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

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:

  1. To present Einstein’s 1916 model for blackbody radiation.
  2. 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.

Einstein

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

  • Amn is 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.
  • Bmn is 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:

Feynman on Einstein

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

formula blackbody

To get the eħω/kT –1, Bmn must be equal to Bnm, 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 Amn, taking into account that Bmn = Bnm, we find that Amn/Bmn =ħω3/π2c2. So we have a ratio here. What about calculating the individual values for Amn and Bmn? 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.

466px-Lasercons

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.

Capture

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?

Ruby_laser_pumping_cavity_assembled_and_disassembled

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:

396px-LaserLevels1

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.

402px-YAG2

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.

Powerlite_NdYAGQ: 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. 🙂

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Planck’s constant (II)

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’ ħωapart from each other. That’s what’s illustrated below.

Equally space energy levels

Now, I don’t want to make too many digressions from the main story, but this En = 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 ω(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 ħωor, what amounts to the same (ħω = (h/2π)(2πν) = hν), hν0. Just to give an example, the frequency of sodium light (ν) is 500×1012 Hz, 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(EαeE/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.

graph

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/Pe−(E1–E0)k= e−ΔE/kT. Now, Pis n1/N and Pis n0/N and, hence, we find that nmust 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 eP.E./kT or eK.E./kT in classical mechanics is that the probability goes down as eΔE/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 N=N0e−2ħω/kT. To simplify the algebra, let us call e−ħω/k= x. Then we simply have N1 = N0x, N2 = N0x2, …, 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 N22ħω=2ħωN0x2 is how much energy we get, and so on. Then we add it all together to get Etot = 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: Ntot = N0(1+x+x2+x3+…). Thus the average energy is

formula

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—
energy

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 → .”

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:

formula blackbody

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 ωin 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/[eħω/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… –x22x3–3x4… = 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 eħω/kis 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)

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:

  1. Equal volumes of different gases, at the same pressure and temperature, will have the same number of molecules.
  2. 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 1.38×10–23 joule 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 3rNow, 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:

  1. The kinetic energy for the CM motion of each molecule is, and will always be, (3/2)kT.
  2. 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 = (2/3)n〈mv2/2〉 formula for gas, and we know that the momentum p is defined as p = mv. So we can substitute mvby (mv)v = pv. So we get = (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 PVγ (i.e. the pressure times the volume raised to the power γ) must equal some constant, so we write:

PVγ = C

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

  1. 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).
  2. 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:

  1. Oscillators have a natural frequency, usually denoted by the (angular) frequency ω0.
  2. 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) = γ0, with γ the damping constant (not to be confused with the γ we used in that PVγ = 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 (1.38×10–23 J/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/ω(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:

blackbody radiation formula

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 (Tand 2T0), as derived above (see the solid lines), and then the actual curves for those two temperatures (see the dotted lines).

Blackbody radation graph

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.

Equally space energy levels

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= 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:

  1. The gas consists of a large number of atomic oscillators, each with their own natural frequency ω0.
  2. The permitted energy levels of these harmonic oscillator are equally spaced and ħωapart.
  3. The probability of occupying a level of energy E is P(Eαe–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:

energy

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 PVγ = C relation, did I? And why all the fuss about photon gas? Well… That was just to introduce that PVγ = 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= 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)

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.

geometry

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

two point sourcesRayleigh criterion

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.

gammaray

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.

Heisenberg_Microscope

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. px, 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= 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:

  1. 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 pwill be distributed over the electron and the photon such that p= p’–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.
  2. The scattered photon goes to the right edge of the lens. In that case, we write p= p”+ 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”– p’= p+ h(ε/2)/λ” – p+ 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”– p’= 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:

  1. 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.
  2. 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.
  3. 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)

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 Line of oscillators

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 θout = 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.

Interference two dipole radiatorsThe 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: 2= 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 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.

5wavelength=slitwidthsprectrum

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?

Six-dipole antenna

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 Few 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.

Iridescence

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 py as p0Δθ, with pthe horizontal momentum. So we have Δpy = 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:

Δpy = 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 Δpy = p0Δθ = p0λ/B = p0(h/p0)/B. Of course, the pfactor vanishes and, hence, bringing B to the other side and substituting for Δy = B yields the following grand result:

Δy·Δp= 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?

Six-dipole antenna Huygens_Fresnel_Principle

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:

n oscillatorsThis, 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 n2I(Iis 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:

  1. 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.
  2. 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): 

geometry

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 = λWhat 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 AR, 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.]    

Adding waves

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 = 2λFor 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 = 3λAnd 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.

283px-Airy-pattern 600px-Laser_Interference

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

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.

Photon wave

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 500×1012 oscillations 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 (600×10–9 meter).

Electromagnetic spectrum

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.

Electron double-slit set-up

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 = h= 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.

Photon wave

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.

Circular.Polarization.Circularly.Polarized.Light_Right.Handed.Animation.305x190.255Colors

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

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 can only sit on top of each other 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’.]

StationaryStatesAnimation

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: i= –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.

Hydrogen_eigenstate_n5_l2_m1

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 :-):

σx·σħ/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 σand σp, 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 6.58×10−16 eV·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 10−16 doesn’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 (6.6×10−16 eV·s)(1/3×108 ‘meter’/second) = 2.2×10−8 eV·’m’. Now, 2.2×10−8 is 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 (6.6×10−16 eV·s)(1/3×108 ‘m’/s)(1×1010 ‘Å’/m) = 220 eV·’Å’, or 22 220 eV·’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 σx·σħ/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:

  1. 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).
  2. 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.

Uncertainty relations

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 = 3×10m/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:

  1. The mathematically (and physically) correct answer involves distinguishing the group and phase velocity of a wave.
  2. 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 Aei(ωtkx, where ω is the frequency, which is related to the classical idea of the energy through ħω, and k is the wave number, which is related to the momentum through 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 Aei(ωtkxequation [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 Aei(ωtkxfunction 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:

Sequential_superposition_of_plane_waves

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.

Photon wave

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.

graph (1)

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. 🙂

decay time

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.

Energy Level Diagrams

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 Eto Ewill be written as ν31 = (E– 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 = (E– 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:

  1. 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.]
  2. 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/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.

Photon - E

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. Wave train

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.

Photon wave

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:

EMR law

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.

graph (1)

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(tr/cas 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.

wave in space

So we’d have something like below.

Photon wave

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. 🙂

800px-Bullwhip_effect

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 10(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 10–8 seconds (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 (500×1012 oscillations per second) and a wavelength of 600 nm (600×10–9 meter), the radiation will lasts about 3.2×10–8 seconds. [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 (3×10m/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)

600px-Hydrogen_transitions

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 10–8 seconds 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 waveSo 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.

Circular.Polarization.Circularly.Polarized.Light_Right.Handed.Animation.305x190.255Colors

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 m(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.

Thomson_scattering_geometry

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).]

compton4

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:

  1. Real numbers are complex numbers too: it’s just that their imaginary part is zero.
  2. 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 Aei(ωtr), 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!