joigus

Yes, thank you for your appreciation of this point. That's what I meant when I said, The very moment you trim down a quantum state from the idealisation of an infinite monochromatic wave to a short pulse, you are introducing infinitely many frequencies, and you need the group velocity to describe its motion. No other velocity makes any sense. In the relativistic case, it's actually superluminal, as I proved from simple relativistic-dynamical and quantum-dynamical constraints.

I'm afraid we have different understandings of what "explicit" means, at least for this case. "Explicit" as in "He gave me very explicit directions on how to get there." (Taken from Oxford's dictionary). Yes, I'm sure about that. https://www.ippp.dur.ac.uk/~krauss/Lectures/QuarksLeptons/QED/GaugeInvariance_2.html Read under the heading "Local Gauge Invariance". IOW: You can re-define the local phase at will as long as you accompany such change by the corresponding gradient in the gauge field. What is "the" phase now? No wonder phase velocity is not an observable. You can concoct situations in which phase and group velocity are essentially the same. I'd venture to say that for those cases you can "measure" the phase velocity. What you're doing (secretly) is, of course, measuring the group velocity, and using the theory to deduce a value of the phase velocity consistent with it and with the choice of quantum state (or "ray") that you have obtained in that particular gauge. No. You got this totally wrong. Quantum waves are not sinusoidal. For starters, they are not real functions. They typically go like complex exponentials. No. I can write A=XY/Z, the relation be totally right, and neither A, nor X, Y or Z be measurable. In fact, what you both are saying contradicts the principles of mainstream quantum mechanics --except for free particles. If the particle is interacting, the so-called Hamiltonian has a potential energy term, and the momentum (the inverse wavelength) does not commute with it, so the principles say no, you can't measure both at the same time. They're called "incompatible." You both are confusing a component (among infinitely many) of the quantum state with the whole quantum state. More on this: https://www.mathpages.com/home/kmath210/kmath210.htm And so on, and so on, and so on...

Again. How? Be explicit, please. That would contradict gauge invariance, which we know to be exact. In a nutshell, gauge invariance tells us that quantum states don't have "a" phase, meaning an unambiguous local phase. Let alone a phase velocity. The phase velocity of ripples on a pond can be measured. The phase of the wave function cannot. You can measure interference patterns. But those have nothing to do with phase velocity. You talk about frequency and wavelength as if there was just one frequency and one wavelength for matter waves. It is precisely because all "realistic" matter waves package many frequencies and wavelenths that the phase velocity is rendered all but meaningless in QM. Only the group velocity makes physical sense. I've just had a déja vu. Didn't I say something like this before? It's either right or wrong. What's sure not to be is irrelevant to the question at hand.

I don't think that's what the text means. I think it's a reference to the group velocity. Whenever group velocity is different from phase velocity, we call the medium "disperssive", and that's because every monochromatic component travels at a slightly different speed. v=v(f)

This sounds all very reasonable, and you do have a point. Nevertheless... We don't have a handle on the wave function itself. X-crystallography and the like is based on phase differences. Interference positions --and thereby wavelengths--... pretty much the same. No way to see the phase itself though. In fact, I'm working with some advantage here. I happen to know there is a very robust principle of physics --the gauge invariance principle-- that tells us it is impossible to know what that phase might be, as I can always gauge away any phase prescription that you take by locally re-defining the phase. So I have a pretty solid understanding of why what you claim cannot be true. Gimme any phase you like and I will "gauge it away" without breaking any known rules of quantum field theory, or classical electromagnetism, etc.

The particle's velocity is the group velocity. What laboratory measurement gives you the phase velocity? Didn't I say this before? And didn't you?

So if I have a "theory" that relates the size of the universe with the size of my nose --never mind how crazy that theory is--, when I'm measuring the size of my nose, can I claim to be measuring the size of the universe? Electron diffraction also measures electron wavelengths. How do you measure both wavelenght and frequency? --Thereby phase velocity. Huh? I'll be patiently waiting for any of your answers. This should be fun.

How?

Really? Please, tell me of an experiment to measure the phase velocity of the wave function of an electron.

Monochromatic-wave solutions of the non-relativistic Schrödinger equation.

They are incompatible, only if you interpret they are both exact. They are compatible, if you interpret one is approximate and the other is exact. If you just say, Then you're misrepresenting what I said. If you juxtapose an approximate statement and an exact one and you take both at face value, of course you would be expected to find contradictions! x-posted with @swansont

I think you are misrepresenting what I said. One is fully relativistic (vp=c²/v); the other (vp=v/2) is a non-relativistic approximation. I see now I must have done a very poor job of explaining that. I thought it was clear for you when you said, Is ex=1+x for small x incompatible with exe-x=1? No, if instead of ex=1+x you write \( e^{x} \simeq 1+x\) instead of just saying they are equal.

You can do as you please, of course. I will just finish by telling you that a theory that fixes a problem that doesn't exist is not a better theory. The phase velocity is not an observable quantity, so any theory that "fixes" the phase velocity to conform to a particular theoretical prejudice of what it should look like is deeply misguided. And please understand that I mean well. I don't want you to waste your time in something that I think is hopeless, and that's why I'm telling you.

Ok, let's take it from there then. This \( \frac{1}{2}mv^{2}=p^{2}/2m \) is what should be identified with the Einstein quantum relation ℏω (Planck's constant times the angular frequency), as the time-independent Schrödinger equation says, \( \left(P^{2}/2m\right)\psi=E\psi \). Then, it's a matter of doing some algebra after plugin in De Broglie's \( p=\hbar k \), \[ \hbar\omega\simeq\sqrt{\left(\hbar k\right)^{2}c^{2}+m^{2}c^{4}}-mc^{2} \] \[ \hbar\omega\simeq mc^{2}\sqrt{\frac{\hbar^{2}k^{2}}{m^{2}c^{2}}+1}-mc^{2} \] Taylor expanding the square root in powers of the small quantity \( \frac{\hbar^{2}k^{2}}{m^{2}c^{2}} \) (as \( mc\gg\hbar k \)), \[ \hbar\omega\simeq mc^{2}\left(1+\frac{1}{2}\frac{\hbar^{2}k^{2}}{m^{2}c^{2}}\right)-mc^{2} \] So that, \[ \hbar\omega\simeq mc^{2}\frac{1}{2}\frac{\hbar^{2}k^{2}}{m^{2}c^{2}}\Rightarrow v_{p}=\frac{\omega}{k}\simeq\frac{1}{2}\frac{\hbar k}{m} \] while, \[ v_{g}=\frac{d\omega}{dk}\simeq\frac{d}{dk}\left(\frac{\hbar k^{2}}{2m}\right)=\frac{\hbar k}{m} \] So indeed, \[ v_{p}=\frac{v_{g}}{2} \] Please, mind the \( \simeq \) in the next-to-the-last line. What about \( v_{g}=c²/v_{p} \) Mind you, this is an exact identity that comes from, \[ v_{p}v_{f}=\frac{\omega}{k}\frac{d\omega}{dk}=c\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\frac{c}{\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}}=c^{2} \] Of course you cannot get that from the Schrödinger dispersion relation, as is clear from the fact that it has no \( c \) in it. So you're mixing an exact identity there with an approximation, which could be part of the reason of your confusion. Does any of that address any of your concerns? Does it make sense?

Ok. So the quantity that's "under the obligation" of giving us back the non-relativistic kinetic energy \( \frac{1}{2}mv^{2} \) is not \( E \), but \( E - mc^{2} \). That is, when \( v \) is very small as compared to \( c \), \[ \frac{1}{2}mv^{2}\simeq\sqrt{p^{2}c^{2}+m^{2}c^{4}}-mc^{2} \] Do you agree so far? I just want to make it very clear that I'm not trying to pull wool over your eyes in any way. This technique of "incremental agreement" I learned from @studiot, by the way.

It's a Taylor-series expansion. Remember?: \[ \frac{mc^{2}}{\sqrt{1-v^{2}/c^{2}}}\simeq mc^{2}\left(1+\frac{v^{2}}{2c^{2}}-\cdots\right)=mc^{2}+\frac{1}{2}mv^{2}-\ldots \] Didn't you understand when I said it, or don't you understand now? It is neither strange nor non-strange. It's a Taylor-series expansion. A Taylor-series expansion is... well. A Taylor-series expansion. What's strange about that?

Then \( E=mc^{2}+\frac{p^{2}}{2m} \) once one corrects for the zero of energies by adding the rest energy, right?

Ok, @martillo, you must agree that the description of a mechanical problem doesn't change if we just re-define the energy by a constant shift E0. Right? E-->E+E0 Equivalent problem. Same solutions. Everything the same. Right? Do we agree on that?

I'm in the middle of answering you, but first please clarify this: vp=mc2/v How is that a velocity?

And it is, if only you just admit that in non-relativistic mechanics you can redefine the spectrum of energy to include this new "energy at zero velocity". It is. It is a particular case. What the non-relativistic approach cannot give you is the mc2 term, for obvious reasons. It cannot give you a formula that's wrong. It only slightly generalises it to be right, ie, to include the rest energy. Haven't you noticed that the second term exactly coincides with Schrödinger's term? Am I not being clear?

I insist: There's a physical reason not to expect the Taylor expansion to recover the Schrödinger dispersion relation. And that's the humongous rest mass factor in the phase of the wave function. That is, \[ e^{⁻imc^{2}t/\hbar} \] which the Schrödinger theory cannot fathom. This "shifts" the Schrödinger dispersion relation. IOW, you should not get the dispersion relation, \[ v_{p}=\frac{\omega}{k}=\frac{\hbar k}{2m} \] But the one corrected for the rest mass, \[ \hbar\omega=mc^{2}+\frac{1}{2}mv^{2}-\ldots=mc^{2}+\frac{\left(\hbar k\right)^{2}}{2m}-\ldots \] \[ v_{p}=\frac{\omega}{k}=\frac{mc^{2}}{\hbar k}+\frac{\hbar k}{2m}-\ldots \] And indeed, that's what you get, if you Taylor expand what I showed you before, \[ v_{p}=c\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\simeq c\frac{mc}{\hbar k}=\frac{mc^{2}}{\hbar k} \] keeping up to terms linear in \( \hbar k \) That is what I can live with. Namely: an otherwise Schrödinger-compatible dispersion relation that only corrects for the rest-mass term. Can't you?

I can live with that. The phase velocity of a Schrödinger wave is not an observable.

Why? Unless you're doing gravitation, the zero of energy is immaterial, and all that matters is differences of energy.

LOL. And again I made a mistake when LateXing the answer. I guess you caught me. The results are OK I think. Don't worry too much about the phase velocity. It's not very substantial (physical). It's very telling that in the relativistic case you get a phase velocity > c. Think about them as vacuum effects that don't do anything but shifting the ground state. The group velocity is a more serious matter. It's related to the particle's actual velocity.

Ok. First of all, sorry I wrote c-1 on a couple of lines where I should have written c. If you go over the equations I wrote you will see that, for v<<c, \[ \sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\simeq\frac{mc}{\hbar k} \] With this approximation, \[ v_{p}=c\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\simeq c\frac{mc}{\hbar k}=\frac{mc^{2}}{\hbar k} \] \[ v_{g}=\frac{c}{\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}}\simeq\frac{c\hbar k}{mc}=\frac{\hbar k}{m} \] So there seems to be a mismatch with the non-relativistic expressions directly obtained from \( E= \frac{1}{2}mv^{2} \). The latter being, \[ v_{p}=\frac{\omega}{k}=\frac{\hbar k}{2m} \] \[ v_{g}=\frac{d\omega}{dk}=\frac{\hbar k}{m} \] In particular, it's the phase velocity that is to blame. I understand that's your point. Is it. not? Well there's a physical reason for that. All these expressions come from the assumption that it's a free particle we're talking about. Einstein's celebrated formula for the kinetic energy, \[ mc^{2}\sqrt{1-v^{2}/c^{2}}\simeq mc^{2}\left(1+\frac{v^{2}}{2c^{2}}-\cdots\right)=mc^{2}+\frac{1}{2}mv^{2}-\ldots \] gives a humongous rest-energy term mc2. This enormous energy shift is a constant, so it is of no effect when it comes to energy considerations, but produces a contribution to the frequency that kicks the Schrödinger dispersion relation out of whack. Is something like that what has you worried?