# The velocity of a quantum mechanical plane wave of a photon

There is a famous "derivation" or "demonstration" that is often presented in introductory classes in quantum mechanics. I find it deeply unsettling and I feel like key information are being left out.

It starts by assuming that a photon is somehow represented by a plane wave:

$$\psi = \exp{(i(kx-\omega t))}$$

where $$\omega$$ is a function of $$k$$ obtained from the relativistic relation $$p=\hbar k$$ subbed in the energy of the photon $$E=\hbar \omega=p^2/2m$$.

Giving us, by relativistic necessity,

$$\omega = p^2 /2m\hbar = (\hbar k)^2 /2m\hbar = \hbar k^2 / 2m$$

therefore,

$$v_p = \omega / k = \hbar k / 2m = p / 2m = v/2$$

and of course by finding $$v_g = d \omega/dk$$ , we obtain that $$v_g = v$$.

Now here a list of things that are bothering me, that seem out of nowhere, and no resource in the entire internet seems to address:

• What does the group velocity even mean for a photon represented by a single plane wave? Why is the calculation for the group velocity meaningful/valid here?
• For a wave-packet consisting of a single phase-wave, don't the phase and group velocities become identical, why does the math not seem to allow this here?
• Why is the "electromagnetic wavelength" treated as unequivocally equal to the "wavefunction wavelength" associated with the de Broglie wavelength of the photon's momentum, is this something that is asserted by the relativistic theory, or is it another postulate or assumption we make that "just works"?

I am failing to find proper, rigorous questions to this matter mainly due to the lack of rigor involved throughout the argument. It feels quite hand-wavy to me and I am not sure what to make of it.

• It is not clear what this supposed proof is about. In any case, the $\omega(p)$ relation is not relativistic at all. Commented Mar 10 at 6:17
• The simplest way Commented Mar 10 at 6:26
• The energy-momentum relation of a massless particle (like the photon) is $E=c|\vec{p}|$. Commented Mar 10 at 6:29
• Commented Mar 12 at 14:07

First of all your energy momentum dispersion for the photon is wrong as $$E = \frac{p^2}{m}$$ is a classical relation. What would one even use as photon mass in that formula? In the case of a relativistic particle with no mass $$E = c |p|$$, as mentioned by Hyperon, should be used instead. In consequence both phase and group velocity become c.

As for your first question: The phase velocity tells you the speed, at which your single plane wave is propagating threw space, the group velocity tells you at what speed the envelope is propagating threw space. Since the envelope of single plane wave is constant in space, propagating it threw space does not do anything and therefore the concept does indeed not feel particularly meaningful in that case.

Regarding your second Question: No, phase and group velocity do not necessarily become identical for a plane wave, in fact they will not if the dispersion relation is not a proportional one, as you show in your question. It’s just that as stated above the group velocity becomes a physically irrelevant mathematical concept, when looking at that special case.

Concerning your last Question: As I am aware the wavelength of the wave function being the same as for the EM-fields in cases where they can be measured macroscopically is less of something asserted by theory(, until you start dealing with relativistic quantum electrodynamics I would guess,) but rather a basic observation, seen when doing slit experiments with singular photons vs slit experiments with larger scale EM-waves for example. Usually assuming a plane wave solution in free space, de-Broglie- and Planck-relations and an energy momentum relation, which all can be motivated from experiments, is a good way to come up with a guess for the QM equation of motion, by just assuming a wave equation and plugging in your frequency wavelength relation. That’s what we did, when I took the introductory QM-course. If you use a classical energy momentum relation there, you will end up with Schrödingers equation for free particles. If you use the relativistic dispersion relation I think you end up with the Klein-Gordon equation.

So the main problem in your example is, that it is talking about photons, but using the classical dispersion relation. If either talking about electrons or using the right relativistic relations I can see it as decent exercise to get some contact with the calculations done in the motivation of the QM equation of motion discussed above. I am not exactly sure what the didactic value is of calculating group velocities for a single plane wave here, but then again calculating such properties can always be seen as a value on itself I guess. Comparing group and phase velocities could also be used to show some vivid differences for the consequences of using the relativistic or nonrelativistic energy-momentum-relations.

Edit: As “The Feadow” pointed out in his comment, the phase velocity is only half the classical movement speed. That is in my opinion indeed an interesting observation. My answer regarding that irritation would be, that a single plane wave is dislocated over the entirety of your position space. That means, that moving it, doesn’t really do anything. Moving the electron does only get interesting, if you have some kind of semi-located wave package or looking at the classical approximation of a hard ball (or the classical limit of a $$\delta$$ function in position-space). But then the group velocity is what gives you the movement of that wave package. So contrary to what I said before, one might claim that phase velocity is physically irrelevant and only group velocity matters. But group velocity also is irrelevant as long as you only look at a single plane wave.

• Aha I see, yeah this calculation was used (in an MIT lecture, no less) to demonstrate and familiarize the idea of phase and group velocities for QM. I took it as a serious derivation/calculation but now that I realize that the classical Energy-momentum relation cannot be used for a photon, I understand that it was just a "white-lie" to familiarize students with certain ideas and terminologies. Commented Mar 10 at 16:00
• Out of curiousity, what would change if I posed the same question for an electron instead of a photon? Commented Mar 10 at 21:00
• @TheFeadow If talking about a non relativistic free electron in a plane wave state, I think your calculations for group and phase velocity would be correct, group velocity would still be physically irrelevant and the question about the EM-wavelength would raze “ERROR: UndefVarError”.
– Zaph
Commented Mar 11 at 10:03
• in this case (non relativistic free electron) the phase velocity calculation yields half the classical velocity (v/2), how does one see through this issue? Commented Mar 12 at 16:53