Does an electron have a quantum phase frequency?

In summary: So clearly, adding a constant does change the physics in some ways.And clearly, any "spectroscopic" phenomenon (like the absorption or emission of light) that depends only on the difference between two energy levels is necessarily invariant under such a transformation.This is not correct. For example, if you add a constant to the potential, the energy of the photon will not be the same, and the photon will be absorbed or emitted.
  • #1
Stationary solutions to the Schrödinger equation factor into a spatial part, e.g. atomic and molecular orbitals, and a temporal part that gives the phase rotation frequency. It is often assumed that adding a constant to the potential leaves the physics unchanged. And clearly, any "spectroscopic" phenomenon (like the absorption or emission of light) that depends only on the difference between two energy levels is necessarily invariant under such a transformation. But a change in energy implies a change in frequency; this is forced by E = h𝜈. So allowing a redefinition of the potential energy means that one can change the phase frequency to whatever one likes.

For example, one could change it to be so high that the energy associated with the electron is sufficient to cause a black hole to form. Since we don't see electrons spontaneously turning into black holes, doesn't this imply some practical upper limit?

Also, we can choose the potential so that the phase frequency is zero. In that case, wouldn't interference effects go away? But we don't see that happening either, I think.

So the question is, does a real electron have a single definite quantum phase frequency? Or is that just something that has no physical reality at all, and can be assumed to be anything you want?

(Note that I am NOT asking about absolute quantum phase, just the frequency. It's easy to prove that one can add constants to potentials in classical mechanics and classical EM with zero effect, so I accept that; the question is about to what extent that is also true in QM. The Aharonov-Bohm effect shows that potentials can have SOME effect in QM, but perhaps only in ways that are themselves gauge-invariant. So my question could also be viewed as being about the limits (if any) of such invariance.)
 
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  • #2
H_A_Landman said:
Stationary solutions to the Schrödinger equation factor into a spatial part, e.g. atomic and molecular orbitals, and a temporal part that gives the phase rotation frequency. It is often assumed that adding a constant to the potential leaves the physics unchanged. And clearly, any "spectroscopic" phenomenon (like the absorption or emission of light) that depends only on the difference between two energy levels is necessarily invariant under such a transformation. But a change in energy implies a change in frequency; this is forced by E = h𝜈.
This is the de Broglie frequency, which is something very different.
H_A_Landman said:
So allowing a redefinition of the potential energy means that one can change the phase frequency to whatever one likes.
This is not correct.
H_A_Landman said:
For example, one could change it to be so high that the energy associated with the electron is sufficient to cause a black hole to form.
This is wrong. You can't turn an electron into a black hole simply by changing your zero-point energy.
H_A_Landman said:
Since we don't see electrons spontaneously turning into black holes, doesn't this imply some practical upper limit?
No. It means you are seriously misinterpreting the physics.
H_A_Landman said:
Also, we can choose the potential so that the phase frequency is zero. In that case, wouldn't interference effects go away? But we don't see that happening either, I think.
This is also wrong. The phase frequency, generally, depends on the state of the electron.
H_A_Landman said:
So the question is, does a real electron have a single definite quantum phase frequency? Or is that just something that has no physical reality at all, and can be assumed to be anything you want?
An electron in a stationary state effectively has a well-defined phase frequency associated with the energy of that state. A free electron is more likely to be a wave-packet and not have a single well-defined energy.
H_A_Landman said:
(Note that I am NOT asking about absolute quantum phase, just the frequency. It's easy to prove that one can add constants to potentials in classical mechanics and classical EM with zero effect, so I accept that; the question is about to what extent that is also true in QM. The Aharonov-Bohm effect shows that potentials can have SOME effect in QM, but perhaps only in ways that are themselves gauge-invariant. So my question could also be viewed as being about the limits (if any) of such invariance.)
I'm not sure what you mean by this. In general, I believe you have seriously misunderstood the physics you have read.
 
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  • #3
H_A_Landman said:
It is often assumed that adding a constant to the potential leaves the physics unchanged.
I think you need to give more specific references. I don't think you will find one that makes this claim as you state it. For one thing, there are common scenarios in which it is obviously false: for example, if you are using QM to model the hydrogen atom, you can't add an arbitrary constant to the Coulomb potential, because the potential has to vanish at infinity.
 
  • #4
PeterDonis said:
I think you need to give more specific references. I don't think you will find one that makes this claim as you state it. For one thing, there are common scenarios in which it is obviously false: for example, if you are using QM to model the hydrogen atom, you can't add an arbitrary constant to the Coulomb potential, because the potential has to vanish at infinity.
I'm not sure what you mean here. Taking the ground state of H as the zero of energy instead of infinitely separated proton + electron shouldn't change anything to the calculation.
 
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  • #5
H_A_Landman said:
For example, one could change it to be so high that the energy associated with the electron is sufficient to cause a black hole to form. Since we don't see electrons spontaneously turning into black holes, doesn't this imply some practical upper limit?

Also, we can choose the potential so that the phase frequency is zero. In that case, wouldn't interference effects go away? But we don't see that happening either, I think.

So the question is, does a real electron have a single definite quantum phase frequency? Or is that just something that has no physical reality at all, and can be assumed to be anything you want?
The complex phase of the wave function is arbitrary, ##\psi## and ##e^{i \alpha} \psi## represent the same physical state, so shifting ##E_0## will have no physical effect on ##e^{-i (E-E_0) t/\hbar} \phi## (for ##\phi## an eigenstate of the Hamiltonian). All interference effects will come from differences in phases, so ##E_0## disappears from any calculation.
 
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  • #6
DrClaude said:
Taking the ground state of H as the zero of energy instead of infinitely separated proton + electron shouldn't change anything to the calculation.
It will change the numerical eigenvalue associated with each eigenstate, correct? If I take the ground state of ##H## as the zero of energy, the eigenvalue of that state is zero; but if I take infinite separation as the zero of energy, the eigenvalue of the ground state is minus whatever the binding energy is (e.g., -13.6 eV for the hydrogen atom).
 
  • #7
Whenever I write down any Hamiltonian I always include my rest mass energy as an additive constant (this is the Hutch Hamiltonian). My calculated results all seem to be equivalent,
although the additive constant continues to change with time.
/
 
  • #8
PeterDonis said:
It will change the numerical eigenvalue associated with each eigenstate, correct? If I take the ground state of ##H## as the zero of energy, the eigenvalue of that state is zero; but if I take infinite separation as the zero of energy, the eigenvalue of the ground state is minus whatever the binding energy is (e.g., -13.6 eV for the hydrogen atom).
It does, but that changes no observable result.
If we are to be completely precise, we wouldn’t say “because the [Coulomb] potential has to vanish at infinity” we would say that its first derivative has to vanish at infinity.
 
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  • #9
Nugatory said:
It does, but that changes no observable result.
It might be helpful to the OP to at least run through a brief explanation of why that is.
 
  • #10
PeterDonis said:
It might be helpful to the OP to at least run through a brief explanation of why that is.
I’m writing that post right now…. I have to admit that I found this digression more interesting than original question :smile: so started in on it first
 
  • #11
PeterDonis said:
I think you need to give more specific references. I don't think you will find one that makes this claim as you state it. For one thing, there are common scenarios in which it is obviously false: for example, if you are using QM to model the hydrogen atom, you can't add an arbitrary constant to the Coulomb potential, because the potential has to vanish at infinity.
Well, it's so obvious that's almost never stated in textbooks. It gets relevant, however, when it comes to relativistic QFT, where you have to get rid of the diverging "zero-point energy" by some choice of the zero-point of energy. That's no physical but a mathematical problem, i.e., by how in the "canonical-quantization procedure" you order the field operators in products, which are ill-defined anyway, because one cannot naively multiply operator-valued distributions.
 
  • #12
PeterDonis said:
It will change the numerical eigenvalue associated with each eigenstate, correct? If I take the ground state of ##H## as the zero of energy, the eigenvalue of that state is zero; but if I take infinite separation as the zero of energy, the eigenvalue of the ground state is minus whatever the binding energy is (e.g., -13.6 eV for the hydrogen atom).
Yes, but this difference in setting the "zero point of energy" doesn't change any observable fact about the hydrogen atom. The binding energy will be always 13.6 eV, no matter whether you make the eigenvalue of the groundstate 0 (an unconventional choice, because usually you set the vacuum state's energy to 0) or -13.6 eV.
 
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  • #13
PeroK said:
This is the de Broglie frequency, which is something very different.
The phase evolution is given by ##\psi(t) = e^{-iEt/\hbar}##. This is pretty much identical to ##E=h\nu##. How is that "very different"?
 
  • #14
PeterDonis said:
I think you need to give more specific references. I don't think you will find one that makes this claim as you state it. For one thing, there are common scenarios in which it is obviously false: for example, if you are using QM to model the hydrogen atom, you can't add an arbitrary constant to the Coulomb potential, because the potential has to vanish at infinity.
I think you're saying that one is required to fix the gauge to the Coulomb/radiation gauge. But isn't that the same thing as saying that there is no gauge invariance for the potential?

Consider an atom, or some simple electron experiment like a 2-slit interference or a charge-mass-ratio one, sitting inside a metal cage that can be charged to various voltages. Most (all?) physics of the system doesn't change. For example, the colors of light emitted by the atom do not change. That's a gauge invariance, yes? But the phase frequency imputed to the electrons has to change, because their potential energy changes with the potential.
 
  • #15
H_A_Landman said:
I think you're saying that one is required to fix the gauge to the Coulomb/radiation gauge.
I'm not talking about QED, I'm talking about the ordinary non-relativistic QM model of the hydrogen atom. The Coulomb potential in that model is just a function ##V(x)## of position; it's not obtained from Maxwell's Equations. There is no gauge choice involved.
 
  • #16
Of course the Coulomb potential is obtained from Maxwell's equations, and also non-relativistic QM is gauge invariant. Otherwise it wouldn't make any physical sense!
 
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  • #17
H_A_Landman said:
The phase evolution is given by ##\psi(t) = e^{-iEt/\hbar}##. This is pretty much identical to ##E=h\nu##. How is that "very different"?
One is QM. The other is the de Broglie matter wave. Very different things.
 
  • #18
Your answer is so terse that I can't even tell which one you claim is which, or how they supposedly differ. Care to elaborate?
 
  • #19
H_A_Landman said:
Care to elaborate?
It should be obvious.
 
  • #20
If you are intentionally trying to be as unhelpful as possible, you are succeeding.
 
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  • #21
H_A_Landman said:
If you are intentionally trying to be as unhelpful as possible, you are succeeding.
See post #2.
 
  • #22
PeroK said:
See post #2.
OK, let's make it simple. Take a hydrogen atom outside of a VDGG sphere, and another one inside the sphere, and charge the sphere to (say) +1 MV. Do you believe that the electrons in the two atoms have the same phase frequency, or different phase frequencies?
 
  • #23
H_A_Landman said:
OK, let's make it simple. Take a hydrogen atom outside of a VDGG sphere, and another one inside the sphere, and charge the sphere to (say) +1 MV. Do you believe that the electrons in the two atoms have the same phase frequency, or different phase frequencies?

Our beliefs here do not matter. Observables do.
Are these two electrons part of the same system (i.e. do they interact in a way that affects the result) ?
  1. If not then it does not matter because the problems separate and they can each have arbitrary additive constants added to their energies.
  2. If these electrons are part of the same system and interact, then they do not have distinguishable eigenenergies and your question is moot.
  3. If you wish to consider the non-interacting particles part of the same system you could probably add arbitrary constants to their potentials but the book-keeping entailed would make this at best ill-advised.
When music is broadcast over the radio by modulating it onto a 100MHz signal, we only recover the music.
 
  • #24
H_A_Landman said:
So the question is, does a real electron have a single definite quantum phase frequency? Or is that just something that has no physical reality at all, and can be assumed to be anything you want?
It looks like the following may be relevant:

https://en.wikipedia.org/wiki/Matter_wave_clock
https://en.wikipedia.org/wiki/Zitterbewegung

So can something like Zitterbewegung frequency observed experimentally? The frequency is very high, so one can have doubts about experimental articles quoted in the Wikipedia articles above ((2008-07-01). "A Search for the de Broglie Particle Internal Clock by Means of Electron Channeling". Foundations of Physics. 38 (7): 659–664. doi:10.1007/s10701-008-9225-1 , "A Clock Directly Linking Time to a Particle's Mass", Science 1 February 2013: Vol. 339 no. 6119 pp. 554–557 doi:10.1126/science.1230767 )

In my articles (see references below) I propose a model of quantum particles as plasma-like collections of a large number of particles and antiparticles. In this model, the Zitterbewegung frequency plays a role of a “natural frequency”, rather than a frequency of some “internal clock”. A pendulum at rest can be an illustration: it has a natural frequency, but this frequency is not manifested at rest.

Eur. Phys. J. C (2013) 73: 2371

Entropy 2022, 24(2), 261

Quantum Rep. 2022, 4(4), 486-508
 
  • #25
hutchphd said:
the problems separate and they can each have arbitrary additive constants added to their energies.
OK, so your answer to my original question would be no, an electron does not have a definite phase frequency, because we can add an arbitrary constant to the energy? PeroK (in post #2) says that is not correct.
 
  • #26
I will not argue with @PeroK by proxy. Read carefully what he says. DeBroglie is not Schrodinger.
Your original question makes no sense to me because the Energy Eigenvalue is associated with the atomic system and not the electron itself. Similarly for the spherical conductor at high potential. The free electron is slightly more cumbersome because of normalization but this is an artifact.
Suppose (for clarity) you are correct about the frequency, exactly how does this make the universe different?
 
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  • #27
hutchphd said:
DeBroglie is not Schrodinger.
Everyone keeps saying that, but I don't see the math. De Broglie gives E=h𝜈. Schrödinger gives an angular frequency which also reduces to E=h𝜈, except the E includes the potential energy. So those seem nearly identical to me. What equations justify your statement?
hutchphd said:
Suppose (for clarity) you are correct about the frequency, exactly how does this make the universe different?
Well, first of all, I haven't stated a position on the answer to the original question, so I can't be correct or incorrect yet (unless the question itself is total nonsense). I'm more trying to understand whether there is a mainstream consensus on this point, and if so what its justifications are.

But yeah, the central question is how could you probe this experimentally. As far as I know, there is no conclusive evidence either way.

One proposal was that you could test for the zitterbewegung frequency by shooting an electron through a channel in a crystal at various speeds; if the speed of the electron was such that the crystal period length (along the channel) was an integer multiple of the zitterbewegung wavelength, you might see a resonance effect. Knowing the wavelength and the speed would give you the frequency. Some experiments of this nature have been done (as in M. Gouanère et al., Annales de la Fondation Louis de Broglie, 30, 109 (2005)), and maybe even some resonances found, but it's not clear (to me) whether they give a definitive answer yet.

Then there's a class of unified theories that predict alterations in unstable elementary particle lifetimes by EM potentials. In those, the phase frequency acts like the particle's "local clock", so the decay rate is proportional to the frequency. Some aspects of this are easily testable with current technology; for example it predicts that a muon in a static electric potential of 1 MV will have its lifetime altered by about 0.95%. That experiment was first proposed by David Apsel around 1978-79, but it hasn't been performed yet. (I am pretty sure of that because I have been trying to be the first one to perform it; see my proposal for details, and an analysis of the various theories in this class, and a discussion of the gauge invariance issues they raise. We don't need to get into that here.)

Most experiments depend on a difference of energies or frequencies. This automatically gives them an invariance to absolute energy or frequency, whether or not the universe itself has that invariance. Finding tests that depend on absolute energy or frequency is hard, but, as the above examples demonstrate, perhaps not impossible.
 
  • #28
I would posit that any measurement is in the end relative and ratiometric.
To measure a force you compare it to another force. Same with energy. Resonance from periodicity usually compares two lengths. I know nothing of the predictions you mention but neither one is very new and I find them each very implausible and know of no general active interest.
I am expert on scattering resonances from periodicity but not so much Zitterbewegung..Sorry
 
  • #29
The total energy, represented by the Hamiltonian, is defined only up to an arbitrary additive constant, which is in no way observable. So the phase velocity of a Schrödinger wave for a free particle depends in this arbitrarily chosen additive constant, but this is without any physical consequences. I've no clue, why everybody is pointing out that de Broglie were different from Schrödinger. Of course, it was Schrödinger who discovered wave mechanics as one representation of modern quantum theory, while de Broglie didn't.
 
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  • #30
vanhees71 said:
The total energy, represented by the Hamiltonian, is defined only up to an arbitrary additive constant, which is in no way observable.
Sounds reasonable.
vanhees71 said:
usually you set the vacuum state's energy to 0
Also sounds reasonable but seems to be in tension with the previous statement.

One could argue that the "arbitrary additive constant" is actually observable, as the maximum energy that can be extracted from a system is the energy of the system as measured from the vacuum state's energy.

Previously, I quoted "A Clock Directly Linking Time to a Particle's Mass", Science 1 February 2013: Vol. 339 no. 6119 pp. 554–557 doi:10.1126/science.1230767 . Among other things, they mention the Compton frequency [itex]\omega_0=mc^2/\hbar[/itex] and note: "In principle, a Compton-frequency clock could be built by annihilating a particle-antiparticle pair and counting the frequencies of the generated photons." This remark suggests that there is some preferred choice of the "additive constant". Indeed, it seems natural to assign the energy of [itex]\hbar \omega[/itex] to a photon.

The above is not meant to criticize anybody. It seems to me that the issue of reality of the Compton frequency is controversial.
.
 
  • #31
H_A_Landman said:
Everyone keeps saying that, but I don't see the math. De Broglie gives E=h𝜈. Schrödinger gives an angular frequency which also reduces to E=h𝜈, except the E includes the potential energy. So those seem nearly identical to me. What equations justify your statement?

Let's say we have a wave packet representing the probability amplitude of finding an electron within some region. The evolution of the probability and of any expected value is independent of any constant you want to add to the Hamiltonian. That's what non-relativistic quantum mechanics actually says.
 
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  • #32
akhmeteli said:
Sounds reasonable.

Also sounds reasonable but seems to be in tension with the previous statement.
It's not. It's just pointing out the usual "arbitrary choice" of the additive constant.
akhmeteli said:
One could argue that the "arbitrary additive constant" is actually observable, as the maximum energy that can be extracted from a system is the energy of the system as measured from the vacuum state's energy.
The arbitary constant is not observable. All physics is invariant under the arbitrary choice of the constant.
akhmeteli said:
Previously, I quoted "A Clock Directly Linking Time to a Particle's Mass", Science 1 February 2013: Vol. 339 no. 6119 pp. 554–557 doi:10.1126/science.1230767 . Among other things, they mention the Compton frequency [itex]\omega_0=mc^2/\hbar[/itex] and note: "In principle, a Compton-frequency clock could be built by annihilating a particle-antiparticle pair and counting the frequencies of the generated photons." This remark suggests that there is some preferred choice of the "additive constant". Indeed, it seems natural to assign the energy of [itex]\hbar \omega[/itex] to a photon.

The above is not meant to criticize anybody. It seems to me that the issue of reality of the Compton frequency is controversial.
.
The frequencies of the generated photons are given by energy-momentum conservation, and thus the arbitrary additive constant simply cancels, because it's on both sides of the "balance equation".
 
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  • #33
vanhees71 said:
It's not. It's just pointing out the usual "arbitrary choice" of the additive constant.
So we disagree.
vanhees71 said:
The arbitary constant is not observable. All physics is invariant under the arbitrary choice of the constant.
Again, one can have a different opinion, arguing that observations indicate what maximum energy one can extract.
vanhees71 said:
The frequencies of the generated photons are given by energy-momentum conservation, and thus the arbitrary additive constant simply cancels, because it's on both sides of the "balance equation".
This is one way to look at this matter. However, one can argue that it would be natural to assign the same energy to each photon with the same frequency and to assign a small energy to a photon with a low frequency. However, such approach is not irrefutable either: there is always the problem of zero-point energy.

Another thing. The OP did not ask if a real electron has a definite energy, he asked if it has a definite frequency. I would say the two photons after electron-positron annihilation have definite frequencies, which seems to imply that an electron has a definite frequency too.

Again, I am not saying that your opinion is wrong, and the opposite opinion is correct, there are arguments in favor of both opinions. This is why I believe this issue is controversial.
 
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  • #34
akhmeteli said:
Again, I am not saying that your opinion is wrong, and the opposite opinion is correct, there are arguments in favor of both opinions.
Physics is about experimental results, not about arguments and opinions.
 
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  • #35
PeroK said:
Physics is about experimental results, not about arguments and opinions.
With all due respect, this sounds like an opinion without arguments.

There are always areas of physics where there is not enough clarity at the moment, so there is place for debates, opinions, and arguments. Sometimes it happens because there are not enough experimental results. For example, at the beginning of the XX century there were a lot of debates on whether atoms exist. There is a dearth of experimental results on the issue discussed in this thread (and the reason is the Compton frequencies are very high). Should we rely on the results of "A Search for the de Broglie Particle Internal Clock by Means of Electron Channeling". Foundations of Physics. 38 (7): 659–664. doi:10.1007/s10701-008-9225-1 ? I don't know.

And sometimes there is a place for debates, opinions, and arguments even when there are a lot of experimental results, as we see in the case of interpretations of quantum theory.
 
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