The Strange Theory of Light and Matter (II)

If we limit our attention to the interaction between light and matter (i.e. the behavior of photons and electrons only—so we we’re not talking quarks and gluons here), then the ‘crazy ideas’ of quantum mechanics can be summarized as follows:

At the atomic or sub-atomic scale, we can no longer look at light as an electromagnetic wave. It consists of photons, and photons come in blobs. Hence, to some extent, photons are ‘particle-like’.
At the atomic or sub-atomic scale, electrons don’t behave like particles. For example, if we send them through a slit that’s small enough, we’ll observe a diffraction pattern. Hence, to some extent, electrons are ‘wave-like’.

In short, photons aren’t waves, but they aren’t particles either. Likewise, electrons aren’t particles, but they aren’t waves either. They are neither. The weirdest thing of all, perhaps, is that, while light and matter are two very different things in our daily experience – light and matter are opposite concepts, I’d say, just like particles and waves are opposite concepts) – they look pretty much the same in quantum physics: they are both represented by a wavefunction.

Let me immediately make a little note on terminology here. The term ‘wavefunction’ is a bit ambiguous, in my view, because it makes one think of a real wave, like a water wave, or an electromagnetic wave. Real waves are described by real-valued wave functions describing, for example, the motion of a ball on a spring, or the displacement of a gas (e.g. air) as a sound wave propagates through it, or – in the case of an electromagnetic wave – the strength of the electric and magnetic field.

You may have questions about the ‘reality’ of fields, but electromagnetic waves – i.e. the classical description of light – are quite ‘real’ too, even if:

Light doesn’t travel in a medium (like water or air: there is no aether), and
The magnitude of the electric and magnetic field (they are usually denoted by E and B) depend on your reference frame: if you calculate the fields using a moving coordinate system, you will get a different mixture of E and B. Therefore, E and B may not feel very ‘real’ when you look at them separately, but they are very real when we think of them as representing one physical phenomenon: the electromagnetic interaction between particles. So the E and B mix is, indeed, a dual representation of one reality. I won’t dwell on that, as I’ve done that in another post of mine.

How ‘real’ is the quantum-mechanical wavefunction?

The quantum-mechanical wavefunction is not like any of these real waves. In fact, I’d rather use the term ‘probability wave’ but, apparently, that’s used only by bloggers like me 🙂 and so it’s not very scientific. That’s for a good reason, because it’s not quite accurate either: the wavefunction in quantum mechanics represents probability amplitudes, not probabilities. So we should, perhaps, be consistent and term it a ‘probability amplitude wave’ – but then that’s too cumbersome obviously, so the term ‘probability wave’ may be confusing, but it’s not so bad, I think.

Amplitudes and probabilities are related as follows:

Probabilities are real numbers between 0 and 1: they represent the probability of something happening, e.g. a photon moves from point A to B, or a photon is absorbed (and emitted) by an electron (i.e. a ‘junction’ or ‘coupling’, as you know).
Amplitudes are complex numbers, or ‘arrows’ as Feynman calls them: they have a length (or magnitude) and a direction.
We get the probabilities by taking the (absolute) square of the amplitudes.

So photons aren’t waves, but they aren’t particles either. Likewise, electrons aren’t particles, but they aren’t waves either. They are neither. So what are they? We don’t have words to describe what they are. Some use the term ‘wavicle’ but that doesn’t answer the question, because who knows what a ‘wavicle’ is? So we don’t know what they are. But we do know how they behave. As Feynman puts it, when comparing the behavior of light and then of electrons in the double-slit experiment—struggling to find language to describe what’s going on: “There is one lucky break: electrons behave just like light.”

He says so because of that wave function: the mathematical formalism is the same, for photons and for electrons. Exactly the same? […] But that’s such a weird thing to say, isn’t it? We can’t help thinking of light as waves, and of electrons as particles. They can’t be the same. They’re different, aren’t they? They are.

Scales and senses

To some extent, the weirdness can be explained because the scale of our world is not atomic or sub-atomic. Therefore, we ‘see’ things differently. Let me say a few words about the instrument we use to look at the world: our eye.

Our eye is particular. The retina has two types of receptors: the so-called cones are used in bright light, and distinguish color, but when we are in a dark room, the so-called rods become sensitive, and it is believed that they actually can detect a single photon of light. However, neural filters only allow a signal to pass to the brain when at least five photons arrive within less than a tenth of a second. A tenth of a second is, roughly, the averaging time of our eye. So, as Feynman puts it: “If we were evolved a little further so we could see ten times more sensitively, we wouldn’t have this discussion—we would all have seen very dim light of one color as a series of intermittent little flashes of equal intensity.” In other words, the ‘particle-like’ character of light would have been obvious to us.

Let me make a few more remarks here, which you may or may not find useful. The sense of ‘color’ is not something ‘out there’: colors, like red or brown, are experiences in our eye and our brain. There are ‘pigments’ in the cones (cones are the receptors that work only if the intensity of the light is high enough) and these pigments absorb the light spectrum somewhat differently, as a result of which we ‘see’ color. Different animals see different things. For example, a bee can distinguish between white paper using zinc white versus lead white, because they reflect light differently in the ultraviolet spectrum, which the bee can see but we don’t. Bees can also tell the direction of the sun without seeing the sun itself, because they are sensitive to polarized light, and the scattered light of the sky (i.e. the blue sky as we see it) is polarized. The bee can also notice flicker up to 200 oscillations per second, while we see it only up to 20, because our averaging time is like a tenth of a second, which is short for us, but so the averaging time of the bee is much shorter. So we cannot see the quick leg movements and/or wing vibrations of bees, but the bee can!

Sometimes we can’t see any color. For example, we see the night sky in ‘black and white’ because the light intensity is very low, and so it’s our rods, not the cones, that process the signal, and so these rods can’t ‘see’ color. So those beautiful color pictures of nebulae are not artificial (although the pictures are often enhanced). It’s just that the camera that is used to take those pictures (film or, nowadays, digital) is much more sensitive than our eye.

Regardless, color is a quality which we add to our experience of the outside world ourselves. What’s out there are electromagnetic waves with this or that wavelength (or, what amounts to the same, this or that frequency). So when critics of the exact sciences say so much is lost when looking at (visible) light as an electromagnetic wave in the range of 430 to 790 teraherz, they’re wrong. Those critics will say that physics reduces reality. That is not the case.

What’s going on is that our senses process the signal that they are receiving, especially when it comes to vision. As Feynman puts it: “None of the other senses involves such a large amount of calculation, so to speak, before the signal gets into a nerve that one can make measurements on. The calculations for all the rest of the senses usually happen in the brain itself, where it is very difficult to get at specific places to make measurements, because there are so many interconnections. Here, with the visual sense, we have the light, three layers of cells making calculations, and the results of the calculations being transmitted through the optic nerve.”

Hence, things like color and all of the other sensations that we have are the object of study of other sciences, including biochemistry and neurobiology, or physiology. For all we know, what’s ‘out there’ is, effectively, just ‘boring’ stuff, like electromagnetic radiation, energy and ‘elementary particles’—whatever they are. No colors. Just frequencies. 🙂

Light versus matter

If we accept the crazy ideas of quantum mechanics, then the what and the how become one and the same. Hence we can say that photons and electrons are a wavefunction somewhere in space. Photons, of course, are always traveling, because they have energy but no rest mass. Hence, all their energy is in the movement: it’s kinetic, not potential. Electrons, on the other hand, usually stick around some nucleus. And, let’s not forget, they have an electric charge, so their energy is not only kinetic but also potential.

But, otherwise, it’s the same type of ‘thing’ in quantum mechanics: a wavefunction, like those below.

Why diagram A and B? It’s just to emphasize the difference between a real-valued wave function and those ‘probability waves’ we’re looking at here (diagram C to H). A and B represent a mass on a spring, oscillating at more or less the same frequency but a different amplitude. The amplitude here means the displacement of the mass. The function describing the displacement of a mass on a spring (so that’s diagram A and B) is an example of a real-valued wave function: it’s a simple sine or cosine function, as depicted below. [Note that a sine and a cosine are the same function really, except for a phase difference of 90°.]

Let’s now go back to our ‘probability waves’. Photons and electrons, light and matter… The same wavefunction? Really? How can the sunlight that warms us up in the morning and makes trees grow be the same as our body, or the tree? The light-matter duality that we experience must be rooted in very different realities, isn’t it?

Well… Yes and no. If we’re looking at one photon or one electron only, it’s the same type of wavefunction indeed. The same type… OK, you’ll say. So they are the same family or genus perhaps, as they say in biology. Indeed, both of them are, obviously, being referred to as ‘elementary particles’ in the so-called Standard Model of physics. But so what makes an electron and a photon specific as a species? What are the differences?

There’re quite a few, obviously:

1. First, as mentioned above, a photon is a traveling wave function and, because it has no rest mass, it travels at the ultimate speed, i.e. the speed of light (c). An electron usually sticks around or, if it travels through a wire, it travels at very low speeds. Indeed, you may find it hard to believe, but the drift velocity of the free electrons in a standard copper wire is measured in cm per hour, so that’s very slow indeed—and while the electrons in an electron microscope beam may be accelerated up to 70% of the speed of light, and close to c in those huge accelerators, you’re not likely to find an electron microscope or accelerator in Nature. In fact, you may want to remember that a simple thing like electricity going through copper wires in our houses is a relatively modern invention. 🙂

So, yes, those oscillating wave functions in those diagrams above are likely to represent some electron, rather than a photon. To be precise, the wave functions above are examples of standing (or stationary) waves, while a photon is a traveling wave: just extend that sine and cosine function in both directions if you’d want to visualize it or, even better, think of a sine and cosine function in an envelope traveling through space, such as the one depicted below.

Indeed, while the wave function of our photon is traveling through space, it is likely to be limited in space because, when everything is said and done, our photon is not everywhere: it must be somewhere.

At this point, it’s good to pause and think about what is traveling through space. It’s the oscillation. But what’s the oscillation? There is no medium here, and even if there would be some medium (like water or air or something like aether—which, let me remind you, isn’t there!), the medium itself would not be moving, or – I should be precise here – it would only move up and down as the wave propagates through space, as illustrated below. To be fully complete, I should add we also have longitudinal waves, like sound waves (pressure waves): in that case, the particles oscillate back and forth along the direction of wave propagation. But you get the point: the medium does not travel with the wave.

When talking electromagnetic waves, we have no medium. These E and B vectors oscillate but is very wrong to assume they use ‘some core of nearby space’, as Feynman puts it. They don’t. Those field vectors represent a condition at one specific point (admittedly, a point along the direction of travel) in space but, for all we know, an electromagnetic wave travels in a straight line and, hence, we can’t talk about its diameter or so.

Still, as mentioned above, we can imagine, more or less, what E and B stand for (we can use field line to visualize them, for instance), even if we have to take into account their relativity (calculating their values from a moving reference frame results in different mixtures of E and B). But what are those amplitudes? How should we visualize them?

The honest answer is: we can’t. They are what they are: two mathematical quantities which, taken together, form a two-dimensional vector, which we square to find a value for a real-life probability, which is something that – unlike the amplitude concept – does make sense to us. Still, that representation of a photon above (i.e. the traveling envelope with a sine and cosine inside) may help us to ‘understand’ it somehow. Again, you absolute have to get rid of the idea that these ‘oscillations’ would somehow occupy some physical space. They don’t. The wave itself has some definite length, for sure, but that’s a measurement in the direction of travel, which is often denoted as x when discussing uncertainty in its position, for example—as in the famous Uncertainty Principle (ΔxΔp > h).

You’ll say: Oh!—but then, at the very least, we can talk about the ‘length’ of a photon, can’t we? So then a photon is one-dimensional at least, not zero-dimensional! The answer is yes and no. I’ve talked about this before and so I’ll be short(er) on it now. A photon is emitted by an atom when an electron jumps from one energy level to another. It thereby emits a wave train that lasts about 10^–8seconds. That’s not very long but, taking into account the rather spectacular speed of light (3×10⁸m/s), that still makes for a wave train with a length of not less than 3 meter. […] That’s quite a length, you’ll say. You’re right. But you forget that light travels at the speed of light and, hence, we will see this length as zero because of the relativistic length contraction effect. So… Well… Let me get back to the question: if photons and electrons are both represented by a wavefunction, what makes them different?

2. A more fundamental difference between photons and electrons is how they interact with each other.

From what I’ve written above, you understand that probability amplitudes are complex numbers, or ‘arrows’, or ‘two-dimensional vectors’. [Note that all of these terms have precise mathematical definitions and so they’re actually not the same, but the difference is too subtle to matter here.] Now, there are two ways of combining amplitudes, which are referred to as ‘positive’ and ‘negative’ interference respectively. I should immediately note that there’s actually nothing ‘positive’ or ‘negative’ about the interaction: we’re just putting two arrows together, and there are two ways to do that. That’s all.

The diagrams below show you these two ways. You’ll say: there are four! However, remember that we square an arrow to get a probability. Hence, the direction of the final arrow doesn’t matter when we’re taking the square: we get the same probability. It’s the direction of the individual amplitudes that matters when combining them. So the square of A+B is the same as the square of –(A+B) = –A+(–B) = –A–B. Likewise, the square of A–B is the same as the square of –(A–B) = –A+B.

These are the only two logical possibilities for combining arrows. I’ve written ad nauseam about this elsewhere: see my post on amplitudes and statistics, and so I won’t go into too much detail here. Or, in case you’d want something less than a full mathematical treatment, I can refer you to my previous post also, where I talked about the ‘stopwatch’ and the ‘phase’: the convention for the stopwatch is to have its hand turn clockwise (obviously!) while, in quantum physics, the phase of a wave function will turn counterclockwise. But so that’s just convention and it doesn’t matter, because it’s the phase difference between two amplitudes that counts. To use plain language: it’s the difference in the angles of the arrows, and so that difference is just the same if we reverse the direction of both arrows (which is equivalent to putting a minus sign in front of the final arrow).

OK. Let me get back to the lesson. The point is: this logical or mathematical dichotomy distinguishes bosons (i.e. force-carrying ‘particles’, like photons, which carry the electromagnetic force) from fermions (i.e. ‘matter-particles’, such as electrons and quarks, which make up protons and neutrons). Indeed, the so-called ‘positive’ and ‘negative’ interference leads to two very different behaviors:

The probability of getting a boson where there are already n present, is n+1 times stronger than it would be if there were none before.
In contrast, the probability of getting two electrons into exactly the same state is zero.

The behavior of photons makes lasers possible: we can pile zillions of photon on top of each other, and then release all of them in one powerful burst. [The ‘flickering’ of a laser beam is due to the quick succession of such light bursts. If you want to know how it works in detail, check my post on lasers.]

The behavior of electrons is referred to as Fermi’s exclusion principle: it is only because real-life electrons can have one of two spin polarizations (i.e. two opposite directions of angular momentum, which are referred to as ‘up’ or ‘down’, but they might as well have been referred to as ‘left’ or ‘right’) that we find two electrons (instead of just one) in any atomic or molecular orbital.

So, yes, while both photons and electrons can be described by a similar-looking wave function, their behavior is fundamentally different indeed. How is that possible? Adding and subtracting ‘arrows’ is a very similar operation, isn’it?

It is and it isn’t. From a mathematical point of view, I’d say: yes. From a physics point of view, it’s obviously not very ‘similar’, as it does lead to these two very different behaviors: the behavior of photons allows for laser shows, while the behavior of electrons explain (almost) all the peculiarities of the material world, including us walking into doors. 🙂 If you want to check it out for yourself, just check Feynman’s Lectures for more details on this or, else, re-read my posts on it indeed.

3. Of course, there are even more differences between photons and electrons than the two key differences I mentioned above. Indeed, I’ve simplified a lot when I wrote what I wrote above. The wavefunctions of electrons in orbit around a nucleus can take very weird shapes, as shown in the illustration below—and please do google a few others if you’re not convinced. As mentioned above, they’re so-called standing waves, because they occupy a well-defined position in space only, but standing waves can look very weird. In contrast, traveling plane waves, or envelope curves like the one above, are much simpler.

In short: yes, the mathematical representation of photons and electrons (i.e. the wavefunction) is very similar, but photons and electrons are very different animals indeed.

Potentiality and interconnectedness

I guess that, by now, you agree that quantum theory is weird but, as you know, quantum theory does explain all of the stuff that couldn’t be explained before: “It works like a charm”, as Feynman puts it. In fact, he’s often quoted as having said the following:

“It is often stated that of all the theories proposed in this century, the silliest is quantum theory. Some say the the only thing that quantum theory has going for it, in fact, is that it is unquestionably correct.”

Silly? Crazy? Uncommon-sensy? Truth be told, you do get used to thinking in terms of amplitudes after a while. And, when you get used to them, those ‘complex’ numbers are no longer complicated. 🙂 Most importantly, when one thinks long and hard enough about it (as I am trying to do), it somehow all starts making sense.

For example, we’ve done away with dualism by adopting a unified mathematical framework, but the distinction between bosons and fermions still stands: an ‘elementary particle’ is either this or that. There are no ‘split personalities’ here. So the dualism just pops up at a different level of description, I’d say. In fact, I’d go one step further and say it pops up at a deeper level of understanding.

But what about the other assumptions in quantum mechanics. Some of them don’t make sense, do they? Well… I struggle for quite a while with the assumption that, in quantum mechanics, anything is possible really. For example, a photon (or an electron) can take any path in space, and it can travel at any speed (including speeds that are lower or higher than light). The probability may be extremely low, but it’s possible.

Now that is a very weird assumption. Why? Well… Think about it. If you enjoy watching soccer, you’ll agree that flying objects (I am talking about the soccer ball here) can have amazing trajectories. Spin, lift, drag, whatever—the result is a weird trajectory, like the one below:

But, frankly, a photon taking the ‘southern’ route in the illustration below? What are the ‘wheels and gears’ there? There’s nothing sensible about that route, is there?

In fact, there’s at least three issues here:

First, you should note that strange curved paths in the real world (such as the trajectories of billiard or soccer balls) are possible only because there’s friction involved—between the felt of the pool table cloth and the ball, or between the balls, or, in the case of soccer, between the ball and the air. There’s no friction in the vacuum. Hence, in empty space, all things should go in a straight line only.
While it’s quite amazing what’s possible, in the real world that is, in terms of ‘weird trajectories’, even the weirdest trajectories of a billiard or soccer ball can be described by a ‘nice’ mathematical function. We obviously can’t say the same of that ‘southern route’ which a photon could follow, in theory that is. Indeed, you’ll agree the function describing that trajectory cannot be ‘nice’. So even we’d allow all kinds of ‘weird’ trajectories, shouldn’t we limit ourselves to ‘nice’ trajectories only? I mean: it doesn’t make sense to allow the photons traveling from your computer screen to your retina take some trajectory to the Sun and back, does it?
Finally, and most fundamentally perhaps, even when we would assume that there’s some mechanism combining (a) internal ‘wheels and gears’ (such as spin or angular momentum) with (b) felt or air or whatever medium to push against, what would be the mechanism determining the choice of the photon in regard to these various paths? In Feynman’s words: How does the photon ‘make up its mind’?

Feynman answers these questions, fully or partially (I’ll let you judge), when discussing the double-slit experiment with photons:

“Saying that a photon goes this or that way is false. I still catch myself saying, “Well, it goes either this way or that way,” but when I say that, I have to keep in mind that I mean in the sense of adding amplitudes: the photon has an amplitude to go one way, and an amplitude to go the other way. If the amplitudes oppose each other, the light won’t get there—even though both holes are open.”

It’s probably worth re-calling the results of that experiment here—if only to help you judge whether or not Feynman fully answer those questions above!

The set-up is shown below. We have a source S, two slits (A and B), and a detector D. The source sends photons out, one by one. In addition, we have two special detectors near the slits, which may or may not detect a photon, depending on whether or not they’re switched on as well as on their accuracy.

First, we close one of the slits, and we find that 1% of the photons goes through the other (so that’s one photon for every 100 photons that leave S). Now, we open both slits to study interference. You know the results already:

If we switch the detectors off (so we have no way of knowing where the photon went), we get interference. The interference pattern depends on the distance between A and B and varies from 0% to 4%, as shown in diagram (a) below. That’s pretty standard. As you know, classical theory can explain that too assuming light is an electromagnetic wave. But so we have blobs of energy – photons – traveling one by one. So it’s really that double-slit experiment with electrons, or whatever other microscopic particles (as you know, they’ve done these interference electrons with large molecules as well—and they get the same result!). We get the interference pattern by using those quantum-mechanical rules to calculate probabilities: we first add the amplitudes, and it’s only when we’re finished adding those amplitudes, that we square the resulting arrow to the final probability.
If we switch those special detectors on, and if they are 100% reliable (i.e. all photons going through are being detected), then our photon suddenly behaves like a particle, instead of as a wave: they will go through one of the slits only, i.e. either through A, or, alternatively, through B. So the two special detectors never go off together. Hence, as Feynman puts it: we shouldn’t think there is “sneaky way that the photon divides in two and then comes back together again.” It’s one or the other way and, and there’s no interference: the detector at D goes off 2% of the time, which is the simple sum of the probabilities for A and B (i.e. 1% + 1%).
When the special detectors near A and B are not 100% reliable (and, hence, do not detect all photons going through), we have three possible final conditions: (i) A and D go off, (ii) B and D go off, and (iii) D goes off alone (none of the special detectors went off). In that case, we have a final curve that’s a mixture, as shown in diagram (c) and (d) below. We get it using the same quantum-mechanical rules: we add amplitudes first, and then we square to get the probabilities.

Now, I think you’ll agree with me that Feynman doesn’t answer my (our) question in regard to the ‘weird paths’. In fact, all of the diagrams he uses assume straight or nearby paths. Let me re-insert two of those diagrams below, to show you what I mean.

So where are all the strange non-linear paths here? Let me, in order to make sure you get what I am saying here, insert that illustration with the three crazy routes once again. What we’ve got above (Figure 33 and 34) is not like that. Not at all: we’ve got only straight lines there! Why? The answer to that question is easy: the crazy paths don’t matter because their amplitudes cancel each other out, and so that allows Feynman to simplify the whole situation and show all the relevant paths as straight lines only.

Now, I struggled with that for quite a while. Not because I can’t see the math or the geometry involved. No. Feynman does a great job showing why those amplitudes cancel each other out indeed (if you want a summary, see my previous post once again). My ‘problem’ is something else. It’s hard to phrase it, but let me try: why would we even allow for the logical or mathematical possibility of ‘weird paths’ (and let me again insert that stupid diagram below) if our ‘set of rules’ ensures that the truly ‘weird’ paths (like that photon traveling from your computer screen to your eye doing a detour taking it to the Sun and back) cancel each other out anyway? Does that respect Occam’s Razor? Can’t we devise some theory including ‘sensible’ paths only?

Of course, I am just an autodidact with limited time, and I know hundreds (if not thousands) of the best scientists have thought long and hard about this question and, hence, I readily accept the answer is quite simply: no. There is no better theory. I accept that answer, ungrudgingly, not only because I think I am not so smart as those scientists but also because, as I pointed out above, one can’t explain any path that deviates from a straight line really, as there is no medium, so there are no ‘wheels and gears’. The only path that makes sense is the straight line, and that’s only because…

Well… Thinking about it… We think the straight path makes sense because we have no good theory for any of the other paths. Hmm… So, from a logical point of view, assuming that the straight line is the only reasonable path is actually pretty random too. When push comes to shove, we have no good theory for the straight line either!

You’ll say I’ve just gone crazy. […] Well… Perhaps you’re right. 🙂 But… Somehow, it starts to make sense to me. We allow for everything to, then, indeed weed out the crazy paths using our interference theory, and so we do end up with what we’re ending up with: some kind of vague idea of “light not really traveling in a straight line but ‘smelling’ all of the neighboring paths around it and, hence, using a small core of nearby space“—as Feynman puts it.

Hmm… It brings me back to Richard Feynman’s introduction to his wonderful little book, in which he says we should just be happy to know how Nature works and not aspire to know why it works that way. In fact, he’s basically saying that, when it comes to quantum mechanics, the ‘how’ and the ‘why’ are one and the same, so asking ‘why’ doesn’t make sense, because we know ‘how’. He compares quantum theory with the system of calculation used by the Maya priests, which was based on a system of bars and dots, which helped them to do complex multiplications and divisions, for example. He writes the following about it: “The rules were tricky, but they were a much more efficient way of getting an answer to complicated questions (such as when Venus would rise again) than by counting beans.”

When I first read this, I thought the comparison was flawed: if a common Maya Indian did not want to use the ‘tricky’ rules of multiplication and what have you (or, more likely, if he didn’t understand them), he or she could still resort to counting beans. But how do we count beans in quantum mechanics? We have no ‘simpler’ rules than those weird rules about adding amplitudes and taking the (absolute) square of complex numbers so… Well… We actually are counting beans here then:

We allow for any possibility—any path: straight, curved or crooked. Anything is possible.
But all those possibilities are inter-connected. Also note that every path has a mirror image: for every route ‘south’, there is a similar route ‘north’, so to say, except for the straight line, which is a mirror image of itself.
And then we have some clock ticking. Time goes by. It ensures that the paths that are too far removed from the straight line cancel each other. [Of course, you’ll ask: what is too far? But I answered that question – convincingly, I hope – in my previous post: it’s not about the ‘number of arrows’ (as suggested in the caption under that Figure 34 above), but about the frequency and, hence, the ‘wavelength’ of our photon.]
And so… Finally, what’s left is a limited number of possibilities that interfere with each other, which results in what we ‘see’: light seems to use a small core of space indeed–a limited number of nearby paths.

You’ll say… Well… That still doesn’t ‘explain’ why the interference pattern disappears with those special detectors or – what amounts to the same – why the special detectors at the slits never click simultaneously.

You’re right. How do we make sense of that? I don’t know. You should try to imagine what happens for yourself. Everyone has his or her own way of ‘conceptualizing’ stuff, I’d say, and you may well be content and just accept all of the above without trying to ‘imagine’ what’s happening really when a ‘photon’ goes through one or both of those slits. In fact, that’s the most sensible thing to do. You should not try to imagine what happens and just follow the crazy calculus rules.

However, when I think about it, I do have some image in my head. The image is of one of those ‘touch-me-not’ weeds. I quickly googled one of these images, but I couldn’t quite find what I am looking for: it would be more like something that, when you touch it, curls up in a little ball. Any case… You know what I mean, I hope.

You’ll shake your head now and solemnly confirm that I’ve gone mad. Touch-me-not weeds? What’s that got to do with photons?

Well… It’s obvious you and I cannot really imagine how a photon looks like. But I think of it as a blob of energy indeed, which is inseparable, and which effectively occupies some space (in three dimensions that is). I also think that, whatever it is, it actually does travel through both slits, because, as it interferes with itself, the interference pattern does depend on the space between the two slits as well as the width of those slits. In short, the whole ‘geometry’ of the situation matters, and so the ‘interaction’ is some kind of ‘spatial’ thing. [Sorry for my awfully imprecise language here.]

Having said that, I think it’s being detected by one detector only because only one of them can sort of ‘hook’ it, somehow. Indeed, because it’s interconnected and inseparable, it’s the whole blob that gets hooked, not just one part of it. [You may or may not imagine that the detectors that’s got the best hold of it gets it, but I think that’s pushing the description too much.] In any case, the point is that a photon is surely not like a lizard dropping its tail while trying to escape. Perhaps it’s some kind of unbreakable ‘string’ indeed – and sorry for summarizing string theory so unscientifically here – but then a string oscillating in dimensions we can’t imagine (or in some dimension we can’t observe, like the Kaluza-Klein theory suggests). It’s something, for sure, and something that stores energy in some kind of oscillation, I think.

What it is, exactly, we can’t imagine, and we’ll probably never find out—unless we accept that the how of quantum mechanics is not only the why, but also the what. 🙂

Does this make sense? Probably not but, if anything, I hope it fired your imagination at least. 🙂

The Strange Theory of Light and Matter (I)

I am of the opinion that Richard Feynman’s wonderful little common-sense introduction to the ‘uncommon-sensy‘ theory of quantum electrodynamics (The Strange Theory of Light and Matter), which were published a few years before his death only, should be mandatory reading for high school students.

I actually mean that: it should just be part of the general education of the first 21st century generation. Either that or, else, the Education Board should include a full-fledged introduction to complex analysis and quantum physics in the curriculum. 🙂

Having praised it (just now, as well as in previous posts), I re-read it recently during a trek in Nepal with my kids – I just grabbed the smallest book I could find the morning we left 🙂 – and, frankly, I now think Ralph Leighton, who transcribed and edited these four short lectures, could have cross-referenced it better. Moreover, there are two or three points where Feynman (or Leighton?) may have sacrificed accuracy for readability. Let me recapitulate the key points and try to improve here and there.

Amplitudes and arrows

The booklet avoids scary mathematical terms and formulas but doesn’t avoid the fundamental concepts behind, and it doesn’t avoid the kind of ‘deep’ analysis one needs to get some kind of ‘feel’ for quantum mechanics either. So what are the simplifications?

A probability amplitude (i.e. a complex number) is, quite simply, an arrow, with a direction and a length. Thus Feynman writes: “Arrows representing probabilities from 0% to 16% [as measured by the surface of the square which has the arrow as its side] have lengths from 0 to 0.4.” That makes sense: such geometrical approach does away, for example, with the need to talk about the absolute square (i.e. the square of the absolute value, or the squared norm) of a complex number – which is what we need to calculate probabilities from probability amplitudes. So, yes, it’s a wonderful metaphor. We have arrows and surfaces now, instead of wave functions and absolute squares of complex numbers.

The way he combines these arrows make sense too. He even notes the difference between photons (bosons) and electrons (fermions): for bosons, we just add arrows; for fermions, we need to subtract them (see my post on amplitudes and statistics in this regard).

There is also the metaphor for the phase of a wave function, which is a stroke of genius really (I mean it): the direction of the ‘arrow’ is determined by a stopwatch hand, which starts turning when a photon leaves the light source, and stops when it arrives, as shown below.

OK. Enough praise. What are the drawbacks?

The illustration above accompanies an analysis of how light is either reflected from the front surface of a sheet of a glass or, else, from the back surface. Because it takes more time to bounce off the back surface (the path is associated with a greater distance), the front and back reflection arrows point in different directions indeed (the stopwatch is stopped somewhat later when the photon reflects from the back surface). Hence, the difference in phase (but that’s a term that Feynman also avoids) is determined by the thickness of the glass. Just look at it. In the upper part of the illustration above, the thickness is such that the chance of a photon reflecting off the front or back surface is 5%: we add two arrows, each with a length of 0.2, and then we square the resulting (aka final) arrow. Bingo! We get a surface measuring 0.05, or 5%.

Huh? Yes. Just look at it: if the angle between the two arrows would be 90° exactly, it would be 0.08 or 8%, but the angle is a bit less. In the lower part of the illustration, the thickness of the glass is such that the two arrows ‘line up’ and, hence, they form an arrow that’s twice the length of either arrow alone (0.2 + 0.2 = 0.4), with a square four times as large (0.16 = 16%). So… It all works like a charm, as Feynman puts it.

[…]

But… Hey! Look at the stopwatch for the front reflection arrows in the upper and lower diagram: they point in the opposite direction of the stopwatch hand! Well… Hmm… You’re right. At this point, Feynman just notes that we need an extra rule: “When we are considering the path of a photon bouncing off the front surface of the glass, we reverse the direction of the arrow.”

He doesn’t say why. He just adds this random rule to the other rules – which most readers who read this book already know. But why this new rule? Frankly, this inconsistency – or lack of clarity – would wake me up at night. This is Feynman: there must be a reason. Why?

Initially, I suspected it had something to do with the two types of ‘statistics’ in quantum mechanics (i.e. those different rules for combining amplitudes of bosons and fermions respectively, which I mentioned above). But… No. Photons are bosons anyway, so we surely need to add, not subtract. So what is it?

[…] Feynman explains it later, much later – in the third of the four chapters of this little book, to be precise. It’s, quite simply, the result of the simplified model he uses in that first chapter. The photon can do anything really, and so there are many more arrows than just two. We actually should look at an infinite number of arrows, representing all possible paths in spacetime, and, hence, the two arrows (i.e. the one for the reflection from the front and back surface respectively) are combinations of many other arrows themselves. So how does that work?

An analysis of partial reflection (I)

The analysis in Chapter 3 of the same phenomenon (i.e. partial reflection by glass) is a simplified analysis too, but it’s much better – because there are no ‘random’ rules here. It is what Leighton promises to the reader in his introduction: “A complete description, accurate in every detail, of a framework onto which more advanced concepts can be attached without modification. Nothing has to be ‘unlearned’ later.”

Well… Accurate in every detail? Perhaps not. But it’s good, and I still warmly recommend a reading of this delightful little book to anyone who’d ask me what to read as a non-mathematical introduction to quantum mechanics. I’ll limit myself here to just some annotations.

The first drawing (a) depicts the situation:

A photon from a light source is being reflected by the glass. Note that it may also go straight through, but that’s a possibility we’ll analyze separately. We first assume that the photon is effectively being reflected by the glass, and so we want to calculate the probability of that event using all these ‘arrows’, i.e. the underlying probability amplitudes.
As for the geometry of the situation: while the light source and the detector seem to be positioned at some angle from the normal, that is not the case: the photon travels straight down (and up again when reflected). It’s just a limitation of the drawing. It doesn’t really matter much for the analysis: we could look at a light beam coming in at some angle, but so we’re not doing that. It’s the simplest situation possible, in terms of experimental set-up that is. I just want to be clear on that.

Now, rather than looking at the front and back surface only (as Feynman does in Chapter 1), the glass sheet is now divided into a number of very thin sections: five, in this case, so we have six points from which the photon can be scattered into the detector at A: X₁ to X₆. So that makes six possible paths. That’s quite a simplification but it’s easy to see it doesn’t matter: adding more sections would result in many more arrows, but these arrows would also be much smaller, and so the final arrow would be the same.

The more significant simplification is that the paths are all straight paths, and that the photon is assumed to travel at the speed of light, always. If you haven’t read the booklet, you’ll say that’s obvious, but it’s not: a photon has an amplitude to go faster or slower than c but, as Feynman points out, these amplitudes cancel out over longer distances. Likewise, a photon can follow any path in space really, including terribly crooked paths, but these paths also cancel out. As Feynman puts it: “Only the paths near the straight-line path have arrows pointing in nearly the same direction, because their timings are nearly the same, and only these arrows are important, because it is from them that we accumulate a large final arrow.” That makes perfect sense, so there’s no problem with the analysis here either.

So let’s have a look at those six arrows in illustration (b). They point in a slightly different direction because the paths are slightly different and, hence, the distances (and, therefore, the timings) are different too. Now, Feynman (but I think it’s Leighton really) loses himself here in a digression on monochromatic light sources. A photon is a photon: it will have some wave function with a phase that varies in time and in space and, hence, illustration (b) makes perfect sense. [I won’t quote what he writes on a ‘monochromatic light source’ because it’s quite confusing and, IMHO, not correct.]

The stopwatch metaphor has only one minor shortcoming: the hand of a stopwatch rotates clockwise (obviously!), while the phase of an actual wave function goes counterclockwise with time. That’s just convention, and I’ll come back to it when I discuss the mathematical representation of the so-called wave function, which gives you these amplitudes. However, it doesn’t change the analysis, because it’s the difference in the phase that matters when combining amplitudes, so the clock can turn in either way indeed, as long as we’re agreed on it.

At this point, I can’t resist: I’ll just throw the math in. If you don’t like it, you can just skip the section that follows.

Feynman’s arrows and the wave function

The mathematical representation of Feynman’s ‘arrows’ is the wave function:

f = f(x–ct)

Is that the wave function? Yes. It is: it’s a function whose argument is x – ct, with x the position in space, and t the time variable. As for c, that’s the speed of light. We throw it in to make the units in which we measure time and position compatible.

Really? Yes: f is just a regular wave function. To make it look somewhat more impressive, I could use the Greek symbol Φ (phi) or Ψ (psi) for it, but it’s just what it is: a function whose value depends on position and time indeed, so we write f = f(x–ct). Let me explain the minus sign and the c in the argument.

Time and space are interchangeable in the argument, provided we measure time in the ‘right’ units, and so that’s why we multiply the time in seconds with c, so the new unit of time becomes the time that light needs to travel a distance of one meter. That also explains the minus sign in front of ct: if we add one distance unit (i.e. one meter) to the argument, we have to subtract one time unit from it – the new time unit of course, so that’s the time that light needs to travel one meter – in order to get the same value for f. [If you don’t get that x–ct thing, just think a while about this, or make some drawing of a wave function. Also note that the spacetime diagram in illustration (b) above assumes the same: time is measured in an equivalent unit as distance, so the 45% line from the south-west to the north-east, that bounces back to the north-west, represents a photon traveling at speed c in space indeed: one unit of time corresponds to one meter of travel.]

Now I want to be a bit more aggressive. I said f is a simple function. That’s true and not true at the same time. It’s a simple function, but it gives you probability amplitudes, which are complex numbers – and you may think that complex numbers are, perhaps, not so simple. However, you shouldn’t be put off. Complex numbers are really like Feynman’s ‘arrows’ and, hence, fairly simple things indeed. They have two dimensions, so to say: an a– and a b-coordinate. [I’d say an x– and y-coordinate, because that’s what you usually see, but then I used the x symbol already for the position variable in the argument of the function, so you have to switch to a and b for a while now.]

This a– and b-coordinate are referred to as the real and imaginary part of a complex number respectively. The terms ‘real’ and ‘imaginary’ are confusing because both parts are ‘real’ – well… As real as numbers can be, I’d say. 🙂 They’re just two different directions in space: the real axis is the a-axis in coordinate space, and the imaginary axis is the b-axis. So we could write it as an ordered pair of numbers (a, b). However, we usually write it as a number itself, and we distinguish the b-coordinate from the a-coordinate by writing an i in front: (a, b) = a + ib. So our function f = f(x–ct) is a complex-valued function: it will give you two numbers (an a and a b) instead of just one when you ‘feed’ it with specific values for x and t. So we write:

f = f(x–ct) = (a, b) = a + ib

So what’s the shape of this function? Is it linear or irregular or what? We’re talking a very regular wave function here, so it’s shape is ‘regular’ indeed. It’s a periodic function, so it repeats itself again and again. The animations below give you some idea of such ‘regular’ wave functions. Animation A and B shows a real-valued ‘wave’: a ball on a string that goes up and down, for ever and ever. Animations C to H are – believe it or not – basically the same thing, but so we have two numbers going up and down. That’s all.

The wave functions above are, obviously, confined in space, and so the horizontal axis represents the position in space. What we see, then, is how the real and imaginary part of these wave functions varies as time goes by. [Think of the blue graph as the real part, and the imaginary part as the pinkish thing – or the other way around. It doesn’t matter.] Now, our wave function – i.e. the one that Feynman uses to calculate all those probabilities – is even more regular than those shown above: its real part is an ordinary cosine function, and it’s imaginary part is a sine. Let me write this in math:

f = f(x–ct) = a + ib = r(cosφ + isinφ)

It’s really the most regular wave function in the world: the very simple illustration below shows how the two components of f vary as a function in space (i.e. the horizontal axis) while we keep the time fixed, or vice versa: it could also show how the function varies in time at one particular point in space, in which case the horizontal axis would represent the time variable. It is what it is: a sine and a cosine function, with the angle φ as its argument.

Note that a sine function is the same as a cosine function, but it just lags a bit. To be precise, the phase difference is 90°, or π/2 in radians (the radian (i.e. the length of the arc on the unit circle) is a much more natural unit to express angles, as it’s fully compatible with our distance unit and, hence, most – if not all – of our other units). Indeed, you may or may not remember the following trigonometric identities: sinφ = cos(π/2–φ) = cos(φ–π/2).

In any case, now we have some r and φ here, instead of a and b. You probably wonder where I am going with all of this. Where are the x and t variables? Be patient! You’re right. We’ll get there. I have to explain that r and φ first. Together, they are the so-called polar coordinates of Feynman’s ‘arrow’ (i.e. the amplitude). Polar coordinates are just as good as coordinates as these Cartesian coordinates we’re used to (i.e. a and b). It’s just a different coordinate system. The illustration below shows how they are related to each other. If you remember anything from your high school trigonometry course, you’ll immediately agree that a is, obviously, equal to rcosφ, and b is rsinφ, which is what I wrote above. Just as good? Well… The polar coordinate system has some disadvantages (all of those expressions and rules we learned in vector analysis assume rectangular coordinates, and so we should watch out!) but, for our purpose here, polar coordinates are actually easier to work with, so they’re better.

Feynman’s wave function is extremely simple because his ‘arrows’ have a fixed length, just like the stopwatch hand. They’re just turning around and around and around as time goes by. In other words, r is constant and does not depend on position and time. It’s the angle φ that’s turning and turning and turning as the stopwatch ticks while our photon is covering larger and larger distances. Hence, we need to find a formula for φ that makes it explicit how φ changes as a function in spacetime. That φ variable is referred to as the phase of the wave function. That’s a term you’ll encounter frequently and so I had better mention it. In fact, it’s generally used as a synonym for any angle, as you can see from my remark on the phase difference between a sine and cosine function.

So how do we express φ as a function of x and t? That’s where Euler’s formula comes in. Feynman calls it the most remarkable formula in mathematics – our jewel! And he’s probably right: of all the theorems and formulas, I guess this is the one we can’t do without when studying physics. I’ve written about this in another post, and repeating what I wrote there would eat up too much space, so I won’t do it and just give you that formula. A regular complex-valued wave function can be represented as a complex (natural) exponential function, i.e. an exponential function with Euler’s number e (i.e. 2.728…) as the base, and the complex number iφ as the (variable) exponent. Indeed, according to Euler’s formula, we can write:

f = f(x–ct) = a + ib = r(cosφ + isinφ) = r·e^iφ

As I haven’t explained Euler’s formula (you should really have a look at my posts on it), you should just believe me when I say that r·e^iφ is an ‘arrow’ indeed, with length r and angle φ (phi), as illustrated above, with a and b coordinates a = rcosφ and b = rsinφ. What you should be able to do now, is to imagine how that φ angle goes round and round as time goes by, just like Feynman’s ‘arrow’ goes round and round – just like a stopwatch hand indeed, but note our φ angle turns counterclockwise indeed.

Fine, you’ll say – but so we need a mathematical expression, don’t we? Yes,we do. We need to know how that φ angle (i.e. the variable in our r·e^iφfunction) changes as a function of x and t indeed. It turns out that the φ in r·e^iφcan be substituted as follows:

r·e^iφ= r·e^i(ωt–kx)= r·e^{–ik(x–ct)}

Huh? Yes. The phase (φ) of the probability amplitude (i.e. the ‘arrow’) is a simple linear function of x and t indeed: φ = ωt–kx = –k(x–ct). What about all these new symbols, k and ω? The ω and k in this equation are the so-called angular frequency and the wave number of the wave. The angular frequency is just the frequency expressed in radians, and you should think of the wave number as the frequency in space. [I could write some more here, but I can’t make it too long, and you can easily look up stuff like this on the Web.] Now, the propagation speed c of the wave is, quite simply, the ratio of these two numbers: c = ω/k. [Again, it’s easy to show how that works, but I won’t do it here.]

Now you know it all, and so it’s time to get back to the lesson.

An analysis of partial reflection (II)

Why did I digress? Well… I think that what I write above makes much more sense than Leighton’s rather convoluted description of a monochromatic light source as he tries to explain those arrows in diagram (b) above. Whatever it is, a monochromatic light source is surely not “a device that has been carefully arranged so that the amplitude for a photon to be emitted at a certain time can be easily calculated.” That’s plain nonsense. Monochromatic light is light of a specific color, so all photons have the same frequency (or, to be precise, their wave functions have all the same well-defined frequency), but these photons are not in phase. Photons are emitted by atoms, as an electron moves from one energy level to the other. Now, when a photon is emitted, what actually happens is that the atom radiates a train of waves only for about 10^–8sec, so that’s about 10 billionths of a second. After 10^–8sec, some other atom takes over, and then another atom, and so on. Each atom emits one photon, whose energy is the difference between the two energy levels that the electron is jumping to or from. So the phase of the light that is being emitted can really only stay the same for about 10^–8sec. Full stop.

Now, what I write above on how atoms actually emit photons is a paraphrase of Feynman’s own words in his much more serious series of Lectures on Mechanics, Radiation and Heat. Therefore, I am pretty sure it’s Leighton who gets somewhat lost when trying to explain what’s happening. It’s not photons that interfere. It’s the probability amplitudes associated with the various paths that a photon can take. To be fully precise, we’re talking the photon here, i.e. the one that ends up in the detector, and so what’s going on is that the photon is interfering with itself. Indeed, that’s exactly what the ‘craziness’ of quantum mechanics is all about: we sent electrons, one by one, through two slits, and we observe an interference pattern. Likewise, we got one photon here, that can go various ways, and it’s those amplitudes that interfere, so… Yes: the photon interferes with itself.

OK. Let’s get back to the lesson and look at diagram (c) now, in which the six arrows are added. As mentioned above, it would not make any difference if we’d divide the glass in 10 or 20 or 1000 or a zillion ‘very thin’ sections: there would be many more arrows, but they would be much smaller ones, and they would cover the same circular segment: its two endpoints would define the same arc, and the same chord on the circle that we can draw when extending that circular segment. Indeed, the six little arrows define a circle, and that’s the key to understanding what happens in the first chapter of Feynman’s QED, where he adds two arrows only, but with a reversal of the direction of the ‘front reflection’ arrow. Here there’s no confusion – Feynman (or Leighton) eloquently describe what they do:

“There is a mathematical trick we can use to get the same answer [i.e. the same final arrow]: Connecting the arrows in order from 1 to 6, we get something like an arc, or part of a circle. The final arrow forms the chord of this arc. If we draw arrows from the center of the ‘circle’ to the tail of arrow 1 and to the head of arrow 6, we get two radii. If the radius arrow from the center to arrow 1 is turned 180° (“subtracted”), then it can be combined with the other radius arrow to give us the same final arrow! That’s what I was doing in the first lecture: these two radii are the two arrows I said represented the ‘front surface’ and ‘back surface’ reflections. They each have the famous length of 0.2.”

That’s what’s shown in part (d) of the illustration above and, in case you’re still wondering what’s going on, the illustration below should help you to make your own drawings now.

So… That explains the phenomenon Feynman wanted to explain, which is a phenomenon that cannot be explained in classical physics. Let me copy the original here:

Partial reflection by glass—a phenomenon that cannot be explained in classical physics? Really?

You’re right to raise an objection: partial reflection by glass can, in fact, be explained by the classical theory of light as an electromagnetic wave. The assumption then is that light is effectively being reflected by both the front and back surface and the reflected waves combine or cancel out (depending on the thickness of the glass and the angle of reflection indeed) to match the observed pattern. In fact, that’s how the phenomenon was explained for hundreds of years! The point to note is that the wave theory of light collapsed as technology advanced, and experiments could be made with very weak light hitting photomultipliers. As Feynman writes: “As the light got dimmer and dimmer, the photomultipliers kept making full-sized clicks—there were just fewer of them. Light behaved as particles!”

The point is that a photon behaves like an electron when going through two slits: it interferes with itself! As Feynman notes, we do not have any ‘common-sense’ theory to explain what’s going on here. We only have quantum mechanics, and quantum mechanics is an “uncommon-sensy” theory: a “strange” or even “absurd” theory, that looks “cockeyed” and incorporates “crazy ideas”. But… It works.

Now that we’re here, I might just as well add a few more paragraphs to fully summarize this lovely publication – if only because summarizing stuff like this helps me to come to terms with understanding things better myself!

Calculating amplitudes: the basic actions

So it all boils down to calculating amplitudes: an event is divided into alternative ways of how the event can happen, and the arrows for each way are ‘added’. Now, every way an event can happen can be further subdivided into successive steps. The amplitudes for these steps are then ‘multiplied’. For example, the amplitude for a photon to go from A to C via B is the ‘product’ of the amplitude to go from A to B and the amplitude to go from B to C.

I marked the terms ‘multiplied’ and ‘product’ with apostrophes, as if to say it’s not a ‘real’ product. But it is an actual multiplication: it’s the product of two complex numbers. Feynman does not explicitly compare this product to other products, such as the dot (•) or cross (×) product of two vectors, but he uses the ∗ symbol for multiplication here, which clearly distinguishes V∗W from V•W or V×W indeed or, more simply, from the product of two ordinary numbers. [Ordinary numbers? Well… With ‘ordinary’ numbers, I mean real numbers, of course, but once you get used to complex numbers, you won’t like that term anymore, because complex numbers start feeling just as ‘real’ as other numbers – especially when you get used to the idea of those complex-valued wave functions underneath reality.]

Now, multiplying complex numbers, or ‘arrows’ using QED’s simpler language, consists of adding their angles and multiplying their lengths. That being said, the arrows here all have a length smaller than one (because their square cannot be larger than one, because that square is a probability, i.e. a (real) number between 0 and 1), Feynman defines successive multiplication as successive ‘shrinks and turns’ of the unit arrow. That all makes sense – very much sense.

But what’s the basic action? As Feynman puts the question: “How far can we push this process of splitting events into simpler and simpler subevents? What are the smallest possible bits and pieces? Is there a limit?” He immediately answers his own question. There are three ‘basic actions’:

A photon goes from one point (in spacetime) to another: this amplitude is denoted by P(A to B).
An electron goes from one point to another: E(A to B).
An electron emits and/or absorbs a photon: this is referred to as a ‘junction’ or a ‘coupling’, and the amplitude for this is denoted by the symbol j, i.e. the so-called junction number.

How do we find the amplitudes for these?

The amplitudes for (1) and (2) are given by a so-called propagator functions, which give you the probability amplitude for a particle to travel from one place to another in a given time indeed, or to travel with a certain energy and momentum. Judging from the Wikipedia article on these functions, the subject-matter is horrendously complicated, and the formulas are too, even if Feynman says it’s ‘very simple’ – for a photon, that is. The key point to note is that any path is possible. Moreover, there are also amplitudes for photons to go faster or slower than the speed of light (c)! However, these amplitudes make smaller contributions, and cancel out over longer distances. The same goes for the crooked paths: the amplitudes cancel each other out as well.

What remains are the ‘nearby paths’. In my previous post (check the section on electromagnetic radiation), I noted that, according to classical wave theory, a light wave does not occupy any physical space: we have electric and magnetic field vectors that oscillate in a direction that’s perpendicular to the direction of propagation, but these do not take up any space. In quantum mechanics, the situation is quite different. As Feynman puts it: “When you try to squeeze light too much [by forcing it to go through a small hole, for example, as illustrated below], it refuses to cooperate and begins to spread out.” He explains this in the text below the second drawing: “There are not enough arrows representing the paths to Q to cancel each other out.”

Not enough arrows? We can subdivide space in as many paths as we want, can’t we? Do probability amplitudes take up space? And now that we’re asking the tougher questions, what’s a ‘small’ hole? What’s ‘small’ and what’s ‘large’ in this funny business?

Unfortunately, there’s not much of an attempt in the booklet to try to answer these questions. One can begin to formulate some kind of answer when doing some more thinking about these wave functions. To be precise, we need to start looking at their wavelength. The frequency of a typical photon (and, hence, of the wave function representing that photon) is astronomically high. For visible light, it’s in the range of 430 to 790 teraherz, i.e. 430–790×10¹² Hz. We can’t imagine such incredible numbers. Because the frequency is so high, the wavelength is unimaginably small. There’s a very simple and straightforward relation between wavelength (λ) and frequency (ν) indeed: c = λν. In words: the speed of a wave is the wavelength (i.e. the distance (in space) of one cycle) times the frequency (i.e. the number of cycles per second). So visible light has a wavelength in the range of 390 to 700 nanometer, i.e. 390–700 billionths of a meter. A meter is a rather large unit, you’ll say, so let me express it differently: it’s less than one thousandth of a micrometer, and a micrometer itself is one thousandth of a millimeter. So, no, we can’t imagine that distance either.

That being said, that wavelength is there, and it does imply that some kind of scale is involved. A wavelength covers one full cycle of the oscillation: it means that, if we travel one wavelength in space, our ‘arrow’ will point in the same direction again. Both drawings above (Figure 33 and 34) suggest the space between the two blocks is less than one wavelength. It’s a bit hard to make sense of the direction of the arrows but note the following:

The phase difference between (a) the ‘arrow’ associated with the straight route (i.e. the ‘middle’ path) and (b) the ‘arrow’ associated with the ‘northern’ or ‘southern’ route (i.e. the ‘highest’ and ‘lowest’ path) in Figure 33 is like quarter of a full turn, i.e. 90°. [Note that the arrows for the northern and southern route to P point in the same direction, because they are associated with the same timing. The same is true for the two arrows in-between the northern/southern route and the middle path.]
In Figure 34, the phase difference between the longer routes and the straight route is much less, like 10° only.

Now, the calculations involved in these analyses are quite complicated but you can see the explanation makes sense: the gap between the two blocks is much narrower in Figure 34 and, hence, the geometry of the situation does imply that the phase difference between the amplitudes associated with the ‘northern’ and ‘southern’ routes to Q is much smaller than the phase difference between those amplitudes in Figure 33. To be precise,

The phase difference between (a) the ‘arrow’ associated with the ‘northern route’ to Q and (b) the ‘arrow’ associated with the ‘southern’ route to Q (i.e. the ‘highest’ and ‘lowest’ path) in Figure 33 is like three quarters of a full turn, i.e. 270°. Hence, the final arrow is very short indeed, which means that the probability of the photon going to Q is very low indeed. [Note that the arrows for the northern and southern route no longer point in the same direction, because they are associated with very different timings: the ‘southern route’ is shorter and, hence, faster.]
In Figure 34, we have a phase difference between the shortest and longest route that is like 60° only and, hence, the final arrow is very sizable and, hence, the probability of the photon going to Q is, accordingly, quite substantial.

OK… What did I say here about P(A to B)? Nothing much. I basically complained about the way Feynman (or Leighton, more probably) explained the interference or diffraction phenomenon and tried to do a better job before tacking the subject indeed: how do we get that P(A to B)?

A photon can follow any path from A to B, including the craziest ones (as shown below). Any path? Good players give a billiard ball extra spin that may make the ball move in a curved trajectory, and will also affect its its collision with any other ball – but a trajectory like the one below? Why would a photon suddenly take a sharp turn left, or right, or up, or down? What’s the mechanism here? What are the ‘wheels and gears inside’ of the photon that (a) make a photon choose this path in the first place and (b) allow it to whirl, swirl and twirl like that?

We don’t know. In fact, the question may make no sense, because we don’t know what actually happens when a photon travels through space. We know it leaves as a lump of energy, and we know it arrives as a similar lump of energy. When we actually put a detector to check which path is followed – by putting special detectors at the slits in the famous double-slit experiment, for example – the interference pattern disappears. So… Well… We don’t know how to describe what’s going on: a photon is not a billiard ball, and it’s not a classical electromagnetic wave either. It is neither. The only thing that we know is that we get probabilities that match with the results of experiment if we accept this nonsensical assumptions and do all of the crazy arithmetic involved. Let me get back to the lesson.

Photons can also travel faster or slower than the speed of light (c is some 3×10⁸meter per second but, in our special time unit, it’s equal to one). Does that violate relativity? It doesn’t, apparently, but for the reasoning behind I must, once again, refer you to more sophisticated writing.

In any case, if the mathematicians and physicists have to take into account both of these assumptions (any path is possible, and speeds higher or lower than c are possible too!), they must be looking at some kind of horrendous integral, don’t they?

They are. When everything is said and done, that propagator function is some monstrous integral indeed, and I can’t explain it to you in a couple of words – if only because I am struggling with it myself. 🙂 So I will just believe Feynman when he says that, when the mathematicians and physicists are finished with that integral, we do get some simple formula which depends on the value of the so-called spacetime interval between two ‘points’ – let’s just call them 1 and 2 – in space and time. You’ve surely heard about it before: it’s denoted by s²or I (or whatever) and it’s zero if an object moves at the speed of light, which is what light is supposed to do – but so we’re dealing with a different situation here. 🙂 To be precise, I consists of two parts:

The distance d between the two points (1 and 2), i.e. Δr, which is just the square root of d²= Δr²= (x₂–x₂)²+(y₂–y₁)²+(z₂–z₁)². [This formula is just a three-dimensional version of the Pythagorean Theorem.]
The ‘distance’ (or difference) in time, which is usually expressed in those ‘equivalent’ time units that we introduced above already, i.e. the time that light – traveling at the speed of light 🙂 – needs to travel one meter. We will usually see that component of I in a squared version too: Δt²= (t₂–t₁)², or, if time is expressed in the ‘old’ unit (i.e. seconds), then we write c²Δt² = c²(t₂–t₁)².

Now, the spacetime interval itself is defined as the excess of the squared distance (in space) over the squared time difference:

s²= I = Δr²– Δt²= (x₂–x₂)²+(y₂–y₁)²+(z₂–z₁)²– (t₂–t₁)²

You know we can then define time-like, space-like and light-like intervals, and these, in turn, define the so-called light cone. The spacetime interval can be negative, for example. In that case, Δt²will be greater than Δr², so there is no ‘excess’ of distance over time: it means that the time difference is large enough to allow for a cause–effect relation between the two events, and the interval is said to be time-like. In any case, that’s not the topic of this post, and I am sorry I keep digressing.

The point to note is that the formula for the propagator favors light-like intervals: they are associated with large arrows. Space- and time-like intervals, on the other hand, will contribute much smaller arrows. In addition, the arrows for space- and time-like intervals point in opposite directions, so they will cancel each other out. So, when everything is said and done, over longer distances, light does tend to travel in a straight line and at the speed of light. At least, that’s what Feynman tells us, and I tend to believe him. 🙂

But so where’s the formula? Feynman doesn’t give it, probably because it would indeed confuse us. Just google ‘propagator for a photon’ and you’ll see what I mean. He does integrate the above conclusions in that illustration (b) though. What illustration?

Oh… Sorry. You probably forgot what I am trying to do here, but so we’re looking at that analysis of partial reflection of light by glass. Let me insert it once again so you don’t have to scroll all the way up.

You’ll remember that Feynman divided the glass sheet into five sections and, hence, there are six points from which the photon can be scattered into the detector at A: X₁ to X₆. So that makes six possible paths: these paths are all straight (so Feynman makes abstraction of all of the crooked paths indeed), and the other assumption is that the photon effectively traveled at the speed of light, whatever path it took (so Feynman also assumes the amplitudes for speeds higher or lower than c cancel each other out). So that explains the difference in time at emission from the light source. The longest path is the path to point X₆ and then back up to the detector. If the photon would have taken that path, it would have to be emitted earlier in time – earlier as compared to the other possibilities, which take less time. So it would have to be emitted at T = T₆. The direction of the ‘arrow’ is like one o’clock. The shorter paths are associated with shorter times (the difference between the time of arrival and departure is shorter) and so T₅ is associated with an arrow in the 12 o’clock direction, T₅ is 11 o’clock, and so on, till T₅, which points at the 9 o’clock direction.

But… What? These arrows also include the reflection, i.e. the interaction between the photon and some electron in the glass, don’t they? […] Right you are. Sorry. So… Yes. The action above involves four ‘basic actions’:

A photon is emitted by the source at a time T = T₁, T₂, T₃, T₄, T₅ or T₆: we don’t know. Quantum-mechanical uncertainty. 🙂
It goes from the source to one of the points X = X₁, X₂, X₃, X₄, X₅ or X₆in the glass: we don’t know which one, because we don’t have a detector there.
The photon interacts with an electron at that point.
It makes it way back up to the detector at A.

Step 1 does not have any amplitude. It’s just the start of the event. Well… We start with the unit arrow pointing north actually, so its length is one and its direction is 12 o’clock. And so we’ll shrink and turn it, i.e. multiply it with other arrows, in the next steps.

Steps 2 and 4 are straightforward and are associated with arrows of the same length. Their direction depends on the distance traveled and/or the time of emission: it amounts to the same because we assume the speed is constant and exactly the same for the six possibilities (that speed is c = 1 obviously). But what length? Well… Some length according to that formula which Feynman didn’t give us. 🙂

So now we need to analyze the third of those three basic actions: a ‘junction’ or ‘coupling’ between an electron and a photon. At this point, Feynman embarks on a delightful story highlighting the difficulties involved in calculating that amplitude. A photon can travel following crooked paths and at devious speeds, but an electron is even worse: it can take what Feynman refers to as ‘one-hop flights’, ‘two-hop flights’, ‘three-hop flights’,… any ‘n-hop flight’ really. Each stop involves an additional amplitude, which is represented by n², with n some number that has been determined from experiment. The formula for E(A to B) then becomes a series of terms: P(A to B) + (P(A to C)∗n²∗(P(C to B) + (P(A to D)∗n²∗P(D to E)∗n²∗P(E to C)+…

P(A to B) is the ‘one-hop flight’ here, while C, D and E are intermediate points, and (P(A to C)∗n²∗(P(C to B) and (P(A to D)∗n²∗P(D to E)∗n²∗P(E to C) are the ‘two-hop’ and ‘three-hop’ flight respectively. Note that this calculation has to be made for all possible intermediate points C, D, E and so on. To make matters worse, the theory assumes that electrons can emit and absorb photons along the way, and then there’s a host of other problems, which Feynman tries to explain in the last and final chapter of his little book. […]

Hey! Stop it!

What?

You’re talking about E(A to B) here. You’re supposed to be talking about that junction number j.

Oh… Sorry. You’re right. Well… That junction number j is about –0.1. I know that looks like an ordinary number, but it’s an amplitude, so you should interpret it as an arrow. When you multiply it with another arrow, it amounts to a shrink to one-tenth, and half a turn. Feynman entertains us also on the difficulties of calculating this number but, you’re right, I shouldn’t be trying to copy him here – if only because it’s about time I finish this post. 🙂

So let me conclude it indeed. We can apply the same transformation (i.e. we multiply with j) to each of the six arrows we’ve got so far, and the result is those six arrows next to the time axis in illustration (b). And then we combine them to get that arc, and then we apply that mathematical trick to show we get the same result as in a classical wave-theoretical analysis of partial reflection.

Done. […] Are you happy now?

[…] You shouldn’t be. There are so many questions that have been left unanswered. For starters, Feynman never gives that formula for the length of P(A to B), so we have no clue about the length of these arrows and, hence, about that arc. If physicists know their length, it seems to have been calculated backwards – from those 0.2 arrows used in the classical wave theory of light. Feynman is actually quite honest about that, and simply writes:

“The radius of the arc [i.e. the arc that determines the final arrow] evidently depends on the length of the arrow for each section, which is ultimately determined by the amplitude S that an electron in an atom of glass scatters a photon. This radius can be calculated using the formulas for the three basic actions. […] It must be said, however, that no direct calculation from first principles for a substance as complex as glass has actually been done. In such cases, the radius is determined by experiment. For glass, it has been determined from experiment that the radius is approximately 0.2 (when the light shines directly onto the glass at right angles).”

Well… OK. I think that says enough. So we have a theory – or first principles at last – but we don’t them to calculate. That actually sounds a bit like metaphysics to me. 🙂 In any case… Well… Bye for now!

But… Hey! You said you’d analyze how light goes straight through the glass as well?

Yes. I did. But I don’t feel like doing that right now. I think we’ve got enough stuff to think about right now, don’t we? 🙂

Applied vector analysis (II)

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. In addition, some of the material was removed by a dark force (that also created problems with the layout, I see now). In any case, we recommend you read our recent papers. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

We’ve covered a lot of ground in the previous post, but we’re not quite there yet. We need to look at a few more things in order to gain some kind of ‘physical’ understanding’ of Maxwell’s equations, as opposed to a merely ‘mathematical’ understanding only. That will probably disappoint you. In fact, you probably wonder why one needs to know about Gauss’ and Stokes’ Theorems if the only objective is to ‘understand’ Maxwell’s equations.

To some extent, your skepticism is justified. It’s already quite something to get some feel for those two new operators we’ve introduced in the previous post, i.e. the divergence (div) and curl operators, denoted by ∇• and ∇× respectively. By now, you understand that these two operators act on a vector field, such as the electric field vector E, or the magnetic field vector B, or, in the example we used, the heat flow h, so we should write ∇•(a vector) and ∇×(a vector. And, as for that del operator – i.e. ∇ without the dot (•) or the cross (×) – if there’s one diagram you should be able to draw off the top of your head, it’s the one below, which shows:

The heat flow vector h, whose magnitude is the thermal energy that passes, per unit time and per unit area, through an infinitesimally small isothermal surface, so we write: h = |h| = ΔJ/ΔA.
The gradient vector ∇T, whose direction is opposite to that of h, and whose magnitude is proportional to h, so we can write the so-called differential equation of heat flow: h = –κ∇T.
The components of the vector dot product ΔT = ∇T•ΔR = |∇T|·ΔR·cosθ.

You should also remember that we can re-write that ΔT = ∇T•ΔR = |∇T|·ΔR·cosθ equation – which we can also write as ΔT/ΔR = |∇T|·cosθ – in a more general form:

Δψ/ΔR = |∇ψ|·cosθ

That equation says that the component of the gradient vector ∇ψ along a small displacement ΔR is equal to the rate of change of ψ in the direction of ΔR. And then we had three important theorems, but I can imagine you don’t want to hear about them anymore. So what can we do without them? Let’s have a look at Maxwell’s equations again and explore some linkages.

Curl-free and divergence-free fields

From what I wrote in my previous post, you should remember that:

The curl of a vector field C (i.e. ∇×C) represents its circulation, i.e. its (infinitesimal) rotation.
Its divergence (i.e. ∇•C) represents the outward flux out of an (infinitesimal) volume around the point we’re considering.

Back to Maxwell’s equations:

Let’s start at the bottom, i.e. with equation (4). It says that a changing electric field (i.e. ∂E/∂t ≠ 0) and/or a (steady) electric current (j/ε₀) will cause some circulation of B, i.e. the magnetic field. It’s important to note that (a) the electric field has to change and/or (b) that electric charges (positive or negative) have to move in order to cause some circulation of B: a steady electric field will not result in any magnetic effects.

This brings us to the first and easiest of all the circumstances we can analyze: the static case. In that case, the time derivatives ∂E/∂t and ∂B/∂t are zero, and Maxwell’s equations reduce to:

∇•E = ρ/ε₀. In this equation, we have ρ, which represents the so-called charge density, which describes the distribution of electric charges in space: ρ = ρ(x, y, z). To put it simply: ρ is the ‘amount of charge’ (which we’ll denote by Δq) per unit volume at a given point. Hence, if we consider a small volume (ΔV) located at point (x, y, z) in space – an infinitesimally small volume, in fact (as usual) –then we can write: Δq = ρ(x, y, z)ΔV. [As for ε₀, you already know this is a constant which ensures all units are ‘compatible’.] This equation basically says we have some flux of E, the exact amount of which is determined by the charge density ρ or, more in general, by the charge distribution in space.
∇×E = 0. That means that the curl of E is zero: everywhere, and always. So there’s no circulation of E. We call this a curl-free field.
∇•B = 0. That means that the divergence of B is zero: everywhere, and always. So there’s no flux of B. None. We call this a divergence-free field.
c²∇×B = j/ε₀. So here we have steady current(s) causing some circulation of B, the exact amount of which is determined by the (total) current j. [What about that c²factor? Well… We talked about that before: magnetism is, basically, a relativistic effect, and so that’s where that factor comes from. I’ll just refer you to what Feynman writes about this in his Lectures, and warmly recommend to read it, because it’s really quite interesting: it gave me at least a much deeper understanding of what it’s all about, and so I hope it will help you as much.]

Now you’ll say: why bother with all these difficult mathematical constructs if we’re going to consider curl-free and divergence-free fields only. Well… B is not curl-free, and E is not divergence-free. To be precise:

E is a field with zero curl and a given divergence, and
B is a field with zero divergence and a given curl.

Yeah, but why can’t we analyze fields that have both curl and divergence? The answer is: we can, and we will, but we have to start somewhere, and so we start with an easier analysis first.

Electrostatics and magnetostatics

The first thing you should note is that, in the static case (i.e. when charges and currents are static), there is no interdependence between E and B. The two fields are not interconnected, so to say. Therefore, we can neatly separate them into two pairs:

Electrostatics: (1) ∇•E = ρ/ε₀ and (2) ∇×E = 0.
Magnetostatics: (1) ∇×B = j/c²ε₀ and (2) ∇•B = 0.

Now, I won’t go through all of the particularities involved. In fact, I’ll refer you to a real physics textbook on that (like Feynman’s Lectures indeed). My aim here is to use these equations to introduce some more math and to gain a better understanding of vector calculus – an understanding that goes, in fact, beyond the math (i.e. a ‘physical’ understanding, as Feynman terms it).

At this point, I have to introduce two additional theorems. They are nice and easy to understand (although not so easy to prove, and so I won’t):

Theorem 1: If we have a vector field – let’s denote it by C – and we find that its curl is zero everywhere, then C must be the gradient of something. In other words, there must be some scalar field ψ (psi) such that C is equal to the gradient of ψ. It’s easier to write this down as follows:

If ∇×C = 0, there is a ψ such that C = ∇ψ.

Theorem 2: If we have a vector field – let’s denote it by D, just to introduce yet another letter – and we find that its divergence is zero everywhere, then D must be the curl of some vector field A. So we can write:

If ∇•D = 0, there is an A such that D = ∇×A.

We can apply this to the situation at hand:

For E, there is some scalar potential Φ such that E = –∇Φ. [Note that we could integrate the minus sign in Φ, but we leave it there as a reminder that the situation is similar to that of heat flow. It’s a matter of convention really: E ‘flows’ from higher to lower potential.]
For B, there is a so-called vector potential A such that B = ∇×A.

The whole game is then to compute Φ and A everywhere. We can then take the gradient of Φ, and the curl of A, to find the electric and magnetic field respectively, at every single point in space. In fact, most of Feynman’s second Volume of his Lectures is devoted to that, so I’ll refer you that if you’d be interested. As said, my goal here is just to introduce the basics of vector calculus, so you gain a better understanding of physics, i.e. an understanding which goes beyond the math.

Electrodynamics

We’re almost done. Electrodynamics is, of course, much more complicated than the static case, but I don’t have the intention to go too much in detail here. The important thing is to see the linkages in Maxwell’s equations. I’ve highlighted them below:

I know this looks messy, but it’s actually not so complicated. The interactions between the electric and magnetic field are governed by equation (2) and (4), so equation (1) and (3) is just ‘statics’. Something needs to trigger it all, of course. I assume it’s an electric current (that’s the arrow marked by [0]).

Indeed, equation (4), i.e. c²∇×B = ∂E/∂t + j/ε₀, implies that a changing electric current – an accelerating electric charge, for instance – will cause the circulation of B to change. More specifically, we can write: ∂[c²∇×B]/∂t = ∂[j/ε₀]∂t. However, as the circulation of B changes, the magnetic field B itself must be changing. Hence, we have a non-zero time derivative of B (∂B/∂t ≠ 0). But, then, according to equation (2), i.e. ∇×E = –∂B/∂t, we’ll have some circulation of E. That’s the dynamics marked by the red arrows [1].

Now, assuming that ∂B/∂t is not constant (because that electric charge accelerates and decelerates, for example), the time derivative ∂E/∂t will be non-zero too (∂E/∂t ≠ 0). But so that feeds back into equation (4), according to which a changing electric field will cause the circulation of B to change. That’s the dynamics marked by the yellow arrows [2].

The ‘feedback loop’ is closed now: I’ve just explained how an electromagnetic field (or radiation) actually propagates through space. Below you can see one of the fancier animations you can find on the Web. The blue oscillation is supposed to represent the oscillating magnetic vector, while the red oscillation is supposed to represent the electric field vector. Note how the effect travels through space.

This is, of course, an extremely simplified view. To be precise, it assumes that the light wave (that’s what an electromagnetic wave actually is) is linearly (aka as plane) polarized, as the electric (and magnetic field) oscillate on a straight line. If we choose the direction of propagation as the z-axis of our reference frame, the electric field vector will oscillate in the xy-plane. In other words, the electric field will have an x- and a y-component, which we’ll denote as E_x and E_xrespectively, as shown in the diagrams below, which give various examples of linear polarization.

Light is, of course, not necessarily plane-polarized. The animation below shows circular polarization, which is a special case of the more general elliptical polarization condition.

The relativity of magnetic and electric fields

Allow me to make a small digression here, which has more to do with physics than with vector analysis. You’ll have noticed that we didn’t talk about the magnetic field vector anymore when discussing the polarization of light. Indeed, when discussing electromagnetic radiation, most – if not all – textbooks start by noting we have E and B vectors, but then proceed to discuss the E vector only. Where’s the magnetic field? We need to note two things here.

1. First, I need to remind you of the force on any electrically charged particle (and note we only have electric charge: there’s no such thing as a magnetic charge according to Maxwell’s third equation) consists of two components. Indeed, the total electromagnetic force (aka Lorentz force) on a charge q is:

F = q(E + v×B) = qE + q(v×B) = F_E + F_M

The velocity vector v is the velocity of the charge: if the charge is not moving, then there’s no magnetic force. The illustration below shows you the components of the vector cross product that, by now, you’re fully familiar with. Indeed, in my previous post, I gave you the expressions for the x, y and z coordinate of a cross product, but there’s a geometrical definition as well:

v×B = |v||B|sin(θ)n

The magnetic force F_M is q(v×B) = qv×B = q|v||B|sin(θ)n. The unit vector n determines the direction of the force, which is determined by that right-hand rule that, by now, you also are fully familiar with: it’s perpendicular to both v and B (cf. the two 90° angles in the illustration). Just to make sure, I’ve also added the right-hand rule illustration above: check it out, as it does involve a bit of arm-twisting in this case. 🙂

In any case, the point to note here is that there’s only one electromagnetic force on the particle. While we distinguish between an E and a B vector, the E and B vector depend on our reference frame. Huh? Yes. The velocity v is relative: we specify the magnetic field in a so-called inertial frame of reference here. If we’d be moving with the charge, the magnetic force would, quite simply, disappear, because we’d have a v equal to zero, so we’d have v×B = 0×B= 0. Of course, all other charges (i.e. all ‘stationary’ and ‘moving’ charges that were causing the field in the first place) would have different velocities as well and, hence, our E and B vector would look very different too: they would come in a ‘different mixture’, as Feynman puts it. [If you’d want to know in what mixture exactly, I’ll refer you Feynman: it’s a rather lengthy analysis (five rather dense pages, in fact), but I can warmly recommend it: in fact, you should go through it if only to test your knowledge at this point, I think.]

You’ll say: So what? That doesn’t answer the question above. Why do physicists leave out the magnetic field vector in all those illustrations?

You’re right. I haven’t answered the question. This first remark is more like a warning. Let me quote Feynman on it:

“Since electric and magnetic fields appear in different mixtures if we change our frame of reference, we must be careful about how we look at the fields E and B. […] The fields are our way of describing what goes on at a point in space. In particular, E and B tell us about the forces that will act on a moving particle. The question “What is the force on a charge from a moving magnetic field?” doesn’t mean anything precise. The force is given by the values of E and B at the charge, and the F = q(E + v×B) formula is not to be altered if the source of E or B is moving: it is the values of E and B that will be altered by the motion. Our mathematical description deals only with the fields as a function of x, y, z, and t with respect to some inertial frame.”

If you allow me, I’ll take this opportunity to insert another warning, one that’s quite specific to how we should interpret this concept of an electromagnetic wave. When we say that an electromagnetic wave ‘travels’ through space, we often tend to think of a wave traveling on a string: we’re smart enough to understand that what is traveling is not the string itself (or some part of the string) but the amplitude of the oscillation: it’s the vertical displacement (i.e. the movement that’s perpendicular to the direction of ‘travel’) that appears first at one place and then at the next and so on and so on. It’s in that sense, and in that sense only, that the wave ‘travels’. However, the problem with this comparison to a wave traveling on a string is that we tend to think that an electromagnetic wave also occupies some space in the directions that are perpendicular to the direction of travel (i.e. the x and y directions in those illustrations on polarization). Now that’s a huge misconception! The electromagnetic field is something physical, for sure, but the E and B vectors do not occupy any physical space in the x and y direction as they ‘travel’ along the z direction!

Let me conclude this digression with Feynman’s conclusion on all of this:

“If we choose another coordinate system, we find another mixture of E and B fields. However, electric and magnetic forces are part of one physical phenomenon—the electromagnetic interactions of particles. While the separation of this interaction into electric and magnetic parts depends very much on the reference frame chosen for the description, the complete electromagnetic description is invariant: electricity and magnetism taken together are consistent with Einstein’s relativity.”

2. You’ll say: I don’t give a damn about other reference frames. Answer the question. Why are magnetic fields left out of the analysis when discussing electromagnetic radiation?

The answer to that question is very mundane. When we know E (in one or the other reference frame), we also know B, and, while B is as ‘essential’ as E when analyzing how an electromagnetic wave propagates through space, the truth is that the magnitude of B is only a very tiny fraction of that of E.

Huh? Yes. That animation with these oscillating blue and red vectors is very misleading in this regard. Let me be precise here and give you the formulas:

I’ve analyzed these formulas in one of my other posts (see, for example, my first post on light and radiation), and so I won’t repeat myself too much here. However, let me recall the basics of it all. The e_R′ vector is a unit vector pointing in the apparent direction of the charge. When I say ‘apparent’, I mean that this unit vector is not pointing towards the present position of the charge, but at where is was a little while ago, because this ‘signal’ can only travel from the charge to where we are now at the same speed of the wave, i.e. at the speed of light c. That’s why we prime the (radial) vector R also (so we write R′ instead of R). So that unit vector wiggles up and down and, as the formula makes clear, it’s the second-order derivative of that movement which determines the electric field. That second-order derivative is the acceleration vector, and it can be substituted for the vertical component of the acceleration of the charge that caused the radiation in the first place but, again, I’ll refer you my post on that, as it’s not the topic we want to cover here.

What we do want to look at here, is that formula for B: it’s the cross product of that e_R′ vector (the minus sign just reverses the direction of the whole thing) and E divided by c. We also know that the E and e_R′ vectors are at right angles to each, so the sine factor (sinθ) is 1 (or –1) too. In other words, the magnitude of B is |E|/c = E/c, which is a very tiny fraction of E indeed (remember: c ≈ 3×10⁸).

So… Yes, for all practical purposes, B doesn’t matter all that much when analyzing electromagnetic radiation, and so that’s why physicists will note it but then proceed and look at E only when discussing radiation. Poor B! That being said, the magnetic force may be tiny, but it’s quite interesting. Just look at its direction! Huh? Why? What’s so interesting about it? I am not talking the direction of B here: I am talking the direction of the force. Oh… OK… Hmm… Well…

Let me spell it out. Take the force formula: F = q(E + v×B) = qE + q(v×B). When our electromagnetic wave hits something real (I mean anything real, like a wall, or some molecule of gas), it is likely to hit some electron, i.e. an actual electric charge. Hence, the electric and magnetic field should have some impact on it. Now, as we pointed here, the magnitude of the electric force will be the most important one – by far – and, hence, it’s the electric field that will ‘drive’ that charge and, in the process, give it some velocity v, as shown below. In what direction? Don’t ask stupid questions: look at the equation. F_E = qE, so the electric force will have the same direction as E.

But we’ve got a moving charge now and, therefore, the magnetic force comes into play as well! That force is F_M = q(v×B) and its direction is given by the right-hand rule: it’s the F above in the direction of the light beam itself. Admittedly, it’s a tiny force, as its magnitude is F = qvE/c only, but it’s there, and it’s what causes the so-called radiation pressure (or light pressure tout court). So, yes, you can start dreaming of fancy solar sailing ships (the illustration below shows one out of of Star Trek) but… Well… Good luck with it! The force is very tiny indeed and, of course, don’t forget there’s light coming from all directions in space!

Jokes aside, it’s a real and interesting effect indeed, but I won’t say much more about it. Just note that we are really talking the momentum of light here, and it’s a ‘real’ as any momentum. In an interesting analysis, Feynman calculates this momentum and, rather unsurprisingly (but please do check out how he calculates these things, as it’s quite interesting), the same 1/c factor comes into play once: the momentum (p) that’s being delivered when light hits something real is equal to 1/c of the energy that’s being absorbed. So, if we denote the energy by W (in order to not create confusion with the E symbol we’ve used already), we can write: p = W/c.

Now I can’t resist one more digression. We’re, obviously, fully in classical physics here and, hence, we shouldn’t mention anything quantum-mechanical here. That being said, you already know that, in quantum physics, we’ll look at light as a stream of photons, i.e. ‘light particles’ that also have energy and momentum. The formula for the energy of a photon is given by the Planck relation: E = hf. The h factor is Planck’s constant here – also quite tiny, as you know – and f is the light frequency of course. Oh – and I am switching back to the symbol E to denote energy, as it’s clear from the context I am no longer talking about the electric field here.

Now, you may or may not remember that relativity theory yields the following relations between momentum and energy:

E² – p²c² = m₀c⁴and/or pc = Ev/c

In this equations, m₀stands, obviously, for the rest mass of the particle, i.e. its mass at v = 0. Now, photons have zero rest mass, but their speed is c. Hence, both equations reduce to p = E/c, so that’s the same as what Feynman found out above: p = W/c.

Of course, you’ll say: that’s obvious. Well… No, it’s not obvious at all. We do find the same formula for the momentum of light (p) – which is great, of course – but so we find the same thing coming from very different necks parts of the woods. The formula for the (relativistic) momentum and energy of particles comes from a very classical analysis of particles – ‘real-life’ objects with mass, a very definite position in space and whatever other properties you’d associate with billiard balls – while that other p = W/c formula comes out of a very long and tedious analysis of light as an electromagnetic wave. The two analytical frameworks couldn’t differ much more, could they? Yet, we come to the same conclusion indeed.

Physics is wonderful. 🙂

So what’s left?

Lots, of course! For starters, it would be nice to show how these formulas for E and B with e_R′in them can be derived from Maxwell’s equations. There’s no obvious relation, is there? You’re right. Yet, they do come out of the very same equations. However, for the details, I have to refer you to Feynman’s Lectures once again – to the second Volume to be precise. Indeed, besides calculating scalar and vector potentials in various situations, a lot of what he writes there is about how to calculate these wave equations from Maxwell’s equations. But so that’s not the topic of this post really. It’s, quite simply, impossible to ‘summarize’ all those arguments and derivations in a single post. The objective here was to give you some idea of what vector analysis really is in physics, and I hope you got the gist of it, because that’s what needed to proceed. 🙂

The other thing I left out is much more relevant to vector calculus. It’s about that del operator (∇) again: you should note that it can be used in many more combinations. More in particular, it can be used in combinations involving second-order derivatives. Indeed, till now, we’ve limited ourselves to first-order derivatives only. I’ll spare you the details and just copy a table with some key results:

∇•(∇T) = div(grad T) = ∇•∇T = (∇•∇)T = ∇²T = ∂²T/∂x²+ ∂²T/∂y²+ ∂²T/∂z²= a scalar field
(∇•∇)h = ∇²h = a vector field
∇(∇•h) = grad(div h) = a vector field
∇×(∇×h) = curl(curl h) =∇(∇•h) – ∇²h
∇•(∇×h) = div(curl h) = 0 (always)
∇×(∇T) = curl(grad T) = 0 (always)

So we have yet another set of operators here: not less than six, to be precise. You may think that we can have some more, like (∇×∇), for example. But… No. A (∇×∇) operator doesn’t make sense. Just write it out and think about it. Perhaps you’ll see why. You can try to invent some more but, if you manage, you’ll see they won’t make sense either. The combinations that do make sense are listed above, all of them.

Now, while of these combinations make (some) sense, it’s obvious that some of these combinations are more useful than others. More in particular, the first operator, ∇², appears very often in physics and, hence, has a special name: it’s the Laplacian. As you can see, it’s the divergence of the gradient of a function.

Note that the Laplace operator (∇²) can be applied to both scalar as well as vector functions. If we operate with it on a vector, we’ll apply it to each component of the vector function. The Wikipedia article on the Laplace operator shows how and where it’s used in physics, and so I’ll refer to that if you’d want to know more. Below, I’ll just write out the operator itself, as well as how we apply it to a vector:

So that covers (1) and (2) above. What about the other ‘operators’?

Let me start at the bottom. Equations (5) and (6) are just what they are: two results that you can use in some mathematical argument or derivation. Equation (4) is… Well… Similar: it’s an identity that may or may not help one when doing some derivation.

What about (3), i.e. the gradient of the divergence of some vector function? Nothing special. As Feynman puts it: “It is a possible vector field, but there is nothing special to say about it. It’s just some vector field which may occasionally come up.”

So… That should conclude my little introduction to vector analysis, and so I’ll call it a day now. 🙂 I hope you enjoyed it.

Applied vector analysis (I)

Original post:

The relationship between math and physics is deep. When studying physics, one sometimes feels physics and math become one and the same. But they are not. In fact, eminent physicists such as Richard Feynman warn against emphasizing the math side of physics too much: “It is not because you understand the Maxwell equations mathematically inside out, that you understand physics inside out.”

We should never lose sight of the fact that all these equations and mathematical constructs represent physical realities. So the math is nothing but the ‘language’ in which we express physical reality and, as Feynman puts it, one (also) needs to develop a ‘physical’ – as opposed to a ‘mathematical’ – understanding of the equations. Now you’ll ask: what’s a ‘physical’ understanding? Well… Let me quote Feynman once again on that: “A physical understanding is a completely unmathematical, imprecise, and inexact thing, but absolutely necessary for a physicist.”

It’s rather surprising to hear that from him: this is a rather philosophical statement, indeed, and Feynman doesn’t like philosophy (see, for example, what he writes on the philosophical implications of the Uncertainty Principle). Indeed, while most physicists – or scientists in general, I’d say – will admit there is some value in a philosophy of science (that’s the branch of philosophy concerned with the foundations and methods of science), they will usually smile derisively when hearing someone talk about metaphysics. However, if metaphysics is the branch of philosophy that deals with ‘first principles’, then it’s obvious that the Standard Model (SM) in physics is, in fact, also some kind of ‘metaphysical’ model! Indeed, what everything is said and done, physicists assume those complex-valued wave functions are, somehow, ‘real’, but all they can ‘see’ (i.e. measure or verify by experiment) are (real-valued) probabilities: we can’t ‘see’ the probability amplitudes.

The only reason why we accept the SM theory is because its predictions agree so well with experiment. Very well indeed. The agreement between theory and experiment is most perfect in the so-called electromagnetic sector of the SM, but the results for the weak force (which I referred to as the ‘weird force’ in some of my posts) are very good too. For example, using CERN data, researchers could finally, recently, observe an extremely rare decay mode which, once again, confirms that the Standard Model, as complicated as it is, is the best we’ve got: just click on the link if you want to hear more about it. [And please do: stuff like this is quite readable and, hence, interesting.]

As this blog makes abundantly clear, it’s not easy to ‘summarize’ the Standard Model in a couple of sentences or in one simple diagram. In fact, I’d say that’s impossible. If there’s one or two diagrams sort of ‘covering’ it all, then it’s the two diagrams that you’ve seen ad nauseam already: (a) the overview of the three generations of matter, with the gauge bosons for the electromagnetic, strong and weak force respectively, as well as the Higgs boson, next to it, and (b) the overview of the various interactions between them. [And, yes, these two diagrams come from Wikipedia.]

I’ve said it before: the complexity of the Standard Model (it has not less than 61 ‘elementary’ particles taking into account that quarks and gluons come in various ‘colors’, and also including all antiparticles – which we have to include them in out count because they are just as ‘real’ as the particles), and the ‘weirdness’ of the weak force, plus a astonishing range of other ‘particularities’ (these ‘quantum numbers’ or ‘charges’ are really not easy to ‘understand’), do not make for a aesthetically pleasing theory but, let me repeat it again, it’s the best we’ve got. Hence, we may not ‘like’ it but, as Feynman puts it: “Whether we like or don’t like a theory is not the essential question. It is whether or not the theory gives predictions that agree with experiment.” (Feynman, QED – The Strange Theory of Light and Matter, p. 10)

It would be foolish to try to reduce the complexity of the Standard Model to a couple of sentences. That being said, when digging into the subject-matter of quantum mechanics over the past year, I actually got the feeling that, when everything is said and done, modern physics has quite a lot in common with Pythagoras’ ‘simple’ belief that mathematical concepts – and numbers in particular – might have greater ‘actuality’ than the reality they are supposed to describe. To put it crudely, the only ‘update’ to the Pythagorean model that’s needed is to replace Pythagoras’ numerological ideas by the equipartition theorem and quantum-mechanical wave functions, describing probability amplitudes that are represented by complex numbers. Indeed, complex numbers are numbers too, and Pythagoras would have reveled in their beauty. In fact, I can’t help thinking that, if he could have imagined them, he would surely have created a ‘religion’ around Euler’s formula, rather than around the tetrad. 🙂

In any case… Let’s leave the jokes and the silly comparisons aside, as that’s not what I want to write about in this post (if you want to read more about this, I’ll refer you another blog of mine). In this post, I want to present the basics of vector calculus, an understanding of which is absolutely essential in order to gain both a mathematical as well as a ‘physical’ understanding of what fields really are. So that’s classical mechanics once again. However, as I found out, one can’t study quantum mechanics without going through the required prerequisites. So let’s go for it.

Vectors in math and physics

What’s a vector? It may surprise you, but the term ‘vector’, in physics and in math, refers to more than a dozen different concepts, and that’s a major source of confusion for people like us–autodidacts. The term ‘vector’ refers to many different things indeed. The most common definitions are:

The term ‘vector’ often refers to a (one-dimensional) array of numbers. In that case, a vector is, quite simply, an element of Rⁿ, while the array will be referred to as an n-tuple. This definition can be generalized to also include arrays of alphanumerical values, or blob files, or any type of object really, but that’s a definition that’s more relevant for other sciences – most notably computer science. In math and physics, we usually limit ourselves to arrays of numbers. However, you should note that a ‘number’ may also be a complex number, and so we have real as well as complex vector spaces. The most straightforward example of a complex vector space is the set of complex numbers itself: C. In that case, the n-tuple is a ‘1-tuple’, aka as a singleton, but the element in it (i.e. a complex number) will have ‘two dimensions’, so to speak. [Just like the term ‘vector’, the term ‘dimension’ has various definitions in math and physics too, and so it may be quite confusing.] However, we can also have 2-tuples, 3-tuples or, more in general, n-tuples of complex numbers. In that case, the vector space is denoted by Cⁿ. I’ve written about vector spaces before and so I won’t say too much about this.
A vector can also be a point vector. In that case, it represents the position of a point in physical space – in one, two or three dimensions – in relation to some arbitrarily chosen origin (i.e. the zero point). As such, we’ll usually write it as x (in one dimension) or, in three dimensions, as (x, y, z). More generally, a point vector is often denoted by the bold-face symbol R. This definition is obviously ‘related’ to the definition above, but it’s not the same: we’re talking physical space here indeed, not some ‘mathematical’ space. Physical space can be curved, as you obviously know when you’re reading this blog, and I also wrote about that in the above-mentioned post, so you can re-visit that topic too if you want. Here, I should just mention one point which may or may not confuse you: while (two-dimensional) point vectors and complex numbers have a lot in common, they are not the same, and it’s important to understand both the similarities as well as the differences between both. For example, multiplying two vectors and multiplying two complex numbers is definitely not the same. I’ll come back to this.
A vector can also be a displacement vector: in that case, it will specify the change in position of a point relative to its previous position. Again, such displacement vectors may be one-, two-, or three-dimensional, depending on the space we’re envisaging, which may be one-dimensional (a simple line), two-dimensional (i.e. the plane), three-dimensional (i.e. three-dimensional space), or four-dimensional (i.e. space-time). A displacement vector is often denoted by s or ΔR, with the delta indicating we’re talking a a distance or a difference indeed: s = ΔR = R₂ – R₁ = (x₂ – x₁, y₂ – y₁, z₂ – z₁). That’s kids’ stuff, isn’t it?
A vector may also refer to a so-called four-vector: a four-vector obeys very specific transformation rules, referred to as the Lorentz transformation. In this regard, you’ve surely heard of space-time vectors, referred to as events, and noted as X = (ct, r), with r the spatial vector r = (x, y, z) and c the speed of light (which, in this case, is nothing but a proportionality constant ensuring that space and time are measured in compatible units). So we can also write X as X = (ct, x, y, z). However, there is another four-vector which you’ve probably also seen already (see, for example, my post on (special) Relativity Theory): P = (E/c, p), which relates energy and momentum in spacetime. Of course, spacetime can also be curved. In fact, Einstein’s (general) Relativity Theory is about the curvature of spacetime, not of ordinary space. But I should not write more about this here, as it’s about time I get back to the main story line of this post.
Finally, we also have vector operators, like the gradient vector ∇. Now that is what I want to write about in this post. Vector operators are also considered to be ‘vectors’ – to some extent, at least: we use them in a ‘vector products’, for example, as I will show below – but, because they are operators and, as such, “hungry for something to operate on”, they are obviously quite different from any of the ‘vectors’ I defined in point (1) to (4) above. [Feynman attributes this ‘hungry for something to operate on’ expression to the British mathematician Sir James Hopwood Jeans, who’s best known from the infamous Rayleigh-Jeans law, whose inconsistency with observations is known as the ultraviolet catastrophe or ‘black-body radiation problem’. But that’s a fairly useless digression so let me got in with it.]

In a text on physics, the term ‘vector’ may refer to any of the above but it’s often the second and third definition (point and/or displacement vectors) that will be implicit. As mentioned above, I want to write about the fifth ‘type’ of vector: vector operators. Now, the title of this post is ‘vector calculus’, and so you’ll immediately wonder why I say these vector operators may or may not be defined as vectors. Moreover, the fact of the matter is that these operators operate on yet another type of ‘vector’ – so that’s a sixth definition I need to introduce here: field vectors.

Now, funnily enough, the term ‘field vector’, while being the most obvious description of what it is, is actually not widely used: what I call a ‘field vector’ is often referred to as a gradient vector, and the vectors E and B are usually referred to as the electric or magnetic field, tout court. Indeed, if you google the terms ‘electromagnetic vector’ (or electric or magnetic vector), you will usually be redirected. However, when everything is said and done, E and B are vectors: they have a magnitude, and they have a direction. To be even more precise, while they depend on both space and time – so we can write E as E = E(x, y, z, t) and we have four independent variables here – they have three components: one of each direction in space, so we can write E as:

E = E(x, y, z, t) = [E_x, E_y, E_z] = [E_x(x, y, z, t), E_y(x, y, z, t), E_z(x, y, z, t)]

So, truth be told, vector calculus (aka vector analysis) in physics is about (vector) fields and (vector) operators,. While the ‘scene’ for these fields and operators is, obviously, physical space (or spacetime) and, hence, a vector space, it’s good to be clear on terminology and remind oneself that, in physics, vector calculus is not about mathematical vectors: it’s about real stuff. That’s why Feynman prefers a much longer term than vector calculus or vector analysis: he calls it differential calculus of vector fields which, indeed, is what it is – but I am sure you would not have bothered starting reading this post if I would have used that term too. 🙂

Now, this has probably become the longest introduction ever to a blog post, and so I had better get on with it. 🙂

Vector fields and scalar fields

Let’s dive straight into it. Vector fields like E and B behave like h, which is the symbol used in a number of textbooks for the heat flow in some body or block of material: E, B and h are all vector fields derived from a scalar field.

Huh? Scalar field? Aren’t we talking vectors? We are. If I say we can derive the vector field h (i.e. the heat flow) from a scalar field, I am basically saying that the relationship between h and the temperature T (i.e. the scalar field) is direct and very straightforward. Likewise, the relationship between E and the scalar field Φ is also direct and very straightforward.

[To be fully precise and complete, I should qualify the latter statement: it’s only true in electrostatics, i.e. when we’re considering charges that don’t move. When we have moving charges, magnetic effects come into play, and then we have a more complicated relationship between (i) two scalar fields, namely A (the magnetic potential – i.e. the ‘magnetic scalar field’) and Φ (the electrostatic potential, or ‘electric scalar field’), and (ii) two vector fields, namely B and E. The relationships between the two are then a bit more complicated than the relationship between T and h. However, the math involved is the same. In fact, the complication arises from the fact that magnetism is actually a relativistic effect. However, at this stage, this statement will only confuse you, and so I will write more about that in my next post.]

Let’s look at h and T. As you know, the temperature is a measure for energy. In a block of material, the temperature T will be a scalar: some real number that we can measure in Kelvin, Fahrenheit or Celsius but – whatever unit we use – any observer using the same unit will measure the same at any given point. That’s what distinguishes a ‘scalar’ quantity from ‘real numbers’ in general: a scalar field is something real. It represents something physical. A real number is just… Well… A real number, i.e. a mathematical concept only.

The same is true for a vector field: it is something real. As Feynman puts it: “It is not true that any three numbers form a vector [in physics]. It is true only if, when we rotate the coordinate system, the components of the vector transform among themselves in the correct way.” What’s the ‘correct way’? It’s a way that ensures that any observer using the same unit will measure the same at any given point.

How does it work?

In physics, we associate a point in space with physical realities, such as:

Temperature, the ‘height‘ of a body in a gravitational field, or the pressure distribution in a gas or a fluid, are all examples of scalar fields: they are just (real) numbers from a math point of view but, because they do represent a physical reality, these ‘numbers’ respect certain mathematical conditions: in practice, they will be a continuous or continuously differentiable function of position.
Heat flow (h), the velocity (v) of the molecules/atoms in a rotating object, or the electric field (E), are examples of vector fields. As mentioned above, the same condition applies: any observer using the same unit should measure the same at any given point.
Tensors, which represent, for example, stress or strain at some point in space (in various directions), or the curvature of space (or spacetime, to be fully correct) in the general theory of relativity.
Finally, there are also spinors, which are often defined as a “generalization of tensors using complex numbers instead of real numbers.” They are very relevant in quantum mechanics, it is said, but I don’t know enough about them to write about them, and so I won’t.

How do we derive a vector field, like h, from a scalar field (i.e. T in this case)? The two illustrations below (taken from Feynman’s Lectures) illustrate the ‘mechanics’ behind it: heat flows, obviously, from the hotter to the colder places. At this point, we need some definitions. Let’s start with the definition of the heat flow: the (magnitude of the) heat flow (h) is the amount of thermal energy (ΔJ) that passes, per unit time and per unit area, through an infinitesimal surface element at right angles to the direction of flow.

A vector has both a magnitude and a direction, as defined above, and, hence, if we define e_f as the unit vector in the direction of flow, we can write:

h = h·e_f = (ΔJ/Δa)·e_f

ΔJ stands for the thermal energy flowing through an area marked as Δa in the diagram above per unit time. So, if we incorporate the idea that the aspect of time is already taken care of, we can simplify the definition above, and just say that the heat flow is the flow of thermal energy per unit area. Simple trigonometry will then yield an equally simple formula for the heat flow through any surface Δa₂ (i.e. any surface that is not at right angles to the heat flow h):

ΔJ/Δa₂ = (ΔJ/Δa₁)cosθ = h·n

When I say ‘simple’, I must add that all is relative, of course, Frankly, I myself did not immediately understand why the heat flow through the Δa₁ and Δa₂ areas below must be the same. That’s why I added the blue square in the illustration above (which I took from Feynman’s Lectures): it’s the same area as Δa₁, but it shows more clearly – I hope! – why the heat flow through the two areas is the same indeed, especially in light of the fact that we are looking at infinitesimally small areas here (so we’re taking a limit here).

As for the cosine factor in the formula above, you should note that, in that ΔJ/Δa₂ = (ΔJ/Δa₁)cosθ = h·n equation, we have a dot product (aka as a scalar product) of two vectors: (1) h, the heat flow and (2) n, the unit vector that is normal (orthogonal) to the surface Δa₂. So let me remind you of the definition of the scalar (or dot) product of two vectors. It yields a (real) number:

A·B = |A||B|cosθ, with θ the angle between A and B

In this case, h·n = |h||n|cosθ = |h|·1·cosθ = |h|cosθ = h·cosθ. What we are saying here is that we get the component of the heat flow that’s perpendicular (or normal, as physicists and mathematicians seem to prefer to say) to the surface Δa₂ by taking the dot product of the heat flow h and the unit normal n. We’ll use this formula later, and so it’s good to take note of it here.

OK. Let’s get back to the lesson. The only thing that we need to do to prove that ΔJ/Δa₂ = (ΔJ/Δa₁)cosθ formula is show that Δa₂ = Δa₁/cosθ or, what amounts to the same, that Δa₁ = Δa₂cosθ. Now that is something you should be able to figure out yourself: it’s quite easy to show that the angle between h and n is equal to the angle between the surfaces Δa₁ and Δa₂. The rest is just plain triangle trigonometry.

For example, when the surfaces coincide, the angle θ will be zero and then h·n is just equal to |h|cosθ = |h| = h·1 = h = ΔJ/Δa₁. The other extreme is that orthogonal surfaces: in that case, the angle θ will be 90° and, hence, h·n = |h||n|cos(π/2) = |h|·1·0 = 0: there is no heat flow normal to the direction of heat flow.

OK. That’s clear enough. The point to note is that the vectors h and n represent physical entities and, therefore, they do not depend on our reference frame (except for the units we use to measure things). That allows us to define vector equations.

The ∇ (del) operator and the gradient

Let’s continue our example of temperature and heat flow. In a block of material, the temperature (T) will vary in the x, y and z direction and, hence, the partial derivatives ∂T/∂x, ∂T/∂y and ∂T/∂z make sense: they measure how the temperature varies with respect to position. Now, the remarkable thing is that the 3-tuple (∂T/∂x, ∂T/∂y, ∂T/∂z) is a physical vector itself: it is independent, indeed, of the reference frame (provided we measure stuff in the same unit) – so we can do a translation and/or a rotation of the coordinate axes and we get the same value. This means this set of three numbers is a vector indeed:

(∂T/∂x, ∂T/∂y, ∂T/∂z) = a vector

If you like to see a formal proof of this, I’ll refer you to Feynman once again – but I think the intuitive argument will do: if temperature and space are real, then the derivatives of temperature in regard to the x-, y- and z-directions should be equally real, isn’t it? Let’s go for the more intricate stuff now.

If we go from one point to another, in the x-, y- or z-direction, then we can define some (infinitesimally small) displacement vector ΔR = (Δx, Δy, Δz), and the difference in temperature between two nearby points (ΔT) will tend to the (total) differential of T – which we denote by ΔT – as the two point get closer and closer. Hence, we write:

ΔT = (∂T/∂x)Δx + (∂T/∂y)Δy + (∂T/∂z)Δz

The two equations above combine to yield:

ΔT = (∂T/∂x, ∂T/∂y, ∂T/∂z)(Δx, Δy, Δz) = ∇T·ΔR

In this equation, we used the ∇ (del) operator, i.e. the vector differential operator. It’s an operator like the differential operator ∂/∂x (i.e. the derivative) but, unlike the derivative, it returns not one but three values, i.e. a vector, which is usually referred to as the gradient, i.e. ∇T in this case. More in general, we can write ∇f(x, y, z), ∇ψ or ∇ followed by whatever symbol for the function we’re looking at.

In other words, the ∇ operator acts on a scalar-valued function (T), aka a scalar field, and yields a vector:

∇T = (∂T/∂x, ∂T/∂y, ∂T/∂z)

That’s why we write ∇ in bold-type too, just like the vector R. Indeed, using bold-type (instead of an arrow or so) is a convenient way to mark a vector, and the difference (in print) between ∇ and ∇ is subtle, but it’s there – and for a good reason as you can see!

[To be precise, I should add that we do not write all of the operators that return three components in bold-type. The most simple example is the common derivative ∂E/∂t = [∂E_x/∂t, ∂E_y/∂t, ∂E_z/∂t]. We have a lot of other possible combinations. Some make sense, and some don’t, like ∂h/∂y = [∂h_x/∂y, ∂h_y/∂y, ∂h_z/∂y], for example.]

If ∇T is a vector, what’s its direction? Think about it. […] The rate of change of T in the x-, y- and z-direction are the x-, y- and z-component of our ∇T vector respectively. In fact, the rate of change of T in any direction will be the component of the ∇T vector in that direction. Now, the magnitude of a vector component will always be smaller than the magnitude of the vector itself, except if it’s the component in the same direction as the vector, in which case the component is the vector. [If you have difficulty understanding this, read what I write once again, but very slowly and attentively.] Therefore, the direction of ∇T will be the direction in which the (instantaneous) rate of change of T is largest. In Feynman’s words: “The gradient of T has the direction of the steepest uphill slope in T.” Now, it should be quite obvious what direction that really is: it is the opposite direction of the heat flow h.

That’s all you need to know to understand our first real vector equation:

h = –κ∇T

Indeed, you don’t need too much math to understand this equation in the way we want to understand it, and that’s in some kind of ‘physical‘ way (as opposed to just the math side of it). Let me spell it out:

The direction of heat flow is opposite to the direction of the gradient vector ∇T. Hence, heat flows from higher to lower temperature (i.e. ‘downhill’), as we would expect, of course!). So that’s the minus sign.
The magnitude of h is proportional to the magnitude of the gradient ∇T, with the constant of proportionality equal to κ (kappa), which is called the thermal conductivity. Now, in case you wonder what this means (again: do go beyond the math, please!), remember that the heat flow is the flow of thermal energy per unit area (and per unit time, of course): |h| = h = ΔJ/Δa.

But… Yes? Why would it be proportional? Why don’t we have some exponential relation or something? Good question, but the answer is simple, and it’s rooted in physical reality – of course! The heat flow between two places – let’s call them 1 and 2 – is proportional to the temperature difference between those two places, so we have: ΔJ ∼ T₂ – T₁. In fact, that’s where the factor of proportionality comes in. If we imagine a very small slab of material (infinitesimally small, in fact) with two faces, parallel to the isothermals, with a surface area ΔA and a tiny distance Δs between them, we can write:

ΔJ = κ(T₂ – T₁)ΔA/Δs = ΔJ = κ·ΔT·ΔA/Δs ⇔ ΔJ/ΔA = κΔT/Δs

Now, we defined ΔJ/ΔA as the magnitude of h. As for its direction, it’s obviously perpendicular (not parallel) to the isothermals. Now, as Δs tends to zero, ΔT/Δs is nothing but the rate of change of T with position. We know it’s the maximum rate of change, because the position change is also perpendicular to the isotherms (if the faces are parallel, that tiny distance Δs is perpendicular). Hence, ΔT/Δs must be the magnitude of the gradient vector (∇T). As its direction is opposite to that of h, we can simply pop in a minus sign and switch from magnitudes to vectors to write what we wrote already: h = –κ∇T.

But let’s get back to the lesson. I think you ‘get’ all of the above. In fact, I should probably not have introduced that extra equation above (the ΔJ expression) and all the extra stuff (i.e. the ‘infinitesimally small slab’ explanation), as it probably only confuses you. So… What’s the point really? Well… Let’s look, once again, at that equation h = –κ∇T and let us generalize the result:

We have a scalar field here, the temperature T – but it could be any scalar field really!
When we have the ‘formula’ for the scalar field – it’s obviously some function T(x, y, z) – we can derive the heat flow h from it, i.e. a vector quantity, which has a property which we can vaguely refer to as ‘flow’.
We do so using this brand-new operator ∇. That’s a so-called vector differential operator aka the del operator. We just apply it to the scalar field and we’re done! The only thing left is to add some proportionality factor, but so that’s just because of our units. [In case you wonder about the symbol it self, ∇ is the so-called nabla symbol: the name comes from the Hebrew word for a harp, which has a similar shape indeed.]

This truly is a most remarkable result, and we’ll encounter the same equation elsewhere. For example, if the electric potential is Φ, then we can immediately calculate the electric field using the following formula:

E = –∇Φ

Indeed, the situation is entirely analogous from a mathematical point of view. For example, we have the same minus sign, so E also ‘flows’ from higher to lower potential. Where’s the factor of proportionality? Well… We don’t have one, as we assume that the units in which we measure E and Φ are ‘fully compatible’ (so don’t worry about them now). Of course, as mentioned above, this formula for E is only valid in electrostatics, i.e. when there are no moving charges. When moving charges are involved, we also have the magnetic force coming into play, and then equations become a bit more complicated. However, this extra complication does not fundamentally alter the logic involved, and I’ll come back to this so you see how it all nicely fits together.

Note: In case you feel I’ve skipped some of the ‘explanation’ of that vector equation h = –κ∇T… Well… You may be right. I feel that it’s enough to simply point out that ∇T is a vector with opposite direction to h, so that explains the minus sign in front of the ∇T factor. The only thing left to ‘explain’ then is the magnitude of h, but so that’s why we pop in that kappa factor (κ), and so we’re done, I think, in terms of ‘understanding’ this equation. But so that’s what I think indeed. Feynman offers a much more elaborate ‘explanation‘, and so you can check that out if you think my approach to it is a bit too much of a shortcut.

Interim summary

So far, we have only have shown two things really:

[I] The first thing to always remember is that h·n product: it gives us the component of ‘flow’ (per unit time and per unit area) of h perpendicular through any surface element Δa. Of course, this result is valid for any other vector field, or any vector for that matter: the scalar product of a vector and a unit vector will always yield the component of that vector in the direction of that unit vector. [But note the second vector needs to be a unit vector: it is not generally true that the dot product of one vector with another yields the component of the first vector in the direction of the second: there’s a scale factor that comes into play.]

Now, you should note that the term ‘component’ (of a vector) usually refers to a number (not to a vector) – and surely in this case, because we calculate it using a scalar product! I am just highlighting this because it did confuse me for quite a while. Why? Well… The concept of a ‘component’ of a vector is, obviously, intimately associated with the idea of ‘direction’: we always talk about the component in some direction, e.g. in the x-, y- or z-direction, or in the direction of any combination of x, y and z. Hence, I think it’s only natural to think of a ‘component’ as a vector in its own right. However, as I note here, we shouldn’t do that: a ‘component’ is just a magnitude, i.e. a number only. If we’d want to include the idea of direction, it’s simple: we can just multiply the component with the normal vector n once again, and then we have a vector quantity once again, instead of just a scalar. So then we just write (h·n)·n = (h·n)n. Simple, isn’t it? 🙂

[As I am smiling here, I should quickly say something about this dot (·) symbol: we use the same symbol here for (i) a product between scalars (i.e. real or complex numbers), like 3·4; (ii) a product between a scalar and a vector, like 3·v – but then I often omit the dot to simply write 3v; and, finally, (iii) a scalar product of two vectors, like h·n indeed. We should, perhaps, introduce some new symbol for multiplying numbers, like ∗ for example, but then I hope you’re smart enough to see from the context what’s going on really.]

Back to the lesson. Let me jot down the formula once again: h·n = |h||n|cosθ = h·cosθ. Hence, the number we get here is h (i.e. the amount of heat flow in the direction of flow) multiplied by cosθ, with θ the angle between (i) the surface we’re looking at (which, as mentioned above, is any surface really) and (ii) the surface that’s perpendicular to the direction of flow.

Hmm… […] The direction of flow? Let’s take a moment to think about what we’re saying here. Is there any particular or unique direction really? Heat flows in all directions from warmer to colder areas, and not just in one direction, doesn’t it? You’re right. Once again, the terminology may confuse you – which is yet another reason why math is so much better as a language to express physical ideas 🙂 – and so we should be precise: the direction of h is the direction in which the amount of heat flow (i.e. h·cosθ) is largest (hence, the angle θ is zero). As we pointed out above, that’s the direction in which the temperature T changes the fastest. In fact, as Feynman notes: “We can, if we wish, consider that this statement defines h.”

That brings me to the second thing you should – always and immediately – remember from all of that I’ve written above.

[II] If we write the infinitesimal (i.e. the differential) change in temperature (in whatever direction) as ΔT, then we know that

ΔT = (∂T/∂x, ∂T/∂y, ∂T/∂z)(Δx, Δy, Δz) = ∇T·ΔR

Now, what does this say really? ΔR is an (infinitesimal) displacement vector: ΔR = (Δx, Δy, Δz). Hence, it has some direction. To be clear: that can be any direction in space really. So that’s simple. What about the second factor in this dot product, i.e. that gradient vector ∇T?

The direction of the gradient (i.e. ∇T) is not just ‘any direction’: it’s the direction in which the rate of change of T is largest, and we know what direction that is: it’s the opposite direction of the heat flow h, as evidenced by the minus sign in our vector equations h = –κ∇T or E = –∇Φ. So, once again, we have a (scalar) product of two vectors here, ∇T·ΔR, which yields… Hmm… Good question. That ∇T·ΔR expression is very similar to that h·n expression above, but it’s not quite the same. It’s also a vector dot product – or a scalar product, in other words, but, unlike that n vector, the ΔR vector is not a unit vector: it’s an infinitesimally small displacement vector. So we do not get some ‘component’ of ∇T. More in general, you should note that the dot product of two vectors A and B does not, in general, yield the component of A in the direction of B, unless B is a unit vector – which, again, is not the case here. So if we don’t have that here, what do we have?

Let’s look at the (physical) geometry of the situation once again. Heat obviously flows in one direction only: from warmer to colder places – not in the opposite direction. Therefore, the θ in the h·n = h·cosθ expression varies from –90° to +90° only. Hence, the cosine factor (cosθ) is always positive. Always. Indeed, we do not have any right- or left-hand rule here to distinguish between the ‘front’ side and the ‘back’ side of our surface area. So when we’re looking at that h·n product, we should remember that that normal unit vector n is a unit vector that’s normal to the surface but which is oriented, generally, towards the direction of flow. Therefore, that h·n product will always yield some positive value, because θ varies from –90° to +90° only indeed.

When looking at that ΔT = ∇T·ΔR product, the situation is quite different: while ∇T has a very specific direction (I really mean unique) – which, as mentioned above is opposite to that of h – that ΔR vector can point in any direction – and then I mean literally any direction, including directions ‘uphill’. Likewise, it’s obvious that the temperature difference ΔT can be both positive or negative (or zero, when we’re moving on a isotherm itself). In fact, it’s rather obvious that, if we go in the direction of flow, we go from higher to lower temperatures and, hence, ΔT will, effectively, be negative: ΔT = T₂ – T₁ < 0, as shown below.

Now, because |∇T| and |ΔR| are absolute values (or magnitudes) of vectors, they are always positive (always!). Therefore, if ΔT has a minus sign, it will have to come from the cosine factor in the ΔT = ∇T·ΔR = |∇T|·|ΔR|·cosθ expression. [Again, if you wonder where this expression comes from: it’s just the definition of a vector dot product.] Therefore, ΔT and cosθ will have the same sign, always, and θ can have any value between –180° to +180°. In other words, we’re effectively looking at the full circle here. To make a long story short, we can write the following:

ΔT = |∇T|·|ΔR|·cosθ = |∇T|·ΔR·cosθ ⇔ ΔT/ΔR = |∇T|cosθ

As you can see, θ is the angle between ∇T and ΔR here and, as mentioned above, it can take on any value – well… Any value between –180° to +180°, that is. ΔT/ΔR is, obviously, the rate of change of T in the direction of ΔR and, from the expression above, we can see it is equal to the component of ∇T in the direction of ΔR:

ΔT/ΔR = |∇T|cosθ

So we have a negative component here? Yes. The rate of change is negative and, therefore, we have a negative component. Indeed, any vector has components in all directions, including directions that point away from it. However, in the directions that point away from it, the component will be negative. More in general, we have the following interesting result: the rate of change of a scalar field ψ in the direction of a (small) displacement ΔR is the component of the gradient ∇ψ along that displacement. We write that result as:

Δψ/ΔR = |∇ψ|cosθ

[Note the (not so) subtle difference between ΔR (i.e. a vector) and ΔR (some real number). It’s quite important. ]

We’ve said a lot of (not so) interesting things here, but we still haven’t answered the original question: what’s ∇T·ΔR? Well… We can’t say more than what we said already: it’s equal to ΔT, which is a differential: ΔT = (∂T/∂x)Δx + (∂T/∂y)Δy + (∂T/∂z)Δz. A differential and a derivative (i.e. a rate of change) are not the same, but they are obviously closely related, as evidenced from the equations above: the rate of change is the change per unit distance. [Likewise, note that |∇ψ|cosθ is just a product of two real numbers, while ∇T·ΔR is a vector dot product, i.e. a (scalar) product of two vectors!]

In any case, this is enough of a recapitulation. In fact, this ‘interim summary’ has become longer than the preceding text! We’re now ready to discuss what I’ll call the First Theorem of vector calculus in physics. Of course, never mind the term: what’s first or second or third doesn’t matter really: you’ll need all of the theorems below to understand vector calculus.

The First Theorem

Let’s assume we have some scalar field ψ in space: ψ might be the temperature, but it could be any scalar field really. Now, if we go from one point (1) to another (2) in space, as shown below, we’ll follow some arbitrary path, which is denoted by the curve Γ in the illustrations below. Each point along the curve can then be associated with a gradient ∇ψ (think of the h = –κ∇T and E = –∇Φ expressions above if you’d want examples). Its tangential component is obviously equal to (∇ψ)_t·Δs= ∇ψ·Δs. [Please note, once again, the subtle difference between Δs(with the s in bold-face) and Δs: Δsis a vector, and Δsis its magnitude.]

As shown in the illustrations above, we can mark off the curve at a number of points (a, b, c, etcetera) and join these points by straight-line segments Δs_i. Now let’s consider the first line segment, i.e. Δs₁. It’s obvious that the change in ψ from point 1 to point a is equal to Δψ₁= ψ(a) – ψ(1). Now, we have that general Δψ = (∂ψ/∂x, ∂ψ/∂y, ∂ψ/∂z)(Δx, Δy, Δz) = ∇ψ·Δs equation. [If you find it difficult to interpret what I am writing here, just substitute ψ for T and Δs for ΔR.] So we can write:

Δψ₁= ψ(a) – ψ(1) = (∇ψ)₁·Δs₁

Likewise, we can write:

ψ(b) – ψ(a) = (∇ψ)₂·Δs₁

In these expressions, (∇ψ)₁and (∇ψ)₂mean the gradient evaluated at segment Δs₁ and point Δs₂ respectively, not at point 1 and 2 – obviously. Now, if we add the two equations above, we get:

ψ(b) – ψ(1) = (∇ψ)₁·Δs₁+ (∇ψ)₂·Δs₁

To make a long story short, we can keep adding such terms to get:

ψ(2) – ψ(1) = ∑(∇ψ)_i·Δs_i

We can add more and more segments and, hence, take a limit: as Δs_i tends to zero, ∑ becomes a sum of an infinite number of terms – which we denote using the integral sign ∫ – in which ds is – quite simply – just the infinitesimally small displacement vector. In other words, we get the following line integral along that curve Γ:

This is a gem, and our First Theorem indeed. It’s a remarkable result, especially taking into account the fact that the path doesn’t matter: we could have chosen any curve Γ indeed, and the result would be the same. So we have:

You’ll say: so what? What do we do with this? Well… Nothing much for the moment, but we’ll need this result later. So I’d say: just hang in there, and note this is the first significant use of our del operator in a mathematical expression that you’ll encounter very often in physics. So just let it sink in, and allow me to proceed with the rest of the story.

Before doing so, however, I should note that even Feynman sins when trying to explain this theorem in a more ‘intuitive’ way. Indeed, in his Lecture on the topic, he writes the following: “Since the gradient represents the rate of change of a field quantity, if we integrate that rate of change, we should get the total change.” Now, from that Δψ/ΔR = |∇ψ|cosθ formula, it’s obvious that the gradient is the rate of change in a specific direction only. To be precise, in this particular case – with the field quantity ψ equal to the temperature T – it’s the direction in which T changes the fastest.

You should also note that the integral above is not the type of integral you known from high school. Indeed, it’s not of the rather straightforward ∫f(x)dx type, with f(x) the integrand and dx the variable of integration. That type of integral, we knew how to solve. A line integral is quite different. Look at it carefully: we have a vector dot product after the ∫ sign. So, unlike what Feynman suggests, it’s not just a matter of “integrating the rate of change.”

Now, I’ll refer you to Wikipedia for a good discussion of what a line integral really is, but I can’t resist the temptation to copy the animation in that article, because it’s very well made indeed. While it shows that we can think of a line integral as the two- or three-dimensional equivalent of the standard type of integral we learned to solve in high school (you’ll remember the solution was also the area under the graph of the function that had to be integrated), the way to go about it is quite different. Solving them will, in general, involve some so-called parametrization of the curve C.

However, this post is becoming way too long and, hence, I really need to move on now.

Operations with ∇: divergence and curl

You may think we’ve covered a lot of ground already, and we did. At the same time, everything I wrote above is actually just the start of it. I emphasized the physics of the situation so far. Let me now turn to the math involved. Let’s start by dissociating the del operator from the scalar field, so we just write:

∇ = (∂/∂x, ∂/∂y, ∂/∂z)

This doesn’t mean anything, you’ll say, because the operator has nothing to operate on. And, yes, you’re right. However, in math, it doesn’t matter: we can combine this ‘meaningless’ operator (which looks like a vector, because it has three components) with something else. For example, we can do a vector dot product:

∇·(a vector)

As mentioned above, we can ‘do’ this product because ∇ has three components, so it’s a ‘vector’ too (although I find such name-giving quite confusing), and so we just need to make sure that the vector we’re operating on has three components too. To continue with our heat flow example, we can write, for example:

∇·h = (∂/∂x, ∂/∂y, ∂/∂z)·(h_x, h_y, h_z) = ∂h_x/∂x + ∂h_y/∂y, ∂h_z/∂z

This del operator followed by a dot, and acting on a vector – i.e. ∇·(vector) – is, in fact, a new operator. Note that we use two existing symbols, the del (∇) and the dot (·), but it’s one operator really. [Inventing a new symbol for it would not be wise, because we’d forget where it comes from and, hence, probably scratch our head when we’d see it.] It’s referred to as a vector operator, just like the del operator, but don’t worry about the terminology here because, once again, the terminology here might confuse you. Indeed, our del operator acted on a scalar to yield a vector, and now it’s the other way around: we have an operator acting on a vector to return a scalar. In a few minutes, we’ll define yet another operator acting on a vector to return a vector. Now, all of these operators are so-called vector operators, not because there’s some vector involved, but because they all involve the del operator. It’s that simple. So there’s no such thing as a scalar operator. 🙂 But let me get back to the main line of the story. This ∇· operator is quite important in physics, and so it has a name (and an abbreviated notation) of its own:

∇·h = div h = the divergence of h

The physical significance of the divergence is related to the so-called flux of a vector field: it measures the magnitude of a field’s source or sink at a given point. Continuing our example with temperature, consider air as it is heated or cooled. The relevant vector field is now the velocity of the moving air at a point. If air is heated in a particular region, it will expand in all directions such that the velocity field points outward from that region. Therefore the divergence of the velocity field in that region would have a positive value, as the region is a source. If the air cools and contracts, the divergence has a negative value, as the region is a sink.

A less intuitive but more accurate definition is the following: the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point.

Phew! That sounds more serious, doesn’t it? We’ll come back to this definition when we’re ready to define the concept of flux somewhat accurately. For now, just note two of Maxwell’s famous equations involve the divergence operator:

∇·E = ρ/ε₀ and ∇·B = 0

In my previous post, I gave a verbal description of those two equations:

The flux of E through a closed surface = (the net charge inside)/ε₀
The flux of B through a closed surface = zero

The first equation basically says that electric charges cause an electric field. The second equation basically says there is no such thing as a magnetic charge: the magnetic force only appears when charges are moving and/or when electric fields are changing. Note that we’re talking closed surface here, so they define a volume indeed. We can also look at the flux through a non-closed surface (and we’ll do that shortly) but, in the context of Maxwell’s equations, we’re talking volumes and, hence, closed surfaces.

Let me quickly throw in some remarks on the units in which we measure stuff. Electric field strength (so the unit we use to measure the magnitude of E) is measured in Newton per Coulomb, so force divided by charge. That makes sense, because E is defined as the force on the unit charge: E = F/q, and so the unit is N/C. Please do think about why we have q in the denominator: if we’d have the same force on an electric charge that is twice as big, then we’d have a field strength that’s only half, so we have an inverse proportionality here. Conversely, if we’d have twice the force on the same electric charge, the field strength would also double.

Now, flux and field strength are obviously related, but not the same. The flux is obviously proportional to the field strength (expressed in N/C), but then we also know it’s some number expressed per unit area. Hence, you might think that the unit of flux is field strength per square meter, i.e. N/C/m². It’s not. It’s a stupid mistake, but one that is commonly made. Flux is expressed in N/C times m², so that’s the product (N/C)·m²= (N·m/C)·m= (J/C)·m. Why is that? Think about the common graphical representation of a field: we just draw lines, all tangent to the direction of the field vector at every point, and the density of the lines (i.e. the number of lines per unit area) represents the magnitude of our electric field vector. Now, the flux through some area is the number of lines we count in that area. Hence, if you double the area, you should get twice the flux. Halve the area, and you should get half the flux. So we have a direct proportionality here. In fact, assuming the electric field is uniform, we can write the (electric) flux as the product of the field strength E and the (vector) area S, so we write Φ_E = E·S = E·S·cosθ.

Huh? Yes. The origin of the mistake is that we, somehow, think the ‘per unit area’ qualification comes with the flux. It doesn’t: it’s in the idea of field strength itself. Indeed, an alternative to the presentation above is just to draw arrows representing the same field strength, as illustrated below. However, instead of drawing more arrows (of some standard length) to represent increasing field strength, we’d just draw longer arrows—not more of them. So then the idea of the number of lines per unit area is no longer valid.

[…] OK. I realize I am probably just confusing you here. Just one more thing, perhaps. We also have magnetic flux, denoted as Φ_B, and it’s defined in the same way: Φ_B = B·S = B·S·cosθ. However, because the unit of magnetic field strength is different, the unit of magnetic flux is different too. It’s the weber, and I’ll let you look up its definition yourself. 🙂 Note that it’s a bit of a different beast, because the magnetic force is a bit of a different beast. 🙂

So let’s get back to our operators. You’ll anticipate the second new operator now, because that’s the one that appears in the other two equations in Maxwell’s set of equations. It’s the cross product:

∇×E = (∂/∂x, ∂/∂y, ∂/∂z)×(E_x, E_y, E_z) = … What?

Well… The cross product is not as straightforward to write down as the dot product. We get a vector indeed, not a scalar, and its three components are:

(∇×E)_z = ∇_xE_y– ∇_yE_x= ∂E_y/∂x – ∂E_x/∂y

(∇×E)_x = ∇_yE_z– ∇_zE_y= ∂E_z/∂y – ∂E_y/∂z

(∇×E)_y = ∇_zE_x– ∇_xE_z= ∂E_x/∂z – ∂E_z/∂x

I know this looks pretty monstrous, but so that’s how cross products work. Please do check it out: you have to play with the order of the x, y and z subscripts. I gave the geometric formula for a dot product above, so I should also give you the same for a cross product:

A×B = |A||B|sin(θ)n

In this formula, we once again have θ, the angle between A and B, but note that, this time around, it’s the sine, not the cosine, that pops up when calculating the magnitude of this vector. In addition, we have n at the end: n is a unit vector at right angles to both A and B. It’s what makes the cross product a vector. Indeed, as you can see, multiplying by n will not alter the magnitude (|A||B|sinθ) of this product, but it gives the whole thing a direction, so we get a new vector indeed. Of course, we have two unit vectors at right angles to A and B, and so we need a rule to choose one of these: the direction of the n vector we want is given by that right-hand rule which we encountered a couple of times already.

Again, it’s two symbols but one operator really, and we also have a special name (and notation) for it:

∇×h = curl h = the curl of h

The curl is, just like the divergence, a so-called vector operator but, as mentioned above, that’s just because it involves the del operator. Just note that it acts on a vector and that its result is a vector too. What’s the geometric interpretation of the curl? Well… It’s a bit hard to describe that but let’s try. The curl describes the ‘rotation’ or ‘circulation’ of a vector field:

The direction of the curl is the axis of rotation, as determined by the right-hand rule.
The magnitude of the curl is the magnitude of rotation.

I know. This is pretty abstract, and I’ll probably have to come back to it in another post. Let’s first ask some basic question: should we associate some unit with the curl? In fact, when you google, you’ll find lots of units used in electromagnetic theory (like the weber, for example), but nothing for circulation. I am not sure why, because if flux is related to some density, the idea of curl (or circulation) is pretty much the same. It’s just that it isn’t used much in actual engineering problems, and surely not those you may have encountered in your high school physics course!

In any case, just note we defined three new operators in this ‘introduction’ to vector calculus:

∇T = grad T = a vector
∇·h = div h = a scalar
∇×h = curl h = a vector

That’s all. It’s all we need to understand Maxwell’s famous equations:

Huh? Hmm… You’re right: understanding the symbols, to some extent, doesn’t mean we ‘understand’ these equations. What does it mean to ‘understand’ an equation? Let me quote Feynman on that: “What it means really to understand an equation—that is, in more than a strictly mathematical sense—was described by Dirac. He said: “I understand what an equation means if I have a way of figuring out the characteristics of its solution without actually solving it.” So if we have a way of knowing what should happen in given circumstances without actually solving the equations, then we “understand” the equations, as applied to these circumstances.”

We’re surely not there yet. In fact, I doubt we’ll ever reach Dirac’s understanding of Maxwell’s equations. But let’s do what we can.

In order to ‘understand’ the equations above in a more ‘physical’ way, let’s explore the concepts of divergence and curl somewhat more. We said the divergence was related to the ‘flux’ of a vector field, and the curl was related to its ‘circulation’. In my previous post, I had already illustrated those two concepts copying the following diagrams from Feynman’s Lectures:

flux = (average normal component)·(surface area)

So that’s the flux (through a non-closed surface).

To illustrate the concept of circulation, we have not one but three diagrams, shown below. Diagram (a) gives us the vector field, such as the velocity field in a liquid. In diagram (b), we imagine a tube (of uniform cross section) that follows some arbitrary closed curve. Finally, in diagram (c), we imagine we’d suddenly freeze the liquid everywhere except inside the tube. Then the liquid in the tube would circulate as shown in (c), and so that’s the concept of circulation.

We have a similar formula as for the flux:

circulation = (the average tangential component)·(the distance around)

In both formulas (flux and circulation), we have a product of two scalars: (i) the average normal component and the average tangential component (for the flux and circulation respectively) and (ii) the surface area and the distance around (again, for the flux and circulation respectively). So we get a scalar as a result. Does that make sense? When we related the concept of flux to the divergence of a vector field, we said that the flux would have a positive value if the region is a source, and a negative value if the region is a sink. So we have a number here (otherwise we wouldn’t be talking ‘positive’ or ‘negative’ values). So that’s OK. But are we talking about the same number? Yes. I’ll show they are the same in a few minutes.

But what about circulation? When we related the concept of circulation of the curl of a vector field, we introduced a vector cross product, so that yields a vector, not a scalar. So what’s the relation between that vector and the number we get when multiplying the ‘average tangential component’ and the ‘distance around’. The answer requires some more mathematical analysis, and I’ll give you what you need in a minute. Let me first make a remark about conventions here.

From what I write above, you see that we use a plus or minus sign for the flux to indicate the direction of flow: the flux has a positive value if the region is a source, and a negative value if the region is a sink. Now, why don’t we do the same for circulation? We said the curl is a vector, and its direction is the axis of rotation as determined by the right-hand rule. Why do we need a vector here? Why can’t we have a scalar taking on positive or negative values, just like we do for the flux?

The intuitive answer to this question (i.e. the ‘non-mathematical’ or ‘physical’ explanation, I’d say) is the following. Although we can calculate the flux through a non-closed surface, from a mathematical point of view, flux is effectively being defined by referring to the infinitesimal volume around some point and, therefore, we can easily, and unambiguously, determine whether we’re inside or outside of that volume. Therefore, the concepts of positive and negative values make sense, as we can define them referring to some unique reference point, which is either inside or outside of the region.

When talking circulation, however, we’re talking about some curve in space. Now it’s not so easy to find some unique reference point. We may say that we are looking at some curve from some point ‘in front of’ that curve, but some other person whose position, from our point of view, would be ‘behind’ the curve, would not agree with our definition of ‘in front of’: in fact, his definition would be exactly the opposite of ours. In short, because of the geometry of the situation involved, our convention in regard to the ‘sign’ of circulation (positive or negative) becomes somewhat more complicated. It’s no longer a simple matter of ‘inward’ or ‘outward’ flow: we need something like a ‘right-hand rule’ indeed. [We could, of course, also adopt a left-hand rule but, by now, you know that, in physics, there’s not much use for a left hand. :-)]

That also ‘explains’ why the vector cross product is non-commutative: A×B ≠ B×A. To be fully precise, A×B and B×A have the same magnitude but opposite direction: A×B = |A||B|sin(θ)n = –|A||B|sin(θ)(–n) = –(B×A) = –B×A. The dot product, on the other hand, is fully commutative: A·B = B·A.

In fact, the concept of circulation is very much related to the concept of angular momentum which, as you’ll remember from a previous post, also involves a vector cross product.

[…]

I’ve confused you too much already. The only way out is the full mathematical treatment. So let’s go for that.

Flux

Some of the confusion as to what flux actually means in electromagnetism is probably caused by the fact that the illustration above is not a closed surface and, from my previous post, you should remember that Maxwell’s first and third equation define the flux of E and B through closed surfaces. It’s not that the formula above for the flux through a non-closed surface is wrong: it’s just that, in electromagnetism, we usually talk about the flux through a closed surface.

A closed surface has no boundary. In contrast, the surface area above does have a clear boundary and, hence, it’s not a closed surface. A sphere is an example of a closed surface. A cube is an example as well. In fact, an infinitesimally small cube is what’s used to prove a very convenient theorem, referred to as Gauss’ Theorem. We will not prove it here, but just try to make sure you ‘understand’ what it says.

Suppose we have some vector field C and that we have some closed surface S – a sphere, for example, but it may also be some very irregular volume. Its shape doesn’t matter: the only requirement is that it’s defined by a closed surface. Let’s then denote the volume that’s enclosed by this surface by V. Now, the flux through some (infinitesimal) surface element da will, effectively, be given by that formula above:

flux = (average normal component)·(surface area)

What’s the average normal component in this case? It’s given by that ΔJ/Δa₂ = (ΔJ/Δa₁)cosθ = h·n formula, except that we just need to substitute h for C here, so we have C·n instead of h·n. To get the flux through the closed surface S, we just need to add all the contributions. Adding those contributions amounts to taking the following surface integral:

Now, I talked about Gauss’ Theorem above, and I said I would not prove it, but this is what Gauss’ Theorem says:

Huh? Don’t panic. Just try to ‘read’ what’s written here. From all that I’ve said so far, you should ‘understand’ the surface integral on the left-hand side. So that should be OK. Let’s now look at the right-hand side. The right-hand side uses the divergence operator which I introduced above: ∇·(vector). In this case, ∇·C. That’s a scalar, as we know, and it represents the outward flux from an infinitesimally small cube inside the surface indeed. The volume integral on the right-hand side adds all of the fluxes out of each part (think of it as zillions of infinitesimally small cubes) of the volume V that is enclosed by the (closed) surface S. So that’s what Gauss’ Theorem is all about. In words, we can state Gauss’ Theorem as follows:

Gauss’ Theorem: The (surface) integral of the normal component of a vector (field) over a closed surface is the (volume) integral of the divergence of the vector over the volume enclosed by the surface.

Again, I said I would not prove Gauss’ Theorem, but its proof is actually quite intuitive: to calculate the flux out of a large volume, we can ‘cut it up’ in smaller volumes, and then calculate the flux out of these volumes. If we add it up, we’ll get the total flux. In any case, I’ll refer you to Feynman in case you’d want to see how it goes exactly. So far, I did what I promised to do, and that’s to relate the formula for flux (i.e. that (average normal component)·(surface area) formula) to the divergence operator. Let’s now do the same for the curl.

Curl

For non-native English speakers (like me), it’s always good to have a look at the common-sense definition of ‘curl’: as a verb (to curl), it means ‘to form or cause to form into a curved or spiral shape’. As a noun (e.g. a curl of hair), it means ‘something having a spiral or inwardly curved form’. It’s clear that, while not the same, we can indeed relate this common-sense definition to the concept of circulation that we introduced above:

circulation = (the average tangential component)·(the distance around)

So that’s the (scalar) product we already mentioned above. How do we relate it to that curl operator?

Patience, please ! The illustration below shows what we actually have to do to calculate the circulation around some loop Γ: we take an infinite number of vector dot products C·ds. Take a careful look at the notation here: I use bold-face for s and, hence, ds is some little vector indeed. Going to the limit, ds becomes a differential indeed. The fact that we’re talking a vector dot product here ensures that only the tangential component of C ‘enters the equation’, so to speak. I’ll come back to that in a moment. Just have a good look at the illustration first.

Such infinite sum of vector dot products C·ds is, once again, an integral. It’s another ‘special’ integral, in fact. To be precise, it’s a line integral. Moreover, it’s not just any line integral: we have to go all around the (closed) loop to take it. We cannot stop somewhere halfway. That’s why Feynman writes it with a little loop (ο) through the integral sign (∫):

Note the subtle difference between the two products in the integrands of the integrals above: C_tds versus C·ds. The first product is just a product of two scalars, while the second is a dot product of two vectors. Just check it out using the definition of a dot product (A·B = |A||B|cosθ) and substitute A and B by C and ds respectively, noting that the tangential component C_t equals C times cosθ indeed.

Now, once again, we want to relate this integral with that dot product inside to one of those vector operators we introduced above. In this case, we’ll relate the circulation with the curl operator. The analysis involves infinitesimal squares (as opposed to those infinitesimal cubes we introduced above), and the result is what is referred to as Stokes’ Theorem. I’ll just write it down:

Again, the integral on the left was explained above: it’s a line integral taking around the full loop Γ. As for the integral on the right-hand side, that’s a surface integral once again but, instead of a div operator, we have the curl operator inside and, moreover, the integrand is the normal component of the curl only. Now, remembering that we can always find the normal component of a vector (i.e. the component that’s normal to the surface) by taking the dot product of that vector and the unit normal vector (n), we can write Stokes’s Theorem also as:

That doesn’t look any ‘nicer’, but it’s the form in which you’ll usually see it. Once again, I will not give you any formal proof of this. Indeed, if you’d want to see how it goes, I’ll just refer you to Feynman’s Lectures. However, the philosophy behind is the same. The first step is to prove that we can break up the surface bounded by the loop Γ into a number of smaller areas, and that the circulation around Γ will be equal to the sum of the circulations around the little loops. The idea is illustrated below:

Of course, we then go to the limit and cut up the surface into an infinite number of infinitesimally small squares. The next step in the proof then shows that the circulation of C around an infinitesimal square is, indeed, (i) the component of the curl of C normal to the surface enclosed by that square multiplied by (ii) the area of that (infinitesimal) square. The diagram and formula below do not give you the proof but just illustrate the idea:

OK, you’ll say, so what? Well… Nothing much. I think you have enough to digest as for now. It probably looks very daunting, but so that’s all we need to know – for the moment that is – to arrive at a better ‘physical’ understanding of Maxwell’s famous equations. I’ll come back to them in my next post. Before proceeding to the summary of this whole post, let me just write down Stokes’ Theorem in words:

Stokes’ Theorem: The line integral of the tangential component of a vector (field) around a closed loop is equal to the surface integral of the normal component of the curl of that vector over any surface which is bounded by the loop.

Summary

We’ve defined three so-called vector operators, which we’ll use very often in physics:

∇T = grad T = a vector
∇·h = div h = a scalar
∇×h = curl h = a vector

Moreover, we also explained three important theorems, which we’ll use as least as much:

[1] The First Theorem:

[2] Gauss Theorem:

[3] Stokes Theorem:

As said, we’ll come back to them in my next post. As for now, just try to familiarize yourself with these div and curl operators. Try to ‘understand’ them as good as you can. Don’t look at them as just some weird mathematical definition: try to understand them in a ‘physical’ way, i.e. in a ‘completely unmathematical, imprecise, and inexact way’, remembering that’s what it takes to understand to truly understand physics. 🙂

Back to tedious stuff: an introduction to electromagnetism

Original post:

It seems I skipped too many chapters in Feynman’s second volume of Lectures (on electromagnetism) and so I have to return to that before getting back to quantum physics. So let me just do that in the next couple of posts. I’ll have to start with the basics: Maxwell’s equations.

Indeed, electromagnetic phenomena are described by a set of four equations known as Maxwell’s equations. They relate two fields: the electric field (E) and the magnetic field (B). The electric field appears when we have electric charges: positive (e.g. protons or positively charged ions) or negative (e.g. electrons or negatively charged ions). That’s obvious.

In contrast, there is no such thing as ‘magnetic charges’. The magnetic field appears only when the electric field changes, or when charges move. In turn, the change in the magnetic field causes an electric field, and that’s how electromagnetic radiation basically works: a changing electric field causes a magnetic field, and the build-up of that magnetic field (so that’s a changing magnetic field) causes a build-up of an electric field, and so on and so on.

OK. That’s obvious too. But how does it work exactly? Before explaining this, I need to point out some more ‘obvious’ things:

1. From Maxwell’s equations, we can calculate the magnitude of E and B. Indeed, a specific functional form for E and B is what we get when we solve Maxwell’s set of equations, and we’ll jot down that solution in a moment–even if I am afraid you will shake your head when you see it. The point to note is that what we get as a solution for E and B is a solution in a particular frame of reference only: if we switch to another reference frame, E and B will look different.

Huh? Yes. According to the principle of relativity, we cannot say which charges are ‘stationary’ and which charges are ‘moving’ in any absolute sense: it all depends on our frame our reference.

But… Yes? Then if we put an electric charge in these fields, the force on it will also be different?

Yes. Forces also look different when moving from one reference to another.

But… Yes? The physical effect surely has to be the same, regardless of the reference frame?

Yes. The point is that, if we look at an electric charge q moving along a current-carrying wire in a coordinate system at rest with respect to the wire, with the same velocity (v₀) as the conduction electrons (v), then the whole force on the electric charge will be ‘magnetic’: F = qv₀×B and E = 0. Now, if we’re looking at the same situation from a frame of reference that is moving with q, then our charge is at rest, and so there can be no magnetic force on it. Hence, the force on it must come from an electric field! But what produces the electric field? Our current-carrying wire is supposed to be neutral!

Well… It turns out that our ‘neutral’ wire appears to be charged when moving. We’ll explain – in very much detail – why this is so later. Now, you should just note that “we should not attach too much reality to E and B, because they appear in different ‘mixtures’ in different coordinate systems”, as Feynman puts it. In fact, you may or may not heard that magnetism is actually nothing but a “relativistic effect” of electricity. Well… That’s true, but we’ll also explain how that works later only. Let’s not jump the gun.

2. The remark above is related to the other ‘obvious’ thing I wanted to say before presenting Maxwell’s equations: fields are very useful to describe what’s going on but, when everything is said and done, what we really want to know is what force will be acting on a charge, because that’s what’s going to tell us how that charge is going to move. In other words, we want to find the equations of motion, and the force determines how the charge’s momentum will change: F = dp/dt = d(mv)/dt (i.e. Newton’s equation of motion).

So how does that work? We’ve given the formula before:

F = q(E + v×B) = qE + q(v×B)

This is a sum of two vectors:

qE is the ‘electric force: that force is in the same direction as the electric field, but with a magnitude equal to q times E. [Note I use a bold letter (E) for a vector (which we may define as some quantity with a direction) and a non-bold letter (E) for its magnitude.]
q(v×B) is the ‘magnetic’ force: that force depends on both v as well as on B. Its direction is given by the so-called right-hand rule for a vector cross-product (as opposed to a dot product, which is denoted by a dot (·) and which yields a scalar instead of a new vector).

That right-hand rule is illustrated below. Note that, if we switch a and b, the b×a vector will point downwards. The magnitude of q(v×B) is given by |v×B| = |v||B|sinθ (with θ the angle between v and B).

We know the direction of v (because we’re talking about some charge that is moving here) but what direction is B? It’s time to be a bit more systematic now.

Flux and circulation

In order to understand Maxwell’s equations, one needs to understand two concepts related to a vector field: flux and circulation. The two concepts are best illustrated referring to a vector field describing the flow of a liquid:

1. If we have a surface, the flux will give us the net amount of fluid going out through the surface per unit time. The illustration below (which I took from Feynman’s Lectures) gives us not only the general idea but a formal definition as well:

2. The concept of circulation is linked to the idea of some net rotational motion around some loop. In fact, that’s exactly what it describes. I’ll again use Feynman’s illustration (and description) because I couldn’t find anything better.

Diagram (a) gives us the velocity field in the liquid. Now, imagine a tube (of uniform cross section) that follows some arbitrary closed curve, like in (b), and then imagine we’d suddenly freeze the liquid everywhere except inside the tube: the liquid in the tube would circulate as shown in (c). Formally, the circulation is defined as:

circulation = (the average tangential component)·(the distance around)

OK. So far, so good. Back to electromagnetism.

E and B

We’re familiar with the electric field E from our high school physics course. Indeed, you’ll probably recognize the two examples below: (a) a (positive) charge near a (neutral) conducting sheet, and (b) two opposite charges next to each other. Note the convention: the field lines emanate from the positive charge. Does that mean that the force is in that direction too? Yes. But remember: if a particle is attracted to another, the latter particle is attracted to the former too! So there’s a force in both directions !

What more can we say about this? Well… It is clear that the field E is directed radially. In terms of our flux and circulation concepts, we say that there’s an outgoing flux from the (positive) point charge. Furthermore, it would seem to be pretty obvious (we’d need to show why, but we won’t do that here: just look at Coulomb’s Law once again) that the flux should be proportional to the charge, and it is: if we double the charge, the flux doubles too. That gives us Maxwell’s first equation:

flux of E through a closed surface = (the net charge inside)/ε₀

Note we’re talking a closed surface here, like a sphere for example–but it does not have to be a nice symmetric shape: Maxwell’s first equation is valid for any closed surface. The expression above is Coulomb’s Law, which you’ll also surely remember from your high school physics course: while it looks very different, it’s the same. It’s just because we’re using that flux concept here that we seem to be getting an entirely different expression. But so we’re not: it’s the same as Coulomb’s Law.

As for the ε₀ factor, that’s just a constant that depends on the units we’re using to measure what we write above, so don’t worry about it. [I am noting it here because you’ll see it pop up later too.]

For B, we’ve got a similar-looking law:

flux of B through a closed surface = 0 (= zero = nil)

That’s not the same, you’ll say. Well… Yes and no. It’s the same really, but the zero on the right-hand side of the expression above says there’s no such thing as a ‘magnetic’ charge.

Hmm… But… If we can’t create any flux of B, because ‘magnetic charges’ don’t exist, so how do we get magnetic fields then?

Well… We wrote that above already, and you should remember it from your high school physics course as well: a magnetic field is created by (1) a moving charge (i.e. a flow or flux of electric current) or (2) a changing electric field.

Situation (1) is illustrated below: the current in the wire creates some circulation of B around the wire. How much? Not much: the magnetic effect is very small as compared to the electric effect (that has to do with magnetism being a relativistic effect of electricity but, as mentioned above, I’ll explain that later only). To be precise, the equation is the following:

c²(circulation of B)= (flux of electric current)/ε₀

That c² factor on the left-hand side becomes 1/c² if we move it to the other side and, yes, c is the speed of light here – so you can see we’re talking a very small amount of circulation only indeed! [As for the ε₀ factor, that’s just the same constant: it’s got to do with the units we’re using to measure stuff.]

One last point perhaps: what’s the direction of the circulation? Well… There’s a so-called right-hand grip rule for that, which is illustrated below.

OK. Enough about this. Let’s go to situation (2): a changing electric field. That effect is usually illustrated with Faraday’s original 1831 experiment, which is shown below with a more modern voltmeter 🙂 : when the wire on one side of the iron ring is connected to the battery, we’ll see a transient current on the other side. It’s transient only, so the current quickly disappears. That’s why transformers don’t work with DC. In fact, it is said that Faraday was quite disappointed to see that the current didn’t last! Likewise, when the wire is disconnected, we’ll briefly see another transient current.

So this effect is due to the changing electric field, which causes a changing magnetic field. But so where is that magnetic field? We’re talking currents here, aren’t we? Yes, you’re right. To understand why we have a transient current in the voltmeter, you need to understand yet another effect: a changing magnetic field causes an electric field, and so that’s what actually generates the transient current. However, what’s going on in the iron ring is the magnetic effect, and so that’s caused by the changing electric field as we connect/disconnect the battery to the wire. Capito?

I guess so… So what’s the equation that captures this situation, i.e. situation (2)? That equation involves both flux and circulation, so we’ll have a surface (S) as well as a curve (C). The equation is the following one: for any surface S (not closed this time because, if the surface was closed, it wouldn’t have an edge!), we have:

c²(circulation of B around C)= d(flux of E through S)/dt

I mentioned above that the reverse is also true. A changing magnetic field causes an electric field, and the equation for that looks very similar, except that we don’t have the c² factor:

circulation of E around C = d(flux of B through S)/dt

Let me quickly mention the presence of absence of that c² or 1/c² factor in the previous equations once again. It is interesting. It’s got nothing to do with the units. It’s really a proportionality factor: any change in E will only cause a little change in B (because of the 1/c² factor in the first equation), but the reverse is not true: there’s no c² in the second equation. Again, it’s got to do with magnetism being a relativistic effect of electricity, so the magnetic effect is, in most cases, tiny as compared to the electric effect, except when we’re talking charges that are moving at relativistic speeds (i.e. speeds close to c). As said, we’ll come back to that–later, much later. Let’s get back to Maxwell’s equations first.

Maxwell’s equations

We can now combine all of the equations above in one set, and so these are Maxwell’s four famous equations:

The flux of E through a closed surface = (the net charge inside)/ε₀
The circulation of E around C = d(flux of B through S)/dt (with C the curve or edge around S)
The flux of B through a closed surface = 0
c²(circulation of B around C)= d(flux of E through S)/dt + (flux of electric current)/ε₀

From a mathematical point of view, this is a set of differential equations, and they are not easy to grasp intuitively. As Feynman puts it: “The laws of Newton were very simple to write down, but they had a lot of complicated consequences and it took us a long time to learn about them all. These laws are not nearly as simple to write down, which means that the consequences are going to be more elaborate and it will take us quite a lot of time to figure them all out.”

Indeed, Feynman needs about twenty (!) Lectures in that second Volume to show what it all implies, as he walks us through electrostatics, magnetostatics and various other ‘special’ cases before giving us the ‘complete’ or ‘general’ solution to the equations. This ‘general’ solution, in mathematical notation, is the following:

Huh? What’s that? Well… The four equations are the equations we explained already, but this time in mathematical notation: flux and circulation can be expressed much more elegantly using the differential operator ∇ indeed. As for the solutions to Maxwell’s set of equations, you can see they are expressed using two other concepts: the scalar potential Φ and the vector potential A.

Now, it is not my intention to summarize two dozen of Feynman’s Lectures in just a few lines, so I’ll have to leave you here for the moment.

[…]

Huh? What? What about my promise to show that magnetism is a relativistic effect of electricity indeed?

Well… I wanted to do that just now, but when I look at it, I realize that I’d end up copying most of Feynman’s little exposé on it and, hence, I’ll just refer you to that particular section. It’s really quite exciting but – as you might expect – it does take a bit of time to wrestle through it.

That being said, it really does give you a kind of an Aha-Erlebnis and, therefore, I really warmly recommend it ! Just click on the link ! 🙂

Amplitudes and statistics

Original post:

When re-reading Feynman’s ‘explanation’ of Bose-Einstein versus Fermi-Dirac statistics (Lectures, Vol. III, Chapter 4), and my own March 2014 post summarizing his argument, I suddenly felt his approach raises as many questions as it answers. So I thought it would be good to re-visit it, which is what I’ll do here. Before you continue reading, however, I should warn you: I am not sure I’ll manage to do a better job now, as compared to a few months ago. But let me give it a try.

Setting up the experiment

The (thought) experiment is simple enough: what’s being analyzed is the (theoretical) behavior of two particles, referred to as particle a and particle b respectively that are being scattered into two detectors, referred to as 1 and 2. That can happen in two ways, as depicted below: situation (a) and situation (b). [And, yes, it’s a bit confusing to use the same letters a and b here, but just note the brackets and you’ll be fine.] It’s an elastic scattering and it’s seen in the center-of-mass reference frame in order to ensure we can analyze it using just one variable, θ, for the angle of incidence. So there is no interaction between those two particles in a quantum-mechanical sense: there is no exchange of spin (spin flipping) nor is there any exchange of energy–like in Compton scattering, in which a photon gives some of its energy to an electron, resulting in a Compton shift (i.e. the wavelength of the scattered photon is different than that of the incoming photon). No, it’s just what it is: two particles deflecting each other. […] Well… Maybe. Let’s fully develop the argument to see what’s going on.

First, the analysis is done for two non-identical particles, say an alpha particle (i.e. a helium nucleus) and then some other nucleus (e.g. oxygen, carbon, beryllium,…). Because of the elasticity of the ‘collision’, the possible outcomes of the experiment are binary: if particle a gets into detector 1, it means particle b will be picked up by detector 2, and vice versa. The first situation (particle a gets into detector 1 and particle b goes into detector 2) is depicted in (a), i.e. the illustration on the left above, while the opposite situation, exchanging the role of the particles, is depicted in (b), i.e. the illustration on the right-hand side. So these two ‘ways’ are two different possibilities which are distinguishable not only in principle but also in practice, for non-identical particles that is (just imagine a detector which can distinguish helium from oxygen, or whatever other substance the other particle is). Therefore, strictly following the rules of quantum mechanics, we should add the probabilities of both events to arrive at the total probability of some particle (and with ‘some’, I mean particle a or particle b) ending up in some detector (again, with ‘some’ detector, I mean detector 1 or detector 2).

Now, this is where Feynman’s explanation becomes somewhat tricky. The whole event (i.e. some particle ending up in some detector) is being reduced to two mutually exclusive possibilities that are both being described by the same (complex-valued) wave function f, which has that angle of incidence as its argument. To be precise: the angle of incidence is θ for the first possibility and it’s π–θ for the second possibility. That being said, it is obvious, even if Feynman doesn’t mention it, that both possibilities actually represent a combination of two separate things themselves:

For situation (a), we have particle a going to detector 1 and particle b going to detector 2. Using Dirac’s so-called bra-ket notation, we should write 〈1|a〉〈2|b〉 = f(θ), with f(θ) a probability amplitude, which should yield a probability when taking its absolute square: P(θ) = |f(θ)|².
For situation (b), we have particle b going to detector 1 and particle a going to 2, so we have 〈1|b〉〈2|a〉, which Feynman equates with f(π–θ), so we write 〈1|b〉〈2|a〉 = 〈2|a〉〈1|b〉 = f(π–θ).

Now, Feynman doesn’t dwell on this–not at all, really–but this casual assumption–i.e. the assumption that situation (b) can be represented by using the same wave function f–merits some more reflection. As said, Feynman is very brief on it: he just says situation (b) is the same situation as (a), but then detector 1 and detector 2 being switched (so we exchange the role of the detectors, I’d say). Hence, the relevant angle is π–θ and, of course, it’s a center-of-mass view again so if a goes to 2, then b has to go to 1. There’s no Third Way here. In short, a priori it would seem to be very obvious indeed to associate only one wave function (i.e. that (complex-valued) f(θ) function) with the two possibilities: that wave function f yields a probability amplitude for θ and, hence, it should also yield some (other) probability amplitude for π–θ, i.e. for the ‘other’ angle. So we have two probability amplitudes but one wave function only.

You’ll say: Of course! What’s the problem? Why are you being fussy? Well… I think these assumptions about f(θ) and f(π–θ) representing the underlying probability amplitudes are all nice and fine (and, yes, they are very reasonable indeed), but I also think we should take them for what they are at this moment: assumptions.

Huh? Yes. At this point, I would like to draw your attention to the fact that the only thing we can measure are real-valued possibilities. Indeed, when we do this experiment like a zillion times, it will give us some real number P for the probability that a goes to 1 and b goes to 2 (let me denote this number as P(θ) = P_a→_{1 and b→}₂), and then, when we change the angle of incidence by switching detector 1 and 2, it will also give us some (other) real number for the probability that a goes to 2 and b goes to 1 (i.e. a number which we can denote as P(π–θ) = P_a→_{2 and b→1}). Now, while it would seem to be very reasonable that the underlying probability amplitudes are the same, we should be honest with ourselves and admit that the probability amplitudes are something we cannot directly measure.

At this point, let me quickly say something about Dirac’s bra-ket notation, just in case you haven’t heard about it yet. As Feynman notes, we have to get away from thinking too much in terms of wave functions traveling through space because, in quantum mechanics, all sort of stuff can happen (e.g. spin flipping) and not all of it can be analyzed in terms of interfering probability amplitudes. Hence, it’s often more useful to think in terms of a system being in some state and then transitioning to some other state, and that’s why that bra-ket notation is so helpful. We have to read these bra-kets from right to left: the part on the right, e.g. |a〉, is the ket and, in this case, that ket just says that we’re looking at some particle referred to as particle a, while the part on the left, i.e. 〈1|, is the bra, i.e. a shorthand for particle a having arrived at detector 1. If we’d want to be complete, we should write:

〈1|a〉 = 〈particle a arrives at detector 1|particle a leaves its source〉

Note that 〈1|a〉 is some complex-valued number (i.e. a probability amplitude) and so we multiply it here with some other complex number, 〈2|b〉, because it’s two things happening together. As said, don’t worry too much about it. Strictly speaking, we don’t need wave functions and/or probability amplitudes to analyze this situation because there is no interaction in the quantum-mechanical sense: we’ve got a scattering process indeed (implying some randomness in where those particles end up, as opposed to what we’d have in a classical analysis of two billiard balls colliding), but we do not have any interference between wave functions (probability amplitudes) here. We’re just introducing the wave function f because we want to illustrate the difference between this situation (i.e. the scattering of non-identical particles) and what we’d have if we’d be looking at identical particles being scattered.

At this point, I should also note that this bra-ket notation is more in line with Feynman’s own so-called path integral formulation of quantum mechanics, which is actually implicit in his line of argument: rather than thinking about the wave function as representing the (complex) amplitude of some particle to be at point x in space at point t in time, we think about the amplitude as something that’s associated with a path, i.e. one of the possible itineraries from the source (its origin) to the detector (its destination). That explains why this f(θ) function doesn’t mention the position (x) and space (t) variables. What x and t variables would we use anyway? Well… I don’t know. It’s true the position of the detectors is fully determined by θ, so we don’t need to associate any x or t with them. Hence, if we’d be thinking about the space-time variables, then we should be talking the position in space and time of both particle a and particle b. Indeed, it’s easy to see that only a slight change in the horizontal (x) or vertical position (y) of either particle would ensure that both particles do not end up in the detectors. However, as mentioned above, Feynman doesn’t even mention this. Hence, we must assume that any randomness in any x or t variable is captured by that wave function f, which explains why this is actually not a classical analysis: so, in short, we do not have two billiard balls colliding here.

Hmm… You’ll say I am a nitpicker. You’ll say that, of course, any uncertainty is indeed being incorporated in the fact that we represent what’s going on by a wave function f which we cannot observe directly but whose absolute square represents a probability (or, to use precise statistical terminology, a probability density), which we can measure: P = |f(θ)|² = f(θ)·f*(θ), with f* the complex conjugate of the complex number f. So… […] What? Well… Nothing. You’re right. This thought experiment describes a classical situation (like two billiard balls colliding) and then it doesn’t, because we cannot predict the outcome (i.e. we can’t say where the two billiard balls are going to end up: we can only describe the likely outcome in terms of probabilities P_a→_{1 and b→}₂ = |f(θ)|²and P_a→2_{and b→}₁ = |f(π–θ)|². Of course, needless to say, the normalization condition should apply: if we add all probabilities over all angles, then we should get 1, we can write: ∫|f(θ)|²dθ = ∫f(θ)·f*(θ)dθ = 1. So that’s it, then?

No. Let this sink in for a while. I’ll come back to it. Let me first make a bit of a detour to illustrate what this thought experiment is supposed to yield, and that’s a more intuitive explanation of Bose-Einstein statistics and Fermi-Dirac statistics, which we’ll get out of the experiment above if we repeat it using identical particles. So we’ll introduce the terms Bose-Einstein statistics and Fermi-Dirac statistics. Hence, there should also be some term for the reference situation described above, i.e a situation in which we non-identical particles are ‘interacting’, so to say, but then with no interference between their wave functions. So, when everything is said and done, it’s a term we should associate with classical mechanics. It’s called Maxwell-Boltzmann statistics.

Huh? Why would we need ‘statistics’ here? Well… We can imagine many particles engaging like this–just colliding elastically and, thereby, interacting in a classical sense, even if we don’t know where exactly they’re going to end up, because of uncertainties in initial positions and what have you. In fact, you already know what this is about: it’s the behavior of particles as described by the kinetic theory of gases (often referred to as statististical mechanics) which, among other things, yields a very elegant function for the distribution of the velocities of gas molecules, as shown below for various gases (helium, neon, argon and xenon) at one specific temperature (25º C), i.e. the graph on the left-hand side, or for the same gas (oxygen) at different temperatures (–100º C, 20º C and 600º C), i.e. the graph on the right-hand side.

Now, all these density functions and what have you are, indeed, referred to as Maxwell-Boltzmann statistics, by physicists and mathematicians that is (you know they always need some special term in order to make sure other people (i.e. people like you and me, I guess) have trouble understanding them).

In fact, we get the same density function for other properties of the molecules, such as their momentum and their total energy. It’s worth elaborating on this, I think, because I’ll later compare with Bose-Einstein and Fermi-Dirac statistics.

Maxwell-Boltzmann statistics

Kinetic gas theory yields a very simple and beautiful theorem. It’s the following: in a gas that’s in thermal equilibrium (or just in equilibrium, if you want), the probability (P) of finding a molecule with energy E is proportional to e^–E/kT, so we have:

P ∝ e^–E/kT

Now that’s a simple function, you may think. If we treat E as just a continuous variable, and T as some constant indeed – hence, if we just treat (the probability) P as a function of (the energy) E – then we get a function like the one below (with the blue, red and green using three different values for T).

So how do we relate that to the nice bell-shaped curves above? The very simple graphs above seem to indicate the probability is greatest for E = 0, and then just goes down, instead of going up initially to reach some maximum around some average value and then drop down again. Well… The fallacy here, of course, is that the constant of proportionality is itself dependent on the temperature. To be precise, the probability density function for velocities is given by:

The function for energy is similar. To be precise, we have the following function:

This (and the velocity function too) is a so-called chi-squared distribution, and ϵ is the energy per degree of freedom in the system. Now these functions will give you such nice bell-shaped curves, and so all is alright. In any case, don’t worry too much about it. I have to get back to that story of the two particles and the two detectors.

However, before I do so, let me jot down two (or three) more formulas. The first one is the formula for the expected number 〈N_i〉 of particles occupying energy level ε_i(and the brackets here, 〈N_i〉, have nothing to do with the bra-ket notation mentioned above: it’s just a general notation for some expected value):

This formula has the same shape as the ones above but we brought the exponential function down, into the denominator, so the minus sign disappears. And then we also simplified it by introducing that g_i factor, which I won’t explain here, because the only reason why I wanted to jot this down is to allow you to compare this formula with the equivalent formula when (a) Fermi-Dirac and (b) Bose-Einstein statistics apply:

Do you see the difference? The only change in the formula is the ±1 term in the denominator: we have a minus one (–1) for Fermi-Dirac statistics and a plus one (+1) for Bose-Einstein statistics indeed. That’s all. That’s the difference with Maxwell-Boltzmann statistics.

Huh? Yes. Think about it, but don’t worry too much. Just make a mental note of it, as it will be handy when you’d be exploring related articles. [And, of course, please don’t think I am bagatellizing the difference between Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac statistics here: that ±1 term in the denominator is, obviously, a very important difference, as evidenced by the consequences of formulas like the one above: just think about the crowding-in effect in lasers as opposed to the Pauli exclusion principle, for example. :-)]

Setting up the experiment (continued)

Let’s get back to our experiment. As mentioned above, we don’t really need probability amplitudes in the classical world: ordinary probabilities, taking into account uncertainties about initial conditions only, will do. Indeed, there’s a limit to the precision with which we can measure the position in space and time of any particle in the classical world as well and, hence, we’d expect some randomness (as captured in the scattering phenomenon) but, as mentioned above, ordinary probabilities would do to capture that. Nevertheless, we did associate probability amplitudes with the events described above in order to illustrate the difference with the quantum-mechanical world. More specifically, we distinguished:

Situation (a): particle a goes to detector 1 and b goes to 2, versus
Situation (b): particle a goes to 2 and b goes to 1.

In our bra-ket notation:

〈1|a〉〈2|b〉 = f(θ), and
〈1|b〉〈2|a〉 = f(π–θ).

The f(θ) function is a quantum-mechanical wave function. As mentioned above, while we’d expect to see some space (x) and time (t) variables in it, these are, apparently, already captured by the θ variable. What about f(π–θ)? Well… As mentioned above also, that’s just the same function as f(θ) but using the angle π–θ as the argument. So, the following remark is probably too trivial to note but let me do it anyway (to make sure you understand what we’re modeling here really): while it’s the same function f, the values f(θ) and f(π–θ) are, of course, not necessarily equal and, hence, the corresponding probabilities are also not necessarily the same. Indeed, some angles of scattering may be more likely than others. However, note that we assume that the function f itself is exactly the same for the two situations (a) and (b), as evidenced by that normalization condition we assume to be respected: if we add all probabilities over all angles, then we should get 1, so ∫|f(θ)|²dθ = ∫f(θ)·f*(θ)dθ = 1.

So far so good, you’ll say. However, let me ask the same critical question once again: why would we use the same wave function f for the second situation?

Huh? You’ll say: why wouldn’t we? Well… Think about it. Again, how do we find that f(θ) function? The assumption here is that we just do the experiment a zillion times while varying the angle θ and, hence, that we’ll find some average corresponding to P(θ), i.e. the probability. Now, the next step then is to equate that average value to |f(θ)|²obviously, because we have this quantum-mechanical theory saying probabilities are the absolute square of probability amplitudes. And, so… Well… Yes. We then just take the square root of the P function to find the f(θ) function, isn’t it?

Well… No. That’s where Feynman is not very accurate when it comes to spelling out all of the assumptions underpinning this thought experiment. We should obviously watch out here, as there’s all kinds of complications when you do something like that. To a large extent (perhaps all of it), the complications are mathematical only.

First, note that any number (real or complex, but note that |f(θ)|²is a real number) has two distinct real square roots: a positive and a negative one: x = ± √x². Secondly, we should also note that, if f(θ) is a regular complex-valued wave function of x and t and θ (and with ‘regular’, we mean, of course, that’s it’s some solution to a Schrödinger (or Schrödinger-like) equation), then we can multiply it with some random factor shifting its phase Θ (usually written as Θ = kx–ωt+α) and the square of its absolute value (i.e. its squared norm) will still yield the same value. In mathematical terms, such factor is just a complex number with a modulus (or length or norm–whatever terminology you prefer) equal to one, which we can write as a complex exponential: e^iα, for example. So we should note that, from a mathematical point of view, any function e^iαf(θ) will yield the same probabilities as f(θ). Indeed,

|f(θ)|²= |e^iαf(θ)|²= (|e^iα||f(θ)|)²= |e^iα|²|f(θ)|²= 1²|f(θ)|²

Likewise, while we assume that this function f(π–θ) is the same function f as that f(θ) function, from a mathematical point of view, the function e^iβf(π–θ) would do just as well, because its absolute square yields the very same (real) probability |f(π–θ)|². So the question as to what wave function we should take for the probability amplitude is not as easy to answer as you may think. Huh? So what function should we take then? Well… We don’t know. Fortunately, it doesn’t matter, for non-identical particles that is. Indeed, when analyzing the scattering of non-identical particles, we’re interested in the probabilities only and we can calculate the total probability of particle a ending up in detector 1 or 2 (and, hence, particle b ending up in detector 2 or 1) as the following sum:

|e^iαf(θ)|²+|e^iβf(π–θ)|²= |f(θ)|²+|f(π–θ)|².

In other words, for non-identical particles, these phase factors (e^iα or e^iβ) don’t matter and we can just forget about them.

However, and that’s the crux of the matter really, we should mention them, of course, in case we’d have to add the probability amplitudes–which is exactly what we’ll have to do when we’re looking at identical particles, of course. In fact, in that case (i.e. when these phase factors e^iα and e^iβwill actually matter), you should note that what matters really is the phase difference, so we could replace α and β with some δ (which is what we’ll do below).

However, let’s not put the cart before the horse and conclude our analysis of what’s going on when we’re considering non-identical parties: in that case, this phase difference doesn’t matter. And the remark about the positive and negative square root doesn’t matter either. In fact, if you want, you can subsume it under the phase difference story by writing e^iαas e^iα = ± 1. To be more explicit: we could say that –f(θ) is the probability amplitude, as |–f(θ)|²is also equal to that very same real number |f(θ)|². OK. Done.

Bose-Einstein and Fermi-Dirac statistics

As I mentioned above, the story becomes an entirely different one when we’re doing the same experiment with identical particles. At this point, Feynman’s argument becomes rather fuzzy and, in my humble opinion, that’s because he refused to be very explicit about all of those implicit assumptions I mentioned above. What I can make of it, is the following:

1. We know that we’ll have to add probability amplitudes, instead of probabilities, because we’re talking one event that can happen in two indistinguishable ways. Indeed, for non-identical particles, we can, in principle (and in practice) distinguish situation (a) and (b) – and so that’s why we only have to add some real-valued numbers representing probabilities – but so we cannot do do that for identical particles.

2. Situation (a) is still being described by some probability amplitude f(θ). We don’t know what function exactly, but we assume there is some unique wave function f(θ) out there that accurately describes the probability amplitude of particle a going to 1 (and, hence, particle b going to 2), even if we can’t tell which is a and which is b. What about the phase factor? Well… We just assume we’ve chosen our t such that α = 0. In short, the assumption is that situation (a) is represented by some probability amplitude (or wave function, if you prefer that term) f(θ).

3. However, a (or some) particle (i.e. particle a or particle b) ending up in a (some) detector (i.e. detector 1 or detector 2) may come about in two ways that cannot be distinguished one from the other. One is the way described above, by that wave function f(θ). The other way is by exchanging the role of the two particles. Now, it would seem logical to associate the amplitude f(π–θ) with the second way. But we’re in the quantum-mechanical world now. There’s uncertainty, in position, in momentum, in energy, in time, whatever. So we can’t be sure about the phase. That being said, the wave function will still have the same functional form, we must assume, as it should yield the same probability when squaring. To account for that, we will allow for a phase factor, and we know it will be important when adding the amplitudes. So, while the probability for the second way (i.e. the square of its absolute value) should be the same, its probability amplitude does not necessarily have to be the same: we have to allow for positive and negative roots or, more generally, a possible phase shift. Hence, we’ll write the probability amplitude as e^iδf(π–θ) for the second way. [Why do I use δ instead of β? Well… Again: note that it’s the phase difference that matters. From a mathematical point of view, it’s the same as inserting an e^iβ factor: δ can take on any value.]

4. Now it’s time for the Big Trick. Nature doesn’t matter about our labeling of particles. If we have to multiply the wave function (i.e. f(π–θ), or f(θ)–it’s the same: we’re talking a complex-valued function of some variable (i.e. the angle θ) here) with a phase factor e^iδ when exchanging the roles of the particles (or, what amounts to the same, exchanging the role of the detectors), we should get back to our point of departure (i.e. no exchange of particles, or detectors) when doing that two times in a row, isn’t it? So we exchange the role of particle a and b in this analysis (or the role of the detectors), and then we’d exchange their roles once again, then there’s no exchange of roles really and we’re back at the original situation. So we must have e^iδe^iδf(θ) = f(θ) (and e^iδe^iδf(π–θ) = f(π–θ) of course, which is exactly the same statement from a mathematical point of view).

5. However, that means (e^iδ)²= +1, which, in turn, implies that e^iδis plus or minus one: e^iδ= ± 1. So that means the phase difference δ must be equal to 0 or π (or –π, which is the same as +π).

In practical terms, that means we have two ways of combining probability amplitudes for identical particles: we either add them or, else, we subtract them. Both cases exist in reality, and lead to the dichotomy between Bose and Fermi particles:

For Bose particles, we find the total probability amplitude for this scattering event by adding the two individual amplitudes: f(θ) + f(π–θ).
For Fermi particles, we find the total probability amplitude for this scattering event by subtracting the two individual amplitudes: f(θ) – f(π–θ).

As compared to the probability for non-identical particles which, you’ll remember, was equal to |f(θ)|²+|f(π–θ)|², we have the following Bose-Einstein and Fermi-Dirac statistics:

For Bose particles: the combined probability is equal to |f(θ)+ f(π–θ)|². For example, if θ is 90°, then we have a scattering probability that is exactly twice the probability for non-identical particles. Indeed, if θ is 90°, then f(θ) = f(π–θ), and then we have |f(π/2)+ f(π/2)|²= |2f(π/2)|² = 4|f(π/2)|². Now, that’s two times |f(π/2)|²+|f(π/2)|² = 2|f(π/2)|² indeed.
For Fermi particles (e.g. electrons), we have a combined probability equal to |f(θ)– f(π–θ)|². Again, if θ is 90°, f(θ) = f(π–θ), and so it would mean that we have a combined probability which is equal to zero ! Now, that‘s a strange result, isn’t it? It is. Fortunately, the strange result has to be modified because electrons will also have spin and, hence, in half of the cases, the two electrons will actually not be identical but have opposite spin. That changes the analysis substantially (see Feynman’s Lectures, III-3-12). To be precise, if we take the spin factor into, we’ll find a total probability (for θ = 90°) equal to |f(π/2)|², so that’s half of the probability for non-identical particles.

Hmm… You’ll say: Now that was a complicated story! I fully agree. Frankly, I must admit I feel like I still don’t quite ‘get‘ the story with that phase shift e^iδ, in an intuitive way that is (and so that’s the reason for going through the trouble of writing out this post). While I think it makes somewhat more sense now (I mean, more than when I wrote a post on this in March), I still feel I’ve only brought some of the implicit assumptions to the fore. In essence, what we’ve got here is a mathematical dichotomy (or a mathematical possibility if you want) corresponding to what turns out to be an actual dichotomy in Nature: in quantum-mechanics, particles are either bosons or fermions. There is no Third Way, in quantum-mechanics that is (there is a Third Way in reality, of course: that’s the classical world!).

I guess it will become more obvious as I’ll get somewhat more acquainted with the real arithmetic involved in quantum-mechanical calculations over the coming weeks. In short, I’ve analyzed this thing over and over again, but it’s still not quite clear me. I guess I should just move on and accept that:

This explanation ‘explains’ the experimental evidence, and that’s different probabilities for identical particles as compared to non-identical particles.
This explanation ‘complements’ analyses such as that 1916 analysis of blackbody radiation by Einstein (see my post on that), which approaches interference from an angle that’s somewhat more intuitive.

A numerical example

I’ve learned that, when some theoretical piece feels hard to read, an old-fashioned numerical example often helps. So let’s try one here. We can experiment with many functional forms but let’s keep things simple. From the illustration (which I copy below for your convenience), that angle θ can take any value between −π and +π, so you shouldn’t think detector 1 can only be ‘north’ of the collision spot: it can be anywhere.

Now, it may or may not make sense (and please work out other examples than this one here), but let’s assume particle a and b are more likely to go in a line that’s more or less straight. In other words, the assumption is that both particles deflect each other only slightly, or even not at all. After all, we’re talking ‘point-like’ particles here and so, even when we try hard, it’s hard to make them collide really.

That would amount to a typical bell-shaped curve for that probability density curve P(θ): one like the blue curve below. That one shows that the probability of particle a and b just bouncing back (i.e. θ ≈ ±π) is (close to) zero, while it’s highest for θ ≈ 0, and some intermediate value for anything angle in-between. The red curve shows P(π–θ), which can be found by mirroring the P(θ) around the vertical axis, which yields the same function because the function is symmetrical: P(θ) = P(–θ), and then shifting it by adding the vertical distance π. It should: it’s the second possibility, remember? Particle a ending up in detector 2. But detector 2 is positioned at the angle π–θ and, hence, if π–θ is close to ±π (so if θ ≈ 0), that means particle 1 is basically bouncing back also, which we said is unlikely. On the other hand, if detector 2 is positioned at an angle π–θ ≈ 0, then we have the highest probability of particle a going right to it. In short, the red curve makes sense too, I would think. [But do think about yourself: you’re the ultimate judge!]

The harder question, of course, concerns the choice of some wave function f(θ) to match those P curves above. Remember that these probability densities P are real numbers and any real number is the absolute square (aka the squared norm) of an infinite number of complex numbers! So we’ve got l’embarras du choix, as they say in French. So… What do to? Well… Let’s keep things simple and stupid and choose a real-valued wave function f(θ), such as the blue function below. Huh? You’ll wonder if that’s legitimate. Frankly, I am not 100% sure, but why not? The blue f(θ) function will give you the blue P(θ) above, so why not go along with it? It’s based on a cosine function but it’s only half of a full cycle. Why? Not sure. I am just trying to match some sinusoidal function with the probability density function here, so… Well… Let’s take the next step.

The red graph above is the associated f(π–θ) function. Could we choose another one? No. There’s no freedom of choice here, I am afraid: if we choose a functional form for f(θ), then our f(π–θ) function is fixed too. So it is what it is: negative between –π and 0, and positive between 0 and +π and 0. Now that is definitely not good, because f(π–θ) for θ = –π is not equal to f(π–θ) for θ = +π: they’re opposite values. That’s nonsensical, isn’t it? Both the f(θ) and the f(π–θ) should be something cyclical… But, again, let’s go along with it as for now: note that the green horizontal line is the sum of the squared (absolute) values of f(θ) and f(π–θ), and note that it’s some constant.

Now, that’s a funny result, because I assumed both particles were more likely to go in some straight line, rather than recoil with some sharp angle θ. It again indicates I must be doing something wrong here. However, the important thing for me here is to compare with the Bose-Einstein and Fermi-Dirac statistics. What’s the total probability there if we take that blue f(θ) function? Well… That’s what’s shown below. The horizontal blue line is the same as the green line in the graph above: a constant probability for some particle (a or b) ending up in some detector (1 or 2). Note that the surface, when added, of the two rectangles above the x-axis (i.e. the θ-axis) should add up to 1. The red graph gives the probability when the experiment is carried out for (identical) bosons (or Bose particles as I like to call them). It’s weird: it makes sense from a mathematical point of view (the surface under the curve adds up to the same surface under the blue line, so it adds up to 1) but, from a physics point of view, what does this mean? A maximum at θ = π/2 and a minimum at θ = –π/2? Likewise, how to interpret the result for fermions?

Is this OK? Well… To some extent, I guess. It surely matches the theoretical results I mentioned above: we have twice the probability for bosons for θ = 90° (red curve), and a probability equal to zero for the same angle when we’re talking fermions (green curve). Still, this numerical example triggers more questions than it answers. Indeed, my starting hypothesis was very symmetrical: both particle a and b are likely to go in a straight line, rather than being deflected in some sharp(er) angle. Now, while that hypothesis gave a somewhat unusual but still understandable probability density function in the classical world (for non-identical particles, we got a constant for P(θ) + P(π–θ)), we get this weird asymmetry in the quantum-mechanical world: we’re much more likely to catch a boson in a detector ‘north’ of the line of firing than ‘south’ of it, and vice versa for fermions.

That’s weird, to say the least. So let’s go back to the drawing board and take another function for f(θ) and, hence, for f(π–θ). This time, the two graphs below assume that (i) f(θ) and f(π–θ) have a real as well as an imaginary part and (ii) that they go through a full cycle, instead of a half-cycle only. This is done by equating the real part of the two functions with cos(θ) and cos(π–θ) respectively, and their imaginary part with sin(θ) and sin(π–θ) respectively. [Note that we conveniently forget about the normalization condition here.]

What do we see? Well… The imaginary part of f(θ) and f(π–θ) is the same, because sin(π–θ) = sin(θ). We also see that the real part of f(θ) and f(π–θ) are the same except for a phase difference equal to π: cos(π–θ) = cos[–(θ–π)] = cos(θ–π). More importantly, we see that the absolute square of both f(θ) and f(π–θ) yields the same constant, and so their sum P = |f(θ)|²+|f(π–θ)|²= 2|f(θ)|²= 2|f(π–θ)|²= 2P(θ) = 2P(π–θ). So that’s another constant. That’s actually OK because, this time, I did not favor one angle over the other (so I did not assume both particles were more likely to go in some straight line rather than recoil).

Now, how does this compare to Bose-Einstein and Fermi-Dirac statistics? That’s shown below. For Bose-Einstein (left-hand side), the sum of the real parts of f(θ) and f(π–θ) yields zero (blue line), while the sum of their imaginary parts (i.e. the red graph) yields a sine-like function but it has double the amplitude of sin(θ). That’s logical: sin(θ) + sin(π–θ) = 2sin(θ). The green curve is the more interesting one, because that’s the total probability we’re looking for. It has two maxima now, at +π/2 and at –π/2. That’s good, as it does away with that ‘weird asymmetry’ we got when we used a ‘half-cycle’ f(θ) function.

Likewise, the Fermi-Dirac probability density function looks good as well (right-hand side). We have the imaginary parts of f(θ) and f(π–θ) that ‘add’ to zero: sin(θ) – sin(π–θ) = 0 (I put ‘add’ between brackets because, with Fermi-Dirac, we’re subtracting of course), while the real parts ‘add’ up to a double cosine function: cos(θ) – cos(π–θ) = cos(θ) – [–cos(θ)] = 2cos(θ). We now get a minimum at +π/2 and at –π/2, which is also in line with the general result we’d expect. The (final) graph below summarizes our findings. It gives the three ‘types’ of probabilities, i.e. the probability of finding some particle in some detector as a function of the angle –π < θ < +π using:

Maxwell-Boltzmann statistics: that’s the green constant (non-identical particles, and probability does not vary with the angle θ).
Bose-Einstein: that’s the blue graph below. It has two maxima, at +π/2 and at –π/2, and two minima, at 0 and at ±π (+π and –π are the same angle obviously), with the maxima equal to twice the value we get under Maxwell-Boltzmann statistics.
Finally, the red graph gives the Fermi-Dirac probabilities. Also two maxima and minima, but at different places: the maxima are at θ = 0 and θ = ±π, while the minima are at at +π/2 and at –π/2.

Funny, isn’t it? These probability density functions are all well-behaved, in the sense that they add up to the same total (which should be 1 when applying the normalization condition). Indeed, the surfaces under the green, blue and red lines are obviously the same. But so we get these weird fluctuations for Bose-Einstein and Fermi-Dirac statistics, favoring two specific angles over all others, while there’s no such favoritism when the experiment involves non-identical particles. This, of course, just follows from our assumption concerning f(θ). What if we double the frequency of f(θ), i.e. from one cycle to two cycles between –π and +π? Well… Just try it: take f(θ) = cos(2·θ) + isin(2·θ) and do the calculations. You should get the following probability graphs: we have the same green line for non-identical particles, but interference with four maxima (and four minima) for the Bose-Einstein and Fermi-Dirac probabilities.

Again… Funny, isn’t it? So… What to make of this? Frankly, I don’t know. But one last graph makes for an interesting observation: if the angular frequency of f(θ) takes on larger and larger values, the Bose-Einstein and Fermi-Dirac probability density functions also start oscillating wildly. For example, the graphs below are based on a f(θ) function equal to f(θ) = cos(25·θ) + isin(25·θ). The explosion of color hurts the eye, doesn’t it? 🙂 But, apart from that, do you now see why physicists say that, at high frequencies, the interference pattern gets smeared out? Indeed, if we move the detector just a little bit (i.e. we change the angle θ just a little bit) in the example below, we hit a maximum instead of a minimum, and vice versa. In short, the granularity may be such that we can only measure that green line, in which case we’d think we’re dealing with Maxwell-Boltzmann statistics, while the underlying reality may be different.

That explains another quote in Feynman’s famous introduction to quantum mechanics (Lectures, Vol. III, Chapter 1): “If the motion of all matter—as well as electrons—must be described in terms of waves, what about the bullets in our first experiment? Why didn’t we see an interference pattern there? It turns out that for the bullets the wavelengths were so tiny that the interference patterns became very fine. So fine, in fact, that with any detector of finite size one could not distinguish the separate maxima and minima. What we saw was only a kind of average, which is the classical curve. In the Figure below, we have tried to indicate schematically what happens with large-scale objects. Part (a) of the figure shows the probability distribution one might predict for bullets, using quantum mechanics. The rapid wiggles are supposed to represent the interference pattern one gets for waves of very short wavelength. Any physical detector, however, straddles several wiggles of the probability curve, so that the measurements show the smooth curve drawn in part (b) of the figure.”

But that should really conclude this post. It has become way too long already. One final remark, though: the ‘smearing out’ effect also explains why those three equations for 〈N_i〉 sometimes do amount to more or less the same thing: the Bose-Einstein and Fermi-Dirac formulas may approximate the Maxwell-Boltzmann equation. In that case, the ±1 term in the denominator does not make much of a difference. As we said a couple of times already, it all depends on scale. 🙂

Concluding remarks

1. The best I can do in terms of interpreting the above, is to tell myself that we cannot fully ‘fix’ the functional form of the wave function for the second or ‘other’ way the event can happen if we’re ‘fixing’ the functional form for the first of the two possibilities. We have to allow for a phase shift e^iδ indeed, which incorporates all kinds of considerations of uncertainty in regard to both time and position and, hence, in regard to energy and momentum also (using both the ΔEΔt = ħ/2 and ΔxΔp = ħ/2 expressions)–I assume (but that’s just a gut instinct). And then the symmetry of the situation then implies e^iδ can only take on one of two possible values: –1 or +1 which, in turn, implies that δ is equal to 0 or π.

2. For those who’d think I am basically doing nothing but re-write a chapter out of Feynman’s Lectures, I’d refute that. One point to note is that Feynman doesn’t seem to accept that we should introduce a phase factor in the analysis for non-identical particles as well. To be specific: just switching the detectors (instead of the particles) also implies that one should allow for the mathematical possibility of the phase of that f function being shifted by some random factor δ. The only difference with the quantum-mechanical analysis (i.e. the analysis for identical particles) is that the phase factor doesn’t make a difference as to the final result, because we’re not adding amplitudes but their absolute squares and, hence, a phase shift doesn’t matter.

3. I think all of the reasoning above makes not only for a very fine but also a very beautiful theoretical argument, even I feel like I don’t fully ‘understand’ it, in an intuitive way that is. I hope this post has made you think. Isn’t it wonderful to see that the theoretical or mathematical possibilities of the model actually correspond to realities, both in the classical as well as in the quantum-mechanical world? In fact, I can imagine that most physicists and mathematicians would shrug this whole reflection off like… Well… Like: “Of course! It’s obvious, isn’t it?” I don’t think it’s obvious. I think it’s deep. I would even qualify it as mysterious, and surely as beautiful. 🙂

Relativity paradoxes

Original post:

My son, who’s fifteen, said he liked my post on lasers. That’s good, because I effectively wrote it thinking of him as part of the audience. He also said it stimulated him to considering taking on studies in engineering later. That’s great. I hope he does, so he doesn’t have to go through what I am going through right now. Indeed, when everything is said and done, you do want your kids to take on as much math and science they can handle when they’re young because, afterwards, it’s tough to catch up.

Now, I struggled quite a bit with bringing relativity into the picture while pondering the ‘essence’ of a photon in my previous post. Hence, I’d thought it would be good to return to the topic of (special) relativity and write another post to (1) refresh my knowledge on the topic and (2) try to stimulate him even more. Indeed, regardless of whether one does or doesn’t understand any of what I write below here, relativity theory sounds fascinating, doesn’t it? 🙂 So, this post intends to present, in a nutshell, what (special) relativity theory is all about.

What relativity does

The thing that’s best known about Einstein’s (special) theory of relativity is the following: the mass of an object, as measured by the (inertial) observer, increases with its speed. The formula for this is m = γm₀, and the γ factor here is the so-called Lorentz factor: γ = (1–u²/c²)^–1/2. Let me give you that diagram of the Lorentz factor once again, which shows that very considerable speeds are required before relativity effects kick in. However, when they do, they kick in with a vengeance, it seems, which makes c the limit !

Now, you may or may not be familiar with two other things that come out of relativity theory as well:

The first is length contraction: objects are measured to be shortened in the direction of motion with respect to the (inertial) observer. The formula to be used incorporates the reciprocal of the Lorentz factor: L = (1/γ)L₀. For example, a stick of one meter in a space ship moving at a velocity v = 0.6c will appear to be only 80 cm to the external/inertial observer seeing it whizz past… That is if he can see anything at all of course: he’d have to take like a photo-finish picture as it zooms past ! 🙂
The second is time dilation, which is also rather well known – just like the mass increase effect – because of the so-called twin paradox: time will appear to be slower in that space ship and, hence, if you send one of two twins away on a space journey, traveling at relativistic speeds (i.e. a velocity sufficiently close to c to make the relativistic effect significant), he will come back younger than his brother. The formula here is equally simple: t = γt₀. Hence, one second in the space ship will be measured as 1.25 seconds by the external observer. Hence, the moving clock will appear to run slower – again: to the external (inertial) observer that is.

These simple rules, which comes out of Einsteins’ special relativity theory, give rise to all kinds of paradoxes. You know what a paradox is: a paradox (in physics) is something that, at first sight, does not make sense but that, when the issue is examined more in detail, does get resolved and actually helps us to better understand what’s going on.

You know the twin paradox already: only of the two twins can be the younger (or the older) when they meet again. However, because one can also say it’s the guy staying on Earth that’s moving (and, hence, is ‘traveling’ at relativistic speed) – so then the reference frame of the guy in the spaceship is the so-called inertial frame, one can say the guy who stayed behind (on Earth) should be the youngest when they meet after the journey. I am not ashamed to say that this actually is a paradox that is difficult to understand. So let me first start with another.

The ladder paradox

While the twin paradox examines the time dilation effect, the ladder paradox examines the length contraction effect. The situation is similar as the one for the twin paradox. However, because we don’t have accelerating and decelerating rockets and all that (cf. the twin paradox), I find this paradox not only more straightforward but also more amusing. Look at the left-hand side first. We have a garage which has both a front and back door. A ladder passes through it, and it seems to fit in the garage as you can see. Now, that may or may not be because of the length contraction effect, of course. Whatever. In any case, it seems we can (very) quickly close both doors of the garage to prove that it fits. Now look at the right-hand side. Here we are moving the garage over the ladder (I know, not very convenient, but just go along with the story). So now the ladder frame is the inertial reference frame and the garage is the moving frame. So, according to that length contraction ‘law’, it’s the garage that gets shorter and it turns out the ladder doesn’t fit any more. Hence, the paradox: does the ladder fit or not? The answer must be unambiguous, no? Yes or no. So what is it?

The paradox pushes us to consider all kinds of important questions which are usually just glossed over. How does we decide if the ladder fits? Well… By closing both the front and back door of course, you’ll say. But then you mean closing them simultaneously, and absolute simultaneity does not exist: two events that appear to happen at the same time in one reference frame may not happen at the same time in another. Only the space-time interval between two events is absolute, in the sense that it’s the same in whatever reference frame we’re measuring it, not the individual space and individual time intervals. Hence, if you’re in the garage shutting those doors at the same time, then that’s your time, but if I am moving with the ladder, I will not see those two doors shutting as something that’s simultaneous. More formally, and using the definition of space-time intervals (and assuming only one space dimension x), we have:

cΔt²– Δx²= cΔt’²– Δx’².

In this equation, we’ll take the x and t coordinates to be those of the inertial frame (so that’s the garage on the left-hand side), while the the primed coordinates (x’ and t’) are the coordinates as measured in the other reference frame, i.e. the reference frame that moves from the perspective of the inertial frame. Indeed, note that we cannot say that one reference frame moves while the other stands still as we we’re talking relative speeds here: one reference frame moves in respect to the other, and vice versa. In any case, the equation with the space-time intervals above implies that:

c(Δt²– Δt’²) – (Δx²– Δx’²) = 0

However, that does not imply that the two terms on the left-hand side of the above equation are zero individually. In fact, they aren’t. Hence, while it must be true that c(Δt²– Δt’²) = Δx²– Δx’², we have:

Δt²– Δt’²≠ 0 and Δx²– Δx’² ≠ 0 or Δt²≠ Δt’²and Δx²≠ Δx’²

To put it simply, if you’re in the garage, and I am moving with the ladder (we’re talking the left-hand side situation) now, you’ll claim that you were able to shut both doors momentarily, so that Δt²= 0. I’ll say: bollocks! Which is rude. I should say: my Δt’²is not equal to zero. Hence, from my point of view, I always saw one of the two doors open and, hence, I don’t think the ladder fits. Hence, what I am seeing, effectively, is the situation on the right-hand side: your garage looks too short for my ladder.

You’ll say: what is this? The ladder fits or it doesn’t, does it? The answer is: no. It is ambiguous. It does depend on your reference frame. It fits in your reference frame but it does not fit in mine. In order to get a non-ambiguous answer you have to stop moving, or I have to stop moving– whatever: the point is that we need to merge our reference frames.

Hence, paradox solved. In fact, now that I think of it, it’s kinda funny that we don’t have such paradoxes for the relativistic mass formula. No one seems to wonder about the apparent contradiction that, if you’re moving away from me, you look heavier than me but that, vice versa, I also look heavier to you. So we both look heavier as seen from our own respective reference frames. So who’s heavier then? Perhaps no one developed a paradox because it is kinda impolite to compare personal weights? 🙂

Of course, I am joking, but think of it: it has to do with our preconceived notions of time and space. Things like inertia (mass is a measure for inertia) don’t grab our attention as much. In any case, now it’s time to discuss time dilation.

Oh ! And do think about that photo-finish picture ! It’s related to the problem of defining what constitutes a length really. 🙂

The twin paradox

I find the twin paradox much more difficult to analyze, and I guess many people do because it’s the one that usually receives all of the attention. [Frankly, I hadn’t heard of this ladder paradox before I started studying physics.] Feynman hardly takes the time to look at it. He basically notes that the situation is not unlike an unstable particle traveling at relativistic speeds: when it does, it lasts (much) longer that its lifetime (measured in the inertial reference frame) suggests. Let me actually just quote Feynman’s account of it:

“Peter and Paul are supposed to be twins, born at the same time. When they are old enough to drive a space ship, Paul flies away at very high speed. Because Peter, who is left on the ground, sees Paul going so fast, all of Paul’s clocks appear to go slower, his heart beats go slower, his thoughts go slower, everything goes slower, from Peter’s point of view. Of course, Paul notices nothing unusual, but if he travels around and about for a while and then comes back, he will be younger than Peter, the man on the ground! That is actually right; it is one of the consequences of the theory of relativity which has been clearly demonstrated. Just as the mu-mesons last longer when they are moving, so also will Paul last longer when he is moving. This is called a “paradox” only by the people who believe that the principle of relativity means that all motion is relative; they say, “Heh, heh, heh, from the point of view of Paul, can’t we say that Peter was moving and should therefore appear to age more slowly? By symmetry, the only possible result is that both should be the same age when they meet.” But in order for them to come back together and make the comparison, Paul must either stop at the end of the trip and make a comparison of clocks or, more simply, he has to come back, and the one who comes back must be the man who was moving, and he knows this, because he had to turn around. When he turned around, all kinds of unusual things happened in his space ship—the rockets went off, things jammed up against one wall, and so on—while Peter felt nothing.

So the way to state the rule is to say that the man who has felt the accelerations, who has seen things fall against the walls, and so on, is the one who would be the younger; that is the difference between them in an “absolute” sense, and it is certainly correct. When we discussed the fact that moving mu-mesons live longer, we used as an example their straight-line motion in the atmosphere. But we can also make mu-mesons in a laboratory and cause them to go in a curve with a magnet, and even under this accelerated motion, they last exactly as much longer as they do when they are moving in a straight line. Although no one has arranged an experiment explicitly so that we can get rid of the paradox, one could compare a mu-meson which is left standing with one that had gone around a complete circle, and it would surely be found that the one that went around the circle lasted longer. Although we have not actually carried out an experiment using a complete circle, it is really not necessary, of course, because everything fits together all right. This may not satisfy those who insist that every single fact be demonstrated directly, but we confidently predict the result of the experiment in which Paul goes in a complete circle.”

[…] Well… I am not sure I am “among those who insist that every single fact be demonstrated directly”, but you’ll admit that Feynman is quite terse here (or more terse than usual, I should say). That being said, I understand why: the calculations involved in demonstrating that the paradox is what it is, i.e. an apparent contradiction only, are not straightforward. I’ve googled a bit but it’s all quite confusing. Good explanations usually involve the so-called Minkowski diagram, also known as the spacetime diagram. You’ve surely seen it before–when the light cone was being discussed and what it implies for the concepts of past, present and future. It’s a way to represent those spacetime intervals. The Minkowski diagram–from the perspective of the twin brother on Earth (hence, we only have unprimed coordinates x and (c)t)– is shown below. Don’t worry about those simultaneity planes as for now. Just try to understand the diagram. The twin brother that stays just moves along the vertical axis: x = 0. His space-traveling brother travels out to some point and then turns back, so he first travels northeast on this diagram and then takes a turn northwest, to meet up again with his brother on Earth.

The point to note is that the twin brother is not traveling along one straight line, but along two. Hence, the argument that we can just as well say his frame of reference is inertial and that of his brother is the moving one is not correct. As Wikipedia notes (from which I got this diagram): “The trajectory of the ship is equally divided between two different inertial frames, while the Earth-based twin stays in the same inertial frame.”

Still, the situation is essentially symmetric and so we could draw a similar-looking spacetime diagram for the primed coordinates, i.e. x’ and ct’, and wonder what’s the difference. That’s where these planes of simultaneity come in. Look at the wonderful animation below: A, B, C are simultaneous events when I am standing still (v = 0). However, when I move at considerable speed (v = 0.3c), that’s no longer the case: it takes more time for news to reach me from ‘point’ A and, hence, assuming news travels at the speed of light, event A appears to happen later. Conversely, event C (in spacetime) appears to have happened before event B. Now that explains these blue so-called simultaneity planes on the diagram above: they’re the white lines traveling from the past to the future on the animation below, but for the trip out only (v > 0). For the trip back, we have the red lines, which correspond to the v = –0.5c situation below. So that’s the return trip (v < 0).

What you see is that, “during the U-turn, the plane of simultaneity jumps from blue to red and very quickly sweeps over a large segment of the world line of the Earth-based twin.” Hence, “when one transfers from the outgoing frame to the incoming frame there is a jump discontinuity in the age of the Earth-based twin.” [I took the quotes taken from Wikipedia here, where you can find the original references.] Now, you will say, that is also symmetric if we switch the reference frames. Yes… Except for the sign. So, yes, it is the traveling brother who effectively skips some time. Paradox solved.

Now… For some real fun…

Now, for some real fun, I’d like to ask you how the world would look like when you were traveling through it riding a photon. So… Think about it. Think hard. I didn’t google at first and I must admit the question really started wracking my brain. There are some many effects to take into account. One basic property, of course, must be that time stands still around you. You see the world as it was when you reached v = c. Well… Yes and no. The fact of the matter is that, because of all the relativistic effects (e.g. aberration, Doppler shift, intensity shifts,…), you actually don’t see a whole lot. One visualization of it (visual effects of relativistic speeds) seems to indicate that (most) science fiction movies actually present the correct picture (if the animation shows the correct visualization, that is): we’re staring into one bright flash of light ahead of us as we’re getting close to v = c. Interesting…

Finally, you should also try to find out what actually happens to the clocks during the deceleration and acceleration as the space ship of that twin brother turns. You’re going to find it fascinating. At the same time, the math behind is, quite simply, daunting and, hence, I won’t even try go into the math of this thing. 🙂

Conclusion

So… Well… That’s it really. I now realize why I never quite got this as a kid. These paradoxes do require some deep thinking and imagination and, most of all, some tools that one just couldn’t find as easily as today.

The Web definitely does make it easier to study without the guidance of professors and the material environment of a university, although I don’t think it can be a substitute for discipline. When everything is said and done, it’s still hard work. Very hard work. But I hope you get there, Vincent ! 🙂 And please do look at that Youtube video by clicking the link above. 🙂

Post scriptum: Because the resolution of the video above is quite low, I looked for others, for example one that describes the journey from the Sun to the Earth, which–as expected–takes about 8 minutes. While it has higher resolution, it is far less informative. I’ll let you google some more. Please tell me if you found something nice. 🙂

The Complementarity Principle

Pre-script (dated 26 June 2020): This post has become less relevant because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. Hence, we recommend you read our recent papers. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

Unlike what you might think when seeing the title of this post, it is not my intention to enter into philosophical discussions here: many authors have been writing about this ‘principle’, most of which–according to eminent physicists–don’t know what they are talking about. So I have no intention to make a fool of myself here too. However, what I do want to do here is explore, in an intuitive way, how the classical and quantum-mechanical explanations of the phenomenon of the diffraction of light are different from each other–and fundamentally so–while, necessarily, having to yield the same predictions. It is in that sense that the two explanations should be ‘complementary’.

The classical explanation

I’ve done a fairly complete analysis of the classical explanation in my posts on Diffraction and the Uncertainty Principle (20 and 21 September), so I won’t dwell on that here. Let me just repeat the basics. The model is based on the so-called Huygens-Fresnel Principle, according to which each point in the slit becomes a source of a secondary spherical wave. These waves then interfere, constructively or destructively, and, hence, by adding them, we get the form of the wave at each point of time and at each point in space behind the slit. The animation below illustrates the idea. However, note that the mathematical analysis does not assume that the point sources are neatly separated from each other: instead of only six point sources, we have an infinite number of them and, hence, adding up the waves amounts to solving some integral (which, as you know, is an infinite sum).

We know what we are supposed to get: a diffraction pattern. The intensity of the light on the screen at the other side depends on (1) the slit width (d), (2) the frequency of the light (λ), and (3) the angle of incidence (θ), as shown below.

$Single_Slit_Diffraction_(english)$

One point to note is that we have smaller bumps left and right. We don’t get that if we’d treat the slit as a single point source only, like Feynman does when he discusses the double-slit experiment for (physical) waves. Indeed, look at the image below: each of the slits acts as one point source only and, hence, the intensity curves I₁and I₂ do not show a diffraction pattern. They are just nice Gaussian “bell” curves, albeit somewhat adjusted because of the angle of incidence (we have two slits above and below the center, instead of just one on the normal itself). So we have an interference pattern on the screen and, now that we’re here, let me be clear on terminology: I am going along with the widespread definition of diffraction being a pattern created by one slit, and the definition of interference as a pattern created by two or more slits. I am noting this just to make sure there’s no confusion.

That should be clear enough. Let’s move on the quantum-mechanical explanation.

The quantum-mechanical explanation

There are several formulations of quantum mechanics: you’ve heard about matrix mechanics and wave mechanics. Roughly speaking, in matrix mechanics “we interpret the physical properties of particles as matrices that evolve in time”, while the wave mechanics approach is primarily based on these complex-valued wave functions–one for each physical property (e.g. position, momentum, energy). Both approaches are mathematically equivalent.

There is also a third approach, which is referred to as the path integral formulation, which “replaces the classical notion of a single, unique trajectory for a system with a sum, or functional integral, over an infinity of possible trajectories to compute an amplitude” (all definitions here were taken from Wikipedia). This approach is associated with Richard Feynman but can also be traced back to Paul Dirac, like most of the math involved in quantum mechanics, it seems. It’s this approach which I’ll try to explain–again, in an intuitive way only–in order to show the two explanations should effectively lead to the same predictions.

The key to understanding the path integral formulation is the assumption that a particle–and a ‘particle’ may refer to both bosons (e.g. photons) or fermions (e.g. electrons)–can follow any path from point A to B, as illustrated below. Each of these paths is associated with a (complex-valued) probability amplitude, and we have to add all these probability amplitudes to arrive at the probability amplitude for the particle to move from A to B.

You can find great animations illustrating what it’s all about in the relevant Wikipedia article but, because I can’t upload video here, I’ll just insert two illustrations from Feynman’s 1985 QED, in which he does what I try to do, and that is to approach the topic intuitively, i.e. without too much mathematical formalism. So probability amplitudes are just ‘arrows’ (with a length and a direction, just like a complex number or a vector), and finding the resultant or final arrow is a matter of just adding all the little arrows to arrive at one big arrow, which is the probability amplitude, which he denotes as P(A, B), as shown below.

This intuitive approach is great and actually goes a very long way in explaining complicated phenomena, such as iridescence for example (the wonderful patterns of color on an oil film!), or the partial reflection of light by glass (anything between 0 and 16%!). All his tricks make sense. For example, different frequencies are interpreted as slower or faster ‘stopwatches’ and, as such, they determine the final direction of the arrows which, in turn, explains why blue and red light are reflected differently. And so on and son. It all works. […] Up to a point.

Indeed, Feynman does get in trouble when trying to explain diffraction. I’ve reproduced his explanation below. The key to the argument is the following:

If we have a slit that’s very wide, there are a lot of possible paths for the photon to take. However, most of these paths cancel each other out, and so that’s why the photon is likely to travel in a straight line. Let me quote Feynman: “When the gap between the blocks is wide enough to allow many neighboring paths to P and Q, the arrows for the paths to P add up (because all the paths to P take nearly the same time), while the paths to Q cancel out (because those paths have a sizable difference in time). So the photomultiplier at Q doesn’t click.” (QED, p.54)
However, “when the gap is nearly closed and there are only a few neighboring paths, the arrows to Q also add up, because there is hardly any difference in time between them, either (see Fig. 34). Of course, both final arrows are small, so there’s not much light either way through such a small hole, but the detector at Q clicks almost as much as the one at P! So when you try to squeeze light too much to make sure it’s going only in a straight line, it refuses to cooperate and begins to spread out.” (QED, p. 55)

This explanation is as simple and intuitive as Feynman’s ‘explanation’ of diffraction using the Uncertainty Principle in his introductory chapter on quantum mechanics (Lectures, I-38-2), which is illustrated below. I won’t go into the detail (I’ve done that before) but you should note that, just like the explanation above, such explanations do not explain the secondary, tertiary etc bumps in the diffraction pattern.

$Diffraction of electrons$

So what’s wrong with these explanations? Nothing much. They’re simple and intuitive, but essentially incomplete, because they do not incorporate all of the math involved in interference. Incorporating the math means doing these integrals for

Electromagnetic waves in classical mechanics: here we are talking ‘wave functions’ with some real-valued amplitude representing the strength of the electric and magnetic field; and
Probability waves: these are complex-valued functions, with the complex-valued amplitude representing probability amplitudes.

The two should, obviously, yield the same result, but a detailed comparison between the approaches is quite complicated, it seems. Now, I’ve googled a lot of stuff, and I duly note that diffraction of electromagnetic waves (i.e. light) is conveniently analyzed by summing up complex-valued waves too, and, moreover, they’re of the same familiar type: ψ = Ae^i(kx–ωt). However, these analyses also duly note that it’s only the real part of the wave that has an actual physical interpretation, and that it’s only because working with natural exponentials (addition, multiplication, integration, derivation, etc) is much easier than working with sine and cosine waves that such complex-valued wave functions are used (also) in classical mechanics. In fact, note the fine print in Feynman’s illustration of interference of physical waves (Fig. 37-2): he calculates the intensities I₁and I₂ by taking the square of the absolute amplitudes ĥ₁ and ĥ₂, and the hat indicates that we’re also talking some complex-valued wave function here.

Hence, we must be talking the same mathematical waves in both explanations, aren’t we? In other words, we should get the same psi functions ψ = Ae^i(kx–ωt) in both explanations, don’t we? Well… Maybe. But… Probably not. As far as I know–but I must be wrong–we cannot just re-normalize the E and B vectors in these electromagnetic waves in order to establish an equivalence with probability waves. I haven’t seen that being done (but I readily admit I still have a lot of reading to do) and so I must assume it’s not very clear-cut at all.

So what? Well… I don’t know. So far, I did not find a ‘nice’ or ‘intuitive’ explanation of a quantum-mechanical approach to the phenomenon of diffraction yielding the same grand diffraction equation, referred to as the Fresnel-Kirchoff diffraction formula (see below), or one of its more comprehensible (because simplified) representations, such as the Fraunhofer diffraction formula, or the even easier formula which I used in my own post (you can google them: they’re somewhat less monstrous and–importantly–they work with real numbers only, which makes them easier to understand).

[…] That looks pretty daunting, isn’t it? You may start to understand it a bit better by noting that (n, r) and (n, s) are angles, so that’s OK in a cosine function. The other variables also have fairly standard interpretations, as shown below, but… Admit it: ‘easy’ is something else, isn’t it?

So… Where are we here? Well… As said, I trust that both explanations are mathematically equivalent – just like matrix and wave mechanics 🙂 –and, hence, that a quantum-mechanical analysis will indeed yield the same formula. However, I think I’ll only understand physics truly if I’ve gone through all of the motions here.

Well then… I guess that should be some kind of personal benchmark that should guide me on this journey, isn’t it? 🙂 I’ll keep you posted.

Post scriptum: To be fair to Feynman, and demonstrating his talent as a teacher once again, he actually acknowledges that the double-slit thought experiment uses simplified assumptions that do not include diffraction effects when the electrons go through the slit(s). He does so, however, only in one of the first chapters of Vol. III of the Lectures, where he comes back to the experiment to further discuss the first principles of quantum mechanics. I’ll just quote him: “Incidentally, we are going to suppose that the holes 1 and 2 are small enough that when we say an electron goes through the hole, we don’t have to discuss which part of the hole. We could, of course, split each hole into pieces with a certain amplitude that the electron goes to the top of the hole and the bottom of the hole and so on. We will suppose that the hole is small enough so that we don’t have to worry about this detail. That is part of the roughness involved; the matter can be made more precise, but we don’t want to do so at this stage.” So here he acknowledges that he omitted the intricacies of diffraction. I noted this only later. Sorry.

A Royal Road to quantum physics?

Pre-script (dated 26 June 2020): This post has become less relevant because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. In addition, some of the material was removed by a dark force. Hence, we recommend you read our recent papers. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

It is said that, when Ptolemy asked Euclid to quickly explain him geometry, Euclid told the King that there was no ‘Royal Road’ to it, by which he meant it’s just difficult and takes a lot of time to understand.

Physicists will tell you the same about quantum physics. So, I know that, at this point, I should just study Feynman’s third Lectures Volume and shut up for a while. However, before I get lost while playing with state vectors, S-matrices, eigenfunctions, eigenvalues and what have you, I’ll try that Royal Road anyway, building on my previous digression on Hamiltonian mechanics.

So… What was that about? Well… If you understood anything from my previous post, it should be that both the Lagrangian and Hamiltonian function use the equations for kinetic and potential energy to derive the equations of motion for a system. The key difference between the Lagrangian and Hamiltonian approach was that the Lagrangian approach yields one differential equation–which had to be solved to yield a functional form for x as a function of time, while the Hamiltonian approach yielded two differential equations–which had to be solved to yield a functional form for both position (x) and momentum (p). In other words, Lagrangian mechanics is a model that focuses on the position variable(s) only, while, in Hamiltonian mechanics, we also keep track of the momentum variable(s). Let me briefly explain the procedure again, so we’re clear on it:

1. We write down a function referred to as the Lagrangian function. The function is L = T – V with T and V the kinetic and potential energy respectively. T has to be expressed as a function of velocity (v) and V has to be expressed as a function of position (x). You’ll say: of course! However, it is an important point to note, otherwise the following step doesn’t make sense. So we take the equations for kinetic and potential energy and combine them to form a function L = L(x, v).

2. We then calculate the so-called Lagrangian equation, in which we use that function L. To be precise: what we have to do is calculate its partial derivatives and insert these in the following equation:

It should be obvious now why I stressed we should write L as a function of velocity and position, i.e. as L = L(x, v). Otherwise those partial derivatives don’t make sense. As to where this equation comes from, don’t worry about it: I did not explain why this works. I didn’t do that here, and I also didn’t do it in my previous post. What we’re doing here is just explaining how it goes, not why.

3. If we’ve done everything right, we should get a second-order differential equation which, as mentioned above, we should then solve for x(t). That’s what ‘solving’ a differential equation is about: find a functional form that satisfies the equation.

Let’s now look at the Hamiltonian approach.

1. We write down a function referred to as the Hamiltonian function. It looks similar to the Lagrangian, except that we sum kinetic and potential energy, and that T has to be expressed as a function of the momentum p. So we have a function H = T + V = H(x, p).

2. We then calculate the so-called Hamiltonian equations, which is a set of two equations, rather than just one equation. [We have two for the one-dimensional situation that we are modeling here: it’s a different story (i.e. we will have more equations) if we’d have more degrees of freedom of course.] It’s the same as in the Lagrangian approach: it’s just a matter of calculating partial derivatives, and insert them in the equations below. Again, note that I am not explaining why this Hamiltonian hocus-pocus actually works. I am just saying how it works.

3. If we’ve done everything right, we should get two first-order differential equations which we should then solve for x(t) and p(t). Now, solving a set of equations may or may not be easy, depending on your point of view. If you wonder how it’s done, there’s excellent stuff on the Web that will show you how (such as, for instance, Paul’s Online Math Notes).

Now, I mentioned in my previous post that the Hamiltonian approach to modeling mechanics is very similar to the approach that’s used in quantum mechanics and that it’s therefore the preferred approach in physics. I also mentioned that, in classical physics, position and momentum are also conjugate variables, and I also showed how we can calculate the momentum as a conjugate variable from the Lagrangian: p = ∂L/∂v. However, I did not dwell on what conjugate variables actually are in classical mechanics. I won’t do that here either. Just accept that conjugate variables, in classical mechanics, are also defined as pairs of variables. They’re not related through some uncertainty relation, like in quantum physics, but they’re related because they can both be obtained as the derivatives of a function which I haven’t introduced as yet. That function is referred to as the action, but… Well… Let’s resist the temptation to digress any further here. If you really want to know what action is–in physics, that is… 🙂 Well… Google it, I’d say. What you should take home from this digression is that position and momentum are also conjugate variables in classical mechanics.

Let’s now move on to quantum mechanics. You’ll see that the ‘similarity’ in approach is… Well… Quite relative, I’d say. 🙂

Position and momentum in quantum mechanics

As you know by now (I wrote at least a dozen posts on this), the concept of position and momentum in quantum mechanics is very different from that in classical physics: we do not have x(t) and p(t) functions which give a unique, precise and unambiguous value for x and p when we assign a value to the time variable and plug it in. No. What we have in quantum physics is some weird wave function, denoted by the Greek letters φ (phi) or ψ (psi) or, using Greek capitals, Φ and Ψ. To be more specific, the psi usually denotes the wave function in the so-called position space (so we write ψ = ψ(x)), and the phi will usually denote the wave function in the so-called momentum space (so we write φ = φ(p)). That sounds more complicated than it is, obviously, but I just wanted to respect terminology here. Finally, note that the ψ(x) and φ(p) wave functions are related through the Uncertainty Principle: they’re conjugate variables, and we have this ΔxΔp = ħ/2 equation, in which the Δ is some standard deviation from some mean value. I should not go into more detail here: you know that by now, don’t you?

While the argument of these functions is some real number, the wave functions themselves are complex-valued, so they have a real and complex amplitude. I’ve also illustrated that a couple of times already but, just to make sure, take a look at the animation below, so you know what we are sort of talking about:

The A and B situations represent a classical oscillator: we know exactly where the red ball is at any point in time.
The C to H situations give us a complex-valued amplitude, with the blue oscillation as the real part, and the pink oscillation as the imaginary part.

So we have such wave function both for x and p. Note that the animation above suggests we’re only looking at the wave function for x but–trust me–we have a similar one for p, and they’re related indeed. [To see how exactly, I’d advise you to go through the proof of the so-called Kennard inequality.] So… What do we do with that?

The position and momentum operators

When we want to know where a particle actually is, or what its momentum is, we need to do something with this wave function ψ or φ. Let’s focus on the position variable first. While the wave function itself is said to have ‘no physical interpretation’ (frankly, I don’t know what that means: I’d think everything has some kind of interpretation (and what’s physical and non-physical?), but let’s not get lost in philosophy here), we know that the square of the absolute value of the probability amplitude yields a probability density. So |ψ(x)|²gives us a probability density function or, to put it simply, the probability to find our ‘particle’ (or ‘wavicle’ if you want) at point x. Let’s now do something more sophisticated and write down the expected value of x, which is usually denoted by 〈x〉 (although that invites confusion with Dirac’s bra-ket notation, but don’t worry about it):

Don’t panic. It’s just an integral. Look at it. ψ* is just the complex conjugate (i.e. a – ib if ψ = a + ib) and you will (or should) remember that the product of a complex number with its (complex) conjugate gives us the square of its absolute value: ψ*ψ = |ψ(x)|². What about that x? Can we just insert that there, in-between ψ* and ψ ? Good question. The answer is: yes, of course! That x is just some real number and we can put it anywhere. However, it’s still a good question because, while multiplication of complex numbers is commutative (hence, z₁z₂ = z₂z₁), the order of our operators – which we will introduce soon – can often not be changed without consequences, so it is something to note.

For the rest, that integral above is quite obvious and it should really not puzzle you: we just multiply a value with its probability of occurring and integrate over the whole domain to get an expected value 〈x〉. Nothing wrong here. Note that we get some real number. [You’ll say: of course! However, I always find it useful to check that when looking at those things mixing complex-valued functions with real-valued variables or arguments. A quick check on the dimensions of what we’re dealing helps greatly in understanding what we’re doing.]

So… You’ve surely heard about the position and momentum operators already. Is that, then, what it is? Doing some integral on some function to get an expected value? Well… No. But there’s a relation. However, let me first make a remark on notation, because that can be quite confusing. The position operator is usually written with a hat on top of the variable – like ẑ – but so I don’t find a hat with every letter with the editor tool for this blog and, hence, I’ll use a bold letter x and p to denote the operator. Don’t confuse it with me using a bold letter for vectors though ! Now, back to the story.

Let’s first give an example of an operator you’re already familiar with in order to understand what an operator actually is. To put it simply: an operator is an instruction to do something with a function. For example: ∂/∂t is an instruction to differentiate some function with regard to the variable t (which usually stands for time). The ∂/∂t operator is obviously referred to as a differentiation operator. When we put a function behind, e.g. f(x, t), we get ∂f(x, t)/∂t, which is just another function in x and t.

So we have the same here: x in itself is just an instruction: you need to put a function behind in order to get some result. So you’ll see it as xψ. In fact, it would be useful to use brackets probably, like x[ψ], especially because I can’t put those hats on the letters here, but I’ll stick to the usual notation, which does not use brackets.

Likewise, we have a momentum operator: p = –iħ∂/∂x. […] Let it sink in. [..]

What’s this? Don’t worry about it. I know: that looks like a very different animal than that x operator. I’ll explain later. Just note, for the moment, that the momentum operator (also) involves a (partial) derivative and, hence, we refer to it as a differential operator (as opposed to differentiation operator). The instruction p = –iħ∂/∂x basically means: differentiate the function with regard to x and multiply with iħ (i.e. the product of Planck’s constant and the imaginary unit i). Nothing wrong with that. Just calculate a derivative and multiply with a tiny imaginary (complex) number.

Now, back to the position operator x. As you can see, that’s a very simple operator–much simpler than the momentum operator in any case. The position operator applied to ψ yields, quite simply, the xψ(x) factor in the integrand above. So we just get a new function xψ(x) when we apply x to ψ, of which the values are simply the product of x and ψ(x). Hence, we write xψ = xψ.

Really? Is it that simple? Yes. For now at least. 🙂

Back to the momentum operator. Where does that come from? That story is not so simple. [Of course not. It can’t be. Just look at it.] Because we have to avoid talking about eigenvalues and all that, my approach to the explanation will be quite intuitive. [As for ‘my’ approach, let me note that it’s basically the approach as used in the Wikipedia article on it. :-)] Just stay with me for a while here.

Let’s assume ψ is given by ψ = eⁱ^(kx^–ωt). So that’s a nice periodic function, albeit complex-valued. Now, we know that functional form doesn’t make all that much sense because it corresponds to the particle being everywhere, because the square of its absolute value is some constant. In fact, we know it doesn’t even respect the normalization condition: all probabilities have to add up to 1. However, that being said, we also know that we can superimpose an infinite number of such waves (all with different k and ω) to get a more localized wave train, and then re-normalize the result to make sure the normalization condition is met. Hence, let’s just go along with this idealized example and see where it leads.

We know the wave number k (i.e. its ‘frequency in space’, as it’s often described) is related to the momentum p through the de Broglie relation: p = ħk. [Again, you should think about a whole bunch of these waves and, hence, some spread in k corresponding to some spread in p, but just go along with the story for now and don’t try to make it even more complicated.] Now, if we differentiate with regard to x, and then substitute, we get ∂ψ/∂x = ∂eⁱ^(kx^–ωt)/∂x = ikeⁱ^(kx^–ωt)= ikψ, or

So what is this? Well… On the left-hand side, we have the (partial) derivative of a complex-valued function (ψ) with regard to x. Now, that derivative is, more likely than not, also some complex-valued function. And if you don’t believe me, just look at the right-hand side of the equation, where we have that i and ψ. In fact, the equation just shows that, when we take that derivative, we get our original function ψ but multiplied by ip/ħ. Hey! We’ve got a differential equation here, don’t we? Yes. And the solution for it is… Well… The natural exponential. Of course! That should be no surprise because we started out with a natural exponential as functional form! So that’s not the point. What is the point, then? Well… If we bring that i/ħ factor to the other side, we get:

(–i/ħ)(∂ψ/∂x) = pψ

[If you’re confused about the –i, remember that i^–1= 1/i = –i.] So… We’ve got pψ on the right-hand side now. So… Well… That’s like xψ, isn’t it? Yes. 🙂 If we define the momentum operator as p = (–i/ħ)(∂/∂x), then we get pψ = pψ. So that’s the same thing as for the position operator. It’s just that p is… Well… A more complex operator, as it has that –i/ħ factor in it. And, yes, of course it also involves an instruction to differentiate, which also sets it apart from the position operator, which is just an instruction to multiply the function with its argument.

I am sure you’ll find this funny–perhaps even fishy–business. And, yes, I have the same questions: what does it all mean? I can’t answer that here. As for now, just accept that this position and momentum operator are what they are, and that I can’t do anything about that. But… I hear you sputter: what about their interpretation? Well… Sorry… I could say that the functions xψ and pψ are so-called linear maps but that is not likely to help you much in understanding what these operators really do. You – and I for sure 🙂 – will indeed have to go through that story of eigenvalues to a somewhat deeper understanding of what these operators actually are. That’s just how it is. As for now, I just have to move on. Sorry for letting you down here. 🙂

Energy operators

Now that we sort of ‘understand’ those position and momentum operators (or their mathematical form at least), it’s time to introduce the energy operators. Indeed, in quantum mechanics, we’ve also got an operator for (a) kinetic energy, and for (b) potential energy. These operators are also denoted with a hat above the T and V symbol. All quantum-mechanical operators are like that, it seems. However, because of the limitations of the editor tool here, I’ll also use a bold T and V respectively. Now, I am sure you’ve had enough of this operators, so let me just jot them down:

V = V, so that’s just an instruction to multiply a function with V = V(x, t). That’s easy enough because that’s just like the position vector.
As for T, that’s more complicated. It involves that momentum operator p, which was also more complicated, remember? Let me just give you the formula:

T = p·p/2m = p²/2m.

So we multiply the operator p with itself here. What does that mean? Well… Because the operator involves a derivative, it means we have to take the derivative twice and… No ! Well… Let me correct myself: yes and no. 🙂 That p·p product is, strictly speaking, a dot product between two vectors, and so it’s not just a matter of differentiating twice. Now that we are here, we may just as well extend the analysis a bit and assume that we also have a y and z coordinate, so we’ll have a position vector r = (x, y, z). [Note that r is a vector here, not an operator. !?! Oh… Well…] Extending the analysis to three (or more) dimensions means that we should replace the differentiation operator by the so-called gradient or del operator: ∇ = (∂/∂x, ∂/∂y, ∂/∂z). And now that dot product p·p will, among other things, yield another operator which you’re surely familiar with: the Laplacian. Let me remind you of it:

Hence, we can write the kinetic energy operator T as:

I quickly copied this formula from Wikipedia, which doesn’t have the limitation of the WordPress editor tool, and so you see it now the way you should see it, i.e. with the hat notation. 🙂

[…]

In case you’re despairing, hang on ! We’re almost there. 🙂 We can, indeed, now define the Hamiltonian operator that’s used in quantum mechanics. While the Hamiltonian function was the sum of the potential and kinetic energy functions in classical physics, in quantum mechanics we add the two energy operators. You’ll grumble and say: that’s not the same as adding energies. And you’re right: adding operators is not the same as adding energy functions. Of course it isn’t. 🙂 But just stick to the story, please, and stop criticizing. [Oh – just in case you wonder where that minus sign comes from: i²= –1, of course.]

Adding the two operators together yields the following:

So. Yes. That’s the famous Hamiltonian operator.

OK. So what?

Yes…. Hmm… What do we do with that operator? Well… We apply it to the function and so we write Hψ = … Hmm…

Well… What?

Well… I am not writing this post just to give some definitions of the type of operators that are used in quantum mechanics and then just do obvious stuff by writing it all out. No. I am writing this post to illustrate how things work.

OK. So how does it work then?

Well… It turns out that, in quantum mechanics, we have similar equations as in classical mechanics. Remember that I just wrote down the set of (two) differential equations when discussing Hamiltonian mechanics? Here I’ll do the same. The Hamiltonian operator appears in an equation of which you’ve surely heard of and which, just like me, you’d love to understand–and then I mean: understand it fully, completely, and intuitively. […] Yes. It’s the Schrödinger equation:

Note, once again, I am not saying anything about where this equation comes from. It’s like jotting down that Lagrange equation, or the set of Hamiltonian equations: I am not saying anything about the why of all this hocus pocus. I am just saying how it goes. So we’ve got another differential equation here, and we have to solve it. If we all write it out using the above definition of the Hamiltonian operator, we get:

If you’re still with me, you’ll immediately wonder about that μ. Well… Don’t. It’s the mass really, but the so-called reduced mass. Don’t worry about it. Just google it if you want to know more about this concept of a ‘reduced’ mass: it’s a fine point which doesn’t matter here really. The point is the grand result.

But… So… What is the grand result? What are we looking at here? Well… Just as I said above: that Schrödinger equation is a differential equation, just like those equations we got when applying the Lagrangian and Hamiltonian approach to modeling a dynamic system in classical mechanics, and, hence, just like what we (were supposed to) do there, we have to solve it. 🙂 Of course, it looks much more daunting than our Lagrangian or Hamiltonian differential equations, because we’ve got complex-valued functions here, and you’re probably scared of that iħ factor too. But you shouldn’t be. When everything is said and done, we’ve got a differential equation here that we need to solve for ψ. In other words, we need to find functional forms for ψ that satisfy the above equation. That’s it. Period.

So how do these solutions look like? Well, they look like those complex-valued oscillating things in the very first animation above. Let me copy them again:

So… That’s it then? Yes. I won’t say anything more about it here, because (1) this post has become way too long already, and so I won’t dwell on the solutions of that Schrödinger equation, and because (2) I do feel it’s about time I really start doing what it takes, and that’s to work on all of the math that’s necessary to actually do all that hocus-pocus. 🙂

Post scriptum: As for understanding the Schrödinger equation “fully, completely, and intuitively”, I am not sure that’s actually possible. But I am trying hard and so let’s see. 🙂 I’ll tell you after I mastered the math. But something inside of me tells me there’s indeed no Royal Road to it. 🙂

Post scriptum 2 (dated 16 November 2015): I’ve added this post scriptum, more than a year later after writing all of the above, because I now realize how immature it actually is. If you really want to know more about quantum math, then you should read my more recent posts, like the one on the Hamiltonian matrix. It’s not that anything that I write above is wrong—it isn’t. But… Well… It’s just that I feel that I’ve jumped the gun. […] But then that’s probably not a bad thing. 🙂

Newtonian, Lagrangian and Hamiltonian mechanics

Post scriptum (dated 16 November 2015): You’ll smile because… Yes, I am starting this post with a post scriptum, indeed. 🙂 I’ve added it, a year later or so, because, before you continue to read, you should note I am not going to explain the Hamiltonian matrix here, as it’s used in quantum physics. That’s the topic of another post, which involves far more advanced mathematical concepts. If you’re here for that, don’t read this post. Just go to my post on the matrix indeed. 🙂 But so here’s my original post. I wrote it to tie up some loose end. 🙂

As an economist, I thought I knew a thing or two about optimization. Indeed, when everything is said and done, optimization is supposed to an economist’s forte, isn’t it? 🙂 Hence, I thought I sort of understood what a Lagrangian would represent in physics, and I also thought I sort of intuitively understood why and how it could be used it to model the behavior of a dynamic system. In short, I thought that Lagrangian mechanics would be all about optimizing something subject to some constraints. Just like in economics, right?

[…] Well… When checking it out, I found that the answer is: yes, and no. And, frankly, the honest answer is more no than yes. 🙂 Economists (like me), and all social scientists (I’d think), learn only about one particular type of Lagrangian equations: the so-called Lagrange equations of the first kind. This approach models constraints as equations that are to be incorporated in an objective function (which is also referred to as a Lagrangian–and that’s where the confusion starts because it’s different from the Lagrangian that’s used in physics, which I’ll introduce below) using so-called Lagrange multipliers. If you’re an economist, you’ll surely remember it: it’s a problem written as “maximize f(x, y) subject to g(x, y) = c”, and we solve it by finding the so-called stationary points (i.e. the points for which the derivative is zero) of the (Lagrangian) objective function f(x, y) + λ[g(x, y) – c].

Now, it turns out that, in physics, they use so-called Lagrange equations of the second kind, which incorporate the constraints directly by what Wikipedia refers to as a “judicious choice of generalized coordinates.”

Generalized coordinates? Don’t worry about it: while generalized coordinates are defined formally as “parameters that describe the configuration of the system relative to some reference configuration”, they are, in practice, those coordinates that make the problem easy to solve. For example, for a particle (or point) that moves on a circle, we’d not use the Cartesian coordinates x and y but just the angle that locates the particles (or point). That simplifies matters because then we only need to find one variable. In practice, the number of parameters (i.e. the number of generalized coordinates) will be defined by the number of degrees of freedom of the system, and we know what that means: it’s the number of independent directions in which the particle (or point) can move. Now, those independent directions may or may not include the x, y and z directions (they may actually exclude one of those), and they also may or may not include rotational and/or vibratory movements. We went over that when discussing kinetic gas theory, so I won’t say more about that here.

So… OK… That was my first surprise: the physicist’s Lagrangian is different from the social scientist’s Lagrangian.

The second surprise was that all physics textbooks seem to dislike the Lagrangian approach. Indeed, they opt for a related but different function when developing a model of a dynamic system: it’s a function referred to as the Hamiltonian. The modeling approach which uses the Hamiltonian instead of the Lagrangian is, of course, referred to as Hamiltonian mechanics. We may think the preference for the Hamiltonian approach has to do with William Rowan Hamilton being Anglo-Irish, while Joseph-Louis Lagrange (born as Giuseppe Lodovico Lagrangia) was Italian-French but… No. 🙂

And then we have good old Newtonian mechanics as well, obviously. In case you wonder what that is: it’s the modeling approach that we’ve been using all along. 🙂 But I’ll remind you of what it is in a moment: it amounts to making sense of some situation by using Newton’s laws of motion only, rather than a more sophisticated mathematical argument using more abstract concepts, such as energy, or action.

Introducing Lagrangian and Hamiltonian mechanics is quite confusing because the functions that are involved (i.e. the so-called Lagrangian and Hamiltonian functions) look very similar: we write the Lagrangian as the difference between the kinetic and potential energy of a system (L = T – V), while the Hamiltonian is the sum of both (H = T + V). Now, I could make this post very simple and just ask you to note that both approaches are basically ‘equivalent’ (in the sense that they lead to the same solutions, i.e. the same equations of motion expressed as a function of time) and that a choice between them is just a matter of preference–like choosing between an English versus a continental breakfast. 🙂 Of course, an English breakfast has usually some extra bacon, or a sausage, so you get more but… Well… Not necessarily something better. 🙂 So that would be the end of this digression then, and I should be done. However, I must assume you’re a curious person, just like me, and, hence, you’ll say that, while being ‘equivalent’, they’re obviously not the same. So how do the two approaches differ exactly?

Let’s try to get a somewhat intuitive understanding of it all by taking, once again, the example of a simple harmonic oscillator, as depicted below. It could be a mass on a spring. In fact, our example will, in fact, be that of an oscillating mass on a spring. Let’s also assume there’s no damping, because that makes the analysis soooooooo much easier.

Of course, we already know all of the relevant equations for this system just from applying Newton’s laws (so that’s Newtonian mechanics). We did that in a previous post. [I can’t remember which one, but I am sure I’ve done this already.] Hence, we don’t really need the Lagrangian or Hamiltonian. But, of course, that’s the point of this post: I want to illustrate how these other approaches to modeling a dynamic system actually work, and so it’s good we have the correct answer already so we can make sure we’re not going off track here. So… Let’s go… 🙂

I. Newtonian mechanics

Let me recapitulate the basics of a mass on a spring which, in jargon, is called a harmonic oscillator. Hooke’s law is there: the force on the mass is proportional to its distance from the zero point (i.e. the displacement), and the direction of the force is towards the zero point–not away from it, and so we have a minus sign. In short, we can write:

F = –kx (i.e. Hooke’s law)

Now, Newton‘s Law (Newton’s second law to be precise) says that F is equal to the mass times the acceleration: F = ma. So we write:

F = ma = m(d²x/dt²) = –kx

So that’s just Newton’s law combined with Hooke’s law. We know this is a differential equation for which there’s a general solution with the following form:

x(t) = A·cos(ωt + α)

If you wonder why… Well… I can’t digress on that here again: just note, from that differential equation, that we apparently need a function x(t) that yields itself when differentiated twice. So that must be some sinusoidal function, like sine or cosine, because these do that. […] OK… Sorry, but I must move on.

As for the new ‘variables’ (A, ω and α), A depends on the initial condition and is the (maximum) amplitude of the motion. We also already know from previous posts (or, more likely, because you already know a lot about physics) that A is related to the energy of the system. To be precise: the energy of the system is proportional to the square of the amplitude: E ∝ A². As for ω, the angular frequency, that’s determined by the spring itself and the oscillating mass on it: ω = (k/m)^1/2= 2π/T = 2πf (with T the period, and f the frequency expressed in oscillations per second, as opposed to the angular frequency, which is the frequency expressed in radians per second). Finally, I should note that α is just a phase shift which depends on how we define our t = 0 point: if x(t) is zero at t = 0, then that cosine function should be zero and then α will be equal to ±π/2.

OK. That’s clear enough. What about the ‘operational currency of the universe’, i.e. the energy of the oscillator? Well… I told you already/ We don’t need the energy concept here to find the equation of motion. In fact, that’s what distinguishes this ‘Newtonian’ approach from the Lagrangian and Hamiltonian approach. But… Now that we’re at it, and we have to move to a discussion of these two animals (I mean the Lagrangian and Hamiltonian), let’s go for it.

We have kinetic versus potential energy. Kinetic energy (T) is what it always is. It depends on the velocity and the mass: K.E. = T = mv²/2 = m(dx/dt)²/2 = p²/2m. Huh? What’s this expression with p in it? […] It’s momentum: p = mv. Just check it: it’s an alternative formula for T really. Nothing more, nothing less. I am just noting it here because it will pop up again in our discussion of the Hamiltonian modeling approach. But that’s for later. Onwards!

What about potential energy (V)? We know that’s equal to V = kx²/2. And because energy is conserved, potential energy (V) and kinetic energy (T) should add up to some constant. Let’s check it: dx/dt = d[Acos(ωt + α)]/dt = –Aωsin(ωt + α). [Please do the derivation: don’t accept things at face value. :-)] Hence, T = mA²ω²sin²(ωt + α)/2 = mA²(k/m)sin²(ωt + α)/2 = kA²sin²(ωt + α)/2. Now, V is equal to V = kx²/2 = k[Acos(ωt + α)]²/2 = k[Acos(ωt + α)]²/2 = kA²cos²(ωt + α)/2. Adding both yields:

T + V = kA²sin²(ωt + α)/2 + kA²cos²(ωt + α)/2

= (1/2)kA²[sin²(ωt + α) + cos²(ωt + α)] = kA²/2.

Ouff! Glad that worked out: the total energy is, indeed, proportional to the square of the amplitude and the constant of proportionality is equal to k/2. [You should now wonder why we do not have m in this formula but, if you’d think about it, you can answer your own question: the amplitude will depend on the mass (bigger mass, smaller amplitude, and vice versa), so it’s actually in the formula already.]

The point to note is that this Hamiltonian function H = T + V is just a constant, not only for this particular case (an oscillation without damping), but in all cases where H represents the total energy of a (closed) system.

OK. That’s clear enough. How does our Lagrangian look like? That’s not a constant obviously. Just so you can visualize things, I’ve drawn the graph below:

The red curve represents kinetic energy (T) as a function of the displacement x: T is zero at the turning points, and reaches a maximum at the x = 0 point.
The blue curve is potential energy (V): unlike T, V reaches a maximum at the turning points, and is zero at the x = 0 point. In short, it’s the mirror image of the red curve.
The Lagrangian is the green graph: L = T – V. Hence, L reaches a minimum at the turning points, and a maximum at the x = 0 point.

While that green function would make an economist think of some Lagrangian optimization problem, it’s worth noting we’re not doing any such thing here: we’re not interested in stationary points. We just want the equation(s) of motion. [I just thought that would be worth stating, in light of my own background and confusion in regard to it all. :-)]

OK. Now that we have an idea of what the Lagrangian and Hamiltonian functions are (it’s probably worth noting also that we do not have a ‘Newtonian function’ of some sort), let us now show how these ‘functions’ are used to solve the problem. What problem? Well… We need to find some equation for the motion, remember? [I find that, in physics, I often have to remind myself of what the problem actually is. Do you feel the same? 🙂 ] So let’s go for it.

II. Lagrangian mechanics

As this post should not turn into a chapter of some math book, I’ll just describe the how, i.e. I’ll just list the steps one should take to model and then solve the problem, and illustrate how it goes for the oscillator above. Hence, I will not try to explain why this approach gives the correct answer (i.e. the equation(s) of motion). So if you want to know why rather than how, then just check it out on the Web: there’s plenty of nice stuff on math out there.

The steps that are involved in the Lagrangian approach are the following:

Compute (i.e. write down) the Lagrangian function L = T – V. Hmm? How do we do that? There’s more than one way to express T and V, isn’t it? Right you are! So let me clarify: in the Lagrangian approach, we should express T as a function of velocity (v) and V as a function of position (x), so your Lagrangian should be L = L(x, v). Indeed, if you don’t pick the right variables, you’ll get nowhere. So, in our example, we have L = mv²/2 – kx²/2.
Compute the partial derivatives ∂L/∂x and ∂L/∂v. So… Well… OK. Got it. Now that we’ve written L using the right variables, that’s a piece of cake. In our example, we have: ∂L/∂x = – kx and ∂L/∂v = mv. Please note how we treat x and v as independent variables here. It’s obvious from the use of the symbol for partial derivatives: ∂. So we’re not taking any total differential here or so. [This is an important point, so I’d rather mention it.]
Write down (‘compute’ sounds awkward, doesn’t it?) Lagrange’s equation: d(∂L/∂v)/dt = ∂L/∂x. […] Yep. That’s it. Why? Well… I told you I wouldn’t tell you why. I am just showing the how here. This is Lagrange’s equation and so you should take it for granted and get on with it. 🙂 In our example: d(∂L/∂v)/dt = d(mv)/dt = –k(dx/dt) = ∂L/∂x = – kx. We can also write this as m(dv/dt) = m(d²x/dt²) = –kx.
Finally, solve the resulting differential equation. […] ?! Well… Yes. […] Of course, we’ve done that already. It’s the same differential equation as the one we found in our ‘Newtonian approach’, i.e. the equation we found by combining Hooke’s and Newton’s laws. So the general solution is x(t) = Acos(ωt + α), as we already noted above.

So, yes, we’re solving the same differential equation here. So you’ll wonder what’s the difference then between Newtonian and Lagrangian mechanics? Yes, you’re right: we’re indeed solving the same second-order differential equation here. Exactly. Fortunately, I’d say, because we don’t want any other equation(s) of motion because we’re talking the same system. The point is: we got that differential equation using an entirely different procedure, which I actually didn’t explain at all: I just said to compute this and then that and… – Surprise, surprise! – we got the same differential equation in the end. 🙂 So, yes, the Newtonian and Lagrangian approach to modeling a dynamic system yield the same equations, but the Lagrangian method is much more (very much more, I should say) convenient when we’re dealing with lots of moving bits and if there’s more directions (i.e. degrees of freedom) in which they can move.

In short, Lagrange could solve a problem more rapidly than Newton with his modeling approach and so that’s why his approach won out. 🙂 In fact, you’ll usually see the spatial variables noted as q_j. In this notation, j = 1, 2,… n, and n is the number of degrees of freedom, i.e. the directions in which the various particles can move. And then, of course, you’ll usually see a second subscript i = 1, 2,… m to keep track of every q_jfor each and every particle in the system, so we’ll have n×m q_ij‘s in our model and so, yes, good to stick to Lagrange in that case.

OK. You get that, I assume. Let’s move on to Hamiltonian mechanics now.

III. Hamiltonian mechanics

The steps here are the following. [Again, I am just explaining the how, not the why. You can find mathematical proofs of why this works in handbooks or, better still, on the Web.]

The first step is very similar as the one above. In fact, it’s exactly the same: write T and V as a function of velocity (v) and position (x) respectively and construct the Lagrangian. So, once again, we have L = L(x, v). In our example: L(x, v) = mv²/2 – kx²/2.
The second step, however, is different. Here, the theory becomes more abstract, as the Hamiltonian approach does not only keep track of the position but also of the momentum of the particles in a system. Position (x) and momentum (p) are so-called canonical variables in Hamiltonian mechanics, and the relation with Lagrangian mechanics is the following: p = ∂L/∂v. Huh? Yeah. Again, don’t worry about the why. Just check it for our example: ∂(mv²/2 – kx²/2)/∂v = 2mv/2 = mv. So, yes, it seems to work. Please note, once again, how we treat x and v as independent variables here, as is evident from the use of the symbol for partial derivatives. Let me get back to the lesson, however. The second step is: calculate the conjugate variables. In more familiar wording: compute the momenta.
The third step is: write down (or ‘build’ as you’ll see it, but I find that wording strange too) the Hamiltonian function H = T + V. We’ve got the same problem here as the one I mentioned with the Lagrangian: there’s more than one way to express T and V. Hence, we need some more guidance. Right you are! When writing your Hamiltonian, you need to make sure you express the kinetic energy as a function of the conjugate variable, i.e. as a function of momentum, rather than velocity. So we have H = H(x, p), not H = H(x, v)! In our example, we have H = T + V = p²/2m + kx²/2.
Finally, write and solve the following set of equations: (I) ∂H/∂p = dx/dt and (II) –∂H/∂x = dp/dt. [Note the minus sign in the second equation.] In our example: (I) p/m = dx/dt and (II) –kx = dp/dt. The first equation is actually nothing but the definition of p: p = mv, and the second equation is just Hooke’s law: F = –kx. However, from a formal-mathematical point of view, we have two first-order differential equations here (as opposed to one second-order equation when using the Lagrangian approach), which should be solved simultaneously in order to find position and momentum as a function of time, i.e. x(t) and p(t). The end result should be the same: x(t) = Acos(ωt + α) and p(t) = … Well… I’ll let you solve this: time to brush up your knowledge about differential equations. 🙂

You’ll say: what the heck? Why are you making things so complicated? Indeed, what am I doing here? Am I making things needlessly complicated?

The answer is the usual one: yes, and no. Yes. If we’d want to do stuff in the classical world only, the answer seems to be: yes! In that case, the Lagrangian approach will do and may actually seem much easier, because we don’t have a set of equations to solve. And why would we need to keep track of p(t)? We’re only interested in the equation(s) of motion, aren’t we? Well… That’s why the answer to your question is also: no! In classical mechanics, we’re usually only interested in position, but in quantum mechanics that concept of conjugate variables (like x and p indeed) becomes much more important, and we will want to find the equations for both. So… Yes. That means a set of differential equations (one for each variable (x and p) in the example above) rather than just one. In short, the real answer to your question in regard to the complexity of the Hamiltonian modeling approach is the following: because the more abstract Hamiltonian approach to mechanics is very similar to the mathematics used in quantum mechanics, we will want to study it, because a good understanding of Hamiltonian mechanics will help us to understand the math involved in quantum mechanics. And so that’s the reason why physicists prefer it to the Lagrangian approach.

[…] Really? […] Well… At least that’s what I know about it from googling stuff here and there. Of course, another reason for physicists to prefer the Hamiltonian approach may well that they think social science (like economics) isn’t real science. Hence, we – social scientists – would surely expect them to develop approaches that are much more intricate and abstract than the ones that are being used by us, wouldn’t we?

[…] And then I am sure some of it is also related to the Anglo-French thing. 🙂

Post scriptum 1 (dated 21 March 2016): I hate to write about stuff and just explain the how—rather than the why. However, in this case, the why is really rather complicated. The math behind is referred to as calculus of variations – which is a rather complicated branch of mathematics – but the physical principle behind is the Principle of Least Action. Just click the link, and you’ll see how the Master used to explain stuff like this. It’s an easy and difficult piece at the same time. Near the end, however, it becomes pretty complicated, as he applies the theory to quantum mechanics, indeed. In any case, I’ll let you judge for yourself. 🙂

Post scriptum 2 (dated 13 September 2017): I started a blog on the Exercises on Feynman’s Lectures, and the posts on the exercises on Chapter 4 have a lot more detail, and basically give you all the math you’ll ever want on this. Just click the link. However, let me warn you: the math is not easy. Not at all, really.

Complex Fourier analysis: an introduction

Pre-script (dated 26 June 2020): This post has become less relevant (almost irrelevant, I would say) because my views on the nature of the concept of uncertainty in the context of quantum mechanics has evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. Hence, we recommend you read our recent papers. I keep blog posts like these to see where I came from. I might review them one day, but I currently don’t have the time or energy for it. It is still interesting, though—in particular because I start by pointing out yet another error or myth in quantum mechanics that gets repeated all too often.

Original post:

One of the most confusing sentences you’ll read in an introduction to quantum mechanics – not only in those simple (math-free) popular books but also in Feynman’s Lecture introducing the topic – is that we cannot define a unique wavelength for a short wave train. In Feynman’s words: “Such a wave train does not have a definite wavelength; there is an indefiniteness in the wave number that is related to the finite length of the train, and thus there is an indefiniteness in the momentum.” (Feynman’s Lectures, Vol. I, Ch. 38, section 1).

That is not only confusing but, in some way, actually wrong. In fact, this is an oft-occurring statement which has effectively hampered my own understanding of quantum mechanics for a long time, and it was only when I had a closer look at what a Fourier analysis really is that I understood what Feynman, and others, wanted to say. In short, it’s a classic example of where a ‘simple’ account of things can lead you astray.

Indeed, we can all imagine a short wave train with a very definite frequency. Just take any sinusoidal function and multiply it with a so-called envelope function in order to shape it into a short pulse. Transients have that shape, and I gave an example in previous posts. Another example is given below. I copied it from the Wikipedia article on Fourier analysis: f(t) is a product of two factors:

The first factor in the product is a cosine function: cos[2π(3t)] to be precise.
The second factor is an exponential function: exp(–πt²).

The frequency of this ‘product function’ is quite precise: cos[2π(3t)] = cos[6πt] = cos[6π(t + 1/3)] for all values t, and so its period is equal to 1/3. [If f(x) is a function with period P, then f(ax+b), where a is a positive constant, is periodic with period P/a.] The only thing that the second factor, i.e. exp(–πt²), does is to shape this cosine function into a nice wave train, as it quickly tends to zero on both sides of the t = 0 point. So that second function is a nice simple bell curve (just plot the graph with a graph plotter) and it doesn’t change the period (or frequency) of the product. In short, the oscillation below–which we should imagine as the representation of ‘something’ traveling through space–has a very definite frequency. So what’s Feynman saying above? There’s no Δf or Δλ here, is there?

The point to note is that these Δ concepts – Δf, Δλ, and so on – actually have very precise mathematical definitions, as one would expect in physics: they usually refer to the standard deviation of the distribution of a variable around the mean.

[…] OK, you’ll say. So what?

Well… That f(t) function above can – and, more importantly, should – be written as the sum of a potentially infinite number of waves in order to make sense of the Δf and Δλ factors in those uncertainty relations. Each of these component waves has a very specific frequency indeed, and each one of them makes its own contribution to the resultant wave. Hence, there is a distribution function for these frequencies, and so that is what Δf refers to. In other words, unlike what you’d think when taking a quick look at that graph above, Δf is not zero. So what is it then?

Well… It’s tempting to get lost in the math of it all now but I don’t want this blog to be technical. The basic ideas, however, are the following. We have a real-valued function here, f(t), which is defined from –∞ to +∞, i.e. over its so-called time domain. Hence, t ranges from –∞ to +∞ (the definition of the zero point is a matter of convention only, and we can easily change the origin by adding or subtracting some constant). [Of course, we could – and, in fact, we should – also define it over a spatial domain, but we’ll keep the analysis simple by leaving out the spatial variable (x).]

Now, the so-called Fourier transform of this function will map it to its so-called frequency domain. The animation below (for which the credit must, once again, go to Wikipedia, from which I borrow most of the material here) clearly illustrates the idea. I’ll just copy the description from the same article: “In the first frames of the animation, a function f is resolved into Fourier series: a linear combination of sines and cosines (in blue). The component frequencies of these sines and cosines spread across the frequency spectrum, are represented as peaks in the frequency domain, as shown shown in the last frames of the animation). The frequency domain representation of the function, $\hat{f}$ , is the collection of these peaks at the frequencies that appear in this resolution of the function.”

[…] OK. You sort of get this (I hope). Now we should go a couple of steps further. In quantum mechanics, we’re talking not real-valued waves but complex-valued waves adding up to give us the resultant wave. Also, unlike what’s shown above, we’ll have a continuous distribution of frequencies. Hence, we’ll not have just six discrete values for the frequencies (and, hence, just six component waves), but an infinite number of them. So how does that work? Well… To do the Fourier analysis, we need to calculate the value of the following integral for each possible frequency, which I’ll denote with the Greek letter nu (ν), as we’ve used the f symbol already–not for the frequency but to denote the function itself! Let me just jot down that integral:

Huh? Don’t be scared now. Just try to understand what it actually represents. So just relax and take a long hard look at it. Note, first, that the integrand (i.e. the function that is to be integrated, between the integral sign and the dt, so that’s f(t)e^–i2πtν) is a complex-valued function (that should be very obvious from the i in the exponent of e). Secondly, note that we need to do such integral for each value of ν. So, for each possible value of ν, we have t ranging from –∞ to +∞ in that integral. Hmm… OK. So… How does that work? Well… The illustration below shows the real and imaginary part respectively of the integrand for ν = 3. [Just in case you still don’t get it: we fix ν here (ν = 3), and calculate the value of the real and imaginary part of the integrand for each possible value of t, so t ranges from –∞ to +∞ indeed.]

So what do we see here? The first thing you should note is that the value of both the real and imaginary part of the integrand quickly tends to zero on both sides of the t = 0 point. That’s because of the shape of f(t), which does exactly the same. However, in-between those ‘zero or close-to-zero values’, the integrand does take on very specific non-zero values. As for the real part of the integrand, which is denoted by Re[e^−2πi(3t)f(t)], we see that’s always positive, with a peak value equal to one at t = 0. Indeed, the real part of the integrand is always positive because f(t) and the real part of e^−2πi(3t)oscillate at the same rate. Hence, when f(t) is positive, so is the real part of e^−2πi(3t), and when f(t) is negative, so is the real part of e^−2πi(3t). However, the story is obviously different for the imaginary part of the integrand, denoted by Im[e^−2πi(3t)f(t)]. That’s because, in general, e^iθ = cosθ + isinθ and the sine and cosine function are essentially the same functions except for a phase difference of π/2 (remember: sin(θ+π/2) = cosθ).

Capito? No? Hmm… Well… Try to read what I am writing above once again. Else, just give up. 🙂

I know this is getting complicated but let me try to summarize what’s going on here. The bottom line is that the integral above will yield a positive real number, 0.5 to be precise (as noted in the margin of the illustration), for the real part of the integrand, but it will give you a zero value for its imaginary part (also as noted in the margin of the illustration). [As for the math involved in calculating an integral of a complex-valued function (with a real-valued argument), just note that we should indeed just separate the real and imaginary parts and integrate separately. However, I don’t want you to get lost in the math so don’t worry about it too much. Just try to stick to the main story line here.]

In short, what we have here is a very significant contribution (the associated density is 0.5) of the frequency ν = 3.

Indeed, let’s compare it to the contribution of the wave with frequency ν = 5. For ν = 5, we get, once again, a value of zero when integrating the imaginary part of the integral above, because the positive and negative values cancel out. As for the real part, we’d think they would do the same if we look at the graph below, but they don’t: the integral does yield, in fact, a very tiny positive value: 1.7×10^–6(so we’re talking 1.7 millionths here). That means that the contribution of the component wave with frequency ν = 5 is close to nil but… Well… It’s not nil: we have some contribution here (i.e. some density in other words).

You get the idea (I hope). We can, and actually should, calculate the value of that integral for each possible value of ν. In other words, we should calculate the integral over the entire frequency domain, so that’s for ν ranging from –∞ to +∞. However, I won’t do that. 🙂 What I will do is just show you the grand general result (below), with the particular results (i.e. the values of 0.5 and 1.7×10^–6for ν = 3 and ν = 5) as a green and red dot respectively. [Note that the graph below uses the ξ symbol instead of ν: I used ν because that’s a more familiar symbol, but so it doesn’t change the analysis.]

Now, if you’re still with me – probably not 🙂 – you’ll immediately wonder why there are two big bumps instead of just one, i.e. two peaks in the density function instead of just one. [You’re used to these Gauss curves, aren’t you?] And you’ll also wonder what negative frequencies actually are: the first bump is a density function for negative frequencies indeed, and… Well… Now that you think of it: why the hell would we do such integral for negative values of ν? I won’t say too much about that: it’s a particularity which results from the fact that e^2πiθ and e^−2πiθ both complete a cycle per second (if θ is measured in seconds, that is) so… Well… Hmm… […] Yes. The fact of the matter is that we do have a mathematical equivalent of the bump for positive frequencies on the negative side of the frequency domain, so… Well… […] Don’t worry about it, I’d say. As mentioned above, we shouldn’t get lost in the math here. For our purpose here, which is just to illustrate what a complex Fourier transform actually is (rather than present all of the mathematical intricacies of it), we should just focus on the second bump of that density function, i.e. the density function for positive frequencies only. 🙂

So what? You’re probably tired by now, and wondering what I want to get at. Well… Nothing much. I’ve done what I wanted to do. I started with a real-valued wave train (think of a transient electric field working its way through space, for example), and I then showed how such wave train can (and should) be analyzed as consisting of an infinite number of complex-valued component waves, which each make their own contribution to the combined wave (which consists of the sum of all component waves) and, hence, can be represented by a graph like the one above, i.e. a real-valued density function around some mean, usually denoted by μ, and with some standard deviation, usually denoted by σ. So now I hope that, when you think of Δf or Δλ in the context of a so-called ‘probability wave’ (i.e. a de Broglie wave), then you’ll think of all this machinery behind.

In other words, it is not just a matter of drawing a simple figure like the one below and saying: “You see: those oscillations represent three photons being emitted one after the other by an atomic oscillator. You can see that’s quite obvious, can’t you?”

No. It is not obvious. Why not? Because anyone that’s somewhat critical will immediately say: “But how does it work really? Those wave trains seem to have a pretty definite frequency (or wavelength), even if their amplitude dies out, and, hence, the Δf factor (or Δλ factor) in that uncertainty relation must be close or, more probably, must be equal to zero. So that means we cannot say these particles are actually somewhere, because Δx must be close or equal to infinity.”

Now you know that’s a very valid remark. Because now you understand that one actually has to go through the tedious exercise of doing that Fourier transform, and so now you understand what those Δ symbols actually represent. I hope you do because of this post, and despite the fact my approach has been very superficial and intuitive. In other words, I didn’t say what physicists would probably say, and that is: “Take a good math course before you study physics!” 🙂

The Uncertainty Principle for energy and time

Pre-script (dated 26 June 2020): My views on the nature of the concept of uncertainty in the context of quantum mechanics has evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. Hence, we recommend you read our recent papers. I keep blog posts like these to see where I came from. I might review them one day, but I currently don’t have the time or energy for it.

Original post:

In all of my posts on the Uncertainty Principle, I left a few points open or rather vague, and that was usually because I didn’t have a clear understanding of them. As I’ve read some more in the meanwhile, I think I sort of ‘get’ these points somewhat better now. Let me share them with you in this and my next posts. This post will focus on the Uncertainty Principle for time and energy.

Indeed, most (if not all) experiments illustrating the Uncertainty Principle (such as the double-slit experiment with electrons for example) focus on the position (x) and momentum (p) variables: Δx·Δp = h. But there is also a similar relationship between time and energy:

ΔE·Δt = h

These pairs of variables (position and momentum, and energy and time) are so-called conjugate variables. I think I said enough about the Δx·Δp = h equation, but what about the ΔE·Δt = h equation? Indeed, we can sort of imagine what ΔE stands for, but what about Δt? It must also be some uncertainty: about time obviously–but what time are we talking about?

I found one particularly appealing explanation in a small booklet that I bought–long time ago– in Berlin: the dtv-Atlas zur Atomphysik. First, note that the uncertainty about the position (Δx) of our ‘wavicle’ (let’s say an electron) is to be related to the length of the (complex-valued) wave-train that represents the ‘particle’ (or ‘wavicle’ if you prefer that term) in space (and in time). In turn, the length of that wave-train is determined by the spread in the frequencies of the component waves that make up that wave-train, as illustrated below. [However, note that the illustration assumes the amplitudes are real-valued only, so there’s no imaginary part. I’ll come back to this point in my next post.]

Now, we can use the de Broglie relation (λ = h/p) to relate the uncertainty about the position to the spread in the wavelengths (and, hence, the frequencies) of the component waves:

p = h/λ and, hence, Δp = Δ(h/λ) = hΔ(1/λ)

In case you wonder why I can simply take h out of the brackets, i.e. why I can write Δ(h/λ) = hΔ(1/λ), just remember that the delta symbol here (Δ) refers to a measure like the standard deviation of a variable, so Δx represents σ_x. Now, one can prove the following:

The standard deviation of some constant function is 0: Δ(k) = 0
The standard deviation is invariant under changes of location: Δ(x + k) = Δx
The standard deviation scales with the scale of the variable: Δ(kx) = |k |Δ(x)

It’s obviously the last rule that we’re using here.

Now, Δx equals h/Δp according to the Uncertainty Principle—if we take it as an equality, rather than as an inequality, that is. Therefore, Δx must equal:

Δx = h/Δp = h/[Δ(h/λ)] =h/[hΔ(1/λ)] = 1/Δ(1/λ)

That’s obvious, but so what? We cannot write Δx = Δλ, because there’s no rule that says that Δ(1/λ) = 1/Δλ and, therefore, h/Δp ≠ Δλ. Indeed, suppose we define Δλ as an interval or a length defined by the difference between its upper bound and its lower bound. Then we can write Δλ as Δλ = λ₂ – λ₁ and, hence, we can then write Δp as Δp = Δ(h/λ) = h/λ₁ – h/λ₂= h(1/λ₁ – 1/λ₂) = h[λ₂ – λ₁]/λ₁λ₂. Now, that’s obviously something very different than h/Δλ = h/(λ₂ – λ₁). So we should surely not write that Δp = h/Δλ. Never ever. Having said that, the Δx = 1/Δ(1/λ) = λ₁λ₂/(λ₂ – λ₁) relationship that emerges here is quite interesting. I encourage you to explore it yourself, as I need to move on here.

So… We’re kinda stuck. What to do? How do we get that energy-time relationship? The de Broglie relation tells us that E = hν, so we can write that ΔE = Δ(hν) = hΔν. But we need to get ΔE = Δ(hν) = hΔν = h/Δt. How do we get Δν = 1/Δt, which is – obviously – the relationship that we need to get ΔE = h/Δt?

To get the answer to that question, we need to ask ourselves another one: what’s Δt here? What are we talking about?

The answer is remarkably mundane: Δt is the measurement time. What measurement time? Relax. You’ll understand in a moment. Let’s go through it.

We know there’s a universal relationship between the propagation speed of a wave (which I’ll denote by c for the time being, but don’t confuse this variable with the speed of light: it can be any speed) and the wavelength and frequency. More specifically, c = λν, and hence, 1/λ = ν/c. So we can now write Δ(1/λ) as Δ(ν/c) = Δ(ν)/c. We also know that the frequency of the wave is the reciprocal of the so-called period of the wave, i.e. the time that’s needed to go through one oscillation: τ = 1/ν and, hence, ν = 1/τ. Hence, we can write Δ(ν) = Δ(1/τ).

OK. That’s stating the obvious. So what? Where do we go from here?

First, note that, for a wavetrain, there’s no precise frequency or period, nor is there any precise number of oscillations. That’s the essence of the Uncertainty Principle in its most ubiquitous form (Δx = h/Δp). But so we can try to measure. Now, to measure something, we need some time. More in particular, to measure the frequency of a wave, we’ll need to look at that wave and register (i.e. measure) at least a few oscillations, as shown below.

I took the image from the above-mentioned German booklet and, hence, the illustration incorporates some German. However, that should not deter you from following the remarkably simple argument, which is the following:

The error in our measurement of the frequency (i.e. the Meβfehler, denoted by Δν) is related to the measurement time (i.e. the Meβzeit, denoted by Δt in the diagram above). Indeed, if τ represents the actual period of the oscillation – which is the reciprocal of the frequency: τ = 1/ν) (both τ and ν are obviously unknown to us: otherwise we wouldn’t be trying to measure the frequency), then we can write Δt as some multiple of τ. More specifically, in the example above we assume that Δt ≈ 4τ = 4/ν. [Note that we use an almost equal to sign (≈) rather than an equality sign (=) because we don’t know τ (or ν). That’s the whole point about it, indeed.]
During that time, we measure four oscillations in our example and, hence, we are tempted to write that ν = 4/Δt. However, because of the measurement error, we should interpret the value for our measurement not as 4 exactly but as 4 plus or minus one: 4 ± 1. Indeed, it’s like measuring the length of something: if our yardstick has millimeter marks, then we’ll measure someone’s length as some number plus or minus 1 mm. Here we are counting the number of oscillations. Hence, the result of our measurement should be written as ν ± Δν = (4 ± 1)/Δt = 4/Δt ± 1/Δt. If you have trouble following the argument, just put in some numbers in order to gain a better understanding. For example, imagine an oscillation of 100 Hz (i.e. 100 oscillations per second), and a measurement time of four hundredths of a second (i.e. Δt = 4×10^–2 s). Suppose, then, we do indeed measure 4 ± 1 oscillations during that time. Then the frequency of this wave must be equal to ν ± Δν = (4 ± 1)/Δt = 4/(4×10^–2 s) ± 1/(4×10^–2 s) = 100 ± 25 Hz. In other words, we here accept that we have a measurement error of Δν/ν = 25/100 = 25%. That’s a relatively large error because the measurement time was relatively short, [Note that ‘relatively short’ means ‘short as compared to the actual period of the oscillation’. Indeed, 4×10^–2 s is obviously not short in any absolute sense: in fact, it is like an eternity when we’re talking light waves, which have frequencies measured in terahertz.]
The example makes it clear that Δν, i.e. the error in our measurement of the frequency, is related to the measurement time as follows: Δν = 1/Δt. Hence, if we double the measurement time, we halve the error in the measurement of the frequency. The relationship is quite straightforward indeed: let’s take the example of that 100 Hz wave once again and assume that our measurement time Δt is equal to Δt = 10τ = 10×10^–2 s = 10^–1 s. In that case, we get Δν = 1/10^–1 s = 10 Hz. Hence, the measurement error is now Δν/ν = 10/100 = 10%.
How long should the measurement time be in order to get a 1% error only? Let’s write the error as a percentage first: Δν/ν = x % = x/100. But Δν = 1/Δt. Hence, we have Δν/ν = (1/Δt)/ν = 1/(Δt·ν) = x/100 or Δt = 100/(x·ν). So, for x = 1 (i.e. an error of 1%), we get Δt = 100/(1·100) = 1 second; for x = 5 (i.e. an error of 5%), we get Δt = 100/(5·100) = 0.2 seconds. Finally, for x = 25 (i.e. an error of 25%), we get Δt = 100/(25·100) = 0.04 seconds, or 4×10^–2 s, which is what this example started out with.

You’ll say: so what? We’re still nowhere… Well… No. We’ve got a formula with the frequency variable here, so we can now derive the Uncertainty Principle for time and energy from the other de Broglie relation (E = hν), which relates the energy of a ‘wavicle’ to the de Broglie frequency. Hence, the uncertainty about the energy about the energy must be related to the measurement time as follows:

E = hν ⇒ ΔE = Δ(hν) = hΔν = h(1/Δt) = h/Δt ⇔ ΔE·Δt = h

So, what this expression of the Uncertainty Principle says is the following: if we increase the measurement time, we’ll reduce the uncertainty in our knowledge of the energy of our ‘wavicle’. Conversely, if we only have a very short measurement time, we’ll not be able to say much about its energy.

A final note needs to be made on the value of h: it’s very tiny. Indeed, a value of (about) 6.6×10⁻³⁴J·s or, using the smaller eV unit for energy, some 4.1×10⁻¹⁵ eV·s is unimaginably small, especially because we need to take into account that the energy concept as used in the de Broglie equation includes the rest mass of a particle. Now, anything that has any rest mass has enormous energy according to Einstein’s mass-energy equivalence relationship: E = mc². Let’s consider, for example, a hydrogen atom. Its atomic mass can be expressed in eV/c², using the same E = mc²but written as m = E/c², although you will usually find it expressed in so-called unified atomic mass units (u). The mass of our hydrogen atom is approximately 1 u ≈ 931.5×10⁶ eV/c². That means its energy is about 931.5×10⁶ eV. In plain language, that’s 931.5 million eV. Hence, if we’d be happy with an uncertainty of plus or minus one million eV, then it’s obvious that even very small values for Δt (i.e. very short measurements) will give us what we want. However, it is likely that we’ll want to reduce the measurement error to much less than plus or minus one million eV, so that means that our measurement time Δt will have to go up. Having said that, the point is still quite clear: we don’t need much time to measure the mass (or the energy) of this hydrogen atom very accurately.

The corollary of this is that the de Broglie frequency f = E/h of such particle is very high. To be precise, the frequency will be in the order of (931.5×10⁶ eV)/(4.1×10⁻¹⁵ eV·s) = 0.2×10²⁴ Hz. In practice, this means that the wavelength is so tiny that there’s no detector which will actually measure the ‘oscillation’: any physical detector will straddle most – in fact, I should say: all – of the wiggles of the probability curve. All these facts basically state the same: a hydrogen atom occupies a very precisely determined position in time and space. Hence, we will see it as a ‘hard’ particle, not as a ‘wavicle’.

That’s why the interference experiment mentions electrons, rather than hydrogen atoms or other ‘big stuff’, even if I should immediately add that interference patterns have been observed using much larger particles as well. However, I wrote about that before, so I won’t repeat myself here. The point was to make that energy-time relationship somewhat more explicit, and I hope I’ve been successful at that at least. You can play with some more numbers yourself now. 🙂

Post scriptum: The Breit-Wigner distribution

The Uncertainty Principle applied to time and energy has an interesting application: it’s used to assign a lifetime to very short-lived particles. In essence, the ‘spread’ around their mean energy (ΔE) is used to calculate their lifetime through the ΔEΔt = ħ/2 equation. I won’t say much about this, because Georgia University’s Hyperphysics website gives an excellent quick explanation of this, and so I just copied that below.

A post for my kids: About Einstein’s laws of radiation, and lasers

Pre-scriptum (dated 26 June 2020): These posts on elementary math and physics have not suffered much the attack by the dark force—which is good because I still like them. While my views on the true nature of light, matter and the force or forces that act on them have evolved significantly as part of my explorations of a more realist (classical) explanation of quantum mechanics, I think most (if not all) of the analysis in this post remains valid and fun to read. In fact, I find the simplest stuff is often the best. 🙂

Original post:

I wrapped up my previous post, which gave Planck’s solution for the blackbody radiation problem, wondering whether or not one could find the same equation using some other model, not involving the assumption that atomic oscillators have discrete energy levels.

I still don’t have an answer to that question but, sure enough, Feynman introduces another model a few pages further in his Lectures. It’s a model developed by Einstein, in 1916, and it’s much ‘richer’ in the sense that it takes into account what we know to be true: unlike matter-particles (fermions), photons like to crowd together. In more advanced quantum-mechanical parlance, their wave functions obey Bose-Einstein statistics. Now, Bose-Einstein statistics are what allows a laser to focus so much energy in one beam, and so I am writing this post for two reasons–one serious and the other not-so-serious:

To present Einstein’s 1916 model for blackbody radiation.
For my kids, so they understand how a laser works.

Let’s start with Einstein’s model first because, if I’d start with the laser, my kids would only read about that and nothing else. [That being said, I am sure my kids will go straight to the second part and, hence, skip Einstein anyway. :-)]

Einstein’s model of blackbody radiation

Einstein’s model is based on Planck’s and, hence, also assumes that the energy of atomic oscillators can also only take on one value of a set of permitted energy levels. However, unlike Planck, he assumes two types of emission. The first is spontaneous, and that’s basically just Planck’s model. The second is induced emission: that’s emission when light is alrady present, and Einstein’s hypothesis was that an atomic oscillator is more likely to emit a photon when there’s light of the same frequency is shining on it.

The basics of the model are shown above, and the two new variables are the following:

A_mnis the probability for the oscillator to have its energy drop from energy level m to energy level n, independent of whether light is shining on the atom or not. So that’s the probability of spontaneous emission and it only depends on m and n.
B_mnis not a probability but a proportionality constant that, together with the intensity of the light shining on the oscillator–denoted by I(ω), co-determines the probability of of induced emission.

Now, as mentioned above, in this post, I basically want to explain how a laser works, and so let me be as brief as possibly by just copying Feynman here, who says it all:

Of course, this result must match Planck’s equation for blackbody radiation, because Planck’s equation matched experiment:

To get the e^ħω/kT–1, B_mnmust be equal to B_nm, and you should not think that’s an obvious result, because it isn’t: this equality says that the induced emission probability and the absorption probability must be equal. Good to know: this keeps the numbers of atoms in the various levels constant through what is referred to as detailed balancing: in thermal equilibrium, every process is balanced by its exact opposite. While that’s nice, and the way it actually works, it’s not obvious. It shows that the process is fully time-reversible. That’s not obvious in a situation involving statistical mechanics, which is what we’re talking about there. In any case, that’s a different topic.

As for A_mn, taking into account that B_mn= B_nm, we find that A_mn/B_mn =ħω³/π²c². So we have a ratio here. What about calculating the individual values for A_mnand B_mn? Can we calculate the absolute spontaneous and induced emission rates? Feynman says: No. Not with what Einstein had at the time. That was possible only a decade or so later, it seems, when Werner Heisenberg, Max Born, Pascual Jordan, Erwin Schrödinger, Paul Dirac and John von Neumann developed a fully complete theory, in the space of just five years (1925-1930), but that’s the subject of the history of science.

The point is: we have got everything here now to sort of understand how lasers work, so let’s try that to do that now.

Lasers

Laser is an acronym which stands for Light Amplification by Stimulated Emission of Radiation. It’s based on the mechanism described above, which I am sure you’ve studied in very much detail. 🙂

The trick is to find a method to get a gas in a state in which the number of atomic oscillators with energy level m is much and much greater than the number with energy level n. So we’re talking a situation that is not in equilibrium. On the contrary: it’s far out of equilibrium. And then, suddenly, we induce emission from this upper state, which creates a sort of chain reaction that makes “the whole lot of them dump down together”, as Feynman puts it.

The diagram below is taken from the Wikipedia article on lasers. It shows a so-called Nd:YAG laser. Huh? Yes. Nd:YAG stands for neodymium-doped yttrium aluminium garnet, and an Nd:YAG laser is a pretty common type of laser. A garnet is a precious stone: a crystal composed of a silicate mineral. And that’s what it is here, and why the laser is so-called solid-state laser, because the so-called laser medium (see the diagram) may also be a gas or even a liquid, as in dye lasers). I could also take a ruby laser, which uses ruby as the laser medium. But let’s go along with this one as for now.

In the set-up as shown above, a simple xenon flash lamp (yes, that’s a ‘neon’ lamp) provides the energy exciting the atomic oscillators in the crystal. It’s important that the so-called pumping source emits light of a higher frequency than the laser light, as shown below. In fact, the light from xenon gas, or any source, will be a spectrum but so it should (also) have light in the blue or violet range (as shown below). The important thing is that it should not have the red laser frequency, because that’s what would trigger the laser, of course.

The diagram above shows how it actually works.The trick is to get the atoms to a higher state (that’s h in the diagram above, but it’s got nothing to do with the Planck constant) from where they trickle down (and, yes, they do emit other photons while doing that), until they all get stuck in the state m, which is referred to as metastable but which is, in effect, unstable. And so then they are all dumped down together by induced emissions. So the source ‘pumps’ the crystal indeed, leading to that ‘metastable’ state which is referred to as population inversion in statistical mechanics: a lot of atoms (i.e. the members of the ‘population’) are in an excited state, rather than in a lower energy state.

And then we have a so-called optical resonator (aka as a cavity) which, in its simplest form, consists of just two mirrors around the gain medium (i.e. the crystal): these mirrors reflect the light so, once the dump starts, the induced effect is enhanced: the light which is emitted gets a chance to induce more emission, and then another chance, and another, and so on. However, although the mirrors are almost one hundred percent reflecting, light does get out because one of the mirrors is only a partial reflector, which is referred to as the output coupler, and which produces the laser’s output beam.

So… That’s all there is to it.

Really? Is it that simple? Yep. I googled a few questions to increase my understanding but so that’s basically it. Perhaps they’ll help you too and so I copied them hereunder. Before you go through that, however, have a look at how they really look like. The image below (from Wikipedia again) shows a disassembled (and assembled) ruby laser head. You can clearly see the crystal rod in the middle, and the two flashlamps that are used for pumping. I am just inserting it here because, in engineering, I found that a diagram of something and the actual thing often have not all that much in common. 🙂 As you can see, it’s not the case here: it looks amazingly simple, doesn’t it?

Q: We have crystal here. What’s the atomic oscillator in the crystal? A: It is the neodymium ion which provides the lasing activity in the crystal, in the same fashion as red chromium ion in ruby lasers.

Q: But how does it work exactly? A: Well… The diagram is a bit misleading. The distance between h and m should not be too big of course, because otherwise half of the energy goes into these photons that are being emitted as the oscillators ‘trickle down’. Also, if these ‘in-between’ emissions would have the same frequency as the laser light, they would induce the emission, which is not what we want. So the actual distances should look more like this:

For an actual Nd:YAG laser, we have absorption mostly in the bands between 730–760 nm and 790–820 nm, and emitted light with a wavelength with a wavelength of 1064 nm. Huh? Yes. Remember: shorter wavelength (λ) is higher frequency (ν = c/λ) and, hence, higher energy (E = hν = hc/λ). So that’s what’s shown below.

Q: But… You’re talking bullsh**. Wavelengths in the 700–800 nm range are infrared (IR) and, hence, not even visible. And light of 1064 nm even less. A: Now you are a smart-ass! You’re right. What actually happens is a bit more complicated, as you might expect. There’s something else going on as well, a process referred to as frequency doubling or second harmonic generation (SHG). It’s a process in which photons with the same frequency (1064 nm) interact with some material to effectively ‘combine’ into new photons with twice the energy, twice the frequency and, therefore, half the wavelength of the initial photons. And so that’s light with a wavelength of 532 nm. We actuall also have so-called higher harmonics, with wavelengths at 355 and 266 nm.

Q: But… That’s green? A: Sure. A Nd:YAG laser produces a green laser beam, as shown below. If you want the red color, buy a ruby laser, which produces pulses of light with a wavelength of 694.3 nm: that’s the deep red color you’d associate with lasers. In fact, the first operational laser, produced by Hughes Research Laboratories back in 1960 (the research arm of Hughes Aircraft, now part of the Raytheon), was a ruby laser.

Q: Pulses? That reminds me of something: lasers pulsate indeed, don’t they? How does that work? A: They do. Lasers have a so-called continuous wave output mode. However, there’s a technique called Q-switching. Here, an optical switch is added to the system. It’s inserted into laser cavity, and it waits for a maximum population inversion before it opens. Then the light wave runs through the cavity, depopulating the excited laser medium at maximum population inversion. It allows to produce light pulses with extremely high peak power, much higher than would be produced by the same laser if it were operating in constant output mode.

Q: What’s the use of lasers? A: Because of their ability to focus, they’re used a surgical knives, in eye surgery, or to remove tumors in the brain and treat skin cancer. Lasers are also widely used for engraving, etching, and marking of metals and plastics. When they pack more power, they can also be used to cut or weld steel. Their ability to focus is why these tiny pocket lasers can damage your eye: it’s not like a flashlight. It’s a really focused beam and so it can really blind you–not for a while but permanently.

Q: Lasers can also be used as weapons, can’t they? A: Yes. As mentioned above, techniques like Q-switching allow to produce pulses packing enormous amounts of energy into one single pulse, and you hear a lot about lasers being used as directed-energy weapons (DEWs). However, they won’t replace explosives anytime soon. Lasers were already widely used for sighting, ranging and targeting for guns, but so they’re not the source of the weapon’s firepower. That being said, the pulse of a megajoule laser would deliver the same energy as 200 grams of high explosive, but all focused on a tiny little spot. Now that’s firepower obviously, and such lasers are now possible. However, their power is more likely to be used for more benign purposes, notably igniting a nuclear fusion reaction. There’s nice stuff out there if you’d want to read more.

Q: No. I think I’ve had it. But what are those pocket lasers? A: They are what they are: handheld lasers. It just shows how technology keeps evolving. The Nano costs a hundred dollars only. I wonder if Einstein would ever have imagined that what he wrote back in 1916 would, ultimately, lead to us manipulating light with little handheld tools. We live in amazing times. 🙂

Planck’s constant (II)

Pre-script (dated 26 June 2020): This post suffered from the removal of material by the the dark force. Its layout also got tampered with and I don’t have the time or energy to put everything back in order. It remains relevant, I think. Among other things, it shows how Planck’s constant was actually discovered—historically and experimentally. If anything, the removal of material will help you to think things through for yourself. 🙂

Original post:

My previous post was tough. Tough for you–if you’ve read it. But tough for me too. 🙂

The blackbody radiation problem is complicated but, when everything is said and done, what the analysis says is that the the ‘equipartition theorem’ in the kinetic theory of gases ‘theorem (or the ‘theorem concerning the average energy of the center-of-mass motion’, as Feynman terms it), is not correct. That equipartition theorem basically states that, in thermal equilibrium, energy is shared equally among all of its various forms. For example, the average kinetic energy per degree of freedom in the translation motion of a molecule should equal that of its rotational motions. That equipartition theorem is also quite precise: it also states that the mean energy, for each atom or molecule, for each degree of freedom, is kT/2. Hence, that’s the (average) energy the 19th century scientists also assigned to the atomic oscillators in a gas.

However, the discrepancy between the theoretical and empirical result of their work shows that adding atomic oscillators–as radiators and absorbers of light–to the system (a box of gas that’s being heated) is not just a matter of adding additional ‘degree of freedom’ to the system. It can’t be analyzed in ‘classical’ terms: the actual spectrum of blackbody radiation shows that these atomic oscillators do not absorb, on average, an amount of energy equal to kT/2. Hence, they are not just another ‘independent direction of motion’.

So what are they then? Well… Who knows? I don’t. But, as I didn’t quite go through the full story in my previous post, the least I can do is to try to do that here. It should be worth the effort. In Feynman’s words: “This was the first quantum-mechanical formula ever known, or discussed, and it was the beautiful culmination of decades of puzzlement.” And then it does not involve complex numbers or wave functions, so that’s another reason why looking at the detail is kind of nice. 🙂

Discrete energy levels and the nature of h

To solve the blackbody radiation problem, Planck assumed that the permitted energy levels of the atomic harmonic oscillator were equally spaced, at ‘distances’ ħω₀apart from each other. That’s what’s illustrated below.

Now, I don’t want to make too many digressions from the main story, but this E_n = nħω₀ formula obviously deserves some attention. First note it immediately shows why the dimension of ħ is expressed in joule-seconds (J·s), or electronvolt-seconds (J·s): we’re multiplying it with a frequency indeed, so that’s something expressed per second (hence, its dimension is s^–1) in order to get a measure of energy: joules or, because of the atomic scale, electronvolts. [The eV is just a (much) smaller measure than the joule, but it amounts to the same: 1 eV ≈ 1.6×10⁻¹⁹ J.]

One thing to note is that the equal spacing consists of distances equal to ħω₀, not of ħ. Hence, while h, or ħ (ħ is the constant to be used when the frequency is expressed in radians per second, rather than oscillations per second, so ħ = h/2π) is now being referred to as the quantum of action (das elementare Wirkungsquantum in German), Planck referred to it as as a Hilfsgrösse only (that’s why he chose the h as a symbol, it seems), so that’s an auxiliary constant only: the actual quantum of action is, of course, ΔE, i.e. the difference between the various energy levels, which is the product of ħ and ω₀(or of h and ν₀ if we express frequency in oscillations per second, rather than in angular frequency). Hence, Planck (and later Einstein) did not assume that an atomic oscillator emits or absorbs packets of energy as tiny as ħ or h, but packets of energy as big as ħω₀or, what amounts to the same (ħω = (h/2π)(2πν) = hν), hν₀. Just to give an example, the frequency of sodium light (ν) is 500×10¹²Hz, and so its energy is E = hν. That’s not a lot–about 2 eV only– but it still packs 500×10¹² ‘quanta of action’ !

Another thing is that ω (or ν) is a continuous variable: hence, the assumption of equally spaced energy levels does not imply that energy itself is a discrete variable: light can have any frequency and, hence, we can also imagine photons with any energy level: the only thing we’re saying is that the energy of a photon of a specific color (i.e. a specific frequency ν) will be a multiple of hν.

Probability assumptions

The second key assumption of Planck as he worked towards a solution of the blackbody radiation problem was that the probability (P) of occupying a level of energy E is P(E) = αe^−E/kT. OK… Why not? But what is this assumption really? You’ll think of some ‘bell curve’, of course. But… No. That wouldn’t make sense. Remember that the energy has to be positive. The general shape of this P(E) curve is shown below.

The highest probability density is near E = 0, and then it goes down as E gets larger, with kT determining the slope of the curve (just take the derivative). In short, this assumption basically states that higher energy levels are not so likely, and that very high energy levels are very unlikely. Indeed, this formula implies that the relative chance, i.e. the probability of being in state E₁ relative to the chance of being in state E₀, is P₁/P₀= e^{−(E₁–E₀)kT}= e^−ΔE/kT. Now, P₁is n₁/N and P₀is n₀/N and, hence, we find that n₁must be equal to n₀e^−ΔE/kT. What this means is that the atomic oscillator is less likely to be in a higher energy state than in a lower one.

That makes sense, doesn’t it? I mean… I don’t want to criticize those 19th century scientists but… What were they thinking? Did they really imagine that infinite energy levels were as likely as… Well… More down-to-earth energy levels? I mean… A mechanical spring will break when you overload it. Hence, I’d think it’s pretty obvious those atomic oscillators cannot be loaded with just about anything, can they? Garbage in, garbage out: of course, that theoretical spectrum of blackbody radiation didn’t make sense!

Let me copy Feynman now, as the rest of the story is pretty straightforward:

Now, we have a lot of oscillators here, and each is a vibrator of frequency w₀. Some of these vibrators will be in the bottom quantum state, some will be in the next one, and so forth. What we would like to know is the average energy of all these oscillators. To find out, let us calculate the total energy of all the oscillators and divide by the number of oscillators. That will be the average energy per oscillator in thermal equilibrium, and will also be the energy that is in equilibrium with the blackbody radiation and that should go in the equation for the intensity of the radiation as a function of the frequency, instead of kT. [See my previous post: that equation is I(ω) = (ω²kt)/(π²c²).]

Thus we let N₀ be the number of oscillators that are in the ground state (the lowest energy state); N₁ the number of oscillators in the state E₁; N₂ the number that are in state E₂; and so on. According to the hypothesis (which we have not proved) that in quantum mechanics the law that replaced the probability e^−P.E./kT or e^−K.E./kT in classical mechanics is that the probability goes down as e^−ΔE/kT, where ΔE is the excess energy, we shall assume that the number N₁ that are in the first state will be the number N₀ that are in the ground state, times e^−ħω/kT. Similarly, N₂, the number of oscillators in the second state, is N₂=N₀e^−2ħω/kT. To simplify the algebra, let us call e^−ħω/kT= x. Then we simply have N₁ = N₀x, N₂ = N₀x², …, N_n= N₀xⁿ.

The total energy of all the oscillators must first be worked out. If an oscillator is in the ground state, there is no energy. If it is in the first state, the energy is ħω, and there are N₁ of them. So N₁ħω, or ħωN₀x is how much energy we get from those. Those that are in the second state have 2ħω, and there are N₂ of them, so N₂⋅2ħω=2ħωN₀x² is how much energy we get, and so on. Then we add it all together to get E_tot= N₀ħω(0+x+2x²+3x³+…).

And now, how many oscillators are there? Of course, N₀ is the number that are in the ground state, N₁ in the first state, and so on, and we add them together: N_tot= N₀(1+x+x²+x³+…). Thus the average energy is

Now the two sums which appear here we shall leave for the reader to play with and have some fun with. When we are all finished summing and substituting for x in the sum, we should get—if we make no mistakes in the sum—

Feynman concludes as follows: “This, then, was the first quantum-mechanical formula ever known, or ever discussed, and it was the beautiful culmination of decades of puzzlement. Maxwell knew that there was something wrong, and the problem was, what was right? Here is the quantitative answer of what is right instead of kT. This expression should, of course, approach kT as ω → 0 or as T → ∞.”

It does, of course. And so Planck’s analysis does result in a theoretical I(ω) curve that matches the observed I(ω) curve as a function of both temperature (T) and frequency (ω). But so what it is, then? What’s the equation describing the dotted curves? It’s given below:

I’ll just quote Feynman once again to explain the shape of those dotted curves: “We see that for a large ω, even though we have ω³in the numerator, there is an e raised to a tremendous power in the denominator, so the curve comes down again and does not “blow up”—we do not get ultraviolet light and x-rays where we do not expect them!”

Is the analysis necessarily discrete?

One question I can’t answer, because I just am not strong enough in math, is the question or whether or not there would be any other way to derive the actual blackbody spectrum. I mean… This analysis obviously makes sense and, hence, provides a theory that’s consistent and in accordance with experiment. However, the question whether or not it would be possible to develop another theory, without having recourse to the assumption that energy levels in atomic oscillators are discrete and equally spaced with the ‘distance’ between equal to hν₀, is not easy to answer. I surely can’t, as I am just a novice, but I can imagine smarter people than me have thought about this question. The answer must be negative, because I don’t know of any other theory: quantum mechanics obviously prevailed. Still… I’d be interested to see the alternatives that must have been considered.

Post scriptum: The “playing with the sums” is a bit confusing. The key to the formula above is the substitution of (0+x+2x²+3x³+…)/(1+x+x²+x³+…) by 1/[(1/x)–1)] = 1/[e^ħω/kT–1]. Now, the denominator 1+x+x²+x³+… is the Maclaurin series for 1/(1–x). So we have:

(0+x+2x²+3x³+…)/(1+x+x²+x³+…) = (0+x+2x²+3x³+…)(1–x)

= x+2x²+3x³… –x²–2x³–3x⁴… = x+x²+x³+x⁴…

= –1+(1+x+x²+x³…) = –1 + 1/(1–x) = –(1–x)+1/(1–x) = x/(1–x).

Note the tricky bit: if x = e^−ħω/kT, then e^ħω/kTis x⁻¹= 1/x, and so we have (1/x)–1 in the denominator of that (mean) energy formula, not 1/(x–1). Now 1/[(1/x)–1)] = 1/[(1–x)/x] = x/(1–x), indeed, and so the formula comes out alright.

Planck’s constant (I)

Pre-script (dated 26 June 2020): This post did not suffer too much from the attack on this blog by the the dark force. It remains relevant. If anything, the removal of material will help you to think things through for yourself. 🙂

Original post:

If you made it here, it means you’re totally fed up with all of the easy stories on quantum mechanics: diffraction, double-slit experiments, imaginary gamma-ray microscopes,… You’ve had it! You now know what quantum mechanics is all about, and you’ve realized all these thought experiments never answer the tough question: where did Planck find that constant (h) which pops up everywhere? And how did he find that Planck relation which seems to underpin all and everything in quantum mechanics?

If you don’t know, that’s because you’ve skipped the blackbody radiation story. So let me give it to you here. What’s blackbody radiation?

Thermal equilibrium of radiation

That’s what the blackbody radiation problem is about: thermal equilibrium of radiation.

Huh?

Yes. Imagine a box with gas inside. You’ll often see it’s described as a furnace, because we heat the box. Hence, the box, and everything inside, acquires a certain temperature, which we then assume to be constant. The gas inside will absorb energy and start emitting radiation, because the gas atoms or molecules are atomic oscillators. Hence, we have electrons getting excited and then jumping up and down from higher to lower energy levels, and then again and again and again, thereby emitting photons with a certain energy and, hence, light of a certain frequency. To put it simply: we’ll find light with various frequencies in the box and, in thermal equilibrium, we should have some distribution of the intensity of the light according to the frequency: what kind of radiation do we find in the furnace? Well… Let’s find out.

The assumption is that the box walls send light back, or that the box has mirror walls. So we assume that all the radiation keeps running around in the box. Now that implies that the atomic oscillators not only radiate energy, but also receive energy, because they’re constantly being illuminated by radiation that comes straight back at them. If the temperature of the box is kept constant, we arrive at a situation which is referred to as thermal equilibrium. In Feynman’s words: “After a while there is a great deal of light rushing around in the box, and although the oscillator is radiating some, the light comes back and returns some of the energy that was radiated.”

OK. That’s easy enough to understand. However, the actual analysis of this equilibrium situation is what gave rise to the ‘problem’ of blackbody radiation in the 19th century which, as you know, led Planck and Einstein to develop a quantum-mechanical view of things. It turned out that the classical analysis predicted a distribution of the intensity of light that didn’t make sense, and no matter how you looked at it, it just didn’t come out right. Theory and experiment did not agree. Now, that is something very serious in science, as you know, because it means your theory isn’t right. In this case, it was disastrous, because it meant the whole of classical theory wasn’t right.

To be frank, the analysis is not all that easy. It involves all that I’ve learned so far: the math behind oscillators and interference, statistics, the so-called kinetic theory of gases and what have you. I’ll try to summarize the story but you’ll see it requires quite an introduction.

Kinetic energy and temperature

The kinetic theory of gases is part of what’s referred to as statistical mechanics: we look at a gas as a large number of inter-colliding atoms and we describe what happens in terms of the collisions between them. As Feynman puts it: “Fundamentally, we assert that the gross properties of matter should be explainable in terms of the motion of its parts.” Now, we can do a lot of intellectual gymnastics, analyzing one gas in one box, two gases in one box, two gases in one box with a piston between them, two gases in two boxes with a hole in the wall between them, and so on and so on, but that would only distract us here. The rather remarkable conclusion of such exercises, which you’ll surely remember from your high school days, is that:

Equal volumes of different gases, at the same pressure and temperature, will have the same number of molecules.
In such view of things, temperature is actually nothing but the mean kinetic energy of those molecules (or atoms if it’s a monatomic gas).

So we can actually measure temperature in terms of the kinetic energy of the molecules of the gas, which, as you know, equals mv²/2, with m the mass and v the velocity of the gas molecules. Hence, we’re tempted to define some absolute measure of temperature T and simply write:

T = 〈mv²/2〉

The 〈 and 〉 brackets denote the mean here. To be precise, we’re talking the root mean square here, aka as the quadratic mean, because we want to average some magnitude of a varying quantity. Of course, the mass of different gases will be different – and so we have 〈m₁v₁²/2〉 for gas 1 and 〈m₂v₂²/2〉 for gas 2 – but that doesn’t matter: we can, actually, imagine measuring temperature in joule, the unit of energy, including kinetic energy. Indeed, the units come out alright: 1 joule = 1 kg·(m²/s²). For historical reasons, however, T is measured in different units: degrees Kelvin, centigrades (i.e. degrees Celsius) or, in the US, in Fahrenheit. Now, we can easily go from one measure to the other as you know and, hence, here I should probably just jot down the so-called ideal gas law–because we need that law for the subsequent analysis of blackbody radiation–and get on with it:

PV = NkT

However, now that we’re here, let me give you an inkling of how we derive that law. A classical (Newtonian) analysis of the collisions (you can find the detail in Feynman’s Lectures, I-39-2) will yield the following equation: P = (2/3)n〈mv²/2〉, with n the number of atoms or molecules per unit volume. So the pressure of a gas (which, as you know, is the force (of a gas on a piston, for example) per unit area: P = F/A) is also equal to the mean kinetic energy of the gas molecules multiplied by (2/3)n. If we multiply that equation by V, we get PV = N(2/3)〈mv²/2〉. However, we know that equal volumes of volumes of different gases, at the same pressure and temperature, will have the same number of molecules, so we have PV = N(2/3)〈m₁v₁²/2〉 = N(2/3)〈m₂v₂²/2〉, which we write as PV = NkT with kT = (2/3)〈m₁v₁²/2〉 = (2/3)〈m₂v₂²/2〉.

In other words, that factor of proportionality k is the one we have to use to convert the temperature as measured by 〈mv²/2〉 (i.e. the mean kinetic energy expressed in joules) to T (i.e. the temperature expressed in the measure we’re used to, and that’s degrees Kelvin–or Celsius or Fahrenheit, but let’s stick to Kelvin, because that’s what’s used in physics). Vice versa, we have 〈mv²/2〉 = (3/2)kT. Now, that constant of proportionality k is equal to k = 1.38×10^–23joule per Kelvin (J/K). So if T is (absolute) temperature, expressed in Kelvin (K), our definition says that the mean molecular kinetic energy is (3/2)kT.

That k factor is a physical constant referred to as the Boltzmann constant. If it’s one of these constants, you may wonder why we don’t integrate that 3/2 factor in it? Well… That’s just how it is, I guess. In any case, it’s rather convenient because we’ll have 2/3 factors in other equations and so these will cancel out with that 3/2 term. However, I am digressing way too much here. I should get back to the main story line. However, before I do that, I need to expand on one more thing, and that’s a small lecture on how things look like when we also allow for internal motion, i.e. the rotational and vibratory motions of the atoms within the gas molecule. Let me first re-write that PV = NkT equation as

PV = NkT = N(2/3)〈m₁v₁²/2〉 = (2/3)U = 2U/3

For monatomic gas, that U would only be the kinetic energy of the atoms, and so we can write it as U = (2/3)NkT. Hence, we have the grand result that the kinetic energy, for each atom, is equal to (3/2)kT, on average that is.

What about non-monatomic gas? Well… For complex molecules, we’d also have energy going into the rotational and vibratory motion of the atoms within the molecule, separate from what is usually referred to as the center-of-mass (CM) motion of the molecules themselves. Now, I’ll again refer you to Feynman for the detail of the analysis, but it turns out that, if we’d have, for example, a diatomic molecule, consisting of an A and B atom, the internal rotational and vibratory motion would, indeed, also absorb energy, and we’d have a total energy equal to (3/2)kT + (3/2)kT = 2×(3/2)kT = 3kT. Now, that amount (3kT) can be split over (i) the energy related to the CM motion, which must still be equal to (3/2)kT, and (ii) the average kinetic energy of the internal motions of the diatomic molecule excluding the bodily motion of the CM. Hence, the latter part must be equal to 3kT – (3/2)kT = (3/2)kT. So, for the diatomic molecule, the total energy happens to consist of two equal parts.

Now, there is a more general theorem here, for which I have to introduce the notion of the degrees of freedom of a system. Each atom can rotate or vibrate or oscillate or whatever in three independent directions–namely the three spatial coordinates x, y and z. These spatial dimensions are referred to as the degrees of freedom of the atom (in the kinetic theory of gases, that is), and if we have two atoms, we have 2×3 = 6 degrees of freedom. More in general, the number of degrees of freedom of a molecule composed of r atoms is equal to 3r. Now, it can be shown that the total energy of an r-atom molecule, including all internal energy as well as the CM motion, will be 3r×kT/2 = 3rkT/2 joules. Hence, for every independent direction of motion that there is, the average kinetic energy for that direction will be kT/2. [Note that ‘independent direction of motion’ is used, somewhat confusingly, as a synonym for degree of freedom, so we don’t have three but six ‘independent directions of motion’ for the diatomic molecule. I just wanted to note that because I do think it causes confusion when reading a textbook like Feynman’s.] Now, that total amount of energy, i.e. 3r(kT/2), will be split as follows according to the “theorem concerning the average energy of the CM motion”, as Feynman terms it:

The kinetic energy for the CM motion of each molecule is, and will always be, (3/2)kT.
The remainder, i.e. r(3/2)kT – (3/2)kT = (3/2)(r–1)kt, is internal vibrational and rotational kinetic energy, i.e. the sum of all vibratory and rotational kinetic energy but excluding the energy of the CM motion of the molecule.

Phew! That’s quite something. And we’re not quite there yet.

The analysis for photon gas

Photon gas? What’s that? Well… Imagine our box is the gas in a very hot star, hotter than the sun. As Feynman writes it: “The sun is not hot enough; there are still too many atoms, but at still higher temperatures in certain very hot stars, we may neglect the atoms and suppose that the only objects that we have in the box are photons.” Well… Let’s just go along with it. We know that photons have no mass but they do have some very tiny momentum, which we related to the magnetic field vector, as opposed to the electric field. It’s tiny indeed. Most of the energy of light goes into the electric field. However, we noted that we can write p as p = E/c, with c the speed of light (3×10⁸). Now, we had that P = (2/3)n〈mv²/2〉 formula for gas, and we know that the momentum p is defined as p = mv. So we can substitute mv²by (mv)v = pv. So we get P = (2/3)n〈pv/2〉 = (1/3)n〈pv〉.

Now, the energy of photons is not quite the same as the kinetic energy of an atom or an molecule, i.e. mv²/2. In fact, we know that, for photons, the speed v is equal to c, and pc = E. Hence, multiplying by the volume V, we get

PV = U/3

So that’s a formula that’s very similar to the one we had for gas, for which we wrote: PV = NkT = 2U/3. The only thing is that we don’t have a factor 2 in the equation but so that’s because of the different energy concepts involved. Indeed, the concept of the energy of a photon (E = pc) is different than the concept of kinetic energy. But so the result is very nice: we have a similar formula for the compressibility of gas and radiation. In fact, both PV = 2U/3 and PV = U/3 will usually be written, more generally, as:

PV = (γ – 1)U

Hence, this γ would be γ = 5/3 ≈ 1.667 for gas and 4/3 ≈ 1.333 for photon gas. Now, I’ll skip the detail (it involves a differential analysis) but it can be shown that this general formula, PV = (γ – 1)U, implies that PV^γ(i.e. the pressure times the volume raised to the power γ) must equal some constant, so we write:

PV^γ= C

So far so good. Back to our problem: blackbody radiation. What you should take away from this introduction is the following:

Temperature is a measure of the average kinetic energy of the atoms or molecules in a gas. More specifically, it’s related to the mean kinetic energy of the CM motion of the atoms or molecules, which is equal to (3/2)kT, with k the Boltzmann constant and T the temperature expressed in Kelvin (i.e. the absolute temperature).
If gas atoms or molecules have additional ‘degrees of freedom’, aka ‘independent directions of motion’, then each of these will absorb additional energy, namely kT/2.

Energy and radiation

The atoms in the box are atomic oscillators, and we’ve analyzed them before. What the analysis above added was that average kinetic energy of the atoms going around is (3/2)kT and that, if we’re talking molecules consisting of r atoms, we have a formula for their internal kinetic energy as well. However, as an oscillator, they also have energy separate from that kinetic energy we’ve been talking about alrady. How much? That’s a tricky analysis. Let me first remind you of the following:

Oscillators have a natural frequency, usually denoted by the (angular) frequency ω₀.
The sum of the potential and kinetic energy stored in an oscillator is a constant, unless there’s some damping constant. In that case, the oscillation dies out. Here, you’ll remember the concept of the Q of an oscillator. If there’s some damping constant, the oscillation will die out and the relevant formula is 1/Q = (dW/dt)/(ω₀W) = γ/ω₀, with γ the damping constant (not to be confused with the γ we used in that PV^γ= C formula).

Now, for gases, we said that, for every independent direction of motion there is, the average kinetic energy for that direction will be kT/2. I admit it’s a bit of a stretch of the imagination but so that’s how the blackbody radiation analysis starts really: our atomic oscillators will have an average kinetic energy equal to kT/2 and, hence, their total energy (kinetic and potential) should be twice that amount, according to the second remark I made above. So that’s kT. We’ll note the total energy as W below, so we can write:

W = kT

Just to make sure we know what we’re talking about (one would forget, wouldn’t one?), kT is the product of the Boltzmann constant (1.38×10^–23J/K) and the temperature of the gas (so note that the product is expressed in joule indeed). Hence, that product is the average energy of our atomic oscillators in the gas in our furnace.

Now, I am not going to repeat all of the detail we presented on atomic oscillators (I’ll refer you, once again, to Feynman) but you may or may not remember that atomic oscillators do have a Q indeed and, hence, some damping constant γ. So we can use and re-write that formula above as

dW/dt = (1/Q)(ω₀W) = (ω₀W)(γ/ω₀) = γW, which implies γ = (dW/dt)/W

What’s γ? Well, we’ve calculated the Q of an atomic oscillator already: Q = 3λ/4πr₀. Now, λ = 2πc/ω₀(we just convert the wavelength into (angular) frequency using λν = c) and γ = ω₀/Q, so we get γ = 4πr₀ω₀/[3(2πc/ω₀)] = (2/3)r₀ω₀²/c. Now, plugging that result back into the equation above, we get

dW/dt = γW = (2/3)(r₀ω₀²kT)/c

Just in case you’d have difficulty following – I admit I did 🙂 – dW/dt is the average rate of radiation of light of (or near) frequency ω₀². I’ll let Feynman take over here:

Next we ask how much light must be shining on the oscillator. It must be enough that the energy absorbed from the light (and thereupon scattered) is just exactly this much. In other words, the emitted light is accounted for as scattered light from the light that is shining on the oscillator in the cavity. So we must now calculate how much light is scattered from the oscillator if there is a certain amount—unknown—of radiation incident on it. Let I(ω)dω be the amount of light energy there is at the frequency ω, within a certain range dω (because there is no light at exactly a certain frequency; it is spread all over the spectrum). So I(ω) is a certain spectral distribution which we are now going to find—it is the color of a furnace at temperature T that we see when we open the door and look in the hole. Now how much light is absorbed? We worked out the amount of radiation absorbed from a given incident light beam, and we calculated it in terms of a cross section. It is just as though we said that all of the light that falls on a certain cross section is absorbed. So the total amount that is re-radiated (scattered) is the incident intensity I(ω)dω multiplied by the cross section σ.

OK. That makes sense. I’ll not copy the rest of his story though, because this is a post in a blog, not a textbook. What we need to find is that I(ω). So I’ll refer you to Feynman for the details (these ‘details’ involve fairly complicated calculations, which are less important than the basic assumptions behind the model, which I presented above) and just write down the result:

This formula is Rayleigh’s law. [And, yes, it’s the same Rayleigh – Lord Rayleigh, I should say respectfully – as the one who invented that criterion I introduced in my previous post, but so this law and that criterion have nothing to do with each other.] This ‘law’ gives the intensity, or the distribution, of light in a furnace. Feynman says it’s referred to as blackbody radiation because “the hole in the furnace that we look at is black when the temperature is zero.” […] OK. Whatever. What we call it doesn’t matter. The point is that this function tells us that the intensity goes as the square of the frequency, which means that if we have a box at any temperature at all, and if we look at it, the X- and gamma rays will be burning out eyes out ! The graph below shows both the theoretical curve for two temperatures (T₀and 2T₀), as derived above (see the solid lines), and then the actual curves for those two temperatures (see the dotted lines).

This is the so-called UV catastrophe: according to classical physics, an ideal black body at thermal equilibrium should emit radiation with infinite power. In reality, of course, it doesn’t: Rayleigh’s law is false. Utterly false. And so that’s where Planck came to the rescue, and he did so by assuming radiation is being emitted and/or absorbed in finite quanta: multiples of h, in fact.

Indeed, Planck studied the actual curve and fitted it with another function. That function assumed the average energy of a harmonic oscillator was not just proportional with the temperature (T), but that it was also a function of the (natural) frequency of the oscillators. By fiddling around, he found a simple derivation for it which involved a very peculiar assumption. That assumption was that the harmonic oscillator can take up energies only ħω at the time, as shown below.

Hence, the assumption is that the harmonic oscillators cannot take whatever (continous) energy level. No. The allowable energy levels of the harmonic oscillators are equally spaced: E_n= nħω. Now, the actual derivation is at least as complex as the derivation of Rayleigh’s law, so I won’t do it here. Let me just give you the key assumptions:

The gas consists of a large number of atomic oscillators, each with their own natural frequency ω₀.
The permitted energy levels of these harmonic oscillator are equally spaced and ħω₀apart.
The probability of occupying a level of energy E is P(E) = αe^–E/kT.

All the rest is tedious calculation, including the calculation of the parameters of the model, which include ħ (and, hence, h, because h = 2πħ) and are found by matching the theoretical curves to the actual curves as measured in experiments. I’ll just mention one result, and that’s the average energy of these oscillators:

As you can see, the average energy does not only depend on the temperature T, but also on their (natural) frequency. So… Now you know where h comes from. As I relied so heavily on Feynman’s presentation here, I’ll include the link. As Feynman puts it: “This, then, was the first quantum-mechanical formula ever known, or ever discussed, and it was the beautiful culmination of decades of puzzlement. Maxwell knew that there was something wrong, and the problem was, what was right? Here is the quantitative answer of what is right instead of kT.”

So there you go. Now you know. 🙂 Oh… And in case you’d wonder: why the h? Well… Not sure. It’s said the h stands for Hilfsgrösse, so that’s some constant which was just supposed to help him out with the calculation. At that time, Planck did not suspect it would turn out to be one of the most fundamental physical constants. 🙂

Post scriptum: I went quite far in my presentation of the basics of the kinetic theory of gases. You may wonder now. I didn’t use that theoretical PV^γ= C relation, did I? And why all the fuss about photon gas? Well… That was just to introduce that PV^γ= C relation, so I could note, here, in this post scriptum, that it has a similar problem. The γ exponent is referred to as the specific heat ratio of a gas, and it can be calculated theoretically as well, as we did–well… Sort of, because we skipped the actual derivation. However, their theoretical value also differs substantially from actually measured values, and the problem is the same: one should not assume that a continuous value for 〈E〉. Agreement between theory and experiment can only be reached when the same assumptions as those of Planck are used: discrete energy levels, multiples of ħ and ω: E_n= nħω. Also, the specific functional form which Planck used to resolve the blackbody radiation problem is also to be used here. For more details, I’ll refer to Feynman too. I can’t say this is easy to digest, but then who said it would be easy? 🙂

The point to note is that the blackbody radiation problem wasn’t the only problem in the 19th century. As Feynman puts it: “One often hears it said that physicists at the latter part of the nineteenth century thought they knew all the significant physical laws and that all they had to do was to calculate more decimal places. Someone may have said that once, and others copied it. But a thorough reading of the literature of the time shows they were all worrying about something.” They were, and so Planck came up with something new. And then Einstein took it to the next level and then… Well… The rest is history. 🙂

Diffraction and the Uncertainty Principle (II)

Pre-script (dated 26 June 2020): This post did not suffer too much from the attack on this blog by the the dark force. It remains relevant. 🙂

Original post:

In my previous post, I derived and explained the general formula for the pattern generated by a light beam going through a slit or a circular aperture: the diffraction pattern. For light going through an aperture, this generates the so-called Airy pattern. In practice, diffraction causes a blurring of the image, and may make it difficult to distinguish two separate points, as shown below (credit for the image must go to Wikipedia again, I am afraid).

Airy_disk_spacing_near_Rayleigh_criterion

What’s actually going on is that the lens acts as a slit or, if it’s circular (which is usually the case), as an aperture indeed: the wavefront of the transmitted light is taken to be spherical or plane when it exits the lens and interferes with itself, thereby creating the ring-shaped diffraction pattern that we explained in the previous post.

The spatial resolution is also known as the angular resolution, which is quite appropriate, because it refers to an angle indeed: we know the first minimum (i.e. the first black ring) occurs at an angle θ such that sinθ = λ/L, with λ the wavelength of the light and L the lens diameter. It’s good to remind ourselves of the geometry of the situation: below we picture the array of oscillators, and so we know that the first minimum occurs at an angle such that Δ = λ. The second, third, fourth etc minimum occurs at an angle θ such that Δ = 2λ, 3λ, 4λ, etc. However, these secondary minima do not play any role in determining the resolving power of a lens, or a telescope, or an electron microscope, etc, and so you can just forget about them for the time being.

For small angles (expressed in radians), we can use the so-called small-angle approximation and equate sinθ with θ: the error of this approximation is less than one percent for angles smaller than 0.244 radians (14°), so we have the amazingly simply result that the first minimum occurs at an angle θ such that:

θ = λ/L

Spatial resolution of a microscope: the Rayleigh criterion versus Dawes’ limit

If we have two point sources right next to each other, they will create two Airy disks, as shown above, which may overlap. That may make it difficult to see them, in a telescope, a microscope, or whatever device. Hence, telescopes, microscopes (using light or electron beams or whatever) have a limited resolving power. How do we measure that?

The so-called Rayleigh criterion regards two point sources as just resolved when the principal diffraction maximum of one image coincides with the first minimum of the other, as shown below. If the distance is greater, the two points are (very) well resolved, and if it is smaller, they are regarded as not resolved. This angle is obviously related to the θ = λ/L angle but it’s not the same: in fact, it’s a slightly wider angle. The analysis involved in calculating the angular resolution in terms of angle, and we use the same symbol θ for it, is quite complicated and so I’ll skip that and just give you the result:

θ = 1.22λ/L

Note that, in this equation, θ stands for the angular resolution, λ for the wavelength of the light being used, and L is the diameter of the (aperture of) the lens. In the first of the three images above, the two points are well separated and, hence, the angle between them is well above the angular resolution. In the second, the angle between just meets the Rayleigh criterion, and in the third the angle between them is smaller than the angular resolution and, hence, the two points are not resolved.

Of course, the Rayleigh criterion is, to some extent, a matter of judgment. In fact, an English 19th century astronomer, named William Rutter Dawes, actually tested human observers on close binary stars of equal brightness, and found they could make out the two stars within an angle that was slightly narrower than the one given by the Rayleigh criterion. Hence, for an optical telescope, you’ll also find the simple θ = λ/L formula, so that’s the formula without the 1.22 factor (of course, λ here is, once again, the wavelength of the observed light or radiation, and L is the diameter of the telescope’s primary lens). This very simple formula allows us, for example, to calculate the diameter of the telescope lens we’d need to build to separate (see) objects in space with a resolution of, for example, 1 arcsec (i.e. 1/3600 of a degree or π/648,000 of a radian). Indeed, if we filter for yellow light only, which has a wavelength of 580 nm, we find L = 580×10⁻⁹ m/(π/648,000) = 0.119633×10⁻⁶ m ≈ 12 cm. [Just so you know: that’s about the size of the lens aperture of a good telescope (4 or 6 inches) for amateur astronomers–just in case you’d want one. :-)]

This simplified formula is called Dawes’ limit, and you’ll often see it used instead of Rayleigh’s criterion. However, the fact that it’s exactly the same formula as our formula for the first minimum of the Airy pattern should not confuse you: angular resolution is something different.

Now, after this introduction, let me get to the real topic of this post: Heisenberg’s Uncertainty Principle according to Heisenberg.

Heisenberg’s Uncertainty Principle according to Heisenberg

I don’t know about you but, as a kid, I didn’t know much about waves and fields and all that, and so I had difficulty understanding why the resolving power of a microscope or any other magnifying device depended on the frequency or wavelength. I now know my understanding was limited because I thought the concept of the amplitude of an electromagnetic wave had some spatial meaning, like the amplitude of a water or a sound wave. You know what I mean: this false idea that an electromagnetic wave is something that sort of wriggles through space, just like a water or sound wave wriggle through their medium (water and air respectively). Now I know better: the amplitude of an electromagnetic wave measures field strength and there’s no medium (no aether). So it’s not like a wave going around some object, or making some medium oscillate. I am not ashamed to acknowledge my stupidity at the time: I am just happy I finally got it, because it helps to really understand Heisenberg’s own illustration of his Uncertainty Principle, which I’ll present now.

Heisenberg imagined a gamma-ray microscope, as shown below (I copied this from the website of the American Institute for Physics ). Gamma-ray microscopes don’t exist – they’re hard to produce: you need a nuclear reactor or so 🙂 – but, as Heisenberg saw the development of new microscopes using higher and higher energy beams (as opposed to the 1.5-3 eV light in the visible spectrum) so as to increase the angular resolution and, hence, be able to see smaller things, he imagined one could use, perhaps, gamma-rays for imaging. Gamma rays are the hardest radiation, with frequencies of 10 exaherz and more (or >10¹⁹ Hz) and, hence, energies above 100 keV (i.e. 100,000 more than photons in the visible light spectrum, and 1000 times more than the electrons used in an average electron microscope). Gamma rays are not the result of some electron jumping from a higher to a lower energy level: they are emitted in decay processes of atomic nuclei (gamma decay). But I am digressing. Back to the main story line. So Heisenberg imagined we could ‘shine’ gamma rays on an electron and that we could then ‘see’ that electron in the microscope because some of the gamma photons would indeed end up in the microscope after their ‘collision’ with the electron, as shown below.

The experiment is described in many places elsewhere but I found these accounts often confusing, and so I present my own here. 🙂

What Heisenberg basically meant to show is that this set-up would allow us to gather precise information on the position of the electron–because we would know where it was–but that, as a result, we’d lose information in regard to its momentum. Why? To put it simply: because the electron recoils as a result of the interaction. The point, of course, is to calculate the exact relationship between the two (position and momentum). In other words: what we want to do is to state the Uncertainty Principle quantitatively, not qualitatively.

Now, the animation above uses the symbol L for the γ-ray wavelength λ, which is confusing because I used L for the diameter of the aperture in my explanation of diffraction above. The animation above also uses a different symbol for the angular resolution: A instead of θ. So let me borrow the diagram used in the Wikipedia article and rephrase the whole situation.

From the diagram above, it’s obvious that, to be scattered into the microscope, the γ-ray photon must be scattered into a cone with angle ε. That angle is obviously related to the angular resolution of the microscope, which is θ = ε/2 = λ/D, with D the diameter of the aperture (i.e. the primary lens). Now, the electron could actually be anywhere, and the scattering angle could be much larger than ε, and, hence, relating D to the uncertainty in position (Δx) is not as obvious as most accounts of this thought experiment make it out to be. The thing is: if the scattering angle is larger than ε, it won’t reach the light detector at the end of the microscope (so that’s the flat top in the diagram above). So that’s why we can equate D with Δx, so we write Δx = ± D/2 = D. To put it differently: the assumption here is basically that this imaginary microscope ‘sees’ an area that is approximately as large as the lens. Using the small-angle approximation (so we write sin(2ε) ≈ 2ε), we can write:

Δx = 2λ/ε

Now, because of the recoil effect, the electron receives some momentum from the γ-ray photon. How much? Well… The situation is somewhat complicated (much more complicated than the Wikipedia article on this very same topic suggests), because the photon keeps some but also gives some of its original momentum. In fact, what’s happening really is Compton scattering: the electron first absorbs the photon, and then emits another with a different energy and, hence, also with different frequency and wavelength. However, what we do now is that the photon’s original momentum was equal to E/c= p = h/λ. That’s just the Planck relation or, if you’d want to look at the photon as a particle, the de Broglie equation.

Now, because we’re doing an analysis in one dimension only (x), we’re only going to look at the momentum in this direction only, i.e. p_x, and we’ll assume that all of the momentum of the photon before the interaction (or ‘collision’ if you want) was horizontal. Hence, we can write p_x= h/λ. After the collision, however, this momentum is spread over the electron and the scattered or emitted photon that’s going into the microscope. Let’s now imagine the two extremes:

The scattered photon goes to the left edge of the lens. Hence, its horizontal momentum is negative (because it moves to the left) and the momentum p_xwill be distributed over the electron and the photon such that p_x= p’_x–h(ε/2)/λ’. Why the ε/2 factor? Well… That’s just trigonometry: the horizontal momentum of the scattered photon is obviously only a tiny fraction of its original horizontal momentum, and that fraction is given by the angle ε/2.
The scattered photon goes to the right edge of the lens. In that case, we write p_x= p”_x+ h(ε/2)/λ”.

Now, the spread in the momentum of the electron, which we’ll simply write as Δp, is obviously equal to:

Δp = p”_x– p’_x= p_x+ h(ε/2)/λ” – p_x+ h(ε/2)/λ’ = h(ε/2)/λ” + h(ε/2)/λ’ = h(ε/2)/λ” + h(ε/2)/λ’

That’s a nice formula, but what can we do with it? What we want is a relationship between Δx and Δp, i.e. the position and the momentum of the electron, and of the electron only. That involves another simplification, which is also dealt with very summarily – too summarily in my view – in most accounts of this experiment. So let me spell it out. The angle ε is obviously very small and, hence, we may equate λ’ and λ”. In addition, while these two wavelengths differ from the wavelength of the incoming photon, the scattered photon is, obviously, still a gamma ray and, therefore, we are probably not too far off when substituting both λ’ and λ” for λ, i.e. the frequency of the incoming γ-ray. Now, we can re-write Δx = 2λ/ε as 1/Δx = ε/(2λ). We then get:

Δp = p”_x– p’_x= hε/2λ” + hε/2λ’ = 2hε/2λ = 2h/Δx

Now that yields ΔpΔx = 2h, which is an approximate expression of Heisenberg’s Uncertainty Principle indeed (don’t worry about the factor 2, as that’s something that comes with all of the approximations).

A final moot point perhaps: it is obviously a thought experiment. Not only because we don’t have gamma-ray microscopes (that’s not relevant because we can effectively imagine constructing one) but because the experiment involves only one photon. A real microscope would organize a proper beam, but that would obviously complicate the analysis. In fact, it would defeat the purpose, because the whole point is to analyze one single interaction here.

The interpretation

Now how should we interpret all of this? Is this Heisenberg’s ‘proof’ of his own Principle? Yes and no, I’d say. It’s part illustration, and part ‘proof’, I would say. The crucial assumptions here are:

We can analyze γ-ray photons, or any photon for that matter, as particles having some momentum, and when ‘colliding’, or interacting, with an electron, the photon will impart some momentum to that electron.
Momentum is being conserved and, hence, the total (linear) momentum before and after the collision, considering both particles–i.e. (1) the incoming ray and the electron before the interaction and (2) the emitted photon and the electron that’s getting the kick after the interaction–must be the same.
For the γ-ray photon, we can relate (or associate, if you prefer that term) its wavelength λ with its momentum p through the Planck relation or, what amounts to the same for photons (because they have no mass), the de Broglie relation.

Now, these assumptions are then applied to an analysis of what we know to be true from experiment, and that’s the phenomenon of diffraction, part of which is the observation that the resolving power of a microscope is limited, and that its resolution is given by the θ = λ/D equation.

Bringing it all together, then gives us a theory which is consistent with experiment and, hence, we then assume the theory is true. Why? Well… I could start a long discourse here on the philosophy of science but, when everything is said and done, we should admit we don’t any ‘better’ theory.

But, you’ll say: what’s a ‘better’ theory? Well… Again, the answer to that question is the subject-matter of philosophers. As for me, I’d just refer to what’s known as Occam’s razor: among competing hypotheses, we should select the one with the fewest assumptions. Hence, while more complicated solutions may ultimately prove correct, the fewer assumptions that are made, the better. Now, when I was a kid, I thought quantum mechanics was very complicated and, hence, describing it here as a ‘simple’ theory sounds strange. But that’s what it is in the end: there’s no better (read: simpler) way to describe, for example, why electrons interfere with each other, and with themselves, when sending them through one or two slits, and so that’s what all these ‘illustrations’ want to show in the end, even if you think there must be simpler way to describe reality. As said, as a kid, I thought so too. 🙂

Diffraction and the Uncertainty Principle (I)

Pre-script (dated 26 June 2020): This post got mutilated by the removal of material by the dark force. It should be possible, however, to follow the main story line. If anything, the lack of illustrations will help you think things through for yourself. 🙂

Original post:

In his Lectures, Feynman advances the double-slit experiment with electrons as the textbook example explaining the “mystery” of quantum mechanics. It shows interference–a property of waves–of ‘particles’, electrons: they no longer behave as particles in this experiment. While it obviously illustrates “the basic peculiarities of quantum mechanics” very well, I think the dual behavior of light – as a wave and as a stream of photons – is at least as good as an illustration. And he could also have elaborated the phenomenon of electron diffraction.

Indeed, the phenomenon of diffraction–light, or an electron beam, interfering with itself as it goes through one slit only–is equally fascinating. Frankly, I think it does not get enough attention in textbooks, including Feynman’s, so that’s why I am devoting a rather long post to it here.

To be fair, Feynman does use the phenomenon of diffraction to illustrate the Uncertainty Principle, both in his Lectures as well as in that little marvel, QED: The Strange Theory of Light of Matter–a must-read for anyone who wants to understand the (probability) wave function concept without any reference to complex numbers or what have you. Let’s have a look at it: light going through a slit or circular aperture, illustrated in the left-hand image below, creates a diffraction pattern, which resembles the interference pattern created by an array of oscillators, as shown in the right-hand image.

$Diffraction for particle wave$

Let’s start with the right-hand illustration, which illustrates interference, not diffraction. We have eight point sources of electromagnetic radiation here (e.g. radio waves, but it can also be higher-energy light) in an array of length L. λ is the wavelength of the radiation that is being emitted, and α is the so-called intrinsic relative phase–or, to put it simply, the phase difference. We assume α is zero here, so the array produces a maximum in the direction θ_out = 0, i.e. perpendicular to the array. There are also weaker side lobes. That’s because the distance between the array and the point where we are measuring the intensity of the emitted radiation does result in a phase difference, even if the oscillators themselves have no intrinsic phase difference.

Interference patterns can be complicated. In the set-up below, for example, we have an array of oscillators producing not just one but many maxima. In fact, the array consists of just two sources of radiation, separated by 10 wavelengths.

The explanation is fairly simple. Once again, the waves emitted by the two point sources will be in phase in the east-west (E-W) direction, and so we get a strong intensity there: four times more, in fact, than what we would get if we’d just have one point source. Indeed, the waves are perfectly in sync and, hence, add up, and the factor four is explained by the fact that the intensity, or the energy of the wave, is proportional to the square of the amplitude: 2²= 4. We get the first minimum at a small angle away (the angle from the normal is denoted by ϕ in the illustration), where the arrival times differ by 180°, and so there is destructive interference and the intensity is zero. To be precise, if we draw a line from each oscillator to a distant point and the difference Δ in the two distances is λ/2, half an oscillation, then they will be out of phase. So this first null occurs when that happens. If we move a bit further, to the point where the difference Δ is equal to λ, then the two waves will be a whole cycle out of phase, i.e. 360°, which is the same as being exactly in phase again! And so we get many maxima (and minima) indeed.

But this post should not turn into a lesson on how to construct a radio broadcasting array. The point to note is that diffraction is usually explained using this rather simple theory on interference of waves assuming that the slit itself is an array of point sources, as illustrated below (while the illustrations above were copied from Feynman’s Lectures, the ones below were taken from the Wikipedia article on diffraction). This is referred to as the Huygens-Fresnel Principle, and the math behind is summarized in Kirchhoff’s diffraction formula.

$500px-Refraction_on_an_aperture_-_Huygens-Fresnel_principle$

Now, that all looks easy enough, but the illustration above triggers an obvious question: what about the spacing between those imaginary point sources? Why do we have six in the illustration above? The relation between the length of the array and the wavelength is obviously important: we get the interference pattern that we get with those two point sources above because the distance between them is 10λ. If that distance would be different, we would get a different interference pattern. But so how does it work exactly? If we’d keep the length of the array the same (L = 10λ) but we would add more point sources, would we get the same pattern?

The easy answer is yes, and Kirchhoff’s formula actually assumes we have an infinite number of point sources between those two slits: every point becomes the source of a spherical wave, and the sum of these secondary waves then yields the interference pattern. The animation below shows the diffraction pattern from a slit with a width equal to five times the wavelength of the incident wave. The diffraction pattern is the same as above: one strong central beam with weaker lobes on the sides.

However, the truth is somewhat more complicated. The illustration below shows the interference pattern for an array of length L = 10λ–so that’s like the situation with two point sources above–but with four additional point sources to the two we had already. The intensity in the E–W direction is much higher, as we would expect. Adding six waves in phase yields a field strength that is six times as great and, hence, the intensity (which is proportional to the square of the field) is thirty-six times as great as compared to the intensity of one individual oscillator. Also, when we look at neighboring points, we find a minimum and then some more ‘bumps’, as before, but then, at an angle of 30°, we get a second beam with the same intensity as the central beam. Now, that’s something we do not see in the diffraction patterns above. So what’s going on here?

Before I answer that question, I’d like to compare with the quantum-mechanical explanation. It turns out that this question in regard to the relevance of the number of point sources also pops up in Feynman’s quantum-mechanical explanation of diffraction.

The quantum-mechanical explanation of diffraction

The illustration below (taken from Feynman’s QED, p. 55-56) presents the quantum-mechanical point of view. It is assumed that light consists of a photons, and these photons can follow any path. Each of these paths is associated with what Feynman simply refers to as an arrow, but so it’s a vector with a magnitude and a direction: in other words, it’s a complex number representing a probability amplitude.

Many arrows

In order to get the probability for a photon to travel from the source (S) to a point (P or Q), we have to add up all the ‘arrows’ to arrive at a final ‘arrow’, and then we take its (absolute) square to get the probability. The text under each of the two illustrations above speaks for itself: when we have ‘enough’ arrows (i.e. when we allow for many neighboring paths, as in the illustration on the left), then the arrows for all of the paths from S to P will add up to one big arrow, because there is hardly any difference in time between them, while the arrows associated with the paths to Q will cancel out, because the difference in time between them is fairly large. Hence, the light will not go to Q but travel to P, i.e. in a straight line.

However, when the gap is nearly closed (so we have a slit or a small aperture), then we have only a few neighboring paths, and then the arrows to Q also add up, because there is hardly any difference in time between them too. As I am quoting from Feynman’s QED here, let me quote all of the relevant paragraph: “Of course, both final arrows are small, so there’s not much light either way through such a small hole, but the detector at Q will click almost as much as the one at P ! So when you try to squeeze light too much to make sure it’s going only in a straight line, it refuses to cooperate and begins to spread out. This is an example of the Uncertainty Principle: there is a kind of complementarity between knowledge of where the light goes between the blocks and where it goes afterwards. Precise knowledge of both is impossible.” (Feynman, QED, p. 55-56).

Feynman’s quantum-mechanical explanation is obviously more ‘true’ that the classical explanation, in the sense that it corresponds to what we know is true from all of the 20th century experiments confirming the quantum-mechanical view of reality: photons are weird ‘wavicles’ and, hence, we should indeed analyze diffraction in terms of probability amplitudes, rather than in terms of interference between waves. That being said, Feynman’s presentation is obviously somewhat more difficult to understand and, hence, the classical explanation remains appealing. In addition, Feynman’s explanation triggers a similar question as the one I had on the number of point sources. Not enough arrows !? What do we mean with that? Why can’t we have more of them? What determines their number?

Let’s first look at their direction. Where does that come from? Feynman is a wonderful teacher here. He uses an imaginary stopwatch to determine their direction: the stopwatch starts timing at the source and stops at the destination. But all depends on the speed of the stopwatch hand of course. So how fast does it turn? Feynman is a bit vague about that but notes that “the stopwatch hand turns around faster when it times a blue photon than when it does when timing a red photon.” In other words, the speed of the stopwatch hand depends on the frequency of the light: blue light has a higher frequency (645 THz) and, hence, a shorter wavelength (465 nm) then red light, for which f = 455 THz and λ = 660 nm. Feynman uses this to explain the typical patterns of red, blue, and violet (separated by borders of black), when one shines red and blue light on a film of oil or, more generally,the phenomenon of iridescence in general, as shown below.

As for the size of the arrows, their length is obviously subject to a normalization condition, because all probabilities have to add up to 1. But what about their number? We didn’t answer that question–yet.

The answer, of course, is that the number of arrows and their size are obviously related: we associate a probability amplitude with every way an event can happen, and the (absolute) square of all these probability amplitudes has to add up to 1. Therefore, if we would identify more paths, we would have more arrows, but they would have to be smaller. The end result would be the same though: when the slit is ‘small enough’, the arrows representing the paths to Q would not cancel each other out and, hence, we’d have diffraction.

You’ll say: Hmm… OK. I sort of see the idea, but how do you explain that pattern–the central beam and the smaller side lobes, and perhaps a second beam as well? Well… You’re right to be skeptical. In order to explain the exact pattern, we need to analyze the wave functions, and that requires a mathematical approach rather than the type of intuitive approach which Feynman uses in his little QED booklet. Before we get started on that, however, let me give another example of such intuitive approach.

Diffraction and the Uncertainty Principle

Let’s look at that very first illustration again, which I’ve copied, for your convenience, again below. Feynman uses it (III-2-2) to (a) explain the diffraction pattern which we observe when we send electrons through a slit and (b) to illustrate the Uncertainty Principle. What’s the story?

Well… Before the electrons hit the wall or enter the slit, we have more or less complete information about their momentum, but nothing on their position: we don’t know where they are exactly, and we also don’t know if they are going to hit the wall or go through the slit. So they can be anywhere. However, we do know their energy and momentum. That momentum is horizontal only, as the electron beam is normal to the wall and the slit. Hence, their vertical momentum is zero–before they hit the wall or enter the slit that is. We’ll denote their (horizontal) momentum, i.e. the momentum before they enter the slit, as p₀.

$Diffraction for particle wave$

Now, if an electron happens to go through the slit, and we know because we detected it on the other side, then we know its vertical position (y) at the slit itself with considerable accuracy: that position will be the center of the slit ±B/2. So the uncertainty in position (Δy) is of the order B, so we can write: Δy = B. However, according to the Uncertainty Principle, we cannot have precise knowledge about its position and its momentum. In addition, from the diffraction pattern itself, we know that the electron acquires some vertical momentum. Indeed, some electrons just go straight, but more stray a bit away from the normal. From the interference pattern, we know that the vast majority stays within an angle Δθ, as shown in the plot. Hence, plain trigonometry allows us to write the spread in the vertical momentum p_y as p₀Δθ, with p₀the horizontal momentum. So we have Δp_y = p₀Δθ.

Now, what is Δθ? Well… Feynman refers to the classical analysis of the phenomenon of diffraction (which I’ll reproduce in the next section) and notes, correctly, that the first minimum occurs at an angle such that the waves from one edge of the slit have to travel one wavelength farther than the waves from the other side. The geometric analysis (which, as noted, I’ll reproduce in the next section) shows that that angle is equal to the wavelength divided by the width of the slit, so we have Δθ = λ/B. So now we can write:

Δp_y = p₀Δθ = p₀λ/B

That shows that the uncertainty in regard to the vertical momentum is, indeed, inversely proportional to the uncertainty in regard to its position (Δy), which is the slit width B. But we can go one step further. The de Broglie relation relates wavelength to momentum: λ = h/p. What momentum? Well… Feynman is a bit vague on that: he equates it with the electron’s horizontal momentum, so he writes λ = h/p₀. Is this correct? Well… Yes and no. The de Broglie relation associates a wavelength with the total momentum, but then it’s obvious that most of the momentum is still horizontal, so let’s go along with this. What about the wavelength? What wavelength are we talking about here? It’s obviously the wavelength of the complex-valued wave function–the ‘probability wave’ so to say.

OK. So, what’s next? Well… Now we can write that Δp_y = p₀Δθ = p₀λ/B = p₀(h/p₀)/B. Of course, the p₀factor vanishes and, hence, bringing B to the other side and substituting for Δy = B yields the following grand result:

Δy·Δp_y= h

Wow ! Did Feynman ‘prove’ Heisenberg’s Uncertainty Principle here?

Well… No. Not really. First, the ‘proof’ above actually assumes there’s fundamental uncertainty as to the position and momentum of a particle (so it actually assumes some uncertainty principle from the start), and then it derives it from another fundamental assumption, i.e. the de Broglie relation, which is obviously related to the Uncertainty Principle. Hence, all of the above is only an illustration of the Uncertainty Principle. It’s no proof. As far as I know, one can’t really ‘prove’ the Uncertainty Principle: it’s a fundamental assumption which, if accepted, makes our observations consistent with the theory that is based on it, i.e. quantum or wave mechanics.

Finally, note that everything that I wrote above also takes the diffraction pattern as a given and, hence, while all of the above indeed illustrates the Uncertainty Principle, it’s not an explanation of the phenomenon of diffraction as such. For such explanation, we need a rigorous mathematical analysis, and that’s a classical analysis. Let’s go for it!

Going from six to n oscillators

The mathematics involved in analyzing diffraction and/or interference are actually quite tricky. If you’re alert, then you should have noticed that I used two illustrations that both have six oscillators but that the interference pattern doesn’t match. I’ve reproduced them below. The illustration on the right-hand side has six oscillators and shows a second beam besides the central one–and, of course, there’s such beam also 30° higher, so we have (at least) three beams with the same intensity here–while the animation on the left-hand side shows only one central beam. So what’s going on here?

The answer is that, in the particular example on the left-hand side, the successive dipole radiators (i.e. the point sources) are separated by a distance of two wavelengths (2λ). In that case, it is actually possible to find an angle where the distance δ between successive dipoles is exactly one wavelength (note the little δ in the illustration, as measured from the second point source), so that the effects from all of them are in phase again. So each one is then delayed relative to the next one by 360 degrees, and so they all come back in phase, and then we have another strong beam in that direction! In this case, the other strong beam has an angle of 30 degrees as compared to the E-W line. If we would put in some more oscillators, to ensure that they are all closer than one wavelength apart, then this cannot happen. And so it’s not happening with light. 🙂 But now that we’re here, I’ll just quickly note that it’s an interesting and useful phenomenon used in diffraction gratings, but I’ll refer you to the literature on that, as I shouldn’t be bothering you with all these digressions. So let’s get back at it.

In fact, let me skip the nitty-gritty of the detailed analysis (I’ll refer you to Feynman’s Lectures for that) and just present the grand result for n oscillators, as depicted below:

This, indeed, shows the diffraction pattern we are familiar with: one strong maximum separated from successive smaller ones (note that the dotted curve magnifies the actual curve with a factor 10). The vertical axis shows the intensity, but expressed as a fraction of the maximum intensity, which is n²I₀(I₀is the intensity we would observe if there was only 1 oscillator). As for the horizontal axis, the variable there is ϕ really, although we re-scale the variable in order to get 1, 2, 2 etcetera for the first, second, etcetera minimum. This ϕ is the phase difference. It consists of two parts:

The intrinsic relative phase α, i.e. the difference in phase between one oscillator and the next: this is assumed to be zero in all of the examples of diffraction patterns above but so the mathematical analysis here is somewhat more general.
The phase difference which results from the fact that we are observing the array in a given direction θ from the normal. Now that‘s the interesting bit, and it’s not so difficult to show that this additional phase is equal to 2πdsinθ/λ, with d the distance between two oscillators, λ the wavelength of the radiation, and θ the angle from the normal.

In short, we write:

ϕ = α + 2πdsinθ/λ

Now, because I’ll have to use the variables below in the analysis that follows, I’ll quickly also reproduce the geometry of the set-up (all illustrations here taken from Feynman’s Lectures):

Before I proceed, please note that we assume that d is less than λ, so we only have one great maximum, and that’s the so-called zero-order beam centered at θ = 0. In order to get subsidiary great maxima (referred to as first-order, second-order, etcetera beams in the context of diffraction gratings), we must have the spacing d of the array greater than one wavelength, but so that’s not relevant for what we’re doing here, and that is to move from a discrete analysis to a continuous one.

Before we do that, let’s look at that curve again and analyze where the first minimum occurs. If we assume that α = 0 (no intrinsic relative phase), then the first minimum occurs when ϕ = 2π/n. Using the ϕ = α + 2πdsinθ/λ formula, we get 2πdsinθ/λ = 2π/n or ndsinθ = λ. What does that mean? Well, nd is the total length L of the array, so we have ndsinθ = Lsinθ = Δ = λ. What that means is that we get the first minimum when Δ is equal to one wavelength.

Now why do we get a minimum when Δ = λ? Because the contributions of the various oscillators are then uniformly distributed in phase from 0° to 360°. What we’re doing, once again, is adding arrows in order to get a resultant arrow A_R, as shown below for n = 6. At the first minimum, the arrows are going around a whole circle: we are adding equal vectors in all directions, and such a sum is zero. So when we have an angle θ such that Δ = λ, we get the first minimum. [Note that simple trigonometry rules imply that θ must be equal to λ/L, a fact which we used in that quantum-mechanical analysis of electron diffraction above.]

What about the second minimum? Well… That occurs when ϕ = 4π/n. Using the ϕ = 2πdsinθ/λ formula again, we get 2πdsinθ/λ = 4π/n or ndsinθ = 2λ. So we get ndsinθ = Lsinθ = Δ = 2λ. So we get the second minimum at an angle θ such that Δ = 2λ. For the third minimum, we have ϕ = 6π/n. So we have 2πdsinθ/λ = 6π/n or ndsinθ = 3λ. So we get the third minimum at an angle θ such that Δ = 3λ. And so on and so on.

The point to note is that the diffraction pattern depends only on the wavelength λ and the total length L of the array, which is the width of the slit of course. Hence, we can actually extend the analysis for n going from some fixed value to infinity, and we’ll find that we will only have one great maximum with a lot of side lobes that are much and much smaller, with the minima occurring at angles such that Δ = λ, 2λ, 3λ, etcetera.

OK. What’s next? Well… Nothing. That’s it. I wanted to do a post on diffraction, and so that’s what I did. However, to wrap it all up, I’ll just include two more images from Wikipedia. The one on the left shows the diffraction pattern of a red laser beam made on a plate after passing a small circular hole in another plate. The pattern is quite clear. On the right-hand side, we have the diffraction pattern generated by ordinary white light going through a hole. In fact, it’s a computer-generated image and the gray scale intensities have been adjusted to enhance the brightness of the outer rings, because we would not be able to see them otherwise.

But… Didn’t I say I would write about diffraction and the Uncertainty Principle? Yes. And I admit I did not write all that much about the Uncertainty Principle above. But so I’ll do that in my next post, in which I intend to look at Heisenberg’s own illustration of the Uncertainty Principle. That example involves a good understanding of the resolving power of a lens or a microscope, and such understanding also involves some good mathematical analysis. However, as this post has become way too long already, I’ll leave that to the next post indeed. I’ll use the left-hand image above for that, so have a good look at it. In fact, let me quickly quote Wikipedia as an introduction to my next post:

The diffraction pattern resulting from a uniformly-illuminated circular aperture has a bright region in the center, known as the Airy disk which together with the series of concentric bright rings around is called the Airy pattern.

We’ll need in order to define the resolving power of a microscope, which is essential to understanding Heisenberg’s illustration of the Principle he advanced himself. But let me stop here, as it’s the topic of my next write-up indeed. This post has become way too long already. 🙂

Photons as strings

Pre-script written much later: In the meanwhile, we figured it all out. We found the common-sense interpretation of quantum physics. No ambiguity. No hocus-pocus. I keep posts like the one below online only to, one day, go back to where I went wrong. 🙂

Jean Louis Van Belle, 20 May 2020

In my previous post, I explored, somewhat jokingly, the grey area between classical physics and quantum mechanics: light as a wave versus light as a particle. I did so by trying to picture a photon as an electromagnetic transient traveling through space, as illustrated below. While actual physicists would probably deride my attempt to think of a photon as an electromagnetic transient traveling through space, the idea illustrates the wave-particle duality quite well, I feel.

Understanding light is the key to understanding physics. Light is a wave, as Thomas Young proved to the Royal Society of London in 1803, thereby demolishing Newton’s corpuscular theory. But its constituents, photons, behave like particles. According to modern-day physics, both were right. Just to put things in perspective, the thickness of the note card which Young used to split the light – ordinary sunlight entering his room through a pinhole in a window shutter – was 1/30 of an inch, or approximately 0.85 mm. Hence, in essence, this is a double-slit experiment with the two slits being separated by a distance of almost 1 millimeter. That’s enormous as compared to modern-day engineering tolerance standards: what was thin then, is obviously not considered to be thin now. Scale matters. I’ll come back to this.

Young’s experiment (from www.physicsclassroom.com)

Young experiment

The table below shows that the ‘particle character’ of electromagnetic radiation becomes apparent when its frequency is a few hundred terahertz, like the sodium light example I used in my previous post: sodium light, as emitted by sodium lamps, has a frequency of 500×10¹²oscillations per second and, therefore (the relation between frequency and wavelength is very straightforward: their product is the velocity of the wave, so for light we have the simple λf = c equation), a wavelength of 600 nanometer (600×10^–9meter).

However, whether something behaves like a particle or a wave also depends on our measurement scale: 0.85 mm was thin in Young’s time, and so it was a delicate experiment then but now, it’s a standard classroom experiment indeed. The theory of light as a wave would hold until more delicate equipment refuted it. Such equipment came with another sense of scale. It’s good to remind oneself that Einstein’s “discovery of the law of the photoelectric effect”, which explained the photoelectric effect as the result of light energy being carried in discrete quantized packets of energy, now referred to as photons, goes back to 1905 only, and that the experimental apparatus which could measure it was not much older. So waves behave like particles if we look at them close enough. Conversely, particles behave like waves if we look at them close enough. So there is this zone where they are neither, the zone for which we invoke the mathematical formalism of quantum mechanics or, to put it more precisely, the formalism of quantum electrodynamics: that “strange theory of light and Matter”, as Feynman calls it.

Let’s have a look at how particles became waves. It should not surprise us that the experimental apparatuses needed to confirm that electrons–or matter in general–can actually behave like a wave is more recent than the 19th century apparatuses which led Einstein to develop his ‘corpuscular’ theory of light (i.e. the theory of light as photons). The engineering tolerances involved are daunting. Let me be precise here. To be sure, the phenomenon of electron diffraction (i.e. electrons going through one slit and producing a diffraction pattern on the other side) had been confirmed experimentally already in 1925, in the famous Davisson-Germer experiment. I am saying because it’s rather famous indeed. First, because electron diffraction was a weird thing to contemplate at the time. Second, because it confirmed the de Broglie hypothesis only two years after Louis de Broglie had advanced it. And, third, because Davisson and Germer had never intended to set it up to detect diffraction: it was pure coincidence. In fact, the observed diffraction pattern was the result of a laboratory accident, and Davisson and Germer weren’t aware of other, conscious, attempts of trying to prove the de Broglie hypothesis. 🙂 […] OK. I am digressing. Sorry. Back to the lesson.

The nanotechnology that was needed to confirm Feynman’s 1965 thought experiment on electron interference (i.e. electrons going through two slits and interfering with each other (rather than producing some diffraction pattern as they go through one slit only) – and, equally significant as an experiment result, with themselves as they go through the slit(s) one by one! – was only developed over the past decades. In fact, it was only in 2008 (and again in 2012) that the experiment was carried out exactly the way Feynman describes it in his Lectures.

It is useful to think of what such experiments entail from a technical point of view. Have a look at the illustration below, which shows the set-up. The insert in the upper-left corner shows the two slits which were used in the 2012 experiment: they are each 62 nanometer wide – that’s 50×10^–9 m! – and the distance between them is 272 nanometer, or 0.272 micrometer. [Just to be complete: they are 4 micrometer tall (4×10^–6 m), and the thing in the middle of the slits is just a little support (150 nm) to make sure the slit width doesn’t vary.]

The second inset (in the upper-right corner) shows the mask that can be moved to close one or both slits partially or completely. The mask is 4.5µm wide ×20µm tall. Please do take a few seconds to contemplate the technology behind this feat: a nanometer is a millionth of a millimeter, so that’s a billionth of a meter, and a micrometer is a millionth of a meter. To imagine how small a nanometer is, you should imagine dividing one millimeter in ten, and then one of these tenths in ten again, and again, and once again, again, and again. In fact, you actually cannot imagine that because we live in the world we live in and, hence, our mind is used only to addition (and subtraction) when it comes to comparing sizes and – to a much more limited extent – with multiplication (and division): our brain is, quite simply, not wired to deal with exponentials and, hence, it can’t really ‘imagine’ these incredible (negative) powers. So don’t think you can imagine it really, because one can’t: in our mind, these scales exist only as mathematical constructs. They don’t correspond to anything we can actually make a mental picture of.

The electron beam consisted of electrons with an (average) energy of 600 eV. That’s not an awful lot: 8.5 times more than the energy of an electron in orbit in a atom, whose energy would be some 70 eV, so the acceleration before they went through the slits was relatively modest. I’ve calculated the corresponding de Broglie wavelength of these electrons in another post (Re-Visiting the Matter-Wave, April 2014), using the de Broglie equations: f = E/h or λ = p/h. And, of course, you could just google the article on the experiment and read about it, but it’s a good exercise, and actually quite simple: just note that you’ll need to express the energy in joule (not in eV) to get it right. Also note that you need to include the rest mass of the electron in the energy. I’ll let you try it (or else just go to that post of mine). You should find a de Broglie wavelength of 50 picometer for these electrons, so that’s 50×10^–12 m. While that wavelength is less than a thousandth of the slit width (62 nm), and about 5,500 times smaller than the space between the two slits (272 nm), the interference effect was unambiguous in the experiment. I advice you to google the results yourself (or read that April 2014 post of mine if you want a summary): the experiment was done at the University of Nebraska-Lincoln in 2012.

Electrons and X-rays

To put everything in perspective: 50 picometer is like the wavelength of X-rays, and you can google similar double-slit experiments for X-rays: they also loose their ‘particle behavior’ when we look at them at this tiny scale. In short, scale matters, and the boundary between ‘classical physics’ (electromagnetics) and quantum physics (wave mechanics) is not clear-cut. If anything, it depends on our perspective, i.e. what we can measure, and we seem to be shifting that boundary constantly. In what direction?

Downwards obviously: we’re devising instruments that measure stuff at smaller and smaller scales, and what’s happening is that we can ‘see’ typical ‘particles’, including hard radiation such as gamma rays, as local wave trains. Indeed, the next step is clear-cut evidence for interference between gamma rays.

Energy levels of photons

We would not associate low-frequency electromagnetic waves, such as radio or radar waves, with photons. But light in the visible spectrum, yes. Obviously. […]

Isn’t that an odd dichotomy? If we see that, on a smaller scale, particles start to look like waves, why would the reverse not be true? Why wouldn’t we analyze radio or radar waves, on a much larger scale, as a stream of very (I must say extremely) low-energy photons? I know the idea sounds ridiculous, because the energies involved would be ridiculously low indeed. Think about it. The energy of a photon is given by the Planck relation: E = hf = hc/λ. For visible light, with wavelengths ranging from 800 nm (red) to 400 nm (violet or indigo), the photon energies range between 1.5 and 3 eV. Now, the shortest wavelengths for radar waves are in the so-called millimeter band, i.e. they range from 1 mm to 1 cm. A wavelength of 1 mm corresponds to a photon energy of 0.00124 eV. That’s close to nothing, of course, and surely not the kind of energy levels that we can currently detect.

But you get the idea: there is a grey area between classical physics and quantum mechanics, and it’s our equipment–notably the scale of our measurements–that determine where that grey area begins, and where it ends, and it seems to become larger and larger as the sensitivity of our equipment improves.

What do I want to get at? Nothing much. Just some awareness of scale, as an introduction to the actual topic of this post, and that’s some thoughts on a rather primitive string theory of photons. What !?

Yes. Purely speculative, of course. 🙂

Photons as strings

I think my calculations in the previous post, as primitive as they were, actually provide quite some food for thought. If we’d treat a photon in the sodium light band (i.e. the light emitted by sodium, from a sodium lamp for instance) just like any other electromagnetic pulse, we would find it’s a pulse of some 10 meter long. We also made sense of this incredibly long distance by noting that, if we’d look at it as a particle (which is what we do when analyzing it as a photon), it should have zero size, because it moves at the speed of light and, hence, the relativistic length contraction effect ensures we (or any observer in whatever reference frame really, because light always moves at the speed of light, regardless of the reference frame) will see it as a zero-size particle.

Having said that, and knowing damn well that we have treat the photon as an elementary particle, I would think it’s very tempting to think of it as a vibrating string.

Huh?

Yes. Let me copy that graph again. The assumption I started with is a standard one in physics, and not something that you’d want to argue with: photons are emitted when an electron jumps from a higher to a lower energy level and, for all practical purposes, this emission can be analyzed as the emission of an electromagnetic pulse by an atomic oscillator. I’ll refer you to my previous post – as silly as it is – for details on these basics: the atomic oscillator has a Q, and so there’s damping involved and, hence, the assumption that the electromagnetic pulse resembles a transient should not sound ridiculous. Because the electric field as a function in space is the ‘reversed’ image of the oscillation in time, the suggested shape has nothing blasphemous.

Just go along with it for a while. First, we need to remind ourselves that what’s vibrating here is nothing physical: it’s an oscillating electromagnetic field. That being said, in my previous post, I toyed with the idea that the oscillation could actually also represent the photon’s wave function, provided we use a unit for the electric field that ensures that the area under the squared curve adds up to one, so as to normalize the probability amplitudes. Hence, I suggested that the field strength over the length of this string could actually represent the probability amplitudes, provided we choose an appropriate unit to measure the electric field.

But then I was joking, right? Well… No. Why not consider it? An electromagnetic oscillation packs energy, and the energy is proportional to the square of the amplitude of the oscillation. Now, the probability of detecting a particle is related to its energy, and such probability is calculated from taking the (absolute) square of probability amplitudes. Hence, mathematically, this makes perfect sense.

It’s quite interesting to think through the consequences, and I hope I will (a) understand enough of physics and (b) find enough time for this—one day! One interesting thing is that the field strength (i.e. the magnitude of the electric field vector) is a real number. Hence, if we equate these magnitudes with probability amplitudes, we’d have real probability amplitudes, instead of complex-valued ones. That’s not a very fundamental issue. It probably indicates we should also take into account the fact that the E vector also oscillates in the other direction that’s normal to the direction of propagation, i.e. the y-coordinate (assuming that the z-axis is the direction of propagation). To put it differently, we should take the polarization of the light into account. The figure below–which I took from Wikipedia again (by far the most convenient place to shop for images and animations: what would I do without it?– shows how the electric field vector moves in the xy-plane indeed, as the wave travels along the z-axis. So… Well… I still have to figure it all out, but the idea surely makes sense.

Another interesting thing to think about is how the collapse of the wave function would come about. If we think of a photon as a string, it must have some ‘hooks’ which could cause it to ‘stick’ or ‘collapse’ into a ‘lump’ as it hits a detector. What kind of hook? What force would come into play?

Well… The interaction between the photon and the photodetector is electromagnetic, but we’re looking for some other kind of ‘hook’ here. What could it be? I have no idea. Having said that, we know that the weakest of all fundamental forces—gravity—becomes much stronger—very much stronger—as the distance becomes smaller and smaller. In fact, it is said that, if we go to the Planck scale, the strength of the force of gravity becomes quite comparable with the other forces. So… Perhaps it’s, quite simply, the equivalent mass of the energy involved that gets ‘hooked’, somehow, as it starts interacting with the photon detector. Hence, when thinking about a photon as an oscillating string of energy, we should also think of that string as having some inseparable (equivalent) mass that, once it’s ‘hooked’, has no other option that to ‘collapse into itself’. [You may note there’s no quantum theory for gravity as yet. I am not sure how, but I’ve got a gut instinct that tells me that may help to explain why a photon consists of one single ‘unbreakable’ lump, although I need to elaborate this argument obviously.]

You must be laughing aloud now. A new string theory–really?

I know… I know… I haven’t reach sophomore level and I am already wildly speculating… Well… Yes. What I am talking about here has probably nothing to do with current string theories, although my proposed string would also replace the point-like photon by a one-dimensional ‘string’. However, ‘my’ string is, quite simply, an electromagnetic pulse (a transient actually, for reasons I explained in my previous post). Naive? Perhaps. However, I note that the earliest version of string theory is referred to as bosonic string theory, because it only incorporated bosons, which is what photons are.

So what? Well… Nothing… I am sure others have thought of this too, and I’ll look into it. It’s surely an idea which I’ll keep in the back of my head as I continue to explore physics. The idea is just too simple and beautiful to disregard, even if I am sure it must be pretty naive indeed. Photons as ten-meter long strings? Let’s just forget about it. 🙂 Onwards !!! 🙂

Post Scriptum: The key to ‘closing’ this discussion is, obviously, to be found in a full-blown analysis of the relativity of fields. So, yes, I have not done all of the required ‘homework’ on this and the previous post. I apologize for that. If anything, I hope it helped you to also try to think somewhat beyond the obvious. I realize I wasted a lot of time trying to understand the pre-cooked ready-made stuff that’s ‘on the market’, so to say. I still am, actually. Perhaps I should first thoroughly digest Feynman’s Lectures. In fact, I think that’s what I’ll try to do in the next year or so. Sorry for any inconvenience caused. 🙂

The shape and size of a photon

Important post script (PS) – dated 22 December 2018: Dear readers of this post, this is one of the more popular posts of my blog but − in the meanwhile − I did move on, and quite a bit, actually! The analysis below is not entirely consistent: I got many questions on it, and I have been thinking differently as a result. The Q&A below sums up everything: I do think of the photon as a pointlike particle now, and Chapter VIII of my book sums up the photon model. At the same time, if you are really interested in this question – how should one think of a photon? – then it’s probably good you also read the original post. If anything, it shows you how easy it is to get confused.

The shape and size of a photon

In reply to:

can we say that size of photon is depend upon its wavelength ? https://briankoberlein.com/2015/04/14/thats-about-the-size-of-it/

Hi Brian – see section III of this paper: http://vixra.org/pdf/1812.0273v2.pdf

Feynman’s classical idea of an atomic oscillator is fine in the context of the blackbody radiation problem, but his description of the photon as a long wavetrain does not make any sense. A photon has to pack two things: (1) the energy difference between the Bohr orbitals and (2) Planck’s constant h, which is the (physical) action associated with one cycle of an oscillation (so it’s a force over a distance (the loop or the radius – depending on the force you’re looking at) over a cycle time). See section V of the paper for how the fine-structure constant pops up here – it’s, as usual, a sort of scaling constant, but this time it scales a force. In any case, the idea is that we should think of a photon as one cycle – rather than a long wavetrain. The one cycle makes sense: when you calculate field strength and force you get quite moderate values (not the kind of black-hole energy concentrations some people suggest). It also makes sense from a logical point of view: the wavelength is something real, and so we should think of the photon amplitude (the electric field strength) as being real as well – especially when you think of how that photon is going to interact or be absorbed into another atom.

Sorry for my late reply. It’s been a while since I checked the comments. Please let me know if this makes sense. I’ll have a look at your blog in the coming days. I am working on a new paper on the anomalous magnetic moment – which is not anomalous as all if you start to think about how things might be working in reality. After many years of study, I’ve come to the conclusion that quantum mechanics is a nice way of describing things, but it doesn’t help us in terms of understanding anything. When we want to understand something, we need to push the classical framework a lot further than we currently do. In any case, that’s another discussion.

OK. Now you can move on to the post itself. 🙂 Sorry if this is confusing the reader, but it is necessary to warn him. I think of this post now as still being here to document the history of my search for a ‘basic version of truth’, as someone called it. [For an even more recent update, see Chapter 8 of my book, A Realist Interpretation of Quantum Mechanics.

Original post:

Photons are weird. All elementary particles are weird. As Feynman puts it, in the very first paragraph of his Lectures on Quantum Mechanics : “Historically, the electron, for example, was thought to behave like a particle, and then it was found that in many respects it behaved like a wave. So it really behaves like neither. Now we have given up. We say: “It is like neither. There is one lucky break, however—electrons behave just like light. The quantum behavior of atomic objects (electrons, protons, neutrons, photons, and so on) is the same for all, they are all “particle waves,” or whatever you want to call them. So what we learn about the properties of electrons will apply also to all “particles,” including photons of light.” (Feynman’s Lectures, Vol. III, Chapter 1, Section 1)

I wouldn’t dare to argue with Feynman, of course, but… What? Well… Photons are like electrons, and then they are not. Obviously not, I’d say. For starters, photons do not have mass or charge, and they are also bosons, i.e. ‘force-carriers’ (as opposed to matter-particles), and so they obey very different quantum-mechanical rules, which are referred to as Bose-Einstein statistics. I’ve written about that in other post (see, for example, my post on Bose-Einstein and Fermi-Dirac statistics), so I won’t do that again here. It’s probably sufficient to remind the reader that these rules imply that the so-called Pauli exclusion principle does not apply to them: bosons like to crowd together, thereby occupying the same quantum state—unlike their counterparts, the so-called fermions or matter-particles: quarks (which make up protons and neutrons) and leptons (including electrons and neutrinos), which can’t do that. Two electrons, for example, can only sit on top of each other (or be very near to each other, I should say) if their spins are opposite (so that makes their quantum state different), and there’s no place whatsoever to add a third one because there are only two possible ‘directions’ for the spin: up or down.

From all that I’ve been writing so far, I am sure you have some kind of picture of matter-particles now, and notably of the electron: it’s not really point-like, because it has a so-called scattering cross-section (I’ll say more about this later), and we can find it somewhere taking into account the Uncertainty Principle, with the probability of finding it at point x at time t given by the absolute square of a so-called ‘wave function’ Ψ(x, t).

But what about the photon? Unlike quarks or electrons, they are really point-like, aren’t they? And can we associate them with a psi function too? I mean, they have a wavelength, obviously, which is given by the Planck-Einstein energy-frequency relation: E = hν, with h the Planck constant and ν the frequency of the associated ‘light’. But an electromagnetic wave is not like a ‘probability wave’. So… Do they have a de Broglie wavelength as well?

Before answering that question, let me present that ‘picture’ of the electron once again.

The wave function for electrons

The electron ‘picture’ can be represented in a number of ways but one of the more scientifically correct ones – whatever that means – is that of a spatially confined wave function representing a complex quantity referred to as the probability amplitude. The animation below (which I took from Wikipedia) visualizes such wave functions. As mentioned above, the wave function is usually represented by the Greek letter psi (Ψ), and it is often referred to as a ‘probability wave’ – by bloggers like me, that is 🙂 – but that term is quite misleading. Why? You surely know that by now: the wave function represents a probability amplitude, not a probability. [So, to be correct, we should say a ‘probability amplitude wave’, or an ‘amplitude wave’, but so these terms are obviously too long and so they’ve been dropped and everybody talks about ‘the’ wave function now, although that’s confusing too, because an electromagnetic wave is a ‘wave function’ too, but describing ‘real’ amplitudes, not some weird complex numbers referred to as ‘probability amplitudes’.]

Having said what I’ve said above, probability amplitude and probability are obviously related: if we take the (absolute) square of the psi function – i.e. if we take the (absolute) square of all these amplitudes Ψ(x, t) – then we get the actual probability of finding that electron at point x at time t. So then we get the so-called probability density functions, which are shown on the right-hand side of the illustration above. [As for the term ‘absolute’ square, the absolute square is the squared norm of the associated ‘vector’. Indeed, you should note that the square of a complex number can be negative as evidenced, for example, by the definition of i: i²= –1. In fact, if there’s only an imaginary part, then its square is always negative. Probabilities are real numbers between 0 and 1, and so they can’t be negative, and so that’s why we always talk about the absolute square, rather than the square as such.]

Below, I’ve inserted another image, which gives a static picture (i.e. one that is not varying in time) of the wave function of a real-life electron. To be precise: it’s the wave function for an electron on the 5d orbital of a hydrogen orbital. You can see it’s much more complicated than those easy things above. However, the idea behind is the same. We have a complex-valued function varying in space and in time. I took it from Wikipedia and so I’ll just copy the explanation here: “The solid body shows the places where the electron’s probability density is above a certain value (0.02), as calculated from the probability amplitude.” What about these colors? Well… The image uses the so-called HSL color system to represent complex numbers: each complex number is represented by a unique color, with a different hue (H), saturation (S) and lightness (L). Just google if you want to know how that works exactly.

OK. That should be clear enough. I wanted to talk about photons here. So let’s go for it. Well… Hmm… I realize I need to talk about some more ‘basics’ first. Sorry for that.

The Uncertainty Principle revisited (1)

The wave function is usually given as a function in space and time: Ψ = Ψ(x, t). However, I should also remind you that we have a similar function in the ‘momentum space’: if ψ is a psi function, then the function in the momentum space is a phi function, and we’ll write it as Φ = Φ(p, t). [As for the notation, x and p are written with capital letters and, hence, represent (three-dimensional) vectors. Likewise, we use a capital letter for psi and phi so we don’t confuse it with, for example, the lower-case φ (phi) representing the phase of a wave function.]

The position-space and momentum-space wave functions Ψ and Φ are related through the Uncertainty Principle. To be precise: they are Fourier transforms of each other. Huh? Don’t be put off by that statement. In fact, I shouldn’t have mentioned it, but then it’s how one can actually prove or derive the Uncertainty Principle from… Well… From ‘first principles’, let’s say, instead of just jotting it down as some God-given rule. Indeed, as Feynman puts: “The Uncertainty Principle should be seen in its historical context. If you get rid of all of the old-fashioned ideas and instead use the ideas that I’m explaining in these lectures—adding arrows for all the ways an event can happen—there is no need for an uncertainty principle!” However, I must assume you’re, just like me, not quite used to the new ideas as yet, and so let me just jot down the Uncertainty Principle once again, as some God-given rule indeed :-):

σ_x·σ_p≥ ħ/2

This is the so-called Kennard formulation of the Principle: it measures the uncertainty about the exact position (x) as well as the momentum (p), in terms of the standard deviation (so that’s the σ (sigma) symbol) around the mean. To be precise, the assumption is that we cannot know the real x and p: we can only find some probability distribution for x and p, which is usually some nice “bell curve” in the textbooks. While the Kennard formulation is the most precise (and exact) formulation of the Uncertainty Principle (or uncertainty relation, I should say), you’ll often find ‘other’ formulations. These ‘other’ formulates usually write Δx and Δp instead of σ_xand σ_p, with the Δ symbol indicating some ‘spread’ or a similar concept—surely do not think of Δ as a differential or so! [Sorry for assuming you don’t know this (I know you do!) but I just want to make sure here!] Also, these ‘other’ formulations will usually (a) not mention the 1/2 factor, (b) substitute ħ for h (ħ = h/2π, as you know, so ħ is preferred when we’re talking things like angular frequency or other stuff involving the unit circle), or (c) put an equality (=) sign in, instead of an inequality sign (≥). Niels Bohr’s early formulation of the Uncertainty Principle actually does all of that:

ΔxΔp = h

So… Well… That’s a bit sloppy, isn’t it? Maybe. In Feynman’s Lectures, you’ll find an oft-quoted ‘application’ of the Uncertainty Principle leading to a pretty accurate calculation of the typical size of an atom (the so-called Bohr radius), which Feynman starts with an equally sloppy statement of the Uncertainty Principle, so he notes: “We needn’t trust our answer to within factors like 2, π etcetera.” Frankly, I used to think that’s ugly and, hence, doubt the ‘seriousness’ of such kind of calculations. Now I know it doesn’t really matter indeed, as the essence of the relationship is clearly not a 2, π or 2π factor. The essence is the uncertainty itself: it’s very tiny (and multiplying it with 2, π or 2π doesn’t make it much bigger) but so it’s there.

In this regard, I need to remind you of how tiny that physical constant ħ actually is: about 6.58×10⁻¹⁶eV·s. So that’s a zero followed by a decimal point and fifteen zeroes: only then we get the first significant digits (65812…). And if 10⁻¹⁶doesn’t look tiny enough for you, then just think about how tiny the electronvolt unit is: it’s the amount of (potential) energy gained (or lost) by an electron as it moves across a potential difference of one volt (which, believe me, is nothing much really): if we’d express ħ in Joule, then we’d have to add nineteen more zeroes, because 1 eV = 1.6×10⁻¹⁹ J. As for such phenomenally small numbers, I’ll just repeat what I’ve said many times before: we just cannot imagine such small number. Indeed, our mind can sort of intuitively deal with addition (and, hence, subtraction), and with multiplication and division (but to some extent only), but our mind is not made to understand non-linear stuff, such as exponentials indeed. If you don’t believe me, think of the Richter scale: can you explain the difference between a 4.0 and a 5.0 earthquake? […] If the answer to that question took you more than a second… Well… I am right. 🙂 [The Richter scale is based on the base-10 exponential function: a 5.0 earthquake has a shaking amplitude that is 10 times that of an earthquake that registered 4.0, and because energy is proportional to the square of the amplitude, that corresponds to an energy release that is 31.6 times that of the lesser earthquake.]

A digression on units

Having said what I said above, I am well aware of the fact that saying that we cannot imagine this or that is what most people say. I am also aware of the fact that they usually say that to avoid having to explain something. So let me try to do something more worthwhile here.

1. First, I should note that ħ is so small because the second, as a unit of time, is so incredibly large. All is relative, of course. 🙂 For sure, we should express time in a more natural unit at the atomic or sub-atomic scale, like the time that’s needed for light to travel one meter. Let’s do it. Let’s express time in a unit that I shall call a ‘meter‘. Of course, it’s not an actual meter (because it doesn’t measure any distance), but so I don’t want to invent a new word and surely not any new symbol here. Hence, I’ll just put apostrophes before and after: so I’ll write ‘meter’ or ‘m’. When adopting the ‘meter’ as a unit of time, we get a value for ‘ħ‘ that is equal to (6.6×10⁻¹⁶eV·s)(1/3×10⁸ ‘meter’/second) = 2.2×10⁻⁸eV·’m’. Now, 2.2×10⁻⁸is a number that is still too tiny to imagine. But then our ‘meter’ is still a rather huge unit at the atomic scale: we should take the ‘millimicron’, aka the ‘nanometer’ (1 nm = 1×10⁻⁹ m), or – even better because more appropriate – the ‘angstrom‘: 1 Å = 0.1 nm = 1×10⁻¹⁰ m. Indeed, the smallest atom (hydrogen) has a radius of 0.25 Å, while larger atoms will have a radius of about 1 or more Å. Now that should work, isn’t it? You’re right, we get a value for ‘ħ‘ equal to (6.6×10⁻¹⁶eV·s)(1/3×10⁸ ‘m’/s)(1×10¹⁰ ‘Å’/m) = 220eV·’Å’, or 22 220eV·’nm’. So… What? Well… If anything, it shows ħ is not a small unit at the atomic or sub-atomic level! Hence, we actually can start imagining how things work at the atomic level when using more adequate units.

[Now, just to test your knowledge, let me ask you: what’s the wavelength of visible light in angstrom? […] Well? […] Let me tell you: 400 to 700 nm is 4000 to 7000 Å. In other words, the wavelength of visible light is quite sizable as compared to the size of atoms or electron orbits!]

2. Secondly, let’s do a quick dimension analysis of that ΔxΔp = h relation and/or its more accurate expression σ_x·σ_p≥ ħ/2.

A position (and its uncertainty or standard deviation) is expressed in distance units, while momentum… Euh… Well… What? […] Momentum is mass times velocity, so it’s kg·m/s. Hence, the dimension of the product on the left-hand side of the inequality is m·kg·m/s = kg·m²/s. So what about this eV·s dimension on the right-hand side? Well… The electronvolt is a unit of energy, and so we can convert it to joules. Now, a joule is a newton-meter (N·m), which is the unit for both energy and work: it’s the work done when applying a force of one newton over a distance of one meter. So we now have N·m·s for ħ, which is nice, because Planck’s constant (h or ħ—whatever: the choice for one of the two depends on the variables we’re looking at) is the quantum for action indeed. It’s a Wirkung as they say in German, so its dimension combines both energy as well as time.

To put it simply, it’s a bit like power, which is what we men are interested in when looking at a car or motorbike engine. 🙂 Power is the energy spent or delivered per second, so its dimension is J/s, not J·s. However, your mind can see the similarity in thinking here. Energy is a nice concept, be it potential (think of a water bucket above your head) or kinetic (think of a punch in a bar fight), but it makes more sense to us when adding the dimension of time (emptying a bucket of water over your head is different than walking in the rain, and the impact of a punch depends on the power with which it is being delivered). In fact, the best way to understand the dimension of Planck’s constant is probably to also write the joule in ‘base units’. Again, one joule is the amount of energy we need to move an object over a distance of one meter against a force of one newton. So one J·s is one N·m·s is (1) a force of one newton acting over a distance of (2) one meter over a time period equal to (3) one second.

I hope that gives you a better idea of what ‘action’ really is in physics. […] In any case, we haven’t answered the question. How do we relate the two sides? Simple: a newton is an oft-used SI unit, but it’s not a SI base unit, and so we should deconstruct it even more (i.e. write it in SI base units). If we do that, we get 1 N = 1 kg·m/s²: one newton is the force needed to give a mass of 1 kg an acceleration of 1 m/s per second. So just substitute and you’ll see the dimension on the right-hand side is kg·(m/s²)·m·s = kg·m²/s, so it comes out alright.

Why this digression on units? Not sure. Perhaps just to remind you also that the Uncertainty Principle can also be expressed in terms of energy and time:

ΔE·Δt = h

Here there’s no confusion in regard to the units on both sides: we don’t need to convert to SI base units to see that they’re the same: [ΔE][Δt] = J·s.

The Uncertainty Principle revisited (2)

The ΔE·Δt = h expression is not so often used as an expression of the Uncertainty Principle. I am not sure why, and I don’t think it’s a good thing. Energy and time are also complementary variables in quantum mechanics, so it’s just like position and momentum indeed. In fact, I like the energy-time expression somewhat more than the position-momentum expression because it does not create any confusion in regard to the units on both sides: it’s just joules (or electronvolts) and seconds on both sides of the equation. So what?

Frankly, I don’t want to digress too much here (this post is going to become awfully long) but, personally, I found it hard, for quite a while, to relate the two expressions of the very same uncertainty ‘principle’ and, hence, let me show you how the two express the same thing really, especially because you may or may not know that there are even more pairs of complementary variables in quantum mechanics. So, I don’t know if the following will help you a lot, but it helped me to note that:

The energy and momentum of a particle are intimately related through the (relativistic) energy-momentum relationship. Now, that formula, E² = p²c² – m₀²c⁴, which links energy, momentum and intrinsic mass (aka rest mass), looks quite monstrous at first. Hence, you may prefer a simpler form: pc = Ev/c. It’s the same really as both are based on the relativistic mass-energy equivalence: E = mc² or, the way I prefer to write it: m = E/c². [Both expressions are the same, obviously, but we can ‘read’ them differently: m = E/c² expresses the idea that energy has a equivalent mass, defined as inertia, and so it makes energy the primordial concept, rather than mass.] Of course, you should note that m is the total mass of the object here, including both (a) its rest mass as well as (b) the equivalent mass it gets from moving at the speed v. So m, not m₀, is the concept of mass used to define p, and note how easy it is to demonstrate the equivalence of both formulas: pc = Ev/c ⇔ mvc = Ev/c ⇔ E = mc². In any case, the bottom line is: don’t think of the energy and momentum of a particle as two separate things; they are two aspects of the same ‘reality’, involving mass (a measure of inertia, as you know) and velocity (as measured in a particular (so-called inertial) reference frame).
Time and space are intimately related through the universal constant c, i.e. the speed of light, as evidenced by the fact that we will often want to express distance not in meter but in light-seconds (i.e. the distance that light travels (in a vacuum) in one second) or, vice versa, express time in meter (i.e. the time that light needs to travel a distance of one meter).

These relationships are interconnected, and the following diagram shows how.

The easiest way to remember it all is to apply the Uncertainty Principle, in both its ΔE·Δt = h as well as its Δp·Δx = h expressions, to a photon. A photon has no rest mass and its velocity v is, obviously, c. So the energy-momentum relationship is a very simple one: p = E/c. We then get both expressions of the Uncertainty Principle by simply substituting E for p, or vice versa, and remember that time and position (or distance) are related in exactly the same way: the constant of proportionality is the very same. It’s c. So we can write: Δx = Δt·c and Δt = Δx/c. If you’re confused, think about it in very practical terms: because the speed of light is what it is, an uncertainty of a second in time amounts, roughly, to an uncertainty in position of some 300,000 km (c = 3×10⁸m/s). Conversely, an uncertainty of some 300,000 km in the position amounts to a uncertainty in time of one second. That’s what the 1-2-3 in the diagram above is all about: please check if you ‘get’ it, because that’s ‘essential’ indeed.

Back to ‘probability waves’

Matter-particles are not the same, but we do have the same relations, including that ‘energy-momentum duality’. The formulas are just somewhat more complicated because they involve mass and velocity (i.e. a velocity less than that of light). For matter-particles, we can see that energy-momentum duality not only in the relationships expressed above (notably the relativistic energy-momentum relation), but also in the (in)famous de Broglie relation, which associates some ‘frequency’ (f) to the energy (E) of a particle or, what amounts to the same, some ‘wavelength’ (λ) to its momentum (p):

λ = h/p and f = E/h

These two complementary equations give a ‘wavelength’ (λ) and/or a ‘frequency’ (f) of a de Broglie wave, or a ‘matter wave’ as it’s sometimes referred to. I am using, once again, apostrophes because the de Broglie wavelength and frequency are a different concept—different than the wavelength or frequency of light, or of any other ‘real’ wave (like water or sound waves, for example). To illustrate the differences, let’s start with a very simple question: what’s the velocity of a de Broglie wave? Well… […] So? You thought you knew, didn’t you?

Let me answer the question:

The mathematically (and physically) correct answer involves distinguishing the group and phase velocity of a wave.
The ‘easy’ answer is: the de Broglie wave of a particle moves with the particle and, hence, its velocity is, obviously, the speed of the particle which, for electrons, is usually non-relativistic (i.e. rather slow as compared to the speed of light).

To be clear on this, the velocity of a de Broglie wave is not the speed of light. So a de Broglie wave is not like an electromagnetic wave at all. They have nothing in common really, except for the fact that we refer to both of them as ‘waves’. 🙂

The second thing to note is that, when we’re talking about the ‘frequency’ or ‘wavelength’ of ‘matter waves’ (i.e. de Broglie waves), we’re talking the frequency and wavelength of a wave with two components: it’s a complex-valued wave function, indeed, and so we get a real and imaginary part when we’re ‘feeding’ the function with some values for x and t.

Thirdly and, perhaps, most importantly, we should always remember the Uncertainty Principle when looking at the de Broglie relation. The Uncertainty Principle implies that we can actually not assign any precise wavelength (or, what amounts to the same, a precise frequency) to a de Broglie wave: if there is a spread in p (and, hence, in E), then there will be a spread in λ (and in f). In fact, I tend to think that it would be better to write the de Broglie relation as an ‘uncertainty relation’ in its own right:

Δλ = Δ(h/p) = hΔp and Δf = ΔE/h = hΔE

Besides from underscoring the fact that we have other ‘pairs’ of complementary variables, this ‘version’ of the de Broglie equation would also remind us continually of the fact that a ‘regular’ wave with an exact frequency and/or an exact wavelength (so a Δλ and/or a Δf equal to zero) would not give us any information about the momentum and/or the energy. Indeed, as Δλ and/or Δf go to zero (Δλ → 0 and/or Δf → 0 ), then Δp and ΔE must go to infinity (Δp → ∞ and ΔE → ∞. That’s just the math involved in such expressions. 🙂

Jokes aside, I’ll admit I used to have a lot of trouble understanding this, so I’ll just quote the expert teacher (Feynman) on this to make sure you don’t get me wrong here:

“The amplitude to find a particle at a place can, in some circumstances, vary in space and time, let us say in one dimension, in this manner: Ψ = Ae^i(ωt−kx), where ω is the frequency, which is related to the classical idea of the energy through E = ħω, and k is the wave number, which is related to the momentum through p = ħk. [These are equivalent formulations of the de Broglie relations using the angular frequency and the wave number instead of wavelength and frequency.] We would say the particle had a definite momentum p if the wave number were exactly k, that is, a perfect wave which goes on with the same amplitude everywhere. The Ψ = Ae^i(ωt−kx)equation [then] gives the [complex-valued probability] amplitude, and if we take the absolute square, we get the relative probability for finding the particle as a function of position and time. This is a constant, which means that the probability to find a [this] particle is the same anywhere.” (Feynman’s Lectures, I-48-5)

You may say or think: What’s the problem here really? Well… If the probability to find a particle is the same anywhere, then the particle can be anywhere and, for all practical purposes, that amounts to saying it’s nowhere really. Hence, that wave function doesn’t serve the purpose. In short, that nice Ψ = Ae^i(ωt−kx)function is completely useless in terms of representing an electron, or any other actual particle moving through space. So what to do?

The Wikipedia article on the Uncertainty Principle has this wonderful animation that shows how we can superimpose several waves, one on top of each other, to form a wave packet. Let me copy it below:

So that’s what the wave we want indeed: a wave packet that travels through space but which is, at the same time, limited in space. Of course, you should note, once again, that it shows only one part of the complex-valued probability amplitude: just visualize the other part (imaginary if the wave above would happen to represent the real part, and vice versa if the wave would happen to represent the imaginary part of the probability amplitude). The animation basically illustrates a mathematical operation. To be precise, it involves a Fourier analysis or decomposition: it separates a wave packet into a finite or (potentially) infinite number of component waves. Indeed, note how, in the illustration above, the frequency of the component waves gradually increases (or, what amounts to the same, how the wavelength gets smaller and smaller) and how, with every wave we ‘add’ to the packet, it becomes increasingly localized. Now, you can easily see that the ‘uncertainty’ or ‘spread’ in the wavelength here (which we’ll denote by Δλ) is, quite simply, the difference between the wavelength of the ‘one-cycle wave’, which is equal to the space the whole wave packet occupies (which we’ll denote by Δx), and the wavelength of the ‘highest-frequency wave’. For all practical purposes, they are about the same, so we can write: Δx ≈ Δλ. Using Bohr’s formulation of the Uncertainty Principle, we can see the expression I used above (Δλ = hΔp) makes sense: Δx = Δλ = h/Δp, so ΔλΔp = h.

[Just to be 100% clear on terminology: a Fourier decomposition is not the same as that Fourier transform I mentioned when talking about the relation between position and momentum in the Kennard formulation of the Uncertainty Principle, although these two mathematical concepts obviously have a few things in common.]

The wave train revisited

All what I’ve said above, is the ‘correct’ interpretation of the Uncertainty Principle and the de Broglie equation. To be frank, it took me quite a while to ‘get’ that—and, as you can see, it also took me quite a while to get ‘here’, of course. 🙂

In fact, I was confused, for quite a few years actually, because I never quite understood whey there had to be a spread in the wavelength of a wave train. Indeed, we can all easily imagine a localized wave train with a fixed frequency and a fixed wavelength, like the one below, which I’ll re-use later. I’ve made this wave train myself: it’s a standard sine and cosine function multiplied with an ‘envelope’ function generating the envelope. As you can see, it’s a complex-valued thing indeed: the blue curve is the real part, and the imaginary part is the red curve.

You can easily make a graph like this yourself. [Just use of one of those online graph tools.] This thing is localized in space and, as mentioned above, it has a fixed frequency and wavelength. So all those enigmatic statements you’ll find in serious or less serious books (i.e. textbooks or popular accounts) on quantum mechanics saying that “we cannot define a unique wavelength for a short wave train” and/or saying that “there is an indefiniteness in the wave number that is related to the finite length of the train, and thus there is an indefiniteness in the momentum” (I am quoting Feynman here, so not one of the lesser gods) are – with all due respect for these authors, especially Feynman – just wrong. I’ve made another ‘short wave train’ below, but this time it depicts the real part of a (possible) wave function only.

Hmm… Now that one has a weird shape, you’ll say. It doesn’t look like a ‘matter wave’! Well… You’re right. Perhaps. [I’ll challenge you in a moment.] The shape of the function above is consistent, though, with the view of a photon as a transient electromagnetic oscillation. Let me come straight to the point by stating the basics: the view of a photon in physics is that photons are emitted by atomic oscillators. As an electron jumps from one energy level to the other, it seems to oscillate back and forth until it’s in equilibrium again, thereby emitting an electromagnetic wave train that looks like a transient.

Huh? What’s a transient? It’s an oscillation like the one above: its amplitude and, hence, its energy, gets smaller and smaller as time goes by. To be precise, its energy level has the same shape as the envelope curve below: E = E₀e^–t/τ. In this expression, we have τ as the so-called decay time, and one can show it’s the inverse of the so-called decay rate: τ = 1/γ with γE = –dE/dt. In case you wonder, check it out on Wikipedia: it’s one of the many applications of the natural exponential function: we’re talking a so-called exponential decay here indeed, involves a quantity (in this case, the amplitude and/or the energy) decreasing at a rate that is proportional to its current value, with the coefficient of proportionality being γ. So we write that as γE = –dE/dt in mathematical notation. 🙂

I need to move on. All of what I wrote above was ‘plain physics’, but so what I really want to explore in this post is a crazy hypothesis. Could these wave trains above – I mean the wave trains with the fixed frequency and wavelength – possible represent a de Broglie wave for a photon?

You’ll say: of course not! But, let’s be honest, you’d have some trouble explaining why. The best answer you could probably come up with is: because no physics textbook says something like that. You’re right. It’s a crazy hypothesis because, when you ask a physicist (believe it or not, but I actually went through the trouble of asking two nuclear scientists), they’ll tell you that photons are not to be associated with de Broglie waves. [You’ll say: why didn’t you try looking for an answer on the Internet? I actually did but – unlike what I am used to – I got very confusing answers on this one, so I gave up trying to find some definite answer on this question on the Internet.]

However, these negative answers don’t discourage me from trying to do some more freewheeling. Before discussing whether or not the idea of a de Broglie wave for a photon makes sense, let’s think about mathematical constraints. I googled a bit but I only see one actually: the amplitudes of a de Broglie wave are subject to a normalization condition. Indeed, when everything is said and done, all probabilities must take a value between 0 and 1, and they must also all add up to exactly 1. So that’s a so-called normalization condition that obviously imposes some constraints on the (complex-valued) probability amplitudes of our wave function.

But let’s get back to the photon. Let me remind you of what happens when a photon is being emitted by inserting the two diagrams below, which gives the energy levels of the atomic orbitals of electrons.

So an electron absorbs or emits a photon when it goes from one energy level to the other, so it absorbs or emits radiation. And, of course, you will also remember that the frequency of the absorbed or emitted light is related to those energy levels. More specifically, the frequency of the light emitted in a transition from, let’s say, energy level E₃to E₁will be written as ν₃₁= (E₃– E₁)/h. This frequency will be one of the so-called characteristic frequencies of the atom and will define a specific so-called spectral emission line.

Now, from a mathematical point of view, there’s no difference between that ν₃₁= (E₃– E₁)/h equation and the de Broglie equation, f = E/h, which assigns a de Broglie wave to a particle. But, of course, from all that I wrote above, it’s obvious that, while these two formulas are the same from a math point of view, they represent very different things. Again, let me repeat what I said above: a de Broglie wave is a matter-wave and, as such, it has nothing to do with an electromagnetic wave.

Let me be even more explicit. A de Broglie wave is not a ‘real’ wave, in a sense (but, of course, that’s a very unscientific statement to make); it’s a psi function, so it represents these weird mathematical quantities–complex probability amplitudes–which allow us to calculate the probability of finding the particle at position x or, if it’s a wave function for the momentum-space, to find a value p for its momentum. In contrast, a photon that’s emitted or absorbed represents a ‘real’ disturbance of the electromagnetic field propagating through space. Hence, that frequency ν is something very different than f, which is why we use another symbol for it (ν is the Greek letter nu, not to be confused with the v symbol we use for velocity). [Of course, you may wonder how ‘real’ or ‘unreal’ an electromagnetic field is but, in the context of this discussion, let me assure you we should look at it as something that’s very real.]

That being said, we also know light is emitted in discrete energy packets: in fact, that’s how photons were defined originally, first by Planck and then by Einstein. Now, when an electron falls from one energy level in an atom to another (lower) energy level, it emits one – and only one – photon with that particular wavelength and energy. The question then is: how should we picture that photon? Does it also have some more or less defined position in space, and some momentum? The answer is definitely yes, on both accounts:

Subject to the constraints of the Uncertainty Principle, we know, more or less indeed, when a photon leaves a source and when it hits some detector. [And, yes, due to the ‘Uncertainty Principle’ or, as Feynman puts it, the rules for adding arrows, it may not travel in a straight line and/or at the speed of light—but that’s a discussion that, believe it or not, is not directly relevant here. If you want to know more about it, check one or more of my posts on it.]
We also know light has a very definite momentum, which I’ve calculated elsewhere and so I’ll just note the result: p = E/c. It’s a ‘pushing momentum’ referred to as radiation pressure, and its in the direction of travel indeed.

In short, it does makes sense, in my humble opinion that is, to associate some wave function with the photon, and then I mean a de Broglie wave. Just think about it yourself. You’re right to say that a de Broglie wave is a ‘matter wave’, and photons aren’t matter but, having said that, photons do behave like like electrons, don’t they? There’s diffraction (when you send a photon through one slit) and interference (when photons go through two slits, altogether or – amazingly – one by one), so it’s the same weirdness as electrons indeed, and so why wouldn’t we associate some kind of wave function with them?

You can react in one of three ways here. The first reaction is: “Well… I don’t know. You tell me.” Well… That’s what I am trying to do here. 🙂

The second reaction may be somewhat more to the point. For example, those who’ve read Feynman’s Strange Theory of Light and Matter, could say: “Of course, why not? That’s what we do when we associate a photon going from point A to B with an amplitude P(A to B), isn’t it?”

Well… No. I am talking about something else here. Not some amplitude associated with a path in spacetime, but a wave function giving an approximate position of the photon.

The third reaction may be the same as the reaction of those two nuclear scientists I asked: “No. It doesn’t make sense. We do not associate photons with a de Broglie wave.” But so they didn’t tell me why because… Well… They didn’t have the time to entertain a guy like me and so I didn’t dare to push the question and continued to explore it more in detail myself.

So I’ve done that, and I thought of one reason why the question, perhaps, may not make all that much sense: a photon travels at the speed of light; therefore, it has no length. Hence, doing what I am doing below, and that’s to associate the electromagnetic transient with a de Broglie wave might not make sense.

Maybe. I’ll let you judge. Before developing the point, I’ll raise two objections to the ‘objection’ raised above (i.e. the statement that a photon has no length). First, if we’re looking at the photon as some particle, it will obviously have no length. However, an electromagnetic transient is just what it is: an electromagnetic transient. I’ve see nothing that makes me think its length should be zero. In fact, if that would be the case, the concept of an electromagnetic wave itself would not make sense, as its ‘length’ would always be zero. Second, even if – somehow – the length of the electromagnetic transient would be reduced to zero because of its speed, we can still imagine that we’re looking at the emission of an electromagnetic pulse (i.e. a photon) using the reference frame of the photon, so that we’re traveling at speed c,’ riding’ with the photon, so to say, as it’s being emitted. Then we would ‘see’ the electromagnetic transient as it’s being radiated into space, wouldn’t we?

Perhaps. I actually don’t know. That’s why I wrote this post and hope someone will react to it. I really don’t know, so I thought it would be nice to just freewheel a bit on this question. So be warned: nothing of what I write below has been researched really, so critical comments and corrections from actual specialists are more than welcome.

The shape of a photon wave

As mentioned above, the answer in regard to the definition of a photon’s position and momentum is, obviously, unambiguous. Perhaps we have to stretch whatever we understand of Einstein’s (special) relativity theory, but we should be able to draw some conclusions, I feel.

Let me say one thing more about the momentum here. As said, I’ll refer you to one of my posts for the detail but, all you should know here is that the momentum of light is related to the magnetic field vector, which we usually never mention when discussing light because it’s so tiny as compared to the electric field vector in our inertial frame of reference. Indeed, the magnitude of the magnetic field vector is equal to the magnitude of the electric field vector divided by c = 3×10⁸, so we write B = E/c. Now, the E here stands for the electric field, so let me use W to refer to the energy instead of E. Using the B = E/c equation and a fairly straightforward calculation of the work that can be done by the associated force on a charge that’s being put into this field, we get that famous equation which we mentioned above already: the momentum of a photon is its total energy divided by c, so we write p = W/c. You’ll say: so what? Well… Nothing. I just wanted to note we get the same p = W/c equation indeed, but from a very different angle of analysis here. We didn’t use the energy-momentum relation here at all! In any case, the point to note is that the momentum of a photon is only a tiny fraction of its energy (p = W/c), and that the associated magnetic field vector is also just a tiny fraction of the electric field vector (B = E/c).

But so it’s there and, in fact, when adopting a moving reference frame, the mix of E and B (i.e. the electric and magnetic field) becomes an entirely different one. One of the ‘gems’ in Feynman’s Lectures is the exposé on the relativity of electric and magnetic fields indeed, in which he analyzes the electric and magnetic field caused by a current, and in which he shows that, if we switch our inertial reference frame for that of the moving electrons in the wire, the ‘magnetic’ field disappears, and the whole electromagnetic effect becomes ‘electric’ indeed.

I am just noting this because I know I should do a similar analysis for the E and B ‘mixture’ involved in the electromagnetic transient that’s being emitted by our atomic oscillator. However, I’ll admit I am not quite comfortably enough with the physics nor the math involved to do that, so… Well… Please do bear this in mind as I will be jotting down some quite speculative thoughts in what follows.

So… A photon is, in essence, a electromagnetic disturbance and so, when trying to picture a photon, we can think of some oscillating electric field vector traveling through–and also limited in–space. [Note that I am leaving the magnetic field vector out of the analysis from the start, which is not ‘nice’ but, in light of that B = E/c relationship, I’ll assume it’s acceptable.] In short, in the classical world – and in the classical world only of course – a photon must be some electromagnetic wave train, like the one below–perhaps.

But why would it have that shape? I only suggested it because it has the same shape as Feynman’s representation of a particle (see below) as a ‘probability wave’ traveling through–and limited in–space.

So, what about it? Let me first remind you once again (I just can’t stress this point enough it seems) that Feynman’s representation – and most are based on his, it seems – is misleading because it suggests that ψ(x) is some real number. It’s not. In the image above, the vertical axis should not represent some real number (and it surely should not represent a probability, i.e. some real positive number between 0 and 1) but a probability amplitude, i.e. a complex number in which both the real and imaginary part are important. Just to be fully complete (in case you forgot), such complex-valued wave function ψ(x) will give you all the probabilities you need when you take its (absolute) square, but so… Well… We’re really talking a different animal here, and the image above gives you only one part of the complex-valued wave function (either the real or the imaginary part), while it should give you both. That’s why I find my graph below much better. 🙂 It’s the same really, but so it shows both the real as well as the complex part of a wave function.

But let me go back to the first illustration: the vertical axis of the first illustration is not ψ but E – the electric field vector. So there’s no imaginary part here: just a real number, representing the strength–or magnitude I should say– of the electric field E as a function of the space coordinate x. [Can magnitudes be negative? The honest answer is: no, they can’t. But just think of it as representing the field vector pointing in the other way .]

Regardless of the shortcomings of this graph, including the fact we only have some real-valued oscillation here, would it work as a ‘suggestion’ of how a real-life photon could look like?

Of course, you could try to not answer that question by mumbling something like: “Well… It surely doesn’t represent anything coming near to a photon in quantum mechanics.” But… Well… That’s not my question here: I am asking you to be creative and ‘think outside of the box’, so to say. 🙂

So you should say ‘No!’ because of some other reason. What reason? Well… If a photon is an electromagnetic transient – in other words, if we adopt a purely classical point of view – it’s going to be a transient wave indeed, and so then it should walk, talk and even look like a transient. 🙂 Let me quickly jot down the formula for the (vertical) component of E as a function of the acceleration of some charge q:

The charge q (i.e. the source of the radiation) is, of course, our electron that’s emitting the photon as it jumps from a higher to a lower energy level (or, vice versa, absorbing it). This formula basically states that the magnitude of the electric field (E) is proportional to the acceleration (a) of the charge (with t–r/c the retarded argument). Hence, the suggested shape of E as a function of x as shown above would imply that the acceleration of the electron is (a) initially quite small, (b) then becomes larger and larger to reach some maximum, and then (c) becomes smaller and smaller again to then die down completely. In short, it does match the definition of a transient wave sensu stricto (Wikipedia defines a transient as “a short-lived burst of energy in a system caused by a sudden change of state”) but it’s not likely to represent any real transient. So, we can’t exclude it, but a real transient is much more likely to look like something what’s depicted below: no gradual increase in amplitude but big swings initially which then dampen to zero. In other words, if our photon is a transient electromagnetic disturbance caused by a ‘sudden burst of energy’ (which is what that electron jump is, I would think), then its representation will, much more likely, resemble a damped wave, like the one below, rather than Feynman’s picture of a moving matter-particle.

In fact, we’d have to flip the image, both vertically and horizontally, because the acceleration of the source and the field are related as shown below. The vertical flip is because of the minus sign in the formula for E(t). The horizontal flip is because of the minus sign in the (t – r/c) term, the retarded argument: if we add a little time (Δt), we get the same value for a(t−r/c) as we would have if we had subtracted a little distance: Δr=−cΔt. So that’s why E as a function of r (or of x), i.e. as a function in space, is a ‘reversed’ plot of the acceleration as a function of time.

So we’d have something like below.

What does this resemble? It’s not a vibrating string (although I do start to understand the attractiveness of string theory now: vibrating strings are great as energy storage systems, so the idea of a photon being some kind of vibrating string sounds great, doesn’t it?). It’s not resembling a bullwhip effect either, because the oscillation of a whip is confined by a different envelope (see below). And, no, it’s also definitely not a trumpet. 🙂

It’s just what it is: an electromagnetic transient traveling through space. Would this be realistic as a ‘picture’ of a photon? Frankly, I don’t know. I’ve looked at a lot of stuff but didn’t find anything on this really. The easy answer, of course, is quite straightforward: we’re not interested in the shape of a photon because we know it is not an electromagnetic wave. It’s a ‘wavicle’, just like an electron.

[…] Sure. I know that too. Feynman told me. 🙂 But then why wouldn’t we associate some wave function with it? Please tell me, because I really can’t find much of an answer to that question in the literature, and so that’s why I am freewheeling here. So just go along with me for a while, and come up with another suggestion. As I said above, your bet is as good as mine. All that I know is that there’s one thing we need to explain when considering the various possibilities: a photon has a very well-defined frequency (which defines its color in the visible light spectrum) and so our wave train should – in my humble opinion – also have that frequency. At least for ‘quite a while’—and then I mean ‘most of the time’, or ‘on average’ at least. Otherwise the concept of a frequency – or a wavelength – wouldn’t make much sense. Indeed, if the photon has no defined wavelength or frequency, then we could not perceive it as some color (as you may or may not know, the sense of ‘color’ is produced by our eye and brain, but so it’s definitely associated with the frequency of the light). A photon should have a color (in phyics, that means a frequency) because, when everything is said and done, that’s what the Planck relation is all about.

What would be your alternative? I mean… Doesn’t it make sense to think that, when jumping from one energy level to the other, the electron would initially sort of overshoot its new equilibrium position, to then overshoot it again on the other side, and so on and so on, but with an amplitude that becomes smaller and smaller as the oscillation dies out? In short, if we look at radiation as being caused by atomic oscillators, why would we not go all the way and think of them as oscillators subject to some damping force? Just think about it. 🙂

The size of a photon wave

Let’s forget about the shape for a while and think about size. We’ve got an electromagnetic train here. So how long would it be? Well… Feynman calculated the Q of these atomic oscillators: it’s of the order of 10⁸(see his Lectures, I-33-3: it’s a wonderfully simple exercise, and one that really shows his greatness as a physics teacher) and, hence, this wave train will last about 10^–8seconds (that’s the time it takes for the radiation to die out by a factor 1/e). To give a somewhat more precise example, for sodium light, which has a frequency of 500 THz (500×10¹²oscillations per second) and a wavelength of 600 nm (600×10^–9meter), the radiation will lasts about 3.2×10^–8seconds. [In fact, that’s the time it takes for the radiation’s energy to die out by a factor 1/e, so(i.e. the so-called decay time τ), so the wavetrain will actually last longer, but so the amplitude becomes quite small after that time.]

So that’s a very short time, but still, taking into account the rather spectacular frequency (500 THz) of sodium light, that still makes for some 16 million oscillations and, taking into the account the rather spectacular speed of light (3×10⁸m/s), that makes for a wave train with a length of, roughly, 9.6 meter. Huh? 9.6 meter!?

You’re right. That’s an incredible distance: it’s like infinity on an atomic scale!

So… Well… What to say? Such length surely cannot match the picture of a photon as a fundamental particle which cannot be broken up, can it? So it surely cannot be right because, if this would be the case, then there surely must be some way to break this thing up and, hence, it cannot be ‘elementary’, can it?

Well… Maybe. But think it through. First note that we will not see the photon as a 10-meter long string because it travels at the speed of light indeed and so the length contraction effect ensure its length, as measured in our reference frame (and from whatever ‘real-life’ reference frame actually, because the speed of light will always be c, regardless of the speeds we mortals could ever reach (including speeds close to c), is zero.

So, yes, I surely must be joking here but, as far as jokes go, I can’t help thinking this one is fairly robust from a scientific point of view. Again, please do double-check and correct me, but all what I’ve written so far is not all that speculative. It corresponds to all what I’ve read about it: only one photon is produced per electron in any de-excitation, and its energy is determined by the number of energy levels it drops, as illustrated (for a simple hydrogen atom) below. For those who continue to be skeptical about my sanity here, I’ll quote Feynman once again:

“What happens in a light source is that first one atom radiates, then another atom radiates, and so forth, and we have just seen that atoms radiate a train of waves only for about 10^–8 sec; after 10^–8 sec, some atom has probably taken over, then another atom takes over, and so on. So the phases can really only stay the same for about 10^–8 sec. Therefore, if we average for very much more than 10^–8 sec, we do not see an interference from two different sources, because they cannot hold their phases steady for longer than 10^–8 sec. With photocells, very high-speed detection is possible, and one can show that there is an interference which varies with time, up and down, in about 10^–8 sec.” (Feynman’s Lectures, I-34-4)

So… Well… Now it’s up to you. I am going along here with the assumption that a photon in the visible light spectrum, from a classical world perspective, should indeed be something that’s several meters long and packs a few million oscillations. So, while we usually measure stuff in seconds, or hours, or years, and, hence, while we would that think 10^–8seconds is short, a photon would actually be a very stretched-out transient that occupies quite a lot of space. I should also add that, in light of that number of ten meter, the dampening seems to happen rather slowly!

[…]

I can see you shaking your head now, for various reasons.

First because this type of analysis is not appropriate. […] You think so? Well… I don’t know. Perhaps you’re right. Perhaps we shouldn’t try to think of a photon as being something different than a discrete packet of energy. But then we also know it is an electromagnetic wave. So why wouldn’t we go all the way?

Second, I guess you may find the math involved in this post not to your liking, even if it’s quite simple and I am not doing anything spectacular here. […] Well… Frankly, I don’t care. Let me bulldozer on. 🙂

What about the ‘vertical’ dimension, the y and the z coordinates in space? We’ve got this long snaky thing: how thick-bodied is it?

Here, we need to watch our language. While it’s fairly obvious to associate a wave with a cross-section that’s normal to its direction of propagation, it is not obvious to associate a photon with the same thing. Not at all actually: as that electric field vector E oscillates up and down (or goes round and round, as shown in the illustration below, which is an image of a circularly polarized wave), it does not actually take any space. Indeed, the electric and magnetic field vectors E and B have a direction and a magnitude in space but they’re not representing something that is actually taking up some small or larger core in space.

Hence, the vertical axis of that graph showing the wave train does not indicate some spatial position: it’s not a y-coordinate but the magnitude of an electric field vector. [Just to underline the fact that the magnitude E has nothing to do with spatial coordinates: note that its value depends on the unit we use to measure field strength (so that’s newton/coulomb, if you want to know), so it’s really got nothing to do with an actual position in space-time.]

So, what can we say about it? Nothing much, perhaps. But let me try.

Cross-sections in nuclear physics

In nuclear physics, the term ‘cross-section’ would usually refer to the so-called Thompson scattering cross-section of an electron (or any charged particle really), which can be defined rather loosely as the target area for the incident wave (i.e. the photons): it is, in fact, a surface which can be calculated from what is referred to as the classical electron radius, which is about 2.82×10^–15 m. Just to compare: you may or may not remember the so-called Bohr radius of an atom, which is about 5.29×10^–11 m, so that’s a length that’s about 20,000 times longer. To be fully complete, let me give you the exact value for the Thompson scattering cross-section of an electron: 6.62×10^–29 m²(note that this is a surface indeed, so we have m squared as a unit, not m).

Now, let me remind you – once again – that we should not associate the oscillation of the electric field vector with something actually happening in space: an electromagnetic field does not move in a medium and, hence, it’s not like a water or sound wave, which makes molecules go up and down as it propagates through its medium. To put it simply: there’s nothing that’s wriggling in space as that photon is flashing through space. However, when it does hit an electron, that electron will effectively ‘move’ (or vibrate or wriggle or whatever you can imagine) as a result of the incident electromagnetic field.

That’s what’s depicted and labeled below: there is a so-called ‘radial component’ of the electric field, and I would say: that’s our photon! [What else would it be?] The illustration below shows that this ‘radial’ component is just E for the incident beam and that, for the scattered beam, it is, in fact, determined by the electron motion caused by the incident beam through that relation described above, in which a is the normal component (i.e. normal to the direction of propagation of the outgoing beam) of the electron’s acceleration.

Now, before I proceed, let me remind you once again that the above illustration is, once again, one of those illustrations that only wants to convey an idea, and so we should not attach too much importance to it: the world at the smallest scale is best not represented by a billiard ball model. In addition, I should also note that the illustration above was taken from the Wikipedia article on elastic scattering (i.e. Thomson scattering), which is only a special case of the more general Compton scattering that actually takes place. It is, in fact, the low-energy limit. Photons with higher energy will usually be absorbed, and then there will be a re-emission, but, in the process, there will be a loss of energy in this ‘collision’ and, hence, the scattered light will have lower energy (and, hence, lower frequency and longer wavelength). But – Hey! – now that I think of it: that’s quite compatible with my idea of damping, isn’t it? 🙂 [If you think I’ve gone crazy, I am really joking here: when it’s Compton scattering, there’s no ‘lost’ energy: the electron will recoil and, hence, its momentum will increase. That’s what’s shown below (credit goes to the HyperPhysics site).]

So… Well… Perhaps we should just assume that a photon is a long wave train indeed (as mentioned above, ten meter is very long indeed: not an atomic scale at all!) but that its effective ‘radius’ should be of the same order as the classical electron radius. So what’s that order? If it’s more or less the same radius, then it would be in the order of femtometers (1 fm = 1 fermi = 1×10^–15 m). That’s good because that’s a typical length-scale in nuclear physics. For example, it would be comparable with the radius of a proton. So we look at a photon here as something very different – because it’s so incredibly long (at least as measured from its own reference frame) – but as something which does have some kind of ‘radius’ that is normal to its direction of propagation and equal or smaller than the classical electron radius. [Now that I think of it, we should probably think of it as being substantially smaller. Why? Well… An electron is obviously fairly massive as compared to a photon (if only because an electron has a rest mass and a photon hasn’t) and so… Well… When everything is said and done, it’s the electron that absorbs a photon–not the other way around!]

Now, that radius determines the area in which it may produce some effect, like hitting an electron, for example, or like being detected in a photon detector, which is just what this so-called radius of an atom or an electron is all about: the area which is susceptible of being hit by some particle (including a photon), or which is likely to emit some particle (including a photon). What is exactly, we don’t know: it’s still as spooky as an electron and, therefore, it also does not make all that much sense to talk about its exact position in space. However, if we’d talk about its position, then we should obviously also invoke the Uncertainty Principle, which will give us some upper and lower bounds for its actual position, just like it does for any other particle: the uncertainty about its position will be related to the uncertainty about its momentum, and more knowledge about the former, will implies less knowledge about the latter, and vice versa. Therefore, we can also associate some complex wave function with this photon which is – for all practical purposes – a de Broglie wave. Now how should we visualize that wave?

Well… I don’t know. I am actually not going to offer anything specific here. First, it’s all speculation. Second, I think I’ve written too much rubbish already. However, if you’re still reading, and you like this kind of unorthodox application of electromagnetics, then the following remarks may stimulate your imagination.

The first thing to note is that we should not end up with a wave function that, when squared, gives us a constant probability for each and every point in space. No. The wave function needs to be confined in space and, hence, we’re also talking a wave train here, and a very short one in this case. So… Well… What about linking its amplitude to the amplitude of the field for the photon. In other words, the probability amplitude could, perhaps, be proportional to the amplitude of E, with the proportionality factor being determined by (a) the unit in which we measure E (i.e. newton/coulomb) and (b) the normalization condition.

OK. I hear you say it now: “Ha-ha! Got you! Now you’re really talking nonsense! How can a complex number (the probability amplitude) be proportional to some real number (the field strength)?”

Well… Be creative. It’s not that difficult to imagine some linkages. First, the electric field vector has both a magnitude and a direction. Hence, there’s more to E than just its magnitude. Second, you should note that the real and imaginary part of a complex-valued wave function is a simple sine and cosine function, and so these two functions are the same really, except for a phase difference of π/2. In other words, if we have a formula for the real part of a wave function, we have a formula for its imaginary part as well. So… Your remark is to the point and then it isn’t.

OK, you’ll say, but then so how exactly would you link the E vector with the ψ(x, t) function for a photon. Well… Frankly, I am a bit exhausted now and so I’ll leave any further speculation to you. The whole idea of a de Broglie wave of a photon, with the (complex-valued) amplitude having some kind of ‘proportional’ relationship to the (magnitude of) the electric field vector makes sense to me, although we’d have to be innovative about what that ‘proportionality’ exactly is.

Let me conclude this speculative business by noting a few more things about our ‘transient’ electromagnetic wave:

1. First, it’s obvious that the usual relations between (a) energy (W), (b) frequency (f) and (c) amplitude (A) hold. If we increase the frequency of a wave, we’ll have a proportional increase in energy (twice the frequency is twice the energy), with the factor of proportionality being given by the Planck-Einstein relation: W = hf. But if we’re talking amplitudes (for which we do not have a formula, which is why we’re engaging in those assumptions on the shape of the transient wave), we should not forget that the energy of a wave is proportional to the square of its amplitude: W ∼ A². Hence, a linear increase of the amplitudes results in an exponential (quadratic) increase in energy (e.g. if you double all amplitudes, you’ll pack four times more energy in that wave).

2. Both factors come into play when an electron emits a photon. Indeed, if the difference between the two energy levels is larger, then the photon will not only have a higher frequency (i.e. we’re talking light (or electromagnetic radiation) in the upper ranges of the spectrum then) but one should also expect that the initial overshooting – and, hence, the initial oscillation – will also be larger. In short, we’ll have larger amplitudes. Hence, higher-energy photons will pack even more energy upfront. They will also have higher frequency, because of the Planck relation. So, yes, both factors would come into play.

What about the length of these wave trains? Would it make them shorter? Yes. I’ll refer you to Feynman’s Lectures to verify that the wavelength appears in the numerator of the formula for Q. Hence, higher frequency means shorter wavelength and, hence, lower Q. Now, I am not quite sure (I am not sure about anything I am writing here it seems) but this may or may not be the reason for yet another statement I never quite understood: photons with higher and higher energy are said to become smaller and smaller, and when they reach the Planck scale, they are said to become black holes.

Hmm… I should check on that. 🙂

Conclusion

So what’s the conclusion? Well… I’ll leave it to you to think about this. As said, I am a bit tired now and so I’ll just wrap this up, as this post has become way too long anyway. Let me, before parting, offer the following bold suggestion in terms of finding a de Broglie wave for our photon: perhaps that transient above actually is the wave function.

You’ll say: What !? What about normalization? All probabilities have to add up to one and, surely, those magnitudes of the electric field vector wouldn’t add up to one, would they?

My answer to that is simple: that’s just a question of units, i.e. of normalization indeed. So just measure the field strength in some other unit and it will come all right.

[…] But… Yes? What? Well… Those magnitudes are real numbers, not complex numbers.

I am not sure how to answer that one but there’s two things I could say:

Real numbers are complex numbers too: it’s just that their imaginary part is zero.
When working with waves, and especially with transients, we’ve always represented them using the complex exponential function. For example, we would write a wave function whose amplitude varies sinusoidally in space and time as Ae^{i(ωt−k·r)}, with ω the (angular) frequency and k the wave number (so that’s the wavelength expressed in radians per unit distance).

So, frankly, think about it: where is the photon? It’s that ten-meter long transient, isn’t it? And the probability to find it somewhere is the (absolute) square of some complex number, right? And then we have a wave function already, representing an electromagnetic wave, for which we know that the energy which it packs is the square of its amplitude, as well as being proportional to its frequency. We also know we’re more likely to detect something with high energy than something with low energy, don’t we? So… Tell me why the transient itself would not make for a good psi function?

But then what about these probability amplitudes being a function of the y and z coordinates?

Well… Frankly, I’ve started to wonder if a photon actually has a radius. If it doesn’t have a mass, it’s probably the only real point-like particle (i.e. a particle not occupying any space) – as opposed to all other matter-particles, which do have mass.

Why?

I don’t know. Your guess is as good as mine. Maybe our concepts of amplitude and frequency of a photon are not very relevant. Perhaps it’s only energy that counts. We know that a photon has a more or less well-defined energy level (within the limits of the Uncertainty Principle) and, hence, our ideas about how that energy actually gets distributed over the frequency, the amplitude and the length of that ‘transient’ have no relation with reality. Perhaps we like to think of a photon as a transient electromagnetic wave, because we’re used to thinking in terms of waves and fields, but perhaps a photon is just a point-like thing indeed, with a wave function that’s got the same shape as that transient. 🙂

Post scriptum: Perhaps I should apologize to you, my dear reader. It’s obvious that, in quantum mechanics, we don’t think of a photon as having some frequency and some wavelength and some dimension in space: it’s just an elementary particle with energy interacting with other elementary particles with energy, and we use these coupling constants and what have you to work with them. So we don’t usually think of photons as ten-meter long transients moving through space. So, when I write that “our concepts of amplitude and frequency of a photon are maybe not very relevant” when trying to picture a photon, and that “perhaps, it’s only energy that counts”, I actually don’t mean “maybe” or “perhaps“. I mean: Of course! […] In the quantum-mechanical world view, that is.

So I apologize for, perhaps, posting what may or may not amount to plain nonsense. However, as all of this nonsense helps me to make sense of these things myself, I’ll just continue. 🙂 I seem to move very slowly on this Road to Reality, but the good thing about moving slowly, is that it will − hopefully − give me the kind of ‘deeper’ understanding I want, i.e. an understanding beyond the formulas and mathematical and physical models. In the end, that’s all that I am striving for when pursuing this ‘hobby’ of mine. Nothing more, nothing less. 🙂 Onwards!

Light: relating waves to photons

Pre-scriptum (dated 26 June 2020): Some of the relevant illustrations in this post were removed as a result of an attack by the dark force. In any case, my ideas on the nature of light and photons have evolved considerably, so you should probably read my papers instead of these old blog posts.

Original post:

This is a concluding note on my ‘series’ on light. The ‘series’ gave you an overview of the ‘classical’ theory: light as an electromagnetic wave. It was very complete, including relativistic effects (see my previous post). I could have added more – there’s an equivalent for four-vectors, for example, when we’re dealing with frequencies and wave numbers: quantities that transform like space and time under the Lorentz transformations – but you got the essence.

One point we never ever touched upon, was that magnetic field vector though. It is there. It is tiny because of that 1/c factor, but it’s there. We wrote it as

All symbols in bold are vectors, of course. The force is another vector vector cross-product: F = qv×B, and you need to apply the usual right-hand screw rule to find the direction of the force. As it turns out, that force – as tiny as it is – is actually oriented in the direction of propagation, and it is what is responsible for the so-called radiation pressure.

So, yes, there is a ‘pushing momentum’. How strong is it? What power can it deliver? Can it indeed make space ships sail? Well… The magnitude of the unit vector e_r’is obviously one, so it’s the values of the other vectors that we need to consider. If we substitute and average F, the thing we need to find is:

〈F〉 = q〈vE〉/c

But the charge q times the field is the electric force, and the force on the charge times the velocity is the work dW/dt being done on the charge. So that should equal the energy absorbed that is being absorbed from the light per second. Now, I didn’t look at that much. It’s actually one of the very few things I left – but I’ll refer you to Feynman’s Lectures if you want to find out more: there’s a fine section on light scattering, introducing the notion of the Thompson scattering cross section, but – as said – I think you had enough as for now. Just note that 〈F〉 = [dW/dt]/c and, hence, that the momentum that light delivers is equal to the energy that is absorbed (dW/dt) divided by c.

So the momentum carried is 1/c times the energy. Now, you may remember that Planck solved the ‘problem’ of black-body radiation – an anomaly that physicists couldn’t explain at the end of the 19th century – by re-introducing a corpuscular theory of light: he said light consisted of photons. We all know that photons are the kind of ‘particles’ that the Greek and medieval corpuscular theories of light envisaged but, well… They have a particle-like character – just as much as they have a wave-like character. They are actually neither, and they are physically and mathematically being described by these wave functions – which, in turn, are functions describing probability amplitudes. But I won’t entertain you with that here, because I’ve written about that in other posts. Let’s just go along with the ‘corpuscular’ theory of photons for a while.

Photons also have energy (which we’ll write as W instead of E, just to be consistent with the symbols above) and momentum (p), and Planck’s Law says how much:

W = hf and p = W/c

So that’s good: we find the same multiplier 1/c here for the momentum of a photon. In fact, this is more than just a coincidence of course: the “wave theory” of light and Planck’s “corpuscular theory” must of course link up, because they are both supposed to help us understand real-life phenomena.

There’s even more nice surprises. We spoke about polarized light, and we showed how the end of the electric field vector describes a circular or elliptical motion as the wave travels to space. It turns out that we can actually relate that to some kind of angular momentum of the wave (I won’t go into the details though – because I really think the previous posts have really been too heavy on equations and complicated mathematical arguments) and that we could also relate it to a model of photons carrying angular momentum, “like spinning rifle bullets” – as Feynman puts it.

However, he also adds: “But this ‘bullet’ picture is as incomplete as the ‘wave’ picture.” And so that’s true and that should be it. And it will be it. I will really end this ‘series’ now. It was quite a journey for me, as I am making my way through all of these complicated models and explanations of how things are supposed to work. But a fascinating one. And it sure gives me a much better feel for the ‘concepts’ that are hastily explained in all of these ‘popular’ books dealing with science and physics, hopefully preparing me better for what I should be doing, and that’s to read Penrose’s advanced mathematical theories.