Is the Wavefunction Speed in Quantum Mechanics Wrong?

[fox26]

This problem bothered me many years ago when I was taking a university course in
quantum mechanics, but I assumed it was due to an error on my part which I, however,
couldn’t locate, and I didn’t ask my course instructor about it. Recently, when
researching a q.m. question on the Internet, I rediscovered it: Both in my textbooks years
ago and on the Internet now, (non-normalizable) plane-wave solutions to the
non-Relativistic, 1-dimensional, but easily extendable to 3 dimensions, Schrodinger
wave equation for any free particle PRTCL with non-zero rest mass (which always
moves at less than the speed of light), such as an electron, are given as proportional to
Ψ = exp(i[kx-ωt]), where k = 2π/λ, λ being Ψ’s wavelength, ω = 2πf, where f is Ψ’s
frequency, i is the imaginary constant, x is the length coordinate, and t is time. The
problem arises from the identification of h/λ with the classical momentum p = ms of
PRTCL, and hf with its classical kinetic energy KE = p²/2m = ms²/2, where m is PRTCL’s
mass, s is its speed, and h is Planck’s Constant. Then Ψ = exp(2πi[px -KEt]/h) = exp(2π
i[msx - ms² t/2]/h), and if t increases one unit, the term it is in increases by ms² /2, so for
Ψ to have the same phase as before, x must increase by ms²/2ms = s/2 units, so the
phase velocity (speed) of Ψ is s/2. This seems to be at variance with the speed of
PRTCL's being s, but it might be objected that Ψ is not a physically realistic wavefunction,
being non-normalizable. However, Ψ can be modified by smoothly limiting it to be zero
outside of some large but finite region R, but to be the same as before inside of some
region R’ inside but nearly as big as R, so that the modified wave function Ψ' is
normalizable, and so that Ψ' is the wave function of some free particle mPRTCL with
only slightly uncertain momentum and speed, with momentum expectation value p and
speed expectation value s, while the phase velocity of the main Fourier component of Ψ′
and the speed of change of its position expectation value is s/2.

The cause of this discrepancy seems to be that KE = hf, which Louis de Broglie, the
inventor of the idea of the wave function of a particle, is supposed to have assumed to
be true for all particles, is true for photons, which move with the speed of light, and all of
whose energy E = mc² is kinetic energy, none being their rest mass equivalent energy,
so their KE = mc² = ms² (E = hf for a photon was Planck’s quantum hypothesis, which began
quantum mechanics, made to avoid the Ultraviolet Catastrophe of then-current
theory [maybe the UC wasn’t implied by then-current-theory, and I’ll post a thread about
that]), but KE = hf, where f is the true frequency of the wave function of a particle in a
momentum eigenstate (i.e., the value of f in the equation above which is necessary to
make the phase velocity, or the speed of the movement of the expectation value of its
position, equal to the particle’s actual speed), gives a value for the kinetic energy KE of
such a particle which has non-zero rest mass and is moving with a speed small
compared to that of light which is about 2 times larger than its actual kinetic energy, so
setting hf = the actual kinetic energy gives a value for hf which is about 2 times too
small. Indeed, the Relativistic (true) kinetic energy rKE of a particle with rest mass m₀
is given by rKE = total energy (excluding potential energy) - rest mass equivalent energy =
m₀c² [1/√(1-s²/c²) - 1] = mc² -m₀c², where m is the particle’s Relativistic (moving) mass;
then the limit of rKE/ms² as s goes to c is 1, while as s goes to 0 it is ½.

Why hasn’t this obvious theoretical error been seen by others? Maybe the fact that it is
an error is known to most competent physicists, but for some reason the error nevertheless
is made in almost all introductory discussions on the Internet of solutions to
Schrodinger’s equation, and, I imagine, since it is omnipresent in such discussions, it is
still being taught to beginning q.m. students in universities. On the other hand, maybe I
am still making the error that years ago I thought I was making, whatever that is.

 

[mfb]

The phase velocity (what you calculated) is not the group velocity (the propagation of your wave packet, and the change of the expectation value). The phase velocity is nothing you can observe.

For relativistic matter waves, the phase velocity is even faster than the speed of light:
$$v_{phase} = \frac{\omega}{k} = f \lambda = \frac{Eh}{hp} = \frac{\gamma m c^2}{\gamma m v} = c \frac{c}{v} > c$$
For nonrelativistic matter waves, the point of zero energy is arbitrary, and correspondingly the phase velocity is arbitrary as well.

Why hasn’t this obvious theoretical error been seen by others?

If the alternatives are "the theory is wrong" and "I misunderstand the theory", expect the second case. It is not a problem of the theory.

Edit: Typo

 

[fox26]

The phase velocity (what you calculated) is not the group velocity (the propagation of your wave packet, and the change of the expectation value). The phase velocity is nothing you can observe.

For relativistic matter waves, the phase velocity is even faster than the speed of light:
$$v_{phase} = \frac{\omega}{k} = \frac {f}{\lambda} = \frac{Eh}{hp} = \frac{\gamma m c^2}{\gamma m v} = c \frac{c}{v} > c$$
For nonrelativistic matter waves, the point of zero energy is arbitrary, and correspondingly the phase velocity is arbitrary as well.If the alternatives are "the theory is wrong" and "I misunderstand the theory", expect the second case. It is not a problem of the theory.

Your formula v_phase = f/λ is obviously wrong. If you multiply both f and λ by any constant a, your formula gives v_phase as being unchanged. whereas multiplying both the frequency and the wavelength by a of course multiplies the phase velocity ("speed" is better) by a². The correct formula is v_phase = fλ. For s ≈ c, rKE/ms² ≈ 1, so v_phase = fλ ≈ (E/h)(h/p) = E/p ≈ mc²/mc = c.

 

[fox26]

Your formula v_phase = f/λ is obviously wrong. If you multiply both f and λ by any constant a, your formula gives v_phase as being unchanged. whereas multiplying both the frequency and the wavelength by a of course multiplies the phase velocity ("speed" is better) by a². The correct formula is v_phase = fλ. For s ≈ c, rKE/ms² ≈ 1, so v_phase = fλ ≈ (E/h)(h/p) = E/p ≈ mc²/mc = c.

mfb,
I figured out where you got the incorrect v = f/λ, it was from taking k = 2πλ and putting that into v = ω/k. Actually, for that v = ω/k to be correct, k must be what is called, I think, the "wave number", which is = 2π/λ. You did get the v (= fλ) = Eh/hp correct, if f = E/h, the Planck formula, and λ = h/p, which λ is usually defined to be. However, E in the Planck formula is the total energy of a photon, which is also its kinetic energy, since its rest mass energy is zero, as I point out in my thread. The phase velocity in the wave equation I considered (Why do you say the phase velocity (speed) is arbitrary? It is fλ, both of which are defined and figure in the [non-relativistic] wave equation I was talking about--of course, this was in a given inertial reference frame.) is calculated using E = the kinetic energy KE of the particle, whereas you used E = mc², where m is the total mass of the particle, so your E includes the rest mass equivalent energy, which must be subtracted from your mc² to get the kinetic energy, as I did in my thread at the bottom of the second paragraph, thus making f = KE/h less than E/h. Maybe using the total energy as E was why your calculation got the phase velocity for a relativistic matter wave to be greater than the speed of light. Using your E gives an f > 0 for a stationary particle, which is incorrect. (But of course you say that its phase velocity is arbitrary, so maybe you consider that its f is also.) The point of my thread was that the phase velocity of the standard wavefunction for a particle is 1/2 the particles velocity (speed), which might be acceptable if it weren't for the argument, admittedly nonrigorous, that I gave that a particle mPRTCL whose wavefunction's construction I outlined, using the standardly calculated wavefunction for a particle PRTCL in a momentum eigenstate, which wavefunction my thread was criticizing, as a starting point, which mPRTCL had a defined speed of motion of its expectation value of position (which maybe is its group velocity?), identical to the phase velocity of the original PRTCL, which was = only 1/2 the speed s = p/m of the actual particle PRTCL which the wavefunction was supposed to be describing. If you are to effectively criticize my argument, it must be by criticizing my contention that mPRTCL would have a speed s' of motion of the expectation value of its position, as calculated from its wavefunction using the assumed values of m, p, and KE (which s' would be about the expectation value of its speed, as also calculated from that wavefunction), that is s/2, where s is the classically calculated speed of a particle with those values of m, p, and KE (a serious discrepancy for s ≠ 0).

 

[Cthugha]

mfb,
I figured out where you got the incorrect v = f/λ, it was from taking k = 2πλ and putting that into v = ω/k.

It is totally obvious that this was just a typo as the math is correct again afterwards. However, you seem to ignore the main point of mfb's response: The velocity of a moving wave packet and therefore also of a particle is given by the group velocity, not by the phase velocity. This is quite trivial to see in dispersive media or in metamaterials, where the phase velocity may actually be opposite to the Poynting vector of the light field and therefore also to the direction of flow of energy. Look up the definitions of group and phase velocity and you should be able to see instantly, why there is no problem.

 

[mfb]

Copied too much from the previous fraction, it should have been $f \lambda$ of course. Just a typo at this step.

Using your E gives an f > 0 for a stationary particle, which is incorrect. (But of course you say that its phase velocity is arbitrary, so maybe you consider that its f is also.)

It is not incorrect. Nonrelativistic the frequency is arbitrary, relativistic the total energy of the particle is the natural choice and a stationary particle will have a frequency larger than zero.

 

[fox26]

It is totally obvious that this was just a typo as the math is correct again afterwards. However, you seem to ignore the main point of mfb's response: The velocity of a moving wave packet and therefore also of a particle is given by the group velocity, not by the phase velocity. This is quite trivial to see in dispersive media or in metamaterials, where the phase velocity may actually be opposite to the Poynting vector of the light field and therefore also to the direction of flow of energy. Look up the definitions of group and phase velocity and you should be able to see instantly, why there is no problem.

I realized shortly after my first reply to mfb that his was probably, if not exactly a typo, which is a typed error due to just a mechanical typing mistake, at least no more than a temporary lapse in thought, and indicated in my second reply that he had gotten the v = Eh/hp correct, using E in Planck's and de Broglie's sense, as the total energy of the particle, which led to the immediate result he wanted, even though not to disproving my post's thesis. As to my ignoring the main point of mfb's response, concerning group velocity, the last 40% of my second response to mfb was devoted to a reprise of something which was in my original post, an argument which involved something like the group velocity of a particle, more precisely the velocity of the expectation value of position, or the expectation value of the velocity, of a wave packet, instead of just the phase velocity of a plane wave, which argument involving the wave packet I admitted wasn't a rigorous argument, but which strongly supported the main point in my original post, and required rebuttal in order for someone to defeat my main point, all of which you seem to have ignored. It just occurred to me that you wrote your reply quoted above before you had read my second reply to mfb. If so, maybe you have reconsidered, but most of what was in that second reply was in my original post, in somewhat different form. Did you read and understand that? By the way, "phase velocity" of a mathematically defined differentiable wave function, at each space-time point, is reasonably well defined, even if not always physically measurable. "Group velocity", however, while well defined for certain periodic waves, such as periodic EM waves in a waveguide, is not, as far as I know, well defined or even somewhat meaningful in general; that is why I used the above-mentioned expectation values instead of group velocity. I have thought of another line of argument leading to the same conclusion as my original post, and may put that on PF after I get it a little better organized.

 

[PeterDonis]

Thread closed for moderation.

 

[berkeman]

Thread has been cleaned up a bit, and will stay closed. Thank you to all who tried to help the OP.