Of course, it’s not because I am done with Feynman’s Lectures, that I am done with physics. I want to move on to more advanced topics now. One of these is general relativity. In fact, I’ve started a page on that. Onwards! 🙂
This post continues my mini-series on Feynman’s Seminar on Superconductivity. Superconductivity is a state which produces many wondrous phenomena, but… Well… The flux quantization phenomenon may not be part of your regular YouTube feed but, as far as I am concerned, it may well be the most amazing manifestation of a quantum-mechanical phenomenon at a macroscopic scale. I mean… Super currents that keep going, with zero resistance, are weird—they explain how we can trap a magnetic flux in the first place—but the fact that such fluxes are quantized is even weirder.
The key idea is the following. When we cool a ring-shaped piece of superconducting material in a magnetic field, all the way down to the critical temperature that causes the electrons to condense into a superconducting fluid, then a super current will emerge—think of an eddy current, here, but with zero resistance—that will force the magnetic field out of the material, as shown below. This current will permanently trap some of the magnetic field, even when the external field is being removed. As said, that’s weird enough by itself but… Well… If we think of the super current as an eddy current encountering zero resistance, then the idea of a permanently trapped magnetic field makes sense, right? In case you’d doubt the effect… Well… Just watch one of the many videos on the effect on YouTube. 🙂 The amazing thing here is not the permanently trapped magnetic field, but the fact that it’s quantized.
To be precise, the trapped flux will always be an integer times 2πħ/q. In other words, the magnetic field which Feynman denotes by Φ (the capitalized Greek letter phi), will always be equal to:
Φ = n·2πħ/q, with n = 0, 1, 2, 3,…
Hence, the flux can be 0, 2πħ/q, 4πħ/q, 6πħ/q , and so on. The fact that it’s a multiple of 2π shows us it’s got to do with the fact that our piece of material is, effectively, a ring. The nice thing about this phenomenon is that the mathematical analysis is, in fact, fairly easy to follow—or… Well… Much easier than what we discussed before. 🙂 Let’s quickly go through it.
We have a formula for the magnetic flux. It must be equal to the line integral of the vector potential (A) around a closed loop Τ, so we write:
Now, we can choose the loop Τ to be well inside the body of the ring, so that it never gets near the surface, as illustrated below. So we know that the current J is zero there. [In case you doubt this, see my previous post.]
One of the equations we introduced in our previous post, ħ∇θ = m·v + q·A, will then reduce to:
ħ∇θ = q·A
Why? The v in the m·v term (the velocity of the superconducting fluid, really), is zero. Remember the analysis is for this particular loop (well inside the ring) only. So… Well… If we integrate the expression above, we get:
Combining the two expressions with the integrals, we get:
Now, the line integral of a gradient from one point to another (say from point 1 to point 2) is the difference of the values of the function at the two points, so we can write:
Now what constraints are there on the values of θ1 and θ2? Well… You might think that, if they’re associated with the same point (we’re talking a closed loop, right?), then the two values should be the same, but… Well… No. All we can say is that the wavefunction must have the same value. We wrote that wavefunction as:
ψ = ρ(r)1/2eθ(r)
The value of this function at some point r is the same if θ changes by n·2π. Hence, when doing one complete turn around the ring, the ∫∇θ·ds integral in the integral formulas we wrote down must be equal to n·2π. Therefore, the second integral expression above can be re-written as:
That’s the result we wanted to explain so… Well… We’re done. Let me wrap up by quoting Feynman’s account of the 1961 experiment which confirmed London’s prediction of the effect, which goes back to 1950! It’s interesting, because… Well… It shows how up to date Feynman’s Lectures really are—or were, back in 1963, at least!
We didn’t get very far in our first post on Feynman’s Seminar on Superconductivity, and then I shifted my attention to other subjects over the past few months. So… Well… Let me re-visit the topic here.
One of the difficulties one encounters when trying to read this so-called seminar—which, according to Feynman, is ‘for entertainment only’ and, therefore, not really part of the Lectures themselves—is that Feynman throws in a lot of stuff that is not all that relevant to the topic itself but… Well… He apparently didn’t manage to throw all that he wanted to throw into his (other) Lectures on Quantum Mechanics and so he inserted a lot of stuff which he could, perhaps, have discussed elsewhere. So let us try to re-construct the main lines of reasoning here.
The first equation is Schrödinger’s equation for some particle with charge q that is moving in an electromagnetic field that is characterized not only by the (scalar) potential Φ but also by a vector potential A:
This closely resembles Schrödinger’s equation for an electron that is moving in an electric field only, which we used to find the energy states of electrons in a hydrogen atom: i·ħ·∂ψ/∂t = −(1/2)·(ħ2/m)∇2ψ + V·ψ. We just need to note the following:
- On the left-hand side, we can, obviously, replace −1/i by i.
- On the right-hand side, we can replace V by q·Φ, because the potential of a charge in an electric field is the product of the charge (q) and the (electric) potential (Φ).
- As for the other term on the right-hand side—so that’s the −(1/2)·(ħ2/m)∇2ψ term—we can re-write −ħ2·∇2ψ as [(ħ/i)·∇]·[(ħ/i)·∇]ψ because (1/i)·(1/i) = 1/i2 = 1/(−1) = −1. 🙂
- So all that’s left now, is that additional −q·A term in the (ħ/i)∇ − q·A expression. In our post, we showed that’s easily explained because we’re talking magnetodynamics: we’ve got to allow for the possibility of changing magnetic fields, and so that’s what the −q·A term captures.
Now, the latter point is not so easy to grasp but… Well… I’ll refer you that first post of mine, in which I show that some charge in a changing magnetic field will effectively gather some extra momentum, whose magnitude will be equal to p = m·v = −q·A. So that’s why we need to introduce another momentum operator here, which we write as:
OK. Next. But… Then… Well… All of what follows are either digressions—like the section on the local conservation of probabilities—or, else, quite intuitive arguments. Indeed, Feynman does not give us the nitty-gritty of the Bardeen-Cooper-Schrieffer theory, nor is the rest of the argument nearly as rigorous as the derivation of the electron orbitals from Schrödinger’s equation in an electrostatic field. So let us closely stick to what he does write, and try our best to follow the arguments.
The key assumption is that there is some attraction between electrons which, at low enough temperatures, can overcome the Coulomb repulsion. Where does this attraction come from? Feynman does not give us any clues here. He just makes a reference to the BCS theory but notes this theory is “not the subject of this seminar”, and that we should just “accept the idea that the electrons do, in some manner or other, work in pairs”, and that “we can think of thos−e pairs as behaving more or less like particles”, and that “we can, therefore, talk about the wavefunction for a pair.”
So we have a new particle, so to speak, which consists of two electrons who move through the conductor as one. To be precise, the electron pair behaves as a boson. Now, bosons have integer spin. According to the spin addition rule, we have four possibilities here but only three possible values:− 1/2 + 1/2 = 1; −1/2 + 1/2 = 0; +1/2 − 1/2 = 0; −1/2 − 1/2 = − 1. Of course, it is tempting to think these Cooper pairs are just like the electron pairs in the atomic orbitals, whose spin is always opposite because of the Fermi exclusion principle. Feynman doesn’t say anything about this, but the Wikipedia article on the BCS theory notes that the two electrons in a Cooper pair are, effectively, correlated because of their opposite spin. Hence, we must assume the Cooper pairs effectively behave like spin-zero particles.
Now, unlike fermions, bosons can collectively share the same energy state. In fact, they are likely to share the same state into what is referred to as a Bose-Einstein condensate. As Feynman puts it: “Since electron pairs are bosons, when there are a lot of them in a given state there is an especially large amplitude for other pairs to go to the same state. So nearly all of the pairs will be locked down at the lowest energy in exactly the same state—it won’t be easy to get one of them into another state. There’s more amplitude to go into the same state than into an unoccupied state by the famous factor √n, where n−1 is the occupancy of the lowest state. So we would expect all the pairs to be moving in the same state.”
Of course, this only happens at very low temperatures, because even if the thermal energy is very low, it will give the electrons sufficient energy to ensure the attractive force is overcome and all pairs are broken up. It is only at very low temperature that they will pair up and go into a Bose-Einstein condensate. Now, Feynman derives this √n factor in a rather abstruse introductory Lecture in the third volume, and I’d advise you to google other material on Bose-Einstein statistics because… Well… The mentioned Lecture is not among Feynman’s finest. OK. Next step.
Cooper pairs and wavefunctions
We know the probability of finding a Cooper pair is equal to the absolute square of its wavefunction. Now, it is very reasonable to assume that this probability will be proportional to the charge density (ρ), so we can write:
|ψ|2 = ψψ* ∼ ρ(r)
The argument here (r) is just the position vector. The next step, then, is to write ψ as the square root of ρ(r) times some phase factor θ. Abstracting away from time, this phase factor will also depend on r, of course. So this is what Feynman writes:
ψ = ρ(r)1/2eθ(r)
As Feynman notes, we can write any complex function of r like this but… Well… The charge density is, obviously, something real. Something we can measure, so we’re not writing the obvious here. The next step is even less obvious.
In our first post, we spent quite some time on Feynman’s digression on the local conservation of probability and… Well… I wrote above I didn’t think this digression was very useful. It now turns out it’s a central piece in the puzzle that Feynman is trying to solve for us here. The key formula here is the one for the so-called probability current, which—as Feynman shows—we write as:
This current J can also be written as:
Now, Feynman skips all of the math here (he notes “it’s just a change of variables” but so he doesn’t want to go through all of the algebra), and so I’ll just believe him when he says that, when substituting ψ for our wavefunction ψ = ρ(r)1/2eθ(r), then we can express this ‘current’ (J) in terms of ρ and θ. To be precise, he writes J as: So what? Well… It’s really fascinating to see what happens next. While J was some rather abstract concept so far—what’s a probability current, really?—Feynman now suggests we may want to think of it as a very classical electric current—the charge density times the velocity of the fluid of electrons. Hence, we equate J to J = ρ·v. Now, if the equation above holds true, but J is also equal to J = ρ·v, then the equation above is equivalent to:
Now, that gives us a formula for ħ∇θ. We write:
ħ∇θ = m·v + q·A
Now, in my previous post on this Seminar, I noted that Feynman attaches a lot of importance to this m·v + q·A quantity because… Well… It’s actually an invariant quantity. The argument can be, very briefly, summarized as follows. During the build-up of (or a change in) a magnetic flux, a charge will pick up some (classical) momentum that is equal to p = m·v = −q·A. Hence, the m·v + q·A sum is zero, and so… Well… That’s it, really: it’s some quantity that… Well… It has a significance in quantum mechanics. What significance? Well… Think of what we’ve been writing here. The v and the A have a physical significance, obviously. Therefore, that phase factor θ(r) must also have a physical significance.
But the question remains: what physical significance, exactly? Well… Let me quote Feynman here:
“The phase is just as observable as the charge density ρ. It is a piece of the current density J. The absolute phase (θ) is not observable, but if the gradient of the phase (∇θ) is known everywhere, then the phase is known except for a constant. You can define the phase at one point, and then the phase everywhere is determined.”
That makes sense, doesn’t it? But it still doesn’t quite answer the question: what is the physical significance of θ(r). What is it, really? We may be able to answer that question after exploring the equations above a bit more, so let’s do that now.
The phenomenon of superconductivity itself is easily explained by the mentioned condensation of the Cooper pairs: they all go into the same energy state. They form, effectively, a superconducting fluid. Feynman’s description of this is as follows:
“There is no electrical resistance. There’s no resistance because all the electrons are collectively in the same state. In the ordinary flow of current you knock one electron or the other out of the regular flow, gradually deteriorating the general momentum. But here to get one electron away from what all the others are doing is very hard because of the tendency of all Bose particles to go in the same state. A current once started, just keeps on going forever.”
Frankly, I’ve re-read this a couple of times, but I don’t think it’s the best description of what we think is going on here. I’d rather compare the situation to… Well… Electrons moving around in an electron orbital. That’s doesn’t involve any radiation or energy transfer either. There’s just movement. Flow. The kind of flow we have in the wavefunction itself. Here I think the video on Bose-Einstein condensates on the French Tout est quantique site is quite instructive: all of the Cooper pairs join to become one giant wavefunction—one superconducting fluid, really. 🙂
The Meissner effect
Feynman describes the Meissner effect as follows:
“If you have a piece of metal in the superconducting state and turn on a magnetic field which isn’t too strong (we won’t go into the details of how strong), the magnetic field can’t penetrate the metal. If, as you build up the magnetic field, any of it were to build up inside the metal, there would be a rate of change of flux which would produce an electric field, and an electric field would immediately generate a current which, by Lenz’s law, would oppose the flux. Since all the electrons will move together, an infinitesimal electric field will generate enough current to oppose completely any applied magnetic field. So if you turn the field on after you’ve cooled a metal to the superconducting state, it will be excluded.
Even more interesting is a related phenomenon discovered experimentally by Meissner. If you have a piece of the metal at a high temperature (so that it is a normal conductor) and establish a magnetic field through it, and then you lower the temperature below the critical temperature (where the metal becomes a superconductor), the field is expelled. In other words, it starts up its own current—and in just the right amount to push the field out.”
The math here is interesting. Feynman first notes that, in any lump of superconducting metal, the divergence of the current must be zero, so we write: ∇·J = 0. At any point? Yes. The current that goes in must go out. No point is a sink or a source. Now the divergence operator (∇·J) is a linear operator. Hence, that means that, when applying the divergence operator to the J = (ħ/m)·[∇θ − (q/ħ)·A]·ρ equation, we’ll need to figure out what ∇·∇θ = = ∇2θ and ∇·A are. Now, as explained in my post on gauges, we can choose to make ∇·A equal to zero so… Well… We’ll make that choice and, hence, the term with ∇·A in it vanishes. So… Well… If ∇·J equals zero, then the term with ∇2θ has to be zero as well, so ∇2θ has to be zero. That, in turn, implies ∇θ has to be some constant (vector).
Now, there is a pretty big error in Feynman’s Lecture here, as it notes: “Now the only way that ∇2θ can be zero everywhere inside the lump of metal is for θ to be a constant.” It should read: ∇2θ can only be zero everywhere if ∇θ is a constant (vector). So now we need to remind ourselves of the reality of θ, as described by Feynman (quoted above): “The absolute phase (θ) is not observable, but if the gradient of the phase (∇θ) is known everywhere, then the phase is known except for a constant. You can define the phase at one point, and then the phase everywhere is determined.” So we can define, or choose, our constant (vector) ∇θ to be 0.
Hmm… We re-set not one but two gauges here: A and ∇θ. Tricky business, but let’s go along with it. [If we want to understand Feynman’s argument, then we actually have no choice than to go long with his argument, right?] The point is: the (ħ/m)·∇θ term in the J = (ħ/m)·[∇θ − (q/ħ)·A]·ρ vanishes, so the equation we’re left with tells us the current—so that’s an actual as well as a probability current!—is proportional to the vector potential:
Now, we’ve neglected any possible variation in the charge density ρ so far because… Well… The charge density in a superconducting fluid must be uniform, right? Why? When the metal is superconducting, an accumulation of electrons in one region would be immediately neutralized by a current, right? [Note that Feynman’s language is more careful here. He writes: the charge density is almost perfectly uniform.]
So what’s next? Well… We have a more general equation from the equations of electromagnetism:
[In case you’d want to know how we get this equation out of Maxwell’s equations, you can look it up online in one of the many standard textbooks on electromagnetism.] You recognize this as a Poisson equation… Well… Three Poisson equations: one for each component of A and J. We can now combine the two equations above by substituting J in that Poisson equation, so we get the following differential equation, which we need to solve for A:
The λ2 in this equation is, of course, a shorthand for the following constant:
Now, it’s very easy to see that both e−λr as well as e−λr are solutions for that Poisson equation. But what do they mean? In one dimension, r becomes the one-dimensional position variable x. You can check the shapes of these solutions with a graphing tool.
Note that only one half of each graph counts: the vector potential must decrease when we go from the surface into the material, and there is a cut-off at the surface of the material itself, of course. So all depends on the size of λ, as compared to the size of our piece of superconducting metal (or whatever other substance our piece is made of). In fact, if we look at e−λx as as an exponential decay function, then τ = 1/λ is the so-called scaling constant (it’s the inverse of the decay constant, which is λ itself). [You can work this out yourself. Note that for x = τ = 1/λ, the value of our function e−λx will be equal to e−λ(1/λ) = e−1 ≈ 0.368, so it means the value of our function is reduced to about 36.8% of its initial value. For all practical purposes, we may say—as Feynman notes—that the field will, effectively, only penetrate to a thin layer at the surface: a layer of about 1/1/λ in thickness. He illustrates this as follows:
Moreover, he calculates the 1/λ distance for lead. Let me copy him here:
Well… That says it all, right? We’re talking two millionths of a centimeter here… 🙂
So what’s left? A lot, like flux quantization, or the equations of motion for the superconducting electron fluid. But we’ll leave that for the next posts. 🙂
My previous post was awfully long, so I must assume many of my readers may have started to read it, but… Well… Gave up halfway or even sooner. 🙂 I added a footnote, though, which is interesting to reflect upon. Also, I know many of my readers aren’t interested in the math—even if they understand one cannot really appreciate quantum theory without the math. But… Yes. I may have left some readers behind. Let me, therefore, pick up the most interesting bit of all of the stories in my last posts in as easy a language as I can find.
We have that weird 360/720° symmetry in quantum physics or—to be precise—we have it for elementary matter-particles (think of electrons, for example). In order to, hopefully, help you understand what it’s all about, I had to explain the often-confused but substantially different concepts of a reference frame and a representational base (or representation tout court). I won’t repeat that explanation, but think of the following.
If we just rotate the reference frame over 360°, we’re just using the same reference frame and so we see the same thing: some object which we, vaguely, describe by some ei·θ function. Think of some spinning object. In its own reference frame, it will just spin around some center or, in ours, it will spin while moving along some axis in its own reference frame or, seen from ours, as moving in some direction while it’s spinning—as illustrated below.
To be precise, I should say that we describe it by some Fourier sum of such functions. Now, if its spin direction is… Well… In the other direction, then we’ll describe it by by some e−i·θ function (again, you should read: a Fourier sum of such functions). Now, the weird thing is is the following: if we rotate the object itself, over the same 360°, we get a different object: our ei·θ and e−i·θ function (again: think of a Fourier sum, so that’s a wave packet, really) becomes a −e±i·θ thing. We get a minus sign in front of it. So what happened here? What’s the difference, really?
Well… I don’t know. It’s very deep. Think of you and me as two electrons who are watching each other. If I do nothing, and you keep watching me while turning around me, for a full 360° (so that’s a rotation of your reference frame over 360°), then you’ll end up where you were when you started and, importantly, you’ll see the same thing: me. 🙂 I mean… You’ll see exactly the same thing: if I was an e+i·θ wave packet, I am still an an e+i·θ wave packet now. Or if I was an e−i·θ wave packet, then I am still an an e−i·θ wave packet now. Easy. Logical. Obvious, right?
But so now we try something different: I turn around, over a full 360° turn, and you stay where you are and watch me while I am turning around. What happens? Classically, nothing should happen but… Well… This is the weird world of quantum mechanics: when I am back where I was—looking at you again, so to speak—then… Well… I am not quite the same any more. Or… Well… Perhaps I am but you see me differently. If I was e+i·θ wave packet, then I’ve become a −e+i·θ wave packet now.
Not hugely different but… Well… That minus sign matters, right? Or If I was wave packet built up from elementary a·e−i·θ waves, then I’ve become a −e−i·θ wave packet now. What happened?
It makes me think of the twin paradox in special relativity. We know it’s a paradox—so that’s an apparent contradiction only: we know which twin stayed on Earth and which one traveled because of the gravitational forces on the traveling twin. The one who stays on Earth does not experience any acceleration or deceleration. Is it the same here? I mean… The one who’s turning around must experience some force.
Can we relate this to the twin paradox? Maybe. Note that a minus sign in front of the e−±i·θ functions amounts a minus sign in front of both the sine and cosine components. So… Well… The negative of a sine and cosine is the sine and cosine but with a phase shift of 180°: −cosθ = cos(θ ± π) and −sinθ = sin(θ ± π). Now, adding or subtracting a common phase factor to/from the argument of the wavefunction amounts to changing the origin of time. So… Well… I do think the twin paradox and this rather weird business of 360° and 720° symmetries are, effectively, related. 🙂
Post scriptum: Google honors Max Born’s 135th birthday today. 🙂 I think that’s a great coincidence in light of the stuff I’ve been writing about lately (possible interpretations of the wavefunction). 🙂
Ouff ! This title is quite a mouthful, isn’t it? 🙂 So… What’s the topic of the day? Well… In our previous posts, we developed a few key ideas in regard to a possible physical interpretation of the (elementary) wavefunction. It’s been an interesting excursion, and I summarized it in another pre-publication paper on the open arXiv.org site.
In my humble view, one of the toughest issues to deal with when thinking about geometric (or physical) interpretations of the wavefunction is the fact that a wavefunction does not seem to obey the classical 360° symmetry in space. In this post, I want to muse a bit about this and show that… Well… It does and it doesn’t. It’s got to do with what happens when you change from one representational base (or representation, tout court) to another which is… Well… Like changing the reference frame but, at the same time, it is also more than just a change of the reference frame—and so that explains the weird stuff (like that 720° symmetry of the amplitudes for spin-1/2 particles, for example).
I should warn you before you start reading: I’ll basically just pick up some statements from my paper (and previous posts) and develop some more thoughts on them. As a result, this post may not be very well structured. Hence, you may want to read the mentioned paper first.
The reality of directions
Huh? The reality of directions? Yes. I warned you. This post may cause brain damage. 🙂 The whole argument revolves around a thought experiment—but one whose results have been verified in zillions of experiments in university student labs so… Well… We do not doubt the results and, therefore, we do not doubt the basic mathematical results: we just want to try to understand them better.
So what is the set-up? Well… In the illustration below (Feynman, III, 6-3), Feynman compares the physics of two situations involving rather special beam splitters. Feynman calls them modified or ‘improved’ Stern-Gerlach apparatuses. The apparatus basically splits and then re-combines the two new beams along the z-axis. It is also possible to block one of the beams, so we filter out only particles with their spin up or, alternatively, with their spin down. Spin (or angular momentum or the magnetic moment) as measured along the z-axis, of course—I should immediately add: we’re talking the z-axis of the apparatus here.
The two situations involve a different relative orientation of the apparatuses: in (a), the angle is 0°, while in (b) we have a (right-handed) rotation of 90° about the z-axis. He then proves—using geometry and logic only—that the probabilities and, therefore, the magnitudes of the amplitudes (denoted by C+ and C− and C’+ and C’− in the S and T representation respectively) must be the same, but the amplitudes must have different phases, noting—in his typical style, mixing academic and colloquial language—that “there must be some way for a particle to tell that it has turned a corner in (b).”
The various interpretations of what actually happens here may shed some light on the heated discussions on the reality of the wavefunction—and of quantum states. In fact, I should note that Feynman’s argument revolves around quantum states. To be precise, the analysis is focused on two-state systems only, and the wavefunction—which captures a continuum of possible states, so to speak—is introduced only later. However, we may look at the amplitude for a particle to be in the up– or down-state as a wavefunction and, therefore (but do note that’s my humble opinion once more), the analysis is actually not all that different.
We know, from theory and experiment, that the amplitudes are different. For example, for the given difference in the relative orientation of the two apparatuses (90°), we know that the amplitudes are given by C’+ = ei∙φ/2∙C+ = e i∙π/4∙C+ and C’− = e−i∙φ/2∙C+ = e− i∙π/4∙C− respectively (the amplitude to go from the down to the up state, or vice versa, is zero). Hence, yes, we—not the particle, Mr. Feynman!—know that, in (b), the electron has, effectively, turned a corner.
The more subtle question here is the following: is the reality of the particle in the two setups the same? Feynman, of course, stays away from such philosophical question. He just notes that, while “(a) and (b) are different”, “the probabilities are the same”. He refrains from making any statement on the particle itself: is or is it not the same? The common sense answer is obvious: of course, it is! The particle is the same, right? In (b), it just took a turn—so it is just going in some other direction. That’s all.
However, common sense is seldom a good guide when thinking about quantum-mechanical realities. Also, from a more philosophical point of view, one may argue that the reality of the particle is not the same: something might—or must—have happened to the electron because, when everything is said and done, the particle did take a turn in (b). It did not in (a). [Note that the difference between ‘might’ and ‘must’ in the previous phrase may well sum up the difference between a deterministic and a non-deterministic world view but… Well… This discussion is going to be way too philosophical already, so let’s refrain from inserting new language here.]
Let us think this through. The (a) and (b) set-up are, obviously, different but… Wait a minute… Nothing is obvious in quantum mechanics, right? How can we experimentally confirm that they are different?
Huh? I must be joking, right? You can see they are different, right? No. I am not joking. In physics, two things are different if we get different measurement results. [That’s a bit of a simplified view of the ontological point of view of mainstream physicists, but you will have to admit I am not far off.] So… Well… We can’t see those amplitudes and so… Well… If we measure the same thing—same probabilities, remember?—why are they different? Think of this: if we look at the two beam splitters as one single tube (an ST tube, we might say), then all we did in (b) was bend the tube. Pursuing the logic that says our particle is still the same even when it takes a turn, we could say the tube is still the same, despite us having wrenched it over a 90° corner.
Now, I am sure you think I’ve just gone nuts, but just try to stick with me a little bit longer. Feynman actually acknowledges the same: we need to experimentally prove (a) and (b) are different. He does so by getting a third apparatus in (U), as shown below, whose relative orientation to T is the same in both (a) and (b), so there is no difference there.
Now, the axis of U is not the z-axis: it is the x-axis in (a), and the y-axis in (b). So what? Well… I will quote Feynman here—not (only) because his words are more important than mine but also because every word matters here:
“The two apparatuses in (a) and (b) are, in fact, different, as we can see in the following way. Suppose that we put an apparatus in front of S which produces a pure +x state. Such particles would be split into +z and −z into beams in S, but the two beams would be recombined to give a +x state again at P1—the exit of S. The same thing happens again in T. If we follow T by a third apparatus U, whose axis is in the +x direction and, as shown in (a), all the particles would go into the + beam of U. Now imagine what happens if T and U are swung around together by 90° to the positions shown in (b). Again, the T apparatus puts out just what it takes in, so the particles that enter U are in a +x state with respect to S, which is different. By symmetry, we would now expect only one-half of the particles to get through.”
I should note that (b) shows the U apparatus wide open so… Well… I must assume that’s a mistake (and should alert the current editors of the Lectures to it): Feynman’s narrative tells us we should also imagine it with the minus channel shut. In that case, it should, effectively, filter approximately half of the particles out, while they all get through in (a). So that’s a measurement result which shows the direction, as we see it, makes a difference.
Now, Feynman would be very angry with me—because, as mentioned, he hates philosophers—but I’d say: this experiment proves that a direction is something real. Of course, the next philosophical question then is: what is a direction? I could answer this by pointing to the experiment above: a direction is something that alters the probabilities between the STU tube as set up in (a) versus the STU tube in (b). In fact—but, I admit, that would be pretty ridiculous—we could use the varying probabilities as we wrench this tube over varying angles to define an angle! But… Well… While that’s a perfectly logical argument, I agree it doesn’t sound very sensical.
OK. Next step. What follows may cause brain damage. 🙂 Please abandon all pre-conceived notions and definitions for a while and think through the following logic.
You know this stuff is about transformations of amplitudes (or wavefunctions), right? [And you also want to hear about those special 720° symmetry, right? No worries. We’ll get there.] So the questions all revolve around this: what happens to amplitudes (or the wavefunction) when we go from one reference frame—or representation, as it’s referred to in quantum mechanics—to another?
Well… I should immediately correct myself here: a reference frame and a representation are two different things. They are related but… Well… Different… Quite different. Not same-same but different. 🙂 I’ll explain why later. Let’s go for it.
Before talking representations, let us first think about what we really mean by changing the reference frame. To change it, we first need to answer the question: what is our reference frame? It is a mathematical notion, of course, but then it is also more than that: it is our reference frame. We use it to make measurements. That’s obvious, you’ll say, but let me make a more formal statement here:
The reference frame is given by (1) the geometry (or the shape, if that sounds easier to you) of the measurement apparatus (so that’s the experimental set-up) here) and (2) our perspective of it.
If we would want to sound academic, we might refer to Kant and other philosophers here, who told us—230 years ago—that the mathematical idea of a three-dimensional reference frame is grounded in our intuitive notions of up and down, and left and right. [If you doubt this, think about the necessity of the various right-hand rules and conventions that we cannot do without in math, and in physics.] But so we do not want to sound academic. Let us be practical. Just think about the following. The apparatus gives us two directions:
(1) The up direction, which we associate with the positive direction of the z-axis, and
(2) the direction of travel of our particle, which we associate with the positive direction of the y-axis.
Now, if we have two axes, then the third axis (the x-axis) will be given by the right-hand rule, right? So we may say the apparatus gives us the reference frame. Full stop. So… Well… Everything is relative? Is this reference frame relative? Are directions relative? That’s what you’ve been told, but think about this: relative to what? Here is where the object meets the subject. What’s relative? What’s absolute? Frankly, I’ve started to think that, in this particular situation, we should, perhaps, not use these two terms. I am not saying that our observation of what physically happens here gives these two directions any absolute character but… Well… You will have to admit they are more than just some mathematical construct: when everything is said and done, we will have to admit that these two directions are real. because… Well… They’re part of the reality that we are observing, right? And the third one… Well… That’s given by our perspective—by our right-hand rule, which is… Well… Our right-hand rule.
Of course, now you’ll say: if you think that ‘relative’ and ‘absolute’ are ambiguous terms and that we, therefore, may want to avoid them a bit more, then ‘real’ and its opposite (unreal?) are ambiguous terms too, right? Well… Maybe. What language would you suggest? 🙂 Just stick to the story for a while. I am not done yet. So… Yes… What is their reality? Let’s think about that in the next section.
Perspectives, reference frames and symmetries
You’ve done some mental exercises already as you’ve been working your way through the previous section, but you’ll need to do plenty more. In fact, they may become physical exercise too: when I first thought about these things (symmetries and, more importantly, asymmetries in space), I found myself walking around the table with some asymmetrical everyday objects and papers with arrows and clocks and other stuff on it—effectively analyzing what right-hand screw, thumb or grip rules actually mean. 🙂
So… Well… I want you to distinguish—just for a while—between the notion of a reference frame (think of the x–y–z reference frame that comes with the apparatus) and your perspective on it. What’s our perspective on it? Well… You may be looking from the top, or from the side and, if from the side, from the left-hand side or the right-hand side—which, if you think about it, you can only define in terms of the various positive and negative directions of the various axes. 🙂 If you think this is getting ridiculous… Well… Don’t. Feynman himself doesn’t think this is ridiculous, because he starts his own “long and abstract side tour” on transformations with a very simple explanation of how the top and side view of the apparatus are related to the axes (i.e. the reference frame) that comes with it. You don’t believe me? This is the very first illustration of his Lecture on this:
He uses it to explain the apparatus (which we don’t do here because you’re supposed to already know how these (modified or improved) Stern-Gerlach apparatuses work). So let’s continue this story. Suppose that we are looking in the positive y-direction—so that’s the direction in which our particle is moving—then we might imagine how it would look like when we would make a 180° turn and look at the situation from the other side, so to speak. We do not change the reference frame (i.e. the orientation) of the apparatus here: we just change our perspective on it. Instead of seeing particles going away from us, into the apparatus, we now see particles coming towards us, out of the apparatus.
What happens—but that’s not scientific language, of course—is that left becomes right, and right becomes left. Top still is top, and bottom is bottom. We are looking now in the negative y-direction, and the positive direction of the x-axis—which pointed right when we were looking in the positive y-direction—now points left. I see you nodding your head now—because you’ve heard about parity inversions, mirror symmetries and what have you—and I hear you say: “That’s the mirror world, right?”
No. It is not. I wrote about this in another post: the world in the mirror is the world in the mirror. We don’t get a mirror image of an object by going around it and looking at its back side. I can’t dwell too much on this (just check that post, and another one who talks about the same), but so don’t try to connect it to the discussions on symmetry-breaking and what have you. Just stick to this story, which is about transformations of amplitudes (or wavefunctions). [If you really want to know—but I know this sounds counterintuitive—the mirror world doesn’t really switch left for right. Your reflection doesn’t do a 180 degree turn: it is just reversed front to back, with no rotation at all. It’s only your brain which mentally adds (or subtracts) the 180 degree turn that you assume must have happened from the observed front to back reversal. So the left to right reversal is only apparent. It’s a common misconception, and… Well… I’ll let you figure this out yourself. I need to move on.] Just note the following:
- The xyz reference frame remains a valid right-handed reference frame. Of course it does: it comes with our beam splitter, and we can’t change its reality, right? We’re just looking at it from another angle. Our perspective on it has changed.
- However, if we think of the real and imaginary part of the wavefunction describing the electrons that are going through our apparatus as perpendicular oscillations (as shown below)—a cosine and sine function respectively—then our change in perspective might, effectively, mess up our convention for measuring angles.
I am not saying it does. Not now, at least. I am just saying it might. It depends on the plane of the oscillation, as I’ll explain in a few moments. Think of this: we measure angles counterclockwise, right? As shown below… But… Well… If the thing below would be some funny clock going backwards—you’ve surely seen them in a bar or so, right?—then… Well… If they’d be transparent, and you’d go around them, you’d see them as going… Yes… Clockwise. 🙂 [This should remind you of a discussion on real versus pseudo-vectors, or polar versus axial vectors, but… Well… We don’t want to complicate the story here.]
Now, if we would assume this clock represents something real—and, of course, I am thinking of the elementary wavefunction eiθ = cosθ + i·sinθ now—then… Well… Then it will look different when we go around it. When going around our backwards clock above and looking at it from… Well… The back, we’d describe it, naively, as… Well… Think! What’s your answer? Give me the formula! 🙂
We’d see it as e−iθ = cos(−θ) + i·sin(−θ) = cosθ − i·sinθ, right? The hand of our clock now goes clockwise, so that’s the opposite direction of our convention for measuring angles. Hence, instead of eiθ, we write e−iθ, right? So that’s the complex conjugate. So we’ve got a different image of the same thing here. Not good. Not good at all.
You’ll say: so what? We can fix this thing easily, right? You don’t need the convention for measuring angles or for the imaginary unit (i) here. This particle is moving, right? So if you’d want to look at the elementary wavefunction as some sort of circularly polarized beam (which, I admit, is very much what I would like to do, but its polarization is rather particular as I’ll explain in a minute), then you just need to define left- and right-handed angles as per the standard right-hand screw rule (illustrated below). To hell with the counterclockwise convention for measuring angles!
You are right. We could use the right-hand rule more consistently. We could, in fact, use it as an alternative convention for measuring angles: we could, effectively, measure them clockwise or counterclockwise depending on the direction of our particle. But… Well… The fact is: we don’t. We do not use that alternative convention when we talk about the wavefunction. Physicists do use the counterclockwise convention all of the time and just jot down these complex exponential functions and don’t realize that, if they are to represent something real, our perspective on the reference frame matters. To put it differently, the direction in which we are looking at things matters! Hence, the direction is not… Well… I am tempted to say… Not relative at all but then… Well… We wanted to avoid that term, right? 🙂
I guess that, by now, your brain may suffered from various short-circuits. If not, stick with me a while longer. Let us analyze how our wavefunction model might be impacted by this symmetry—or asymmetry, I should say.
The flywheel model of an electron
In our previous posts, we offered a model that interprets the real and the imaginary part of the wavefunction as oscillations which each carry half of the total energy of the particle. These oscillations are perpendicular to each other, and the interplay between both is how energy propagates through spacetime. Let us recap the fundamental premises:
- The dimension of the matter-wave field vector is force per unit mass (N/kg), as opposed to the force per unit charge (N/C) dimension of the electric field vector. This dimension is an acceleration (m/s2), which is the dimension of the gravitational field.
- We assume this gravitational disturbance causes our electron (or a charged mass in general) to move about some center, combining linear and circular motion. This interpretation reconciles the wave-particle duality: fields interfere but if, at the same time, they do drive a pointlike particle, then we understand why, as Feynman puts it, “when you do find the electron some place, the entire charge is there.” Of course, we cannot prove anything here, but our elegant yet simple derivation of the Compton radius of an electron is… Well… Just nice. 🙂
- Finally, and most importantly in the context of this discussion, we noted that, in light of the direction of the magnetic moment of an electron in an inhomogeneous magnetic field, the plane which circumscribes the circulatory motion of the electron should also comprise the direction of its linear motion. Hence, unlike an electromagnetic wave, the plane of the two-dimensional oscillation (so that’s the polarization plane, really) cannot be perpendicular to the direction of motion of our electron.
Let’s say some more about the latter point here. The illustrations below (one from Feynman, and the other is just open-source) show what we’re thinking of. The direction of the angular momentum (and the magnetic moment) of an electron—or, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron is traveling—cannot be parallel to the direction of motion. On the contrary, it must be perpendicular to the direction of motion. In other words, if we imagine our electron as spinning around some center (see the illustration on the left-hand side), then the disk it circumscribes (i.e. the plane of the polarization) has to comprise the direction of motion.
Of course, we need to add another detail here. As my readers will know, we do not really have a precise direction of angular momentum in quantum physics. While there is no fully satisfactory explanation of this, the classical explanation—combined with the quantization hypothesis—goes a long way in explaining this: an object with an angular momentum J and a magnetic moment μ that is not exactly parallel to some magnetic field B, will not line up: it will precess—and, as mentioned, the quantization of angular momentum may well explain the rest. [Well… Maybe… We have detailed our attempts in this regard in various posts on this (just search for spin or angular momentum on this blog, and you’ll get a dozen posts or so), but these attempts are, admittedly, not fully satisfactory. Having said that, they do go a long way in relating angles to spin numbers.]
The thing is: we do assume our electron is spinning around. If we look from the up-direction only, then it will be spinning clockwise if its angular momentum is down (so its magnetic moment is up). Conversely, it will be spinning counterclockwise if its angular momentum is up. Let us take the up-state. So we have a top view of the apparatus, and we see something like this:I know you are laughing aloud now but think of your amusement as a nice reward for having stuck to the story so far. Thank you. 🙂 And, yes, do check it yourself by doing some drawings on your table or so, and then look at them from various directions as you walk around the table as—I am not ashamed to admit this—I did when thinking about this. So what do we get when we change the perspective? Let us walk around it, counterclockwise, let’s say, so we’re measuring our angle of rotation as some positive angle. Walking around it—in whatever direction, clockwise or counterclockwise—doesn’t change the counterclockwise direction of our… Well… That weird object that might—just might—represent an electron that has its spin up and that is traveling in the positive y-direction.
When we look in the direction of propagation (so that’s from left to right as you’re looking at this page), and we abstract away from its linear motion, then we could, vaguely, describe this by some wrenched eiθ = cosθ + i·sinθ function, right? The x- and y-axes of the apparatus may be used to measure the cosine and sine components respectively.
Let us keep looking from the top but walk around it, rotating ourselves over a 180° angle so we’re looking in the negative y-direction now. As I explained in one of those posts on symmetries, our mind will want to switch to a new reference frame: we’ll keep the z-axis (up is up, and down is down), but we’ll want the positive direction of the x-axis to… Well… Point right. And we’ll want the y-axis to point away, rather than towards us. In short, we have a transformation of the reference frame here: z’ = z, y’ = − y, and x’ = − x. Mind you, this is still a regular right-handed reference frame. [That’s the difference with a mirror image: a mirrored right-hand reference frame is no longer right-handed.] So, in our new reference frame, that we choose to coincide with our perspective, we will now describe the same thing as some −cosθ − i·sinθ = −eiθ function. Of course, −cosθ = cos(θ + π) and −sinθ = sin(θ + π) so we can write this as:
−cosθ − i·sinθ = cos(θ + π) + i·sinθ = ei·(θ+π) = eiπ·eiθ = −eiθ.
Sweet ! But… Well… First note this is not the complex conjugate: e−iθ = cosθ − i·sinθ ≠ −cosθ − i·sinθ = −eiθ. Why is that? Aren’t we looking at the same clock, but from the back? No. The plane of polarization is different. Our clock is more like those in Dali’s painting: it’s flat. 🙂 And, yes, let me lighten up the discussion with that painting here. 🙂 We need to have some fun while torturing our brain, right?
So, because we assume the plane of polarization is different, we get an −eiθ function instead of a e−iθ function.
Let us now think about the ei·(θ+π) function. It’s the same as −eiθ but… Well… We walked around the z-axis taking a full 180° turn, right? So that’s π in radians. So that’s the phase shift here. Hey! Try the following now. Go back and walk around the apparatus once more, but let the reference frame rotate with us, as shown below. So we start left and look in the direction of propagation, and then we start moving about the z-axis (which points out of this page, toward you, as you are looking at this), let’s say by some small angle α. So we rotate the reference frame about the z-axis by α and… Well… Of course, our ei·θ now becomes an our ei·(θ+α) function, right? We’ve just derived the transformation coefficient for a rotation about the z-axis, didn’t we? It’s equal to ei·α, right? We get the transformed wavefunction in the new reference frame by multiplying the old one by ei·α, right? It’s equal to ei·α·ei·θ = ei·(θ+α), right?
No. The answer is: no. The transformation coefficient is not ei·α but ei·α/2. So we get an additional 1/2 factor in the phase shift.
Huh? Yes. That’s what it is: when we change the representation, by rotating our apparatus over some angle α about the z-axis, then we will, effectively, get a new wavefunction, which will differ from the old one by a phase shift that is equal to only half of the rotation angle only.
Huh? Yes. It’s even weirder than that. For a spin down electron, the transformation coefficient is e−i·α/2, so we get an additional minus sign in the argument.
I know you are terribly disappointed, but that’s how it is. That’s what hampers an easy geometric interpretation of the wavefunction. Paraphrasing Feynman, I’d say that, somehow, our electron not only knows whether or not it has taken a turn, but it also knows whether or not it is moving away from us or, conversely, towards us.
But… Hey! Wait a minute! That’s it, right?
What? Well… That’s it! The electron doesn’t know whether it’s moving away or towards us. That’s nonsense. But… Well… It’s like this:
Our ei·α coefficient describes a rotation of the reference frame. In contrast, the ei·α/2 and e−i·α/2 coefficients describe what happens when we rotate the T apparatus! Now that is a very different proposition.
Right! You got it! Representations and reference frames are different things. Quite different, I’d say: representations are real, reference frames aren’t—but then you don’t like philosophical language, do you? 🙂 But think of it. When we just go about the z-axis, a full 180°, but we don’t touch that T-apparatus, we don’t change reality. When we were looking at the electron while standing left to the apparatus, we watched the electrons going in and moving away from us, and when we go about the z-axis, a full 180°, looking at it from the right-hand side, we see the electrons coming out, moving towards us. But it’s still the same reality. We simply change the reference frame—from xyz to x’y’z’ to be precise: we do not change the representation.
In contrast, when we rotate the T apparatus over a full 180°, our electron now goes in the opposite direction. And whether that’s away or towards us, that doesn’t matter: it was going in one direction while traveling through S, and now it goes in the opposite direction—relative to the direction it was going in S, that is.
So what happens, really, when we change the representation, rather than the reference frame? Well… Let’s think about that. 🙂
The transformation matrix for the amplitude of a system to be in an up or down state (and, hence, presumably, for a wavefunction) for a rotation about the z-axis is the following one:
Feynman derives this matrix in a rather remarkable intellectual tour de force in the 6th of his Lectures on Quantum Mechanics. So that’s pretty early on. He’s actually worried about that himself, apparently, and warns his students that “This chapter is a rather long and abstract side tour, and it does not introduce any idea which we will not also come to by a different route in later chapters. You can, therefore, skip over it, and come back later if you are interested.”
Well… That’s how I approached it. I skipped it, and didn’t worry about those transformations for quite a while. But… Well… You can’t avoid them. In some weird way, they are at the heart of the weirdness of quantum mechanics itself. Let us re-visit his argument. Feynman immediately gets that the whole transformation issue here is just a matter of finding an easy formula for that phase shift. Why? He doesn’t tell us. Lesser mortals like us must just assume that’s how the instinct of a genius works, right? 🙂 So… Well… Because he knows—from experiment—that the coefficient is ei·α/2 instead of ei·α, he just says the phase shift—which he denotes by λ—must be some proportional to the angle of rotation—which he denotes by φ rather than α (so as to avoid confusion with the Euler angle α). So he writes:
λ = m·φ
Initially, he also tries the obvious thing: m should be one, right? So λ = φ, right? Well… No. It can’t be. Feynman shows why that can’t be the case by adding a third apparatus once again, as shown below.
Let me quote him here, as I can’t explain it any better:
“Suppose C+ and C’− = e−i·πC− = −C−. [Feynman works with states here, instead of the wavefunction of the particle as a whole. I’ll come back to this.] However, this is just the original state all over again. Both amplitudes are just multiplied by which gives back the original physical system. (It is again a case of a common phase change.) This means that if the angle between T and S is increased to 180°, the system would be indistinguishable from the zero-degree situation, and the particles would again go through the (+) state of the U apparatus. At , though, the (+) state of the U apparatus is the state of the original S apparatus. So a (+x) state would become a state. But we have done nothing to change the original state; the answer is wrong. We cannot have We must have the situation that a rotation by no smaller angle reproduces the same physical state. This will happen ifis rotated by ; then, clearly, it is right back at zero degrees, and we should have C’+ = C+ and C’− = C− or, what is the same thing, ei·m·2π . We get m = . [But no!] This argument is wrong! To see that it is, consider that T is rotated by . If m were equal to 1, we would have C’+ = ei·π
The result, of course, is this weird 720° symmetry. While we get the same physics after a 360° rotation of the T apparatus, we do not get the same amplitudes. We get the opposite (complex) number: C’+ = ei·2π/2C+ = −C+ and C’− = e−i·2π/2C− = −C−. That’s OK, because… Well… It’s a common phase shift, so it’s just like changing the origin of time. Nothing more. Nothing less. Same physics. Same reality. But… Well… C’+ ≠ −C+ and C’− ≠ −C−, right? We only get our original amplitudes back if we rotate the T apparatus two times, so that’s by a full 720 degrees—as opposed to the 360° we’d expect.
Now, space is isotropic, right? So this 720° business doesn’t make sense, right?
Well… It does and it doesn’t. We shouldn’t dramatize the situation. What’s the actual difference between a complex number and its opposite? It’s like x or −x, or t and −t. I’ve said this a couple of times already again, and I’ll keep saying it many times more: Nature surely can’t be bothered by how we measure stuff, right? In the positive or the negative direction—that’s just our choice, right? Our convention. So… Well… It’s just like that −eiθ function we got when looking at the same experimental set-up from the other side: our eiθ and −eiθ functions did not describe a different reality. We just changed our perspective. The reference frame. As such, the reference frame isn’t real. The experimental set-up is. And—I know I will anger mainstream physicists with this—the representation is. Yes. Let me say it loud and clear here:
A different representation describes a different reality.
In contrast, a different perspective—or a different reference frame—does not.
While you might have had a lot of trouble going through all of the weird stuff above, the point is: it is not all that weird. We can understand quantum mechanics. And in a fairly intuitive way, really. It’s just that… Well… I think some of the conventions in physics hamper such understanding. Well… Let me be precise: one convention in particular, really. It’s that convention for measuring angles. Indeed, Mr. Leonhard Euler, back in the 18th century, might well be “the master of us all” (as Laplace is supposed to have said) but… Well… He couldn’t foresee how his omnipresent formula—eiθ = cosθ + i·sinθ—would, one day, be used to represent something real: an electron, or any elementary particle, really. If he would have known, I am sure he would have noted what I am noting here: Nature can’t be bothered by our conventions. Hence, if eiθ represents something real, then e−iθ must also represent something real. [Coz I admire this genius so much, I can’t resist the temptation. Here’s his portrait. He looks kinda funny here, doesn’t he? :-)]
Frankly, he would probably have understood quantum-mechanical theory as easily and instinctively as Dirac, I think, and I am pretty sure he would have noted—and, if he would have known about circularly polarized waves, probably agreed to—that alternative convention for measuring angles: we could, effectively, measure angles clockwise or counterclockwise depending on the direction of our particle—as opposed to Euler’s ‘one-size-fits-all’ counterclockwise convention. But so we did not adopt that alternative convention because… Well… We want to keep honoring Euler, I guess. 🙂
So… Well… If we’re going to keep honoring Euler by sticking to that ‘one-size-fits-all’ counterclockwise convention, then I do believe that eiθ and e−iθ represent two different realities: spin up versus spin down.
Yes. In our geometric interpretation of the wavefunction, these are, effectively, two different spin directions. And… Well… These are real directions: we see something different when they go through a Stern-Gerlach apparatus. So it’s not just some convention to count things like 0, 1, 2, etcetera versus 0, −1, −2 etcetera. It’s the same story again: different but related mathematical notions are (often) related to different but related physical possibilities. So… Well… I think that’s what we’ve got here. Think of it. Mainstream quantum math treats all wavefunctions as right-handed but… Well… A particle with up spin is a different particle than one with down spin, right? And, again, Nature surely cannot be bothered about our convention of measuring phase angles clockwise or counterclockwise, right? So… Well… Kinda obvious, right? 🙂
Let me spell out my conclusions here:
1. The angular momentum can be positive or, alternatively, negative: J = +ħ/2 or −ħ/2. [Let me note that this is not obvious. Or less obvious than it seems, at first. In classical theory, you would expect an electron, or an atomic magnet, to line up with the field. Well… The Stern-Gerlach experiment shows they don’t: they keep their original orientation. Well… If the field is weak enough.]
2. Therefore, we would probably like to think that an actual particle—think of an electron, or whatever other particle you’d think of—comes in two variants: right-handed and left-handed. They will, therefore, either consist of (elementary) right-handed waves or, else, (elementary) left-handed waves. An elementary right-handed wave would be written as: ψ(θi) = eiθi = ai·(cosθi + i·sinθi). In contrast, an elementary left-handed wave would be written as: ψ(θi) = e−iθi = ai·(cosθi − i·sinθi). So that’s the complex conjugate.
So… Well… Yes, I think complex conjugates are not just some mathematical notion: I believe they represent something real. It’s the usual thing: Nature has shown us that (most) mathematical possibilities correspond to real physical situations so… Well… Here you go. It is really just like the left- or right-handed circular polarization of an electromagnetic wave: we can have both for the matter-wave too! [As for the differences—different polarization plane and dimensions and what have you—I’ve already summed those up, so I won’t repeat myself here.] The point is: if we have two different physical situations, we’ll want to have two different functions to describe it. Think of it like this: why would we have two—yes, I admit, two related—amplitudes to describe the up or down state of the same system, but only one wavefunction for it? You tell me.
Authors like me are looked down upon by the so-called professional class of physicists. The few who bothered to react to my attempts to make sense of Einstein’s basic intuition in regard to the nature of the wavefunction all said pretty much the same thing: “Whatever your geometric (or physical) interpretation of the wavefunction might be, it won’t be compatible with the isotropy of space. You cannot imagine an object with a 720° symmetry. That’s geometrically impossible.”
Well… Almost three years ago, I wrote the following on this blog: “As strange as it sounds, a spin-1/2 particle needs two full rotations (2×360°=720°) until it is again in the same state. Now, in regard to that particularity, you’ll often read something like: “There is nothing in our macroscopic world which has a symmetry like that.” Or, worse, “Common sense tells us that something like that cannot exist, that it simply is impossible.” [I won’t quote the site from which I took this quotes, because it is, in fact, the site of a very respectable research center!] Bollocks! The Wikipedia article on spin has this wonderful animation: look at how the spirals flip between clockwise and counterclockwise orientations, and note that it’s only after spinning a full 720 degrees that this ‘point’ returns to its original configuration after spinning a full 720 degrees.
So… Well… I am still pursuing my original dream which is… Well… Let me re-phrase what I wrote back in January 2015:
Yes, we can actually imagine spin-1/2 particles, and we actually do not need all that much imagination!
In fact, I am tempted to think that I’ve found a pretty good representation or… Well… A pretty good image, I should say, because… Well… A representation is something real, remember? 🙂
Post scriptum (10 December 2017): Our flywheel model of an electron makes sense, but also leaves many unanswered questions. The most obvious one question, perhaps, is: why the up and down state only?
I am not so worried about that question, even if I can’t answer it right away because… Well… Our apparatus—the way we measure reality—is set up to measure the angular momentum (or the magnetic moment, to be precise) in one direction only. If our electron is captured by some harmonic (or non-harmonic?) oscillation in multiple dimensions, then it should not be all that difficult to show its magnetic moment is going to align, somehow, in the same or, alternatively, the opposite direction of the magnetic field it is forced to travel through.
Of course, the analysis for the spin up situation (magnetic moment down) is quite peculiar: if our electron is a mini-magnet, why would it not line up with the magnetic field? We understand the precession of a spinning top in a gravitational field, but… Hey… It’s actually not that different. Try to imagine some spinning top on the ceiling. 🙂 I am sure we can work out the math. 🙂 The electron must be some gyroscope, really: it won’t change direction. In other words, its magnetic moment won’t line up. It will precess, and it can do so in two directions, depending on its state. 🙂 […] At least, that’s why my instinct tells me. I admit I need to work out the math to convince you. 🙂
The second question is more important. If we just rotate the reference frame over 360°, we see the same thing: some rotating object which we, vaguely, describe by some e+i·θ function—to be precise, I should say: by some Fourier sum of such functions—or, if the rotation is in the other direction, by some e−i·θ function (again, you should read: a Fourier sum of such functions). Now, the weird thing, as I tried to explain above is the following: if we rotate the object itself, over the same 360°, we get a different object: our ei·θ and e−i·θ function (again: think of a Fourier sum, so that’s a wave packet, really) becomes a −e±i·θ thing. We get a minus sign in front of it. So what happened here? What’s the difference, really?
Well… I don’t know. It’s very deep. If I do nothing, and you keep watching me while turning around me, for a full 360°, then you’ll end up where you were when you started and, importantly, you’ll see the same thing. Exactly the same thing: if I was an e+i·θ wave packet, I am still an an e+i·θ wave packet now. Or if I was an e−i·θ wave packet, then I am still an an e−i·θ wave packet now. Easy. Logical. Obvious, right?
But so now we try something different: I turn around, over a full 360° turn, and you stay where you are. When I am back where I was—looking at you again, so to speak—then… Well… I am not quite the same any more. Or… Well… Perhaps I am but you see me differently. If I was e+i·θ wave packet, then I’ve become a −e+i·θ wave packet now. Not hugely different but… Well… That minus sign matters, right? Or If I was wave packet built up from elementary a·e−i·θ waves, then I’ve become a −e−i·θ wave packet now. What happened?
It makes me think of the twin paradox in special relativity. We know it’s a paradox—so that’s an apparent contradiction only: we know which twin stayed on Earth and which one traveled because of the gravitational forces on the traveling twin. The one who stays on Earth does not experience any acceleration or deceleration. Is it the same here? I mean… The one who’s turning around must experience some force.
Can we relate this to the twin paradox? Maybe. Note that a minus sign in front of the e−±i·θ functions amounts a minus sign in front of both the sine and cosine components. So… Well… The negative of a sine and cosine is the sine and cosine but with a phase shift of 180°: −cosθ = cos(θ ± π) and −sinθ = sin(θ ± π). Now, adding or subtracting a common phase factor to/from the argument of the wavefunction amounts to changing the origin of time. So… Well… I do think the twin paradox and this rather weird business of 360° and 720° symmetries are, effectively, related. 🙂
If you haven’t read any of my previous posts on the geometry of the wavefunction (this link goes to the most recent one of them), then don’t attempt to read this one. It brings too much stuff together to be comprehensible. In fact, I am not even sure if I am going to understand what I write myself. 🙂 [OK. Poor joke. Acknowledged.]
Just to recap the essentials, I part ways with mainstream physicists in regard to the interpretation of the wavefunction. For mainstream physicists, the wavefunction is just some mathematical construct. Nothing real. Of course, I acknowledge mainstream physicists have very good reasons for that, but… Well… I believe that, if there is interference, or diffraction, then something must be interfering, or something must be diffracting. I won’t dwell on this because… Well… I have done that too many times already. My hypothesis is that the wavefunction is, in effect, a rotating field vector, so it’s just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).
Of course, it must be different, and it is. First, the (physical) dimension of the field vector of the matter-wave must be different. So what is it? Well… I am tempted to associate the real and imaginary component of the wavefunction with a force per unit mass (as opposed to the force per unit charge dimension of the electric field vector). Of course, the newton/kg dimension reduces to the dimension of acceleration (m/s2), so that’s the dimension of a gravitational field.
Second, I also am tempted to think that this gravitational disturbance causes an electron (or any matter-particle) to move about some center, and I believe it does so at the speed of light. In contrast, electromagnetic waves do not involve any mass: they’re just an oscillating field. Nothing more. Nothing less. Why would I believe there must still be some pointlike particle involved? Well… As Feynman puts it: “When you do find the electron some place, the entire charge is there.” (Feynman’s Lectures, III-21-4) So… Well… That’s why.
The third difference is one that I thought of only recently: the plane of the oscillation cannot be perpendicular to the direction of motion of our electron, because then we can’t explain the direction of its magnetic moment, which is either up or down when traveling through a Stern-Gerlach apparatus. I am more explicit on that in the mentioned post, so you may want to check there. 🙂
I wish I mastered the software to make animations such as the one above (for which I have to credit Wikipedia), but so I don’t. You’ll just have to imagine it. That’s great mental exercise, so… Well… Just try it. 🙂
Let’s now think about rotating reference frames and transformations. If the z-direction is the direction along which we measure the angular momentum (or the magnetic moment), then the up-direction will be the positive z-direction. We’ll also assume the y-direction is the direction of travel of our elementary particle—and let’s just consider an electron here so we’re more real. 🙂 So we’re in the reference frame that Feynman used to derive the transformation matrices for spin-1/2 particles (or for two-state systems in general). His ‘improved’ Stern-Gerlach apparatus—which I’ll refer to as a beam splitter—illustrates this geometry.
So I think the magnetic moment—or the angular momentum, really—comes from an oscillatory motion in the x– and y-directions. One is the real component (the cosine function) and the other is the imaginary component (the sine function), as illustrated below.
So the crucial difference with the animations above (which illustrate left- and a right-handed polarization respectively) is that we, somehow, need to imagine the circular motion is not in the xz-plane, but in the yz-plane. Now what happens if we change the reference frame?
Well… That depends on what you mean by changing the reference frame. Suppose we’re looking in the positive y-direction—so that’s the direction in which our particle is moving—, then we might imagine how it would look like when we would make a 180° turn and look at the situation from the other side, so to speak. Now, I did a post on that earlier this year, which you may want to re-read. When we’re looking at the same thing from the other side (from the back side, so to speak), we will want to use our familiar reference frame. So we will want to keep the z-axis as it is (pointing upwards), and we will also want to define the x– and y-axis using the familiar right-hand rule for defining a coordinate frame. So our new x-axis and our new y-axis will the same as the old x- and y-axes but with the sign reversed. In short, we’ll have the following mini-transformation: (1) z‘ = z, (2) x’ = −x, and (3) y’ = −y.
So… Well… If we’re effectively looking at something real that was moving along the y-axis, then it will now still be moving along the y’-axis, but in the negative direction. Hence, our elementary wavefunction eiθ = cosθ + i·sinθ will transform into −cosθ − i·sinθ = −cosθ − i·sinθ = cosθ − i·sinθ. It’s the same wavefunction. We just… Well… We just changed our reference frame. We didn’t change reality.
Now you’ll cry wolf, of course, because we just went through all that transformational stuff in our last post. To be specific, we presented the following transformation matrix for a rotation along the z-axis:
Now, if φ is equal to 180° (so that’s π in radians), then these eiφ/2 and e−iφ/2/√2 factors are equal to eiπ/2 = +i and e−iπ/2 = −i respectively. Hence, our eiθ = cosθ + i·sinθ becomes…
Hey ! Wait a minute ! We’re talking about two very different things here, right? The eiθ = cosθ + i·sinθ is an elementary wavefunction which, we presume, describes some real-life particle—we talked about an electron with its spin in the up-direction—while these transformation matrices are to be applied to amplitudes describing… Well… Either an up– or a down-state, right?
Right. But… Well… Is it so different, really? Suppose our eiθ = cosθ + i·sinθ wavefunction describes an up-electron, then we still have to apply that eiφ/2 = eiπ/2 = +i factor, right? So we get a new wavefunction that will be equal to eiφ/2·eiθ = eiπ/2·eiθ = +i·eiθ = i·cosθ + i2·sinθ = sinθ − i·cosθ, right? So how can we reconcile that with the cosθ − i·sinθ function we thought we’d find?
We can’t. So… Well… Either my theory is wrong or… Well… Feynman can’t be wrong, can he? I mean… It’s not only Feynman here. We’re talking all mainstream physicists here, right?
Right. But think of it. Our electron in that thought experiment does, effectively, make a turn of 180°, so it is going in the other direction now ! That’s more than just… Well… Going around the apparatus and looking at stuff from the other side.
Hmm… Interesting. Let’s think about the difference between the sinθ − i·cosθ and cosθ − i·sinθ functions. First, note that they will give us the same probabilities: the square of the absolute value of both complex numbers is the same. [It’s equal to 1 because we didn’t bother to put a coefficient in front.] Secondly, we should note that the sine and cosine functions are essentially the same. They just differ by a phase factor: cosθ = sin(θ + π/2) and −sinθ = cos(θ + π/2). Let’s see what we can do with that. We can write the following, for example:
sinθ − i·cosθ = −cos(θ + π/2) − i·sin(θ + π/2) = −[cos(θ + π/2) + i·sin(θ + π/2)] = −ei·(θ + π/2)
Well… I guess that’s something at least ! The ei·θ and −ei·(θ + π/2) functions differ by a phase shift and a minus sign so… Well… That’s what it takes to reverse the direction of an electron. 🙂 Let us mull over that in the coming days. As I mentioned, these more philosophical topics are not easily exhausted. 🙂
Some say it is not possible to fully understand quantum-mechanical spin. Now, I do agree it is difficult, but I do not believe it is impossible. That’s why I wrote so many posts on it. Most of these focused on elaborating how the classical view of how a rotating charge precesses in a magnetic field might translate into the weird world of quantum mechanics. Others were more focused on the corollary of the quantization of the angular momentum, which is that, in the quantum-mechanical world, the angular momentum is never quite all in one direction only—so that explains some of the seemingly inexplicable randomness in particle behavior.
Frankly, I think those explanations help us quite a bit already but… Well… We need to go the extra mile, right? In fact, that’s drives my search for a geometric (or physical) interpretation of the wavefunction: the extra mile. 🙂
Now, in one of these many posts on spin and angular momentum, I advise my readers – you, that is – to try to work yourself through Feynman’s 6th Lecture on quantum mechanics, which is highly abstract and, therefore, usually skipped. [Feynman himself told his students to skip it, so I am sure that’s what they did.] However, if we believe the physical (or geometric) interpretation of the wavefunction that we presented in previous posts is, somehow, true, then we need to relate it to the abstract math of these so-called transformations between representations. That’s what we’re going to try to do here. It’s going to be just a start, and I will probably end up doing several posts on this but… Well… We do have to start somewhere, right? So let’s see where we get today. 🙂
The thought experiment that Feynman uses throughout his Lecture makes use of what Feynman’s refers to as modified or improved Stern-Gerlach apparatuses. They allow us to prepare a pure state or, alternatively, as Feynman puts it, to analyze a state. In theory, that is. The illustration below present a side and top view of such apparatus. We may already note that the apparatus itself—or, to be precise, our perspective of it—gives us two directions: (1) the up direction, so that’s the positive direction of the z-axis, and (2) the direction of travel of our particle, which coincides with the positive direction of the y-axis. [This is obvious and, at the same time, not so obvious, but I’ll talk about that in my next post. In this one, we basically need to work ourselves through the math, so we don’t want to think too much about philosophical stuff.]
The kind of questions we want to answer in this post are variants of the following basic one: if a spin-1/2 particle (let’s think of an electron here, even if the Stern-Gerlach experiment is usually done with an atomic beam) was prepared in a given condition by one apparatus S, say the +S state, what is the probability (or the amplitude) that it will get through a second apparatus T if that was set to filter out the +T state?
The result will, of course, depend on the angles between the two apparatuses S and T, as illustrated below. [Just to respect copyright, I should explicitly note here that all illustrations are taken from the mentioned Lecture, and that the line of reasoning sticks close to Feynman’s treatment of the matter too.]
We should make a few remarks here. First, this thought experiment assumes our particle doesn’t get lost. That’s obvious but… Well… If you haven’t thought about this possibility, I suspect you will at some point in time. So we do assume that, somehow, this particle makes a turn. It’s an important point because… Well… Feynman’s argument—who, remember, represents mainstream physics—somehow assumes that doesn’t really matter. It’s the same particle, right? It just took a turn, so it’s going in some other direction. That’s all, right? Hmm… That’s where I part ways with mainstream physics: the transformation matrices for the amplitudes that we’ll find here describe something real, I think. It’s not just perspective: something happened to the electron. That something does not only change the amplitudes but… Well… It describes a different electron. It describes an electron that goes in a different direction now. But… Well… As said, these are reflections I will further develop in my next post. 🙂 Let’s focus on the math here. The philosophy will follow later. 🙂 Next remark.
Second, we assume the (a) and (b) illustrations above represent the same physical reality because the relative orientation between the two apparatuses, as measured by the angle α, is the same. Now that is obvious, you’ll say, but, as Feynman notes, we can only make that assumption because experiments effectively confirm that spacetime is, effectively, isotropic. In other words, there is no aether allowing us to establish some sense of absolute direction. Directions are relative—relative to the observer, that is… But… Well… Again, in my next post, I’ll argue that it’s not because directions are relative that they are, somehow, not real. Indeed, in my humble opinion, it does matter whether an electron goes here or, alternatively, there. These two different directions are not just two different coordinate frames. But… Well… Again. The philosophy will follow later. We need to stay focused on the math here.
Third and final remark. This one is actually very tricky. In his argument, Feynman also assumes the two set-ups below are, somehow, equivalent.
You’ll say: Huh? If not, say it! Huh? 🙂 Yes. Good. Huh? Feynman writes equivalent—not the same because… Well… They’re not the same, obviously:
- In the first set-up (a), T is wide open, so the apparatus is not supposed to do anything with the beam: it just splits and re-combines it.
- In set-up (b) the T apparatus is, quite simply, not there, so… Well… Again. Nothing is supposed to happen with our particles as they come out of S and travel to U.
The fundamental idea here is that our spin-1/2 particle (again, think of an electron here) enters apparatus U in the same state as it left apparatus S. In both set-ups, that is! Now that is a very tricky assumption, because… Well… While the net turn of our electron is the same, it is quite obvious it has to take two turns to get to U in (a), while it only takes one turn in (b). And so… Well… You can probably think of other differences too. So… Yes. And no. Same-same but different, right? 🙂
Right. That is why Feynman goes out of his way to explain the nitty-gritty behind: he actually devotes a full page in small print on this, which I’ll try to summarize in just a few paragraphs here. [And, yes, you should check my summary against Feynman’s actual writing on this.] It’s like this. While traveling through apparatus T in set-up (a), time goes by and, therefore, the amplitude would be different by some phase factor δ. [Feynman doesn’t say anything about this, but… Well… In the particle’s own frame of reference, this phase factor depend on the energy, the momentum and the time and distance traveled. Think of the argument of the elementary wavefunction here: θ = (E∙t – p∙x)/ħ).] Now, if we believe that the amplitude is just some mathematical construct—so that’s what mainstream physicists (not me!) believe—then we could effectively say that the physics of (a) and (b) are the same, as Feynman does. In fact, let me quote him here:
“The physics of set-up (a) and (b) should be the same but the amplitudes could be different by some phase factor without changing the result of any calculation about the real world.”
Hmm… It’s one of those mysterious short passages where we’d all like geniuses like Feynman (or Einstein, or whomever) to be more explicit on their world view: if the amplitudes are different, can the physics really be the same? I mean… Exactly the same? It all boils down to that unfathomable belief that, somehow, the particle is real but the wavefunction that describes it, is not. Of course, I admit that it’s true that choosing another zero point for the time variable would also change all amplitudes by a common phase factor and… Well… That’s something that I consider to be not real. But… Well… The time and distance traveled in the T apparatus is the time and distance traveled in the T apparatus, right?
Bon… I have to stay away from these questions as for now—we need to move on with the math here—but I will come back to it later. But… Well… Talking math, I should note a very interesting mathematical point here. We have these transformation matrices for amplitudes, right? Well… Not yet. In fact, the coefficient of these matrices are exactly what we’re going to try to derive in this post, but… Well… Let’s assume we know them already. 🙂 So we have a 2-by-2 matrix to go from S to T, from T to U, and then one to go from S to U without going through T, which we can write as RST, RTU, and RSU respectively. Adding the subscripts for the base states in each representation, the equivalence between the (a) and (b) situations can then be captured by the following formula:
So we have that phase factor here: the left- and right-hand side of this equation is, effectively, same-same but different, as they would say in Asia. 🙂 Now, Feynman develops a beautiful mathematical argument to show that the eiδ factor effectively disappears if we convert our rotation matrices to some rather special form that is defined as follows:
I won’t copy his argument here, but I’d recommend you go over it because it is wonderfully easy to follow and very intriguing at the same time. [Yes. Simple things can be very intriguing.] Indeed, the calculation below shows that the determinant of these special rotation matrices will be equal to 1.
So… Well… So what? You’re right. I am being sidetracked here. The point is that, if we put all of our rotation matrices in this special form, the eiδ factor vanishes and the formula above reduces to:
So… Yes. End of excursion. Let us remind ourselves of what it is that we are trying to do here. As mentioned above, the kind of questions we want to answer will be variants of the following basic one: if a spin-1/2 particle was prepared in a given condition by one apparatus (S), say the +S state, what is the probability (or the amplitude) that it will get through a second apparatus (T) if that was set to filter out the +T state?
We said the result would depend on the angles between the two apparatuses S and T. I wrote: angles—plural. Why? Because a rotation will generally be described by the three so-called Euler angles: α, β and γ. Now, it is easy to make a mistake here, because there is a sequence to these so-called elemental rotations—and right-hand rules, of course—but I will let you figure that out. 🙂
The basic idea is the following: if we can work out the transformation matrices for each of these elemental rotations, then we can combine them and find the transformation matrix for any rotation. So… Well… That fills most of Feynman’s Lecture on this, so we don’t want to copy all that. We’ll limit ourselves to the logic for a rotation about the z-axis, and then… Well… You’ll see. 🙂
So… The z-axis… We take that to be the direction along which we are measuring the angular momentum of our electron, so that’s the direction of the (magnetic) field gradient, so that’s the up-axis of the apparatus. In the illustration below, that direction points out of the page, so to speak, because it is perpendicular to the direction of the x– and the y-axis that are shown. Note that the y-axis is the initial direction of our beam.
Now, because the (physical) orientation of the fields and the field gradients of S and T is the same, Feynman says that—despite the angle—the probability for a particle to be up or down with regard to S and T respectively should be the same. Well… Let’s be fair. He does not only say that: experiment shows it to be true. [Again, I am tempted to interject here that it is not because the probabilities for (a) and (b) are the same, that the reality of (a) and (b) is the same, but… Well… You get me. That’s for the next post. Let’s get back to the lesson here.] The probability is, of course, the square of the absolute value of the amplitude, which we will denote as C+, C−, C’+, and C’− respectively. Hence, we can write the following:
Now, the absolute values (or the magnitudes) are the same, but the amplitudes may differ. In fact, they must be different by some phase factor because, otherwise, we would not be able to distinguish the two situations, which are obviously different. As Feynman, finally, admits himself—jokingly or seriously: “There must be some way for a particle to know that it has turned the corner at P1.” [P1 is the midway point between S and T in the illustration, of course—not some probability.]
So… Well… We write:
Now, it shouldn’t you too long to figure out that λ’ is equal to λ’ = λ/2 + μ/2 = −μ’. So… Well… Then we can just adopt the convention that λ = −μ. So our C’+ = eiλ ·C+ and C’− = eiμ ·C− equations can now be written as:
C’+ = eiλ ·C+ and C’− = e−iλ·C−
The absolute values are the same, but the phases are different. Right. OK. Good move. What’s next?
Well… The next assumption is that the phase shift λ is proportional to the angle (α) between the two apparatuses. Hence, λ is equal to λ = m·α, and we can re-write the above as:
C’+ = eimα·C+ and C’− = e−imα·C−
Now, this assumption may or may not seem reasonable. Feynman justifies it with a continuity argument, arguing any rotation can be built up as a sequence of infinitesimal rotations and… Well… Let’s not get into the nitty-gritty here. [If you want it, check Feynman’s Lecture itself.] Back to the main line of reasoning. So we’ll assume we can write λ as λ = m·α. The next question then is: what is the value for m? Now, we obviously do get exactly the same physics if we rotate T by 360°, or 2π radians. So we might conclude that the amplitudes should be the same and, therefore, that eimα = eim·2π has to be equal to one, so C’+ = C+ and C’− = C− . That’s the case if m is equal to 1. But… Well… No. It’s the same thing again: the probabilities (or the magnitudes) have to be the same, but the amplitudes may be different because of some phase factor. In fact, they should be different. If m = 1/2, then we also get the same physics, even if the amplitudes are not the same. They will be each other’s opposite:
Huh? Yes. Think of it. The coefficient of proportionality (m) cannot be equal to 1. If it would be equal to 1, and we’d rotate T by 180° only, then we’d also get those C’+ = −C+ and C’− = −C− equations, and so these coefficients would, therefore, also describe the same physical situation. Now, you will understand, intuitively, that a rotation of the T apparatus by 180° will not give us the same physical situation… So… Well… In case you’d want a more formal argument proving a rotation by 180° does not give us the same physical situation, Feynman has one for you. 🙂
I know that, by now, you’re totally tired and bored, and so you only want the grand conclusion at this point. Well… All of what I wrote above should, hopefully, help you to understand that conclusion, which – I quote Feynman here – is the following:
If we know the amplitudes C+ and C− of spin one-half particles with respect to a reference frame S, and we then use new base states, defined with respect to a reference frame T which is obtained from S by a rotation α around the z-axis, the new amplitudes are given in terms of the old by the following formulas:
[Feynman denotes our angle α by phi (φ) because… He uses the Euler angles a bit differently. But don’t worry: it’s the same angle.]
What about the amplitude to go from the C− to the C’+ state, and from the C+ to the C’− state? Well… That amplitude is zero. So the transformation matrix is this one:
Let’s take a moment and think about this. Feynman notes the following, among other things: “It is very curious to say that if you turn the apparatus 360° you get new amplitudes. [They aren’t really new, though, because the common change of sign doesn’t give any different physics.] But if something has been rotated by a sequence of small rotations whose net result is to return it to the original orientation, then it is possible to define the idea that it has been rotated 360°—as distinct from zero net rotation—if you have kept track of the whole history.”
This is very deep. It connects space and time into one single geometric space, so to speak. But… Well… I’ll try to explain this rather sweeping statement later. Feynman also notes that a net rotation of 720° does give us the same amplitudes and, therefore, cannot be distinguished from the original orientation. Feynman finds that intriguing but… Well… I am not sure if it’s very significant. I do note some symmetries in quantum physics involve 720° rotations but… Well… I’ll let you think about this. 🙂
Note that the determinant of our matrix is equal to a·d − b·c = eiφ/2·e−iφ/2 = 1. So… Well… Our rotation matrix is, effectively, in that special form! How comes? Well… When equating λ = −μ, we are effectively putting the transformation into that special form. Let us also, just for fun, quickly check the normalization condition. It requires that the probabilities, in any given representation, add to up to one. So… Well… Do they? When they come out of S, our electrons are equally likely to be in the up or down state. So the amplitudes are 1/√2. [To be precise, they are ±1/√2 but… Well… It’s the phase factor story once again.] That’s normalized: |1/√2|2 + |1/√2|2 = 1. The amplitudes to come out of the T apparatus in the up or down state are eiφ/2/√2 and eiφ/2/√2 respectively, so the probabilities add up to |eiφ/2/√2|2 + |e−iφ/2/√2|2 = … Well… It’s 1. Check it. 🙂
Let me add an extra remark here. The normalization condition will result in matrices whose determinant will be equal to some pure imaginary exponential, like eiα. So is that what we have here? Yes. We can re-write 1 as 1 = ei·0 = e0, so α = 0. 🙂 Capito? Probably not, but… Well… Don’t worry about it. Just think about the grand results. As Feynman puts it, this Lecture is really “a sort of cultural excursion.” 🙂
Let’s do a practical calculation here. Let’s suppose the angle is, effectively, 180°. So the eiφ/2 and e−iφ/2/√2 factors are equal to eiπ/2 = +i and e−iπ/2 = −i, so… Well… What does that mean—in terms of the geometry of the wavefunction? Hmm… We need to do some more thinking about the implications of all this transformation business for our geometric interpretation of he wavefunction, but so we’ll do that in our next post. Let us first work our way out of this rather hellish transformation logic. 🙂 [See? I do admit it is all quite difficult and abstruse, but… Well… We can do this, right?]
So what’s next? Well… Feynman develops a similar argument (I should say same-same but different once more) to derive the coefficients for a rotation of ±90° around the y-axis. Why 90° only? Well… Let me quote Feynman here, as I can’t sum it up more succinctly than he does: “With just two transformations—about the y-axis, and an arbitrary angle about the z-axis [which we described above]—we can generate any rotation at all.”
So how does that work? Check the illustration below. In Feynman’s words again: “Suppose that we want the angle α around x. We know how to deal with the angle α , but now we want it around x. How do we get it? First, we turn the axis z down onto x—which is a rotation of Then we turn through the angle α around x = . Then we rotate about y”. The net result of the three rotations is the same as turning around x by the angle α. It is a property of space.”around
Besides helping us greatly to derive the transformation matrix for any rotation, the mentioned property of space is rather mysterious and deep. It sort of reduces the degrees of freedom, so to speak. Feynman writes the following about this:
“These facts of the combinations of rotations, and what they produce, are hard to grasp intuitively. It is rather strange, because we live in three dimensions, but it is hard for us to appreciate what happens if we turn this way and then that way. Perhaps, if we were fish or birds and had a real appreciation of what happens when we turn somersaults in space, we could more easily appreciate such things.”
In any case, I should limit the number of philosophical interjections. If you go through the motions, then you’ll find the following elemental rotation matrices:
What about the determinants of the Rx(φ) and Ry(φ) matrices? They’re also equal to one, so… Yes. A pure imaginary exponential, right? 1 = ei·0 = e0. 🙂
What’s next? Well… We’re done. We can now combine the elemental transformations above in a more general format, using the standardized Euler angles. Again, just go through the motions. The Grand Result is:
Does it give us normalized amplitudes? It should, but it looks like our determinant is going to be a much more complicated complex exponential. 🙂 Hmm… Let’s take some time to mull over this. As promised, I’ll be back with more reflections in my next post.
Our previous posts showed how a simple geometric interpretation of the elementary wavefunction yielded the (Compton scattering) radius of an elementary particle—for an electron, at least: for the proton, we only got the order of magnitude right—but then a proton is not an elementary particle. We got lots of other interesting equations as well… But… Well… When everything is said and done, it’s that equivalence between the E = m·a2·ω2 and E = m·c2 relations that we… Well… We need to be more specific about it.
Indeed, I’ve been ambiguous here and there—oscillating between various interpretations, so to speak. 🙂 In my own mind, I refer to my unanswered questions, or my ambiguous answers to them, as the form factor problem. So… Well… That explains the title of my post. But so… Well… I do want to be somewhat more conclusive in this post. So let’s go and see where we end up. 🙂
To help focus our mind, let us recall the metaphor of the V-2 perpetuum mobile, as illustrated below. With permanently closed valves, the air inside the cylinder compresses and decompresses as the pistons move up and down. It provides, therefore, a restoring force. As such, it will store potential energy, just like a spring, and the motion of the pistons will also reflect that of a mass on a spring: it is described by a sinusoidal function, with the zero point at the center of each cylinder. We can, therefore, think of the moving pistons as harmonic oscillators, just like mechanical springs. Of course, instead of two cylinders with pistons, one may also think of connecting two springs with a crankshaft, but then that’s not fancy enough for me. 🙂
At first sight, the analogy between our flywheel model of an electron and the V-twin engine seems to be complete: the 90 degree angle of our V-2 engine makes it possible to perfectly balance the pistons and we may, therefore, think of the flywheel as a (symmetric) rotating mass, whose angular momentum is given by the product of the angular frequency and the moment of inertia: L = ω·I. Of course, the moment of inertia (aka the angular mass) will depend on the form (or shape) of our flywheel:
- I = m·a2 for a rotating point mass m or, what amounts to the same, for a circular hoop of mass m and radius r = a.
- For a rotating (uniformly solid) disk, we must add a 1/2 factor: I = m·a2/2.
How can we relate those formulas to the E = m·a2·ω2 formula? The kinetic energy that is being stored in a flywheel is equal Ekinetic = I·ω2/2, so that is only half of the E = m·a2·ω2 product if we substitute I for I = m·a2. [For a disk, we get a factor 1/4, so that’s even worse!] However, our flywheel model of an electron incorporates potential energy too. In fact, the E = m·a2·ω2 formula just adds the (kinetic and potential) energy of two oscillators: we do not really consider the energy in the flywheel itself because… Well… The essence of our flywheel model of an electron is not the flywheel: the flywheel just transfers energy from one oscillator to the other, but so… Well… We don’t include it in our energy calculations. The essence of our model is that two-dimensional oscillation which drives the electron, and which is reflected in Einstein’s E = m·c2 formula. That two-dimensional oscillation—the a2·ω2 = c2 equation, really—tells us that the resonant (or natural) frequency of the fabric of spacetime is given by the speed of light—but measured in units of a. [If you don’t quite get this, re-write the a2·ω2 = c2 equation as ω = c/a: the radius of our electron appears as a natural distance unit here.]
Now, we were extremely happy with this interpretation not only because of the key results mentioned above, but also because it has lots of other nice consequences. Think of our probabilities as being proportional to energy densities, for example—and all of the other stuff I describe in my published paper on this. But there is even more on the horizon: a follower of this blog (a reader with an actual PhD in physics, for a change) sent me an article analyzing elementary particles as tiny black holes because… Well… If our electron is effectively spinning around, then its tangential velocity is equal to v = a·ω = c. Now, recent research suggest black holes are also spinning at (nearly) the speed of light. Interesting, right? However, in order to understand what she’s trying to tell me, I’ll first need to get a better grasp of general relativity, so I can relate what I’ve been writing here and in previous posts to the Schwarzschild radius and other stuff.
Let me get back to the lesson here. In the reference frame of our particle, the wavefunction really looks like the animation below: it has two components, and the amplitude of the two-dimensional oscillation is equal to a, which we calculated as a = ħ·/(m·c) = 3.8616×10−13 m, so that’s the (reduced) Compton scattering radius of an electron.
In my original article on this, I used a more complicated argument involving the angular momentum formula, but I now prefer a more straightforward calculation:
c = a·ω = a·E/ħ = a·m·c2/ħ ⇔ a = ħ/(m·c)
The question is: what is that rotating arrow? I’ve been vague and not so vague on this. The thing is: I can’t prove anything in this regard. But my hypothesis is that it is, in effect, a rotating field vector, so it’s just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).
There are a number of crucial differences though:
- The (physical) dimension of the field vector of the matter-wave is different: I associate the real and imaginary component of the wavefunction with a force per unit mass (as opposed to the force per unit charge dimension of the electric field vector). Of course, the newton/kg dimension reduces to the dimension of acceleration (m/s2), so that’s the dimension of a gravitational field.
- I do believe this gravitational disturbance, so to speak, does cause an electron to move about some center, and I believe it does so at the speed of light. In contrast, electromagnetic waves do not involve any mass: they’re just an oscillating field. Nothing more. Nothing less. In contrast, as Feynman puts it: “When you do find the electron some place, the entire charge is there.” (Feynman’s Lectures, III-21-4)
- The third difference is one that I thought of only recently: the plane of the oscillation cannot be perpendicular to the direction of motion of our electron, because then we can’t explain the direction of its magnetic moment, which is either up or down when traveling through a Stern-Gerlach apparatus.
I mentioned that in my previous post but, for your convenience, I’ll repeat what I wrote there. The basic idea here is illustrated below (credit for this illustration goes to another blogger on physics). As for the Stern-Gerlach experiment itself, let me refer you to a YouTube video from the Quantum Made Simple site.
The point is: the direction of the angular momentum (and the magnetic moment) of an electron—or, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron is traveling—cannot be parallel to the direction of motion. On the contrary, it is perpendicular to the direction of motion. In other words, if we imagine our electron as spinning around some center, then the disk it circumscribes will comprise the direction of motion.
However, we need to add an interesting detail here. As you know, we don’t really have a precise direction of angular momentum in quantum physics. [If you don’t know this… Well… Just look at one of my many posts on spin and angular momentum in quantum physics.] Now, we’ve explored a number of hypotheses but, when everything is said and done, a rather classical explanation turns out to be the best: an object with an angular momentum J and a magnetic moment μ (I used bold-face because these are vector quantities) that is parallel to some magnetic field B, will not line up, as you’d expect a tiny magnet to do in a magnetic field—or not completely, at least: it will precess. I explained that in another post on quantum-mechanical spin, which I advise you to re-read if you want to appreciate the point that I am trying to make here. That post integrates some interesting formulas, and so one of the things on my ‘to do’ list is to prove that these formulas are, effectively, compatible with the electron model we’ve presented in this and previous posts.
Indeed, when one advances a hypothesis like this, it’s not enough to just sort of show that the general geometry of the situation makes sense: we also need to show the numbers come out alright. So… Well… Whatever we think our electron—or its wavefunction—might be, it needs to be compatible with stuff like the observed precession frequency of an electron in a magnetic field.
Our model also needs to be compatible with the transformation formulas for amplitudes. I’ve been talking about this for quite a while now, and so it’s about time I get going on that.
Last but not least, those articles that relate matter-particles to (quantum) gravity—such as the one I mentioned above—are intriguing too and, hence, whatever hypotheses I advance here, I’d better check them against those more advanced theories too, right? 🙂 Unfortunately, that’s going to take me a few more years of studying… But… Well… I still have many years ahead—I hope. 🙂
Post scriptum: It’s funny how one’s brain keeps working when sleeping. When I woke up this morning, I thought: “But it is that flywheel that matters, right? That’s the energy storage mechanism and also explains how photons possibly interact with electrons. The oscillators drive the flywheel but, without the flywheel, nothing is happening. It is really the transfer of energy—through the flywheel—which explains why our flywheel goes round and round.”
It may or may not be useful to remind ourselves of the math in this regard. The motion of our first oscillator is given by the cos(ω·t) = cosθ function (θ = ω·t), and its kinetic energy will be equal to sin2θ. Hence, the (instantaneous) change in kinetic energy at any point in time (as a function of the angle θ) is equal to: d(sin2θ)/dθ = 2∙sinθ∙d(sinθ)/dθ = 2∙sinθ∙cosθ. Now, the motion of the second oscillator (just look at that second piston going up and down in the V-2 engine) is given by the sinθ function, which 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 again) is equal to 2∙sin(θ − π /2)∙cos(θ − π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ. So here we have our energy transfer: the flywheel organizes the borrowing and returning of energy, so to speak. That’s the crux of the matter.
So… Well… What if the relevant energy formula is E = m·a2·ω2/2 instead of E = m·a2·ω2? What are the implications? Well… We get a √2 factor in our formula for the radius a, as shown below.
Now that is not so nice. For the tangential velocity, we get v = a·ω = √2·c. This is also not so nice. How can we save our model? I am not sure, but here I am thinking of the mentioned precession—the wobbling of our flywheel in a magnetic field. Remember we may think of Jz—the angular momentum or, to be precise, its component in the z-direction (the direction in which we measure it—as the projection of the real angular momentum J. Let me insert Feynman’s illustration here again (Feynman’s Lectures, II-34-3), so you get what I am talking about.
Now, all depends on the angle (θ) between Jz and J, of course. We did a rather obscure post on these angles, but the formulas there come in handy now. Just click the link and review it if and when you’d want to understand the following formulas for the magnitude of the presumed actual momentum:In this particular case (spin-1/2 particles), j is equal to 1/2 (in units of ħ, of course). Hence, J is equal to √0.75 ≈ 0.866. Elementary geometry then tells us cos(θ) = (1/2)/√(3/4) = = 1/√3. Hence, θ ≈ 54.73561°. That’s a big angle—larger than the 45° angle we had secretly expected because… Well… The 45° angle has that √2 factor in it: cos(45°) = sin(45°) = 1/√2.
Hmm… As you can see, there is no easy fix here. Those damn 1/2 factors! They pop up everywhere, don’t they? 🙂 We’ll solve the puzzle. One day… But not today, I am afraid. I’ll call it the form factor problem… Because… Well… It sounds better than the 1/2 or √2 problem, right? 🙂
Note: If you’re into quantum math, you’ll note a = ħ/(m·c) is the reduced Compton scattering radius. The standard Compton scattering radius is equal to a·2π = (2π·ħ)/(m·c) = h/(m·c) = h/(m·c). It doesn’t solve the √2 problem. Sorry. The form factor problem. 🙂
To be honest, I finished my published paper on all of this with a suggestion that, perhaps, we should think of two circular oscillations, as opposed to linear ones. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation – around any axis – will be some combination of a rotation around the two other axes. Hence, we may want to think of our two-dimensional oscillation as an oscillation of a polar and azimuthal angle. It’s just a thought but… Well… I am sure it’s going to keep me busy for a while. 🙂They are oscillations, still, so I am not thinking of two flywheels that keep going around in the same direction. No. More like a wobbling object on a spring. Something like the movement of a bobblehead on a spring perhaps. 🙂
In our previous posts, we interpreted the elementary wavefunction ψ = a·e−i∙θ = a·cosθ − i·a·sinθ as a two-dimensional oscillation in spacetime. In addition to assuming the two directions of the oscillation were perpendicular to each other, we also assumed they were perpendicular to the direction of motion. While the first assumption is essential in our interpretation, the second assumption is solely based on an analogy with a circularly polarized electromagnetic wave. We also assumed the matter wave could be right-handed as well as left-handed (as illustrated below), and that these two physical possibilities corresponded to the angular momentum being equal to plus or minus ħ/2 respectively.
This allowed us to derive the Compton scattering radius of an elementary particle. Indeed, we interpreted the rotating vector as a resultant vector, which we get by adding the sine and cosine motions, which represent the real and imaginary components of our wavefunction. The energy of this two-dimensional oscillation is twice the energy of a one-dimensional oscillator and, therefore, equal to E = m·a2·ω2. Now, the angular frequency is given by ω = E/ħ and E must, obviously, also be equal to E = m·c2. Substitition and re-arranging the terms gives us the Compton scattering radius:
The value given above is the (reduced) Compton scattering radius for an electron. For a proton, we get a value of about 2.1×10−16 m, which is about 1/4 of the radius of a proton as measured in scattering experiments. Hence, for a proton, our formula does not give us the exact (i.e. experimentally verified) value but it does give us the correct order of magnitude—which is fine because we know a proton is not an elementary particle and, hence, the motion of its constituent parts (quarks) is… Well… It complicates the picture hugely.
If we’d presume the electron charge would, effectively, be rotating about the center, then its tangential velocity is given by v = a·ω = [ħ·/(m·c)]·(E/ħ) = c, which is yet another wonderful implication of our hypothesis. Finally, the c = a·ω formula allowed us to interpret the speed of light as the resonant frequency of the fabric of space itself, as illustrated when re-writing this equality as follows:
This gave us a natural and forceful interpretation of Einstein’s mass-energy equivalence formula: the energy in the E = m·c2· equation is, effectively, a two-dimensional oscillation of mass.
However, while toying with this and other results (for example, we may derive a Poynting vector and show probabilities are, effectively, proportional to energy densities), I realize the plane of our two-dimensional oscillation cannot be perpendicular to the direction of motion of our particle. In fact, the direction of motion must lie in the same plane. This is a direct consequence of the direction of the angular momentum as measured by, for example, the Stern-Gerlach experiment. The basic idea here is illustrated below (credit for this illustration goes to another blogger on physics). As for the Stern-Gerlach experiment itself, let me refer you to a YouTube video from the Quantum Made Simple site.
The point is: the direction of the angular momentum (and the magnetic moment) of an electron—or, to be precise, its component as measured in the direction of the (inhomogenous) magnetic field through which our electron is traveling—cannot be parallel to the direction of motion. On the contrary, it is perpendicular to the direction of motion. In other words, if we imagine our electron as some rotating disk or a flywheel, then it will actually comprise the direction of motion.
What are the implications? I am not sure. I will definitely need to review whatever I wrote about the de Broglie wavelength in previous posts. We will also need to look at those transformations of amplitudes once again. Finally, we will also need to relate this to the quantum-mechanical formulas for the angular momentum and the magnetic moment.
Post scriptum: As in previous posts, I need to mention one particularity of our model. When playing with those formulas, we contemplated two different formulas for the angular mass: one is the formula for a rotating mass (I = m·r2/2), and the other is the one for a rotating mass (I = m·r2). The only difference between the two is a 1/2 factor, but it turns out we need it to get a sensical result. For a rotating mass, the angular momentum is equal to the radius times the momentum, so that’s the radius times the mass times the velocity: L = m·v·r. [See also Feynman, Vol. II-34-2, in this regard)] Hence, for our model, we get L = m·v·r = m·c·a = m·c·ħ/(m·c) = ħ. Now, we know it’s equal to ±ħ/2, so we need that 1/2 factor in the formula.
Can we relate this 1/2 factor to the g-factor for the electron’s magnetic moment, which is (approximately) equal to 2? Maybe. We’d need to look at the formula for a rotating charged disk. That’s for a later post, however. It’s been enough for today, right? 🙂
I would just like to signal another interesting consequence of our model. If we would interpret the radius of our disk (a)—so that’s the Compton radius of our electron, as opposed to the Thomson radius—as the uncertainty in the position of our electron, then our L = m·v·r = m·c·a = p·a = ħ/2 formula as a very particular expression of the Uncertainty Principle: p·Δx= ħ/2. Isn’t that just plain nice? 🙂
The geometry of the elementary wavefunction
In the reference frame of the particle itself, the geometry of the wavefunction simplifies to what is illustrated below: an oscillation in two dimensions which, viewed together, form a plane that would be perpendicular to the direction of motion—but then our particle doesn’t move in its own reference frame, obviously. Hence, we could be looking at our particle from any direction and we should, presumably, see a similar two-dimensional oscillation. That is interesting because… Well… If we rotate this circle around its center (in whatever direction we’d choose), we get a sphere, right? It’s only when it starts moving, that it loses its symmetry. Now, that is very intriguing, but let’s think about that later.
Let’s assume we’re looking at it from some specific direction. Then we presumably have some charge (the green dot) moving about some center, and its movement can be analyzed as the sum of two oscillations (the sine and cosine) which represent the real and imaginary component of the wavefunction respectively—as we observe it, so to speak. [Of course, you’ve been told you can’t observe wavefunctions so… Well… You should probably stop reading this. :-)] We write:
ψ = = a·e−i∙θ = a·e−i∙E·t/ħ = a·cos(−E∙t/ħ) + i·a·sin(−E∙t/ħ) = a·cos(E∙t/ħ) − i·a·sin(E∙t/ħ)
So that’s the wavefunction in the reference frame of the particle itself. When we think of it as moving in some direction (so relativity kicks in), we need to add the p·x term to the argument (θ = E·t − p∙x). It is easy to show this term doesn’t change the argument (θ), because we also get a different value for the energy in the new reference frame: Ev = γ·E0 and so… Well… I’ll refer you to my post on this, in which I show the argument of the wavefunction is invariant under a Lorentz transformation: the way Ev and pv and, importantly, the coordinates x and t relativistically transform ensures the invariance.
In fact, I’ve always wanted to read de Broglie‘s original thesis because I strongly suspect he saw that immediately. If you click this link, you’ll find an author who suggests the same. Having said that, I should immediately add this does not imply there is no need for a relativistic wave equation: the wavefunction is a solution for the wave equation and, yes, I am the first to note the Schrödinger equation has some obvious issues, which I briefly touch upon in one of my other posts—and which is why Schrödinger himself and other contemporaries came up with a relativistic wave equation (Oskar Klein and Walter Gordon got the credit but others (including Louis de Broglie) also suggested a relativistic wave equation when Schrödinger published his). In my humble opinion, the key issue is not that Schrödinger’s equation is non-relativistic. It’s that 1/2 factor again but… Well… I won’t dwell on that here. We need to move on. So let’s leave the wave equation for what it is and go back to our wavefunction.
You’ll note the argument (or phase) of our wavefunction moves clockwise—or counterclockwise, depending on whether you’re standing in front of behind the clock. Of course, Nature doesn’t care about where we stand or—to put it differently—whether we measure time clockwise, counterclockwise, in the positive, the negative or whatever direction. Hence, I’ve argued we can have both left- as well as right-handed wavefunctions, as illustrated below (for p ≠ 0). Our hypothesis is that these two physical possibilities correspond to the angular momentum of our electron being either positive or negative: Jz = +ħ/2 or, else, Jz = −ħ/2. [If you’ve read a thing or two about neutrinos, then… Well… They’re kinda special in this regard: they have no charge and neutrinos and antineutrinos are actually defined by their helicity. But… Well… Let’s stick to trying to describing electrons for a while.]
The line of reasoning that we followed allowed us to calculate the amplitude a. We got a result that tentatively confirms we’re on the right track with our interpretation: we found that a = ħ/me·c, so that’s the Compton scattering radius of our electron. All good ! But we were still a bit stuck—or ambiguous, I should say—on what the components of our wavefunction actually are. Are we really imagining the tip of that rotating arrow is a pointlike electric charge spinning around the center? [Pointlike or… Well… Perhaps we should think of the Thomson radius of the electron here, i.e. the so-called classical electron radius, which is equal to the Compton radius times the fine-structure constant: rThomson = α·rCompton ≈ 3.86×10−13/137.]
So that would be the flywheel model.
In contrast, we may also think the whole arrow is some rotating field vector—something like the electric field vector, with the same or some other physical dimension, like newton per charge unit, or newton per mass unit? So that’s the field model. Now, these interpretations may or may not be compatible—or complementary, I should say. I sure hope they are but… Well… What can we reasonably say about it?
Let us first note that the flywheel interpretation has a very obvious advantage, because it allows us to explain the interaction between a photon and an electron, as I demonstrated in my previous post: the electromagnetic energy of the photon will drive the circulatory motion of our electron… So… Well… That’s a nice physical explanation for the transfer of energy. However, when we think about interference or diffraction, we’re stuck: flywheels don’t interfere or diffract. Only waves do. So… Well… What to say?
I am not sure, but here I want to think some more by pushing the flywheel metaphor to its logical limits. Let me remind you of what triggered it all: it was the mathematical equivalence of the energy equation for an oscillator (E = m·a2·ω2) and Einstein’s formula (E = m·c2), which tells us energy and mass are equivalent but… Well… They’re not the same. So what are they then? What is energy, and what is mass—in the context of these matter-waves that we’re looking at. To be precise, the E = m·a2·ω2 formula gives us the energy of two oscillators, so we need a two-spring model which—because I love motorbikes—I referred to as my V-twin engine model, but it’s not an engine, really: it’s two frictionless pistons (or springs) whose direction of motion is perpendicular to each other, so they are in a 90° degree angle and, therefore, their motion is, effectively, independent. In other words: they will not interfere with each other. It’s probably worth showing the illustration just one more time. And… Well… Yes. I’ll also briefly review the math one more time.
If the magnitude of the oscillation is equal to a, then the motion of these piston (or the mass on a spring) will be described by x = a·cos(ω·t + Δ). Needless to say, Δ is just a phase factor which defines our t = 0 point, and ω is the natural angular frequency of our oscillator. Because of the 90° angle between the two cylinders, Δ would be 0 for one oscillator, and –π/2 for the other. Hence, the motion of one piston is given by x = a·cos(ω·t), while the motion of the other is given by x = a·cos(ω·t–π/2) = a·sin(ω·t). The kinetic and potential energy of one oscillator – think of one piston or one spring only – can then be calculated as:
- K.E. = T = m·v2/2 = (1/2)·m·ω2·a2·sin2(ω·t + Δ)
- P.E. = U = k·x2/2 = (1/2)·k·a2·cos2(ω·t + Δ)
The coefficient k in the potential energy formula characterizes the restoring force: F = −k·x. From the dynamics involved, it is obvious that k must be equal to m·ω2. Hence, the total energy—for one piston, or one spring—is equal to:
E = T + U = (1/2)· m·ω2·a2·[sin2(ω·t + Δ) + cos2(ω·t + Δ)] = m·a2·ω2/2
Hence, adding the energy of the two oscillators, we have a perpetuum mobile storing an energy that is equal to twice this amount: E = m·a2·ω2. It is a great metaphor. Somehow, in this beautiful interplay between linear and circular motion, energy is borrowed from one place and then returns to the other, cycle after cycle. However, we still have to prove this engine is, effectively, a perpetuum mobile: we need to prove the energy that is being borrowed or returned by one piston is the energy that is being returned or borrowed by the other. That is easy to do, but I won’t bother you with that proof here: you can double-check it in the referenced post or – more formally – in an article I posted on viXra.org.
It is all beautiful, and the key question is obvious: if we want to relate the E = m·a2·ω2 and E = m·c2 formulas, we need to explain why we could, potentially, write c as c = a·ω = a·√(k/m). We’ve done that already—to some extent at least. The tangential velocity of a pointlike particle spinning around some axis is given by v = r·ω. Now, the radius r is given by a = ħ/(m·c), and ω = E/ħ = m·c2/ħ, so v is equal to to v = [ħ/(m·c)]·[m·c2/ħ] = c. Another beautiful result, but what does it mean? We need to think about the meaning of the ω = √(k/m) formula here. In the mentioned article, we boldly wrote that the speed of light is to be interpreted as the resonant frequency of spacetime, but so… Well… What do we really mean by that? Think of the following.
Einstein’s E = mc2 equation implies the ratio between the energy and the mass of any particle is always the same:
This effectively reminds us of the ω2 = C–1/L or ω2 = k/m formula for harmonic oscillators. The key difference is that the ω2= C–1/L and ω2 = k/m formulas introduce two (or more) degrees of freedom. In contrast, c2= E/m for any particle, always. However, that is exactly the point: we can modulate the resistance, inductance and capacitance of electric circuits, and the stiffness of springs and the masses we put on them, but we live in one physical space only: our spacetime. Hence, the speed of light (c) emerges here as the defining property of spacetime: the resonant frequency, so to speak. We have no further degrees of freedom here.
Let’s think about k. [I am not trying to avoid the ω2= 1/LC formula here. It’s basically the same concept: the ω2= 1/LC formula gives us the natural or resonant frequency for a electric circuit consisting of a resistor, an inductor, and a capacitor. Writing the formula as ω2= C−1/L introduces the concept of elastance, which is the equivalent of the mechanical stiffness (k) of a spring, so… Well… You get it, right? The ω2= C–1/L and ω2 = k/m sort of describe the same thing: harmonic oscillation. It’s just… Well… Unlike the ω2= C–1/L, the ω2 = k/m is directly compatible with our V-twin engine metaphor, because it also involves physical distances, as I’ll show you here.] The k in the ω2 = k/m is, effectively, the stiffness of the spring. It is defined by Hooke’s Law, which states that the force that is needed to extend or compress a spring by some distance x is linearly proportional to that distance, so we write: F = k·x.
Now that is interesting, isn’t it? We’re talking exactly the same thing here: spacetime is, presumably, isotropic, so it should oscillate the same in any direction—I am talking those sine and cosine oscillations now, but in physical space—so there is nothing imaginary here: all is real or… Well… As real as we can imagine it to be. 🙂
We can elaborate the point as follows. The F = k·x equation implies k is a force per unit distance: k = F/x. Hence, its physical dimension is newton per meter (N/m). Now, the x in this equation may be equated to the maximum extension of our spring, or the amplitude of the oscillation, so that’s the radius r = a in the metaphor we’re analyzing here. Now look at how we can re-write the c = a·ω = a·√(k/m) equation:
In case you wonder about the E = F·a substitution: just remember that energy is force times distance. [Just do a dimensional analysis: you’ll see it works out.] So we have a spectacular result here, for several reasons. The first, and perhaps most obvious reason, is that we can actually derive Einstein’s E = m·c2 formula from our flywheel model. Now, that is truly glorious, I think. However, even more importantly, this equation suggests we do not necessarily need to think of some actual mass oscillating up and down and sideways at the same time: the energy in the oscillation can be thought of a force acting over some distance, regardless of whether or not it is actually acting on a particle. Now, that energy will have an equivalent mass which is—or should be, I’d say… Well… The mass of our electron or, generalizing, the mass of the particle we’re looking at.
Huh? Yes. In case you wonder what I am trying to get at, I am trying to convey the idea that the two interpretations—the field versus the flywheel model—are actually fully equivalent, or compatible, if you prefer that term. In Asia, they would say: they are the “same-same but different” 🙂 but, using the language that’s used when discussing the Copenhagen interpretation of quantum physics, we should actually say the two models are complementary.
You may shrug your shoulders but… Well… It is a very deep philosophical point, really. 🙂 As far as I am concerned, I’ve never seen a better illustration of the (in)famous Complementarity Principle in quantum physics because… Well… It goes much beyond complementarity. This is about equivalence. 🙂 So it’s just like Einstein’s equation. 🙂
Post scriptum: If you read my posts carefully, you’ll remember I struggle with those 1/2 factors here and there. Textbooks don’t care about them. For example, when deriving the size of an atom, or the Rydberg energy, even Feynman casually writes that “we need not trust our answer [to questions like this] within factors like 2, π, etcetera.” Frankly, that’s disappointing. Factors like 2, 1/2, π or 2π are pretty fundamental numbers, and so they need an explanation. So… Well… I do loose sleep over them. Let me advance some possible explanation here.
As for Feynman’s model, and the derivation of electron orbitals in general, I think it’s got to do with the fact that electrons do want to pair up when thermal motion does not come into play: think of the Cooper pairs we use to explain superconductivity (so that’s the BCS theory). The 1/2 factor in Schrödinger’s equation also has weird consequences (when you plug in the elementary wavefunction and do the derivatives, you get a weird energy concept: E = m·v2, to be precise). This problem may also be solved when assuming we’re actually calculating orbitals for a pair of electrons, rather than orbitals for just one electron only. [We’d get twice the mass (and, presumably, the charge, so… Well… It might work—but I haven’t done it yet. It’s on my agenda—as so many other things, but I’ll get there… One day. :-)]
So… Well… Let’s get back to the lesson here. In this particular context (i.e. in the context of trying to find some reasonable physical interpretation of the wavefunction), you may or may not remember (if not, check my post on it) ‘ll remember I had to use the I = m·r2/2 formula for the angular momentum, as opposed to the I = m·r2 formula. I = m·r2/2 (with the 1/2 factor) gives us the angular momentum of a disk with radius r, as opposed to a point mass going around some circle with radius r. I noted that “the addition of this 1/2 factor may seem arbitrary”—and it totally is, of course—but so it gave us the result we wanted: the exact (Compton scattering) radius of our electron.
Now, the arbitrary 1/2 factor may or may be explained as follows. In the field model of our electron, the force is linearly proportional to the extension or compression. Hence, to calculate the energy involved in stretching it from x = 0 to x = a, we need to calculate it as the following integral:
So… Well… That will give you some food for thought, I’d guess. 🙂 If it racks your brain too much—or if you’re too exhausted by this point (which is OK, because it racks my brain too!)—just note we’ve also shown that the energy is proportional to the square of the amplitude here, so that’s a nice result as well… 🙂
Talking food for thought, let me make one final point here. The c2 = a2·k/m relation implies a value for k which is equal to k = m·c2/a = E/a. What does this tell us? In one of our previous posts, we wrote that the radius of our electron appeared as a natural distance unit. We wrote that because of another reason: the remark was triggered by the fact that we can write the c/ω ratio as c/ω = a·ω/ω = a. This implies the tangential and angular velocity in our flywheel model of an electron would be the same if we’d measure distance in units of a. Now, the E = a·k = a·F/x (just re-writing…) implies that the force is proportional to the energy— F = (x/a)·E — and the proportionality coefficient is… Well… x/a. So that’s the distance measured in units of a. So… Well… Isn’t that great? The radius of our atom appearing as a natural distance unit does fit in nicely with our geometric interpretation of the wavefunction, doesn’t it? I mean… Do I need to say more?
I hope not because… Well… I can’t explain any better for the time being. I hope I sort of managed to convey the message. Just to make sure, in case you wonder what I was trying to do here, it’s the following: I told you c appears as a resonant frequency of spacetime and, in this post, I tried to explain what that really means. I’d appreciate if you could let me know if you got it. If not, I’ll try again. 🙂 When everything is said and done, one only truly understands stuff when one is able to explain it to someone else, right? 🙂 Please do think of more innovative or creative ways if you can! 🙂
OK. That’s it but… Well… I should, perhaps, talk about one other thing here. It’s what I mentioned in the beginning of this post: this analysis assumes we’re looking at our particle from some specific direction. It could be any direction but… Well… It’s some direction. We have no depth in our line of sight, so to speak. That’s really interesting, and I should do some more thinking about it. Because the direction could be any direction, our analysis is valid for any direction. Hence, if our interpretation would happen to be some true—and that’s a big if, of course—then our particle has to be spherical, right? Why? Well… Because we see this circular thing from any direction, so it has to be a sphere, right?
Well… Yes. But then… Well… While that logic seems to be incontournable, as they say in French, I am somewhat reluctant to accept it at face value. Why? I am not sure. Something inside of me says I should look at the symmetries involved… I mean the transformation formulas for wavefunction when doing rotations and stuff. So… Well… I’ll be busy with that for a while, I guess. 😦
Post scriptum 2: You may wonder whether this line of reasoning would also work for a proton. Well… Let’s try it. Because its mass is so much larger than that of an electron (about 1835 times), the a = ħ/(m·c) formula gives a much smaller radius: 1835 times smaller, to be precise, so that’s around 2.1×10−16 m, which is about 1/4 of the so-called charge radius of a proton, as measured by scattering experiments. So… Well… We’re not that far off, but… Well… We clearly need some more theory here. Having said that, a proton is not an elementary particle, so its mass incorporates other factors than what we’re considering here (two-dimensional oscillations).
One of my readers sent me the following question on the geometric (or even physical) interpretation of the wavefunction that I’ve been offering in recent posts:
“Does this mean that the wave function is merely describing excitations in a matter field; or is this unsupported?“
My reply was very short: “Yes. In fact, we can think of a matter-particle as a tiny flywheel that stores energy.”
However, I realize this answer answers the question only partially. Moreover, I now feel I’ve been quite ambiguous in my description. When looking at the geometry of the elementary wavefunction (see the animation below, which shows us a left- and right-handed wave respectively), two obvious but somewhat conflicting interpretations readily come to mind:
(1) One is that the components of the elementary wavefunction represent an oscillation (in two dimensions) of a field. We may call it a matter field (yes, think of the scalar Higgs field here), but we could also think of it as an oscillation of the spacetime fabric itself: a tiny gravitational wave, in effect. All we need to do here is to associate the sine and cosine component with a physical dimension. The analogy here is the electromagnetic field vector, whose dimension is force per unit charge (newton/coulomb). So we may associate the sine and cosine components of the wavefunction with, say, the force per unit mass dimension (newton/kg) which, using Newton’s Law (F = m·a) reduces to the dimension of acceleration (m/s2), which is the dimension of gravitational fields. I’ll refer to this interpretation as the field interpretation of the matter wave (or wavefunction).
(2) The other interpretation is what I refer to as the flywheel interpretation of the electron. If you google this, you won’t find anything. However, you will probably stumble upon the so-called Zitterbewegung interpretation of quantum mechanics, which is a more elaborate theory based on the same basic intuition. The Zitterbewegung (a term which was coined by Erwin Schrödinger himself, and which you’ll see abbreviated as zbw) is, effectively, a local circulatory motion of the electron, which is presumed to be the basis of the electron’s spin and magnetic moment. All that I am doing, is… Well… I think I do push the envelope of this interpretation quite a bit. 🙂
The flywheel interpretation
A light beam—an electromagnetic wave—consists of a large number of photons. These photons are thought of as being circularly polarized: look at those animations above again. The Planck-Einstein equation tells us the energy of each photon is equal to E = ħ·ω = h·f. [I should, perhaps, quickly note that the frequency f is, obviously, the frequency of the electromagnetic wave. It, therefore, is not to be associated with a matter wave: the de Broglie wavelength and the wavelength of light are very different concepts, even if the Planck-Einstein equation looks the same for both.]
Now, if our beam consists of N photons, the total energy of our beam will be equal to W = N·E = N·ħ·ω. It is crucially important to note that this energy is to be interpreted as the energy that is carried by the beam in a certain time: we should think of the beam as being finite, somehow, in time and in space. Otherwise, our reasoning doesn’t make sense.
The photons carry angular momentum. Just look at those animations (above) once more. It doesn’t matter much whether or not we think of light as particles or as a wave: you can see there is angular momentum there. Photons are spin-1 particles, so the angular momentum will be equal to ± ħ. Hence, then the total angular momentum Jz (the direction of propagation is supposed to be the z-axis here) will be equal to Jz = N·ħ. [This, of course, assumes all photons are polarized in the same way, which may or may not be the case. You should just go along with the argument right now.] Combining the W = N·ħ·ω and Jz = N·ħ equations, we get:
Jz = N·ħ = W/ω
For a photon, we do accept the field interpretation, as illustrated below. As mentioned above, the z-axis here is the direction of propagation (so that’s the line of sight when looking at the diagram). So we have an electric field vector, which we write as ε (epsilon) so as to not cause any confusion with the Ε we used for the energy. [You may wonder if we shouldn’t also consider the magnetic field vector, but then we know the magnetic field vector is, basically, a relativistic effect which vanishes in the reference frame of the charge itself.] The phase of the electric field vector is φ = ω·t.
Now, a charge (so that’s our electron now) will experience a force which is equal to F = q·ε. We use bold letters here because F and ε are vectors. We now need to look at our electron which, in our interpretation of the elementary wavefunction, we think of as rotating about some axis. So that’s what’s represented below. [Both illustrations are Feynman’s, not mine. As for the animations above, I borrowed them from Wikipedia.]
Now, in previous posts, we calculated the radius r based on a similar argument as the one Feynman used to get that Jz = N·ħ = W/ω equation. I’ll refer you those posts and just mention the result here: r is the Compton scattering radius for an electron, which is equal to:
An equally spectacular implication of our flywheel model of the electron was the following: we found that the angular velocity v was equal to v = r·ω = [ħ·/(m·c)]·(E/ħ) = c. Hence, in our flywheel model of an electron, it is effectively spinning around at the speed of light. Note that the angular frequency (ω) in the v = r·ω equation is not the angular frequency of our photon: it’s the frequency of our electron. So we use the same Planck-Einstein equation (ω = E/ħ) but the energy E is the (rest) energy of our electron, so that’s about 0.511 MeV (so that’s an order of magnitude which is 100,000 to 300,000 times that of photons in the visible spectrum). Hence, the angular frequencies of our electron and our photon are very different. Feynman casually reflects this difference by noting the phases of our electron and our photon will differ by a phase factor, which he writes as φ0.
Just to be clear here, at this point, our analysis here diverges from Feynman’s. Feynman had no intention whatsoever to talk about Schrödinger’s Zitterbewegung hypothesis when he wrote what he wrote back in the 1960s. In fact, Feynman is very reluctant to venture into physical interpretations of the wavefunction in all his Lectures on quantum mechanics—which is surprising. Because he comes so tantalizing close at many occasions—as he does here: he describes the motion of the electron here as that of a harmonic oscillator which can be driven by an external electric field. Now that is a physical interpretation, and it is totally consistent with the one I’ve advanced in my recent posts. Indeed, Feynman also describes it as an oscillation in two dimensions—perpendicular to each other and to the direction of motion, as we do— in both the flywheel as well as the field interpretation of the wavefunction!
This point is important enough to quote Feynman himself in this regard:
“We have often described the motion of the electron in the atom as a harmonic oscillator which can be driven into oscillation by an external electric field. We’ll suppose that the atom is isotropic, so that it can oscillate equally well in the x– or y- directions. Then in the circularly polarized light, the x displacement and the y displacement are the same, but one is 90° behind the other. The net result is that the electron moves in a circle.”
Right on! But so what happens really? As our light beam—the photons, really—are being absorbed by our electron (or our atom), it absorbs angular momentum. In other words, there is a torque about the central axis. Let me remind you of the formulas for the angular momentum and for torque respectively: L = r×p and τ = r×F. Needless to say, we have two vector cross-products here. Hence, if we use the τ = r×F formula, we need to find the tangential component of the force (Ft), whose magnitude will be equal to Ft = q·εt. Now, energy is force over some distance so… Well… You may need to think about it for a while but, if you’ve understood all of the above, you should also be able to understand the following formula:
dW/dt = q·εt·v
[If you have trouble, remember v is equal to ds/dt = Δs/Δt for Δt → 0, and re-write the equation above as dW = q·εt·v·dt = q·εt·ds = Ft·ds. Capito?]
Now, you may or may not remember that the time rate of change of angular momentum must be equal to the torque that is being applied. Now, the torque is equal to τ = Ft·r = q·εt·r, so we get:
dJz/dt = q·εt·v
The ratio of dW/dt and dJz/dt gives us the following interesting equation:
Now, Feynman tries to relate this to the Jz = N·ħ = W/ω formula but… Well… We should remind ourselves that the angular frequency of these photons is not the angular frequency of our electron. So… Well… What can we say about this equation? Feynman suggests to integrate dJz and dW over some time interval, which makes sense: as mentioned, we interpreted W as the energy that is carried by the beam in a certain time. So if we integrate dW over this time interval, we get W. Likewise, if we integrate dJz over the same time interval, we should get the total angular momentum that our electron is absorbing from the light beam. Now, because dJz = dW/ω, we do concur with Feynman’s conclusion: the total angular momentum which is being absorbed by the electron is proportional to the total energy of the beam, and the constant of proportionality is equal to 1/ω.
It’s just… Well… The ω here is the angular frequency of the electron. It’s not the angular frequency of the beam. Not in our flywheel model of the electron which, admittedly, is not the model which Feynman used in his analysis. Feynman’s analysis is simpler: he assumes an electron at rest, so to speak, and then the beam drives it so it goes around in a circle with a velocity that is, effectively, given by the angular frequency of the beam itself. So… Well… Fine. Makes sense. As said, I just pushed the analysis a bit further along here. Both analyses raise an interesting question: how and where is the absorbed energy being stored? What is the mechanism here?
In Feynman’s analysis, the answer is quite simple: the electron did not have any motion before but does spin around after the beam hit it. So it has more energy now: it wasn’t a tiny flywheel before, but it is now!
In contrast, in my interpretation of the matter wave, the electron was spinning around already, so where does the extra energy go now? As its energy increases, ω = E/ħ must increase, right? Right. At the same time, the velocity v = r·ω must still be equal to v = r·ω = [ħ·/(m·c)]·(E/ħ) = c, right? Right. So… If ω increases, but r·ω must equal the speed of light, then r must actually decrease somewhat, right?
Right. It’s a weird but inevitable conclusion, it seems. I’ll let you think about it. 🙂
To conclude this post—which, I hope, the reader who triggered it will find interesting—I would like to quote Feynman on an issue on which most textbooks remain silent: the two-state nature of photons. I will just quote him without trying to comment or alter what he writes, because what he writes is clear enough, I think:
“Now let’s ask the following question: If light is linearly polarized in the x-direction, what is its angular momentum? Light polarized in the x-direction can be represented as the superposition of RHC and LHC polarized light. […] The interference of these two amplitudes produces the linear polarization, but it has equal probabilities to appear with plus or minus one unit of angular momentum. [Macroscopic measurements made on a beam of linearly polarized light will show that it carries zero angular momentum, because in a large number of photons there are nearly equal numbers of RHC and LHC photons contributing opposite amounts of angular momentum—the average angular momentum is zero.]
Now, we have said that any spin-one particle can have three values of Jz, namely with zero mass only the states with the components and along the direction of motion exist. For example, light does not have three states, but only two—although a photon is still an object of spin one.”(the three states we saw in the Stern-Gerlach experiment). But light is screwy; it has only two states. It does not have the zero case. This strange lack is related to the fact that light cannot stand still. For a particle of spin j which is standing still, there must be the possible states with values of Jz going in steps of from to . But it turns out that for something of spin
In his typical style and frankness—for which he is revered by some (like me) but disliked by others—he admits this is very puzzling, and not obvious at all! Let me quote him once more:
“How is this consistent with our earlier proofs—based on what happens under rotations in space—that for spin-one particles three states are necessary? For a particle at rest, rotations can be made about any axis without changing the momentum state. Particles with zero rest mass (like photons and neutrinos) cannot be at rest; only rotations about the axis along the direction of motion do not change the momentum state. Arguments about rotations around one axis only are insufficient to prove that three states are required. We have tried to find at least a proof that the component of angular momentum along the direction of motion must for a zero mass particle be an integral multiple of ħ/2—and not something like ħ/3. Even using all sorts of properties of the Lorentz transformation and what not, we failed. Maybe it’s not true. We’ll have to talk about it with Prof. Wigner, who knows all about such things.”
The reference to Eugene Wigner is historically interesting. Feynman probably knew him very well—if only because they had both worked together on the Manhattan Project—and it’s true Wigner was not only a great physicist but a mathematical genius as well. However, Feynman probably quotes him here for the 1963 Nobel Prize he got for… Well… Wigner’s “contributions to the theory of the atomic nucleus and elementary particles, particularly through the discovery and application of fundamental symmetry principles.” 🙂 I’ll let you figure out how what I write about in this post, and symmetry arguments, might be related. 🙂
That’s it for today, folks! I hope you enjoyed this. 🙂
Post scriptum: The main disadvantage of the flywheel interpretation is that it doesn’t explain interference: waves interfere—some rotating mass doesn’t. Ultimately, the wave and flywheel interpretation must, somehow, be compatible. One way to think about it is that the electron can only move as it does—in a “local circulatory motion”—if there is a force on it that makes it move the way it does. That force must be gravitational because… Well… There is no other candidate, is there? [We’re not talking some electron orbital here—some negative charge orbiting around a positive nucleus. We’re just considering the electron itself.] So we just need to prove that our rotating arrow will also represent a force, whose components will make our electron move the way it does. That should not be difficult. The analogy of the V-twin engine should do the trick. I’ll deal with that in my next post. If we’re able to provide such proof (which, as mentioned, should not be difficult), it will be a wonderful illustration of the complementarity principle. 🙂
However, just thinking about it does raise some questions already. Circular motion like this can be explained in two equivalent ways. The most obvious way to think about it is to assume some central field. It’s the planetary model (illustrated below). However, that doesn’t suit our purposes because it’s hard – if possible at all – to relate it to the wavefunction oscillation.
The second model is our two-spring or V-twin engine model (illustrated below), but then what is the mass here? One hypothesis that comes to mind is that we’re constantly accelerating and decelerating an electric charge (the electron charge)—against all other charges in the Universe, so to speak. So that’s a force over a distance—energy. And energy has an equivalent mass.
The question which remains open, then, is the following: what is the nature of this force? In previous posts, I suggested it might be gravitational, but so here we’re back to the drawing board: we’re talking an electrical force, but applied to some mass which acquires mass because of… Well… Because of the force—because of the oscillation (the moving charge) itself. Hmm…. I need to think about this.
After all of the lengthy and speculative excursions into the nature of the wavefunction for an electron, it is time to get back to Feynman’s Lectures and look at photon-electron interactions. So that’s chapter 17 and 18 of Volume III. Of all of the sections in those chapters – which are quite technical here and there – I find the one on the angular momentum of polarized light the most interesting.
Feynman provides an eminently readable explanation of how the electromagnetic energy of a photon may be absorbed by an electron as kinetic energy. It is entirely compatible with our physical interpretation of the wavefunction of an electron as… Well… We’ve basically been looking at the electron as a little flywheel, right? 🙂 I won’t copy Feynman here, except the illustration, which speaks for itself.
However, I do recommend you explore these two Lectures for yourself. Among other interesting passages, Feynman notes that, while photons are spin-1 particles and, therefore, are supposed to be associated with three possible values for the angular momentum (Jz = +ħ, 0 or −ħ), there are only two states: the zero case doesn’t exist. As Feynman notes: “This strange lack is related to the fact that light cannot stand still.” But I will let you explore this for yourself. 🙂
While browsing for something else, I stumbled on an article which derides Feynman’s qualities as a teacher, and the Caltech Feynman Lectures themselves. It is an interesting read. Let me quote (part of) the conclusion:
“Richard Feynman constructed an “introductory” physics course at Caltech suitable primarily for perhaps imaginary extreme physics prodigies like himself or how he pictured himself as an eighteen year old. It is an open question how well the actual eighteen year old Feynman would have done in the forty-three year old Feynman’s “introductory” physics course. Like many adults had Feynman lost touch with what it had been like to be eighteen? In any case, such extreme physics prodigies made up only a small fraction of the highly qualified undergraduate students at Caltech either in the 1960’s or 1980’s. An educational system designed by extreme prodigies for extreme prodigies, often from academic families, extremely wealthy families, or other unusual backgrounds rare even among most top students as conventionally defined, is a prescription for disaster for the vast majority of students and society at large.”
The article actually reacts to a blog post from Bill Gates, who extols Feynman’s virtues as a teacher. So… Was or wasn’t he a great teacher?
It all depends on your definition of a great teacher. I respect the views in the mentioned article mentioned above—if only because the author, John F. McGowan, is not just anyone: he is a B.S. from Caltech itself, and he has a Ph.D. in physics. I don’t, so… Well… He is an authority, obviously. Frankly, I must agree I struggled with Feynman’s Lectures too, and I will probably continue to do so as I read and re-read them time after time. On the other hand, below I copy one of those typical Feynman illustrations you will not find in any other textbook. Feynman tries to give us a physical explanation of the photon-electron interaction here. Most introductory physics textbooks just don’t bother: they’ll give you the mathematical formalism and then some exercises, and that’s it. Worse, those textbooks will repeatedly tell you you can’t really ‘understand’ quantum math. Just go through the math and apply the rules. That’s the general message.
I find that very disappointing. I must admit that Feynman has racked my brain—but in a good way. I still feel I do not quite understand quantum physics “the way we would like to”. It is still “peculiar and mysterious”, but then that’s just how Richard Feynman feels about it too—and he’s humble enough to admit that in the very first paragraph of his very first Lecture on QM.
I have spent a lot of my free time over the past years thinking about a physical or geometric interpretation of the wavefunction—half of my life, in a way—and I think I found it. The article I recently published on it got downloaded for the 100th time today, and this blog – as wordy, nerdy and pedantic as it is – attracted 5,000 visitors last month alone. People like me: people who want to understand physics beyond the equations.
So… Well… Feynman himself admits he was mainly interested in the “one or two dozen students who — very surprisingly — understood almost everything in all of the lectures, and who were quite active in working with the material and worrying about the many points in an excited and interested way.” I think there are many people like those students. People like me: people who want to understand but can’t afford to study physics on a full-time basis.
For those, I think Feynman’s Lectures are truly inspirational. At the very least, they’ve provided me with many wonderful evenings of self-study—some productive, in the classical sense of the word (moving ahead) and… Some… Well… Much of what I read did—and still does—keep me awake at night. 🙂
My previous post on the speed of light as an angular velocity was rather cryptic. This post will be a bit more elaborate. Not all that much, however: this stuff is and remains quite dense, unfortunately. 😦 But I’ll do my best to try to explain what I am thinking of. Remember the formula (or definition) of the elementary wavefunction:
ψ = a·e−i[E·t − p∙x]/ħ = a·cos(p∙x/ħ − E∙t/ħ) + i·a·sin(p∙x/ħ − E∙t/ħ)
How should we interpret this? We know an actual particle will be represented by a wave packet: a sum of wavefunctions, each with its own amplitude ak and its own argument θk = (Ek∙t − pk∙x)/ħ. But… Well… Let’s see how far we get when analyzing the elementary wavefunction itself only.
According to mathematical convention, the imaginary unit (i) is a 90° angle in the counterclockwise direction. However, Nature surely cannot be bothered about our convention of measuring phase angles – or time itself – clockwise or counterclockwise. Therefore, both right- as well as left-handed polarization may be possible, as illustrated below.
The left-handed elementary wavefunction would be written as:
ψ = a·ei[E·t − p∙x]/ħ = a·cos(p∙x/ħ − E∙t/ħ) − i·a·sin(p∙x/ħ − E∙t/ħ)
In my previous posts, I hypothesized that the two physical possibilities correspond to the angular momentum of our particle – say, an electron – being either positive or negative: J = +ħ/2 or, else, J = −ħ/2. I will come back to this in a moment. Let us first further examine the functional form of the wavefunction.
We should note that both the direction as well as the magnitude of the (linear) momentum (p) are relative: they depend on the orientation and relative velocity of our reference frame – which are, in effect, relative to the reference frame of our object. As such, the wavefunction itself is relative: another observer will obtain a different value for both the momentum (p) as well as for the energy (E). Of course, this makes us think of the relativity of the electric and magnetic field vectors (E and B) but… Well… It’s not quite the same because – as I will explain in a moment – the argument of the wavefunction, considered as a whole, is actually invariant under a Lorentz transformation.
Let me elaborate this point. If we consider the reference frame of the particle itself, then the idea of direction and momentum sort of vanishes, as the momentum vector shrinks to the origin itself: p = 0. Let us now look at how the argument of the wavefunction transforms. The E and p in the argument of the wavefunction (θ = ω∙t – k∙x = (E/ħ)∙t – (p/ħ)∙x = (E∙t – p∙x)/ħ) are, of course, the energy and momentum as measured in our frame of reference. Hence, we will want to write these quantities as E = Ev and p = pv = pv∙v. If we then use natural time and distance units (hence, the numerical value of c is equal to 1 and, hence, the (relative) velocity is then measured as a fraction of c, with a value between 0 and 1), we can relate the energy and momentum of a moving object to its energy and momentum when at rest using the following relativistic formulas:
Ev = γ·E0 and pv = γ·m0∙v = γ·E0∙v/c2
The argument of the wavefunction can then be re-written as:
θ = [γ·E0/ħ]∙t – [(γ·E0∙v/c2)/ħ]∙x = (E0/ħ)·(t − v∙x/c2)·γ = (E0/ħ)∙t’
The γ in these formulas is, of course, the Lorentz factor, and t’ is the proper time: t’ = (t − v∙x/c2)/√(1−v2/c2). Two essential points should be noted here:
1. The argument of the wavefunction is invariant. There is a primed time (t’) but there is no primed θ (θ’): θ = (Ev/ħ)·t – (pv/ħ)·x = (E0/ħ)∙t’.
2. The E0/ħ coefficient pops up as an angular frequency: E0/ħ = ω0. We may refer to it as the frequency of the elementary wavefunction.
Now, if you don’t like the concept of angular frequency, we can also write: f0 = ω0/2π = (E0/ħ)/2π = E0/h. Alternatively, and perhaps more elucidating, we get the following formula for the period of the oscillation:
T0 = 1/f0 = h/E0
This is interesting, because we can look at the period as a natural unit of time for our particle. This period is inversely proportional to the (rest) energy of the particle, and the constant of proportionality is h. Substituting E0 for m0·c2, we may also say it’s inversely proportional to the (rest) mass of the particle, with the constant of proportionality equal to h/c2. The period of an electron, for example, would be equal to about 8×10−21 s. That’s very small, and it only gets smaller for larger objects ! But what does all of this really tell us? What does it actually mean?
We can look at the sine and cosine components of the wavefunction as an oscillation in two dimensions, as illustrated below.
Look at the little green dot going around. Imagine it is some mass going around and around. Its circular motion is equivalent to the two-dimensional oscillation. Indeed, instead of saying it moves along a circle, we may also say it moves simultaneously (1) left and right and back again (the cosine) while also moving (2) up and down and back again (the sine).
Now, a mass that rotates about a fixed axis has angular momentum, which we can write as the vector cross-product L = r×p or, alternatively, as the product of an angular velocity (ω) and rotational inertia (I), aka as the moment of inertia or the angular mass: L = I·ω. [Note we write L and ω in boldface here because they are (axial) vectors. If we consider their magnitudes only, we write L = I·ω (no boldface).]
We can now do some calculations. We already know the angular velocity (ω) is equal to E0/ħ. Now, the magnitude of r in the L = r×p vector cross-product should equal the magnitude of ψ = a·e−i∙E·t/ħ, so we write: r = a. What’s next? Well… The momentum (p) is the product of a linear velocity (v) – in this case, the tangential velocity – and some mass (m): p = m·v. If we switch to scalar instead of vector quantities, then the (tangential) velocity is given by v = r·ω.
So now we only need to think about what formula we should use for the angular mass. If we’re thinking, as we are doing here, of some point mass going around some center, then the formula to use is I = m·r2. However, we may also want to think that the two-dimensional oscillation of our point mass actually describes the surface of a disk, in which case the formula for I becomes I = m·r2/2. Of course, the addition of this 1/2 factor may seem arbitrary but, as you will see, it will give us a more intuitive result. This is what we get:
L = I·ω = (m·r2/2)·(E/ħ) = (1/2)·a2·(E/c2)·(E/ħ) = a2·E2/(2·ħ·c2)
Note that our frame of reference is that of the particle itself, so we should actually write ω0, m0 and E0 instead of ω, m and E. The value of the rest energy of an electron is about 0.510 MeV, or 8.1871×10−14 N∙m. Now, this momentum should equal J = ±ħ/2. We can, therefore, derive the (Compton scattering) radius of an electron:Substituting the various constants with their numerical values, we find that a is equal 3.8616×10−13 m, which is the (reduced) Compton scattering radius of an electron. The (tangential) velocity (v) can now be calculated as being equal to v = r·ω = a·ω = [ħ·/(m·c)]·(E/ħ) = c. This is an amazing result. Let us think about it.
In our previous posts, we introduced the metaphor of two springs or oscillators, whose energy was equal to E = m·ω2. Is this compatible with Einstein’s E = m·c2 mass-energy equivalence relation? It is. The E = m·c2 implies E/m = c2. We, therefore, can write the following:
ω = E/ħ = m·c2/ħ = m·(E/m)·/ħ ⇔ ω = E/ħ
Hence, we should actually have titled this and the previous post somewhat differently: the speed of light appears as a tangential velocity. Think of the following: the ratio of c and ω is equal to c/ω = a·ω/ω = a. Hence, the tangential and angular velocity would be the same if we’d measure distance in units of a. In other words, the radius of an electron appears as a natural distance unit here: if we’d measure ω in units of a per second, rather than in radians (which are expressed in the SI unit of distance, i.e. the meter) per second, the two concepts would coincide.
More fundamentally, we may want to look at the radius of an electron as a natural unit of velocity. Huh? Yes. Just re-write the c/ω = a as ω = c/a. What does it say? Exactly what I said, right? As such, the radius of an electron is not only a norm for measuring distance but also for time. 🙂
If you don’t quite get this, think of the following. For an electron, we get an angular frequency that is equal to ω = E/ħ = (8.19×10−14 N·m)/(1.05×10−34 N·m·s) ≈ 7.76×1020 radians per second. That’s an incredible velocity, because radians are expressed in distance units—so that’s in meter. However, our mass is not moving along the unit circle, but along a much tinier orbit. The ratio of the radius of the unit circle and a is equal to 1/a ≈ (1 m)/(3.86×10−13 m) ≈ 2.59×1012. Now, if we divide the above-mentioned velocity of 7.76×1020 radians per second by this factor, we get… Right ! The speed of light: 2.998×1082 m/s. 🙂
Post scriptum: I have no clear answer to the question as to why we should use the I = m·r2/2 formula, as opposed to the I = m·r2 formula. It ensures we get the result we want, but this 1/2 factor is actually rather enigmatic. It makes me think of the 1/2 factor in Schrödinger’s equation, which is also quite enigmatic. In my view, the 1/2 factor should not be there in Schrödinger’s equation. Electron orbitals tend to be occupied by two electrons with opposite spin. That’s why their energy levels should be twice as much. And so I’d get rid of the 1/2 factor, solve for the energy levels, and then divide them by two again. Or something like that. 🙂 But then that’s just my personal opinion or… Well… I’ve always been intrigued by the difference between the original printed edition of the Feynman Lectures and the online version, which has been edited on this point. My printed edition is the third printing, which is dated July 1966, and – on this point – it says the following:
“Don’t forget that meff 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 2 to 20 times the free-space mass of the electron.”
Two to twenty times. Not 1 or 0.5 to 20 times. No. Two times. As I’ve explained a couple of times, if we’d define a new effective mass which would be twice the old concept – so meffNEW = 2∙meffOLD – then such re-definition would not only solve a number of paradoxes and inconsistencies, but it will also justify my interpretation of energy as a two-dimensional oscillation of mass.
However, the online edition has been edited here to reflect the current knowledge about the behavior of an electron in a medium. Hence, if you click on the link above, you will read that the effective mass can be “about 0.1 to 30 times” the free-space mass of the electron. Well… This is another topic altogether, and so I’ll sign off here and let you think about it all. 🙂
Over the weekend, I worked on a revised version of my paper on a physical interpretation of the wavefunction. However, I forgot to add the final remarks on the speed of light as an angular velocity. I know… This post is for my faithful followers only. It is dense, but let me add the missing bits here:
Post scriptum (29 October): Einstein’s view on aether theories probably still holds true: “We may say that according to the general theory of relativity space is endowed with physical qualities; in this sense, therefore, there exists an aether. According to the general theory of relativity, space without aether is unthinkable – for in such space there not only would be no propagation of light, but also no possibility of existence for standards of space and time (measuring-rods and clocks), nor therefore any space-time intervals in the physical sense. But this aether may not be thought of as endowed with the quality characteristic of ponderable media, as consisting of parts which may be tracked through time. The idea of motion may not be applied to it.”
The above quote is taken from the Wikipedia article on aether theories. The same article also quotes Robert Laughlin, the 1998 Nobel Laureate in Physics, who said this about aether in 2005: “It is ironic that Einstein’s most creative work, the general theory of relativity, should boil down to conceptualizing space as a medium when his original premise [in special relativity] was that no such medium existed. […] The word ‘aether’ has extremely negative connotations in theoretical physics because of its past association with opposition to relativity. This is unfortunate because, stripped of these connotations, it rather nicely captures the way most physicists actually think about the vacuum. […]The modern concept of the vacuum of space, confirmed every day by experiment, is a relativistic aether. But we do not call it this because it is taboo.”
I really love this: a relativistic aether. My interpretation of the wavefunction is very consistent with that.
My last posts were all about a possible physical interpretation of the quantum-mechanical wavefunction. To be precise, we have been interpreting the wavefunction as a gravitational wave. In this interpretation, the real and imaginary component of the wavefunction get a physical dimension: force per unit mass (newton per kg). The inspiration here was the structural similarity between Coulomb’s and Newton’s force laws. They both look alike: it’s just that one gives us a force per unit charge (newton per coulomb), while the other gives us a force per unit mass.
So… Well… Many nice things came out of this – and I wrote about that at length – but last night I was thinking this interpretation may also offer an explanation of relativistic length contraction. Before we get there, let us re-visit our hypothesis.
The geometry of the wavefunction
The elementary wavefunction is written as:
ψ = a·e−i(E·t − p∙x)/ħ = a·cos(p∙x/ħ – E∙t/ħ) + i·a·sin(p∙x/ħ – E∙t/ħ)
Nature should not care about our conventions for measuring the phase angle clockwise or counterclockwise and, therefore, the ψ = a·ei[E·t − p∙x]/ħ function may also be permitted. We know that cos(θ) = cos(–θ) and sinθ = –sin(–θ), so we can write:
ψ = a·ei(E·t − p∙x)/ħ = a·cos(E∙t/ħ – p∙x/ħ) + i·a·sin(E∙t/ħ – p∙x/ħ)
= a·cos(p∙x/ħ – E∙t/ħ) – i·a·sin(p∙x/ħ – E∙t/ħ)
The vectors p and x are the the momentum and position vector respectively: p = (px, py, pz) and x = (x, y, z). However, if we assume there is no uncertainty about p – not about the direction nor the magnitude – then we may choose an x-axis which reflects the direction of p. As such, x = (x, y, z) reduces to (x, 0, 0), and p∙x/ħ reduces to p∙x/ħ. This amounts to saying our particle is traveling along the x-axis or, if p = 0, that our particle is located somewhere on the x-axis. Hence, the analysis is one-dimensional only.
The geometry of the elementary wavefunction is illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.
Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. You can check this as follows: if the origin is the (x, t) = (0, 0) point, then cos(θ) = cos(0) = 1 and sin(θ) = sin(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively. We speculated this should correspond to the two possible values for the quantum-mechanical spin of the wave: +ħ/2 or −ħ/2. The cosine and sine components for the left-handed wave are shown below. Needless to say, the cosine and sine function are the same, except for a phase difference of π/2: sin(θ) = cos(θ − π/2).
As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any wave F(kx – ωt) is given by vp = ω/k = (E/ħ)/(p/ħ) = E/p. Of course, the momentum might also be in the negative x-direction, in which case k would be equal to -p and, therefore, we would get a negative phase velocity: vp = ω/k = –E/p.
The de Broglie relations
E/ħ = ω gives the frequency in time (expressed in radians per second), while p/ħ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). Of course, we may write: f = ω/2π and λ = 2π/k, which gives us the two de Broglie relations:
- E = ħ∙ω = h∙f
- p = ħ∙k = h/λ
The frequency in time is easy to interpret. The wavefunction of a particle with more energy, or more mass, will have a higher density in time than a particle with less energy.
In contrast, the second de Broglie relation is somewhat harder to interpret. According to the p = h/λ relation, the wavelength is inversely proportional to the momentum: λ = h/p. The velocity of a photon, or a (theoretical) particle with zero rest mass (m0 = 0), is c and, therefore, we find that p = mv∙v = mc∙c = m∙c (all of the energy is kinetic). Hence, we can write: p∙c = m∙c2 = E, which we may also write as: E/p = c. Hence, for a particle with zero rest mass, the wavelength can be written as:
λ = h/p = hc/E = h/mc
Hence, if p goes to zero, then the wavelength becomes infinitely long: if p → 0 then λ → ∞. How should we interpret this inverse proportionality between λ and p? To answer this question, let us first see what this wavelength λ actually represents.
If we look at the ψ = a·cos(p∙x/ħ – E∙t/ħ) – i·a·sin(p∙x/ħ – E∙t/ħ) once more, and if we write p∙x/ħ as Δ, then we can look at p∙x/ħ as a phase factor, and so we will be interested to know for what x this phase factor Δ = p∙x/ħ will be equal to 2π. So we write:
Δ =p∙x/ħ = 2π ⇔ x = 2π∙ħ/p = h/p = λ
So now we get a meaningful interpretation for that wavelength. It is the distance between the crests (or the troughs) of the wave, so to speak, as illustrated below. Of course, this two-dimensional wave has no real crests or troughs: we measure crests and troughs against the y-axis here. Hence, our definition depend on the frame of reference.
Now we know what λ actually represents for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/f = 2π·(ħ/E). Hence, we can now calculate the wave velocity:
v = λ/T = (h/p)/[2π·(ħ/E)] = E/p
Unsurprisingly, we just get the phase velocity that we had calculated already: v = vp = E/p. The question remains: what if p is zero? What if we are looking at some particle at rest? It is an intriguing question: we get an infinitely long wavelength, and an infinite wave velocity.
Now, re-writing the v = E/p as v = m∙c2/m∙vg = c/βg, in which βg is the relative classical velocity of our particle βg = vg/c) tells us that the phase velocities will effectively be superluminal (βg < 1 so 1/ βg > 1), but what if βg approaches zero? The conclusion seems unavoidable: for a particle at rest, we only have a frequency in time, as the wavefunction reduces to:
ψ = a·e−i·E·t/ħ = a·cos(E∙t/ħ) – i·a·sin(E∙t/ħ)
How should we interpret this?
A physical interpretation of relativistic length contraction?
In my previous posts, we argued that the oscillations of the wavefunction pack energy. Because the energy of our particle is finite, the wave train cannot be infinitely long. If we assume some definite number of oscillations, then the string of oscillations will be shorter as λ decreases. Hence, the physical interpretation of the wavefunction that is offered here may explain relativistic length contraction.
Yep. Think about it. 🙂
 Even neutrinos have some (rest) mass. This was first confirmed by the US-Japan Super-Kamiokande collaboration in 1998. Neutrinos oscillate between three so-called flavors: electron neutrinos, muon neutrinos and tau neutrinos. Recent data suggests that the sum of their masses is less than a millionth of the rest mass of an electron. Hence, they propagate at speeds that are very near to the speed of light.
 Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as KE = E − E0 = mvc2 − m0c2 = m0γc2 − m0c2 = m0c2(γ − 1). As v approaches c, γ approaches infinity and, therefore, the kinetic energy would become infinite as well.
 Because our particle will be represented by a wave packet, i.e. a superimposition of elementary waves with different E and p, the classical velocity of the particle becomes the group velocity of the wave, which is why we denote it by vg.
This post further builds on the rather remarkable results we got in our previous posts. Let us start with the basics once again. The elementary wavefunction is written as:
ψ = a·e−i[E·t − p∙x]/ħ = a·cos(p∙x/ħ − E∙t/ħ) + i·a·sin(p∙x/ħ − E∙t/ħ)
Of course, Nature (or God, as Einstein would put it) does not care about our conventions for measuring an angle (i.e. the phase of our wavefunction) clockwise or counterclockwise and, therefore, the ψ = a·ei[E·t − p∙x]/ħ function is also permitted. We know that cos(θ) = cos(−θ) and sinθ = −sin(−θ), so we can write:
ψ = a·ei[E·t − p∙x]/ħ = a·cos(E∙t/ħ − p∙x/ħ) + i·a·sin(E∙t/ħ − p∙x/ħ)
= a·cos(p∙x/ħ − E∙t/ħ) − i·a·sin(p∙x/ħ − E∙t/ħ)
The vectors p and x are the momentum and position vector respectively: p = (px, py, pz) and x = (x, y, z). However, if we assume there is no uncertainty about p – not about the direction, and not about the magnitude – then the direction of p can be our x-axis. In this reference frame, x = (x, y, z) reduces to (x, 0, 0), and p∙x/ħ reduces to p∙x/ħ. This amounts to saying our particle is traveling along the x-axis or, if p = 0, that our particle is located somewhere on the x-axis. So we have an analysis in one dimension only then, which facilitates our calculations. The geometry of the wavefunction is then as illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.
Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. [You can check this as follows: if the origin is the (x, t) = (0, 0) point, then cos(θ) = cos(0) = 1 and sin(θ) = sin(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively.]
Now, you will remember that we speculated the two polarizations (left- versus right-handed) should correspond to the two possible values for the quantum-mechanical spin of the wave (+ħ/2 or −ħ/2). We will come back to this at the end of this post. Just focus on the essentials first: the cosine and sine components for the left-handed wave are shown below. Look at it carefully and try to understand. Needless to say, the cosine and sine function are the same, except for a phase difference of π/2: sin(θ) = cos(θ − π/2).
As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any waveform F(kx − ωt) is given by vp = ω/k. In our case, we find that vp = ω/k = (E/ħ)/(p/ħ) = E/p. Of course, the momentum might also be in the negative x-direction, in which case k would be equal to −p and, therefore, we would get a negative phase velocity: vp = ω/k = (E/ħ)/(−p/ħ) = −E/p.
As you know, E/ħ = ω gives the frequency in time (expressed in radians per second), while p/ħ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). [If in doubt, check my post on essential wave math.] Now, you also know that f = ω/2π and λ = 2π/k, which gives us the two de Broglie relations:
- E = ħ∙ω = h∙f
- p = ħ∙k = h/λ
The frequency in time (oscillations or radians per second) is easy to interpret. A particle will always have some mass and, therefore, some energy, and it is easy to appreciate the fact that the wavefunction of a particle with more energy (or more mass) will have a higher density in time than a particle with less energy.
However, the second de Broglie relation is somewhat harder to interpret. Note that the wavelength is inversely proportional to the momentum: λ = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long, so we write:
If p → 0 then λ → ∞.
For the limit situation, a particle with zero rest mass (m0 = 0), the velocity may be c and, therefore, we find that p = mv∙v = mc∙c = m∙c (all of the energy is kinetic) and, therefore, p∙c = m∙c2 = E, which we may also write as: E/p = c. Hence, for a particle with zero rest mass (m0 = 0), the wavelength can be written as:
λ = h/p = hc/E = h/mc
Of course, we are talking a photon here. We get the zero rest mass for a photon. In contrast, all matter-particles should have some mass and, therefore, their velocity will never equal c. The question remains: how should we interpret the inverse proportionality between λ and p?
Let us first see what this wavelength λ actually represents. If we look at the ψ = a·cos(p∙x/ħ − E∙t/ħ) − i·a·sin(p∙x/ħ – E∙t/ħ) once more, and if we write p∙x/ħ as Δ, then we can look at p∙x/ħ as a phase factor, and so we will be interested to know for what x this phase factor Δ = p∙x/ħ will be equal to 2π. So we write:
Δ =p∙x/ħ = 2π ⇔ x = 2π∙ħ/p = h/p = λ
So now we get a meaningful interpretation for that wavelength. It is the distance between the crests (or the troughs) of the wave, so to speak, as illustrated below. Of course, this two-dimensional wave has no real crests or troughs: they depend on your frame of reference.
So now we know what λ actually represent for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/f = 2π·(ħ/E). Hence, we can now calculate the wave velocity:
v = λ/T = (h/p)/[2π·(ħ/E)] = E/p
Unsurprisingly, we just get the phase velocity that we had calculated already: v = vp = E/p. It does not answer the question: what if p is zero? What if we are looking at some particle at rest? It is an intriguing question: we get an infinitely long wavelength, and an infinite phase velocity. Now, we know phase velocities can be superluminal, but they should not be infinite. So what does the mathematical inconsistency tell us? Do these infinitely long wavelengths and infinite wave velocities tell us that our particle has to move? Do they tell us our notion of a particle at rest is mathematically inconsistent?
Maybe. But maybe not. Perhaps the inconsistency just tells us our elementary wavefunction – or the concept of a precise energy, and a precise momentum – does not make sense. This is where the Uncertainty Principle comes in: stating that p = 0, implies zero uncertainty. Hence, the σp factor in the σp∙σx ≤ ħ/2 would be zero and, therefore, σp∙σx would be zero which, according to the Uncertainty Principle, it cannot be: it can be very small, but it cannot be zero.
It is interesting to note here that σp refers to the standard deviation from the mean, as illustrated below. Of course, the distribution may be or may not be normal – we don’t know – but a normal distribution makes intuitive sense, of course. Also, if we assume the mean is zero, then the uncertainty is basically about the direction in which our particle is moving, as the momentum might then be positive or negative.
The question of natural units may pop up. The Uncertainty Principle suggests a numerical value of the natural unit for momentum and distance that is equal to the square root of ħ/2, so that’s about 0.726×10−17 m for the distance unit and 0.726×10−17 N∙s for the momentum unit, as the product of both gives us ħ/2. To make this somewhat more real, we may note that 0.726×10−17 m is the attometer scale (1 am = 1×10−18 m), so that is very small but not unreasonably small.
Hence, we need to superimpose a potentially infinite number of waves with energies and momenta centered on some mean value. It is only then that we get meaningful results. For example, the idea of a group velocity – which should correspond to the classical idea of the velocity of our particle – only makes sense in the context of wave packet. Indeed, the group velocity of a wave packet (vg) is calculated as follows:
vg = ∂ωi/∂ki = ∂(Ei/ħ)/∂(pi/ħ) = ∂(Ei)/∂(pi)
This assumes the existence of a dispersion relation which gives us ωi as a function of ki – what amounts to the same – Ei as a function of pi. How do we get that? Well… There are a few ways to go about it but one interesting way of doing it is to re-write Schrödinger’s equation as the following pair of equations:
- Re(∂ψ/∂t) = −[ħ/(2meff)]·Im(∇2ψ) ⇔ ω·cos(kx − ωt) = k2·[ħ/(2meff)]·cos(kx − ωt)
- Im(∂ψ/∂t) = [ħ/(2meff)]·Re(∇2ψ) ⇔ ω·sin(kx − ωt) = k2·[ħ/(2meff)]·sin(kx − ωt)
These equations imply the following dispersion relation:
ω = ħ·k2/(2m)
Of course, we need to think about the subscripts now: we have ωi, ki, but… What about meff or, dropping the subscript, about m? Do we write it as mi? If so, what is it? Well… It is the equivalent mass of Ei obviously, and so we get it from the mass-energy equivalence relation: mi = Ei/c2. It is a fine point, but one most people forget about: they usually just write m. However, if there is uncertainty in the energy, then Einstein’s mass-energy relation tells us we must have some uncertainty in the (equivalent) mass too, and the two will, obviously, be related as follows: σm = σE/c2. We are tempted to do a few substitutions here. Let’s first check what we get when doing the mi = Ei/c2 substitution:
ωi = ħ·ki2/(2mi) = (1/2)∙ħ·ki2∙c2/Ei = (1/2)∙ħ·ki2∙c2/(ωi∙ħ) = (1/2)∙ħ·ki2∙c2/ωi
⇔ ωi2/ki2 = c2/2 ⇔ ωi/ki = vp = c/2 !?
We get a very interesting but nonsensical condition for the dispersion relation here. I wonder what mistake I made. 😦
Let us try another substitution. The group velocity is what it is, right? It is the velocity of the group, so we can write: ki = p/ħ = mi ·vg. This gives us the following result:
ωi = ħ·(mi ·vg)2/(2mi) = ħ·mi·vg2/2
It is yet another interesting condition for the dispersion relation. Does it make any more sense? I am not so sure. That factor 1/2 troubles us. It only makes sense when we drop it. Now you will object that Schrödinger’s equation gives us the electron orbitals – and many other correct descriptions of quantum-mechanical stuff – so, surely, Schrödinger’s equation cannot be wrong. You’re right. It’s just that… Well… When we are splitting in up in two equations, as we are doing here, then we are looking at one of the two dimensions of the oscillation only and, therefore, it’s only half of the mass that counts. Complicated explanation but… Well… It should make sense, because the results that come out make sense. Think of it. So we write this:
- Re(∂ψ/∂t) = −(ħ/meff)·Im(∇2ψ) ⇔ ω·cos(kx − ωt) = k2·(ħ/meff)·cos(kx − ωt)
- Im(∂ψ/∂t) = (ħ/meff)·Re(∇2ψ) ⇔ ω·sin(kx − ωt) = k2·(ħ/meff)·sin(kx − ωt)
We then get the dispersion relation without that 1/2 factor:
ωi = ħ·ki2/mi
The mi = Ei/c2 substitution then gives us the result we sort of expected to see:
ωi = ħ·ki2/mi = ħ·ki2∙c2/Ei = ħ·ki2∙c2/(ωi∙ħ) ⇔ ωi/ki = vp = c
Likewise, the other calculation also looks more meaningful now:
ωi = ħ·(mi ·vg)2/mi = ħ·mi·vg2
Sweet ! 🙂
Let us put this aside for the moment and focus on something else. If you look at the illustrations above, you see we can sort of distinguish (1) a linear velocity – the speed with which those wave crests (or troughs) move – and (2) some kind of circular or tangential velocity – the velocity along the red contour line above. We’ll need the formula for a tangential velocity: vt = a∙ω.
Now, if λ is zero, then vt = a∙ω = a∙E/ħ is just all there is. We may double-check this as follows: the distance traveled in one period will be equal to 2πa, and the period of the oscillation is T = 2π·(ħ/E). Therefore, vt will, effectively, be equal to vt = 2πa/(2πħ/E) = a∙E/ħ. However, if λ is non-zero, then the distance traveled in one period will be equal to 2πa + λ. The period remains the same: T = 2π·(ħ/E). Hence, we can write:
For an electron, we did this weird calculation. We had an angular momentum formula (for an electron) which we equated with the real-life +ħ/2 or −ħ/2 values of its spin, and we got a numerical value for a. It was the Compton radius: the scattering radius for an electron. Let us write it out:
Using the right numbers, you’ll find the numerical value for a: 3.8616×10−13 m. But let us just substitute the formula itself here:
This is fascinating ! And we just calculated that vp is equal to c. For the elementary wavefunction, that is. Hence, we get this amazing result:
vt = 2c
This tangential velocity is twice the linear velocity !
Of course, the question is: what is the physical significance of this? I need to further look at this. Wave velocities are, essentially, mathematical concepts only: the wave propagates through space, but nothing else is really moving. However, the geometric implications are obviously quite interesting and, hence, need further exploration.
One conclusion stands out: all these results reinforce our interpretation of the speed of light as a property of the vacuum – or of the fabric of spacetime itself. 🙂
 Even neutrinos should have some (rest) mass. In fact, the mass of the known neutrino flavors was estimated to be smaller than 0.12 eV/c2. This mass combines the three known neutrino flavors.
 Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as KE = E − E0 = mvc2 − m0c2 = m0γc2 − m0c2 = m0c2(γ − 1). As v approaches c, γ approaches infinity and, therefore, the kinetic energy would become infinite as well.
 It is, of course, extremely small, but 1 am is the current sensitivity of the LIGO detector for gravitational waves. It is also thought of as the upper limit for the length of an electron, for quarks, and for fundamental strings in string theory. It is, in any case, 1,000,000,000,000,000,000 times larger than the order of magnitude of the Planck length (1.616229(38)×10−35 m).
 The meff is the effective mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write meff = m. As for the equations, they are easily derived from noting 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. Now, the ∂ψ/∂t = i∙(ħ/meff)∙∇2ψ equation amounts to writing something like this: a + i∙b = i∙(c + i∙d). Now, remembering that i2 = −1, you can easily figure out that i∙(c + i∙d) = i∙c + i2∙d = − d + i∙c.
My posts and article on the wavefunction as a gravitational wave are rather short on the exact geometry of the wavefunction, so let us explore that a bit here. By now, you know the formula for the elementary wavefunction by heart:
ψ = a·e−i[E·t − p∙x]/ħ = a·cos(p∙x/ħ − E∙t/ħ) + i·a·sin(p∙x/ħ − E∙t/ħ)
If we assume the momentum p is all in the x-direction, then the p and x vectors will have the same direction, and p∙x/ħ reduces to p∙x/ħ. This amounts to saying our particle is traveling along the x-axis. The geometry of the wavefunction is illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.
Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. If the origin is the (x, t) = (0, 0) point, then cos(θ) = cos(0) = 1 and sin(θ) = sin(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively. I am convinced these correspond to the two possible values for the quantum-mechanical spin of the wave: +ħ/2 or −ħ/2. But… Well… Who am I? The cosine and sine components are shown below. Needless to say, the cosine and sine function are the same, except for a phase difference of π/2: sin(θ) = cos(θ − π/2)
Surely, Nature doesn’t care a hoot about our conventions for measuring the phase angle clockwise or counterclockwise and therefore, the ψ = a·ei[E·t − p∙x]/ħ function should, effectively, also be permitted. We know that cos(θ) = cos(–θ) and sinθ = –sin(–θ), so we can write:
ψ = a·ei[E·t − p∙x]/ħ = a·cos(E∙t/ħ − p∙x/ħ) + i·a·sin(E∙t/ħ − p∙x/ħ)
= a·cos(p∙x/ħ − E∙t/ħ) − i·a·sin(p∙x/ħ − E∙t/ħ)
E/ħ = ω gives the frequency in time (expressed in radians per second), while p/ħ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). Of course, we may write: f = ω/2π and λ = 2π/k, which gives us the two de Broglie relations:
- E = ħ∙ω = h∙f
- p = ħ∙k = h/λ
The frequency in time is easy to interpret (a particle will always have some mass and, therefore, some energy), but the wavelength is inversely proportional to the momentum: λ = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long: if p → 0, then λ → ∞. For the limit situation, a particle with zero rest mass (m0 = 0), the velocity may be c and, therefore, we find that p = mv∙v = m∙c and, therefore, p∙c = m∙c2 = E, which we may also write as: E/p = c. Hence, for a particle with zero rest mass, the wavelength can be written as:
λ = h/p = hc/E = h/mc
However, we argued that the physical dimension of the components of the wavefunction may be usefully expressed in N/kg units (force per unit mass), while the physical dimension of the electromagnetic wave are expressed in N/C (force per unit charge). This, in fact, explains the dichotomy between bosons (photons) and fermions (spin-1/2 particles). Hence, all matter-particles should have some mass. But how we interpret the inverse proportionality between λ and p?
We should probably first ask ourselves what wavelength we are talking about. The wave only has a phase velocity here, which is equal to vp = ω/k = (E/ħ)/(p/ħ) = E/p. Of course, we know that, classically, the momentum will be equal to the group velocity times the mass: p = m·vg. However, when p is zero, we have a division by zero once more: if p → 0, then vp = E/p → ∞. Infinite wavelengths and infinite phase velocities probably tell us that our particle has to move: our notion of a particle at rest is mathematically inconsistent. If we associate this elementary wavefunction with some particle, and if we then imagine it to move, somehow, then we get an interesting relation between the group and the phase velocity:
vp = ω/k = E/p = E/(m·vg) = (m·c2)/(m·vg) = c2/vg
We can re-write this as vp·vg = c2, which reminds us of the relationship between the electric and magnetic constant (1/ε0)·(1/μ0) = c2. But what is the group velocity of the elementary wavefunction? Is it a meaningful concept?
The phase velocity is just the ratio of ω/k. In contrast, the group velocity is the derivative of ω with respect to k. So we need to write ω as a function of k. Can we do that even if we have only one wave? We do not have a wave packet here, right? Just some hypothetical building block of a real-life wavefunction, right? Right. So we should introduce uncertainty about E and p and build up the wave packet, right? Well… Yes. But let’s wait with that, and see how far we can go in our interpretation of this elementary wavefunction. Let’s first get that ω = ω(k) relation. You’ll remember we can write Schrödinger’s equation – the equation that describes the propagation mechanism for matter-waves – as 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)
This tells us that ω = ħ·k2/(2m). Therefore, we can calculate ∂ω/∂k as:
∂ω/∂k = ħ·k/m = p/m = vg
We learn nothing new. We are going round and round in circles here, and we always end up with a tautology: as soon as we have a non-zero momentum, we have a mathematical formula for the group velocity – but we don’t know what it represents – and a finite wavelength. In fact, using the p = ħ∙k = h/λ relation, we can write one as a function of the other:
λ = h/p = h/mvg ⇔ vg = h/mλ
What does this mean? It resembles the c = h/mλ relation we had for a particle with zero rest mass. Of course, it does: the λ = h/mc relation is, once again, a limit for vg going to c. By the way, it is interesting to note that the vp·vg = c2 relation implies that the phase velocity is always superluminal. That’ easy to see when you re-write the equation in terms of relative velocities: (vp/c)·(vg/c) = βphase·βgroup = 1. Hence, if βgroup < 1, then βphase > 1.
So what is the geometry, really? Let’s look at the ψ = a·cos(p∙x/ħ – E∙t/ħ) – i·a·sin(p∙x/ħ – E∙t/ħ) formula once more. If we write p∙x/ħ as Δ, then we will be interested to know for what x this phase factor will be equal to 2π. So we write:
Δ =p∙x/ħ = 2π ⇔ x = 2π∙ħ/p = h/p = λ
So now we get a meaningful interpretation for that wavelength: it’s that distance between the crests of the wave, so to speak, as illustrated below.
Can we now find a meaningful (i.e. geometric) interpretation for the group and phase velocity? If you look at the illustration above, you see we can sort of distinguish (1) a linear velocity (the speed with which those wave crests move) and (2) some kind of circular or tangential velocity (the velocity along the red contour line above). We’ll probably need the formula for the tangential velocity: v = a∙ω. If p = 0 (so we have that weird infinitesimally long wavelength), then we have two velocities:
- The tangential velocity around the a·ei·E·t circle, so to speak, and that will just be equal to v = a∙ω = a∙E/ħ.
- The red contour line sort of gets stretched out, like infinitely long, and the velocity becomes… What does it do? Does it go to ∞ , or to c?
Let’s think about this. For a particle at rest, we had this weird calculation. We had an angular momentum formula (for an electron) which we equated with the real-life +ħ/2 or −ħ/2 values of its spin. And so we got a numerical value for a. It was the Compton radius: the scattering radius for an electron. Let me copy it once again:
Just to bring this story a bit back to Earth, you should note the calculated value: a = 3.8616×10−13 m. We did then another weird calculation. We said all of the energy of the electron had to be packed in this cylinder that might of might not be there. The point was: the energy is finite, so that elementary wavefunction cannot have an infinite length in space. Indeed, assuming that the energy was distributed uniformly, we jotted down this formula, which reflects the formula for the volume of a cylinder:
E = π·a2·l ⇔ l = E/(π·a2)
Using the value we got for the Compton scattering radius (a = 3.8616×10−13 m), we got an astronomical value for l. Let me write it out:
l = (8.19×10−14)/(π·14.9×10−26) ≈ 0.175×1012 m
It is, literally, an astronomical value: 0.175×1012 m is 175 million kilometer, so that’s like the distance between the Sun and the Earth. We wrote, jokingly, that such space is too large to look for an electron and, hence, that we should really build a proper packet by making use of the Uncertainty Principle: allowing for uncertainty in the energy should, effectively, reduce the uncertainty in position.
But… Well… What if we use that value as the value for λ? We’d get that linear velocity, right? Let’s try it. The period is equal to T = T = 2π·(ħ/E) = h/E and λ = E/(π·a2), so we write:We can write this as a function of m and the c and ħ constants only:
A weird formula but not necessarily nonsensical: we get a finite speed. Now, if the wavelength becomes somewhat less astronomical, we’ll get different values of course. I have a strange feeling that, with these formula, we should, somehow, be able to explain relativistic length contraction. But I will let you think about that as for now. Here I just wanted to show the geometry of the wavefunction a bit more in detail.
 The discussions on the mass of neutrinos are interesting in this regard. Scientists all felt the neutrino had to have some (rest) mass, so my instinct on this is theirs. In fact, only recently experimental confirmation came in, and the mass of the known neutrino flavors was estimated to be something like 0.12 eV/c2. This mass combines the three known neutrino flavors. To understand this number, you should note it is the same order of magnitude of the equivalent mass of low-energy photons, like infrared or microwave radiation.
One of my beloved brothers just sent me the news on this year’s Nobel Prize for Physics. Of course, it went to the MIT/Caltech LIGO scientists – who confirmed the reality of gravitational waves. That’s exactly the topic that I am exploring when trying to digest all this quantum math and stuff. Brilliant !
I actually sent the physicists a congratulatory message – and my paper ! I can’t believe I actually did that.
In the best case, I just made a fool of myself. In the worst case… Well… I just made a fool of myself. 🙂
In my previous posts, I’ve been playing with… Well… At the very least, a new didactic approach to understanding the quantum-mechanical wavefunction. I just boldly assumed the matter-wave is a gravitational wave. I did so by associating its components with the dimension of gravitational field strength: newton per kg, which is the dimension of acceleration (N/kg = m/s2). Why? When you remember the physical dimension of the electromagnetic field is N/C (force per unit charge), then that’s kinda logical, right? 🙂 The math is beautiful. Key consequences include the following:
- Schrodinger’s equation becomes an energy diffusion equation.
- Energy densities give us probabilities.
- The elementary wavefunction for the electron gives us the electron radius.
- Spin angular momentum can be interpreted as reflecting the right- or left-handedness of the wavefunction.
- Finally, the mysterious boson-fermion dichotomy is no longer “deep down in relativistic quantum mechanics”, as Feynman famously put it.
It’s all great. Every day brings something new. 🙂 Today I want to focus on our weird electron model and how we get God’s number (aka the fine-structure constant) out of it. Let’s recall the basics of it. We had the elementary wavefunction:
ψ = a·e−i[E·t − p∙x]/ħ = a·e−i[E·t − p∙x]/ħ = a·cos(p∙x/ħ − E∙t/ħ) + i·a·sin(p∙x/ħ − E∙t/ħ)
In one-dimensional space (think of a particle traveling along some line), the vectors (p and x) become scalars, and so we simply write:
ψ = a·e−i[E·t − p∙x]/ħ = a·e−i[E·t − p∙x]/ħ = a·cos(p∙x/ħ − E∙t/ħ) + i·a·sin(p∙x/ħ − E∙t/ħ)
This wavefunction comes with constant probabilities |ψ|2 = a2, so we need to define a space outside of which ψ = 0. Think of the particle-in-a-box model. This is obvious oscillations pack energy, and the energy of our particle is finite. Hence, each particle – be it a photon or an electron – will pack a finite number of oscillations. It will, therefore, occupy a finite amount of space. Mathematically, this corresponds to the normalization condition: all probabilities have to add up to one, as illustrated below.Now, all oscillations of the elementary wavefunction have the same amplitude: a. [Terminology is a bit confusing here because we use the term amplitude to refer to two very different things here: we may say a is the amplitude of the (probability) amplitude ψ. So how many oscillations do we have? What is the size of our box? Let us assume our particle is an electron, and we will reduce its motion to a one-dimensional motion only: we’re thinking of it as traveling along the x-axis. We can then use the y- and z-axes as mathematical axes only: they will show us how the magnitude and direction of the real and imaginary component of ψ. The animation below (for which I have to credit Wikipedia) shows how it looks like.Of course, we can have right- as well as left-handed particle waves because, while time physically goes by in one direction only (we can’t reverse time), we can count it in two directions: 1, 2, 3, etcetera or −1, −2, −3, etcetera. In the latter case, think of time ticking away. 🙂 Of course, in our physical interpretation of the wavefunction, this should explain the (spin) angular momentum of the electron, which is – for some mysterious reason that we now understand 🙂 – always equal to J = ± ħ/2.
Now, because a is some constant here, we may think of our box as a cylinder along the x-axis. Now, the rest mass of an electron is about 0.510 MeV, so that’s around 8.19×10−14 N∙m, so it will pack some 1.24×1020 oscillations per second. So how long is our cylinder here? To answer that question, we need to calculate the phase velocity of our wave. We’ll come back to that in a moment. Just note how this compares to a photon: the energy of a photon will typically be a few electronvolt only (1 eV ≈ 1.6 ×10−19 N·m) and, therefore, it will pack like 1015 oscillations per second, so that’s a density (in time) that is about 100,000 times less.
Back to the angular momentum. The classical formula for it is L = I·ω, so that’s angular frequency times angular mass. What’s the angular velocity here? That’s easy: ω = E/ħ. What’s the angular mass? If we think of our particle as a tiny cylinder, we may use the formula for its angular mass: I = m·r2/2. We have m: that’s the electron mass, right? Right? So what is r? That should be the magnitude of the rotating vector, right? So that’s a. Of course, the mass-energy equivalence relation tells us that E = mc2, so we can write:
L = I·ω = (m·r2/2)·(E/ħ) = (1/2)·a2·m·(mc2/ħ) = (1/2)·a2·m2·c2/ħ
Does it make sense? Maybe. Maybe not. You can check the physical dimensions on both sides of the equation, and that works out: we do get something that is expressed in N·m·s, so that’s action or angular momentum units. Now, we know L must be equal to J = ± ħ/2. [As mentioned above, the plus or minus sign depends on the left- or right-handedness of our wavefunction, so don’t worry about that.] How do we know that? Because of the Stern-Gerlach experiment, which has been repeated a zillion times, if not more. Now, if L = J, then we get the following equation for a: This is the formula for the radius of an electron. To be precise, it is the Compton scattering radius, so that’s the effective radius of an electron as determined by scattering experiments. You can calculate it: it is about 3.8616×10−13 m, so that’s the picometer scale, as we would expect.
This is a rather spectacular result. As far as I am concerned, it is spectacular enough for me to actually believe my interpretation of the wavefunction makes sense.
Let us now try to think about the length of our cylinder once again. The period of our wave is equal to T = 1/f = 1/(ω/2π) = 1/[(E/ħ)·2π] = 1/(E/h) = h/E. Now, the phase velocity (vp) will be given by:
vp = λ·f = (2π/k)·(ω/2π) = ω/k = (E/ħ)/(p/ħ) = E/p = E/(m·vg) = (m·c2)/(m·vg) = c2/vg
This is very interesting, because it establishes an inverse proportionality between the group and the phase velocity of our wave, with c2 as the coefficient of inverse proportionality. In fact, this equation looks better if we write as vp·vg = c2. Of course, the group velocity (vg) is the classical velocity of our electron. This equation shows us the idea of an electron at rest doesn’t make sense: if vg = 0, then vp times zero must equal c2, which cannot be the case: electrons must move in space. More generally, speaking, matter-particles must move in space, with the photon as our limiting case: it moves at the speed of light. Hence, for a photon, we find that vp = vg = E/p = c.
How can we calculate the length of a photon or an electron? It is an interesting question. The mentioned orders or magnitude of the frequency (1015 or 1020) gives us the number of oscillations per second. But how many do we have in one photon, or in one electron?
Let’s first think about photons, because we have more clues here. Photons are emitted by atomic oscillators: atoms going from one state (energy level) to another. We know how to calculate to calculate the Q of these atomic oscillators (see, for example, Feynman I-32-3): it is of the order of 108, which means the wave train will last about 10–8 seconds (to be precise, that is the time it takes for the radiation to die out by a factor 1/e). Now, the frequency of sodium light, for example, is 0.5×1015 oscillations per second, and the decay time is about 3.2×10–8 seconds, so that makes for (0.5×1015)·(3.2×10–8) = 16 million oscillations. Now, the wavelength is 600 nanometer (600×10–9) m), so that gives us a wavetrain with a length of (600×10–9)·(16×106) = 9.6 m.
These oscillations may or may not have the same amplitude and, hence, each of these oscillations may pack a different amount of energies. However, if the total energy of our sodium light photon (i.e. about 2 eV ≈ 3.3×10–19 J) are to be packed in those oscillations, then each oscillation would pack about 2×10–26 J, on average, that is. We speculated in other posts on how we might imagine the actual wave pulse that atoms emit when going from one energy state to another, so we don’t do that again here. However, the following illustration of the decay of a transient signal dies out may be useful.
This calculation is interesting. It also gives us an interesting paradox: if a photon is a pointlike particle, how can we say its length is like 10 meter or more? Relativity theory saves us here. We need to distinguish the reference frame of the photon – riding along the wave as it is being emitted, so to speak – and our stationary reference frame, which is that of the emitting atom. Now, because the photon travels at the speed of light, relativistic length contraction will make it look like a pointlike particle.
What about the electron? Can we use similar assumptions? For the photon, we can use the decay time to calculate the effective number of oscillations. What can we use for an electron? We will need to make some assumption about the phase velocity or, what amounts to the same, the group velocity of the particle. What formulas can we use? The p = m·v is the relativistically correct formula for the momentum of an object if m = mv, so that’s the same m we use in the E = mc2 formula. Of course, v here is, obviously, the group velocity (vg), so that’s the classical velocity of our particle. Hence, we can write:
p = m·vg = (E/c2)·vg ⇔ vg = p/m = p·c2/E
This is just another way of writing that vg = c2/vp or vp = c2/vg so it doesn’t help, does it? Maybe. Maybe not. Let us substitute in our formula for the wavelength:
λ = vp/f = vp·T = vp⋅(h/E) = (c2/vg)·(h/E) = h/(m·vg) = h/p
This gives us the other de Broglie relation: λ = h/p. This doesn’t help us much, although it is interesting to think about it. The f = E/h relation is somewhat intuitive: higher energy, higher frequency. In contrast, what the λ = h/p relation tells us that we get an infinite wavelength if the momentum becomes really small. What does this tell us? I am not sure. Frankly, I’ve look at the second de Broglie relation like a zillion times now, and I think it’s rubbish. It’s meant to be used for the group velocity, I feel. I am saying that because we get a non-sensical energy formula out of it. Look at this:
- E = h·f and p = h/λ. Therefore, f = E/h and λ = p/h.
- v = f·λ = (E/h)∙(p/h) = E/p
- p = m·v. Therefore, E = v·p = m·v2
E = m·v2? This formula is only correct if v = c, in which case it becomes the E = mc2 equation. So it then describes a photon, or a massless matter-particle which… Well… That’s a contradictio in terminis. 🙂 In all other cases, we get nonsense.
Let’s try something differently. If our particle is at rest, then p = 0 and the p·x/ħ term in our wavefunction vanishes, so it’s just:
ψ = a·e−i·E·t/ħ = a·cos(E∙t/ħ) − i·a·sin(E∙t/ħ)
Hence, our wave doesn’t travel. It has the same amplitude at every point in space at any point in time. Both the phase and group velocity become meaningless concepts. The amplitude varies – because of the sine and cosine – but the probability remains the same: |ψ|2 = a2. Hmm… So we need to find another way to define the size of our box. One of the formulas I jotted down in my paper in which I analyze the wavefunction as a gravitational wave was this one:
It was a physical normalization condition: the energy contributions of the waves that make up a wave packet need to add up to the total energy of our wave. Of course, for our elementary wavefunction here, the subscripts vanish and so the formula reduces to E = (E/c2)·a2·(E2/ħ2), out of which we get our formula for the scattering radius: a = ħ/mc. Now how do we pack that energy in our cylinder? Assuming that energy is distributed uniformly, we’re tempted to write something like E = a2·l or, looking at the geometry of the situation:
E = π·a2·l ⇔ l = E/(π·a2)
It’s just the formula for the volume of a cylinder. Using the value we got for the Compton scattering radius (a = 3.8616×10−13 m), we find an l that’s equal to (8.19×10−14)/(π·14.9×10−26) =≈ 0.175×1012… Meter? Yes. We get the following formula:
0.175×1012 m is 175 million kilometer. That’s – literally – astronomic. It corresponds to 583 light-seconds, or 9.7 light-minutes. So that’s about 1.17 times the (average) distance between the Sun and the Earth. You can see that we do need to build a wave packet: that space is a bit too large to look for an electron, right? 🙂
Could we possibly get some less astronomic proportions? What if we impose that l should equal a? We get the following condition:We find that m would have to be equal to m ≈ 1.11×10−36 kg. That’s tiny. In fact, it’s equivalent to an energy of about equivalent to 0.623 eV (which you’ll see written as 623 milli-eV. This corresponds to light with a wavelength of about 2 micro-meter (μm), so that’s in the infrared spectrum. It’s a funny formula: we find, basically, that the l/a ratio is proportional to m4. Hmm… What should we think of this? If you have any ideas, let me know !
Post scriptum (3 October 2017): The paper is going well. Getting lots of downloads, and the views on my blog are picking up too. But I have been vicious. Substituting B for (1/c)∙i∙E or for −(1/c)∙i∙E implies a very specific choice of reference frame. The imaginary unit is a two-dimensional concept: it only makes sense when giving it a plane view. Literally. Indeed, my formulas assume the i (or −i) plane is perpendicular to the direction of propagation of the elementary quantum-mechanical wavefunction. So… Yes. The need for rotation matrices is obvious. But my physical interpretation of the wavefunction stands. 🙂