Re-visiting electron orbitals (II)

Pre-script (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or classical) interpretation of all things quantum-mechanical. In addition, I note the dark force has amused himself by removing some material. So no use to read this. Read my recent papers instead. 🙂

Original post:

I’ve talked about electron orbitals in a couple of posts already – including a fairly recent one, which is why I put the (II) after the title. However, I just wanted to tie up some loose ends here – and do some more thinking about the concept of a definite energy state. What is it really? We know the wavefunction for a definite energy state can always be written as:

ψ(x, t) = e^{−i·(E/ħ)·t}·ψ(x)

Well… In fact, we should probably formally prove that but… Well… Let us just explore this formula in a more intuitive way – for the time being, that is – using those electron orbitals we’ve derived.

First, let me note that ψ(x, t) and ψ(x) are very different functions and, therefore, the choice of the same symbol for both (the Greek psi) is – in my humble opinion – not very fortunate, but then… Well… It is the choice of physicists – as copied in textbooks all over – and so we’ll just have to live with it. Of course, we can appreciate why they choose to use the same symbol – ψ(x) is like a time-independent wavefunction now, so that’s nice – but… Well… You should note that it is not so obvious to write some function as the product of two other functions. To be complete, I’ll be a bit more explicit here: if some function in two variables – say F(x, y) – can be written as the product of two functions in one variable – say f(x) and g(y), so we can write F as F(x, y) = f(x)·g(y) – then we say F is a separable function. For a full overview of what that means, click on this link. And note mathematicians do choose a different symbol for the functions F and g. It would probably be interesting to explore what the conditions for separability actually imply in terms of properties of… Well… The wavefunction and its argument, i.e. the space and time variables. But… Well… That’s stuff for another post. 🙂

Secondly, note that the momentum variable (p) – i.e. the p in our elementary wavefunction a·e^{i·(p·x−E·t)/ħ} has sort of vanished: ψ(x) is a function of the position only. Now, you may think it should be somewhere there – that, perhaps, we can write something like ψ(x) = ψ[x), p(x)]. But… No. The momentum variable has effectively vanished. Look at Feynman’s solutions for the electron orbitals of a hydrogen atom: Grand Equation The Y_l,m(θ, φ) and F_n,l(ρ) functions here are functions of the (polar) coordinates ρ, θ, φ only. So that’s the position only (these coordinates are polar or spherical coordinates, so ρ is the radial distance, θ is the polar angle, and φ is the azimuthal angle). There’s no idea whatsoever of any momentum in one or the other spatial direction here. I find that rather remarkable. Let’s see how it all works with a simple example.

The functions below are the Y_l,m(θ, φ) for l = 1. Note the symmetry: if we swap θ and φ for -θ and -φ respectively, we get the other function: 2^-1/2·sin(-θ)·e^–i(-φ) = -2^-1/2·sinθ·e^iφ.

example

To get the probabilities, we need to take the absolute square of the whole thing, including e^−i·(E/ħ), but we know |e^i·δ|² = 1 for any value of δ. Why? Because the absolute square of any complex number is the product of the number with its complex conjugate, so |e^i·δ|² = e^i·δ·e^–i·δ= e^i·0= 1. So we only have to look at the absolute square of the Y_l,m(θ, φ) and F_n,l(ρ) functions here. The F_n,l(ρ) function is a real-valued function, so its absolute square is just what it is: some real number (I gave you the formula for the a_k coefficients in my post on it, and you shouldn’t worry about them: they’re real too). In contrast, the Y_l,m(θ, φ) functions are complex-valued – most of them are, at least. Unsurprisingly, we find the probabilities are also symmetric:

P = |-2^-1/2·sinθ·e^iφ|² = (-2^-1/2·sinθ·e^iφ)·(-2^-1/2·sinθ·e^–iφ)

= (2^-1/2·sinθ·e^–iφ)·(2^-1/2·sinθ·e^iφ) = |2^-1/2·sinθ·e^–iφ|² = (1/2)·sin²θ

Of course, for m = 0, the probability is just cos²θ. The graphs below are the polar graphs for the cos²θ and (1/2)·sin²θ functions respectively.

These polar graphs are not so easy to interpret, so let me say a few words about them. The points that are plotted combine (a) some radial distance from the center – which I wrote as P because this distance is, effectively, a probability – with (b) the polar angle θ (so that’s one of the three coordinates). To be precise, the plot gives us, for a given ρ, all of the (θ, P) combinations. It works as follows. To calculate the probability for some ρ and θ (note that φ can be any angle), we must take the absolute square of that ψ_n,l,m, = Y_l,m(θ, φ)·F_n,l(ρ) product. Hence, we must calculate |Y_l,m(θ, φ)·F_n,l(ρ)|² = |F_n,l(ρ)|²·cos²θ for m = 0, and (1/2)·|F_n,l(ρ)|²·sin²θ for m = ±1. Hence, the value of ρ determines the value of F_n,l(ρ), and that F_n,l(ρ) value then determines the shape of the polar graph. The three graphs below – P = cos²θ, P = (1/2)·cos²θ and P = (1/4)·cos²θ – illustrate the idea. polar comparative Note that we’re measuring θ from the z-axis here, as we should. So that gives us the right orientation of this volume, as opposed to the other polar graphs above, which measured θ from the x-axis. So… Well… We’re getting there, aren’t we? 🙂

Now you’ll have two or three – or even more – obvious questions. The first one is: where is the third lobe? That’s a good question. Most illustrations will represent the p-orbitals as follows: p orbitals Three lobes. Well… Frankly, I am not quite sure here, but the equations speak for themselves: the probabilities only depend on ρ and θ. Hence, the azimuthal angle φ can be anything. So you just need to rotate those P = (1/2)·sin²θ and P = cos²θ curves about the the z-axis. In case you wonder how to do that, the illustration below may inspire you. spherical The second obvious question is about the size of those lobes. That 1/2 factor must surely matter, right? Well… We still have that F_n,l(ρ) factor, of course, but you’re right: that factor does not depend on the value for m: it’s the same for m = 0 or ± 1. So… Well… Those representations above – with the three lobes, all of the same volume – may not be accurate. I found an interesting site – Atom in a Box – with an app that visualizes the atomic orbitals in a fun and exciting way. Unfortunately, it’s for Mac and iPhone only – but this YouTube video shows how it works. I encourage you to explore it. In fact, I need to explore it – but what I’ve seen on that YouTube video (I don’t have a Mac nor an iPhone) suggests the three-lobe illustrations may effectively be wrong: there’s some asymmetry here – which we’d expect, because those p-orbitals are actually supposed to be asymmetric! In fact, the most accurate pictures may well be the ones below. I took them from Wikimedia Commons. The author explains the use of the color codes as follows: “The depicted rigid body is where the probability density exceeds a certain value. The color shows the complex phase of the wavefunction, where blue means real positive, red means imaginary positive, yellow means real negative and green means imaginary negative.” I must assume he refers to the sign of a and b when writing a complex number as a + i·b

The third obvious question is related to the one above: we should get some cloud, right? Not some rigid body or some surface. Well… I think you can answer that question yourself now, based on what the author of the illustration above wrote: if we change the cut-off value for the probability, then we’ll give a different shape. So you can play with that and, yes, it’s some cloud, and that’s what the mentioned app visualizes. 🙂

The fourth question is the most obvious of all. It’s the question I started this post with: what are those definite energy states? We have uncertainty, right? So how does that play out? Now that is a question I’ll try to tackle in my next post. Stay tuned ! 🙂

Post scriptum: Let me add a few remarks here so as to – hopefully – contribute to an even better interpretation of what’s going on here. As mentioned, the key to understanding is, obviously, the following basic functional form:

ψ(r, t) = e^{−i·(E/ħ)·t}·ψ(r)

Wikipedia refers to the e^{−i·(E/ħ)·t}factor as a time-dependent phase factor which, as you can see, we can separate out because we are looking at a definite energy state here. Note the minus sign in the exponent – which reminds us of the minus sign in the exponent of the elementary wavefunction, which we wrote as:

a·e^−i·θ = a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]} = a·e^{i·[(p/ħ)∙x − (E/ħ)·t]} = a·e^{−i·(E/ħ)·t}·e^{i·(p/ħ)∙x}

We know this elementary wavefunction is problematic in terms of interpretation because its absolute square gives us some constant probability P(x, t) = |a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]}|²= a². In other words, at any point in time, our electron is equally likely to be anywhere in space. That is not consistent with the idea of our electron being somewhere at some point in time.

The other question is: what reference frame do we use to measure E and p? Indeed, the value of E and p = (p_x, p_y, p_z) depends on our reference frame: from the electron’s own point of view, it has no momentum whatsoever: p = 0. Fortunately, we do have a point of reference here: the nucleus of our hydrogen atom. And our own position, of course, because you should note, indeed, that both the subject and the object of the observation are necessary to define the Cartesian x = x, y, z – or, more relevant in this context – the polar r = ρ, θ, φ coordinates.

This, then, defines some finite or infinite box in space in which the (linear) momentum (p) of our electron vanishes, and then we just need to solve Schrödinger’s diffusion equation to find the solutions for ψ(r). These solutions are more conveniently written in terms of the radial distance ρ, the polar angle θ, and the azimuthal angle φ: Grand Equation

The functions below are the Y_l,m(θ, φ) functions for l = 1.

example

The interesting thing about these Y_l,m(θ, φ) functions is the e^i·φ and/or e^−i·φ factor. Indeed, note the following:

Because the sinθ and cosθ factors are real-valued, they only define some envelope for the ψ(r) function.
In contrast, the e^i·φ and/or e^−i·φ factor define some phase shift.

Let’s have a look at the physicality of the situation, which is depicted below.

spherical

The nucleus of our hydrogen atom is at the center. The polar angle is measured from the z-axis, and we know we only have an amplitude there for m = 0, so let’s look at what that cosθ factor does. If θ = 0°, the amplitude is just what it is, but when θ > 0°, then |cosθ| < 1 and, therefore, the probability P = |F_n,l(ρ)|²·cos²θ will diminish. Hence, for the same radial distance (ρ), we are less likely to find the electron at some angle θ > 0° than on the z-axis itself. Now that makes sense, obviously. You can work out the argument for m = ± 1 yourself, I hope. [The axis of symmetry will be different, obviously!] angle 2 In contrast, the e^i·φ and/or e^−i·φ factor work very differently. These just give us a phase shift, as illustrated below. A re-set of our zero point for measuring time, so to speak, and the e^i·φ and/or e^−i·φ factor effectively disappears when we’re calculating probabilities, which is consistent with the fact that this angle clearly doesn’t influence the magnitude of the amplitude fluctuations. phase shift So… Well… That’s it, really. I hope you enjoyed this ! 🙂

Some more on symmetries…

In our previous post, we talked a lot about symmetries in space – in a rather playful way. Let’s try to take it further here by doing some more thinking on symmetries in spacetime. This post will pick up some older stuff – from my posts on states and the related quantum math in November 2015, for example – but that shouldn’t trouble you too much. On the contrary, I actually hope to tie up some loose ends here.

Let’s first review some obvious ideas. Think about the direction of time. On a time axis, time goes from left to right. It will usually be measured from some zero point – like when we started our experiment or something 🙂 – to some +t point but we may also think of some point in time before our zero point, so the minus (−t) points – the left side of the axis – make sense as well. So the direction of time is clear and intuitive. Now, what does it mean to reverse the direction of time? We need to distinguish two things here: the convention, and… Well… Reality. If we would suddenly decide to reverse the direction in which we measure time, then that’s just another convention. We don’t change reality: trees and kids would still grow the way they always did. 🙂 We would just have to change the numbers on our clocks or, alternatively, the direction of rotation of the hand(s) of our clock, as shown below. [I only showed the hour hand because… Well… I don’t want to complicate things by introducing two time units. But adding the minute hand doesn’t make any difference.]

Now, imagine you’re the dictator who decided to change our time measuring convention. How would you go about it? Would you change the numbers on the clock or the direction of rotation? Personally, I’d be in favor of changing the direction of rotation. Why? Well… First, we wouldn’t have to change expressions such as: “If you are looking north right now, then west is in the 9 o’clock direction, so go there.” 🙂 More importantly, it would align our clocks with the way we’re measuring angles. On the other hand, it would not align our clocks with the way the argument (θ) of our elementary wavefunction ψ = a·e^−iθ = e^–i·^{(E·t – p·x)/ħ} is measured, because that’s… Well… Clockwise.

So… What are the implications here? We would need to change t for −t in our wavefunction as well, right? Yep. Good point. So that’s another convention that would change: we should write our elementary wavefunction now as ψ = a·eⁱ^·^{(E·t – p·x)/ħ}. So we would have to re-define θ as θ = –E·t + p·x = p·x –E·t. So… Well… Done!

So… Well… What’s next? Nothing. Note that we’re not changing reality here. We’re just adapting our formulas to a new dictatorial convention according to which we should count time from positive to negative – like 2, 1, 0, -1, -2 etcetera, as shown below. Fortunately, we can fix all of our laws and formulas in physics by swapping t for -t. So that’s great. No sweat. time reversal

Is that all? Yes. We don’t need to do anything else. We’ll still measure the argument of our wavefunction as an angle, so that’s… Well… After changing our convention, it’s now clockwise. 🙂 Whatever you want to call it: it’s still the same direction. Our dictator can’t change physical reality 🙂

Hmm… But so we are obviously interested in changing physical reality. I mean… Anyone can become a dictator, right? In contrast, we – enlightened scientists – want to really change the world, don’t we? 🙂 So what’s a time reversal in reality? Well… I don’t know… You tell me. 🙂 We may imagine some movie being played backwards, or trees and kids shrinking instead of growing, or some bird flying backwards – and I am not talking the hummingbird here. 🙂

Hey! The latter illustration – that bird flying backwards – is probably the better one: if we reverse the direction of time – in reality, that is – then we should also reverse all directions in space. But… Well… What does that mean, really? We need to think in terms of force fields here. A stone that’d be falling must now go back up. Two opposite charges that were going towards each other, should now move away from each other. But… My God! Such world cannot exist, can it?

No. It cannot. And we don’t need to invoke the second law of thermodynamics for that. 🙂 None of what happens in a movie that’s played backwards makes sense: a heavy stone does not suddenly fly up and decelerate upwards. So it is not like the anti-matter world we described in our previous post. No. We can effectively imagine some world in which all charges have been replaced by their opposite: we’d have positive electrons (positrons) around negatively charged nuclei consisting of antiprotons and antineutrons and, somehow, negative masses. But Coulomb’s law would still tell us two opposite charges – q₁ and –q₂ , for example – don’t repel but attract each other, with a force that’s proportional to the product of their charges, i.e. q₁·(-q₂) = –q₁·q₂. Likewise, Newton’s law of gravitation would still tell us that two masses m₁ and m₂ – negative or positive – will attract each other with a force that’s proportional to the product of their masses, i.e. m₁·m₂= (-m₁)·(-m₂). If you’d make a movie in the antimatter world, it would look just like any other movie. It would definitely not look like a movie being played backwards.

In fact, the latter formula – m₁·m₂= (-m₁)·(-m₂) – tells us why: we’re not changing anything by putting a minus sign in front of all of our variables, which are time (t), position (x), mass (m) and charge (q). [Did I forget one? I don’t think so.] Hence, the famous CPT Theorem – which tells us that a world in which (1) time is reversed, (2) all charges have been conjugated (i.e. all particles have been replaced by their antiparticles), and (3) all spatial coordinates now have the opposite sign, is entirely possible (because it would obey the same Laws of Nature that we, in our world, have discovered over the past few hundred years) – is actually nothing but a tautology. Now, I mean that literally: a tautology is a statement that is true by necessity or by virtue of its logical form. Well… That’s the case here: if we flip the signs of all of our variables, we basically just agreed to count or measure everything from positive to negative. That’s it. Full stop. Such exotic convention is… Well… Exotic, but it cannot change the real world. Full stop.

Of course, this leaves the more intriguing questions entirely open. Partial symmetries. Like time reversal only. 🙂 Or charge conjugation only. 🙂 So let’s think about that.

We know that the world that we see in a mirror must be made of anti-matter but, apart from that particularity, that world makes sense: if we drop a stone in front of the mirror, the stone in the mirror will drop down too. Two like charges will be seen as repelling each other in the mirror too, and concepts such as kinetic or potential energy look just the same. So time just seems to tick away in both worlds – no time reversal here! – and… Well… We’ve got two CP-symmetrical worlds here, don’t we? We only flipped the sign of the coordinate frame and of the charges. Both are possible, right? And what’s possible must exist, right? Well… Maybe. That’s the next step. Let’s first see if both are possible. 🙂

Now, when you’ve read my previous post, you’ll note that I did not flip the z-coordinate when reflecting my world in the mirror. That’s true. But… Well… That’s entirely beside the point. We could flip the z-axis too and so then we’d have a full parity inversion. [Or parity transformation – sounds more serious, doesn’t it? But it’s only a simple inversion, really.] It really doesn’t matter. The point is: axial vectors have the opposite sign in the mirror world, and so it’s not only about whether or not an antimatter world is possible (it should be, right?): it’s about whether or not the sign reversal of all of those axial vectors makes sense in each and every situation. The illustration below, for example, shows how a left-handed neutrino should be a right-handed antineutrino in the mirror world. right-handed antineutrino I hope you understand the left- versus right-handed thing. Think, for example, of how the left-circularly polarized wavefunction below would look like in the mirror. Just apply the customary right-hand rule to determine the direction of the angular momentum vector. You’ll agree it will be right-circularly polarized in the mirror, right? That’s why we need the charge conjugation: think of the magnetic moment of a circulating charge! So… Well… I can’t dwell on this too much but – if Maxwell’s equations are to hold – then that world in the mirror must be made of antimatter. animation

Now, we know that some processes – in our world – are not entirely CP-symmetrical. I wrote about this at length in previous posts, so I won’t dwell on these experiments here. The point is: these experiments – which are not easy to understand – lead physicists, philosophers, bloggers and what have you to solemnly state that the world in the mirror cannot really exist. And… Well… They’re right. However, I think their observations are beside the point. Literally.

So… Well… I would just like to make a very fundamental philosophical remark about all those discussions. My point is quite simple:

We should realize that the mirror world and our world are effectively separated by the mirror. So we should not be looking at stuff in the mirror from our perspective, because that perspective is well… Outside of the mirror. A different world. 🙂 In my humble opinion, the valid point of reference would be the observer in the mirror, like the photographer in the image below. Now note the following: if the real photographer, on this side of the mirror, would have a left-circularly polarized beam in front of him, then the imaginary photographer, on the other side of the mirror, would see the mirror image of this left-circularly polarized beam as a left-circularly polarized beam too. 🙂 I know that sounds complicated but re-read it a couple of times and – I hope – you’ll see the point. If you don’t… Well… Let me try to rephrase it: the point is that the observer in the mirror would be seeing our world – just the same laws and what have you, all makes sense! – but he would see our world in his world, so he’d see it in the mirror world. 🙂

Mirror

Capito? If you would actually be living in the mirror world, then all the things you would see in the mirror world would make perfectly sense. But you would be living in the mirror world. You would not look at it from outside, i.e. from the other side of the mirror. In short, I actually think the mirror world does exist – but in the mirror only. 🙂 […] I am, obviously, joking here. Let me be explicit: our world is our world, and I think those CP violations in Nature are telling us that it’s the only real world. The other worlds exist in our mind only – or in some mirror. 🙂

Post scriptum: I know the Die Hard philosophers among you will now have an immediate rapid-backfire question. [Hey – I just invented a new word, didn’t I? A rapid-backfire question. Neat.] How would the photographer in the mirror look at our world? The answer to that question is simple: symmetry! He (or she) would think it’s a mirror world only. His world and our world would be separated by the same mirror. So… What are the implications here?

Well… That mirror is only a piece of glass with a coating. We made it. Or… Well… Some man-made company made it. 🙂 So… Well… If you think that observer in the mirror – I am talking about that image of the photographer in that picture above now – would actually exist, then… Well… Then you need to be aware of the consequences: the corollary of his existence is that you do not exist. 🙂 And… Well… No. I won’t say more. If you’re reading stuff like this, then you’re smart enough to figure it out for yourself. We live in one world. Quantum mechanics tells us the perspective on that world matters very much – amplitudes are different in different reference frames – but… Well… Quantum mechanics – or physics in general – does not give us many degrees of freedoms. None, really. It basically tells us the world we live in is the only world that’s possible, really. But… Then… Well… That’s just because physics… Well… When everything is said and done, it’s just mankind’s drive to ensure our perception of the Universe lines up with… Well… What we perceive it to be. 😦 or 🙂 Whatever your appreciation of it. Those Great Minds did an incredible job. 🙂

Symmetries and transformations

In my previous post, I promised to do something on symmetries. Something simple but then… Well… You know how it goes: one question always triggers another one. 🙂

Look at the situation in the illustration on the left below. We suppose we have something real going on there: something is moving from left to right (so that’s in the 3 o’clock direction), and then something else is going around clockwise (so that’s not the direction in which we measure angles (which also include the argument θ of our wavefunction), because that’s always counter-clockwise, as I note at the bottom of the illustration). To be precise, we should note that the angular momentum here is all about the y-axis, so the angular momentum vector L points in the (positive) y-direction. We get that direction from the familiar right-hand rule, which is illustrated in the top right corner.

mirror Now, suppose someone else is looking at this from the other side – or just think of yourself going around a full 180° to look at the same thing from the back side. You’ll agree you’ll see the same thing going from right to left (so that’s in the 9 o’clock direction now – or, if our clock is transparent, the 3 o’clock direction of our reversed clock). Likewise, the thing that’s turning around will now go counter-clockwise.

Note that both observers – so that’s me and that other person (or myself after my walk around this whole thing) – use a regular coordinate system, which implies the following:

We’ve got regular 90° degree angles between our coordinates axes.
Our x-axis goes from negative to positive from left to right, and our y-axis does the same going away from us.
We also both define our z-axis using, once again, the ubiquitous right-hand rule, so our z-axis points upwards.

So we have two observers looking at the same reality – some linear as well as some angular momentum – but from opposite sides. And so we’ve got a reversal of both the linear as well as the angular momentum. Not in reality, of course, because we’re looking at the same thing. But we measure it differently. Indeed, if we use the subscripts 1 and 2 to denote the measurements in the two coordinate systems, we find that p₂ = –p₁. Likewise, we also find that L₂ = –L₁.

Now, when you see these two equations, you will probably not worry about that p₂ = –p₁ equation – although you should, because it’s actually only valid for this rather particular orientation of the linear momentum (I’ll come back to that in a moment). It’s the L₂ = –L₁ equation which should surprise you most. Why? Because you’ve always been told there is a big difference between (1) real vectors (aka polar vectors), like the momentum p, or the velocity v, or the force F, and (2) pseudo-vectors (aka axial vectors), like the angular momentum L. You may also remember how to distinguish between the two: if you change the direction of the axes of your reference frame, polar vectors will change sign too, as opposed to axial vectors: axial vectors do not swap sign if we swap the coordinate signs.

So… Well… How does that work here? In fact, what we should ask ourselves is: why does that not work here? Well… It’s simple, really. We’re not changing the direction of the axes here. Or… Well… Let me be more precise: we’re only swapping the sign of the x– and y-axis. We did not flip the z-axis. So we turned things around, but we didn’t turn them upside down. It makes a huge difference. Note, for example, that if all of the linear momentum would have been in the z-direction only (so our p vector would have been pointing in the z-direction, and in the z-direction only), it would not swap sign. The illustration below shows what really happens with the coordinates of some vector when we’re doing a rotation. It’s, effectively, only the x– and y-coordinates that flip sign. reflection symmetry

It’s easy to see that this rotation about the z-axis here preserves our deep sense of ‘up’ versus ‘down’, but that it swaps ‘left’ for ‘right’, and vice versa. Note that this is not a reflection. We are not looking at some mirror world here. The difference between a reflection (a mirror world) and a rotation (the real world seen from another angle) is illustrated below. It’s quite confusing but, unlike what you might think, a reflection does not swap left for right. It does turn things inside out, but that’s what a rotation does as well: near becomes far, and far becomes near. difference between reflection and rotation

Before we move on, let me say a few things about the mirror world and, more in particular, about the obvious question: could it possibly exist? Well… What do you think? Your first reaction might well be: “Of course! What nonsense question! We just walk around whatever it is that we’re seeing – or, what amounts to the same, we just turn it around – and there it is: that’s the mirror world, right? So of course it exists!” Well… No. That’s not the mirror world. That’s just the real world seen from the opposite direction, and that world… Well… That’s just the real world. 🙂 The mirror world is, literally, the world in the mirror – like the photographer in the illustration below. We don’t swap left for right here: some object going from left to right in the real world is still going from left to right in the mirror world! Mirror Of course, you may now involve the photographer in the picture above and observe – note that you’re now an observer of the observer of the mirror 🙂 – that, if he would move his left arm in the real world, the photographer in the mirror world would be moving his right arm. But… Well… No. You’re saying that because you’re now imaging that you’re the photographer in the mirror world yourself now, who’s looking at the real world from inside, so to speak. So you’ve rotated the perspective in your mind and you’re saying it’s his right arm because you imagine yourself to be the photographer in the mirror. We usually do that because… Well… Because we look in a mirror every day, right? So we’re used to seeing ourselves that way and we always think it’s us we’re seeing. 🙂 However, the illustration above is correct: the mirror world only swaps near for far, and far for near, so it only swaps the sign of the y-axis.

So the question is relevant: could the mirror world actually exist? What we’re really asking here is the following: can we swap the sign of one coordinate axis only in all of our physical laws and equations and… Well… Do we then still get the same laws and equations? Do we get the same Universe – because that’s what those laws and equations describe? If so, our mirror world can exist. If not, then not.

Now, I’ve done a post on that, in which I explain that mirror world can only exist if it would consist of anti-matter. So if our real world and the mirror world would actually meet, they would annihilate each other. 🙂 But that post is quite technical. Here I want to keep it very simple: I basically only want to show what the rotation operation implies for the wavefunction. There is no doubt whatsoever that the rotated world exists. In fact, the rotated world is just our world. We walk around some object, or we turn it around, but so we’re still watching the same object. So we’re not thinking about the mirror world here. We just want to know how things look like when adopting some other perspective.

So, back to the starting point: we just have two observers here, who look at the same thing but from opposite directions. Mathematically, this corresponds to a rotation of our reference frame about the z-axis of 180°. Let me spell out – somewhat more precisely – what happens to the linear and angular momentum here:

The direction of the linear momentum in the xy-plane swaps direction.
The angular momentum about the y-axis, as well as about the x-axis, swaps direction too.

Note that the illustration only shows angular momentum about the y-axis, but you can easily verify the statement about the angular momentum about the x-axis. In fact, the angular momentum about any line in the xy-plane will swap direction.

Of course, the x-, y-, z-axes in the other reference frame are different than mine, and so I should give them a subscript, right? Or, at the very least, write something like x’, y’, z’, so we have a primed reference frame here, right? Well… Maybe. Maybe not. Think about it. 🙂 A coordinate system is just a mathematical thing… Only the momentum is real… Linear or angular… Equally real… And then Nature doesn’t care about our position, does it? So… Well… No subscript needed, right? Or… Well… What do you think? 🙂

It’s just funny, isn’t it? It looks like we can’t really separate reality and perception here. Indeed, note how our p₂ = –p₁and L₂ = –L₁ equations already mix reality with how we perceive it. It’s the same thing in reality but the coordinates of p₁ and L₁are positive, while the coordinates of p₂ and L₂are negative. To be precise, these coordinates will look like this:

p₁ = (p, 0, 0) and L₁= (0, L, 0)
p₂ = (−p, 0, 0) and L₁= (0, −L, 0)

So are they two different things or are they not? 🙂 Think about it. I’ll move on in the meanwhile. 🙂

Now, you probably know a thing or two about parity symmetry, or P-symmetry: if if we flip the sign of all coordinates, then we’ll still find the same physical laws, like F = m·a and what have you. [It works for all physical laws, including quantum-mechanical laws – except those involving the weak force (read: radioactive decay processes).] But so here we are talking rotational symmetry. That’s not the same as P-symmetry. If we flip the signs of all coordinates, we’re also swapping ‘up’ for ‘down’, so we’re not only turning around, but we’re also getting upside down. The difference between rotational symmetry and P-symmetry is shown below. up and down swap

As mentioned, we’ve talked about P-symmetry at length in other posts, and you can easily google a lot more on that. The question we want to examine here – just as a fun exercise – is the following:

How does that rotational symmetry work for a wavefunction?

The very first illustration in this post gave you the functional form of the elementary wavefunction eⁱ^θ = eⁱ^·^{(E·t – p·x)/ħ}. We should actually use a bold type x = (x, y, z) in this formula but we’ll assume we’re talking something similar to that p vector: something moving in the x-direction only – or in the xy-plane only. The z-component doesn’t change. Now, you know that we can reduce all actual wavefunctions to some linear combination of such elementary wavefunctions by doing a Fourier decomposition, so it’s fine to look at the elementary wavefunction only – so we don’t make it too complicated here. Now think of the following.

The energy E in the eⁱ^θ = eⁱ^·^{(E·t – p·x)/ħ}function is a scalar, so it doesn’t have any direction and we’ll measure it the same from both sides – as kinetic or potential energy or, more likely, by adding both. But… Well… Writing eⁱ^·^{(E·t – p·x)/ħ}or eⁱ^·^{(E·t + p·x)/ħ}is not the same, right? No, it’s not. However, think of it as follows: we won’t be changing the direction of time, right? So it’s OK to not change the sign of E. In fact, we can re-write the two expressions as follows:

eⁱ^·^{(E·t – p·x)/ħ}= eⁱ^·^(E/ħ)·t·e^–ⁱ^·(^p/ħ)·x
eⁱ^·^{(E·t + p·x)/ħ}= eⁱ^·^(E/ħ)·t·eⁱ^·(^p/ħ)·x

The first wavefunction describes some particle going in the positive x-direction, while the second wavefunction describes some particle going in the negative x-direction, so… Well… That’s exactly what we see in those two reference frames, so there is no issue whatsoever. 🙂 It’s just… Well… I just wanted to show the wavefunction does look different too when looking at something from another angle.

So why am I writing about this? Why am I being fussy? Well.. It’s just to show you that those transformations are actually quite natural – just as natural as it is to see some particle go in one direction in one reference frame and see it go in the other in the other. 🙂 It also illustrates another point that I’ve been trying to make: the wavefunction is something real. It’s not just a figment of our imagination. The real and imaginary part of our wavefunction have a precise geometrical meaning – and I explained what that might be in my more speculative posts, which I’ve brought together in the Deep Blue page of this blog. But… Well… I can’t dwell on that here because… Well… You should read that page. 🙂

The point to note is the following: we do have different wavefunctions in different reference frames, but these wavefunctions describe the same physical reality, and they also do respect the symmetries we’d expect them to respect, except… Well… The laws describing the weak force don’t, but I wrote about that a very long time ago, and it was not in the context of trying to explain the relatively simple basic laws of quantum mechanics. 🙂 If you’re interested, you should check out my post(s) on that or, else, just google a bit. It’s really exciting stuff, but not something that will help you much to understand the basics, which is what we’re trying to do here. 🙂

The second point to note is that those transformations of the wavefunction – or of quantum-mechanical states – which we go through when rotating our reference frame, for example – are really quite natural. There’s nothing special about them. We had such transformations in classical mechanics too! But… Well… Yes, I admit they do look complicated. But then that’s why you’re so fascinated and why you’re reading this blog, isn’t it? 🙂

Post scriptum: It’s probably useful to be somewhat more precise on all of this. You’ll remember we visualized the wavefunction in some of our posts using the animation below. It uses a left-handed coordinate system, which is rather unusual but then it may have been made with a software which uses a left-handed coordinate system (like RenderMan, for example). Now the rotating arrow at the center moves with time and gives us the polarization of our wave. Applying our customary right-hand rule,you can see this beam is left-circularly polarized. [I know… It’s quite confusing, but just go through the motions here and be consistent.] Animation Now, you know that e^–ⁱ^·(^p/ħ)·xand e^–ⁱ^·(^p/ħ)·x are each other’s complex conjugate:

e^–ⁱ^·k^·x= cos(k·x) + i·sin(k·x)
e^–ⁱ^·k^·x= cos(-k·x) + i·sin(-k·x) = cos(k·x) − i·sin(k·x)

Their real part – the cosine function – is the same, but the imaginary part – the sine function – has the opposite sign. So, assuming the direction of propagation is, effectively, the x-direction, then what’s the polarization of the mirror image? Well… The wave will now go from right to left, and its polarization… Hmm… Well… What?

Well… If you can’t figure it out, then just forget about those signs and just imagine you’re effectively looking at the same thing from the backside. In fact, if you have a laptop, you can push the screen down and go around your computer. 🙂 There’s no shame in that. In fact, I did that just to make sure I am not talking nonsense here. 🙂 If you look at this beam from the backside, you’ll effectively see it go from right to left – instead of from what you see on this side, which is a left-to-right direction. And as for its polarization… Well… The angular momentum vector swaps direction too but the beam is still left-circularly polarized. So… Well… That’s consistent with what we wrote above. 🙂 The real world is real, and axial vectors are as real as polar vectors. This real beam will only appear to be right-circularly polarized in a mirror. Now, as mentioned above, that mirror world is not our world. If it would exist – in some other Universe – then it would be made up of anti-matter. 🙂

So… Well… Might it actually exist? Is there some other world made of anti-matter out there? I don’t know. We need to think about that reversal of ‘near’ and ‘far’ too: as mentioned, a mirror turns things inside out, so to speak. So what’s the implication of that? When we walk around something – or do a rotation – then the reversal between ‘near’ and ‘far’ is something physical: we go near to what was far, and we go away from what was near. But so how would we get into our mirror world, so to speak? We may say that this anti-matter world in the mirror is entirely possible, but then how would we get there? We’d need to turn ourselves, literally, inside out – like short of shrink to the zero point and then come back out of it to do that parity inversion along our line of sight. So… Well… I don’t see that happen, which is why I am a fan of the One World hypothesis. 🙂 So I think the mirror world is just what it is: the mirror world. Nothing real. But… Then… Well… What do you think? 🙂

Quantum-mechanical magnitudes

As I was writing about those rotations in my previous post (on electron orbitals), I suddenly felt I should do some more thinking on (1) symmetries and (2) the concept of quantum-mechanical magnitudes of vectors. I’ll write about the first topic (symmetries) in some other post. Let’s first tackle the latter concept. Oh… And for those I frightened with my last post… Well… This should really be an easy read. More of a short philosophical reflection about quantum mechanics. Not a technical thing. Something intuitive. At least I hope it will come out that way. 🙂

First, you should note that the fundamental idea that quantities like energy, or momentum, may be quantized is a very natural one. In fact, it’s what the early Greek philosophers thought about Nature. Of course, while the idea of quantization comes naturally to us (I think it’s easier to understand than, say, the idea of infinity), it is, perhaps, not so easy to deal with it mathematically. Indeed, most mathematical ideas – like functions and derivatives – are based on what I’ll loosely refer to as continuum theory. So… Yes, quantization does yield some surprising results, like that formula for the magnitude of some vector J: Magnitude formulas The J·J in the classical formula above is, of course, the equally classical vector dot product, and the formula itself is nothing but Pythagoras’ Theorem in three dimensions. Easy. I just put a + sign in front of the square roots so as to remind you we actually always have two square roots and that we should take the positive one. 🙂

I will now show you how we get that quantum-mechanical formula. The logic behind it is fairly straightforward but, at the same time… Well… You’ll see. 🙂 We know that a quantum-mechanical variable – like the spin of an electron, or the angular momentum of an atom – is not continuous but discrete: it will have some value m = j, j-1, j-2, …, -(j-2), -(j-1), –j. Our j here is the maximum value of the magnitude of the component of our vector (J) in the direction of measurement, which – as you know – is usually written as J_z. Why? Because we will usually choose our coordinate system such that our z-axis is aligned accordingly. 🙂 Those values j, j-1, j-2, …, -(j-2), -(j-1), –j are separated by one unit. That unit would be Planck’s quantum of action ħ ≈ 1.0545718×10⁻³⁴ N·m·s – by the way, isn’t it amazing we can actually measure such tiny stuff in some experiment? 🙂 – if J would happen to be the angular momentum, but the approach here is more general – action can express itself in various ways 🙂 – so the unit doesn’t matter: it’s just the unit, so that’s just one. 🙂 It’s easy to see that this separation implies j must be some integer or half-integer. [Of course, now you might think the values of a series like 2.4, 1.4, 0.4, -0.6, -1.6 are also separated by one unit, but… Well… That would violate the most basic symmetry requirement so… Well… No. Our j has to be an integer or a half-integer. Please also note that the number of possible values for m is equal to 2j+1, as we’ll use that in a moment.]

OK. You’re familiar with this by now and so I should not repeat the obvious. To make things somewhat more real, let’s assume j = 3/2, so m = 3/2, 1/2, -1/2 or +3/2. Now, we don’t know anything about the system and, therefore, these four values are all equally likely. Now, you may not agree with this assumption but… Well… You’ll have to agree that, at this point, you can’t come up with anything else that would make sense, right? It’s just like a classical situation: J might point in any direction, so we have to give all angles an equal probability. [In fact, I’ll show you – in a minute or so – that you actually have a point here: we should think some more about this assumption – but so that’s for later. I am asking you to just go along with this story as for now.]

So the expected value of J_z is E[J_z] is equal to E[J_z] = (1/4)·(3/2)+(1/4)·(1/2)+(1/4)·(-1/2)+(1/4)·(-3/2) = 0. Nothing new here. We just multiply probabilities with all of the possible values to get an expected value. So we get zero here because our values are distributed symmetrically around the zero point. No surprise. Now, to calculate a magnitude, we don’t need J_zbut J_z². In case you wonder, that’s what this squaring business is all about: we’re abstracting away from the direction and so we’re going to square both positive as well as negative values to then add it all up and take a square root. Now, the expected value of J_z² is equal to E[J_z] = (1/4)·(3/2)²+(1/4)·(1/2)²+(1/4)·(-1/2)²+(1/4)·(-3/2)² = 5/4 = 1.25. Some positive value.

You may note that it’s a bit larger than the average of the absolute value of our variable, which is equal to (|3/2|+|1/2|+|-1/2|+|-3/2|)/4 = 1, but that’s just because the squaring favors larger values 🙂 Also note that, of course, we’d also get some positive value if J_zwould be a continuous variable over the [-3/2, +3/2] interval, but I’ll let you think about what positive value we’d get for J_z²assuming J_z is uniform distributed over the [-3/2, +3/2] interval, because that calculation is actually not so straightforward as it may seem at first. In any case, these considerations are not very relevant to our story here, so let’s move on.

Of course, our z-direction was random, and so we get the same thing for whatever direction. More in particular, we’ll also get it for the x– and y-directions: E[J_x] = E[J_y] = E[J_z] = 5/4. Now, at this point it’s probably good to give you a more generalized formula for these quantities. I think you’ll easily agree to the following one: magnitude squared formula So now we can apply our classical J·J = J_x²+ J_y²+ J_z²formula to these quantities by calculating the expected value of J = J·J, which is equal to:

E[J·J] = E[J_x²] + E[J_y²] + E[J_z²] = 3·E[J_x²] = 3·E[J_y²] = 3·E[J_z²]

You should note we’re making use of the E[X + Y] = E[X]+ E[Y] property here: the expected value of the sum of two variables is equal to the sum of the expected values of the variables, and you should also note this is true even if the individual variables would happen to be correlated – which might or might not be the case. [What do you think is the case here?]

For j = 3/2, it’s easy to see we get E[J·J] = 3·E[J_x] = 3·5/4 = (3/2)·(3/2+1) = j·(j+1). We should now generalize this formula for other values of j, which is not so easy… Hmm… It obviously involves some formula for a series, and I am not good at that… So… Well… I just checked if it was true for j = 1/2 and j = 1 (please check that at least for yourself too!) and then I just believe the authorities on this for all other values of j. 🙂

Now, in a classical situation, we know that J·J product will be the same for whatever direction J would happen to have, and so its expected value will be equal to its constant value J·J. So we can write: E[J·J] = J·J. So… Well… That’s why we write what we wrote above: Magnitude formulas

Makes sense, no? E[J·J] = E[J_x²+J_y²+J_z²] = E[J_x²]+E[J_y²]+E[J_z²] = j·(j+1) = J·J = J², so J = +√[j(j+1)], right?

Hold your horses, man! Think! What are we doing here, really? We didn’t calculate all that much above. We only found that E[J_x²]+E[J_y²]+E[J_z²] = E[J_x²+J_y²+J_z²] = j·(j+1). So what? Well… That’s not a proof that the J vector actually exists.

Huh?

Yes. That J vector might just be some theoretical concept. When everything is said and done, all we’ve been doing – or at least, we imagined we did – is those repeated measurements of J_x, J_y and J_z here – or whatever subscript you’d want to use, like J_θ,φ, for example (the example is not random, of course) – and so, of course, it’s only natural that we assume these things are the magnitude of the component (in the direction of measurement) of some real vector that is out there, but then… Well… Who knows? Think of what we wrote about the angular momentum in our previous post on electron orbitals. We imagine – or do like to think – that there’s some angular momentum vector J out there, which we think of as being “cocked” at some angle, so its projection onto the z-axis gives us those discrete values for m which, for j = 2, for example, are equal to 0, 1 or 2 (and -1 and -2, of course) – like in the illustration below. 🙂 cocked angle 2 But… Well… Note those weird angles: we get something close to 24.1° and then another value close to 54.7°. No symmetry here. 😦 The table below gives some more values for larger j. They’re easy to calculate – it’s, once again, just Pythagoras’ Theorem – but… Well… No symmetries here. Just weird values. [I am not saying the formula for these angles is not straightforward. That formula is easy enough: θ = sin^-1(m/√[j(j+1)]). It’s just… Well… No symmetry. You’ll see why that matters in a moment.] Capture I skipped the half-integer values for j in the table above so you might think they might make it easier to come up with some kind of sensible explanation for the angles. Well… No. They don’t. For example, for j = 1/2 and m = ± 1/2, the angles are ±35.2644° – more or less, that is. 🙂 As you can see, these angles do not nicely cut up our circle in equal pieces, which triggers the obvious question: are these angles really equally likely? Equal angles do not correspond to equal distances on the z-axis (in case you don’t appreciate the point, look at the illustration below). angles distance

So… Well… Let me summarize the issue on hand as follows: the idea of the angle of the J vector being randomly distributed is not compatible with the idea of those J_z values being equally spaced and equally likely. The latter idea – equally spaced and equally likely J_z values – relates to different possible states of the system being equally likely, so… Well… It’s just a different idea. 😦

Now there is another thing which we should mention here. The maximum value of the z-component of our J vector is always smaller than that quantum-mechanical magnitude, and quite significantly so for small j, as shown in the table below. It is only for larger values of j that the ratio of the two starts to converge to 1. For example, for j = 25, it is about 1.02, so that’s only 2% off. convergence That’s why physicists tell us that, in quantum mechanics, the angular momentum is never “completely along the z-direction.” It is obvious that this actually challenges the idea of a very precise direction in quantum mechanics, but then that shouldn’t surprise us, does it? After, isn’t this what the Uncertainty Principle is all about?

Different states, rather than different directions… And then Uncertainty because… Well… Because of discrete variables that won’t split in the middle. Hmm… 😦

Perhaps. Perhaps I should just accept all of this and go along with it… But… Well… I am really not satisfied here, despite Feynman’s assurance that that’s OK: “Understanding of these matters comes very slowly, if at all. Of course, one does get better able to know what is going to happen in a quantum-mechanical situation—if that is what understanding means—but one never gets a comfortable feeling that these quantum-mechanical rules are ‘natural’.”

I do want to get that comfortable feeling – on some sunny day, at least. 🙂 And so I’ll keep playing with this, until… Well… Until I give up. 🙂 In the meanwhile, if you’d feel you’ve got some better or some more intuitive explanation for all of this, please do let me know. I’d be very grateful to you. 🙂

Post scriptum: Of course, we would all want to believe that J somehow exists because… Well… We want to explain those states somehow, right? I, for one, am not happy with being told to just accept things and shut up. So let me add some remarks here. First, you may think that the narrative above should distinguish between polar and axial vectors. You’ll remember polar vectors are the real vectors, like a radius vector r, or a force F, or velocity or (linear) momentum. Axial vectors (also known as pseudo-vectors) are vectors like the angular momentum vector: we sort of construct them from… Well… From real vectors. The angular momentum L, for example, is the vector cross product of the radius vector r and the linear momentum vector p: we write L = r×p. In that sense, they’re a figment of our imagination. But then… What’s real and unreal? The magnitude of L, for example, does correspond to something real, doesn’t it? And its direction does give us the direction of circulation, right? You’re right. Hence, I think polar and axial vectors are both real – in whatever sense you’d want to define real. Their reality is just different, and that’s reflected in their mathematical behavior: if you change the direction of the axes of your reference frame, polar vectors will change sign too, as opposed to axial vectors: they don’t swap sign. They do something else, which I’ll explain in my next post, where I’ll be talking symmetries.

But let us, for the sake of argument, assume whatever I wrote about those angles applies to axial vectors only. Let’s be even more specific, and say it applies to the angular momentum vector only. If that’s the case, we may want to think of a classical equivalent for the mentioned lack of a precise direction: free nutation. It’s a complicated thing – even more complicated than the phenomenon of precession, which we should be familiar with by now. Look at the illustration below (which I took from an article of a physics professor from Saint Petersburg), which shows both precession as well as nutation. Think of the movement of a spinning top when you release it: its axis will, at first, nutate around the axis of precession, before it settles in a more steady precession.The nutation is caused by the gravitational force field, and the nutation movement usually dies out quickly because of dampening forces (read: friction). Now, we don’t think of gravitational fields when analyzing angular momentum in quantum mechanics, and we shouldn’t. But there is something else we may want to think of. There is also a phenomenon which is referred to as free nutation, i.e. a nutation that is not caused by an external force field. The Earth, for example, nutates slowly because of a gravitational pull from the Sun and the other planets – so that’s not a free nutation – but, in addition to this, there’s an even smaller wobble – which is an example of free nutation – because the Earth is not exactly spherical. In fact, the Great Mathematician, Leonhard Euler, had already predicted this, back in 1765, but it took another 125 years or so before an astronomist, Seth Chandler, could finally experimentally confirm and measure it. So they named this wobble the Chandler wobble (Euler already has too many things named after him). 🙂

Now I don’t have much backup here – none, actually 🙂 – but why wouldn’t we imagine our electron would also sort of nutate freely because of… Well… Some symmetric asymmetry – something like the slightly elliptical shape of our Earth. 🙂 We may then effectively imagine the angular momentum vector as continually changing direction between a minimum and a maximum angle – something like what’s shown below, perhaps, between 0 and 40 degrees. Think of it as a rotation within a rotation, or an oscillation within an oscillation – or a standing wave within a standing wave. 🙂 wobbling I am not sure if this approach would solve the problem of our angles and distances – the issue of whether we should think in equally likely angles or equally likely distances along the z-axis, really – but… Well… I’ll let you play with this. Please do send me some feedback if you think you’ve found something. 🙂

Whatever your solution is, it is likely to involve the equipartition theorem and harmonics, right? Perhaps we can, indeed, imagine standing waves within standing waves, and then standing waves within standing waves. How far can we go? 🙂

Post scriptum 2: When re-reading this post, I was thinking I should probably do something with the following idea. If we’ve got a sphere, and we’re thinking of some vector pointing to some point on the surface of that sphere, then we’re doing something which is referred to as point picking on the surface of a sphere, and the probability distributions – as a function of the polar and azimuthal angles θ and φ – are quite particular. See the article on the Wolfram site on this, for example. I am not sure if it’s going to lead to some easy explanation of the ‘angle problem’ we’ve laid out here but… Well… It’s surely an element in the explanation. The key idea here is shown in the illustration below: if the direction of our momentum in three-dimensional space is really random, there may still be more of a chance of an orientation towards the equator, rather than towards the pole. So… Well… We need to study the math of this. 🙂 But that’s for later. density

Re-visiting electron orbitals

Pre-script (dated 26 June 2020): This post got mutilated by the removal of some material by the dark force. You should be able to follow the main story line, however. If anything, the lack of illustrations might actually help you to think things through for yourself. In any case, we now have different views on these concepts as part of our realist interpretation of quantum mechanics, so we recommend you read our recent papers instead of these old blog posts.

Original post:

One of the pieces I barely gave a glance when reading Feynman’s Lectures over the past few years, was the derivation of the non-spherical electron orbitals for the hydrogen atom. It just looked like a boring piece of math – and I thought the derivation of the s-orbitals – the spherically symmetrical ones – was interesting enough already. To some extent, it is – but there is so much more to it. When I read it now, the derivation of those p-, d-, f– etc. orbitals brings all of the weirdness of quantum mechanics together and, while doing so, also provides for a deeper understanding of all of the ideas and concepts we’re trying to get used to. In addition, Feynman’s treatment of the matter is actually much shorter than what you’ll find in other textbooks, because… Well… As he puts it, he takes a shortcut. So let’s try to follow the bright mind of our Master as he walks us through it.

You’ll remember – if not, check it out again – that we found the spherically symmetric solutions for Schrödinger’s equation for our hydrogen atom. Just to be make sure, Schrödinger’s equation is a differential equation – a condition we impose on the wavefunction for our electron – and so we need to find the functional form for the wavefunctions that describe the electron orbitals. [Quantum math is so confusing that it’s often good to regularly think of what it is that we’re actually trying to do. :-)] In fact, that functional form gives us a whole bunch of solutions – or wavefunctions – which are defined by three quantum numbers: n, l, and m. The parameter n corresponds to an energy level (E_n), l is the orbital (quantum) number, and m is the z-component of the angular momentum. But that doesn’t say much. Let’s go step by step.

First, we derived those spherically symmetric solutions – which are referred to as s-states – assuming this was a state with zero (orbital) angular momentum, which we write as l = 0. [As you know, Feynman does not incorporate the spin of the electron in his analysis, which is, therefore, approximative only.] Now what exactly is a state with zero angular momentum? When everything is said and done, we are effectively trying to describe some electron orbital here, right? So that’s an amplitude for the electron to be somewhere, but then we also know it always moves. So, when everything is said and done, the electron is some circulating negative charge, right? So there is always some angular momentum and, therefore, some magnetic moment, right?

Well… If you google this question on Physics Stack Exchange, you’ll get a lot of mumbo jumbo telling you that you shouldn’t think of the electron actually orbiting around. But… Then… Well… A lot of that mumbo jumbo is contradictory. For example, one of the academics writing there does note that, while we shouldn’t think of an electron as some particle, the orbital is still a distribution which gives you the probability of actually finding the electron at some point (x,y,z). So… Well… It is some kind of circulating charge – as a point, as a cloud or as whatever. The only reasonable answer – in my humble opinion – is that l = 0 probably means there is no net circulating charge, so the movement in this or that direction must balance the movement in the other. One may note, in this regard, that the phenomenon of electron capture in nuclear reactions suggests electrons do travel through the nucleus for at least part of the time, which is entirely coherent with the wavefunctions for s-states – shown below – which tell us that the most probable (x, y, z) position for the electron is right at the center – so that’s where the nucleus is. There is also a non-zero probability for the electron to be at the center for the other orbitals (p, d, etcetera).In fact, now that I’ve shown this graph, I should quickly explain it. The three graphs are the spherically symmetric wavefunctions for the first three energy levels. For the first energy level – which is conventionally written as n = 1, not as n = 0 – the amplitude approaches zero rather quickly. For n = 2 and n = 3, there are zero-crossings: the curve passes the r-axis. Feynman calls these zero-crossing radial nodes. To be precise, the number of zero-crossings for these s-states is n − 1, so there’s none for n = 1, one for n = 2, two for n = 3, etcetera.

Now, why is the amplitude – apparently – some real-valued function here? That’s because we’re actually not looking at ψ(r, t) here but at the ψ(r) function which appears in the following break-up of the actual wavefunction ψ(r, t):

ψ(r, t) = e^{−i·(E/ħ)·t}·ψ(r)

So ψ(r) is more of an envelope function for the actual wavefunction, which varies both in space as well as in time. It’s good to remember that: I would have used another symbol, because ψ(r, t) and ψ(r) are two different beasts, really – but then physicists want you to think, right? And Mr. Feynman would surely want you to do that, so why not inject some confusing notation from time to time? 🙂 So for n = 3, for example, ψ(r) goes from positive to negative and then to positive, and these areas are separated by radial nodes. Feynman put it on the blackboard like this:I am just inserting it to compare this concept of radial nodes with the concept of a nodal plane, which we’ll encounter when discussing p-states in a moment, but I can already tell you what they are now: those p-states are symmetrical in one direction only, as shown below, and so we have a nodal plane instead of a radial node. But so I am getting ahead of myself here… 🙂 nodal planes Before going back to where I was, I just need to add one more thing. 🙂 Of course, you know that we’ll take the square of the absolute value of our amplitude to calculate a probability (or the absolute square – as we abbreviate it), so you may wonder why the sign is relevant at all. Well… I am not quite sure either but there’s this concept of orbital parity which you may have heard of. The orbital parity tells us what will happen to the sign if we calculate the value for ψ for −r rather than for r. If ψ(−r) = ψ(r), then we have an even function – or even orbital parity. Likewise, if ψ(−r) = −ψ(r), then we’ll the function odd – and so we’ll have an odd orbital parity. The orbital parity is always equal to (-1)^l = ±1. The exponent l is that angular quantum number, and +1, or + tout court, means even, and -1 or just − means odd. The angular quantum number for those p-states is l = 1, so that works with the illustration of the nodal plane. 🙂 As said, it’s not hugely important but I might as well mention in passing – especially because we’ll re-visit the topic of symmetries a few posts from now. 🙂

OK. I said I would talk about states with some angular momentum (so l ≠ 0) and so it’s about time I start doing that. As you know, our orbital angular momentum l is measured in units of ħ (just like the total angular momentum J, which we’ve discussed ad nauseam already). We also know that if we’d measure its component along any direction – any direction really, but physicists will usually make sure that the z-axis of their reference frame coincides with, so we call it the z-axis 🙂 – then we will find that it can only have one of a discrete set of values m·ħ = l·ħ, (l-1)·ħ, …, -(l-1)·ħ, –l·ħ. Hence, l just takes the role of our good old quantum number j here, and m is just J_z. Likewise, I’d like to introduce l as the equivalent of J, so we can easily talk about the angular momentum vector. And now that we’re here, why not write m in bold type too, and say that m is the z-component itself – i.e. the whole vector quantity, so that’s the direction and the magnitude.

Now, we do need to note one crucial difference between j and l, or between J and l: our j could be an integer or a half-integer. In contrast, l must be some integer. Why? Well… If l can be zero, and the values of l must be separated by a full unit, then l must be 1, 2, 3 etcetera. 🙂 If this simple answer doesn’t satisfy you, I’ll refer you to Feynman’s, which is also short but more elegant than mine. 🙂 Now, you may or may not remember that the quantum-mechanical equivalent of the magnitude of a vector quantity such as l is to be calculated as √[l·(l+1)]·ħ, so if l = 1, that magnitude will be √2·ħ ≈ 1.4142·ħ, so that’s – as expected – larger than the maximum value for m, which is +1. As you know, that leads us to think of that z-component m as a projection of l. Paraphrasing Feynman, the limited set of values for m imply that the angular momentum is always “cocked” at some angle. For l = 1, that angle is either +45° or, else, −45°, as shown below. cocked angle What if l = 2? The magnitude of l is then equal to √[2·(2+1)]·ħ = √6·ħ ≈ 2.4495·ħ. How do we relate that to those “cocked” angles? The values of m now range from -2 to +2, with a unit distance in-between. The illustration below shows the angles. [I didn’t mention ħ any more in that illustration because, by now, we should know it’s our unit of measurement – always.]

cocked angle 2 Note we’ve got a bigger circle here (the radius is about 2.45 here, as opposed to a bit more than 1.4 for m = 0). Also note that it’s not a nice cake with perfectly equal pieces. From the graph, it’s obvious that the formula for the angle is the following: angle formula It’s simple but intriguing. Needless to say, the sin ⁻¹function is the inverse sine, also known as the arcsine. I’ve calculated the values for all m for l = 1, 2, 3, 4 and 5 below. The most interesting values are the angles for m = 1 and m = l. As the graphs underneath show, for m = 1, the values start approaching the zero angle for very large l, so there’s not much difference any more between m = ±1 and m = 1 for large values of l. What about the m = l case? Well… Believe it or not, if l becomes really large, then these angles do approach 90°. If you don’t remember how to calculate limits, then just calculate θ for some huge value for l and m. For l = m = 1,000,000, for example, you should find that θ = 89.9427…°. 🙂 angles

graph Isn’t this fascinating? I’ve actually never seen this in a textbook – so it might be an original contribution. 🙂 OK. I need to get back to the grind: Feynman’s derivation of non-symmetrical electron orbitals. Look carefully at the illustration below. If m is really the projection of some angular momentum that’s “cocked”, either at a zero-degree or, alternatively, at ±45º (for the l = 1 situation we show here) – a projection on the z-axis, that is – then the value of m (+1, 0 or -1) does actually correspond to some idea of the orientation of the space in which our electron is circulating. For m = 0, that space – think of some torus or whatever other space in which our electron might circulate – would have some alignment with the z-axis. For m = ±1, there is no such alignment. m = 0

The interpretation is tricky, however, and the illustration on the right-hand side above is surely too much of a simplification: an orbital is definitely not like a planetary orbit. It doesn’t even look like a torus. In fact, the illustration in the bottom right corner, which shows the probability density, i.e. the space in which we are actually likely to find the electron, is a picture that is much more accurate – and it surely does not resemble a planetary orbit or some torus. However, despite that, the idea that, for m = 0, we’d have some alignment of the space in which our electron moves with the z-axis is not wrong. Feynman expresses it as follows:

“Suppose m is zero, then there can be some non-zero amplitude to find the electron on the z-axis at some distance r. We’ll call this amplitude F_l(r).”

You’ll say: so what? And you’ll also say that illustration in the bottom right corner suggests the electron is actually circulating around the z-axis, rather than through it. Well… No. That illustration does not show any circulation. It only shows a probability density. No suggestion of any actual movement or circulation. So the idea is valid: if m = 0, then the implication is that, somehow, the space of circulation of current around the direction of the angular momentum vector (J), as per the well-known right-hand rule, will include the z-axis. So the idea of that electron orbiting through the z-axis for m = 0 is essentially correct, and the corollary is… Well… I’ll talk about that in a moment.

But… Well… So what? What’s so special about that F_l(r) amplitude? What can we do with that? Well… If we would find a way to calculate F_l(r), then we know everything. Huh? Everything? Yes. The reasoning here is quite complicated, so please bear with me as we go through it.

The first thing you need to accept, is rather weird. The thing we said about the non-zero amplitudes to find the electron somewhere on the z-axis for the m = 0 state – which, using Dirac’s bra-ket notation, we’ll write as |l, m = 0〉 – has a very categorical corollary:

The amplitude to find an electron whose state m is not equal to zero on the z-axis (at some non-zero distance r) is zero. We can only find an electron on the z-axis unless the z-component of its angular momentum (m) is zero.

Now, I know this is hard to swallow, especially when looking at those 45° angles for J in our illustrations, because these suggest the actual circulation of current may also include at least part of the z-axis. But… Well… No. Why not? Well… I have no good answer here except for the usual one which, I admit, is quite unsatisfactory: it’s quantum mechanics, not classical mechanics. So we have to look at the m and −m vectors, which are pointed along the z-axis itself for m = ±1 and, hence, the circulation we’d associate with those momentum vectors (even if they’re the z–component only) is around the z-axis. Not through or on it. I know it’s a really poor argument, but it’s consistent with our picture of the actual electron orbitals – that picture in terms of probability densities, which I copy below. For m = −1, we have the yz-plane as the nodal plane between the two lobes of our distribution, so no amplitude to find the electron on the z-axis (nor would we find it on the y-axis, as you can see). Likewise, for m = +1, we have the xz-plane as the nodal plane. Both nodal planes include the z-axis and, therefore, there’s zero probability on that axis. p orbitals

In addition, you may also want to note the 45° angle we associate with m = ±1 does sort of demarcate the lobes of the distribution by defining a three-dimensional cone and… Well… I know these arguments are rather intuitive, and so you may refuse to accept them. In fact, to some extent, I refuse to accept them. 🙂 Indeed, let me say this loud and clear: I really want to understand this in a better way!

But… Then… Well… Such better understanding may never come. Feynman’s warning, just before he starts explaining the Stern-Gerlach experiment and the quantization of angular momentum, rings very true here: “Understanding of these matters comes very slowly, if at all. Of course, one does get better able to know what is going to happen in a quantum-mechanical situation—if that is what understanding means—but one never gets a comfortable feeling that these quantum-mechanical rules are “natural.” Of course they are, but they are not natural to our own experience at an ordinary level.” So… Well… What can I say?

It is now time to pull the rabbit out of the hat. To understand what we’re going to do next, you need to remember that our amplitudes – or wavefunctions – are always expressed with regard to a specific frame of reference, i.e. some specific choice of an x-, y– and z-axis. If we change the reference frame – say, to some new set of x’-, y’– and z’-axes – then we need to re-write our amplitudes (or wavefunctions) in terms of the new reference frame. In order to do so, one should use a set of transformation rules. I’ve written several posts on that – including a very basic one, which you may want to re-read (just click the link here).

Look at the illustration below. We want to calculate the amplitude to find the electron at some point in space. Our reference frame is the x, y, z frame and the polar coordinates (or spherical coordinates, I should say) of our point are the radial distance r, the polar angle θ (theta), and the azimuthal angle φ (phi). [The illustration below – which I copied from Feynman’s exposé – uses a capital letter for phi, but I stick to the more usual or more modern convention here.]

change of reference frame

In case you wonder why we’d use polar coordinates rather than Cartesian coordinates… Well… I need to refer you to my other post on the topic of electron orbitals, i.e. the one in which I explain how we get the spherically symmetric solutions: if you have radial (central) fields, then it’s easier to solve stuff using polar coordinates – although you wouldn’t think so if you think of that monster equation that we’re actually trying to solve here:

new de

It’s really Schrödinger’s equation for the situation on hand (i.e. a hydrogen atom, with a radial or central Coulomb field because of its positively charged nucleus), but re-written in terms of polar coordinates. For the detail, see the mentioned post. Here, you should just remember we got the spherically symmetric solutions assuming the derivatives of the wavefunction with respect to θ and φ – so that’s the ∂ψ/∂θ and ∂ψ/∂φ in the equation above – were zero. So now we don’t assume these partial derivatives to be zero: we’re looking for states with an angular dependence, as Feynman puts it somewhat enigmatically. […] Yes. I know. This post is becoming very long, and so you are getting impatient. Look at the illustration with the (r, θ, φ) point, and let me quote Feynman on the line of reasoning now:

“Suppose we have the atom in some |l, m〉 state, what is the amplitude to find the electron at the angles θ and φ and the distance r from the origin? Put a new z-axis, say z’, at that angle (see the illustration above), and ask: what is the amplitude that the electron will be at the distance r along the new z’-axis? We know that it cannot be found along z’ unless its z’-component of angular momentum, say m’, is zero. When m’ is zero, however, the amplitude to find the electron along z’ is F_l(r). Therefore, the result is the product of two factors. The first is the amplitude that an atom in the state |l, m〉 along the z-axis will be in the state |l, m’ = 0〉 with respect to the z’-axis. Multiply that amplitude by F_l(r) and you have the amplitude ψ_l,m(r) to find the electron at (r, θ, φ) with respect to the original axes.”

So what is he telling us here? Well… He’s going a bit fast here. 🙂 Worse, I think he may actually not have chosen the right words here, so let me try to rephrase it. We’ve introduced the F_l(r) function above: it was the amplitude, for m = 0, to find the electron on the z-axis at some distance r. But so here we’re obviously in the x’, y’, z’ frame and so F_l(r) is the amplitude for m’ = 0, it’s the amplitude to find the electron on the z-axis at some distance r along the z’-axis. Of course, for this amplitude to be non-zero, we must be in the |l, m’ = 0〉 state, but are we? Well… |l, m’ = 0〉 actually gives us the amplitude for that. So we’re going to multiply two amplitudes here:

F_l(r)·|l, m’ = 0〉

So this amplitude is the product of two amplitudes as measured in the the x’, y’, z’ frame. Note it’s symmetric: we may also write it as |l, m’ = 0〉·F_l(r). We now need to sort of translate that into an amplitude as measured in the x, y, z frame. To go from x, y, z to x’, y’, z’, we first rotated around the z-axis by the angle φ, and then rotated around the new y’-axis by the angle θ. Now, the order of rotation matters: you can easily check that by taking a non-symmetrical object in your hand and doing those rotations in the two different sequences: check what happens to the orientation of your object. Hence, to go back we should first rotate about the y’-axis by the angle −θ, so our z’-axis folds into the old z-axis, and then rotate about the z-axis by the angle −φ.

Now, we will denote the transformation matrices that correspond to these rotations as R_y’(−θ) and R_z(−φ) respectively. These transformation matrices are complicated beasts. They are surely not the easy rotation matrices that you can use for the coordinates themselves. You can click this link to see how they look like for l = 1. For larger l, there are other formulas, which Feynman derives in another chapter of his Lectures on quantum mechanics. But let’s move on. Here’s the grand result:

The amplitude for our wavefunction ψ_l,m(r) – which denotes the amplitude for (1) the atom to be in the state that’s characterized by the quantum numbers l and m and – let’s not forget – (2) find the electron at r – note the bold type: r = (x, y, z) – would be equal to:

ψ_l,m(r) = 〈l, m|R_z(−φ) R_y’(−θ)|l, m’ = 0〉·F_l(r)

Well… Hmm… Maybe. […] That’s not how Feynman writes it. He writes it as follows:

ψ_l,m(r) = 〈l, 0|R_y(θ) R_z(φ)|l, m〉·F_l(r)

I am not quite sure what I did wrong. Perhaps the two expressions are equivalent. Or perhaps – is it possible at all? – Feynman made a mistake? I’ll find out. [P.S: I re-visited this point in the meanwhile: see the P.S. to this post. :-)] The point to note is that we have some combined rotation matrix R_y(θ) R_z(φ). The elements of this matrix are algebraic functions of θ and φ, which we will write as Y_l,m(θ, φ), so we write:

a·Y_l,m(θ, φ) = 〈l, 0|R_y(θ) R_z(φ)|l, m〉

Or a·Y_l,m(θ, φ) = 〈l, m|R_z(−φ) R_y’(−θ)|l, m’ = 0〉, if Feynman would have it wrong and my line of reasoning above would be correct – which is obviously not so likely. Hence, the ψ_l,m(r) function is now written as:

ψ_l,m(r) = a·Y_l,m(θ, φ)·F_l(r)

The coefficient a is, as usual, a normalization coefficient so as to make sure the surface under the probability density function is 1. As mentioned above, we get these Y_l,m(θ, φ) functions from combining those rotation matrices. For l = 1, and m = -1, 0, +1, they are: spherical harmonics A more complete table is given below: spherical harmonics 2 So, yes, we’re done. Those equations above give us those wonderful shapes for the electron orbitals, as illustrated below (credit for the illustration goes to an interesting site of the UC Davis school).But… Hey! Wait a moment! We only have these Y_l,m(θ, φ) functions here. What about F_l(r)?

You’re right. We’re not quite there yet, because we don’t have a functional form for F_l(r). Not yet, that is. Unfortunately, that derivation is another lengthy development – and that derivation actually is just tedious math only. Hence, I will refer you to Feynman for that. 🙂 Let me just insert one more thing before giving you The Grand Equation, and that’s a explanation of how we get those nice graphs. They are so-called polar graphs. There is a nice and easy article on them on the website of the University of Illinois, but I’ll summarize it for you. Polar graphs use a polar coordinate grid, as opposed to the Cartesian (or rectangular) coordinate grid that we’re used to. It’s shown below.

The origin is now referred to as the pole – like in North or South Pole indeed. 🙂 The straight lines from the pole (like the diagonals, for example, or the axes themselves, or any line in-between) measure the distance from the pole which, in this case, goes from 0 to 10, and we can connect the equidistant points by a series of circles – as shown in the illustration also. These lines from the pole are defined by some angle – which we’ll write as θ to make things easy 🙂 – which just goes from 0 to 2π = 0 and then round and round and round again. The rest is simple: you’re just going to graph a function, or an equation – just like you’d graph y = ax + b in the Cartesian plane – but it’s going to be a polar equation. Referring back to our p-orbitals, we’ll want to graph the cos²θ = ρ equation, for example, because that’s going to show us the shape of that probability density function for l = 1 and m = 0. So our graph is going to connect the (θ, ρ) points for which the angle (θ) and the distance from the pole (ρ) satisfies the cos²θ = ρ equation. There is a really nice widget on the WolframAlpha site that produces those graphs for you. I used it to produce the graph below, which shows the 1.1547·cos²θ = ρ graph (the 1.1547 coefficient is the normalization coefficient a). Now, you’ll wonder why this is a curve, or a curved line. That widget even calculates its length: it’s about 6.374743 units long. So why don’t we have a surface or a volume here? We didn’t specify any value for ρ, did we? No, we didn’t. The widget calculates those values from the equation. So… Yes. It’s a valid question: where’s the distribution? We were talking about some electron cloud or something, right?

Right. To get that cloud – those probability densities really – we need that F_l(r) function. Our cos²θ = ρ is, once again, just some kind of envelope function: it marks a space but doesn’t fill it, so to speak. 🙂 In fact, I should now give you the complete description, which has all of the possible states of the hydrogen atom – everything! No separate pieces anymore. Here it is. It also includes n. It’s The Grand Equation:The a_k coefficients in the formula for ρF_n,l(ρ) are the solutions to the equation below, which I copied from Feynman’s text on it all. I’ll also refer you to the same text to see how you actually get solutions out of it, and what they then actually represent. 🙂We’re done. Finally!

I hope you enjoyed this. Look at what we’ve achieved. We had this differential equation (a simple diffusion equation, really, albeit in the complex space), and then we have a central Coulomb field and the rather simple concept of quantized (i.e. non-continuous or discrete) angular momentum. Now see what magic comes out of it! We literally constructed the atomic structure out of it, and it’s all wonderfully elegant and beautiful.

Now I think that’s amazing, and if you’re reading this, then I am sure you’ll find it as amazing as I do.

Note: I did a better job in explaining the intricacies of actually representing those orbitals in a later post. I recommend you have a look at it by clicking the link here.

Post scriptum on the transformation matrices:

You must find the explanation for that 〈l, 0|R_y(θ) R_z(φ)|l, m〉·F_l(r) product highly unsatisfactory, and it is. 🙂 I just wanted to make you think – rather than just superficially read through it. First note that F_l(r)·|l, m’ = 0〉 is not a product of two amplitudes: it is the product of an amplitude with a state. A state is a vector in a rather special vector space – a Hilbert space (just a nice word to throw around, isn’t it?). The point is: a state vector is written as some linear combination of base states. Something inside of me tells me we may look at the three p-states as base states, but I need to look into that.

Let’s first calculate the R_y(θ) R_zmatrix to see if we get those formulas for the angular dependence of the amplitudes. It’s the product of the R_y(θ) and R_zmatrices, which I reproduce below.

Note that this product is non-commutative because… Well… Matrix products generally are non-commutative. 🙂 So… Well… There they are: the second row gives us those functions, so I am wrong, obviously, and Dr. Feynman is right. Of course, he is. He is always right – especially because his Lectures have gone through so many revised editions that all errors must be out by now. 🙂

However, let me – just for fun – also calculate my R_z(−φ) R_y’(−θ) product. I can do so in two steps: first I calculate R_z(φ) R_y’(θ), and then I substitute the angles φ and θ for –φ and –θ, remembering that cos(–α) = cos(α) and sin(–α) = –sin(α). I might have made a mistake, but I got this:The functions look the same but… Well… No. The e^iφ and e^−iφ are in the wrong place (it’s just one minus sign – but it’s crucially different). And then these functions should not be in a column. That doesn’t make sense when you write it all out. So Feynman’s expression is, of course, fully correct. But so how do we interpret that 〈l, 0|R_y(θ) R_z(φ)|l, m〉 expression then? This amplitude probably answers the following question:

Given that our atom is in the |l, m〉 state, what is the amplitude for it to be in the 〈l, 0| state in the x’, y’, z’ frame?

That makes sense – because we did start out with the assumption that our atom was in the the |l, m〉 state, so… Yes. Think about it some more and you’ll see it all makes sense: we can – and should – multiply this amplitude with the F_l(r) amplitude.

OK. Now we’re really done with this. 🙂

Note: As for the 〈 | and | 〉 symbols to denote a state, note that there’s not much difference: both are state vectors, but a state vector that’s written as an end state – so that’s like 〈 Φ | – is a 1×3 vector (so that’s a column vector), while a vector written as | Φ 〉 is a 3×1 vector (so that’s a row vector). So that’s why 〈l, 0|R_y(θ) R_z(φ)|l, m〉 does give us some number. We’ve got a (1×3)·(3×3)·(3×1) matrix product here – but so it gives us what we want: a 1×1 amplitude. 🙂

The state(s) of a photon

Original post:

While hurrying to try to understand the things I wanted to understand most – like Schrödinger’s equation and, equally important, its solutions explaining the weird shapes of electron orbitals – I skipped some interesting bits and pieces. Worse, I skipped two or three of Feynman’s Lectures on quantum mechanics entirely. These include Chapter 17 – on symmetry and conservation laws – and Chapter 18 – on angular momentum. With the benefit of hindsight, that was not the right thing to do. If anything, doing all of the Lectures would, at the very least, ensure I would have more than an ephemeral grasp of it all. So… In this and the next post, I want to tidy up and go over everything I skipped so far. 🙂

We’ve written a lot on how quantum mechanics applies to both bosons as well as fermions. For example, we pointed out – in very much detail – that the mathematical structure of the electromagnetic wave – light! 🙂 – is quite similar to that of the ubiquitous wavefunction. Equally fundamental – if not more – is the fact that light also arrives in lumps – little light-particles which we call photons. It’s the photoelectric effect, which Einstein explained in 1905 by… Well… By telling us that light consists of quanta – photons – whose energy must be high enough so as to be able to dislodge an electron. It’s what got him his Nobel Prize. [Einstein never got a Nobel Prize for his relativity theory, which is – arguably – at least as important. There’s a lot of controversy around that but, in any case, that’s history.]

So it shouldn’t surprise you that there’s an equivalent to the spin of an electron. With spin, we refer to the angular momentum of a quantum-mechanical system – an atom, a nucleus, an electron, whatever – which, as you know, can only be one of a set of discrete values when measured along some direction, which we usually refer to as the z-direction. More formally, we write that the z-component of the angular moment J is equal to

J_z = j·ħ, (j-1)·ħ, (j-2)·ħ, …, -(j-2)·ħ, -(j-1)·ħ, –j·ħ

The j in this expression is the so-called spin of the system. For an electron, it’s equal to ±1/2, which we referred to as “up” and “down” states respectively because of obvious reasons: one state points upwards – more or less, that is (we know the angular momentum will actually precess around the direction of the magnetic field) – while the other points downwards.

We also know that the magnetic energy of an electron in a (weak) magnetic field – which, as you know, we conveniently assume to be pointing in the same z-direction, so B_z= B – will be equal to:

U_mag = g·μ_z·B·j = ± 2·μ_z·B·(1/2) = ± μ_z·B = ± B·(q_e·ħ)/(2m)

In short, the magnetic energy is proportional to the magnetic field, and the constant of proportionality is the so-called Bohr magneton q_e·ħ/2m. So far, so good. What’s the analog for a photon?

Well… Let’s first discuss the equivalent of a Stern-Gerlach apparatus for photons. That would be a polarizing material, like a piece of calcite, for example. Now, it is, unfortunately, much more difficult to explain how a polarizing material works than to explain how a Stern-Gerlach apparatus works. [If you thought the workings of that (hypothetical) Stern-Gerlach filter were difficult to understand, think again.] We actually have different types of polarizers – some complicated, some easy. We’ll take the easy ones: linear ones. In addition, the phenomenon of polarization itself is a bit intricate. The phenomenon is well described in Chapter 33 of Feynman’s first Volume of Lectures, out of which I copied the two illustrations below the next paragraph.

Of course, to make sure you think about whatever is that you’re reading, Feynman now chooses the z-direction such that it coincides with the direction of propagation of the electromagnetic radiation. So it’s now the x– and y-direction that we’re looking at. Not the z-direction any more. As usual, we forget about the magnetic field vector B and so we think of the oscillating electric field vector E only. Why can we forget about B? Well… If we have E, we know B. Full stop. As you know, I think B is pretty essential in the analysis too but… Well… You’ll see all textbooks on physics quickly forget about B when describing light. I don’t want to do that, but… Well… I need to move on. [I’ll come back to the matter – sideways – at the end of this post. :-)]

So we know the electric field vector E may oscillate in a plane (so that’s up and down and back again) but – interestingly enough – its direction may also rotate around the z-axis (again, remember the z-axis is the direction of propagation). Why? Well… Because E has an x– and a y-component (no z-component!), and these two components may oscillate in phase or out of phase, and so all of the combinations below are possible. Linear polarization Elliptical polarization To make a long story short, light comes in two varieties: linearly polarized and elliptically polarized. Of course, elliptically may be circularly – if you’re lucky! 🙂

Now, a (linear) polarizer has an optical axis, and only light whose E vector is oscillating along that axis will go through. […] OK. That’s not true: the component along the optical axis of some E pointing in some other direction will go through too! I’ll show how that works in a moment. But so all the rest is absorbed, and the absorbed energy just heats up the polarizer (which, of course, then radiates heat back out).

In any case, if the optical axis happens to be our x-axis, then we know that the light that comes through will be x-polarized, so that corresponds to the rather peculiar E_x = 1 and E_y = 0 notation. [This notation refers to coefficients we’ll use later to resolve states into base states – but don’t worry about it now.] Needless to say, you shouldn’t confuse the electric field vector E with the energy of our photon, which we denote as E. No bold letter here. No subscript. 🙂

Pfff… This introduction is becoming way too long. What about our photon? We want to talk about one photon only and we’ve already written over a page and haven’t started yet. 🙂

Well… First, we must note that we’ll assume the light is perfectly monochromatic, so all photons will have an energy that’s equal to E = h·f, so the energy is proportional to the frequency of our light, and the constant of proportionality is Planck’s constant. That’s Einstein’s relation, not a de Broglie relation. Just remember: we’re talking definite energy states here.

Second – and much more importantly – we may define two base states for our photon, |x〉 and |y〉 respectively, which correspond to the classical linear x– and y-polarization. So a photon can be in state |x〉 or |y〉 but, as usual, it is much more likely to be in some state that is some linear combination of these two base states.

OK. Now we can start playing with these ideas. Imagine a polarizer – or polaroid, as Feynman calls it – whose optical axis is tilted – say, it’s at an angle θ from the x-axis, as shown below. Classically, the light that comes through will be polarized in the x’-direction, which we associate with that angle θ. So we say the photons will be in the |x‘〉 state. So far, so good. But what happens if we have two polarizers, set up as shown below, with the optical axis of the first one at an angle θ, which is, say, equal to 30°? Will any light get through? two polarizers

Well? No answer? […] Think about it. What happens classically? […] No answer? Let me tell you. In a classical analysis, we’d say that only the x-component of the light that comes through the first polarizer would get through the second one. Huh? Yes. It is not all or nothing in a classical analysis. This is where the magnitude of E comes in, which we’ll write as E₀, so as to not confuse it with the energy E. [I know you’ll confuse it anyway but… Well… I need to move on or I won’t get anywhere with this story.] So if E₀ is the (maximum) magnitude (or amplitude – in the classical sense of the word, that is) of E as the light leaves the first polarizer, then its x-component will be equal to E₀·cosθ. [I don’t need to make a drawing here, do I?] Of course, you know that the intensity of the light will be proportional to the square of the (maximum) field, which is equal to E₀²·cos²θ = 0.75·E₀² for θ = 30°.

So our classical theory says that only 3/4 of the energy that we were sending in will get through. The rest (1/4) will be absorbed. So how do we model that quantum-mechanically? It’s amazingly simple. We’ve already associated the |x‘〉 state with the photons coming out of the first polaroid, and so now we’ll just say that this |x‘〉 state is equal to the following linear combination of the |x〉 and |y〉 base states:

|x‘〉 = cosθ·|x〉 + sinθ·|y〉

〈x|x‘〉 = cosθ·〈x|x〉 + sinθ·〈x|y〉

〈x|x‘〉 = cosθ

Lo and behold! The absolute square of that is equal to cos²θ, so each of these photons have an (average) probability of 3/4 to get through. So if we were to have like 10 billion photons, then some 7.5 billion of them would get through. As these photons are all associated with a definite energy – and they go through as one whole, of course (no such thing as a 3/4 photon!) – we find that 3/4 of all of the energy goes through. The quantum-mechanical theory gives the same result as the classical theory – as it should, in this case at least!

Now that’s all good for linear polarization. What about elliptical or circular polarization? Hmm… That’s a bit more complicated, but equally feasible. If we denote the state of a photon with a right-hand circular polarization (RHC) as |R〉 and, likewise, the state of a photon with a left-hand circular polarization (LHC) as |L〉, then we can write these as the following linear combinations of our base states |x〉 and |y〉: linear combination RHC and LHC That’s where those coefficients under illustrations (c) and (g) come in, although I think they’ve got the sign of i (the imaginary unit) wrong. 🙂 So how does it work? Well… That 1/√2 factor is – obviously – just there to make sure everything’s normalized, so all probabilities over all states add up to 1. So that is taken care of and now we just need to explain how and why we’re adding |x〉 and |y〉. For |R〉, the amplitudes must be the same but with a phase difference of 90°. That corresponds to the sine and cosine function, which are the same except for a phase difference of π/2 (90°), indeed: sin(φ + π/2) = cosφ. Now, a phase shift of 90° corresponds to a multiplication with the imaginary unit i. Indeed, i = e^i·π/2 and, therefore, it is obvious that e^i·π/2·e^i·φ = e^{i·(φ + π/2)}.

Of course, if we can write RHC and LHC states as a linear combination of the base states |x〉 and |y〉, then you’ll believe me if I say that we can write any polarization state – including non-circular elliptical ones – as a linear combination of these base states. Now, there are two or three other things I’d like to point out here:

1. The RHC and LHC states can be used as base states themselves – so they satisfy all of the conditions for a set of base states. Indeed, it’s easy to add and then subtract the two equations above to get the following: new base set As an exercise, you should verify the right and left polarization states effectively satisfy the conditions for a set of base states.

2. We can also rotate the xy-plane around the z-axis (as mentioned, that’s the direction of propagation of our beam) and use the resulting |x‘〉 and |y‘〉 states as base states. In short, we can effectively, as Feynman puts it, “You can resolve light into x– and y– polarizations, or into x’– and y’-polarizations, or into right and left polarizations as a basis.” These pairs are always orthogonal and also satisfy the other conditions we’d impose on a set of base states.

3. The last point I want to make here is much more enigmatic but, as far as I am concerned – by far – the most interesting of all of Feynman’s Lecture on this topic. It’s actually just a footnote, but I am very excited about it. So… Well… What is it?

Well… Feynman does the calculations to show how a circularly polarized photon looks like when we rotate the coordinates around the z-axis, and shows the phase of the right and left polarized states effectively keeps track of the x– and y-axes, so all of our “right-hand” rules don’t get lost somehow. He compares this analysis to an analysis he actually did – in a much earlier Lecture (in Chapter 5) – for spin-one particles. But, of course, here we’ve been analyzing the photon as a two-state system, right?

So… Well… Don’t we have a contradiction here? If photons are spin-one particles, then they’re supposed to be analyzed in terms of three base states, right? Well… I guess so… But then Feynman adds a footnote – with a very important remark:

“The photon is a spin-one particle which has, however, no ‘zero’-state.”

Why I am noting that? Because it confirms my theory about photons – force-particles – being different from matter-particles not only because of the different rules for adding amplitudes, but also because we get two wavefunctions for the price of one and, therefore, twice the energy for every oscillation! And so we’ll also have a distance of two Planck units between the equivalent of the “up” and “down” states of the photon, rather than one Planck unit, like what we have for the angular momentum for an electron.

I described the gist of my argument in my e-book, which you’ll find under another tab of this blog, and so I’ll refer you there. However, in case you’re interested, the summary of the summary is as follows:

We can think of a photon having some energy that’s equal to E = p = m (assuming we choose our time and distance units such that c = 1), but that energy would be split up in an electric and a magnetic wavefunction respectively: ψ_Eand ψ_B.
Now, Schrödinger’s equation would then apply to both wavefunctions, but the E, p and m in those two wavefunctions are the same and not the same: their numerical value is the same (p_E =E_E = m_E = p_B =E_B = m_B), but they’re conceptually different. [They must be: I showed that, if they aren’t, then we get a phase and group velocity for the wave that doesn’t make sense.]

It is then easy to show that – using the B = i·E relation between the magnetic and the electric field vectors – we find a composite wavefunction for our photon which we can write as:

E + B = ψ_E+ ψ_B= E + i·E = √2·eⁱ^{(p·x/2 − E·t/2 + π/4)}= √2·eⁱ^(π/4)·eⁱ^{(p·x/2 − E·t/2)}= √2·eⁱ^(π/4)·E

The whole thing then becomes:

ψ = ψ_E+ ψ_B= √2·eⁱ^{(p·x/2 − E·t/2 + π/4)} = √2·eⁱ^(π/4)·eⁱ^{(p·x/2 − E·t/2)}

So we’ve got a √2 factor here in front of our combined wavefunction for our photon which, knowing that the energy is proportional to the square of the amplitude gives us twice the energy we’d associate with a regular amplitude… [With “regular”, I mean the wavefunction for matter-particles – fermions, that is.] So… Well… That little footnote of Feynman seems to confirm I really am on to something. Nice! Very nice, actually! 🙂

Comments on the MIT’s Stern-Gerlach lab experiment

In my previous post, I noted that I’d go through the MIT’s documentation on the Stern-Gerlach experiment that their undergrad students have to do, because we should now – after 175 posts on quantum physics 🙂 – be ready to fully understand what is said in there. So this post is just going to be a list of comments. I’ll organize it section by section.

Theory of atomic beam experiments

The theory is known – and then it isn’t, of course. The key idea is that individual atoms behave like little magnets. Why? In the simplest and most naive of models, it’s because the electrons somehow circle around the nucleus. You’ve seen the illustration below before. Note that current is, by convention, the flow of positive charge, which is, of course, opposite to the electron flow. You can check the direction by applying the right-hand rule: if you curl the fingers of your right hand in the direction of the current in the loop (so that’s opposite to v), your thumb will point in the direction of the magnetic moment (μ). orbital-angular-momentum So the electron orbit – in whatever way we’d want to visualize it – gives us L, which we refer to as the orbital angular momentum. We know the electron is also supposed to spin about its own axis – even if we know this planetary model of an electron isn’t quite correct. So that gives us a spin angular momentum S. In the so-called vector model of the atom, we simply add the two to get the total angular momentum J = L + S.

Of course, now you’ll say: only hydrogen has one electron, so how does it work with multiple electrons? Well… Then we have multiple orbital angular momentum l_i which are to be added to give a total orbital angular momentum L. Likewise, the electrons spins s_i can also be added to give some total spin angular momentum S. So we write:

J = L + S with L = Σ_i l_i and S = Σ_i s_i

Really? Well… If you’d google this to double-check – check the Wikipedia article on it, for example – then you’ll find this additivity property is valid only for relatively light atoms (Z ≤ 30) and only if any external magnetic field is weak enough. The way individual orbital and spin angular momenta have to be combined so as to arrive at some total L, S and J is referred to a coupling scheme: the additivity rule above is referred to as LS coupling, but one may also encounter LK coupling, or jj coupling, or other stuff. The US National Institute of Standards and Technology (NIST) has a nice article on all these models – but we need to move on here. Just note that we do assume the LS coupling scheme applies to our potassium beam – because its atomic number (Z) is 19, and the external magnetic field is assumed to be weak enough.

The vector model of the atom describes the atom using angular momentum vectors. Of course, we know that a magnetic field will cause our atomic magnet to precess – rather than line up. At this point, the classical analogy between a spinning top – or a gyroscope – and our atomic magnet becomes quite problematic. First, think about the implications for L and S when assuming, as we usually do, that J precesses nicely about an axis that is parallel to the magnetic field – as shown in the illustration below, which I took from Feynman’s treatment of the matter. If J is the sum of two other vectors L and S, then this has rather weird implications for the precession of L and S, as shown in the illustration below – which I took from the Wikipedia article on LS coupling. Think about: if L and S are independent, then the axis of precession for these two vectors should be just the same as for J, right? So their axis of precession should also be parallel to the magnetic field (B), so that’s the direction of the J_z component, which is just the z-axis of our reference frame here. 375px-ls_coupling More importantly, our classical model also gets into trouble when actually measuring the magnitude of J_z: repeated measurements will not yield some randomly distributed continuous variable, as one would classically expect. No. In fact, that’s what this experiment is all about: it shows that J_z will take only certain quantized values. That is what is shown in the illustration below (which once again assumes the magnetic field (B) is along the z-axis). vector-model I copied the illustration above from the HyperPhysics site, because I found it to be enlightening and intriguing at the same time. First, it also shows this rather weird implication of the vector model: if J continually changes direction because of its precession in a weak magnetic field, then L and S must, obviously, also continually change direction. However, this illustration is even more intriguing than the Wikipedia illustration because it assumes the axis of precession of L and S and L actually the same!

So what’s going on here? To better understand what’s going on, I started to read the whole HyperPhysics article on the vector model, which also includes the illustration below, with the following comments: “When orbital angular momentum L and electron spin S are combined to produce the total angular momentum of an atomic electron, the combination process can be visualized in terms of a vector model. Both the orbital and spin angular momenta are seen as precessing about the direction of the total angular momentum J. This diagram can be seen as describing a single electron, or multiple electrons for which the spin and orbital angular momenta have been combined to produce composite angular momenta S and L respectively. In so doing, one has made assumptions about the coupling of the angular momenta which are described by the LS coupling scheme which is appropriate for light atoms with relatively small external magnetic fields.” vector-model-2 Hmm… What about those illustrations on the right-hand side – with the vector sums and those values for j and m_j? I guess the idea may also be illustrated by the table below: combining different values for l (±1) and s (±1/2) gives four possible values, ranging from +3/2 to -1/2, for j = l + s. table Having said that, the illustration raises a very fundamental question: the length of the sum of two vectors is definitely not the same as the sum of the length of the two vectors! So… Well… Hmm… Something doesn’t make sense here! However, I can’t dwell any longer on this. I just wanted to note you should not take all that’s published on those oft-used sites on quantum mechanics for granted. But so I need to move on. Back to the other illustration – copied once more below. vector-model We have that very special formula for the magnitude (J) of the angular momentum J:

│J│= J = √(J·J) = √[j·(j+1)·ħ²] = √[j·(j+1)]·ħ

So if j = 3/2, then J is equal to √[(3/2)·(3/2+1)]·ħ = √[(15/4)·ħ ≈ 1.9635·ħ, so that’s almost 2ħ. 🙂 At the same time, we know that for j = 3/2, the possible values of J_z can only be +3ħ/2, +ħ/2, -ħ/2, and -3ħ/2. So that’s what’s shown in that half-circular diagram: the magnitude of J is larger than its z-component – always!

OK. Next. What’s that 3p_3/2 notation? Excellent question! Don’t think this 3p denotes an electron orbital, like 1s or 3d – i.e. the orbitals we got from solving Schrödinger’s equation. No. In fact, the illustration above is somewhat misleading because the correct notation is not 3p_3/2 but ³P_3/2. So we have a capital P which is preceded by a superscript 3. This is the notation for the so-called term symbol for a nuclear, atomic or molecular (ground) state which – assuming our LS coupling model is valid – because we’ve got other term symbols for other coupling models – we can write, more generally, as:

^2S+1L_J

The J, L and S in this thing are the following:

1. The J is the total angular momentum quantum number, so it is – the notation gets even more confusing now – the j in the │J│= J = √(J·J) = √[j·(j+1)·ħ²] = √[j·(j+1)]·ħ expression. We know that number is 1/2 for electrons, but it may take on other values for nuclei, atoms or molecules. For example, it is 3/2 for nitrogen, and 2 for oxygen, for which the corresponding terms are ⁴S_3/2 and ³P₂ respectively.

2. The S in the term symbol is the total spin quantum number, and 2S+1 itself is referred to as the fine-structure multiplicity. It is not an easy concept. Just remember that the fine structure describes the splitting of the spectral lines of atoms due to electron spin. In contrast, the gross structure energy levels are those we effectively get from solving Schrödinger’s equation assuming our electrons have no spin. We also have a hyperfine structure, which is due to the existence of a (small) nuclear magnetic moment, which we do not take into consideration here, which is why the ⁴S_3/2 and ³P₂ terms are sometimes being referred to as describing electronic ground states. In fact, the MIT lab document, which we are studying here, refers to the ground state of the potassium atoms in the beam as an electronic ground state, which is written up as ²S_1/2. So S is, effectively, equal to 1/2. [Are you still there? If so, just write it down: 2S+1 = 2 ⇒ S = 1/2. That means the following: our potassium atom behaves like an electron: its spin is either ‘up’ or, else, it is ‘down’. There is no in-between.]

3. Finally, the L in the term symbol is the total orbital angular momentum quantum number but, rather than using a number, the values of L are often represented as S, P, D, F, etcetera. This L number is very confusing because – as mentioned above – one would think it represents those s, p, d, f, g,… orbitals. However, that is not the case. The difference may easily be illustrated by observing that a carbon atom, for example, has six electrons, which are distributed over the 1s, 2s and 2p orbitals (one pair each). However, its ground state only gets one L number: L = P. Hence, its value is 1. Of course, now you will wonder how we get that number.

Well… I wish I could give you an easy answer, but I can’t. For two p electrons – think of our carbon atom once again – we can have L = 0, 1 or 2, or S, P and D. They effectively correspond to different energy levels, which are related to the way these two electrons interact with each other. The phenomenon is referred to as angular momentum coupling. In fact, all of the numbers we discussed so far – J, S and L – are numbers resulting from angular momentum coupling. As Wikipedia puts it: “Angular momentum coupling refers to the construction of eigenstates of total angular momentum out of eigenstates of separate angular momentum.” [As you know, each eigenstate corresponds to an energy level, of course.]

Now that should clear some of the confusion on the ^2S+1L_J notation: the capital letters J, S and L refer to some total, as opposed to the quantum numbers you are used to, i.e. n, l, m and s, i.e. the so-called principal, orbital, magnetic and spin quantum number respectively. The lowercase letters are quantum numbers that describe an electron in an atom, while those capital letters denote quantum numbers describing the atom – or a molecule – itself.

OK. Onwards. But where were we? 🙂 Oh… Yes. That J = L + S formula gives us some total electronic angular momentum, but we’ll also have some nuclear angular momentum, which our MIT paper denotes as I. Our vector model of our potassium atom allows us, once again, to simply add the two to get the total angular momentum, which is written as F = J + I = L + S + I. This, then, explains why the MIT experiment writes the magnitude of the total angular momentum as:

magnitude-of-f

Of course, here I don’t need to explain – or so I hope – why this quantum-mechanical formula for the calculation of the magnitude is what it is (or, equivalently, why the usual Euclidean metric – i.e. √(x² + y² + z²) – is not to be used here. If you do need an explanation, you’ll need to go through the basics once again.

Now, the whole point, of course, is that the z-component of F can have only the discrete values that are specified by the F_z = m_f·ħ equation, with m_f – i.e. the (total) magnetic quantum number – having an equally discrete value equal to m_f = −f, −(f−1), …, +(f+1), f.

For the rest, I probably shouldn’t describe the experiment itself: you know it. But let me just copy the set-up below, so it’s clear what it is that we’re expecting to happen. In addition, you’ll also need the illustration below because I’ll refer to those d₁ and d₂ distances shown in what follows. set-up

Note the MIT documentation does spell out some additional assumptions. Most notably, it says that the potassium atoms that emerge from the oven (at a temperature of 200°) will be:

(1) almost exclusively in the ground electronic state,

(2) nearly equally distributed among the two (magnetic) sub-states characterized by f, and, finally,

(3) very nearly equally distributed among the hyperfine states, i.e. the states with the same f but with different m_f.

I am just noting these assumptions because it is interesting to note that – according to the man or woman who wrote this paper – we would actually have states within states here. The paper states that the hyperfine splitting of the two sub-beams we expect to come out of the magnet can only be resolved by very advanced atomic beam techniques, so… Well… That’s not the apparatus that’s being used for this experiment.

However, it’s all a bit weird, because the paper notes that the rules for combining the electronic and nuclear angular momentum – using that F = J + I = L + S + I formula – imply that our quantum number f = i ± j can be either 1 or 2. These two values would be associated with the following m_f and m_fforce values:

f = 1 ⇒ F_z = m_f·ħ = −ħ, 0 or +ħ (so we’d have three beams here)

f = 2 ⇒ F_z = m_f·ħ = −2ħ, −ħ, 0, +ħ or +2ħ (so we’d have five beams here)

Neither of the two possibilities relates to the situation at hand – which assumes two beams only. In short, I think the man or women who wrote the theoretical introduction – an assistant professor, most likely (no disrespect here: that’s how far I progressed in economics – nothing more, nothing less) – might have made a mistake. Or perhaps he or she may have wanted to confuse us.

I’ll look into it over the coming days. As for now, all you need to know – please jot it down! – is that our potassium atom is fully described by ²S_1/2. That shorthand notation has all the quantum number we need to know. Most importantly, it tells us S is, effectively, equal to 1/2. So… Well… That ²S_1/2 notation tells us our potassium atom should behave like an electron: its spin is either ‘up’ or ‘down’. No in-between. 🙂 So we should have two beams. Not three or five. No fine or hyperfine sub-structures! 🙂 In any case, the rest of the paper makes it clear the assumption is, effectively, that the angular momentum number is equal to j = 1/2. So… Two beams only. 🙂

How to calculate the expected deflection

We know that the inhomogeneous magnetic field (B), whose direction is the z-axis, will result in a force, which we have calculated a couple of times already as being equal to:In case you’d want to check this, you can check one of my posts on this. I just need to make one horrifying remark on notation here: while the same symbol is used, the force F_zis, obviously, not to be confused with the z-component of the angular momentum F = J + I = L + S + I that we described above. Frankly, I hope that the MIT guys have corrected that in the meanwhile, because it’s really terribly confusing notation! In any case… Let’s move on.

Now, we assume the deflecting force is constant because of the rather particular design of the magnet pole pieces (see Appendix I of the paper). We can then use Newton’s Second Law (F = m·a) to calculate the velocity in the z-direction, which is denoted by V_z (I am not sure why a capital letter is used here, but that’s not important, of course). That velocity is assumed to go from 0 to its final value V_z while our potassium atom travels between the two magnet poles but – to be clear – at any point in time, V_z will increase linearly – not exponentially – so we can write: V_z = a·t₁, with t₁ the time that is needed to travel through the magnet. Now, the relevant mass is the mass of the atom, of course, which is denoted by M. Hence, it is easy to see that a = F_z/M = V_z/t₁. Hence, we find that V_z= F_z·t₁/M.

Now, the vertical distance traveled (z) can be calculated by solving the usual integral: z = ∫₀^t₁v(t)·dt = ∫₀^t₁a·t·dt = a·t₁²/2 = (V_z/t₁)·t₁²/2 = V_z·t₁/2. Of course, once our potassium atom comes out of the magnetic field, it will continue to travel upward or downward with the same velocity V_z, which adds V_z·t₂ to the total distance traveled along the z-direction. Hence, the formula for the deflection is, effectively, the one that you’ll find in the paper:

z = V_z·t₁/2 + V_z·t₂= V_z·(t₁/2 + t₂)

Now, the travel times depend on the velocity of our potassium atom along the y-axis, which is approximated by equating it with │V│= V, because the y-component of the velocity is easily the largest – by far! Hence, t₁ = d₁/V and t₂ = d₂/V. Some more manipulation will then give you the expression we need, which is a formula for the deflection in terms of variables that we actually know:

z-formula

Statistical mechanics

We now need to combine this with the Maxwell-Boltzmann distribution for the velocities we gave you in our previous post: formula-a The next step is to use this formula so as to be able to calculate a distribution which would describe the intensity of the beam. Now, it’s easy to understand such intensity will be related to the flux of potassium atoms, and it’s equally easy to get that a flux is defined as the rate of flow per unit area. Hmm… So how does this get us the formula below? density-formula-2 The tricky thing – of course – is the use of those normalized velocities because… Well… It’s easy to see that the right-hand side of the equation above – just forget about the d(V/V₀ ) bit for a second, as we have it on both sides of the equation and so it cancels out anyway – is just density times velocity. We do have a product of the density of particles and the velocity with which they emerge here – albeit a normalized velocity. But then… Who cares? The normalization is just a division by V₀– or a multiplication by 1/V₀, which is some constant. From a math point of view, it doesn’t make any difference: our variable is V/V₀ instead of V. It’s just like using some other unit. No worries here – as long as you use the new variable consistently everywhere. 🙂

Alright. […] What’s next? Well… Nothing much. The only thing that we still need to explain now is that factor 2. It’s easy to see that’s just a normalization factor – just like that 4/√π factor in the first formula. So we get it from imposing the usual condition: density So… What’s next… Well… We’re almost there. 🙂 As the MIT paper notes, the f(V) and I(V/V₀) functions can be mapped to each other: the related transformation maps a velocity distribution to an intensity distribution – i.e. a distribution of the deflection – and vice versa.

Now, the rest of the paper is just a lot of algebraic manipulations – distinguishing the case of a quantized F_zversus a continuous F_z. Here again, I must admit I am a bit shocked by the mix-up of concepts and symbols. The paper talks about a quantized deflecting force – while it’s obvious we should be talking a quantized angular momentum. The two concepts – and their units – are fundamentally different: the unit in which angular momentum is measured is the action unit: newton·meter·second (N·m·s). Force is just force: x newton.

Having said that, the mix-up does trigger an interesting philosophical question: what is quantized really? Force (expressed in N)? Energy (expressed in N·m)? Momentum (expressed in N·s)? Action (expressed in N·m·s, i.e. the unit of angular momentum)? Space? Time? Or space-time – related through the absolute speed of light (c)? Three factors (force, distance and time), six possibilities. What’s your guess?

[…]

What’s my guess? Well… The formulas tell us the only thing that’s quantized is action: Nature itself tells us we have to express it in terms of Planck units. However, because action is a product involving all of these factors, with different dimensions, the quantum-mechanical quantization of action can, obviously, express itself in various ways. 🙂

The quantization of magnetic moments

Original post:

You may not have many questions after a first read of Feynman’s Lecture on the Stern-Gerlach experiment and his more general musings on the quantization of the magnetic moment of an elementary particle. [At least I didn’t have all that many after my first reading, which I summarized in a previous post.]

However, a second, third or fourth reading should trigger some, I’d think. My key question is the following: what happens to that magnetic moment of a particle – and its spin [1] – as it travels through a homogeneous or inhomogeneous magnetic field? We know – or, to be precise, we assume – its spin is either “up” (J_z = +ħ/2) or “down” (J_z = −ħ/2) when it enters the Stern-Gerlach apparatus, but then – when it’s moving in the field itself – we would expect that the magnetic field would, somehow, line up the magnetic moment, right?

Feynman says that it doesn’t: from all of the schematic drawings – and the subsequent discussion of Stern-Gerlach filters – it is obvious that the magnetic field – which we denote as B, and which we assume to be inhomogeneous [2] – should not result in a change of the magnetic moment. Feynman states it as follows: “The magnetic field produces a torque. Such a torque you would think is trying to line up the (atomic) magnet with the field, but it only causes its precession.”

[…] OK. That’s too much information already, I guess. Let’s start with the basics. The key to a good understanding of this discussion is the force formula:

We should first explain this formula before discussing the obvious question: over what time – or over what distance – should we expect this force to pull the particle up or down in the magnetic field? Indeed, if the force ends up aligning the moment, then the force will disappear!

So let’s first explain the formula. We start by explaining the energy U. U is the potential energy of our particle, which it gets from its magnetic moment μ and its orientation in the magnetic field B. To be precise, we can write the following:

Of course, μ and B are the magnitudes of μ and B respectively, and θ is the angle between μ and B: if the angle θ is zero, then U_mag will be negative. Hence, the total energy of our particle (U) will actually be less than what it would be without the magnetic field: it is the energy when the magnetic moment of our particle is fully lined up with the magnetic field. When the angle is a right angle (θ = ±π/2), then the energy doesn’t change (U_mag = 0). Finally, when θ is equal to π or −π, then its energy will be more than what it would be outside of the magnetic field. [Note that the angle θ effectively varies between –π and π – not between 0 and 2π!] angles Of course, we may already note that, in quantum mechanics, U_mag will only take on a very limited set of values. To be precise, for a particle with spin number j = 1/2, the possible values of U_mag will be limited to two values only. We will come back to that in a moment. First that force formula.

Energy is force over a distance. To be precise, when a particle is moved from point a to point b, then its change in energy can be written as the following line integral:

Note that the minus sign is there because of the convention that we’re doing work against the force when increasing the (potential) energy of that what we’re moving. Also note that F∙ds product is a vector (dot) product: it is, obviously, equal to F_t times ds, with F_t the magnitude of the tangential component of the force. The equation above gives us that force formula:

Feynman calls it the principle of virtual work, which sounds a bit mysterious – but so you get it by taking the derivative of both sides of the energy formula.

Let me now get back to the real mystery of quantum mechanics, which tells us that the magnetic moment – as measured along our z-axis – will only take one of two possible values. To be precise, we have the following formula for μ_z:

This is a formula you just have to accept for the moment. It needs a bit of interpretation, and you need to watch out for the sign. The g-factor is the so-called Landé g-factor: it is equal to 1 for a so-called pure orbital moment, 2 for a so-called pure spin moment, and some number in-between in reality, which is always some mixture of the two: both the electron’s orbit around the nucleus as well as the electron’s rotation about its own axis contribute to the total angular momentum and, hence, to the total magnetic moment of our electron. As for the other factors, m and q_e are, of course, the mass and the charge of our electron, and J_z is either +ħ/2 or −ħ/2. Hence, if we know g, we can easily calculate the two possible values for μ_z.

Now, that also means we could – theoretically – calculate the two possible values of that angle θ. For some reason, no handbook in physics ever does that. The reason is probably a good one: electron orbits, and the concept of spin itself, are not like the orbit and the spin of some planet in a planetary system. In fact, we know that we should not think of electrons like that at all: quantum physicists tell us we may only think of it as some kind of weird cloud around a center. That cloud has a density which is to be calculated by taking the absolute square of the quantum-mechanical amplitude of our electron.

In fact, when thinking about the two possible values for θ, we may want to remind ourselves of another peculiar consequence of the fact that the angular momentum – and, hence, the magnetic moment – is not continuous but quantized: the magnitude of the angular momentum J is not J = √(J·J) = √J² in quantum mechanics but J = √(J·J) = √[j·(j+1)·ħ²] = √[j·(j+1)]·ħ. For our electron, j = 1/2 and, hence, the magnitude of J is equal to J = √[(1/2)∙(3/2)]∙ ħ = √(3/4)∙ħ ≈ 0.866∙ħ. Hence, the magnitude of the angular momentum is larger than the maximum value of J_z – and not just a little bit, because the maximum value of ħ is ħ/2! That leads to that weird conclusion: in quantum mechanics, we find that the angular momentum is never completely along any one direction [3]! In fact, this conclusion basically undercuts the very idea of the angular momentum – and, hence, the magnetic moment – of having any precise direction at all! [This may sound spectacular, but there is actually a classical equivalent to the idea of the angular momentum having no precisely defined direction: gyroscopes may not only precess, but nutate as well. Nutation refers to a kind of wobbling around the direction of the angular momentum. For more details, see the post I wrote after my first reading of Feynman’s Lecture on the quantization of magnetic moments. :-)]

Let’s move on. So if, in quantum mechanics, we cannot associate the magnetic moment – or the angular momentum – with some specific direction, then how should we imagine it? Well… I won’t dwell on that here, but you may want to have a look at another post of mine, where I develop a metaphor for the wavefunction which may help you to sort of understand what it might be. The metaphor may help you to think of some oscillation in two directions – rather than in one only – with the two directions separated by a right angle. Hence, the whole thing obviously points in some direction but it’s not very precise. In any case, I need to move on here.

We said that the magnetic moment will take one of two values only, in any direction along which we’d want to measure it. We also said that the (maximum) value along that direction – any direction, really – will be smaller than the magnitude of the moment. [To be precise, we said that for the angular momentum, but the formulas above make it clear the conclusions also hold for the magnetic moment.] So that means that the magnetic moment is, in fact, never fully aligned with the magnetic field. Now, if it is not aligned – and, importantly, if it also does not line up – then it should precess. Now, precession is a difficult enough concept in classical mechanics, so you may think it’s going to be totally abstruse in quantum mechanics. Well… That is true – to some extent. At the same time, it is surely not unintelligible. I will not repeat Feynman’s argument here, but he uses the classical formulas once more to calculate an angular velocity and a precession frequency – although he doesn’t explain what they might actually physically represent. Let me just jot down the formula for the precession frequency:

We get the same factors: g, q_e and m. In addition, you should also note that the precession frequency is directly proportional to the strength of the magnetic field, which makes sense. Now, you may wonder: what is the relevance of this? Can we actually measure any of this?

We can. In fact, you may wonder about the if I inserted above: if we can measure the Landé g-factor… Can we? We can. It’s done in a resonance experiment, which is referred to as the Rabi molecular-beam method – but then it might also be just an atomic beam, of course!

The experiment is interesting, because it shows the precession is – somehow – real. It also illustrates some other principles we have been describing above.

The set-up looks pretty complicated. We have a series of three magnets. The first magnet is just a Stern-Gerlach apparatus: a magnet with a very sharp edge on one of the pole tips so as to produce an inhomogeneous magnetic field. Indeed, a homogeneous magnetic field implies that ∂B/∂z = 0 and, hence, the force along the z-direction would be zero and our atomic magnets would not be displaced.

The second magnet is more complicated. Its magnetic field is uniform, so there are no vertical forces on the atoms and they go straight through. However, the magnet includes an extra set of coils that can produce an alternating horizontal field as well. I’ll come back to that in a moment. Finally, the third magnet is just like the first one, but with the field inverted. Have a look at it:

rabi-apparatus

It may not look very obvious but, after some thinking, you’ll agree that the atoms can only arrive at the detector if they follow the trajectories a and/or b. In fact, these trajectories are the only possible ones because of the slits S₁ and S₂.

Now what’s the idea of that horizontal field B’ in magnet 2? In a classical situation, we could change the angular momentum – and the magnetic moment – by applying some torque about the z-axis. The idea is shown in Figure (a) and (b) below.

changing-angular-momentum

Figure (a) shows – or tries to show – some rotating field B’ – one that is always at right angles to both the angular momentum as well as to the (uniform) B field. That would be effective. However, Figure (b) shows another arrangement that is almost equally effective: an oscillating field that sort of pulls and pushes at some frequency ω. Classically, such fields would effectively change the angle of our gyroscope with respect to the z-axis. Is it also the case quantum-mechanically?

It turns out it sort of works the same in quantum mechanics. There is a big difference though. Classically, μ_z would change gradually, but in quantum mechanics it cannot: in quantum mechanics, it must jump suddenly from one value to the other, i.e. from +ħ/2 to −ħ/2, or the other way around. In other words, it must flip up or down. Now, if an atom flips, then it will, of course, no longer follow the (a) or (b) trajectories: it will follow some other path, like a’ or b’, which make it crash into the magnet. Now, it turns out that almost all atoms will flip if we get that frequency ω right. The graph below shows this ‘resonance’ phenomenon: there is a sharp drop in the ’current’ of atoms if ω is close or equal to ω_p.

resonance

What’s ω_p? It’s that precession frequency for which we gave you that formula above. To make a long story short, from the experiment, we can calculate the Landé g-factor for that particular beam of atoms – say, silver atoms [4]. So… Well… Now we know it all, don’t we?

Maybe. As mentioned when I started this post, when going through all of this material, I always wonder why there is no magnetization effect: why would an atom remain in the same state when it crosses a magnetic field? When it’s already aligned with the magnetic field – to the maximum extent possible, that is – then it shouldn’t flip, but what if its magnetic moment is opposite? It should lower its energy by flipping, right? And it should flip just like that. Why would it need an oscillating B’ field?

In fact, Feynman does describe how the magnetization phenomenon can be analyzed – classically and quantum-mechanically, but he does that for bulk materials: solids, or liquids, or gases – anything that involves lots of atoms that are kicked around because of the thermal motions. So that involves statistical mechanics – which I am sure you’ve skipped so far. 🙂 It is a beautiful argument – which ends with an equally beautiful formula, which tells us the magnetization (M) of a material – which is defined as the net magnetic moment per unit volume – has the same direction as the magnetic field (B) and a magnitude M that is proportional the magnitude of B:

The μ in this formula is the magnitude of the magnetic moment of the individual atoms and so… Well… It’s just like the formula for the electric polarization P, which we described in some other post. In fact, the formula for P and M are same-same but different, as they would say in Thailand. 🙂 But this wonderful story doesn’t answer our question. The magnetic moment of an individual particle should not stay what it is: if it doesn’t change because of all the kicking around as a result of thermal motions, then… Well… These little atomic magnets should line up. That means atoms with their spin “up” should go into the “spin-down” state.

I don’t have an answer to my own question as for now. I suspect it’s got to do with the strength of the magnetic field: a Stern-Gerlach apparatus involves a weak magnetic field. If it’s too strong, the atomic magnets must flip. Hence, a more advanced analysis should probably include that flipping effect. When quickly googling – just now – I found an MIT lab exercise on it, which also provides a historical account of the Stern-Gerlach experiment itself. I skimmed through it – and will read all of it in the coming days – but let me just quote this from the historical background section:

“Stern predicted that the effect would be be just barely observable. They had difficulty in raising support in the midst of the post war financial turmoil in Germany. The apparatus, which required extremely precise alignment and a high vacuum, kept breaking down. Finally, after a year of struggle, they obtained an exposure of sufficient length to give promise of an observable silver deposit. At first, when they examined the glass plate they saw nothing. Then, gradually, the deposit became visible, showing a beam separation of 0.2 millimeters! Apparently, Stern could only afford cheap cigars with a high sulfur content. As he breathed on the glass plate, sulfur fumes converted the invisible silver deposit into visible black silver sufide, and the splitting of the beam was discovered.”

Isn’t this funny? And great at the same time? 🙂 But… Well… The point is: the paper for that MIT lab exercise makes me realize Feynman does cut corners when explaining stuff – and some corners are more significant than others. I note, for example, that they talk about interference peaks rather than “two distinct spots on the glass plate.” Hence, the analysis is somewhat more sophisticated than Feynman pretends it to be. So, when everything is said and done, Feynman’s Lectures may indeed be reading for undergraduate students only. Is it time to move on?

[1] The magnetic moment – as measured in a particular coordinate system – is equal to μ = −g·[q/(2m)]·J. The factor J in this expression is the angular momentum, and the coordinate system is chosen such that its z-axis is along the direction of the magnetic field B. The component of J along the z-axis is written as J_z. This z-component of the angular momentum is what is, rather loosely, being referred to as the spin of the particle in this context. In most other contexts, spin refers to the spin number j which appears in the formula for the value of J_z, which is J_z = j∙ħ, (j−1)∙ħ, (j−2)∙ħ,…, (−j+2)∙ħ, (−j+1), −j∙ħ. Note the separation between the possible values of J_z is equal to ħ. Hence, j itself must be an integer (e.g. 1 or 2) or a half-integer (e.g. 1/2). We usually look at electrons, whose spin number j is 1/2.

[2] One of the pole tips of the magnet that is used in the Stern-Gerlach experiment has a sharp edge. Therefore, the magnetic field strength varies with z. We write: ∂B/∂z ≠ 0.

[3] The z-direction can be any direction, really.

[4] The original experiment was effectively done with a beam of silver atoms. The lab exercise which MIT uses to show the effect to physics students involves potassium atoms.

Feynman’s Seminar on Superconductivity (1)

Original post:

The ultimate challenge for students of Feynman’s iconic Lectures series is, of course, to understand his final one: A Seminar on Superconductivity. As he notes in his introduction to this formidably dense piece, the text does not present the detail of each and every step in the development and, therefore, we’re not supposed to immediately understand everything. As Feynman puts it: we should just believe (more or less) that things would come out if we would be able to go through each and every step. Well… Let’s see. Feynman throws a lot of stuff in here—including, I suspect, some stuff that may not be directly relevant, but that he sort of couldn’t insert into all of his other Lectures. So where do we start?

It took me one long maddening day to figure out the first formula:It says that the amplitude for a particle to go from a to b in a vector potential (think of a classical magnetic field) is the amplitude for the same particle to go from a to b when there is no field (A = 0) multiplied by the exponential of the line integral of the vector potential times the electric charge divided by Planck’s constant. I stared at this for quite a while, but then I recognized the formula for the magnetic effect on an amplitude, which I described in my previous post, which tells us that a magnetic field will shift the phase of the amplitude of a particle with an amount equal to:

Hence, if we write 〈b|a〉 for A = 0 as 〈b|a〉_{A = 0} = C·eⁱ^θ, then 〈b|a〉 in A will, naturally, be equal to 〈b|a〉 _{in A}= C·eⁱ⁽^θ+φ) = C·eⁱ^θ·eⁱ^φ = 〈b|a〉_{A = 0}·eⁱ^φ, and so that explains it. 🙂 Alright… Next. Or… Well… Let us briefly re-examine the concept of the vector potential, because we’ll need it a lot. We introduced it in our post on magnetostatics. Let’s briefly re-cap the development there. In Maxwell’s set of equations, two out of the four equations give us the magnetic field: ∇•B = 0 and c²∇×B = j/ε₀. We noted the following in this regard:

The ∇•B = 0 equation is true, always, unlike the ∇×E = 0 expression, which is true for electrostatics only (no moving charges). So the ∇•B = 0 equation says the divergence of B is zero, always.
The divergence of the curl of a vector field is always zero. Hence, if A is some vector field, then div(curl A) = ∇•(∇×A) = 0, always.
We can now apply another theorem: if the divergence of a vector field, say D, is zero—so if ∇•D = 0—then $D will be the$ the curl of some other vector field $C, so we can write: D = \nabla \times C . Applying this to \nabla • B = 0, we can write:$

If ∇•B = 0, then there is an A such that B = ∇×A

So, in essence, we’re just re-defining the magnetic field (B) in terms of some other vector field. To be precise, we write it as the curl of some other vector field, which we refer to as the (magnetic) vector potential. The components of the magnetic field vector can then be re-written as:

We need to note an important point here: the equations above suggest that the components of B depend on position only. In other words, we assume static magnetic fields, so they do not change with time. That, in turn, assumes steady currents. We will want to extend the analysis to also include magnetodynamics. It complicates the analysis but… Well… Quantum mechanics is complicated. Let us remind ourselves here of Feynman’s re-formulation of Maxwell’s equations as a set of two equations (expressed in terms of the magnetic (vector) and the electric potential) only:

Wave equation for A

Wave equation for potential

These equations are wave equations, as you can see by writing out the second equation:

It is a wave equation in three dimensions. Note that, even in regions where we do no have any charges or currents, we have non-zero solutions for φ and A. These non-zero solutions are, effectively, representing the electric and magnetic fields as they travel through free space. As Feynman notes, the advantage of re-writing Maxwell’s equations as we do above, is that the two new equations make it immediately apparent that we’re talking electromagnetic waves, really. As he notes, for many practical purposes, it will still be convenient to use the original equations in terms of E and B, but… Well… Not in quantum mechanics, it turns out. As Feynman puts it: “E and B are on the other side of the mountain we have climbed. Now we are ready to cross over to the other side of the peak. Things will look different—we are ready for some new and beautiful views.”

Well… Maybe. Appreciating those views, as part of our study of quantum mechanics, does take time and effort, unfortunately. 😦

The Schrödinger equation in an electromagnetic field

Feynman then jots down Schrödinger’s equation for the same particle (with charge q) moving in an electromagnetic field that is characterized not only by the (scalar) potential Φ but also by a vector potential A:

schrodinger

Now where does that come from? We know the standard formula in an electric field, right? It’s the formula we used to find the energy states of electrons in a hydrogen atom:

i·ħ·∂ψ/∂t = −(1/2)·(ħ²/m)∇²ψ + V·ψ

Of course, it is easy to see that we replaced V by q·Φ, which makes sense: the potential of a charge in an electric field is the product of the charge (q) and the (electric) potential (Φ), because Φ is, obviously, the potential energy of the unit charge. It’s also easy to see we can re-write −ħ²·∇²ψ as [(ħ/i)·∇]·[(ħ/i)·∇]ψ because (1/i)·(1/i) = 1/i² = 1/(−1) = −1. 🙂 Alright. So it’s just that −q·A term in the (ħ/i)∇ − q·A expression that we need to explain now.

Unfortunately, that explanation is not so easy. Feynman basically re-derives Schrödinger’s equation using his trade-mark historical argument – which did not include any magnetic field – with a vector potential. The re-derivation is rather annoying, and I didn’t have the courage to go through it myself, so you should – just like me – just believe Feynman when he says that, when there’s a vector potential – i.e. when there’s a magnetic field – then that (ħ/i)·∇ operator – which is the momentum operator– ought to be replaced by a new momentum operator:

new-momentum-operator

So… Well… There we are… 🙂 So far, so good? Well… Maybe.

While, as mentioned, you won’t be interested in the mathematical argument, it is probably worthwhile to reproduce Feynman’s more intuitive explanation of why the operator above is what it is. In other words, let us try to understand that −qA term. Look at the following situation: we’ve got a solenoid here, and some current I is going through it so there’s a magnetic field B. Think of the dynamics while we turn on this flux. Maxwell’s second equation (∇×E = −∂B/∂t) tells us the line integral of E around a loop will be equal to the time rate of change of the magnetic flux through that loop. The ∇×E = −∂B/∂t equation is a differential equation, of course, so it doesn’t have the integral, but you get the idea—I hope. solenoid

Now, using the B = ∇×A equation we can re-write the ∇×E = −∂B/∂t as ∇×E = −∂(∇×A)/∂t. This allows us to write the following:

∇×E = −∂(∇×A)/∂t = −∇×(∂A/∂t) ⇔ E = −∂A/∂t

This is a remarkable expression. Note its derivation is based on the commutativity of the curl and time derivative operators, which is a property that can easily be explained: if we have a function in two variables—say x and t—then the order of the derivation doesn’t matter: we can first take the derivative with respect to x and then to t or, alternatively, we can first take the time derivative and then do the ∂/∂x operation. So… Well… The curl is, effectively, a derivative with regard to the spatial variables. OK. So what? What’s the point?

Well… If we’d have some charge q, as shown in the illustration above, that would happen to be there as the flux is being switched on, it will experience a force which is equal to F = qE. We can now integrate this over the time interval (t) during which the flux is being built up to get the following:

∫₀^t F = ∫₀^t m·a = ∫₀^t m·dv/dt = m·v_t= ∫₀^t q·E = −∫₀^t q·∂A/∂t = −q·A_t

Assuming v₀ and A₀are zero, we may drop the time subscript and simply write:

m·v = −q·A

The point is: during the build-up of the magnetic flux, our charge will pick up some (classical) momentum that is equal to p = m·v = −q·A. So… Well… That sort of explains the additional term in our new momentum operator.

Note: For some reason I don’t quite understand, Feynman introduces the weird concept of ‘dynamical momentum’, which he defines as the quantity m·v + q·A, so that quantity must be zero in the analysis above. I quickly googled to see why but didn’t invest too much time in the research here. It’s just… Well… A bit puzzling. I don’t really see the relevance of his point here: I am quite happy to go along with the new operator, as it’s rather obvious that introducing changing magnetic fields must, obviously, also have some impact on our wave equations—in classical as well as in quantum mechanics.

Local conservation of probability

The title of this section in Feynman’s Lecture (yes, still the same Lecture – we’re not switching topics here) is the equation of continuity for probabilities. I find it brilliant, because it confirms my interpretation of the wave function as describing some kind of energy flow. Let me quote Feynman on his endeavor here:

“An important part of the Schrödinger equation for a single particle is the idea that the probability to find the particle at a position is given by the absolute square of the wave function. It is also characteristic of the quantum mechanics that probability is conserved in a local sense. When the probability of finding the electron somewhere decreases, while the probability of the electron being elsewhere increases (keeping the total probability unchanged), something must be going on in between. In other words, the electron has a continuity in the sense that if the probability decreases at one place and builds up at another place, there must be some kind of flow between. If you put a wall, for example, in the way, it will have an influence and the probabilities will not be the same. So the conservation of probability alone is not the complete statement of the conservation law, just as the conservation of energy alone is not as deep and important as the local conservation of energy. If energy is disappearing, there must be a flow of energy to correspond. In the same way, we would like to find a “current” of probability such that if there is any change in the probability density (the probability of being found in a unit volume), it can be considered as coming from an inflow or an outflow due to some current.”

This is it, really ! The wave function does represent some kind of energy flow – between a so-called ‘real’ and a so-called ‘imaginary’ space, which are to be defined in terms of directional versus rotational energy, as I try to point out – admittedly: more by appealing to intuition than to mathematical rigor – in that post of mine on the meaning of the wavefunction.

So what is the flow – or probability current as Feynman refers to it? Well… Here’s the formula:

probability-current-2

Huh? Yes. Don’t worry too much about it right now. The essential point is to understand what this current – denoted by J – actually stands for:

probability-current-1

So what’s next? Well… Nothing. I’ll actually refer you to Feynman now, because I can’t improve on how he explains how pairs of electrons start behaving when temperatures are low enough to render Boltzmann’s Law irrelevant: the kinetic energy that’s associated with temperature can no longer break up electron pairs if temperature comes close to the zero point.

Huh? What? Electron pairs? Electrons are not supposed to form pairs, are they? They carry the same charge and are, therefore, supposed to repel each other. Well… Yes and no. In my post on the electron orbitals in a hydrogen atom – which just presented Feynman’s presentation on the subject-matter in a, hopefully, somewhat more readable format – we calculated electron orbitals neglecting spin. In Feynman’s words:

“We make another approximation by forgetting that the electron has spin. […] The non-relativistic Schrödinger equation disregards magnetic effects. [However] Small magnetic effects [do] occur because, from the electron’s point-of-view, the proton is a circulating charge which produces a magnetic field. In this field the electron will have a different energy with its spin up than with it down. [Hence] The energy of the atom will be shifted a little bit from what we will calculate. We will ignore this small energy shift. Also we will imagine that the electron is just like a gyroscope moving around in space always keeping the same direction of spin. Since we will be considering a free atom in space the total angular momentum will be conserved. In our approximation we will assume that the angular momentum of the electron spin stays constant, so all the rest of the angular momentum of the atom—what is usually called “orbital” angular momentum—will also be conserved. To an excellent approximation the electron moves in the hydrogen atom like a particle without spin—the angular momentum of the motion is a constant.”

To an excellent approximation… But… Well… Electrons in a metal do form pairs, because they can give up energy in that way and, hence, they are more stable that way. Feynman does not go into the details here – I guess because that’s way beyond the undergrad level – but refers to the Bardeen-Coopers-Schrieffer (BCS) theory instead – the authors of which got a Nobel Prize in Physics in 1972 (that’s a decade or so after Feynman wrote this particular Lecture), so I must assume the theory is well accepted now. 🙂

Of course, you’ll shout now: Hey! Hydrogen is not a metal! Well… Think again: the latest breakthrough in physics is making hydrogen behave like a metal. 🙂 And I am really talking the latest breakthrough: Science just published the findings of this experiment last month! 🙂 🙂 In any case, we’re not talking hydrogen here but superconducting materials, to which – as far as we know – the BCS theory does apply.

So… Well… I am done. I just wanted to show you why it’s important to work your way through Feynman’s last Lecture because… Well… Quantum mechanics does explain everything – although the nitty-gritty of it (the Meissner effect, the London equation, flux quantization, etc.) are rather hard bullets to bite. 😦

Don’t give up ! I am struggling with the nitty-gritty too ! 🙂

The Aharonov-Bohm effect

Original post:

This title sounds very exciting. It is – or was, I should say – one of these things I thought I would never ever understand, until I started studying physics, that is. 🙂

Having said that, there is – incidentally – nothing very special about the Aharonov-Bohm effect. As Feynman puts it: “The theory was known from the beginning of quantum mechanics in 1926. […] The implication was there all the time, but no one paid attention to it.”

To be fair, he also admits the experiment itself – proving the effect – is “very, very difficult”, which is why the first experiment that claimed to confirm the predicted effect was set up in 1960 only. In fact, some claim the results of that experiment were ambiguous, and that it was only in 1986, with the experiment of Akira Tonomura, that the Aharonov-Bohm effect was unambiguously demonstrated. So what is it about?

In essence, it proves the reality of the vector potential—and of the (related) magnetic field. What do we mean with a real field? To put it simply, a real field cannot act on some particle from a distance through some kind of spooky ‘action-at-a-distance’: real fields must be specified at the position of the particle itself and describe what happens there. Now you’ll immediately wonder: so what’s a non-real field? Well… Some field that does act through some kind of spooky ‘action-at-a-distance.’ As for an example… Well… I can’t give you one because we’ve only been discussing real fields so far. 🙂

So it’s about what a magnetic (or an electric) field does in terms influencing motion and/or quantum-mechanical amplitudes. In fact, we discussed this matter quite a while ago (check my 2015 post on it). Now, I don’t want to re-write that post, but let me just remind you of the essentials. The two equations for the magnetic field (B) in Maxwell’s set of four equations (the two others specify the electric field E) are: (1) ∇•B = 0 and (2) c²∇×B = j/ε₀ + ∂E/ ∂t. Now, you can temporarily forget about the second equation, but you should note that the ∇•B = 0 equation is always true (unlike the ∇×E = 0 expression, which is true for electrostatics only, when there are no moving charges). So it says that the divergence of B is zero, always.

Now, from our posts on vector calculus, you may or may not remember that the divergence of the curl of a vector field is always zero. We wrote: div (curl A) = ∇•(∇×A) = 0, always. Now, there is another theorem that we can now apply, which says the following: if the divergence of a vector field, say D, is zero – so if ∇•D = 0, then $D will be the$ curl of some other vector field $C, so we can write: D = \nabla \times C . When we now apply this to our \nabla • B = 0 equation, we can confidently state the following:$

If ∇•B = 0, then there is an A such that B = ∇×A

We can also write this as follows: ∇·B = ∇·(∇×A) = 0 and, hence, B = ∇×A. Now, it’s this vector field A that is referred to as the (magnetic) vector potential, and so that’s what we want to talk about here. As a start, it may be good to write out all of the components of our B = ∇×A vector:

In that 2015 post, I answered the question as to why we’d need this new vector field in a way that wasn’t very truthful: I just said that, in many situations, it would be more convenient – from a mathematical point of view, that is – to first find A, and then calculate the derivatives above to get B.

Now, Feynman says the following about this argument in his Lecture on the topic: “It is true that in many complex problems it is easier to work with A, but it would be hard to argue that this ease of technique would justify making you learn about one more vector field. […] We have introduced A because it does have an important physical significance: it is a real physical field.” Let us follow his argument here.

Quantum-mechanical interference effects

Let us first remind ourselves of the quintessential electron interference experiment illustrated below. [For a much more modern rendering of this experiment, check out the Tout Est Quantique video on it. It’s much more amusing than my rather dry exposé here, but it doesn’t give you the math.]

interference

We have electrons, all of (nearly) the same energy, which leave the source – one by one – and travel towards a wall with two narrow slits. Beyond the wall is a backstop with a movable detector which measures the rate, which we call I, at which electrons arrive at a small region of the backstop at the distance x from the axis of symmetry. The rate (or intensity) I is proportional to the probability that an individual electron that leaves the source will reach that region of the backstop. This probability has the complicated-looking distribution shown in the illustration, which we understand is due to the interference of two amplitudes, one from each slit. So we associate the two trajectories with two amplitudes, which Feynman writes as A₁eⁱ^Φ₁and A₂eⁱ^Φ₂ respectively.

As usual, Feynman abstracts away from the time variable here because it is, effectively, not relevant: the interference pattern depends on distances and angles only. Having said that, for a good understanding, we should – perhaps – write our two wavefunctions as A₁eⁱ⁽^{ωt +}^Φ₁)and A₂eⁱ⁽^{ωt +}^Φ₂⁾respectively. The point is: we’ve got two wavefunctions – one for each trajectory – even if it’s only one electron going through the slit: that’s the mystery of quantum mechanics. 🙂 We need to add these waves so as to get the interference effect:

R = A₁eⁱ⁽^{ωt +}^Φ₁)+ A₂eⁱ⁽^{ωt +}^Φ₂⁾= [A₁eⁱ^Φ₁+ A₂eⁱ^Φ₂]·eⁱ^ωt

Now, we know we need to take the absolute square of this thing to get the intensity – or probability (before normalization). The absolute square of a product, is the product of the absolute squares of the factors, and we also know that the absolute square of any complex number is just the product of the same number with its complex conjugate. Hence, the absolute square of the eⁱ^ωt factor is equal to |eⁱ^ωt|² = eⁱ^ωt∙e^–iωt = e⁰= 1. So the time-dependent factor doesn’t matter: that’s why we can always abstract away from it. Let us now take the absolute square of the [A₁eⁱ^Φ₁+ A₂eⁱ^Φ₂] factor, which we can write as:

|R|²= |A₁eⁱ^Φ₁+ A₂eⁱ^Φ₂|²= (A₁eⁱ^Φ₁+ A₂eⁱ^Φ₂)·(A₁e^–i^Φ₁+ A₂e^–ⁱ^Φ₂)

= A₁²+ A₂²+ 2·A₁·A₂·cos(Φ₁−Φ₂) = A₁²+ A₂²+ 2·A₁·A₂·cosδ with δ = Φ₁−Φ₂

OK. This is probably going a bit quick, but you should be able to figure it out, especially when remembering that eⁱ^Φ+ e^–i^Φ= 2·cosΦ and cosΦ = cos(−Φ). The point to note is that the intensity is equal to the sum of the intensities of both waves plus a correction factor, which is equal to 2·A₁·A₂·cos(Φ₁−Φ₂) and, hence, ranges from −2·A₁·A₂ to +2·A₁·A₂. Now, it takes a bit of geometrical wizardry to be able to write the phase difference δ = Φ₁−Φ₂as

δ = 2π·a/λ = 2π·(x/L)·d/λ

—but it can be done. 🙂 Well… […] OK. 🙂 Let me quickly help you here by copying another diagram from Feynman – one he uses to derive the formula for the phase difference on arrival between the signals from two oscillators. A₁ and A₂ are equal here (A₁ = A₂ = A) so that makes the situation below somewhat simpler to analyze. However, instead, we have the added complication of a phase difference (α) at the origin – which Feynman refers to as an intrinsic relative phase. triangle

When we apply the geometry shown above to our electron passing through the slits, we should, of course, equate α to zero. For the rest, the picture is pretty similar as the two-slit picture. The distance a in the two-slit – i.e. the difference in the path lengths for the two trajectories of our electron(s) – is, obviously, equal to the d·sinθ factor in the oscillator picture. Also, because L is huge as compared to x, we may assume that trajectory 1 and 2 are more or less parallel and, importantly, that the triangles in the picture – small and large – are rectangular. Now, trigonometry tells us that sinθ is equal to the ratio of the opposite side of the triangle and the hypotenuse (i.e. the longest side of the rectangular triangle). The opposite side of the triangle is x and, because x is very, very small as compared to L, we may approximate the length of the hypotenuse with L. [I know—a lot of approximations here, but… Well… Just go along with it as for now…] Hence, we can equate sinθ to x/L and, therefore, a = d·x/L. Now we need to calculate the phase difference. How many wavelengths do we have in a? That’s simple: a/λ, i.e. the total distance divided by the wavelength. Now these wavelengths correspond to 2π·a/λ radians (one cycle corresponds to one wavelength which, in turn, corresponds to 2π radians). So we’re done. We’ve got the formula: δ = Φ₁−Φ₂= 2π·a/λ = 2π·(x/L)·d/λ.

Huh? Yes. Just think about it. I need to move on. The point is: when x is equal to zero, the two waves are in phase, and the probability will have a maximum. When δ = π, then the waves are out of phase and interfere destructively (cosπ = −1), so the intensity (and, hence, the probability) reaches a minimum.

So that’s pretty obvious – or should be pretty obvious if you’ve understood some of the basics we presented in this blog. We now move to the non-standard stuff, i.e. the Aharonov-Bohm effect(s).

Interference in the presence of an electromagnetic field

In essence, the Aharonov-Bohm effect is nothing special: it is just a law – two laws, to be precise – that tells us how the phase of our wavefunction changes because of the presence of a magnetic and/or electric field. As such, it is not very different from previous analyses and presentations, such as those showing how amplitudes are affected by a potential − such as an electric potential, or a gravitational field, or a magnetic field − and how they relate to a classical analysis of the situation (see, for example, my November 2015 post on this topic). If anything, it’s just a more systematic approach to the topic and – importantly – an approach centered around the use of the vector potential A (and the electric potential Φ). Let me give you the formulas:

The first formula tells us that the phase of the amplitude for our electron (or whatever charged particle) to arrive at some location via some trajectory is changed by an amount that is equal to the integral of the vector potential along the trajectory times the charge of the particle over Planck’s constant. I know that’s quite a mouthful but just read it a couple of times.

The second formula tells us that, if there’s an electrostatic field, it will produce a phase change given by the negative of the time integral of the (scalar) potential Φ.

These two expressions – taken together – tell us what happens for any electromagnetic field, static or dynamic. In fact, they are really the (two) law(s) replacing the F = q(E + v×B) expression in classical mechanics.

So how does it work? Let me further follow Feynman’s treatment of the matter—which analyzes what happens when we’d have some magnetic field in the two-slit experiment (so we assume there’s no electric field: we only look at some magnetic field). We said Φ₁ was the phase of the wave along trajectory 1, and Φ₂ was the phase of the wave along trajectory 2. Without magnetic field, that is, so B = 0. Now, the (first) formula above tells us that, when the field is switched on, the new phases will be the following:

Hence, the phase difference δ = Φ₁−Φ₂will now be equal to:

Now, we can combine the two integrals into one that goes forward along trajectory 1 and comes back along trajectory 2. We’ll denote this path as 1-2 and write the new integral as follows:

Note that we’re using a notation here which suggests that the 1-2 path is closed, which is… Well… Yet another approximation of the Master. In fact, his assumption that the new 1-2 path is closed proves to be essential in the argument that follows the one we presented above, in which he shows that the inherent arbitrariness in our choice of a vector potential function doesn’t matter, but… Well… I don’t want to get too technical here.

Let me conclude this post by noting we can re-write our grand formula above in terms of the flux of the magnetic field B:

So… Well… That’s it, really. I’ll refer you to Feynman’s Lecture on this matter for a detailed description of the 1960 experiment itself, which involves a magnetized iron whisker that acts like a tiny solenoid—small enough to match the tiny scale of the interference experiment itself. I must warn you though: there is a rather long discussion in that Lecture on the ‘reality’ of the magnetic and the vector potential field which – unlike Feynman’s usual approach to discussions like this – is rather philosophical and partially misinformed, as it assumes there is zero magnetic field outside of a solenoid. That’s true for infinitely long solenoids, but not true for real-life solenoids: if we have some A, then we must also have some B, and vice versa. Hence, if the magnetic field (B) is a real field (in the sense that it cannot act on some particle from a distance through some kind of spooky ‘action-at-a-distance’), then the vector potential A is an equally real field—and vice versa. Feynman admits as much as he concludes his rather lengthy philosophical excursion with the following conclusion (out of which I already quoted one line in my introduction to this post):

“This subject has an interesting history. The theory we have described was known from the beginning of quantum mechanics in 1926. The fact that the vector potential appears in the wave equation of quantum mechanics (called the Schrödinger equation) was obvious from the day it was written. That it cannot be replaced by the magnetic field in any easy way was observed by one man after the other who tried to do so. This is also clear from our example of electrons moving in a region where there is no field and being affected nevertheless. But because in classical mechanics A did not appear to have any direct importance and, furthermore, because it could be changed by adding a gradient, people repeatedly said that the vector potential had no direct physical significance—that only the magnetic and electric fields are “real” even in quantum mechanics. It seems strange in retrospect that no one thought of discussing this experiment until 1956, when Bohm and Aharonov first suggested it and made the whole question crystal clear. The implication was there all the time, but no one paid attention to it. Thus many people were rather shocked when the matter was brought up. That’s why someone thought it would be worthwhile to do the experiment to see if it was really right, even though quantum mechanics, which had been believed for so many years, gave an unequivocal answer. It is interesting that something like this can be around for thirty years but, because of certain prejudices of what is and is not significant, continues to be ignored.”

Well… That’s it, folks! Enough for today! 🙂

An interpretation of the wavefunction

This is my umpteenth post on the same topic. 😦 It is obvious that this search for a sensible interpretation is consuming me. Why? I am not sure. Studying physics is frustrating. As a leading physicist puts it:

“The teaching of quantum mechanics these days usually follows the same dogma: firstly, the student is told about the failure of classical physics at the beginning of the last century; secondly, the heroic confusions of the founding fathers are described and the student is given to understand that no humble undergraduate student could hope to actually understand quantum mechanics for himself; thirdly, a deus ex machina arrives in the form of a set of postulates (the Schrödinger equation, the collapse of the wavefunction, etc); fourthly, a bombardment of experimental verifications is given, so that the student cannot doubt that QM is correct; fifthly, the student learns how to solve the problems that will appear on the exam paper, hopefully with as little thought as possible.”

That’s obviously not the way we want to understand quantum mechanics. [With we, I mean, me, of course, and you, if you’re reading this blog.] Of course, that doesn’t mean I don’t believe Richard Feynman, one of the greatest physicists ever, when he tells us no one, including himself, understands physics quite the way we’d like to understand it. Such statements should not prevent us from trying harder. So let’s look for better metaphors. The animation below shows the two components of the archetypal wavefunction – a simple sine and cosine. They’re the same function actually, but their phases differ by 90 degrees (π/2).

circle_cos_sin

It makes me think of a V-2 engine with the pistons at a 90-degree angle. Look at the illustration below, which I took from a rather simple article on cars and engines that has nothing to do with quantum mechanics. Think of the moving pistons as harmonic oscillators, like springs.

two-timer-576-px-photo-369911-s-original

We will also think of the center of each cylinder as the zero point: think of that point as a point where – if we’re looking at one cylinder alone – the internal and external pressure balance each other, so the piston would not move… Well… If it weren’t for the other piston, because the second piston is not at the center when the first is. In fact, it is easy to verify and compare the following positions of both pistons, as well as the associated dynamics of the situation:

Piston 1	Piston 2	Motion of Piston 1	Motion Piston 2
Top	Center	Compressed air will push piston down	Piston moves down against external pressure
Center	Bottom	Piston moves down against external pressure	External air pressure will push piston up
Bottom	Center	External air pressure will push piston up	Piston moves further up and compresses the air
Center	Top	Piston moves further up and compresses the air	Compressed air will push piston down

When the pistons move, their linear motion will be described by a sinusoidal function: a sine or a cosine. In fact, the 90-degree V-2 configuration ensures that the linear motion of the two pistons will be exactly the same, except for a phase difference of 90 degrees. [Of course, because of the sideways motion of the connecting rods, our sine and cosine function describes the linear motion only approximately, but you can easily imagine the idealized limit situation. If not, check Feynman’s description of the harmonic oscillator.]

The question is: if we’d have a set-up like this, two springs – or two harmonic oscillators – attached to a shaft through a crank, would this really work as a perpetuum mobile? We obviously talk energy being transferred back and forth between the rotating shaft and the moving pistons… So… Well… Let’s model this: the total energy, potential and kinetic, in each harmonic oscillator is constant. Hence, the piston only delivers or receives kinetic energy from the rotating mass of the shaft.

Now, in physics, that’s a bit of an oxymoron: we don’t think of negative or positive kinetic (or potential) energy in the context of oscillators. We don’t think of the direction of energy. But… Well… If we’ve got two oscillators, our picture changes, and so we may have to adjust our thinking here.

Let me start by giving you an authoritative derivation of the various formulas involved here, taking the example of the physical spring as an oscillator—but the formulas are basically the same for any harmonic oscillator.

energy harmonic oscillator

The first formula is a general description of the motion of our oscillator. The coefficient in front of the cosine function (a) is the maximum amplitude. Of course, you will also recognize ω₀ as the natural frequency of the oscillator, and Δ as the phase factor, which takes into account our t = 0 point. In our case, for example, we have two oscillators with a phase difference equal to π/2 and, hence, Δ would be 0 for one oscillator, and –π/2 for the other. [The formula to apply here is sinθ = cos(θ – π/2).] Also note that we can equate our θ argument to ω₀·t. Now, if a = 1 (which is the case here), then these formulas simplify to:

K.E. = T = m·v²/2 = m·ω₀²·sin²(θ + Δ) = m·ω₀²·sin²(ω₀·t + Δ)
P.E. = U = k·x²/2 = k·cos²(θ + Δ)

The coefficient k in the potential energy formula characterizes the force: F = −k·x. The minus sign reminds us our oscillator wants to return to the center point, so the force pulls back. From the dynamics involved, it is obvious that k must be equal to m·ω₀²., so that gives us the famous T + U = m·ω₀²/2 formula or, including a once again, T + U = m·a²·ω₀²/2.

Now, if we normalize our functions by equating k to one (k = 1), then the motion of our first oscillator is given by the cosθ function, and its kinetic energy will be equal to sin²θ. Hence, the (instantaneous) change in kinetic energy at any point in time will be equal to:

d(sin²θ)/dθ = 2∙sinθ∙d(sinθ)/dt = 2∙sinθ∙cosθ

Let’s look at the second oscillator now. Just think of the second piston going up and down in our V-twin engine. Its motion is given by the sinθ function which, as mentioned above, is equal to cos(θ−π /2). Hence, its kinetic energy is equal to sin²(θ−π /2), and how it changes – as a function of θ – will be equal to:

2∙sin(θ−π /2)∙cos(θ−π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ

We have our perpetuum mobile! While transferring kinetic energy from one piston to the other, the rotating shaft moves at constant speed. Linear motion becomes circular motion, and vice versa, in a frictionless Universe. We have the metaphor we were looking for!

Somehow, in this beautiful interplay between linear and circular motion, energy is being borrowed from one place to another, and then returned. From what place to what place? I am not sure. We may call it the real and imaginary energy space respectively, but what does that mean? One thing is for sure, however: the interplay between the real and imaginary part of the wavefunction describes how energy propagates through space!

How exactly? Again, I am not sure. Energy is, obviously, mass in motion – as evidenced by the E = m·c² equation, and it may not have any direction (when everything is said and done, it’s a scalar quantity without direction), but the energy in a linear motion is surely different from that in a circular motion, and our metaphor suggests we need to think somewhat more along those lines. Perhaps we will, one day, able to square this circle. 🙂

Schrödinger’s equation

Let’s analyze the interplay between the real and imaginary part of the wavefunction through an analysis of Schrödinger’s equation, which we write as:

i·ħ∙∂ψ/∂t = –(ħ²/2m)∙∇²ψ + V·ψ

We can do a quick dimensional analysis of both sides:

[i·ħ∙∂ψ/∂t] = N∙m∙s/s = N∙m
[–(ħ²/2m)∙∇²ψ] = N∙m³/m² = N∙m
[V·ψ] = N∙m

Note the dimension of the ‘diffusion’ constant ħ²/2m: [ħ²/2m] = N²∙m²∙s²/kg = N²∙m²∙s²/(N·s²/m) = N∙m³. Also note that, in order for the dimensions to come out alright, the dimension of V – the potential – must be that of energy. Hence, Feynman’s description of it as the potential energy – rather than the potential tout court – is somewhat confusing but correct: V must equal the potential energy of the electron. Hence, V is not the conventional (potential) energy of the unit charge (1 coulomb). Instead, the natural unit of charge is used here, i.e. the charge of the electron itself.

Now, Schrödinger’s equation – without the V·ψ term – can be written as the following pair of equations:

Re(∂ψ/∂t) = −(1/2)∙(ħ/m)∙Im(∇²ψ)
Im(∂ψ/∂t) = (1/2)∙(ħ/m)∙Re(∇²ψ)

This closely resembles the propagation mechanism of an electromagnetic wave as described by Maxwell’s equation for free space (i.e. a space with no charges), but E and B are vectors, not scalars. How do we get this result. Well… ψ is a complex function, which we can write as a + i∙b. Likewise, ∂ψ/∂t is a complex function, which we can write as c + i∙d, and ∇²ψ can then be written as e + i∙f. If we temporarily forget about the coefficients (ħ, ħ²/m and V), then Schrödinger’s equation – including V·ψ term – amounts to writing something like this:

i∙(c + i∙d) = –(e + i∙f) + (a + i∙b) ⇔ a + i∙b = i∙c − d + e+ i∙f ⇔ a = −d + e and b = c + f

Hence, we can now write:

V∙Re(ψ) = −ħ∙Im(∂ψ/∂t) + (1/2)∙( ħ²/m)∙Re(∇²ψ)
V∙Im(ψ) = ħ∙Re(∂ψ/∂t) + (1/2)∙( ħ²/m)∙Im(∇²ψ)

This simplifies to the two equations above for V = 0, i.e. when there is no potential (electron in free space). Now we can bring the Re and Im operators into the brackets to get:

V∙Re(ψ) = −ħ∙∂Im (ψ)/∂t + (1/2)∙( ħ²/m)∙∇²Re(ψ)
V∙Im(ψ) = ħ∙∂Re(ψ)/∂t + (1/2)∙( ħ²/m)∙∇²Im(ψ)

This is very interesting, because we can re-write this using the quantum-mechanical energy operator H = –(ħ²/2m)∙∇² + V· (note the multiplication sign after the V, which we do not have – for obvious reasons – for the –(ħ²/2m)∙∇² expression):

H[Re (ψ)] = −ħ∙∂Im(ψ)/∂t
H[Im(ψ)] = ħ∙∂Re(ψ)/∂t

A dimensional analysis shows us both sides are, once again, expressed in N∙m. It’s a beautiful expression because – if we write the real and imaginary part of ψ as r∙cosθ and r∙sinθ, we get:

H[cosθ] = −ħ∙∂sinθ/∂t = E∙cosθ
H[sinθ] = ħ∙∂cosθ/∂t = E∙sinθ

Indeed, θ = (E∙t − p∙x)/ħ and, hence, −ħ∙∂sinθ/∂t = ħ∙cosθ∙E/ħ = E∙cosθ and ħ∙∂cosθ/∂t = ħ∙sinθ∙E/ħ = E∙sinθ. Now we can combine the two equations in one equation again and write:

H[r∙(cosθ + i∙sinθ)] = r∙(E∙cosθ + i∙sinθ) ⇔ H[ψ] = E∙ψ

The operator H – applied to the wavefunction – gives us the (scalar) product of the energy E and the wavefunction itself. Isn’t this strange?

Hmm… I need to further verify and explain this result… I’ll probably do so in yet another post on the same topic… 🙂

Post scriptum: The symmetry of our V-2 engine – or perpetuum mobile – is interesting: its cross-section has only one axis of symmetry. Hence, we may associate some angle with it, so as to define its orientation in the two-dimensional cross-sectional plane. Of course, the cross-sectional plane itself is at right angles to the crankshaft axis, which we may also associate with some angle in three-dimensional space. Hence, its geometry defines two orthogonal directions which, in turn, define a spherical coordinate system, as shown below.

We may, therefore, say that three-dimensional space is actually being implied by the geometry of our V-2 engine. Now that is interesting, isn’t it? 🙂

Quantum-mechanical operators

I wrote a post on quantum-mechanical operators some while ago but, when re-reading it now, I am not very happy about it, because it tries to cover too much ground in one go. In essence, I regret my attempt to constantly switch between the matrix representation of quantum physics – with the | state 〉 symbols – and the wavefunction approach, so as to show how the operators work for both cases. But then that’s how Feynman approaches this.

However, let’s admit it: while Heisenberg’s matrix approach is equivalent to Schrödinger’s wavefunction approach – and while it’s the only approach that works well for n-state systems – the wavefunction approach is more intuitive, because:

Most practical examples of quantum-mechanical systems (like the description of the electron orbitals of an atomic system) involve continuous coordinate spaces, so we have an infinite number of states and, hence, we need to describe it using the wavefunction approach.
Most of us are much better-versed in using derivatives and integrals, as opposed to matrix operations.
A more intuitive statement of the same argument above is the following: the idea of one state flowing into another, rather than being transformed through some matrix, is much more appealing. 🙂

So let’s stick to the wavefunction approach here. So, while you need to remember that there’s a ‘matrix equivalent’ for each of the equations we’re going to use in this post, we’re not going to talk about it.

The operator idea

In classical physics – high school physics, really – we would describe a pointlike particle traveling in space by a function relating its position (x) to time (t): x = x(t). Its (instantaneous) velocity is, obviously, v(t) = dx/dt. Simple. Obvious. Let’s complicate matters now by saying that the idea of a velocity operator would sort of generalize the v(t) = dx/dt velocity equation by making abstraction of the specifics of the x = x(t) function.

Huh? Yes. We could define a velocity ‘operator’ as:

Now, you may think that’s a rather ridiculous way to describe what an operator does, but – in essence – it’s correct. We have some function – describing an elementary particle, or a system, or an aspect of the system – and then we have some operator, which we apply to our function, to extract the information from it that we want: its velocity, its momentum, its energy. Whatever. Hence, in quantum physics, we have an energy operator, a position operator, a momentum operator, an angular momentum operator and… Well… I guess I listed the most important ones. 🙂

It’s kinda logical. Our velocity operator looks at one particular aspect of whatever it is that’s going on: the time rate of change of position. We do refer to that as the velocity. Our quantum-mechanical operators do the same: they look at one aspect of what’s being described by the wavefunction. [At this point, you may wonder what the other properties of our classical ‘system’ – i.e. other properties than velocity – because we’re just looking at a pointlike particle here, but… Well… Think of electric charge and forces acting on it, so it accelerates and decelerates in all kinds of ways, and we have kinetic and potential energy and all that. Or momentum. So it’s just the same: the x = x(t) function may cover a lot of complexities, just like the wavefunction does!]

The Wikipedia article on the momentum operator is, for a change (I usually find Wikipedia quite abstruse on these matters), quite simple – and, therefore – quite enlightening here. It applies the following simple logic to the elementary wavefunction ψ = e^{−i·(ω·t − k∙x)}, with the de Broglie relations telling us that ω = E/ħ and k = p/ħ:

mom op 1

Note we forget about the normalization coefficient a here. It doesn’t matter: we can always stuff it in later. The point to note is that we can sort of forget about ψ (or abstract away from it—as mathematicians and physicists would say) by defining the momentum operator, which we’ll write as:

mom op 2

Its three-dimensional equivalent is calculated in very much the same way:

wiki

So this operator, when operating on a particular wavefunction, gives us the (expected) momentum when we would actually catch our particle there, provided the momentum doesn’t vary in time. [Note that it may – and actually is likely to – vary in space!]

So that’s the basic idea of an operator. However, the comparison goes further. Indeed, a superficial reading of what operators are all about gives you the impression we get all these observables (or properties of the system) just by applying the operator to the (wave)function. That’s not the case. There is the randomness. The uncertainty. Actual wavefunctions are superpositions of several elementary waves with various coefficients representing their amplitudes. So we need averages, or expected values: E[X] Even our velocity operator ∂/∂t – in the classical world – gives us an instantaneous velocity only. To get the average velocity (in quantum mechanics, we’ll be interested in the the average momentum, or the average position, or the average energy – rather than the average velocity), we’re going to have the calculate the total distance traveled. Now, that’s going to involve a line integral:

S = ∫_Lds.

The principle is illustrated below.

line integral

You’ll say: this is kids stuff, and it is. Just note how we write the same integral in terms of the x and t coordinate, and using our new velocity operator:

integral

Kids stuff. Yes. But it’s good to think about what it represents really. For example, the simplest quantum-mechanical operator is the position operator. It’s just x for the x-coordinate, y for the y-coordinate, and z for the z-coordinate. To get the average position of a stationary particle – represented by the wavefunction ψ(r, t) – in three-dimensional space, we need to calculate the following volume integral:

position operator 3D V2

Simple? Yes and no. The r·|ψ(r)|² integrand is obvious: we multiply each possible position (r) by its probability (or likelihood), which is equal to P(r) = |ψ(r)|². However, look at the assumptions: we already omitted the time variable. Hence, the particle we’re describing here must be stationary, indeed! So we’ll need to re-visit the whole subject allowing for averages to change with time. We’ll do that later. I just wanted to show you that those integrals – even with very simple operators, like the position operator – can become very complicated. So you just need to make sure you know what you’re looking at.

One wavefunction—or two? Or more?

There is another reason why, with the immeasurable benefit of hindsight, I now feel that my earlier post is confusing: I kept switching between the position and the momentum wavefunction, which gives the impression we have different wavefunctions describing different aspects of the same thing. That’s just not true. The position and momentum wavefunction describe essentially the same thing: we can go from one to the other, and back again, by a simple mathematical manipulation. So I should have stuck to descriptions in terms of ψ(x, t), instead of switching back and forth between the ψ(x, t) and φ(x, t) representations.

In any case, the damage is done, so let’s move forward. The key idea is that, when we know the wavefunction, we know everything. I tried to convey that by noting that the real and imaginary part of the wavefunction must, somehow, represent the total energy of the particle. The structural similarity between the mass-energy equivalence relation (i.e. Einstein’s formula: E = m·c²) and the energy formulas for oscillators and spinning masses is too obvious:

The energy of any oscillator is given by the E = m·ω₀²/2. We may want to liken the real and imaginary component of our wavefunction to two oscillators and, hence, add them up. The E = m·ω₀² formula we get is then identical to the E = m·c² formula.
The energy of a spinning mass is given by an equivalent formula: E = I·ω²/2 (I is the moment of inertia in this formula). The same 1/2 factor tells us our particle is, somehow, spinning in two dimensions at the same time (i.e. a ‘real’ as well as an ‘imaginary’ space—but both are equally real, because amplitudes interfere), so we get the E = I·ω² formula.

Hence, the formulas tell us we should imagine an electron – or an electron orbital – as a very complicated two-dimensional standing wave. Now, when I write two-dimensional, I refer to the real and imaginary component of our wavefunction, as illustrated below. What I am asking you, however, is to not only imagine these two components oscillating up and down, but also spinning about. Hence, if we think about energy as some oscillating mass – which is what the E = m·c² formula tells us to do, we should remind ourselves we’re talking very complicated motions here: mass oscillates, swirls and spins, and it does so both in real as well as in imaginary space. rising_circular

What I like about the illustration above is that it shows us – in a very obvious way – why the wavefunction depends on our reference frame. These oscillations do represent something in absolute space, but how we measure it depends on our orientation in that absolute space. But so I am writing this post to talk about operators, not about my grand theory about the essence of mass and energy. So let’s talk about operators now. 🙂

In that post of mine, I showed how the position, momentum and energy operator would give us the average position, momentum and energy of whatever it was that we were looking at, but I didn’t introduce the angular momentum operator. So let me do that now. However, I’ll first recapitulate what we’ve learnt so far in regard to operators.

The energy, position and momentum operators

The equation below defines the energy operator, and also shows how we would apply it to the wavefunction:

To the purists: sorry for not (always) using the hat symbol. [I explained why in that post of mine: it’s just too cumbersome.] The others 🙂 should note the following:

E_average is also an expected value: E_av= E[E]
The * symbol tells us to take the complex conjugate of the wavefunction.
As for the integral, it’s an integral over some volume, so that’s what the d³r shows. Many authors use double or triple integral signs (∫∫ or ∫∫∫) to show it’s a surface or a volume integral, but that makes things look very complicated, and so I don’t that. I could also have written the integral as ∫ψ(r)*·H·ψ(r) dV, but then I’d need to explain that the dV stands for dVolume, not for any (differental) potential energy (V).
We must normalize our wavefunction for these formulas to work, so all probabilities over the volume add up to 1.

OK. That’s the energy operator. As you can see, it’s a pretty formidable beast, but then it just reflects Schrödinger’s equation which, as I explained a couple of times already, we can interpret as an energy propagation mechanism, or an energy diffusion equation, so it is actually not that difficult to memorize the formula: if you’re able to remember Schrödinger’s equation, then you’ll also have the operator. If not… Well… Then you won’t pass your undergrad physics exam. 🙂

I already mentioned that the position operator is a much simpler beast. That’s because it’s so intimately related to our interpretation of the wavefunction. It’s the one thing you know about quantum mechanics: the absolute square of the wavefunction gives us the probability density function. So, for one-dimensional space, the position operator is just:

The equivalent operator for three-dimensional space is equally simple:

position operator 3D V2

Note how the operator, for the one- as well as for the three-dimensional case, gets rid of time as a variable. In fact, the idea itself of an average makes abstraction of the temporal aspect. Well… Here, at least—because we’re looking at some box in space, rather than some box in spacetime. We’ll re-visit that rather particular idea of an average, and allow for averages that change with time, in a short while.

Next, we introduced the momentum operator in that post of mine. For one dimension, Feynman shows this operator is given by the following formula:

Now that does not look very simple. You might think that the ∂/∂x operator reflects our velocity operator, but… Well… No: ∂/∂t gives us a time rate of change, while ∂/∂x gives us the spatial variation. So it’s not the same. Also, that ħ/i factor is quite intriguing, isn’t it? We’ll come back to it in the next section of this post. Let me just give you the three-dimensional equivalent which, remembering that 1/i = −i, you’ll understand to be equal to the following vector operator:

Now it’s time to define the operator we wanted to talk about, i.e. the angular momentum operator.

The angular momentum operator

The formula for the angular momentum operator is remarkably simple:

Why do I call this a simple formula? Because it looks like the familiar formula of classical mechanics for the z-component of the classical angular momentum L = r × p. I must assume you know how to calculate a vector cross product. If not, check one of my many posts on vector analysis. I must also assume you remember the L = r × p formula. If not, the following animation might bring it all back. If that doesn’t help, check my post on gyroscopes. 🙂

Now, spin is a complicated phenomenon, and so, to simplify the analysis, we should think of orbital angular momentum only. This is a simplification, because electron spin is some complicated mix of intrinsic and orbital angular momentum. Hence, the angular momentum operator we’re introducing here is only the orbital angular momentum operator. However, let us not get bogged down in all of the nitty-gritty and, hence, let’s just go along with it for the time being.

I am somewhat hesitant to show you how we get that formula for our operator, but I’ll try to show you using an intuitive approach, which uses only bits and pieces of Feynman’s more detailed derivation. It will, hopefully, give you a bit of an idea of how these differential operators work. Think about a rotation of our reference frame over an infinitesimally small angle – which we’ll denote as ε – as illustrated below.

rotation

Now, the whole idea is that, because of that rotation of our reference frame, our wavefunction will look different. It’s nothing fundamental, but… Well… It’s just because we’re using a different coordinate system. Indeed, that’s where all these complicated transformation rules for amplitudes come in. I’ve spoken about these at length when we were still discussing n-state systems. In contrast, the transformation rules for the coordinates themselves are very simple:

rotation

Now, because ε is an infinitesimally small angle, we may equate cos(θ) = cos(ε) to 1, and cos(θ) = sin(ε) to ε. Hence, x’ and y’ are then written as x’ = x + εy and y’ = y − εx, while z‘ remains z. Vice versa, we can also write the old coordinates in terms of the new ones: x = x’ − εy, y = y’ + εx, and z = z‘. That’s obvious. Now comes the difficult thing: you need to think about the two-dimensional equivalent of the simple illustration below.

izvod

If we have some function y = f(x), then we know that, for small Δx, we have the following approximation formula for f(x + Δx): f(x + Δx) ≈ f(x) + (dy/dx)·Δx. It’s the formula you saw in high school: you would then take a limit (Δx → 0), and define dy/dx as the Δy/Δx ratio for Δx → 0. You would this after re-writing the f(x + Δx) ≈ f(x) + (dy/dx)·Δx formula as:

Δy = Δf = f(x + Δx) − f(x) ≈ (dy/dx)·Δx

Now you need to substitute f for ψ, and Δx for ε. There is only one complication here: ψ is a function of two variables: x and y. In fact, it’s a function of three variables – x, y and z – but we keep z constant. So think of moving from x and y to x + εy = x + Δx and to y + Δy = y − εx. Hence, Δx = εy and Δy = −εx. It then makes sense to write Δψ as:

angular momentum operator v2

If you agree with that, you’ll also agree we can write something like this:

formula 2

Now that implies the following formula for Δψ:

repair

This looks great! You can see we get some sort of differential operator here, which is what we want. So the next step should be simple: we just let ε go to zero and then we’re done, right? Well… No. In quantum mechanics, it’s always a bit more complicated. But it’s logical stuff. Think of the following:

1. We will want to re-write the infinitesimally small ε angle as a fraction of i, i.e. the imaginary unit.

Huh? Yes. This little i represents many things. In this particular case, we want to look at it as a right angle. In fact, you know multiplication with i amounts to a rotation by 90 degrees. So we should replace ε by ε·i. It’s like measuring ε in natural units. However, we’re not done.

2. We should also note that Nature measures angles clockwise, rather than counter-clockwise, as evidenced by the fact that the argument of our wavefunction rotates clockwise as time goes by. So our ε is, in fact, a −ε. We will just bring the minus sign inside of the brackets to solve this issue.

Huh? Yes. Sorry. I told you this is a rather intuitive approach to getting what we want to get. 🙂

3. The third modification we’d want to make is to express ε·i as a multiple of Planck’s constant.

Huh? Yes. This is a very weird thing, but it should make sense—intuitively: we’re talking angular momentum here, and its dimension is the same as that of physical action: N·m·s. Therefore, Planck’s quantum of action (ħ = h/2π ≈ 1×10⁻³⁴ J·s ≈ 6.6×10⁻¹⁶ eV·s) naturally appears as… Well… A natural unit, or a scaling factor, I should say.

To make a long story short, we’ll want to re-write ε as −(i/ħ)·ε. However, there is a thing called mathematical consistency, and so, if we want to do such substitutions and prepare for that limit situation (ε → 0), we should re-write that Δψ equation as follows:

final

So now – finally! – we do have the formula we wanted to find for our angular momentum operator:

final 2

The final substitution, which yields the formula we just gave you when commencing this section, just uses the formula for the linear momentum operator in the x– and y-direction respectively. We’re done! 🙂 Finally!

Well… No. 🙂 The question, of course, is the same as always: what does it all mean, really? That’s always a great question. 🙂 Unfortunately, the answer is rather boring: we can calculate the average angular momentum in the z-direction, using a similar integral as the one we used to get the average energy, or the average linear momentum in some direction. That’s basically it.

To compensate for that very boring answer, however, I will show you something that is far less boring. 🙂

Quantum-mechanical weirdness

I’ll shameless copy from Feynman here. He notes that many classical equations get carried over into a quantum-mechanical form (I’ll copy some of his illustrations later). But then there are some that don’t. As Feynman puts it—rather humorously: “There had better be some that don’t come out right, because if everything did, then there would be nothing different about quantum mechanics. There would be no new physics.” He then looks at the following super-obvious equation in classical mechanics:

x·p_x− p_x·x = 0

In fact, this equation is so super-obvious that it’s almost meaningless. Almost. It’s super-obvious because multiplication is commutative (for real as well for complex numbers). However, when we replace x and p_xby the position and momentum operator, we get an entirely different result. You can verify the following yourself:

strange

This is plain weird! What does it mean? I am not sure. Feynman’s take on it is nice but leaves us in the dark on it:

Feynman quote 2

He adds: “If Planck’s constant were zero, the classical and quantum results would be the same, and there would be no quantum mechanics to learn!” Hmm… What does it mean, really? Not sure. Let me make two remarks here:

1. We should not put any dot (·) between our operators, because they do not amount to multiplying one with another. We just apply operators successively. Hence, commutativity is not what we should expect.

2. Note that Feynman forgot to put the subscript in that quote. When doing the same calculations for the equivalent of the x·p_y− p_y·x expression, we do get zero, as shown below:

not strange

These equations – zero or not – are referred to as ‘commutation rules’. [Again, I should not have used any dot between x and p_y, because there is no multiplication here. It’s just a separation mark.] Let me quote Feynman on it, so the matter is dealt with:

quote

OK. So what do we conclude? What are we talking about?

Conclusions

Some of the stuff above was really intriguing. For example, we found that the linear and angular momentum operators are differential operators in the true sense of the word. The angular momentum operator shows us what happens to the wavefunction if we rotate our reference frame over an infinitesimally small angle ε. That’s what’s captured by the formulas we’ve developed, as summarized below:

angular momentum

Likewise, the linear momentum operator captures what happens to the wavefunction for an infinitesimally small displacement of the reference frame, as shown by the equivalent formulas below:

linear momentum

What’s the interpretation for the position operator, and the energy operator? Here we are not so sure. The integrals above make sense, but these integrals are used to calculate averages values, as opposed to instantaneous values. So… Well… There is not all that much I can say about the position and energy operator right now, except… Well… We now need to explore the question of how averages could possibly change over time. Let’s do that now.

Averages that change with time

I know: you are totally quantum-mechanicked out by now. So am I. But we’re almost there. In fact, this is Feynman’s last Lecture on quantum mechanics and, hence, I think I should let the Master speak here. So just click on the link and read for yourself. It’s a really interesting chapter, as he shows us the equivalent of Newton’s Law in quantum mechanics, as well as the quantum-mechanical equivalent of other standard equations in classical mechanics. However, I need to warn you: Feynman keeps testing the limits of our intellectual absorption capacity by switching back and forth between matrix and wave mechanics. Interesting, but not easy. For example, you’ll need to remind yourself of the fact that the Hamiltonian matrix is equal to its own complex conjugate (or – because it’s a matrix – its own conjugate transpose.

Having said that, it’s all wonderful. The time rate of change of all those average values is denoted by using the over-dot notation. For example, the time rate of change of the average position is denoted by:

Once you ‘get’ that new notation, you will quickly understand the derivations. They are not easy (what derivations are in quantum mechanics?), but we get very interesting results. Nice things to play with, or think about—like this identity:

formula2

It takes a while, but you suddenly realize this is the equivalent of the classical dx/dt = v = p/m formula. 🙂

Another sweet result is the following one:

formula3

This is the quantum-mechanical equivalent of Newton’s force law: F = m·a. Huh? Yes. Think of it: the spatial derivative of the (potential) energy is the force. Now just think of the classical dp/dt = d(m·v) = m·dv/dt = m·a formula. […] Can you see it now? Isn’t this just Great Fun?

Note, however, that these formulas also show the limits of our analysis so far, because they treat m as some constant. Hence, we’ll need to relativistically correct them. But that’s complicated, and so we’ll postpone that to another day.

[…]

Well… That’s it, folks! We’re really through! This was the last of the last of Feynman’s Lectures on Physics. So we’re totally done now. Isn’t this great? What an adventure! I hope that, despite the enormous mental energy that’s required to digest all this stuff, you enjoyed it as much as I did. 🙂

Post scriptum 1: I just love Feynman but, frankly, I think he’s sometimes somewhat sloppy with terminology. In regard to what these operators really mean, we should make use of better terminology: an average is something else than an expected value. Our momentum operator, for example, as such returns an expected value – not an average momentum. We need to deepen the analysis here somewhat, but I’ll also leave that for later.

Post scriptum 2: There is something really interesting about that i·ħ or −(i/ħ) scaling factor – or whatever you want to call it – appearing in our formulas. Remember the Schrödinger equation can also be written as:

i·ħ·∂ψ/∂t = −(1/2)·(ħ²/m)∇²ψ + V·ψ = Hψ

This is interesting in light of our interpretation of the Schrödinger equation as an energy propagation mechanism. If we write Schrödinger’s equation like we write it here, then we have the energy on the right-hand side – which is time-independent. How do we interpret the left-hand side now? Well… It’s kinda simple, but we just have the time rate of change of the real and imaginary part of the wavefunction here, and the i·ħ factor then becomes a sort of unit in which we measure the time rate of change. Alternatively, you may think of ‘splitting’ Planck’s constant in two: Planck’s energy, and Planck’s time unit, and then you bring the Planck energy unit to the other side, so we’d express the energy in natural units. Likewise, the time rate of change of the components of our wavefunction would also be measured in natural time units if we’d do that.

I know this is all very abstract but, frankly, it’s crystal clear to me. This formula tells us that the energy of the particle that’s being described by the wavefunction is being carried by the oscillations of the wavefunction. In fact, the oscillations are the energy. You can play with the mass factor, by moving it to the left-hand side too, or by using Einstein’s mass-energy equivalence relation. The interpretation remains consistent.

In fact, there is something really interesting here. You know that we usually separate out the spatial and temporal part of the wavefunction, so we write: ψ(r, t) = ψ(r)·e^{−i·(E/ħ)·t}. In fact, it is quite common to refer to ψ(r) – rather than to ψ(r, t) – as the wavefunction, even if, personally, I find that quite confusing and misleading (see my page onSchrödinger’s equation). Now, we may want to think of what happens when we’d apply the energy operator to ψ(r) rather than to ψ(r, t). We may think that we’d get a time-independent value for the energy at that point in space, so energy is some function of position only, not of time. That’s an interesting thought, and we should explore it. For example, we then may think of energy as an average that changes with position—as opposed to the (average) position and momentum, which we like to think of as averages than change with time, as mentioned above. I will come back to this later – but perhaps in another post or so. Not now. The only point I want to mention here is the following: you cannot use ψ(r) in Schrödinger’s equation. Why? Well… Schrödinger’s equation is no longer valid when substituting ψ for ψ(r), because the left-hand side is always zero, as ∂ψ(r)/∂t is zero – for any r.

There is another, related, point to this observation. If you think that Schrödinger’s equation implies that the operators on both sides of Schrödinger’s equation must be equivalent (i.e. the same), you’re wrong:

i·ħ·∂/∂t ≠ H = −(1/2)·(ħ²/m)∇² + V

It’s a basic thing, really: Schrödinger’s equation is not valid for just any function. Hence, it does not work for ψ(r). Only ψ(r, t) makes it work, because… Well… Schrödinger’s equation gave us ψ(r, t)!

The energy and 1/2 factor in Schrödinger’s equation

Schrödinger’s equation, for a particle moving in free space (so we have no external force fields acting on it, so V = 0 and, therefore, the Vψ term disappears) is written as:

∂ψ(x, t)/∂t = i·(1/2)·(ħ/m_eff)·∇²ψ(x, t)

We already noted and explained the structural similarity with the ubiquitous diffusion equation in physics:

∂φ(x, t)/∂t = D·∇²φ(x, t) with x = (x, y, z)

The big difference between the wave equation and an ordinary diffusion equation is that the wave equation gives us two equations for the price of one: ψ is a complex-valued function, with a real and an imaginary part which, despite their name, are both equally fundamental, or essential. Whatever word you prefer. 🙂 That’s also what the presence of the imaginary unit (i) in the equation tells us. But for the rest it’s the same: the diffusion constant (D) in Schrödinger’s equation is equal to (1/2)·(ħ/m_eff).

Why the 1/2 factor? It’s ugly. Think of the following: If we bring the (1/2)·(ħ/m_eff) to the other side, we can write it as m_eff/(ħ/2). The ħ/2 now appears as a scaling factor in the diffusion constant, just like ħ does in the de Broglie equations: ω = E/ħ and k = p/ħ, or in the argument of the wavefunction: θ = (E·t − p∙x)/ħ. Planck’s constant is, effectively, a physical scaling factor. As a physical scaling constant, it usually does two things:

It fixes the numbers (so that’s its function as a mathematical constant).
As a physical constant, it also fixes the physical dimensions. Note, for example, how the 1/ħ factor in ω = E/ħ and k = p/ħ ensures that the ω·t = (E/ħ)·t and k·x = (p/ħ)·x terms in the argument of the wavefunction are both expressed as some dimensionless number, so they can effectively be added together. Physicists don’t like adding apples and oranges.

The question is: why did Schrödinger use ħ/2, rather than ħ, as a scaling factor? Let’s explore the question.

The 1/2 factor

We may want to think that 1/2 factor just echoes the 1/2 factor in the Uncertainty Principle, which we should think of as a pair of relations: σ_x·σ_p ≥ ħ/2 and σ_E·σ_t≥ ħ/2. However, the 1/2 factor in those relations only makes sense because we chose to equate the fundamental uncertainty (Δ) in x, p, E and t with the mathematical concept of the standard deviation (σ), or the half-width, as Feynman calls it in his wonderfully clear exposé on it in one of his Lectures on quantum mechanics (for a summary with some comments, see my blog post on it). We may just as well choose to equate Δ with the full-width of those probability distributions we get for x and p, or for E and t. If we do that, we get σ_x·σ_p ≥ ħ and σ_E·σ_t≥ ħ.

It’s a bit like measuring the weight of a person on an old-fashioned (non-digital) bathroom scale with 1 kg marks only: do we say this person is x kg ± 1 kg, or x kg ± 500 g? Do we take the half-width or the full-width as the margin of error? In short, it’s a matter of appreciation, and the 1/2 factor in our pair of uncertainty relations is not there because we’ve got two relations. Likewise, it’s not because I mentioned we can think of Schrödinger’s equation as a pair of relations that, taken together, represent an energy propagation mechanism that’s quite similar in its structure to Maxwell’s equations for an electromagnetic wave (as shown below), that we’d insert (or not) that 1/2 factor: either of the two representations below works. It just depends on our definition of the concept of the effective mass.

The 1/2 factor is really a matter of choice, because the rather peculiar – and flexible – concept of the effective mass takes care of it. However, we could define some new effective mass concept, by writing: m_eff^NEW = 2∙m_eff^OLD, and then Schrödinger’s equation would look more elegant:

∂ψ/∂t = i·(ħ/m_eff^NEW)·∇²ψ

Now you’ll want the definition, of course! What is that effective mass concept? Feynman talks at length about it, but his exposé is embedded in a much longer and more general argument on the propagation of electrons in a crystal lattice, which you may not necessarily want to go through right now. So let’s try to answer that question by doing something stupid: let’s substitute ψ in the equation for ψ = a·e^{−i·[E·t − p∙x]/ħ} (which is an elementary wavefunction), calculate the time derivative and the Laplacian, and see what we get. If we do that, the ∂ψ/∂t = i·(1/2)·(ħ/m_eff)·∇²ψ equation becomes:

~~−i·a~~·(E/ħ)·~~e^{−i∙(E·t − p∙x)/ħ}~~ = ~~−i·a~~·(1/2)·(ħ/m_eff)(p²/ħ²)·~~e^{−i∙(E·t − p∙x)/ħ}~~

⇔ E = (1/2)·p²/m_eff = (1/2)·(m·v)²/m_eff ⇔ m_eff = (1/2)·(m/E)·m·v²

⇔ m_eff = (1/c²)·(m·v²/2) = m·β²/2

Hence, the effective mass appears in this equation as the equivalent mass of the kinetic energy (K.E.) of the elementary particle that’s being represented by the wavefunction. Now, you may think that sounds good – and it does – but you should note the following:

1. The K.E. = m·v²/2 formula is only correct for non-relativistic speeds. In fact, it’s the kinetic energy formula if, and only if, if m ≈ m₀. The relativistically correct formula for the kinetic energy calculates it as the difference between (1) the total energy (which is given by the E = m·c² formula, always) and (2) its rest energy, so we write:

K.E. = E − E₀ = m_v·c² − m₀·c² = m₀·γ·c² − m₀·c² = m₀·c²·(γ − 1)

2. The energy concept in the wavefunction ψ = a·e^{−i·[E·t − p∙x]/ħ}is, obviously, the total energy of the particle. For non-relativistic speeds, the kinetic energy is only a very small fraction of the total energy. In fact, using the formula above, you can calculate the ratio between the kinetic and the total energy: you’ll find it’s equal to 1 − 1/γ = 1 − √(1−v²/c²), and its graph goes from 0 to 1.

Now, if we discard the 1/2 factor, the calculations above yield the following:

−i·a·(E/ħ)·e^{−i∙(E·t − p∙x)/ħ} = −i·a·(ħ/m_eff)(p²/ħ²)·e^{−i∙(E·t − p∙x)/ħ}

⇔ E = p²/m_eff = (m·v)²/m_eff ⇔ m_eff = (m/E)·m·v²

⇔ m_eff = m·v²/c²= m·β²

In fact, it is fair to say that both definitions are equally weird, even if the dimensions come out alright: the effective mass is measured in old-fashioned mass units, and the β²or β²/2 factor appears as a sort of correction factor, varying between 0 and 1 (for β²) or between 0 and 1/2 (for β²/2). I prefer the new definition, as it ensures that m_eff becomes equal to m in the limit for the velocity going to c. In addition, if we bring the ħ/m_eff or (1/2)∙ħ/m_eff factor to the other side of the equation, the choice becomes one between a m_eff^NEW/ħ or a 2∙m_eff^OLD/ħ coefficient.

It’s a choice, really. Personally, I think the equation without the 1/2 factor – and, hence, the use of ħ rather than ħ/2 as the scaling factor – looks better, but then you may argue that – if half of the energy of our particle is in the oscillating real part of the wavefunction, and the other is in the imaginary part – then the 1/2 factor should stay, because it ensures that m_eff becomes equal to m/2 as v goes to c (or, what amounts to the same, β goes to 1). But then that’s the argument about whether or not we should have a 1/2 factor because we get two equations for the price of one, like we did for the Uncertainty Principle.

So… What to do? Let’s first ask ourselves whether that derivation of the effective mass actually makes sense. Let’s therefore look at both limit situations.

1. For v going to c (or β = v/c going to 1), we do not have much of a problem: m_eff just becomes the total mass of the particle that we’re looking at, and Schrödinger’s equation can easily be interpreted as an energy propagation mechanism. Our particle has zero rest mass in that case ( we may also say that the concept of a rest mass is meaningless in this situation) and all of the energy – and, therefore, all of the equivalent mass – is kinetic: m = E/c²and the effective mass is just the mass: m_eff = m·c²/c²= m. Hence, our particle is everywhere and nowhere. In fact, you should note that the concept of velocity itself doesn’t make sense in this rather particular case. It’s like a photon (but note it’s not a photon: we’re talking some theoretical particle here with zero spin and zero rest mass): it’s a wave in its own frame of reference, but as it zips by at the speed of light, we think of it as a particle.

2. Let’s look at the other limit situation. For v going to 0 (or β = v/c going to 0), Schrödinger’s equation no longer makes sense, because the diffusion constant goes to zero, so we get a nonsensical equation. Huh? What’s wrong with our analysis?

Well… I must be honest. We started off on the wrong foot. You should note that it’s hard – in fact, plain impossible – to reconcile our simple a·e^{−i·[E·t − p∙x]/ħ} function with the idea of the classical velocity of our particle. Indeed, the classical velocity corresponds to a group velocity, or the velocity of a wave packet, and so we just have one wave here: no group. So we get nonsense. You can see the same when equating p to zero in the wave equation: we get another nonsensical equation, because the Laplacian is zero! Check it. If our elementary wavefunction is equal to ψ = a·e^{−i·(E/ħ)·t}, then that Laplacian is zero.

Hence, our calculation of the effective mass is not very sensical. Why? Because the elementary wavefunction is a theoretical concept only: it may represent some box in space, that is uniformly filled with energy, but it cannot represent any actual particle. Actual particles are always some superposition of two or more elementary waves, so then we’ve got a wave packet (as illustrated below) that we can actually associate with some real-life particle moving in space, like an electron in some orbital indeed. 🙂

wave-packet

I must credit Oregon State University for the animation above. It’s quite nice: a simple particle in a box model without potential. As I showed on my other page (explaining various models), we must add at least two waves – traveling in opposite directions – to model a particle in a box. Why? Because we represent it by a standing wave, and a standing wave is the sum of two waves traveling in opposite directions.

So, if our derivation above was not very meaningful, then what is the actual concept of the effective mass?

The concept of the effective mass

I am afraid that, at this point, I do have to direct you back to the Grand Master himself for the detail. Let me just try to sum it up very succinctly. If we have a wave packet, there is – obviously – some energy in it, and it’s energy we may associate with the classical concept of the velocity of our particle – because it’s the group velocity of our wave packet. Hence, we have a new energy concept here – and the equivalent mass, of course. Now, Feynman’s analysis – which is Schrödinger’s analysis, really – shows we can write that energy as:

E = m_eff·v²/2

So… Well… That’s the classical kinetic energy formula. And it’s the very classical one, because it’s not relativistic. 😦 But that’s OK for relatively small-moving electrons! [Remember the typical (relative) velocity is given by the fine-structure constant: α = β = v/c. So that’s impressive (about 2,188 km per second), but it’s only a tiny fraction of the speed of light, so non-relativistic formulas should work.]

Now, the m_eff factor in this equation is a function of the various parameters of the model he uses. To be precise, we get the following formula out of his model (which, as mentioned above, is a model of electrons propagating in a crystal lattice):

m_eff = ħ²/(2·A·b² )

Now, the b in this formula is the spacing between the atoms in the lattice. The A basically represents an energy barrier: to move from one atom to another, the electron needs to get across it. I talked about this in my post on it, and so I won’t explain the graph below – because I did that in that post. Just note that we don’t need that factor 2: there is no reason whatsoever to write E₀+ 2·A and E₀− 2·A. We could just re-define a new A: (1/2)·A^NEW = A^OLD. The formula for m_eff then simplifies to ħ²/(2·A^OLD·b²) = ħ²/(A^NEW·b²). We then get an E_eff = m_eff·v²formula for the extra energy.

E_eff = m_eff·v²?!? What energy formula is that? Schrödinger must have thought the same thing, and so that’s why we have that ugly 1/2 factor in his equation. However, think about it. Our analysis shows that it is quite straightforward to model energy as a two-dimensional oscillation of mass. In this analysis, both the real and the imaginary component of the wavefunction each store half of the total energy of the object, which is equal to E = m·c². Remember, indeed, that we compared it to the energy in an oscillator, which is equal to the sum of kinetic and potential energy, and for which we have the T + U = m·ω₀²/2 formula. But so we have two oscillators here and, hence, twice the energy. Hence, the E = m·c² corresponds to m·ω₀²and, hence, we may think of c as the natural frequency of the vacuum.

Therefore, the E_eff = m_eff·v² formula makes much more sense. It nicely mirrors Einstein’s E = m·c² formula and, in fact, naturally merges into E = m·c² for v approaching c. But, I admit, it is not so easy to interpret. It’s much easier to just say that the effective mass is the mass of our electron as it appears in the kinetic energy formula, or – alternatively – in the momentum formula. Indeed, Feynman also writes the following formula:

m_eff·v = p = ħ·k

Now, that is something we easily recognize! 🙂

So… Well… What do we do now? Do we use the 1/2 factor or not?

It would be very convenient, of course, to just stick with tradition and use m_eff as everyone else uses it: it is just the mass as it appears in whatever medium we happen to look it, which may be a crystal lattice (or a semi-conductor), or just free space. In short, it’s the mass of the electron as it appears to us, i.e. as it appears in the (non-relativistic) kinetic energy formula (K.E. = m_eff·v²/2), the formula for the momentum of an electron (p = m_eff·v), or in the wavefunction itself (k = p/ħ = (m_eff·v)/ħ. In fact, in his analysis of the electron orbitals, Feynman (who just follows Schrödinger here) drops the eff subscript altogether, and so the effective mass is just the mass: m_eff = m. Hence, the apparent mass of the electron in the hydrogen atom serves as a reference point, and the effective mass in a different medium (such as a crystal lattice, rather than free space or, I should say, a hydrogen atom in free space) will also be different.

The thing is: we get the right results out of Schrödinger’s equation, with the 1/2 factor in it. Hence, Schrödinger’s equation works: we get the actual electron orbitals out of it. Hence, Schrödinger’s equation is true – without any doubt. Hence, if we take that 1/2 factor out, then we do need to use the other effective mass concept. We can do that. Think about the actual relation between the effective mass and the real mass of the electron, about which Feynman writes the following: “The effective mass has nothing to do with the real mass of an electron. It may be quite different—although in commonly used metals and semiconductors it often happens to turn out to be the same general order of magnitude: about 0.1 to 30 times the free-space mass of the electron.” Hence, if we write the relation between m_effand m as m_eff= g(m), then the same relation for our m_eff^NEW = 2∙m_eff^OLD becomes m_eff^NEW= 2·g(m), and the “about 0.1 to 30 times” becomes “about 0.2 to 60 times.”

In fact, in the original 1963 edition, Feynman writes that the effective mass is “about 2 to 20 times” the free-space mass of the electron. Isn’t that interesting? I mean… Note that factor 2! If we’d write m_eff= 2·m, then we’re fine. We can then write Schrödinger’s equation in the following two equivalent ways:

(m_eff/ħ)·∂ψ/∂t = i·∇²ψ
(2m/ħ)·∂ψ/∂t = i·∇²ψ

Both would be correct, and it explains why Schrödinger’s equation works. So let’s go for that compromise and write Schrödinger’s equation in either of the two equivalent ways. 🙂 The question then becomes: how to interpret that factor 2? The answer to that question is, effectively, related to the fact that we get two waves for the price of one here. So we have two oscillators, so to speak. Now that‘s quite deep, and I will explore that in one of my next posts.

Let me now address the second weird thing in Schrödinger’s equation: the energy factor. I should be more precise: the weirdness arises when solving Schrödinger’s equation. Indeed, in the texts I’ve read, there is this constant switching back and forth between interpreting E as the energy of the atom, versus the energy of the electron. Now, both concepts are obviously quite different, so which one is it really?

The energy factor E

It’s a confusing point—for me, at least and, hence, I must assume for students as well. Let me indicate, by way of example, how the confusion arises in Feynman’s exposé on the solutions to the Schrödinger equation. Initially, the development is quite straightforward. Replacing V by −e²/r, Schrödinger’s equation becomes:

As usual, it is then assumed that a solution of the form ψ (r, t) = e^{−(i/ħ)·E·t}·ψ(r) will work. Apart from the confusion that arises because we use the same symbol, ψ, for two different functions (you will agree that ψ (r, t), a function in two variables, is obviously not the same as ψ(r), a function in one variable only), this assumption is quite straightforward and allows us to re-write the differential equation above as:

To get this, you just need to actually to do that time derivative, noting that the ψ in our equation is now ψ(r), not ψ (r, t). Feynman duly notes this as he writes: “The function ψ(r) must solve this equation, where E is some constant—the energy of the atom.” So far, so good. In one of the (many) next steps, we re-write E as E = E_R·ε, with E_R= m·e⁴/2ħ². So we just use the Rydberg energy (E_R≈ 13.6 eV) here as a ‘natural’ atomic energy unit. That’s all. No harm in that.

Then all kinds of complicated but legitimate mathematical manipulations follow, in an attempt to solve this differential equation—attempt that is successful, of course! However, after all these manipulations, one ends up with the grand simple solution for the s-states of the atom (i.e. the spherically symmetric solutions):

E_n = −E_R/n²with 1/n²= 1, 1/4, 1/9, 1/16,…, 1

So we get: E_n = −13.6 eV, −3.4 eV, −1.5 eV, etcetera. Now how is that possible? How can the energy of the atom suddenly be negative? More importantly, why is so tiny in comparison with the rest energy of the proton (which is about 938 mega-electronvolt), or the electron (0.511 MeV)? The energy levels above are a few eV only, not a few million electronvolt. Feynman answers this question rather vaguely when he states the following:

“There is, incidentally, nothing mysterious about negative numbers for the energy. The energies are negative because when we chose to write V = −e²/r, we picked our zero point as the energy of an electron located far from the proton. When it is close to the proton, its energy is less, so somewhat below zero. The energy is lowest (most negative) for n = 1, and increases toward zero with increasing n.”

We picked our zero point as the energy of an electron located far away from the proton? But we were talking the energy of the atom all along, right? You’re right. Feynman doesn’t answer the question. The solution is OK – well, sort of, at least – but, in one of those mathematical complications, there is a ‘normalization’ – a choice of some constant that pops up when combining and substituting stuff – that is not so innocent. To be precise, at some point, Feynman substitutes the ε variable for the square of another variable – to be even more precise, he writes: ε = −α². He then performs some more hat tricks – all legitimate, no doubt – and finds that the only sensible solutions to the differential equation require α to be equal to 1/n, which immediately leads to the above-mentioned solution for our s-states.

The real answer to the question is given somewhere else. In fact, Feynman casually gives us an explanation in one of his very first Lectures on quantum mechanics, where he writes the following:

“If we have a “condition” which is a mixture of two different states with different energies, then the amplitude for each of the two states will vary with time according to an equation like a·e^−iωt, with ħ·ω = E₀ = m·c². Hence, we can write the amplitude for the two states, for example as:

e^{−i(E₁/ħ)·t} and e^{−i(E₂/ħ)·t}

And if we have some combination of the two, we will have an interference. But notice that if we added a constant to both energies, it wouldn’t make any difference. If somebody else were to use a different scale of energy in which all the energies were increased (or decreased) by a constant amount—say, by the amount A—then the amplitudes in the two states would, from his point of view, be

e - i (E 1 +A)\cdott/ħ and e - i (E 2 +A)\cdott/ħ

All of his amplitudes would be multiplied by the same factor e^{−i(A/ħ)·t}, and all linear combinations, or interferences, would have the same factor. When we take the absolute squares to find the probabilities, all the answers would be the same. The choice of an origin for our energy scale makes no difference; we can measure energy from any zero we want. For relativistic purposes it is nice to measure the energy so that the rest mass is included, but for many purposes that aren’t relativistic it is often nice to subtract some standard amount from all energies that appear. For instance, in the case of an atom, it is usually convenient to subtract the energy M_s·c², where M_s is the mass of all the separate pieces—the nucleus and the electrons—which is, of course, different from the mass of the atom. For other problems, it may be useful to subtract from all energies the amount M_g·c², where M_g is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom. So, sometimes we may shift our zero of energy by some very large constant, but it doesn’t make any difference, provided we shift all the energies in a particular calculation by the same constant.”

It’s a rather long quotation, but it’s important. The key phrase here is, obviously, the following: “For other problems, it may be useful to subtract from all energies the amount M_g·c², where M_g is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom.” So that’s what he’s doing when solving Schrödinger’s equation. However, I should make the following point here: if we shift the origin of our energy scale, it does not make any difference in regard to the probabilities we calculate, but it obviously does make a difference in terms of our wavefunction itself. To be precise, its density in time will be very different. Hence, if we’d want to give the wavefunction some physical meaning – which is what I’ve been trying to do all along – it does make a huge difference. When we leave the rest mass of all of the pieces in our system out, we can no longer pretend we capture their energy.

This is a rather simple observation, but one that has profound implications in terms of our interpretation of the wavefunction. Personally, I admire the Great Teacher’s Lectures, but I am really disappointed that he doesn’t pay more attention to this. 😦

The Essence of Reality

Original post:

I know it’s a crazy title. It has no place in a physics blog, but then I am sure this article will go elsewhere. […] Well… […] Let me be honest: it’s probably gonna go nowhere. Whatever. I don’t care too much. My life is happier than Wittgenstein’s. 🙂

My original title for this post was: discrete spacetime. That was somewhat less offensive but, while being less offensive, it suffered from the same drawback: the terminology was ambiguous. The commonly accepted term for discrete spacetime is the quantum vacuum. However, because I am just an arrogant bastard trying to establish myself in this field, I am telling you that term is meaningless. Indeed, wouldn’t you agree that, if the quantum vacuum is a vacuum, then it’s empty. So it’s nothing. Hence, it cannot have any properties and, therefore, it cannot be discrete – or continuous, or whatever. We need to put stuff in it to make it real.

Therefore, I’d rather distinguish mathematical versus physical space. Of course, you are smart, and so you now you’ll say that my terminology is as bad as that of the quantum vacuumists. And you are right. However, this is a story that I am writing, and so I will write it the way I want to write it. 🙂 So where were we? Spacetime! Discrete spacetime.

Yes. Thank you! Because relativity tells us we should think in terms of four-vectors, we should not talk about space but about spacetime. Hence, we should distinguish mathematical spacetime from physical spacetime. So what’s the definitional difference?

Mathematical spacetime is just what it is: a coordinate space – Cartesian, polar, or whatever – which we define by choosing a representation, or a base. And all the other elements of the set are just some algebraic combination of the base set. Mathematical space involves numbers. They don’t – let me emphasize that: they do not!– involve the physical dimensions of the variables. Always remember: math shows us the relations, but it doesn’t show us the stuff itself. Think of it: even if we may refer to the coordinate axes as time, or distance, we do not really think of them as something physical. In math, the physical dimension is just a label. Nothing more. Nothing less.

In contrast, physical spacetime is filled with something – with waves, or with particles – so it’s spacetime filled with energy and/or matter. In fact, we should analyze matter and energy as essentially the same thing, and please do carefully re-read what I wrote: I said they are essentially the same. I did not say they are the same. Energy and mass are equivalent, but not quite the same. I’ll tell you what that means in a moment.

These waves, or particles, come with mass, energy and momentum. There is an equivalence between mass and energy, but they are not the same. There is a twist – literally (only after reading the next paragraphs, you’ll realize how literally): even when choosing our time and distance units such that c is numerically equal to 1 – e.g. when measuring distance in light-seconds (or time in light-meters), or when using Planck units – the physical dimension of the c²factor in Einstein’s E = mc²equation doesn’t vanish: the physical dimension of energy is kg·m²/s².

Using Newton’s force law (1 N = 1 kg·m/s²), we can easily see this rather strange unit is effectively equivalent to the energy unit, i.e. the joule (1 J = 1 kg·m²/s² = 1 (N·s²/m)·m²/s²= 1 N·m), but that’s not the point. The (m/s)² factor – i.e. the square of the velocity dimension – reflects the following:

Energy is nothing but mass in motion. To be precise, it’s oscillating mass. [And, yes, that’s what string theory is all about, but I didn’t want to mention that. It’s just terminology once again: I prefer to say ‘oscillating’ rather than ‘vibrating’. :-)]
The rapidly oscillating real and imaginary component of the matter-wave (or wavefunction, we should say) each capture half of the total energy of the object E = mc².
The oscillation is an oscillation of the mass of the particle (or wave) that we’re looking at.

In the mentioned publication, I explore the structural similarity between:

The oscillating electric and magnetic field vectors (E and B) that represent the electromagnetic wave, and
The oscillating real and imaginary part of the matter-wave.

The story is simple or complicated, depending on what you know already, but it can be told in an abnoxiously easy way. Note that the associated force laws do not differ in their structure:

Coulomb Law

gravitation law

The only difference is the dimension of m versus q: mass – the measure of inertia -versus charge. Mass comes in one color only, so to speak: it’s always positive. In contrast, electric charge comes in two colors: positive and negative. You can guess what comes next, but I won’t talk about that here. Just note the absolute distance between two charges (with the same or the opposite sign) is twice the distance between 0 and 1, which must explains the rather mysterious 2 factor I get for the Schrödinger equation for the electromagnetic wave (but I still need to show how that works out exactly).

The point is: remembering that the physical dimension of the electric field is N/C (newton per coulomb, i.e. force per unit of charge) it should not come as a surprise that we find that the physical dimension of the components of the matter-wave is N/kg: newton per kg, i.e. force per unit of mass. For the detail, I’ll refer you to that article of mine (and, because I know you will not want to work your way through it, let me tell you it’s the last chapter that tells you how to do the trick).

So where were we? Strange. I actually just wanted to talk about discrete spacetime here, but I realize I’ve already dealt with all of the metaphysical questions you could possible have, except the (existential) Who Am I? question, which I cannot answer on your behalf. 🙂

I wanted to talk about physical spacetime, so that’s sanitized mathematical space plus something. A date without logistics. Our mind is a lazy host, indeed.

Reality is the guest that brings all of the wine and the food to the party.

In fact, it’s a guest that brings everything to the party: you – the observer – just need to set the time and the place. In fact, in light of what Kant – and many other eminent philosophers – wrote about space and time being constructs of the mind, that’s another statement which you should interpret literally. So physical spacetime is spacetime filled with something – like a wave, or a field. So how does that look like? Well… Frankly, I don’t know! But let me share my idea of it.

Because of the unity of Planck’s quantum of action (ħ ≈ 1.0545718×10⁻³⁴ N·m·s), a wave traveling in spacetime might be represented as a set of discrete spacetime points and the associated amplitudes, as illustrated below. [I just made an easy Excel graph. Nothing fancy.]

spacetime

The space in-between the discrete spacetime points, which are separated by the Planck time and distance units, is not real. It is plain nothingness, or – if you prefer that term – the space in-between in is mathematical space only: a figment of the mind – nothing real, because quantum theory tells us that the real, physical, space is discontinuous.

Why is that so? Well… Smaller time and distance units cannot exist, because we would not be able to pack Planck’s quantum of action in them: a box of the Planck scale, with ħ in it, is just a black hole and, hence, nothing could go from here to there, because all would be trapped. Of course, now you’ll wonder what it means to ‘pack‘ Planck’s quantum of action in a Planck-scale spacetime box. Let me try to explain this. It’s going to be a rather rudimentary explanation and, hence, it may not satisfy you. But then the alternative is to learn more about black holes and the Schwarzschild radius, which I warmly recommend for two equivalent reasons:

The matter is actually quite deep, and I’d recommend you try to fully understand it by reading some decent physics course.
You’d stop reading this nonsense.

If, despite my warning, you would continue to read what I write, you may want to note that we could also use the logic below to define Planck’s quantum of action, rather than using it to define the Planck time and distance unit. Everything is related to everything in physics. But let me now give the rather naive explanation itself:

Planck’s quantum of action (ħ ≈ 1.0545718×10⁻³⁴ N·m·s) is the smallest thing possible. It may express itself as some momentum (whose physical dimension is N·s) over some distance (Δs), or as some amount of energy (whose dimension is N·m) over some time (Δt).
Now, energy is an oscillation of mass (I will repeat that a couple of times, and show you the detail of what that means in the last chapter) and, hence, ħ must necessarily express itself both as momentum as well as energy over some time and some distance. Hence, it is what it is: some force over some distance over some time. This reflects the physical dimension of ħ, which is the product of force, distance and time. So let’s assume some force ΔF, some distance Δs, and some time Δt, so we can write ħ as ħ = ΔF·Δs·Δt.
Now let’s pack that into a traveling particle – like a photon, for example – which, as you know (and as I will show in this publication) is, effectively, just some oscillation of mass, or an energy flow. Now let’s think about one cycle of that oscillation. How small can we make it? In spacetime, I mean.
If we decrease Δs and/or Δt, then ΔF must increase, so as to ensure the integrity (or unity) of ħ as the fundamental quantum of action. Note that the increase in the momentum (ΔF·Δt) and the energy (ΔF·Δs) is proportional to the decrease in Δt and Δs. Now, in our search for the Planck-size spacetime box, we will obviously want to decrease Δs and Δt simultaneously.
Because nothing can exceed the speed of light, we may want to use equivalent time and distance units, so the numerical value of the speed of light is equal to 1 and all velocities become relative velocities. If we now assume our particle is traveling at the speed of light – so it must be a photon, or a (theoretical) matter-particle with zero rest mass (which is something different than a photon) – then our Δs and Δt should respect the following condition: Δs/Δt = c = 1.
Now, when Δs = 1.6162×10⁻³⁵ m and Δt = 5.391×10⁻⁴⁴ s, we find that Δs/Δt = c, but ΔF = ħ/(Δs·Δt) = (1.0545718×10⁻³⁴ N·m·s)/[(1.6162×10⁻³⁵ m)·(5.391×10⁻⁴⁴ s)] ≈ 1.21×10⁴⁴ N. That force is monstrously huge. Think of it: because of gravitation, a mass of 1 kg in our hand, here on Earth, will exert a force of 9.8 N. Now note the exponent in that 1.21×10⁴⁴ number.
If we multiply that monstrous force with Δs – which is extremely tiny – we get the Planck energy: (1.6162×10⁻³⁵ m)·(1.21×10⁴⁴ N) ≈ 1.956×10⁹ joule. Despite the tininess of Δs, we still get a fairly big value for the Planck energy. Just to give you an idea, it’s the energy that you’d get out of burning 60 liters of gasoline—or the mileage you’d get out of 16 gallons of fuel! In fact, the equivalent mass of that energy, packed in such tiny space, makes it a black hole.
In short, the conclusion is that our particle can’t move (or, thinking of it as a wave, that our wave can’t wave) because it’s caught in the black hole it creates by its own energy: so the energy can’t escape and, hence, it can’t flow. 🙂

Of course, you will now say that we could imagine half a cycle, or a quarter of that cycle. And you are right: we can surely imagine that, but we get the same thing: to respect the unity of ħ, we’ll then have to pack it into half a cycle, or a quarter of a cycle, which just means the energy of the whole cycle is 2·ħ, or 4·ħ. However, our conclusion still stands: we won’t be able to pack that half-cycle, or that quarter-cycle, into something smaller than the Planck-size spacetime box, because it would make it a black hole, and so our wave wouldn’t go anywhere, and the idea of our wave itself – or the particle – just doesn’t make sense anymore.

This brings me to the final point I’d like to make here. When Maxwell or Einstein, or the quantum vacuumists – or I 🙂 – say that the speed of light is just a property of the vacuum, then that’s correct and not correct at the same time. First, we should note that, if we say that, we might also say that ħ is a property of the vacuum. All physical constants are. Hence, it’s a pretty meaningless statement. Still, it’s a statement that helps us to understand the essence of reality. Second, and more importantly, we should dissect that statement. The speed of light combines two very different aspects:

It’s a physical constant, i.e. some fixed number that we will find to be the same regardless of our reference frame. As such, it’s as essential as those immovable physical laws that we find to be the same in each and every reference frame.
However, its physical dimension is the ratio of the distance and the time unit: m/s. We may choose other time and distance units, but we will still combine them in that ratio. These two units represent the two dimensions in our mind that – as Kant noted – structure our perception of reality: the temporal and spatial dimension.

Hence, we cannot just say that c is ‘just a property of the vacuum’. In our definition of c as a velocity, we mix reality – the ‘outside world’ – with our perception of it. It’s unavoidable. Frankly, while we should obviously try – and we should try very hard! – to separate what’s ‘out there’ versus ‘how we make sense of it’, it is and remains an impossible job because… Well… When everything is said and done, what we observe ‘out there’ is just that: it’s just what we – humans – observe. 🙂

So, when everything is said and done, the essence of reality consists of four things:

Nothing
Mass, i.e. something, or not nothing
Movement (of something), from nowhere to somewhere.
Us: our mind. Or God’s Mind. Whatever. Mind.

The first is like yin and yang, or manicheism, or whatever dualistic religious system. As for Movement and Mind… Hmm… In some very weird way, I feel they must be part of one and the same thing as well. 🙂 In fact, we may also think of those four things as:

0 (zero)
1 (one), or as some sine or a cosine, which is anything in-between 0 and 1.
Well… I am not sure! I can’t really separate point 3 and point 4, because they combine point 1 and point 2.

So we’ve don’t have a quadrupality, right? We do have a Trinity here, don’t we? […] Maybe. I won’t comment, because I think I just found Unity here. 🙂

The wavefunction and relativity

When reading about quantum theory, and wave mechanics, you will often encounter the rather enigmatic statement that the Schrödinger equation is not relativistically correct. What does that mean?

In my previous post on the wavefunction and relativity, I boldly claimed that relativity theory had been around for quite a while when the young Comte Louis de Broglie wrote his short groundbreaking PhD thesis, back in 1924. Moreover, it is more than likely that he suggested the θ = ω∙t – k∙x = (E∙t – p∙x)/ħ formula for the argument of the wavefunction exactly because relativity theory had already established the invariance of the four-vector product p_μx_μ = E∙t – p∙x = p_μ‘x_μ‘ = E’∙t’ – p’∙x’. [Note that Planck’s constant, as a physical constant, should obviously not depend on the reference frame either. Hence, if the E∙t – p∙x product is invariant, so is (E∙t – p∙x)/ħ.] However, I didn’t prove that, and I didn’t relate it to Schrödinger’s equation. Hence, let’s explore the matter somewhat further here.

I don’t want to do the academic thing, of course – and that is to prove the invariance of the four-vector dot product. If you want such proof, let me just give you a link to some course material that does just that. Here, I will just summarize the conclusions of such course material:

Four-vector dot products – like x_μx_μ = x_μ², p_μp_μ = p_μ², the spacetime interval s²= (Δr)²– Δt², or our p_μx_μ product here – are invariant under a Lorentz transformation (aka as a Lorentz boost). To be formally correct, I should write x_μx^μ, p_μp^μ, and p_μx^μ, because the product multiplies a row vector with a column vector, which is what the sub- and superscript indicate.
Four-vector dot products are referred to as Lorentz scalars.
When derivatives are involved, we must use the so-called four-gradient, which is denoted by ∂ or ∇_μ and defined as:

∂ = ∇_μ = (∂/∂t, –∇) = (∂/∂t, –∂/∂x, –∂/∂y, –∂/∂z)

Applying the four-gradient vector operator to the wavefunction, we get:

∇_μψ= (∂ψ/∂t, –∇ψ) = (∂ψ/∂t, –∂ψ/∂x, –∂ψ/∂y, –∂ψ/∂z)

We wrote about that in the context of electromagnetic theory (see, for instance, my post on the relativistic transformation of fields), so I won’t dwell on it here. Note, however, that that’s the weak spot in Schrödinger’s equation: it’s good, but not good enough. However, in the context in which it’s being used – i.e. to calculate electron orbitals – the approximation works just fine, so you shouldn’t worry about it. The point to remember is that the wavefunction itself is relativistically correct. 🙂

Of course, it is always good to work through a simple example, so let’s do that here. Let me first remind you of that transformation we presented a couple of times already, and that’s how to calculate the argument of the wavefunction in the reference frame of the particle itself, i.e. the inertial frame. It goes like this: when measuring all variables in Planck units, the physical constants ħ and c are numerically equal to one, then we can then re-write the argument of the wavefunction as follows:

ħ = 1 ⇒ θ = (E∙t – p∙x)/ħ = E∙t – p∙x = E_v∙t − (m_v∙v)∙x
E_v= E₀/√(1−v²) and m_v= m₀/√(1−v²) ⇒ θ = [E₀/√(1−v²)]∙t – [m₀∙v/√(1−v²)]∙x
c = 1 ⇒ m₀ = E₀⇒ θ = [E₀/√(1−v²)]∙t – [E₀∙v/√(1−v²)]∙x = E₀∙(t − v∙x)/√(1−v²)

⇔ θ = E₀∙t’ = E’·t’ with t’ = (t − v∙x)/√(1−v²)

The t’ in the θ = E₀∙t’ expression is, obviously, the proper time as measured in the inertial reference frame. Needless to say, v is the relative velocity, which is usually denoted by β. Note that this derivation uses the numerical m₀ = E₀ identity, which emerges when using natural time and distance units (c = 1). However, while mass and energy are equivalent, they are different physical concepts and, hence, they still have different physical dimensions. It is interesting to spell out what happens with the dimensions here:

The dimension of E_vt and/or E₀∙t’ is (N∙m)∙s, i.e. the dimension of (physical) action.
The dimension of the (m_v∙v)∙x term must be the same, but how is that possible? Despite us using natural units – so the value of v is now some number between 0 and 1 – velocity is what it is: velocity. Hence, its dimension is m/s. Hence, the dimension of the m_v∙v∙x term is kg∙m = (N∙s²/m)∙(m/s)∙m = N∙m∙s.
Hence, the dimension of the [E₀∙v/√(1−v²)]∙x term only makes sense if we remember the m²/s² dimension of the c² factor in the E = m∙c² equivalence relation. We write: [E₀∙v∙x] = [E₀]∙[v]∙[x] = [(N∙m)∙(s²/m²)]∙(m/s)∙m = N∙m∙s. In short, when doing the m_v = E_v and/or m₀ = E₀ substitution, we should not get rid of the physical 1/c² dimension.

That should be clear enough. Let’s now do the example. The rest energy of an electron, expressed in Planck units, E_eP = E_e/E_P = (0.511×10⁶eV)/(1.22×10²⁸eV) = 4.181×10⁻²³. That is a very tiny fraction. However, the numerical value of the Planck time unit is even smaller: about 5.4×10⁻⁴⁴ seconds. Hence, as a frequency is expressed as the number of cycles (or, as an angular frequency, as the number of radians) per time unit, the natural frequency of the wavefunction of the electron is 4.181×10⁻²³ rad per Planck time unit, so that’s a frequency in the order of [4.181×10⁻²³/(2π)]/(5.4×10⁻⁴⁴ s) ≈ 1×10²⁰ cycles per second (or hertz). The relevant calculations are given hereunder.

Electron
Rest energy (in joule)	8.1871E-14
Planck energy (in joule)	1.9562E+09
Rest energy in Planck units	4.1853E-23
Frequency in cycles per second	1.2356E+20

Because of these rather incredible numbers (like 10^–31or 10²⁰), the calculations are not always very obvious, but the logic is clear enough: a higher rest mass increases the (angular) frequency of the real and imaginary part of the wavefunction, and gives them a much higher density in spacetime. How does a frequency like 1.235×10²⁰ Hz compare to, say, the frequency of gamma rays. The answer may surprise you: they are of the same order, as is their energy! 🙂 However, their nature, as a wave ,is obviously very different: gamma rays are an electromagnetic wave, so they involve an E and B vector, rather than the two components of the matter-wave. As an energy propagation mechanism, they are structurally similar, though, as I showed in my previous post.

Now, the typical speed of an electron is given by of the fine-structure constant (α), which is (also) equal to the is the (relative) speed of an electron (for the many interpretations of the fine-structure constant, see my post on it). So we write:

α = β = v/c

More importantly, we can use this formula to calculate it, which is done hereunder. As you can see, while the typical electron speed is quite impressive (about 2,188 km per second), it is only a fraction of the speed of light and, therefore, the Lorentz factor is still equal to one for all practical purposes. Therefore, its speed adds hardly anything to its energy.

Fine-structure constant	0.007297353
Typical speed of the electron (m/s)	2.1877E+06
Typical speed of the electron (km/s)	2,188 km/s
Lorentz factor (γ)	1.0000266267

But I admit it does have momentum now and, hence, the p∙x term in the θ = E∙t – p∙x comes into play. What is its momentum? That’s calculated below. Remember we calculate all in Planck units here!

Electron energy moving at alpha (in Planck units)	4.1854E-23
Electron mass moving at alpha (in Planck units)	4.1854E-23
Planck momentum (p = m·v = m·α )	3.0542E-25

The momentum is tiny, but it’s real. Also note the increase in its energy. Now, when substituting x for x = v·t, we get the following formula for the argument of our wavefunction:

θ = E·t – p·x = E·t − p·v·t = m_v·t − m_v·v·v·t = m_v·(1 − v²)·t

Now, how does that compare to our θ = θ = E₀∙t’ = E’·t’ expression? Well… The value of the two coefficients is calculated below. You can, effectively, see it hardly matters.

m_v·(1 − v²)	4.1852E-23
Rest energy in Planck units	4.1853E-23

With that, we are finally ready to use the non-relativistic Schrödinger equation in a non-relativistic way, i.e. we can start calculating electron orbitals with it now, which is what we did in one of my previous posts, but I will re-visit that post soon – and provide some extra commentary! 🙂

The Poynting vector for the matter-wave

Original post:

In my various posts on the wavefunction – which I summarized in my e-book – I wrote at the length on the structural similarities between the matter-wave and the electromagnetic wave. Look at the following images once more:

Animation 5d_euler_f

Both are the same, and then they are not. The illustration on the right-hand side is a regular quantum-mechanical wavefunction, i.e. an amplitude wavefunction: the x-axis represents time, so we are looking at the wavefunction at some particular point in space. [Of course, we could just switch the dimensions and it would all look the same.] The illustration on the left-hand side looks similar, but it is not an amplitude wavefunction. The animation shows how the electric field vector (E) of an electromagnetic wave travels through space. Its shape is the same. So it is the same function. Is it also the same reality?

Yes and no. The two energy propagation mechanisms are structurally similar. The key difference is that, in electromagnetics, we get two waves for the price of one. Indeed, the animation above does not show the accompanying magnetic field vector (B), which is equally essential. But, for the rest, Schrödinger’s equation and Maxwell’s equation model a similar energy propagation mechanism, as shown below.

They have to, as the force laws are similar too:

Coulomb Law

gravitation law

The only difference is that mass comes in one color only, so to speak: it’s always positive. In contrast, electric charge comes in two colors: positive and negative. You can now guess what comes next: quantum chromodynamics, but I won’t write about that here, because I haven’t studied that yet. I won’t repeat what I wrote elsewhere, but I want to make good on one promise, and that is to develop the idea of the Poynting vector for the matter-wave. So let’s do that now. Let me first remind you of the basic ideas, however.

Basics

The animation below shows the two components of the archetypal wavefunction, i.e. the sine and cosine:

circle_cos_sin

Think of the two oscillations as (each) packing half of the total energy of a particle (like an electron or a photon, for example). Look at how the sine and cosine mutually feed into each other: the sine reaches zero as the cosine reaches plus or minus one, and vice versa. Look at how the moving dot accelerates as it goes to the center point of the axis, and how it decelerates when reaching the end points, so as to switch direction. The two functions are exactly the same function, but for a phase difference of 90 degrees, i.e. a right angle. Now, I love engines, and so it makes me think of a V-2 engine with the pistons at a 90-degree angle. Look at the illustration below. If there is no friction, we have a perpetual motion machine: it would store energy in its moving parts, while not requiring any external energy to keep it going.

two-timer-576-px-photo-369911-s-original

If it is easier for you, you can replace each piston by a physical spring, as I did below. However, I should learn how to make animations myself, because the image below does not capture the phase difference. Hence, it does not show how the real and imaginary part of the wavefunction mutually feed into each other, which is (one of the reasons) why I like the V-2 image much better. 🙂

summary 2

The point to note is: all of the illustrations above are true representations – whatever that means – of (idealized) stationary particles, and both for matter (fermions) as well as for force-carrying particles (bosons). Let me give you an example. The (rest) energy of an electron is tiny: about 8.2×10⁻¹⁴ joule. Note the minus 14 exponent: that’s an unimaginably small amount. It sounds better when using the more commonly used electronvolt scale for the energy of elementary particles: 0.511 MeV. Despite its tiny mass (or energy, I should say, but then mass and energy are directly proportional to each other: the proportionality coefficient is given by the E = m·c² formula), the frequency of the matter-wave of the electron is of the order of 1×10²⁰ = 100,000,000,000,000,000,000 cycles per second. That’s an unimaginably large number and – as I will show when we get there – that’s not because the second is a huge unit at the atomic or sub-atomic scale.

We may refer to this as the natural frequency of the electron. Higher rest masses increase the frequency and, hence, give the wavefunction an even higher density in spacetime. Let me summarize things in a very simple way:

The (total) energy that is stored in an oscillating spring is the sum of the kinetic and potential energy (T and U) and is given by the following formula: E = T + U = a₀²·m·ω₀²/2. The a₀factor is the maximum amplitude – which depends on the initial conditions, i.e. the initial pull or push. The ω₀in the formula is the natural frequency of our spring, which is a function of the stiffness of the spring (k) and the mass on the spring (m): ω₀² = k/m.
Hence, the total energy that’s stored in two springs is equal to a₀²·m·ω₀².
The similarity between the E = a₀²·m·ω₀² and the E = m·c² formula is much more than just striking. It is fundamental: the two oscillating components of the wavefunction each store half of the total energy of our particle.
To emphasize the point: ω₀ = √(k/m) is, obviously, a characteristic of the system. Likewise, c = √(E/m) is just the same: a property of spacetime.

Of course, the key question is: what is that is oscillating here? In our V-2 engine, we have the moving parts. Now what exactly is moving when it comes to the wavefunction? The easy answer is: it’s the same thing. The V-2 engine, or our springs, store energy because of the moving parts. Hence, energy is equivalent only to mass that moves, and the frequency of the oscillation obviously matters, as evidenced by the E = a₀²·m·ω₀²/2 formula for the energy in a oscillating spring. Mass. Energy is moving mass. To be precise, it’s oscillating mass. Think of it: mass and energy are equivalent, but they are not the same. That’s why the dimension of the c² factor in Einstein’s famous E = m·c² formula matters. The equivalent energy of a 1 kg object is approximately 9×10¹⁶ joule. To be precise, it is the following monstrous number:

89,875,517,873,681,764 kg·m²/s²

Note its dimension: the joule is the product of the mass unit and the square of the velocity unit. So that, then, is, perhaps, the true meaning of Einstein’s famous formula: energy is not just equivalent to mass. It’s equivalent to mass that’s moving. In this case, an oscillating mass. But we should explore the question much more rigorously, which is what I do in the next section. Let me warn you: it is not an easy matter and, even if you are able to work your way through all of the other material below in order to understand the answer, I cannot promise you that the answer will satisfy you entirely. However, it will surely help you to phrase the question.

The Poynting vector for the matter-wave

For the photon, we have the electric and magnetic field vectors E and B. The boldface highlights the fact that these are vectors indeed: they have a direction as well as a magnitude. Their magnitude has a physical dimension. The dimension of E is straightforward: the electric field strength (E) is a quantity expressed in newton per coulomb (N/C), i.e. force per unit charge. This follows straight from the F = q·E force relation.

The dimension of B is much less obvious: the magnetic field strength (B) is measured in (N/C)/(m/s) = (N/C)·(s/m). That’s what comes out of the F = q·v×B force relation. Just to make sure you understand: v×B is a vector cross product, and yields another vector, which is given by the following formula:

a×b = |a×b|·n = |a|·|b|·sinφ·n

The φ in this formula is the angle between a and b (in the plane containing them) and, hence, is always some angle between 0 and π. The n is the unit vector that is perpendicular to the plane containing a and b in the direction given by the right-hand rule. The animation below shows it works for some rather special angles:

We may also need the vector dot product, so let me quickly give you that formula too. The vector dot product yields a scalar given by the following formula:

a•b = |a|·|b|·cosφ

Let’s get back to the F = q·v×B relation. A dimensional analysis shows that the dimension of B must involve the reciprocal of the velocity dimension in order to ensure the dimensions come out alright:

[F]= [q·v×B] = [q]·[v]·[B] = C·(m/s)·(N/C)·(s/m) = N

We can derive the same result in a different way. First, note that the magnitude of B will always be equal to E/c (except when none of the charges is moving, so B is zero), which implies the same:

[B] = [E/c] = [E]/[c] = (N/C)/(m/s) = (N/C)·(s/m)

Finally, the Maxwell equation we used to derive the wavefunction of the photon was ∂E/∂t = c²∇×B, which also tells us the physical dimension of B must involve that s/m factor. Otherwise, the dimensional analysis would not work out:

[∂E/∂t] = (N/C)/s = N/(C·s)
[c²∇×B] = [c²]·[∇×B] = (m²/s²)·[(N/C)·(s/m)]/m= N/(C·s)

This analysis involves the curl operator ∇×, which is a rather special vector operator. It gives us the (infinitesimal) rotation of a three-dimensional vector field. You should look it up so you understand what we’re doing here.

Now, when deriving the wavefunction for the photon, we gave you a purely geometric formula for B:

B = e_x×E = i·E

Now I am going to ask you to be extremely flexible: wouldn’t you agree that the B = E/c and the B = e_x×E = i·E formulas, jointly, only make sense if we’d assign the s/m dimension to e_x and/or to i? I know you’ll think that’s nonsense because you’ve learned to think of the e_x× and/or i· operation as a rotation only. What I am saying here is that it also transforms the physical dimension of the vector on which we do the operation: it multiplies it with the reciprocal of the velocity dimension. Don’t think too much about it, because I’ll do yet another hat trick. We can think of the real and imaginary part of the wavefunction as being geometrically equivalent to the E and B vector. Just compare the illustrations below:

e-and-b Rising_circular

Of course, you are smart, and you’ll note the phase difference between the sine and the cosine (illustrated below). So what should we do with that? Not sure. Let’s hold our breath for the moment.

Let’s first think about what dimension we could possible assign to the real part of the wavefunction. We said this oscillation stores half of the energy of the elementary particle that is being described by the wavefunction. How does that storage work for the E vector? As I explained in my post on the topic, the Poynting vector describes the energy flow in a varying electromagnetic field. It’s a bit of a convoluted story (which I won’t repeat here), but the upshot is that the energy density is given by the following formula:

Its shape should not surprise you. The formula is quite intuitive really, even if its derivation is not. The formula represents the one thing that everyone knows about a wave, electromagnetic or not: the energy in it is proportional to the square of its amplitude, and so that’s E•E = E² and B•B = B². You should also note he c²factor that comes with the B•B product. It does two things here:

As a physical constant, with some dimension of its own, it ensures that the dimensions on both sides of the equation come out alright.
The magnitude of B is 1/c of that of E, so cB = E, and so that explains the extra c² factor in the second term: we do get two waves for the price of one here and, therefore, twice the energy.

Speaking of dimensions, let’s quickly do the dimensional analysis:

E is measured in newton per coulomb, so [E•E] = [E²] = N²/C².
B is measured in (N/C)/(m/s), so we get [B•B] = [B²] = (N²/C²)·(s²/m²). However, the dimension of our c² factor is (m²/s²) and so we’re left with N²/C². That’s nice, because we need to add stuff that’s expressed in the same units.
The ε₀ is that ubiquitous physical constant in electromagnetic theory: the electric constant, aka as the vacuum permittivity. Besides ensuring proportionality, it also ‘fixes’ our units, and so we should trust it to do the same thing here, and it does: [ε₀] = C²/(N·m²), so if we multiply that with N²/C², we find that u is expressed in N/m².

Why is N/m² an energy density? The correct answer to that question involves a rather complicated analysis, but there is an easier way to think about it: just multiply N/m²with m/m, and then its dimension becomes N·m/m³= J/m³, so that’s joule per cubic meter. That looks more like an energy density dimension, doesn’t it? But it’s actually the same thing. In any case, I need to move on.

We talked about the Poynting vector, and said it represents an energy flow. So how does that work? It is also quite intuitive, as its formula really speaks for itself. Let me write it down:

Just look at it: u is the energy density, so that’s the amount of energy per unit volume at a given point, and so whatever flows out of that point must represent its time rate of change. As for the –∇•S expression… Well… The ∇• operator is the divergence, and so it give us the magnitude of a (vector) field’s source or sink at a given point. If C is a vector field (any vector field, really), then ∇•C is a scalar, and if it’s positive in a region, then that region is a source. Conversely, if it’s negative, then it’s a sink. To be precise, the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. So, in this case, it gives us the volume density of the flux of S. If you’re somewhat familiar with electromagnetic theory, then you will immediately note that the formula has exactly the same shape as the ∇•j = −∂ρ/∂t formula, which represents a flow of electric charge.

But I need to get on with my own story here. In order to not create confusion, I will denote the total energy by U, rather than E, because we will continue to use E for the magnitude of the electric field. We said the real and the imaginary component of the wavefunction were like the E and B vector, but what’s their dimension? It must involve force, but it should obviously not involve any electric charge. So what are our options here? You know the electric force law (i.e. Coulomb’s Law) and the gravitational force law are structurally similar:

Coulomb Law

gravitation law

So what if we would just guess that the dimension of the real and imaginary component of our wavefunction should involve a newton per kg factor (N/kg), so that’s force per mass unit rather than force per unit charge? But… Hey! Wait a minute! Newton’s force law defines the newton in terms of mass and acceleration, so we can do a substitution here: 1 N = 1 kg·m/s² ⇔ 1 kg = 1 N·s²/m. Hence, our N/kg dimension becomes:

N/kg = N/(N·s²/m)= m/s²

What is this: m/s²? Is that the dimension of the a·cosθ term in the a·e^−i·θ= a·cosθ − i·a·sinθ wavefunction? I hear you. This is getting quite crazy, but let’s see where it leads us. To calculate the equivalent energy density, we’d then need an equivalent for the ε₀ factor, which – replacing the C by kg in the [ε₀] = C²/(N·m²) expression – would be equal to kg²/(N·m²). Because we know what we want (energy is defined using the force unit, not the mass unit), we’ll want to substitute the kg unit once again, so – temporarily using the μ₀ symbol for the equivalent of that ε₀ constant – we get:

[μ₀] = [N·s²/m]²/(N·m²) = N·s⁴/m⁴

Hence, the dimension of the equivalent of that ε₀·E² term becomes:

[(μ₀/2)]·[cosθ]² = (N·s⁴/m⁴)·m²/s⁴= N/m²

Bingo! How does it work for the other component? The other component has the imaginary unit (i) in front. If we continue to pursue our comparison with the E and B vectors, we should assign an extra s/m dimension because of the e_x and/or i factor, so the physical dimension of the i·sinθ term would be (m/s²)·(s/m) = s. What? Just the second? Relax. That second term in the energy density formula has the c² factor, so it all works out:

[(μ₀/2)]·[c²]·[i·sinθ]² = [(μ₀/2)]·[c²]·[i]²·[sinθ]² (N·s⁴/m⁴)·(m²/s²)·(s²/m²)·m²/s⁴= N/m²

As weird as it is, it all works out. We can calculate u and, hence, we can now also calculate the equivalent Poynting vector (S). However, I will let you think about that as an exercise. 🙂 Just note the grand conclusions:

The physical dimension of the argument of the wavefunction is physical action (newton·meter·second) and Planck’s quantum of action is the scaling factor.
The physical dimension of both the real and imaginary component of the elementary wavefunction is newton per kg (N/kg). This allows us to analyze the wavefunction as an energy propagation mechanism that is structurally similar to Maxwell’s equations, which represent the energy propagation mechanism when electromagnetic energy is involved.

As such, all we presented so far was a deep exploration of the mathematical equivalence between the gravitational and electromagnetic force laws:

Coulomb Law

gravitation law

Despite our grand conclusions, you should note we have not answered the most fundamental question of all. What is mass? What is electric charge? We have all these relations and equations, but are we any wiser, really? The answer to that question probably lies in general relativity: mass is that what curves spacetime. Likewise, we may look at electric charge as causing a very special type of spacetime curvature. However, even such answer – which would involve a much more complicated mathematical analysis – may not satisfy you. In any case, I will let you digest this post. I hope you enjoyed it as much as I enjoyed writing it. 🙂

Post scriptum: Of all of the weird stuff I presented here, I think the dimensional analyses were the most interesting. Think of the N/kg = N/(N·s²/m)= m/s²identity, for example. The m/s² dimension is the dimension of physical acceleration (or deceleration): the rate of change of the velocity of an object. The identity comes straight out of Newton’s force law:

F = m·a ⇔ F/m = a

Now look, once again, at the animation, and remember the formula for the argument of the wavefunction: θ = E₀∙t’. The energy of the particle that is being described is the (angular) frequency of the real and imaginary components of the wavefunction.

circle_cos_sin

The relation between (1) the (angular) frequency of a harmonic oscillator (which is what the sine and cosine represent here) and (2) the acceleration along the axis is given by the following equation:

a(x) = −ω₀²·x

I’ll let you think about what that means. I know you will struggle with it – because I did – and, hence, let me give you the following hint:

The energy of an ordinary string wave, like a guitar string oscillating in one dimension only, will be proportional to the square of the frequency.
However, for two-dimensional waves – such as an electromagnetic wave – we find that the energy is directly proportional to the frequency. Think of Einstein’s E = h·f = ħ·ω relation, for example. There is no squaring here!

It is a strange observation. Those two-dimensional waves – the matter-wave, or the electromagnetic wave – give us two waves for the price of one, each carrying half of the total energy but, as a result, we no longer have that square function. Think about it. Solving the mystery will make you feel like you’ve squared the circle, which – as you know – is impossible. 🙂

Quantum Mechanics: The Other Introduction

About three weeks ago, I brought my most substantial posts together in one document: it’s the Deep Blue page of this site. I also published it on Amazon/Kindle. It’s nice. It crowns many years of self-study, and many nights of short and bad sleep – as I was mulling over yet another paradox haunting me in my dreams. It’s been an extraordinary climb but, frankly, the view from the top is magnificent. 🙂

The offer is there: anyone who is willing to go through it and offer constructive and/or substantial comments will be included in the book’s acknowledgements section when I go for a second edition (which it needs, I think). First person to be acknowledged here is my wife though, Maria Elena Barron, as she has given me the spacetime and, more importantly, the freedom to take this bull by its horns.Below I just copy the foreword, just to give you a taste of it. 🙂

Foreword

Another introduction to quantum mechanics? Yep. I am not hoping to sell many copies, but I do hope my unusual background—I graduated as an economist, not as a physicist—will encourage you to take on the challenge and grind through this.

I’ve always wanted to thoroughly understand, rather than just vaguely know, those quintessential equations: the Lorentz transformations, the wavefunction and, above all, Schrödinger’s wave equation. In my bookcase, I’ve always had what is probably the most famous physics course in the history of physics: Richard Feynman’s Lectures on Physics, which have been used for decades, not only at Caltech but at many of the best universities in the world. Plus a few dozen other books. Popular books—which I now regret I ever read, because they were an utter waste of time: the language of physics is math and, hence, one should read physics in math—not in any other language.

But Feynman’s Lectures on Physics—three volumes of about fifty chapters each—are not easy to read. However, the experimental verification of the existence of the Higgs particle in CERN’s LHC accelerator a couple of years ago, and the award of the Nobel prize to the scientists who had predicted its existence (including Peter Higgs and François Englert), convinced me it was about time I take the bull by its horns. While, I consider myself to be of average intelligence only, I do feel there’s value in the ideal of the ‘Renaissance man’ and, hence, I think stuff like this is something we all should try to understand—somehow. So I started to read, and I also started a blog (www.readingfeynman.org) to externalize my frustration as I tried to cope with the difficulties involved. The site attracted hundreds of visitors every week and, hence, it encouraged me to publish this booklet.

So what is it about? What makes it special? In essence, it is a common-sense introduction to the key concepts in quantum physics. However, while common-sense, it does not shy away from the math, which is complicated, but not impossible. So this little book is surely not a Guide to the Universe for Dummies. I do hope it will guide some Not-So-Dummies. It basically recycles what I consider to be my more interesting posts, but combines them in a comprehensive structure.

It is a bit of a philosophical analysis of quantum mechanics as well, as I will – hopefully – do a better job than others in distinguishing the mathematical concepts from what they are supposed to describe, i.e. physical reality.

Last but not least, it does offer some new didactic perspectives. For those who know the subject already, let me briefly point these out:

I. Few, if any, of the popular writers seems to have noted that the argument of the wavefunction (θ = E·t – p·t) – using natural units (hence, the numerical value of ħ and c is one), and for an object moving at constant velocity (hence, x = v·t) – can be written as the product of the proper time of the object and its rest mass:

θ = E·t – p·x = E·t − p·x = m_v·t − m_v·v·x = m_v·(t − v·x)

⇔ θ = m₀·(t − v·x)/√(1 – v²) = m₀·t’

Hence, the argument of the wavefunction is just the proper time of the object with the rest mass acting as a scaling factor for the time: the internal clock of the object ticks much faster if it’s heavier. This symmetry between the argument of the wavefunction of the object as measured in its own (inertial) reference frame, and its argument as measured by us, in our own reference frame, is remarkable, and allows to understand the nature of the wavefunction in a more intuitive way.

While this approach reflects Feynman’s idea of the photon stopwatch, the presentation in this booklet generalizes the concept for all wavefunctions, first and foremost the wavefunction of the matter-particles that we’re used to (e.g. electrons).

II. Few, if any, have thought of looking at Schrödinger’s wave equation as an energy propagation mechanism. In fact, when helping my daughter out as she was trying to understand non-linear regression (logit and Poisson regressions), it suddenly realized we can analyze the wavefunction as a link function that connects two physical spaces: the physical space of our moving object, and a physical energy space.

Re-inserting Planck’s quantum of action in the argument of the wavefunction – so we write θ as θ = (E/ħ)·t – (p/ħ)·x = [E·t – p·x]/ħ – we may assign a physical dimension to it: when interpreting ħ as a scaling factor only (and, hence, when we only consider its numerical value, not its physical dimension), θ becomes a quantity expressed in newton·meter·second, i.e. the (physical) dimension of action. It is only natural, then, that we would associate the real and imaginary part of the wavefunction with some physical dimension too, and a dimensional analysis of Schrödinger’s equation tells us this dimension must be energy.

This perspective allows us to look at the wavefunction as an energy propagation mechanism, with the real and imaginary part of the probability amplitude interacting in very much the same way as the electric and magnetic field vectors E and B. This leads me to the next point, which I make rather emphatically in this booklet: the propagation mechanism for electromagnetic energy – as described by Maxwell’s equations – is mathematically equivalent to the propagation mechanism that’s implicit in the Schrödinger equation.

I am, therefore, able to present the Schrödinger equation in a much more coherent way, describing not only how this famous equation works for electrons, or matter-particles in general (i.e. fermions or spin-1/2 particles), which is probably the only use of the Schrödinger equation you are familiar with, but also how it works for bosons, including the photon, of course, but also the theoretical zero-spin boson!

In fact, I am personally rather proud of this. Not because I am doing something that hasn’t been done before (I am sure many have come to the same conclusions before me), but because one always has to trust one’s intuition. So let me say something about that third innovation: the photon wavefunction.

III. Let me tell you the little story behind my photon wavefunction. One of my acquaintances is a retired nuclear scientist. While he knew I was delving into it all, I knew he had little time to answer any of my queries. However, when I asked him about the wavefunction for photons, he bluntly told me photons didn’t have a wavefunction. I should just study Maxwell’s equations and that’s it: there’s no wavefunction for photons: just this traveling electric and a magnetic field vector. Look at Feynman’s Lectures, or any textbook, he said. None of them talk about photon wavefunctions. That’s true, but I knew he had to be wrong. I mulled over it for several months, and then just sat down and started doing to fiddle with Maxwell’s equations, assuming the oscillations of the E and B vector could be described by regular sinusoids. And – Lo and behold! – I derived a wavefunction for the photon. It’s fully equivalent to the classical description, but the new expression solves the Schrödinger equation, if we modify it in a rather logical way: we have to double the diffusion constant, which makes sense, because E and B give you two waves for the price of one!

[…]

In any case, I am getting ahead of myself here, and so I should wrap up this rather long introduction. Let me just say that, through my rather long journey in search of understanding – rather than knowledge alone – I have learned there are so many wrong answers out there: wrong answers that hamper rather than promote a better understanding. Moreover, I was most shocked to find out that such wrong answers are not the preserve of amateurs alone! This emboldened me to write what I write here, and to publish it. Quantum mechanics is a logical and coherent framework, and it is not all that difficult to understand. One just needs good pointers, and that’s what I want to provide here.

As of now, it focuses on the mechanics in particular, i.e. the concept of the wavefunction and wave equation (better known as Schrödinger’s equation). The other aspect of quantum mechanics – i.e. the idea of uncertainty as implied by the quantum idea – will receive more attention in a later version of this document. I should also say I will limit myself to quantum electrodynamics (QED) only, so I won’t discuss quarks (i.e. quantum chromodynamics, which is an entirely different realm), nor will I delve into any of the other more recent advances of physics.

In the end, you’ll still be left with lots of unanswered questions. However, that’s quite OK, as Richard Feynman himself was of the opinion that he himself did not understand the topic the way he would like to understand it. But then that’s exactly what draws all of us to quantum physics: a common search for a deep and full understanding of reality, rather than just some superficial description of it, i.e. knowledge alone.

So let’s get on with it. I am not saying this is going to be easy reading. In fact, I blogged about much easier stuff than this in my blog—treating only aspects of the whole theory. This is the whole thing, and it’s not easy to swallow. In fact, it may well too big to swallow as a whole. But please do give it a try. I wanted this to be an intuitive but formally correct introduction to quantum math. However, when everything is said and done, you are the only who can judge if I reached that goal.

Of course, I should not forget the acknowledgements but… Well… It was a rather lonely venture, so I am only going to acknowledge my wife here, Maria, who gave me all of the spacetime and all of the freedom I needed, as I would get up early, or work late after coming home from my regular job. I sacrificed weekends, which we could have spent together, and – when mulling over yet another paradox – the nights were often short and bad. Frankly, it’s been an extraordinary climb, but the view from the top is magnificent.

I just need to insert one caution, my site (www.readingfeynman.org) includes animations, which make it much easier to grasp some of the mathematical concepts that I will be explaining. Hence, I warmly recommend you also have a look at that site, and its Deep Blue page in particular – as that page has the same contents, more or less, but the animations make it a much easier read.

Have fun with it!

Jean Louis Van Belle, BA, MA, BPhil, Drs.

Wave functions and equations: a summary

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link. In fact, I actually made some (small) mistakes when writing the post below.

Original post:

Schrödinger’s wave equation for spin-zero, spin-1/2, and spin-one particles in free space differ from each other by a factor two:

For particles with zero spin, we write: ∂ψ/∂t = i·(ħ/m)·∇²ψ. We get this by multiplying the ħ/(2m) factor in Schrödinger’s original wave equation – which applies to spin-1/2 particles (e.g. electrons) only – by two. Hence, the correction that needs to be made is very straightforward.
For fermions (spin-1/2 particles), Schrödinger’s equation is what it is: ∂ψ/∂t = i·[ħ/(2m)]·∇²ψ.
For spin-1 particles (photons), we have ∂ψ/∂t = i·(2ħ/m)·∇²ψ, so here we multiply the ħ/m factor in Schrödinger’s wave equation for spin-zero particles by two, which amounts to multiplying Schrödinger’s original coefficient by four.

Look at the coefficients carefully. It’s a strange succession:

The ħ/m factor (which is just the reciprocal of the mass measured in units of ħ) works for spin-0 particles.
For spin-1/2 particles, we take only half that factor: ħ/(2m) = (1/2)·(ħ/m).
For spin-1 particles, we double that factor: 2ħ/m = 2·(ħ/m).

I describe the detail on my Deep Blue page, so please go there for more detail. What I did there, can be summarized as follows:

The spin-one particle is the photon, and we derived the photon wavefunction from Maxwell’s equations in free space, and found that it solves the ∂ψ/∂t = i·(2ħ/m)·∇²ψ equation, not the ∂ψ/∂t = i·(ħ/m)·∇²ψ or ∂ψ/∂t = i·[ħ/(2m)]·∇²ψ equations.
As for the spin-zero particles, we simplified the analysis by assuming our particle had zero rest mass, and we found that we were basically modeling an energy flow.
The analysis for spin-1/2 particles is just the standard analysis you’ll find in textbooks.

We can speculate how things would look like for spin-3/2 particles, or for spin-2 particles, but let’s not do that here. In any case, we will come back to this. Let’s first focus on the more familiar terrain, i.e. the wave equation for spin-1/2 particles, such as protons or electrons. [A proton is not elementary – as it consists of quarks – but it is a spin-1/2 particle, i.e. a fermion.]

The phase and group velocity of the wavefunction for spin-1/2 particles (fermions)

We’ll start with the very beginning of it all, i.e. the two equations that the young Comte Louis de Broglie presented in his 1924 PhD thesis, which give us the temporal and spatial frequency of the wavefunction, i.e. the ω and k in the θ = ω·t − k·t argument of the a·e^–i·θwavefunction:

ω = E/ħ
k = p/ħ

This allows to calculate the phase velocity of the wavefunction:

v_p = ω/k = (E/ħ)/(p/ħ) = E/p

This is an elementary wavefunction, several of which we would add with appropriate coefficients, with uncertainty in the energy and momentum ensuring our component waves have different frequencies, and, therefore, the concept of a group velocity does not apply. In effect, the a·e^–i·θwavefunction does not describe a localized particle: the probability to find it somewhere is the same everywhere. We may want to think of our wavefunction being confined to some narrow band in space, with us having no prior information about the probability density function, and, therefore, we assume a uniform distribution. Assuming our box in space is defined by Δx = x₂ − x₁, and imposing the normalization condition (all probabilities have to add up to one), we find that the following logic should hold:

(Δx)·a² = (x₂−x₁)·a²= 1 ⇔ Δx = 1/a²

However, we are, of course, interested in the group velocity, as the group velocity should correspond to the classical velocity of the particle. The group velocity of a composite wave is given by the v_g = ∂ω/∂k formula. Of course, that formula assumes an unambiguous relation between the temporal and spatial frequency of the component waves, which we may want to denote as ω_n and k_n, with n = 1, 2, 3,… However, we will not use the index as the context makes it quite clear what we are talking about.

The relation between ω_n and k_n is known as the dispersion relation, and one particularly nice way to calculate ω as a function of kis to distinguish the real and imaginary parts of the ∂ψ/∂t =i·[ħ/(2m)]·∇²ψ wave equation and, hence, re-write it as a pair of two equations:

Re(∂ψ_B/∂t) = −[ħ/(2m)]·Im(∇²ψ_B) ⇔ ω·cos(kx − ωt) = k²·[ħ/(2m)]·cos(kx − ωt)
Im(∂ψ_B/∂t) = [ħ/(2m)]·Re(∇²ψ_B) ⇔ ω·sin(kx − ωt) = k²·[ħ/(2m)]·sin(kx − ωt)

Both equations imply the following dispersion relation:

ω = ħ·k²/(2m)

We can now calculate v_g = ∂ω/∂k as:

v_g = ∂ω/∂k = ∂[ħ·k²/(2m)]/∂k = 2ħk/(2m) = ħ·(p/ħ)/m = p/m = m·v/m = v

That’s nice, because it’s what we wanted to find. If the group velocity would not equal the classical velocity of our particle, then our model would not make sense.

We used the classical momentum formula in our calculation above: p = m·v. To calculate the phase velocity of our wavefunction, we need to calculate that E/p ratio and, hence, we need an energy formula. Here we have a lot of choice, as energy can be defined in many ways: is it rest energy, potential energy, or kinetic energy? At this point, I need to remind you of the basic concepts.

The argument of the wavefunction as the proper time

It is obvious that the energy concept that is to be used in the ω = E/ħ is the total energy. Louis de Broglie himself noted that the energy of a particle consisted of three parts:

The particle’s rest energy m₀c², which de Broglie referred to as internal energy (E_int): it includes the rest mass of the ‘internal pieces’, as de Broglie put it (now we call those ‘internal pieces’ quarks), as well as their binding energy (i.e. the quarks’ interaction energy);
Any potential energy (V) it may have because of some field (so de Broglie was not assuming the particle was traveling in free space): the field(s) can be anything—gravitational, electromagnetic—you name it: whatever changes the energy because of the position of the particle;
The particle’s kinetic energy, which he wrote in terms of its momentum p: K.E. = m·v²/2 = m²·v²/(2m) = (m·v)²/(2m) = p²/(2m).

So the wavefunction, as de Broglie wrote it, can be written as follows:

ψ(θ) = ψ(x, t) = a·e⁻ⁱ^θ = a·e^{−i[(E_int+ p²/(2m) + V)·t − p∙x]/ħ}

This formula allows us to analyze interesting phenomena such as the tunneling effect and, hence, you may want to stop here and start playing with it. However, you should note that the kinetic energy formula that is used here is non-relativistic. The relativistically correct energy formula is E = m_vc², and the relativistically correct formula for the kinetic energy is the difference between the total energy and the rest energy:

K.E. = E − E₀ = m_v·c² − m₀·c² = m₀·γ·c² − m₀·c² = m₀·c²·(γ − 1), with γ the Lorentz factor.

At this point, we should simplify our calculations by adopting natural units, so as to ensure the numerical value of c = 1, and likewise for ħ. Hence, we assume all is described in Planck units, but please note that the physical dimensions of our variables do not change when adopting natural units: time is time, energy is energy, etcetera. But when using natural units, the E = m_vc² reduces to E = m_v. As for our formula for the momentum, this formula remains p = m_v·v, but v is now some relative velocity, i.e. a fraction between 0 and 1. We can now re-write θ = (E/ħ)·t – (p/ħ)·x as:

θ = E·t – p·x = E·t − p·v·t = m_v·t − m_v·v·v·t = m_v·(1 − v²)·t

We can also write this as:

ψ(x, t) = a·e^{−i·(m_v·t − p∙x)}= a·e^{−i·[(m₀/√(1−v²))·t − (m₀·v/√(1−v²)∙x)}= a·e^{−i·m₀·(t − v∙x)/√(1−v²)}

The (t − v∙x)/√(1−v²) factor in the argument is the proper time of the particle as given by the formulas for the Lorentz transformation of spacetime:

relativity

However, both the θ = m_v·(1 − v²)·t and θ = m₀·t’ = m₀·(t − v∙x)/√(1−v²) are relativistically correct. Note that the rest mass of the particle (m₀) acts as a scaling factor as we multiply it with the proper time: a higher m₀ gives the wavefunction a higher density, in time as well as in space.

Let’s go back to our v_p = E/p formula. Using natural units, it becomes:

v_p = E/p = m_v/m_v·v = 1/v

Interesting! The phase velocity is the reciprocal of the classical velocity! This implies it is always superluminal, ranging from v_p = ∞ to v_p = c = 1 for v going from 0 to 1 = c, as illustrated in the simple graph below.

Let me note something here, as you may also want to use the dispersion relation, i.e. ω = ħ·k²/(2m), to calculate the phase velocity. You’d write:

v_p = ω/k = [ħ·k²/(2m)]/k = ħ·k/(2m) = ħ·(p/ħ)/(2m) = m·v/(2m) = v/2

That’s a nonsensical result. Why do we get it? Because we are mixing two different mass concepts here: the mass that’s associated with the component wave, and the mass that’s associated with the composite wave. Think of it. That’s where Schrödinger’s equation is different from all of the other diffusion equations you’ve seen: the mass factor in the ∂ψ/∂t = i·[ħ/(2m)]·∇²ψ equation is the mass of the particle that’s being represented by the wavefunction that solves the equation. Hence, the diffusion constant ħ/(2m) is not a property of the medium. In that sense, it’s different from the κ/k factor in the ∂T/∂t = (κ/k)·∇²T heat diffusion, for example. We don’t have a medium here and, therefore, Schrödinger’s equation and the associated wavefunction are intimately connected.

It’s an interesting point, because if we’re going to be measuring the mass as multiples of ħ/2 (as suggested by the ħ/(2m) = 1/[m/[ħ/2)] factor itself), then its possible values (for ħ = 1) will be 1/2, 1, 3/2, 2, 5/2,… Now that should remind you of a few things—things like harmonics, or allowable spin values, or… Well… So many things. 🙂

Let’s do the exercise for bosons now.

The phase and group velocity of the wavefunction for spin-0 particles

My Deep Blue page explains why we need to drop the 1/2 factor in Schrödinger’s equation to make it fit the wavefunction for bosons. We distinguished two bosons: (1) the (theoretical) zero-mass particle (which has spin zero), and the (actual) photon (which has spin one). Let’s first do the analysis for the spin-zero particle.

A zero-mass particle (i.e. a particle with zero rest mass) should be traveling at the speed of light: both its phase as well as its group velocity should be equal to c = 1. In fact, we’re not talking composite wavefunctions here, so there’s no such thing as a group velocity. We’re not adding waves: there is only one wavefunction. [Note that we don’t need to add waves with different frequencies in order to localize our particle, because quantum mechanics and relativity theory come together here in what might well be the most logical and absurd conclusion ever: as an outside observer, we’re going to see all those zero-mass particles as point objects whizzing by because of the relativistic length contraction. So their wavefunction is only all over spacetime in their proper space and time, but not in ours!]
Now, it’s easy to show that, if we choose our time and distance units such that c = 1, then the energy formula reduces to E = m∙c² = m. Likewise, we find that p = m∙c = m. So we have this strange condition: E = p = m.
Now, this is not consistent with the ω = ħ·k²/(2m) we get out of the ∂ψ/∂t = i·[ħ/(2m)]·∇²ψ equation, because E/ħ = ħ·(p/ħ)²/(2m) ⇔ E = m²/(2m) = m/2. That does not fit the E = p = m condition. The only way out is to drop the 1/2 factor, i.e. to multiply Schrödinger’s coefficient with 2.

Let’s quickly check if it does the trick. We assume E, p and m will be multiples of ħ/2 (E = p = m = n·(ħ/2), so the wavefunction is e^{−i∙[t − x]n·/2}, Schrödinger’s constant becomes 2/n, and the derivatives for ∂ψ/∂t = i·(ħ/m)·∇²ψ are:

∂ψ/∂t = −i·(n/2)·e^{−i∙[t − x]·n/2}
∇²ψ = ∂²[e^{−i∙[t − x]·n/2}]/∂x²= i·(n/2)·∂[e^{−i∙[t − x]·n/2}]/∂x = −(n²/4)·e^{−i∙[t − x]·n/2}

So the Schrödinger equation becomes:

−i·(n/2)·e^{−i∙[t − x]n·/2}) = −i·(2/n)·(n²/4)·e^{−i∙[t − x]·n/2} ⇔ n/2 = n/2 ⇔ 1 = 1

As Feynman would say, it works like a charm, and note that n does not have to be some integer to make this work.

So what makes spin-1/2 particles different? The answer is: they have both linear as well as angular momentum, and the equipartition theorem tells us the energy will be shared equally among both , so they will pick up linear and angular momentum. Hence, the associated condition is not E = p = m, but E = p = 2m. We’ll come back to this.

Let’s now summarize how it works for spin-one particles

The phase and group velocity of the wavefunction for spin-1 particles (photons)

Because of the particularities that characterize an electromagnetic wave, the wavefunction packs two waves, capturing both the electric as well as the magnetic field vector (i.e. E and B). For the detail, I’ll refer you to the mentioned page, because the proof is rather lengthy (but easy to follow, so please do check it out). I will just briefly summarize the logic here.

1. For the spin-zero particle, we measured E, m and p in units of – or as multiples of – the ħ/2 factor. Hence, the elementary wavefunction (i.e. the wavefunction for E = p = m = 1) for the zero-mass particle is eⁱ^{(x/2 − t/2)}.

2. For the spin-1 particle (the photon), one can show that we get two of these elementary wavefunctions (ψ_Eand ψ_B), and one can then prove that we can write the sum of the electric and magnetic field vector as:

E + B = E + B = ψ_E+ ψ_B= E + i·E

= √2·eⁱ^{(x/2 − t/2+ π/4)}= √2·eⁱ^(π/4)·eⁱ^{(x/2 − t/2)}= √2·eⁱ^(π/4)·E = √2·eⁱ^(π/4)·eⁱ^{(x/2 − t/2)}

Hence, the photon has a special wavefunction. Does it solve the Schrödinger equation? It does when we use the 2ħ/m diffusion constant, rather than the ħ/m or ħ/(2m) coefficient. Let us quickly check it. The derivatives are:

∂ψ/∂t = −√2·eⁱ^(π/4)·e^{−i∙[t − x]/2}·(i/2)
∇²ψ = ∂²[√2·eⁱ^(π/4)·e^{−i∙[t − x]/2}]/∂x²= √2·eⁱ^(π/4)·∂[e^{−i∙[t − x]/2}·(i/2)]/∂x = −√2·eⁱ^(π/4)·e^{−i∙[t − x]/2}·(1/4)

Note, however, that we have two mass, energy and momentum concepts here: E_E, p_E, m_E and E_B, p_B, and m_B respectively. Hence, if E_E= p_E= m_E = E_B= p_B= m_B = 1/2, then E = E_E+ E_B, p = p_E+ p_B and m = m_E+ m_Bare all equal to 1. Hence, because E = p = m = 1 and we measure in units of ħ, the 2ħ/m factor is equal to 2 and, therefore, the modified Schrödinger’s equation ∂ψ/∂t = i·(2ħ/m)·∇²ψ becomes:

−i·√2·eⁱ^(π/4)·e^{−i∙[t − x]/2}·(1/2) = −i·√2·2·eⁱ^(π/4)·e^{−i∙[t − x]/2}·(1/4) ⇔ 1/2 = 2/4 = 1/2

It all works out. Let’s quickly check it for E, m and p being multiples of ħ, so we write: E = p = m = n·ħ = n, so the wavefunction is √2·eⁱ^(π/4)·e^{−i∙[t − x]n·/2}, Schrödinger’s 2ħ/m constant becomes 2ħ/m = 2/n, and the derivatives for ∂ψ/∂t = i·(ħ/m)·∇²ψ are:

∂ψ/∂t = −i·(n/2)·√2·eⁱ^(π/4)·e^{−i∙[t − x]·n/2}
∇²ψ = ∂²[e^{−i∙[t − x]·n/2}]/∂x²= i·√2·eⁱ^(π/4)·(n/2)·∂[e^{−i∙[t − x]·n/2}]/∂x = −√2·(n²/4)·eⁱ^(π/4)·e^{−i∙[t − x]·n/2}

So the Schrödinger equation becomes:

−i·√2·eⁱ^(π/4)·(n/2)·e^{−i∙[t − x]·n/2}) = −i·√2·eⁱ^(π/4)·(2/n)·(n²/4)·e^{−i∙[t − x]·n/2} ⇔ n/2 = n/2 ⇔ 1 = 1

It works like a charm again. Note the subtlety of the difference between the ħ/(2m) and 2ħ/m factor: it depends on us measuring the mass (and, hence, the energy and momentum as well) in units of ħ/2 (for spin-0 particles) or, alternatively (for spin-1 particles), in units of ħ. This is very deep—but it does make sense in light of the E_n =n·ħ·ω = n·h·f formula that solves the black-body radiation problem, as illustrated below. [The formula next to the energy levels is the probability of an atomic oscillator occupying that energy level, which is given by Boltzmann’s Law. You can check things in my post on it.]

It is now time to look at something else.

Schrödinger’s equation as an energy propagation mechanism

The Schrödinger equations above are not complete. The complete equation includes force fields, i.e. potential energy:

To write the equation like this, we need to move the i on the right-hand side of our ∂ψ/∂t = i·(2ħ/m)·∇²ψ equation to the other side, and multiply both sides with −1. [Remember: 1/i = −i.] Now, it is very interesting to do a dimensional analysis of this equation. Let’s do the right-hand side first. The ħ²factor in the ħ/(2m) is expressed in J²·s². Now that doesn’t make much sense, but then that mass factor in the denominator makes everything come out alright. Indeed, we can use the mass-equivalence relation to express m in J/(m/s)² units. So we get: (J²·s²)·[(m/s)²/J] = J·m². But so we multiply that with some quantity (the Laplacian) that’s expressed per m². So −(ħ²/2m)·∇²ψ is something expressed in joule, so it’s some amount of energy! Interesting, isn’t it? [Note that it works out fine with the addition Vψ term, which is also expressed in joule.] On the left-hand side, we have ħ, and its dimension is the action dimension: J·s, i.e. force times distance times time (N·m·s). So we multiply that with a time derivative and we get J once again, the unit of energy. So it works out: we have joule units both left and right. But what does it mean?

Well… The Laplacian on the right-hand side works just the same as for our heat diffusion equation: it gives us a flux density, i.e. something expressed per square meter (1/m²). Likewise, the time derivative on the left-hand side gives us a flow per second. But so what is it that is flowing here? Well… My interpretation is that it is energy, and it’s flowing between a real and an imaginary space—but don’t be fooled by the terms here: both spaces are equally real, as both have an actual physical dimension. Let me explain.

Things become somewhat more comprehensible when we remind ourselves that the Schrödinger equation is equivalent to the following pair of equations:

Re(∂ψ/∂t) = −(ħ/2m)·Im(∇²ψ) ⇔ ω·cos(kx − ωt) = k²·(ħ/2m)·cos(kx − ωt)
Im(∂ψ/∂t) = (ħ/2m)·Re(∇²ψ) ⇔ ω·sin(kx − ωt) = k²·(ħ/2m)·sin(kx − ωt)

So what? Let me insert an illustration here. See what happens. The wavefunction acts as a link function between our physical spacetime and some other space whose dimensions – in my humble opinion – are also physical. We have those sines and cosines, which mirror the energy of the system at any point in time, as measured by the proper time of the system.

summary

Let me more precise. The wavefunction, as a link function between two spaces here, associates every point in spacetime with some real as well as some imaginary energy here—but, as mentioned above, that imaginary energy is as real as the real energy. What it embodies really is the energy conservation law: at any point in time (as measured by the proper time) the sum of kinetic and potential energy must be equal to some constant, and so that’s what’s shown here. Indeed, you should note the phase shift between the sine and the cosine function: if one reaches the +1 or −1 value, then the other function reaches the zero point—and vice versa. It’s a beautiful structure.

Of course, the million-dollar question is: is it a physical structure, or a mathematical structure? The answer is: it’s a bit of both. It’s a mathematical structure but, at the same time, its dimension is physical: it’s an energy space. It’s that energy that explains why amplitudes interfere—which, as you know, is what they do. So these amplitudes are something real, and as the dimensional analysis of Schrödinger’s equation reveals their dimension is expressed in joule, then… Well… Then these physical equations say what they say, don’t they? And what they say, is something like the diagram below.

summary 2

Note that the diagram above does not show the phase difference between the two springs. The animation below does a better job here, although you need to realize the hand of the clock will move faster or slower as our object travels through force fields and accelerates or decelerates accordingly.

Circle_cos_sin

We may relate that picture above to the principle of least action, which ensures that the difference between the kinetic energy (KE) and potential energy (PE) in the integrand of the action integral, i.e.

action

is minimized along the path of travel.

The spring metaphor should also make us think of the energy formula for a harmonic oscillator, which tells us that the total energy – kinetic (i.e. the energy related to its momentum) plus potential (i.e. the energy stored in the spring) – is equal to T + U = m·ω₀²/2. The ω₀here is the angular velocity, and we have two springs here, so the total energy would be the sum of both, i.e. m·ω₀², without the 1/2 factor. Does that make sense? It’s like an E = m·v²equation, so that’s twice the (non-relativistic) kinetic energy. Does that formula make any sense?

In the context of what we’re discussing here, it does. Think about the limit situation by trying to imagine a zero-mass particle here (I am talking a zero-mass spin-1/2 particle this time). It would have no rest energy, so it’s only energy is kinetic, which is equal to:

K.E. = E − E₀ = m_v·c² − m₀·c² = m_c·c²

Why is m_vequal to m_c? Zero-mass particles must travel at the speed of light, as the slightest force on them gives them an infinite acceleration. So there we are: the m·ω₀²equation makes sense! But what if we have a non-zero rest mass? In that case, look at that pair of equations again: they give us a dispersion relation, i.e. a relation between ω and k. Indeed, using natural units again, so the numerical value of ħ = 1, we can write:

ω = k²/(2m) = p²/(2m) = (m·v)²/(2m) = m·v²/2

This equation seems to represent the kinetic energy but m is not the rest mass here: it’s the relativistic mass, so that makes it different from the classical kinetic energy formula (K.E. = m₀·v²/2). [It may be useful here to remind you of how we get that classical formula. We basically integrate force over distance, from some start to some final point of a path in spacetime. So we write: ∫ F·ds = ∫ (m·a)·ds = ∫ (m·a)·ds = ∫ [m·(dv/dt)]·ds = ∫ [m·(ds/dt)]·dv = ∫ m·v·ds. So we can solve that using the m·v²/2 primitive but only if m does not vary, i.e. if m = m₀. If velocity are high, we need the relativistic mass concept.]

So we have a new energy concept here: m·v², and it’s split over those two springs. Hmm… The interpretation of all of this is not so easy, so I will need to re-visit this. As for now, however, it looks like the Universe can be represented by a V-twin engine! 🙂

V-Twin engine

Is it real?

You may still doubt whether that new ‘space’ has an actual energy dimension. It’s a figment of our mind, right? Well… Yes and no. Again, it’s a bit of a mixture between a mathematical and a physical space: it’s definitely not our physical space, as it’s not the spacetime we’re living in. But, having said that, I don’t think this energy space is just a figment of our mind. Let me give you some additional reasons, beside the dimensional analysis we did above.

For example, there is the fact that we need to to take the absolute square of the wavefunction to get the probability that our elementary particle is actually right there! Now that’s something real! Hence, let me say a few more things about that. The absolute square gets rid of the time factor. Just write it out to see what happens:

|re^iθ|²= |r|²|e^iθ|² = r²[√(cos²θ + sin²θ)]²= r²(√1)²= r²

Now, the r gives us the maximum amplitude (sorry for the mix of terminology here: I am just talking the wave amplitude here – i.e. the classical concept of an amplitude – not the quantum-mechanical concept of a probability amplitude). Now, we know that the energy of a wave – anywave, really – is proportional to the amplitude of a wave. It would also be logical to expect that the probability of finding our particle at some point x is proportional to the energy densitythere, isn’t it? [I know what you’ll say now: you’re squaring the amplitude, so if the dimension of its square is energy, then its own dimension must be the square root, right? No. Wrong. That’s why this confusion between amplitude and probability amplitude is so bad. Look at the formula: we’re squaring the sine and cosine, to then take the square root again, so the dimension doesn’t change: it’s √J² = J.]

The third reason why I think the probability amplitude represents some energy is that its real and imaginary part also interfere with each other, as is evident when you take the ordinary square (i.e. not the absolute square). Then the i² = –1 rule comes into play and, therefore, the square of the imaginary part starts messing with the square of the real part. Just write it out:

(re^iθ)²= r²(cosθ + isinθ)²= r²(cos²θ – sin²θ + 2icosθsinθ)²= r²(1 – 2sin²θ + 2icosθsinθ)²

As mentioned above, if there’s interference, then something is happening, and so then we’re talking something real. Hence, the real and imaginary part of the wavefunction must have some dimension, and not just any dimension: it must be energy, as that’s the currency of the Universe, so to speak.

Let me add a philosophical note here—or an ontological note, I should say. When you think we should only have one physical space, you’re right. This new physical space, in which we relate energy to time, is not our physical space. It’s not reality—as we know, as we experience it. So, in that sense, you’re right. It’s not physical space. But then… Well… It’s a definitional matter. Any space whose dimensions are physical, is a physical space for me. But then I should probably be more careful. What we have here is some kind of projection of our physical space to a space that lacks… Well… It lacks the spatial dimension. It’s just time – but a special kind of time: relativistic proper time – and energy—albeit energy in two dimensions, so to speak. So… What can I say? Just what I said a couple of times already: it’s some kind of mixture between a physical and mathematical space. But then… Well… Our own physical space – including the spatial dimension – is something like a mixture as well, isn’t it? We can try to disentangle them – which is what I am trying to do – but we’ll never fully succeed.

The Imaginary Energy Space

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link. In addition, I see the dark force has amused himself by removing some material even here!

Original post:

Intriguing title, isn’t it? You’ll think this is going to be highly speculative and you’re right. In fact, I could also have written: the imaginary action space, or the imaginary momentum space. Whatever. It all works ! It’s an imaginary space – but a very real one, because it holds energy, or momentum, or a combination of both, i.e. action. 🙂

So the title is either going to deter you or, else, encourage you to read on. I hope it’s the latter. 🙂

In my post on Richard Feynman’s exposé on how Schrödinger got his famous wave equation, I noted an ambiguity in how he deals with the energy concept. I wrote that piece in February, and we are now May. In-between, I looked at Schrödinger’s equation from various perspectives, as evidenced from the many posts that followed that February post, which I summarized on my Deep Blue page, where I note the following:

The argument of the wavefunction (i.e. θ = ωt – kx = [E·t – p·x]/ħ) is just the proper time of the object that’s being represented by the wavefunction (which, in most cases, is an elementary particle—an electron, for example).
The 1/2 factor in Schrödinger’s equation (∂ψ/∂t = i·(ħ/2m)·∇²ψ) doesn’t make all that much sense, so we should just drop it. Writing ∂ψ/∂t = i·(m/ħ)∇²ψ (i.e. Schrödinger’s equation without the 1/2 factor) does away with the mentioned ambiguities and, more importantly, avoids obvious contradictions.

Both remarks are rather unusual—especially the second one. In fact, if you’re not shocked by what I wrote above (Schrödinger got something wrong!), then stop reading—because then you’re likely not to understand a thing of what follows. 🙂 In any case, I thought it would be good to follow up by devoting a separate post to this matter.

The argument of the wavefunction as the proper time

Frankly, it took me quite a while to see that the argument of the wavefunction is nothing but the t’ = (t − v∙x)/√(1−v²)] formula that we know from the Lorentz transformation of spacetime. Let me quickly give you the formulas (just substitute the u for v):

relativity

In fact, let me be precise: the argument of the wavefunction also has the particle’s rest mass m₀ in it. That mass factor (m₀) appears in it as a general scaling factor, so it determines the density of the wavefunction both in time as well as in space. Let me jot it down:

ψ(x, t) = a·e^{−i·(m_v·t − p∙x)}= a·e^{−i·[(m₀/√(1−v²))·t − (m₀·v/√(1−v²))∙x]}= a·e^{−i·m₀·(t − v∙x)/√(1−v²)}

Huh? Yes. Let me show you how we get from θ = ωt – kx = [E·t – p·x]/ħ to θ = m_v·t − p∙x. It’s really easy. We first need to choose our units such that the speed of light and Planck’s constant are numerically equal to one, so we write: c = 1 and ħ = 1. So now the 1/ħ factor no longer appears.

[Let me note something here: using natural units does not do away with the dimensions: the dimensions of whatever is there remain what they are. For example, energy remains what it is, and so that’s force over distance: 1 joule = 1 newton·meter (1 J = 1 N·m. Likewise, momentum remains what it is: force times time (or mass times velocity). Finally, the dimension of the quantum of action doesn’t disappear either: it remains the product of force, distance and time (N·m·s). So you should distinguish between the numerical value of our variables and their dimension. Always! That’s where physics is different from algebra: the equations actually mean something!]

Now, because we’re working in natural units, the numerical value of both c and c²will be equal to 1. It’s obvious, then, that Einstein’s mass-energy equivalence relation reduces from E = m_vc² to E = m_v. You can work out the rest yourself – noting that p = m_v·v and m_v= m₀/√(1−v²). Done! For a more intuitive explanation, I refer you to the above-mentioned page.

So that’s for the wavefunction. Let’s now look at Schrödinger’s wave equation, i.e. that differential equation of which our wavefunction is a solution. In my introduction, I bluntly said there was something wrong with it: that 1/2 factor shouldn’t be there. Why not?

What’s wrong with Schrödinger’s equation?

When deriving his famous equation, Schrödinger uses the mass concept as it appears in the classical kinetic energy formula: K.E. = m·v²/2, and that’s why – after all the complicated turns – that 1/2 factor is there. There are many reasons why that factor doesn’t make sense. Let me sum up a few.

[I] The most important reason is that de Broglie made it quite clear that the energy concept in his equations for the temporal and spatial frequency for the wavefunction – i.e. the ω = E/ħ and k = p/ħ relations – is the total energy, including rest energy (m₀), kinetic energy (m·v²/2) and any potential energy (V). In fact, if we just multiply the two de Broglie (aka as matter-wave equations) and use the old-fashioned v = f·λ relation (so we write E as E = ω·ħ = (2π·f)·(h/2π) = f·h, and p as p = k·ħ = (2π/λ)·(h/2π) = h/λ and, therefore, we have f = E/h and p = h/p), we find that the energy concept that’s implicit in the two matter-wave equations is equal to E = m∙v², as shown below:

f·λ = (E/h)·(h/p) = E/p
v = f·λ ⇒ f·λ = v = E/p ⇔ E = v·p = v·(m·v) ⇒ E = m·v²

Huh? E = m∙v²? Yes. Not E = m∙c² or m·v²/2 or whatever else you might be thinking of. In fact, this E = m∙v² formula makes a lot of sense in light of the two following points.

Skeptical note: You may – and actually should – wonder whether we can use that v = f·λ relation for a wave like this, i.e. a wave with both a real (cos(-θ)) as well as an imaginary component (i·sin(-θ). It’s a deep question, and I’ll come back to it later. But… Yes. It’s the right question to ask. 😦

[II] Newton told us that force is mass time acceleration. Newton’s law is still valid in Einstein’s world. The only difference between Newton’s and Einstein’s world is that, since Einstein, we should treat the mass factor as a variable as well. We write: F = m_v·a = m_v·a = [m₀/√(1−v²)]·a. This formula gives us the definition of the newton as a force unit: 1 N = 1 kg·(m/s)/s = 1 kg·m/s². [Note that the 1/√(1−v²) factor – i.e. the Lorentz factor (γ) – has no dimension, because v is measured as a relative velocity here, i.e. as a fraction between 0 and 1.]

Now, you’ll agree the definition of energy as a force over some distance is valid in Einstein’s world as well. Hence, if 1 joule is 1 N·m, then 1 J is also equal to 1 (kg·m/s²)·m = 1 kg·(m²/s²), so this also reflects the E = m∙v² concept. [I can hear you mutter: that kg factor refers to the rest mass, no? No. It doesn’t. The kg is just a measure of inertia: as a unit, it applies to both m₀as well as m_v. Full stop.]

Very skeptical note: You will say this doesn’t prove anything – because this argument just shows the dimensional analysis for both equations (i.e. E = m∙v² and E = m∙c²) is OK. Hmm… Yes. You’re right. 🙂 But the next point will surely convince you! 🙂

[III] The third argument is the most intricate and the most beautiful at the same time—not because it’s simple (like the arguments above) but because it gives us an interpretation of what’s going on here. It’s fairly easy to verify that Schrödinger’s equation, ∂ψ/∂t = i·(ħ/2m)·∇²ψ equation (including the 1/2 factor to which I object), is equivalent to the following set of two equations:

Re(∂ψ/∂t) = −(ħ/2m)·Im(∇²ψ)
Im(∂ψ/∂t) = (ħ/2m)·Re(∇²ψ)

[In case you don’t see it immediately, note that two complex numbers a + i·b and c + i·d are equal if, and only if, their real and imaginary parts are the same. However, here we have something like this: a + i·b = i·(c + i·d) = i·c + i²·d = − d + i·c (remember i²= −1).]

Now, before we proceed (i.e. before I show you what’s wrong here with that 1/2 factor), let us look at the dimensions first. For that, we’d better analyze the complete Schrödinger equation so as to make sure we’re not doing anything stupid here by looking at one aspect of the equation only. The complete equation, in its original form, is:

Notice that, to simplify the analysis above, I had moved the i and the ħ on the left-hand side to the right-hand side (note that 1/i = −i, so −(ħ²/2m)/(i·ħ) = ħ/2m). Now, the ħ²factor on the right-hand side is expressed in J²·s². Now that doesn’t make much sense, but then that mass factor in the denominator makes everything come out alright. Indeed, we can use the mass-equivalence relation to express m in J/(m/s)² units. So our ħ²/2m coefficient is expressed in (J²·s²)/[J/(m/s)²] = J·m². Now we multiply that by that Laplacian operating on some scalar, which yields some quantity per square meter. So the whole right-hand side becomes some amount expressed in joule, i.e. the unit of energy! Interesting, isn’t it?

On the left-hand side, we have i and ħ. We shouldn’t worry about the imaginary unit because we can treat that as just another number, albeit a very special number (because its square is minus 1). However, in this equation, it’s like a mathematical constant and you can think of it as something like π or e. [Think of the magical formula: e^iπ= i² = −1.] In contrast, ħ is a physical constant, and so that constant comes with some dimension and, therefore, we cannot just do what we want. [I’ll show, later, that even moving it to the other side of the equation comes with interpretation problems, so be careful with physical constants, as they really mean something!] In this case, its dimension is the action dimension: J·s = N·m·s, so that’s force times distance times time. So we multiply that with a time derivative and we get joule once again (N·m·s/s = N·m = J), so that’s the unit of energy. So it works out: we have joule units both left and right in Schrödinger’s equation. Nice! Yes. But what does it mean? 🙂

Well… You know that we can – and should – think of Schrödinger’s equation as a diffusion equation – just like a heat diffusion equation, for example – but then one describing the diffusion of a probability amplitude. [In case you are not familiar with this interpretation, please do check my post on it, or my Deep Blue page.] But then we didn’t describe the mechanism in very much detail, so let me try to do that now and, in the process, finally explain the problem with the 1/2 factor.

The missing energy

There are various ways to explain the problem. One of them involves calculating group and phase velocities of the elementary wavefunction satisfying Schrödinger’s equation but that’s a more complicated approach and I’ve done that elsewhere, so just click the reference if you prefer the more complicated stuff. I find it easier to just use those two equations above:

Re(∂ψ/∂t) = −(ħ/2m)·Im(∇²ψ)
Im(∂ψ/∂t) = (ħ/2m)·Re(∇²ψ)

The argument is the following: if our elementary wavefunction is equal to eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt), then it’s easy to proof that this pair of conditions is fulfilled if, and only if, ω = k²·(ħ/2m). [Note that I am omitting the normalization coefficient in front of the wavefunction: you can put it back in if you want. The argument here is valid, with or without normalization coefficients.] Easy? Yes. Check it out. The time derivative on the left-hand side is equal to:

∂ψ/∂t = −iω·ieⁱ^{(kx − ωt)}= ω·[cos(kx − ωt) + i·sin(kx − ωt)] = ω·cos(kx − ωt) + iω·sin(kx − ωt)

And the second-order derivative on the right-hand side is equal to:

∇²ψ= ∂²ψ/∂x²= i·k²·eⁱ^{(kx − ωt)}= k²·cos(kx − ωt) + i·k²·sin(kx − ωt)

So the two equations above are equivalent to writing:

Re(∂ψ_B/∂t) = −(ħ/2m)·Im(∇²ψ_B) ⇔ ω·cos(kx − ωt) = k²·(ħ/2m)·cos(kx − ωt)
Im(∂ψ_B/∂t) = (ħ/2m)·Re(∇²ψ_B) ⇔ ω·sin(kx − ωt) = k²·(ħ/2m)·sin(kx − ωt)

So both conditions are fulfilled if, and only if, ω = k²·(ħ/2m). You’ll say: so what? Well… We have a contradiction here—something that doesn’t make sense. Indeed, the second of the two de Broglie equations (always look at them as a pair) tells us that k = p/ħ, so we can re-write the ω = k²·(ħ/2m) condition as:

ω/k = v_p = k²·(ħ/2m)/k = k·ħ/(2m) = (p/ħ)·(ħ/2m) = p/2m ⇔ p = 2m

You’ll say: so what? Well… Stop reading, I’d say. That p = 2m doesn’t make sense—at all! Nope! In fact, if you thought that the E = m·v² is weird—which, I hope, is no longer the case by now—then… Well… This p = 2m equation is much weirder. In fact, it’s plain nonsense: this condition makes no sense whatsoever. The only way out is to remove the 1/2 factor, and to re-write the Schrödinger equation as I wrote it, i.e. with an ħ/m coefficient only, rather than an (1/2)·(ħ/m) coefficient.

Huh? Yes.

As mentioned above, I could do those group and phase velocity calculations to show you what rubbish that 1/2 factor leads to – and I’ll do that eventually – but let me first find yet another way to present the same paradox. Let’s simplify our life by choosing our units such that c = ħ = 1, so we’re using so-called natural units rather than our SI units. [Again, note that switching to natural units doesn’t do anything to the physical dimensions: a force remains a force, a distance remains a distance, and so on.] Our mass-energy equivalence then becomes: E = m·c²= m·1²= m. [Again, note that switching to natural units doesn’t do anything to the physical dimensions: a force remains a force, a distance remains a distance, and so on. So we’d still measure energy and mass in different but equivalent units. Hence, the equality sign should not make you think mass and energy are actually the same: energy is energy (i.e. force times distance), while mass is mass (i.e. a measure of inertia). I am saying this because it’s important, and because it took me a while to make these rather subtle distinctions.]

Let’s now go one step further and imagine a hypothetical particle with zero rest mass, so m₀ = 0. Hence, all its energy is kinetic and so we write: K.E. = m_v·v/2. Now, because this particle has zero rest mass, the slightest acceleration will make it travel at the speed of light. In fact, we would expect it to travel at the speed, so m_v = m_c and, according to the mass-energy equivalence relation, its total energy is, effectively, E = m_v = m_c. However, we just said its total energy is kinetic energy only. Hence, its total energy must be equal to E = K.E. = m_c·c/2 = m_c/2. So we’ve got only half the energy we need. Where’s the other half? Where’s the missing energy? Quid est veritas? Is its energy E = m_c or E = m_c/2?

It’s just a paradox, of course, but one we have to solve. Of course, we may just say we trust Einstein’s E = m·c²formula more than the kinetic energy formula, but that answer is not very scientific. 🙂 We’ve got a problem here and, in order to solve it, I’ve come to the following conclusion: just because of its sheer existence, our zero-mass particle must have some hidden energy, and that hidden energy is also equal to E = m·c²/2. Hence, the kinetic and the hidden energy add up to E = m·c² and all is alright.

Huh? Hidden energy? I must be joking, right?

Well… No. Let me explain. Oh. And just in case you wonder why I bother to try to imagine zero-mass particles. Let me tell you: it’s the first step towards finding a wavefunction for a photon and, secondly, you’ll see it just amounts to modeling the propagation mechanism of energy itself. 🙂

The hidden energy as imaginary energy

I am tempted to refer to the missing energy as imaginary energy, because it’s linked to the imaginary part of the wavefunction. However, it’s anything but imaginary: it’s as real as the imaginary part of the wavefunction. [I know that sounds a bit nonsensical, but… Well… Think about it. And read on!]

Back to that factor 1/2. As mentioned above, it also pops up when calculating the group and the phase velocity of the wavefunction. In fact, let me show you that calculation now. [Sorry. Just hang in there.] It goes like this.

The de Broglie relations tell us that the k and the ω in the eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt) wavefunction (i.e. the spatial and temporal frequency respectively) are equal to k = p/ħ, and ω = E/ħ. Let’s now think of that zero-mass particle once more, so we assume all of its energy is kinetic: no rest energy, no potential! So… If we now use the kinetic energy formula E = m·v²/2 – which we can also write as E = m·v·v/2 = p·v/2 = p·p/2m = p²/2m, with v = p/m the classical velocity of the elementary particle that Louis de Broglie was thinking of – then we can calculate the group velocity of our eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt) wavefunction as:

v_g = ∂ω/∂k = ∂[E/ħ]/∂[p/ħ] = ∂E/∂p = ∂[p²/2m]/∂p = 2p/2m = p/m = v

[Don’t tell me I can’t treat m as a constant when calculating ∂ω/∂k: I can. Think about it.]

Fine. Now the phase velocity. For the phase velocity of our eⁱ^{(kx − ωt)} wavefunction, we find:

v_p = ω/k = (E/ħ)/(p/ħ) = E/p = (p²/2m)/p = p/2m = v/2

So that’s only half of v: it’s the 1/2 factor once more! Strange, isn’t it? Why would we get a different value for the phase velocity here? It’s not like we have two different frequencies here, do we? Well… No. You may also note that the phase velocity turns out to be smaller than the group velocity (as mentioned, it’s only half of the group velocity), which is quite exceptional as well! So… Well… What’s the matter here? We’ve got a problem!

What’s going on here? We have only one wave here—one frequency and, hence, only one k and ω. However, on the other hand, it’s also true that the eⁱ^{(kx − ωt)} wavefunction gives us two functions for the price of one—one real and one imaginary: eⁱ^{(kx − ωt)} = cos(kx−ωt) + i∙sin(kx−ωt). So the question here is: are we adding waves, or are we not? It’s a deep question. If we’re adding waves, we may get different group and phase velocities, but if we’re not, then… Well… Then the group and phase velocity of our wave should be the same, right? The answer is: we are and we aren’t. It all depends on what you mean by ‘adding’ waves. I know you don’t like that answer, but that’s the way it is, really. 🙂

Let me make a small digression here that will make you feel even more confused. You know – or you should know – that the sine and the cosine function are the same except for a phase difference of 90 degrees: sinθ = cos(θ + π/2). Now, at the same time, multiplying something with i amounts to a rotation by 90 degrees, as shown below.

Hence, in order to sort of visualize what our eⁱ^{(kx − ωt)} function really looks like, we may want to super-impose the two graphs and think of something like this:

You’ll have to admit that, when you see this, our formulas for the group or phase velocity, or our v = f·λ relation, do no longer make much sense, do they? 🙂

Having said that, that 1/2 factor is and remains puzzling, and there must be some logical reason for it. For example, it also pops up in the Uncertainty Relations:

Δx·Δp ≥ ħ/2 and ΔE·Δt ≥ ħ/2

So we have ħ/2 in both, not ħ. Why do we need to divide the quantum of action here? How do we solve all these paradoxes? It’s easy to see how: the apparent contradiction (i.e. the different group and phase velocity) gets solved if we’d use the E = m∙v² formula rather than the kinetic energy E = m∙v²/2. But then… What energy formula is the correct one: E = m∙v² or m∙c²? Einstein’s formula is always right, isn’t it? It must be, so let me postpone the discussion a bit by looking at a limit situation. If v = c, then we don’t need to make a choice, obviously. 🙂 So let’s look at that limit situation first. So we’re discussing our zero-mass particle once again, assuming it travels at the speed of light. What do we get?

Well… Measuring time and distance in natural units, so c = 1, we have:

E = m∙c² = m and p = m∙c= m, so we get: E = m = p

Waw ! E = m = p ! What a weird combination, isn’t it? Well… Yes. But it’s fully OK. [You tell me why it wouldn’t be OK. It’s true we’re glossing over the dimensions here, but natural units are natural units and, hence, the numerical value of c and c²is 1. Just figure it out for yourself.] The point to note is that the E = m = p equality yields extremely simple but also very sensible results. For the group velocity of our eⁱ^{(kx − ωt)} wavefunction, we get:

v_g = ∂ω/∂k = ∂[E/ħ]/∂[p/ħ] = ∂E/∂p = ∂p/∂p = 1

So that’s the velocity of our zero-mass particle (remember: the 1 stands for c here, i.e. the speed of light) expressed in natural units once more—just like what we found before. For the phase velocity, we get:

v_p = ω/k = (E/ħ)/(p/ħ) = E/p = p/p = 1

Same result! No factor 1/2 here! Isn’t that great? My ‘hidden energy theory’ makes a lot of sense.

However, if there’s hidden energy, we still need to show where it’s hidden. 🙂 Now that question is linked to the propagation mechanism that’s described by those two equations, which now – leaving the 1/2 factor out, simplify to:

Re(∂ψ/∂t) = −(ħ/m)·Im(∇²ψ)
Im(∂ψ/∂t) = (ħ/m)·Re(∇²ψ)

Propagation mechanism? Yes. That’s what we’re talking about here: the propagation mechanism of energy. Huh? Yes. Let me explain in another separate section, so as to improve readability. Before I do, however, let me add another note—for the skeptics among you. 🙂

Indeed, the skeptics among you may wonder whether our zero-mass particle wavefunction makes any sense at all, and they should do so for the following reason: if x = 0 at t = 0, and it’s traveling at the speed of light, then x(t) = t. Always. So if E = m = p, the argument of our wavefunction becomes E·t – p·x = E·t – E·t = 0! So what’s that? The proper time of our zero-mass particle is zero—always and everywhere!?

Well… Yes. That’s why our zero-mass particle – as a point-like object – does not really exist. What we’re talking about is energy itself, and its propagation mechanism. 🙂

While I am sure that, by now, you’re very tired of my rambling, I beg you to read on. Frankly, if you got as far as you have, then you should really be able to work yourself through the rest of this post. 🙂 And I am sure that – if anything – you’ll find it stimulating! 🙂

The imaginary energy space

Look at the propagation mechanism for the electromagnetic wave in free space, which (for c = 1) is represented by the following two equations:

∂B/∂t = –∇×E
∂E/∂t = ∇×B

[In case you wonder, these are Maxwell’s equations for free space, so we have no stationary nor moving charges around.] See how similar this is to the two equations above? In fact, in my Deep Blue page, I use these two equations to derive the quantum-mechanical wavefunction for the photon (which is not the same as that hypothetical zero-mass particle I introduced above), but I won’t bother you with that here. Just note the so-called curl operator in the two equations above (∇×) can be related to the Laplacian we’ve used so far (∇²). It’s not the same thing, though: for starters, the curl operator operates on a vector quantity, while the Laplacian operates on a scalar (including complex scalars). But don’t get distracted now. Let’s look at the revised Schrödinger’s equation, i.e. the one without the 1/2 factor:

∂ψ/∂t = i·(ħ/m)·∇²ψ

On the left-hand side, we have a time derivative, so that’s a flow per second. On the right-hand side we have the Laplacian and the i·ħ/m factor. Now, written like this, Schrödinger’s equation really looks exactly the same as the general diffusion equation, which is written as: ∂φ/∂t = D·∇²φ, except for the imaginary unit, which makes it clear we’re getting two equations for the price of one here, rather than one only! 🙂 The point is: we may now look at that ħ/m factor as a diffusion constant, because it does exactly the same thing as the diffusion constant D in the diffusion equation ∂φ/∂t = D·∇²φ, i.e:

As a constant of proportionality, it quantifies the relationship between both derivatives.
As a physical constant, it ensures the dimensions on both sides of the equation are compatible.

So the diffusion constant for Schrödinger’s equation is ħ/m. What is its dimension? That’s easy: (N·m·s)/(N·s²/m) = m²/s. [Remember: 1 N = 1 kg·m/s².] But then we multiply it with the Laplacian, so that’s something expressed per square meter, so we get something per second on both sides.

Of course, you wonder: what per second? Not sure. That’s hard to say. Let’s continue with our analogy with the heat diffusion equation so as to try to get a better understanding of what’s being written here. Let me give you that heat diffusion equation here. Assuming the heat per unit volume (q) is proportional to the temperature (T) – which is the case when expressing T in degrees Kelvin (K), so we can write q as q = k·T – we can write it as:

So that’s structurally similar to Schrödinger’s equation, and to the two equivalent equations we jotted down above. So we’ve got T (temperature) in the role of ψ here—or, to be precise, in the role of ψ ‘s real and imaginary part respectively. So what’s temperature? From the kinetic theory of gases, we know that temperature is not just a scalar: temperature measures the mean (kinetic) energy of the molecules in the gas. That’s why we can confidently state that the heat diffusion equation models an energy flow, both in space as well as in time.

Let me make the point by doing the dimensional analysis for that heat diffusion equation. The time derivative on the left-hand side (∂T/∂t) is expressed in K/s (Kelvin per second). Weird, isn’t it? What’s a Kelvin per second? Well… Think of a Kelvin as some very small amount of energy in some equally small amount of space—think of the space that one molecule needs, and its (mean) energy—and then it all makes sense, doesn’t it?

However, in case you find that a bit difficult, just work out the dimensions of all the other constants and variables. The constant in front (k) makes sense of it. That coefficient (k) is the (volume) heat capacity of the substance, which is expressed in J/(m³·K). So the dimension of the whole thing on the left-hand side (k·∂T/∂t) is J/(m³·s), so that’s energy (J) per cubic meter (m³) and per second (s). Nice, isn’t it? What about the right-hand side? On the right-hand side we have the Laplacian operator – i.e. ∇²= ∇·∇, with ∇ = (∂/∂x, ∂/∂y, ∂/∂z) – operating on T. The Laplacian operator, when operating on a scalar quantity, gives us a flux density, i.e. something expressed per square meter (1/m²). In this case, it’s operating on T, so the dimension of ∇²T is K/m². Again, that doesn’t tell us very much (what’s the meaning of a Kelvin per square meter?) but we multiply it by the thermal conductivity (κ), whose dimension is W/(m·K) = J/(m·s·K). Hence, the dimension of the product is the same as the left-hand side: J/(m³·s). So that’s OK again, as energy (J) per cubic meter (m³) and per second (s) is definitely something we can associate with an energy flow.

In fact, we can play with this. We can bring k from the left- to the right-hand side of the equation, for example. The dimension of κ/k is m²/s (check it!), and multiplying that by K/m²(i.e. the dimension of ∇²T) gives us some quantity expressed in Kelvin per second, and so that’s the same dimension as that of ∂T/∂t. Done!

In fact, we’ve got two different ways of writing Schrödinger’s diffusion equation. We can write it as ∂ψ/∂t = i·(ħ/m)·∇²ψ or, else, we can write it as ħ·∂ψ/∂t = i·(ħ²/m)·∇²ψ. Does it matter? I don’t think it does. The dimensions come out OK in both cases. However, interestingly, if we do a dimensional analysis of the ħ·∂ψ/∂t = i·(ħ²/m)·∇²ψ equation, we get joule on both sides. Interesting, isn’t it? The key question, of course, is: what is it that is flowing here?

I don’t have a very convincing answer to that, but the answer I have is interesting—I think. 🙂 Think of the following: we can multiply Schrödinger’s equation with whatever we want, and then we get all kinds of flows. For example, if we multiply both sides with 1/(m²·s) or 1/(m³·s), we get a equation expressing the energy conservation law, indeed! [And you may want to think about the minus sign of the right-hand side of Schrödinger’s equation now, because it makes much more sense now!]

We could also multiply both sides with s, so then we get J·s on both sides, i.e. the dimension of physical action (J·s = N·m·s). So then the equation expresses the conservation of action! Huh? Yes. Let me re-phrase that: then it expresses the conservation of angular momentum—as you’ll surely remember that the dimension of action and angular momentum are the same. 🙂

And then we can divide both sides by m, so then we get N·s on both sides, so that’s momentum. So then Schrödinger’s equation embodies the momentum conservation law.

Isn’t it just wonderful? Schrödinger’s equation packs all of the conservation laws! The only catch is that it flows back and forth from the real to the imaginary space, using that propagation mechanism as described in those two equations.

Now that is really interesting, because it does provide an explanation – as fuzzy as it may seem – for all those weird concepts one encounters when studying physics, such as the tunneling effect, which amounts to energy flowing from the imaginary space to the real space and, then, inevitably, flowing back. It also allows for borrowing time from the imaginary space. Hmm… Interesting! [I know I still need to make these points much more formally, but… Well… You kinda get what I mean, don’t you?]

To conclude, let me re-baptize my real and imaginary ‘space’ by referring to them to what they really are: a real and imaginary energy space respectively. Although… Now that I think of it: it could also be real and imaginary momentum space, or a real and imaginary action space. Hmm… The latter term may be the best. 🙂

Isn’t this all great? I mean… I could go on and on—but I’ll stop here, so you can freewheel around yourself. For example, you may wonder how similar that energy propagation mechanism actually is as compared to the propagation mechanism of the electromagnetic wave? The answer is: very similar. You can check how similar in one of my posts on the photon wavefunction or, if you’d want a more general argument, check my Deep Blue page. Have fun exploring! 🙂

So… Well… That’s it, folks. I hope you enjoyed this post—if only because I really enjoyed writing it. 🙂

[…]

OK. You’re right. I still haven’t answered the fundamental question.

So what about the 1/2 factor?

What about that 1/2 factor? Did Schrödinger miss it? Well… Think about it for yourself. First, I’d encourage you to further explore that weird graph with the real and imaginary part of the wavefunction. I copied it below, but with an added 45º line—yes, the green diagonal. To make it somewhat more real, imagine you’re the zero-mass point-like particle moving along that line, and we observe you from our inertial frame of reference, using equivalent time and distance units.

So we’ve got that cosine (cosθ) varying as you travel, and we’ve also got the i·sinθ part of the wavefunction going while you’re zipping through spacetime. Now, THINK of it: the phase velocity of the cosine bit (i.e. the red graph) contributes as much to your lightning speed as the i·sinθ bit, doesn’t it? Should we apply Pythagoras’ basic r² = x² + y²Theorem here? Yes: the velocity vector along the green diagonal is going to be the sum of the velocity vectors along the horizontal and vertical axes. So… That’s great.

Yes. It is. However, we still have a problem here: it’s the velocity vectors that add up—not their magnitudes. Indeed, if we denote the velocity vector along the green diagonal as u, then we can calculate its magnitude as:

u = √u² = √[(v/2)² + (v/2)²] = √[2·(v²/4) = √[v²/2] = v/√2 ≈ 0.7·v

So, as mentioned, we’re adding the vectors, but not their magnitudes. We’re somewhat better off than we were in terms of showing that the phase velocity of those sine and cosine velocities add up—somehow, that is—but… Well… We’re not quite there.

Fortunately, Einstein saves us once again. Remember we’re actually transforming our reference frame when working with the wavefunction? Well… Look at the diagram below (for which I thank the author)

special relativity

In fact, let me insert an animated illustration, which shows what happens when the velocity u goes up and down from (close to) −c to +c and back again. It’s beautiful, and I must credit the author here too. It sort of speaks for itself, but please do click the link as the accompanying text is quite illuminating. 🙂

Animated_Lorentz_Transformation

The point is: for our zero-mass particle, the x’ and t’ axis will rotate into the diagonal itself which, as I mentioned a couple of times already, represents the speed of light and, therefore, our zero-mass particle traveling at c. It’s obvious that we’re now adding two vectors that point in the same direction and, hence, their magnitudes just add without any square root factor. So, instead of u = √[(v/2)² + (v/2)²], we just have v/2 + v/2 = v! Done! We solved the phase velocity paradox! 🙂

So… I still haven’t answered that question. Should that 1/2 factor in Schrödinger’s equation be there or not? The answer is, obviously: yes. It should be there. And as for Schrödinger using the mass concept as it appears in the classical kinetic energy formula: K.E. = m·v²/2… Well… What other mass concept would he use? I probably got a bit confused with Feynman’s exposé – especially this notion of ‘choosing the zero point for the energy’ – but then I should probably just re-visit the thing and adjust the language here and there. But the formula is correct.

Thinking it all through, the ħ/2m constant in Schrödinger’s equation should be thought of as the reciprocal of m/(ħ/2). So what we’re doing basically is measuring the mass of our object in units of ħ/2, rather than units of ħ. That makes perfect sense, if only because it’s ħ/2, rather than ħthe factor that appears in the Uncertainty Relations Δx·Δp ≥ ħ/2 and ΔE·Δt ≥ ħ/2. In fact, in my post on the wavefunction of the zero-mass particle, I noted its elementary wavefunction should use the m = E = p = ħ/2 values, so it becomes ψ(x, t) = a·e^{−i∙[(ħ/2)∙t − (ħ/2)∙x]/ħ}= a·e^{−i∙[t − x]/2}.

Isn’t that just nice? 🙂 I need to stop here, however, because it looks like this post is becoming a book. Oh—and note that nothing what I wrote above discredits my ‘hidden energy’ theory. On the contrary, it confirms it. In fact, the nice thing about those illustrations above is that it associates the imaginary component of our wavefunction with travel in time, while the real component is associated with travel in space. That makes our theory quite complete: the ‘hidden’ energy is the energy that moves time forward. The only thing I need to do is to connect it to that idea of action expressing itself in time or in space, cf. what I wrote on my Deep Blue page: we can look at the dimension of Planck’s constant, or at the concept of action in general, in two very different ways—from two different perspectives, so to speak:

[Planck’s constant] = [action] = N∙m∙s = (N∙m)∙s = [energy]∙[time]
[Planck’s constant] = [action] = N∙m∙s = (N∙s)∙m = [momentum]∙[distance]

Hmm… I need to combine that with the idea of the quantum vacuum, i.e. the mathematical space that’s associated with time and distance becoming countable variables…. In any case. Next time. 🙂

Before I sign off, however, let’s quickly check if our a·e^{−i∙[t − x]/2}wavefunction solves the Schrödinger equation:

∂ψ/∂t = −a·e^{−i∙[t − x]/2}·(i/2)
∇²ψ = ∂²[a·e^{−i∙[t − x]/2}]/∂x²= ∂[a·e^{−i∙[t − x]/2}·(i/2)]/∂x = −a·e^{−i∙[t − x]/2}·(1/4)

So the ∂ψ/∂t = i·(ħ/2m)·∇²ψ equation becomes:

−a·e^{−i∙[t − x]/2}·(i/2) = −i·(ħ/[2·(ħ/2)])·a·e^{−i∙[t − x]/2}·(1/4)

⇔ 1/2 = 1/4 !?

The damn 1/2 factor. Schrödinger wants it in his wave equation, but not in the wavefunction—apparently! So what if we take the m = E = p = ħ solution? We get:

∂ψ/∂t = −a·i·e^{−i∙[t − x]}
∇²ψ = ∂²[a·e^{−i∙[t − x]}]/∂x²= ∂[a·i·e^{−i∙[t − x]}]/∂x = −a·e^{−i∙[t − x]}

So the ∂ψ/∂t = i·(ħ/2m)·∇²ψ equation now becomes:

−a·i·e^{−i∙[t − x]} = −i·(ħ/[2·ħ])·a·e^{−i∙[t − x]}

⇔ 1 = 1/2 !?

We’re still in trouble! So… Was Schrödinger wrong after all? There’s no difficulty whatsoever with the ∂ψ/∂t = i·(ħ/m)·∇²ψ equation:

−a·e^{−i∙[t − x]/2}·(i/2) = −i·[ħ/(ħ/2)]·a·e^{−i∙[t − x]/2}·(1/4) ⇔ 1 = 1
−a·i·e^{−i∙[t − x]} = −i·(ħ/ħ)·a·e^{−i∙[t − x]}⇔ 1 = 1

What these equations might tell us is that we should measure mass, energy and momentum in terms of ħ (and not in terms of ħ/2) but that the fundamental uncertainty is ± ħ/2. That solves it all. So the magnitude of the uncertainty is ħ but it separates not 0 and ± 1, but −ħ/2 and −ħ/2. Or, more generally, the following series:

…, −7ħ/2, −5ħ/2, −3ħ/2, −ħ/2, +ħ/2, +3ħ/2,+5ħ/2, +7ħ/2,…

Why are we not surprised? The series represent the energy values that a spin one-half particle can possibly have, and ordinary matter – i.e. all fermions – is composed of spin one-half particles.

To conclude this post, let’s see if we can get any indication on the energy concepts that Schrödinger’s revised wave equation implies. We’ll do so by just calculating the derivatives in the ∂ψ/∂t = i·(ħ/m)·∇²ψ equation (i.e. the equation without the 1/2 factor). Let’s also not assume we’re measuring stuff in natural units, so our wavefunction is just what it is: a·e^{−i·[E·t − p∙x]/ħ}. The derivatives now become:

∂ψ/∂t = −a·i·(E/ħ)·e^{−i∙[E·t − p∙x]/ħ}
∇²ψ = ∂²[a·e^{−i∙[E·t − p∙x]/ħ}]/∂x²= ∂[a·i·(p/ħ)·e^{−i∙[E·t − p∙x]/ħ}]/∂x = −a·(p²/ħ²)·e^{−i∙[E·t − p∙x]/ħ}

So the ∂ψ/∂t = i·(ħ/m)·∇²ψ = i·(1/m)·∇²ψ equation now becomes:

−a·i·(E/ħ)·e^{−i∙[E·t − p∙x]/ħ} = −i·(ħ/m)·a·(p²/ħ²)·e^{−i∙[E·t − p∙x]/ħ}⇔ E = p²/m = m·v²

It all works like a charm. Note that we do not assume stuff like E = m = p here. It’s all quite general. Also note that the E = p²/m closely resembles the kinetic energy formula one often sees: K.E. = m·v²/2 = m·m·v²/(2m) = p²/(2m). We just don’t have the 1/2 factor in our E = p²/m formula, which is great—because we don’t want it! Of course, if you’d add the 1/2 factor in Schrödinger’s equation again, you’d get it back in your energy formula, which would just be that old kinetic energy formula which gave us all these contradictions and ambiguities. 😦

Finally, and just to make sure: let me add that, when we wrote that E = m = p – like we did above – we mean their numerical values are the same. Their dimensions remain what they are, of course. Just to make sure you get that subtle point, we’ll do a quick dimensional analysis of that E = p²/m formula:

[E] = [p²/m] ⇔ N·m = N²·s²/kg = N²·s²/[N·m/s²] = N·m = joule (J)

So… Well… It’s all perfect. 🙂

Post scriptum: I revised my Deep Blue page after writing this post, and I think that a number of the ideas that I express above are presented more consistently and coherently there. In any case, the missing energy theory makes sense. Think of it: any oscillator involves both kinetic as well as potential energy, and they both add up to twice the average kinetic (or potential) energy. So why not here? When everything is said and done, our elementary wavefunction does describe an oscillator. 🙂

Schrödinger’s equation in action

Pre-script (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or classical) interpretation of all things quantum-mechanical. So no use to read this. Read my recent papers instead. 🙂

Original post:

This post is about something I promised to write about aeons ago: how do we get those electron orbitals out of Schrödinger’s equation? So let me write it now – for the simplest of atoms: hydrogen. I’ll largely follow Richard Feynman’s exposé on it: this text just intends to walk you through it and provide some comments here and there.

Let me first remind you of what that famous Schrödinger’s equation actually represents. In its simplest form – i.e. not including any potential, so then it’s an equation that’s valid for free space only—no force fields!—it reduces to:

i·ħ∙∂ψ/∂t = –(1/2)∙(ħ²/m_eff)∙∇²ψ

Note the enigmatic concept of the efficient mass in it (m_eff), as well as the rather awkward 1/2 factor, which we may get rid of by re-defining it. We then write: m_eff^NEW = 2∙m_eff^OLD, and Schrödinger’s equation then simplifies to:

∂ψ/∂t + i∙(V/ħ)·ψ = i(ħ/m_eff)·∇²ψ
In free space (no potential): ∂ψ/∂t = i∙(ħ/m_eff)·∇²ψ

In case you wonder where the minus sign went, I just brought the imaginary unit to the other side. Remember 1/i = −i. 🙂

Now, in my post on quantum-mechanical operators, I drew your attention to the fact that this equation is structurally similar to the heat diffusion equation – or to any diffusion equation, really. Indeed, assuming the heat per unit volume (q) is proportional to the temperature (T) – which is the case when expressing T in degrees Kelvin (K), so we can write q as q = k·T – we can write the heat diffusion equation as:

Moreover, I noted the similarity is not only structural. There is more to it: both equations model energy flows. How exactly is something I wrote about in my e-publication on this, so let me refer you to that. Let’s jot down the complete equation once more:

∂ψ/∂t + i∙(V/ħ)·ψ = i(ħ/m_eff)·∇²ψ

In fact, it is rather surprising that Feynman drops the eff subscript almost immediately, so he just writes: schrodinger 5

Let me first remind you that ψ is a function of position in space and time, so we write: ψ = ψ(x, y, z, t) = ψ(r, t), with (x, y, z) = r. And m, on the other side of the equation, is what it always was: the effective electron mass. Now, we talked about the subtleties involved before, so let’s not bother about the definition of the effective electron mass, or wonder where that factor 1/2 comes from here.

What about V? V is the potential energy of the electron: it depends on the distance (r) from the proton. We write: V = −e²/│r│ = −e²/r. Why the minus sign? Because we say the potential energy is zero at large distances (see my post on potential energy). Back to Schrödinger’s equation.

On the left-hand side, we have ħ, and its dimension is J·s (or N·m·s, if you want). So we multiply that with a time derivative and we get J, the unit of energy. On the right-hand side, we have Planck’s constant squared, the mass factor in the denominator, and the Laplacian operator – i.e. ∇²= ∇·∇, with ∇ = (∂/∂x, ∂/∂y, ∂/∂z) – operating on the wavefunction.

Let’s start with the latter. The Laplacian works just the same as for our heat diffusion equation: it gives us a flux density, i.e. something expressed per square meter (1/m²). The ħ²factor gives us J²·s². The mass factor makes everything come out alright, if we use the mass-equivalence relation, which says it’s OK to express the mass in J/(m/s)². [The mass of an electron is usually expressed as being equal to 0.5109989461(31) MeV/c². That unit uses the E = m·c²mass-equivalence formula. As for the eV, you know we can convert that into joule, which is a rather large unit—which is why we use the electronvolt as a measure of energy.] To make a long story short, we’re OK: (J²·s²)·[(m/s)²/J]·(1/m²) = J! Perfect. [As for the Vψ term, that’s obviously expressed in joule too.]

In short, Schrödinger’s equation expresses the energy conservation law too, and we may express it per square meter or per second or per cubic meter as well, if we’d wish: we can just multiply both sides by 1/m² or 1/s or 1/m³or by whatever dimension you want. Again, if you want more detail on the Schrödinger equation as an energy propagation mechanism, read the mentioned e-publication. So let’s get back to our equation, which, taking into account our formula for V, now looks like this:

Eq1

Feynman then injects one of these enigmatic phrases—enigmatic for novices like us, at least!

“We want to look for definite energy states, so we try to find solutions which have the form: ψ (r, t) = e^{−(i/ħ)·E·t}·ψ(r).”

At first, you may think he’s just trying to get rid of the relativistic correction in the argument of the wavefunction. Indeed, as I explain in that little booklet of mine, the –(p/ħ)·x term in the argument of the elementary wavefunction a·e^−i·θ = a·e^{−i·[(E/ħ)·t – (p/ħ)·x]} is there because the young Comte Louis de Broglie, back in 1924, when he wrote his groundbreaking PhD thesis, suggested the θ = ω∙t – k∙x = (E∙t – p∙x)/ħ formula for the argument of the wavefunction, as he knew that relativity theory had already established the invariance of the four-vector (dot) product p_μx_μ = E∙t – p∙x = p_μ‘x_μ‘ = E’∙t’ – p’∙x’. [Note that Planck’s constant, as a physical constant, should obviously not depend on the reference frame either. Hence, if the E∙t – p∙x product is invariant, so is (E∙t – p∙x)/ħ.] So the θ = E∙t – p∙x and the θ = E₀∙t’ = E’·t’ are fully equivalent. Using lingo, we can say that the argument of the wavefunction is a Lorentz scalar and, therefore, invariant under a Lorentz boost. Sounds much better, doesn’t it? 🙂

But… Well. That’s not why Feynman says what he says. He just makes abstraction of uncertainty here, as he looks for states with a definite energy state, indeed. Nothing more, nothing less. Indeed, you should just note that we can re-write the elementary a·e^{−i[(E/ħ)·t – (p/ħ)·x]} function as e^{−(i/ħ)·E·t}·a·e^{i·(p/ħ)·x]}. So that’s what Feynman does here: he just eases the search for functional forms that satisfy Schrödinger’s equation. You should note the following:

Writing the coefficient in front of the complex exponential as ψ(r) = a·e^{i·(p/ħ)·x]} does the trick we want it to do: we do not want that coefficient to depend on time: it should only depend on the size of our ‘box’ in space, as I explained in one of my posts.
Having said that, you should also note that the ψ in the ψ(r, t) function and the ψ in the ψ(r) denote two different beasts: one is a function of two variables (r and t), while the other makes abstraction of the time factor and, hence, becomes a function of one variable only (r). I would have used another symbol for the ψ(r) function, but then the Master probably just wants to test your understanding. 🙂

In any case, the differential equation we need to solve now becomes:

Huh? How does that work? Well… Just take the time derivative of e^{−(i/ħ)·E·t}·ψ(r), multiply with the i·ħ in front of that term in Schrödinger’s original equation and re-arrange the terms. [Just do it: ∂[e^{−(i/ħ)·E·t}·ψ(r)]/∂t = −(i/ħ)·E·e^{−(i/ħ)·E·t}·ψ(r). Now multiply that with i·ħ: the ħ factor cancels and the minus disappears because i²= −1.]

So now we need to solve that differential equation, i.e. we need to find functional forms for ψ – and please do note we’re talking ψ(r) here – not ψ(r, t)! – that satisfy the above equation. Interesting question: is our equation still Schrödinger’s equation? Well… It is and it isn’t. Any linear combination of the definite energy solutions we find will also solve Schrödinger’s equation, but so we limited the solution set here to those definite energy solutions only. Hence, it’s not quite the same equation. We removed the time dependency here – and in a rather interesting way, I’d say.

The next thing to do is to switch from Cartesian to polar coordinates. Why? Well… When you have a central-force problem – like this one (because of the potential) – it’s easier to solve them using polar coordinates. In fact, because we’ve got three dimensions here, we’re actually talking a spherical coordinate system. The illustration and formulas below show how spherical and Cartesian coordinates are related:

x = r·sinθ·cosφ; y = r·sinθ·sinφ; z = r·cosθ

558px-3D_Spherical

As you know, θ (theta) is referred to as the polar angle, while φ (phi) is the azimuthal angle, and the coordinate transformation formulas can be easily derived. The rather simple differential equation above now becomes the following monster:

new de

Huh? Yes, I am very sorry. That’s how it is. Feynman does this to help us. If you think you can get to the solutions by directly solving the equation in Cartesian coordinates, please do let me know. 🙂 To tame the beast, we might imagine to first look for solutions that are spherically symmetric, i.e. solutions that do not depend on θ and φ. That means we could rotate the reference frame and none of the amplitudes would change. That means the ∂ψ/∂θ and ∂ψ/∂φ (partial) derivatives in our formula are equal to zero. These spherically symmetric states, or s-states as they are referred to, are states with zero (orbital) angular momentum, but you may want to think about that statement before accepting it. 🙂 [It’s not that there’s no angular momentum (on the contrary: there’s lots of it), but the total angular momentum should obviously be zero, and so that’s what meant when these states are denoted as l = 0 states.] So now we have to solve:

de 3

Now that looks somewhat less monstrous, but Feynman still fills two rather dense pages to show how this differential equation can be solved. It’s not only tedious but also complicated, so please check it yourself by clicking on the link. One of the steps is a switch in variables, or a re-scaling, I should say. Both E and r are now measured as follows:

The complicated-looking factors are just the Bohr radius (r_B= ħ²/(m·e²) ≈ 0.528 Å) and the Rydberg energy (E_R= m·e⁴/2·ħ² ≈ 13.6 eV). We calculated those long time ago using a rather heuristic model to describe an atom. In case you’d want to check the dimensions, note e²is a rather special animal. It’s got nothing to do with Euler’s number. Instead, e²is equal to k_e·q_e², and the k_e here is Coulomb’s constant: k_e = 1/(4πε₀). This allows to re-write the force between two electrons as a function of the distance: F = e²/r². This, in turn, explains the rather weird dimension of e²: [e²] = N·e²= J·m. But I am digressing too much. The bottom line is: the various energy levels that fit the equation, i.e. the allowable energies, are fractions of the Rydberg energy, i.e. E_R=m·e⁴/2·ħ². To be precise, the formula for the n^thenergy level is:

E_n= − E_R/n².

The interesting thing is that the spherically symmetric solutions yield real-valued ψ(r) functions. The solutions for n = 1, 2, and 3 respectively, and their graph is given below.

graph As Feynman writes, all of the wave functions approach zero rapidly for large r (also, confusingly, denoted as ρ) after oscillating a few times, with the number of ‘bumps’ equal to n. Of course, you should note that you should put the time factor back in in order to correctly interpret these functions. Indeed, remember how we separated them when we wrote:

ψ(r, t) = e^{−i·(E/ħ)·t}·ψ(r)

We might say the ψ(r) function is sort of an envelope function for the whole wavefunction, but it’s not quite as straightforward as that. However, I am sure you’ll figure it out.

States with an angular dependence

So far, so good. But what if those partial derivatives are not zero? Now the calculations become really complicated. Among other things, we need these transformation matrices for rotations, which we introduced a very long time ago. As mentioned above, I don’t have the intention to copy Feynman here, who needs another two or three dense pages to work out the logic. Let me just state the grand result:

We’ve got a whole range of definite energy states, which correspond to orbitals that form an orthonormal basis for the actual wavefunction of the electron.
The orbitals are characterized by three quantum numbers, denoted as l, n and m respectively:
- The l is the quantum number of (total) angular momentum, and it’s equal to 0, 1, 2, 3, etcetera. [Of course, as usual, we’re measuring in units of ħ.] The l = 0 states are referred to as s-states, the l = 1 states are referred to as p-states, and the l = 2 states are d-states. They are followed by f, g, h, etcetera—for no particular good reason. [As Feynman notes: “The letters don’t mean anything now. They did once—they meant “sharp” lines, “principal” lines, “diffuse” lines and “fundamental” lines of the optical spectra of atoms. But those were in the days when people did not know where the lines came from. After f there were no special names, so we now just continue with g, h, and so on.]
- The m is referred to as the ‘magnetic’ quantum number, and it ranges from −l to +l.
- The n is the ‘principle’ quantum number, and it goes from l + 1 to infinity (∞).

How do these things actually look like? Let me insert two illustrations here: one from Feynman, and the other from Wikipedia.

shape

The number in front just tracks the number of s-, p-, d-, etc. orbital. The shaded region shows where the amplitudes are large, and the plus and minus signs show the relative sign of the amplitude. [See my remark above on the fact that the ψ factor is real-valued, even if the wavefunction as a whole is complex-valued.] The Wikipedia image shows the same density plots but, as it was made some 50 years later, with some more color. 🙂

660px-Hydrogen_Density_Plots

This is it, guys. Feynman takes it further by also developing the electron configurations for the next 35 elements in the periodic table but… Well… I am sure you’ll want to read the original here, rather than my summaries. 🙂

Congrats ! We now know all what we need to know. All that remains is lots of practical exercises, so you can be sure you master the material for your exam. 🙂