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Energy, frequency, and Planck's constant

  1. May 14, 2014 #1
    I'm trying to get a deeper understanding of why Planck's constant shows up in so many seemingly unrelated places, and what I'm finding is that in most of the cases I've considered I can reduce it to a relation between some "physical" (depending on interpretation) frequency and an energy. The obvious example is for photons where [itex]E=\hbar\omega[/itex]. Most of the places the constant shows up can be traced back to this relation, with a few exceptions. The two I'm focusing on now are the Hydrogen atom and the electron.

    Using a semi-classical approach and neglecting l > 0 states, you can derive various details about the orbit of the electron in the hydrogen atom. For example,
    r_n = n^2\dfrac{\hbar^2}{me^2}\\
    E_n = \dfrac{e^2}{2a_n}\\
    v_n = \sqrt{\dfrac{2E_n}{m}} = \dfrac{e^2}{\hbar n}\\
    \omega_n = \dfrac{v_n}{r_n} = \dfrac{n\hbar}{ma_0} = \dfrac{me^4}{n^3\hbar^3}\\
    Which gives the result [itex]E_n = \frac{1}{2}n\hbar\omega_n[/itex]. Now this makes a little sense, but it deviates significantly from the usual expression [itex]E=\hbar\omega[/itex]. Where does that factor of 2 come from, and why is there suddenly a dependence on n??

    The next case is just simply with any fermion. The usually quoted formula for the de broglie frequency of a particle used [itex]E=\hbar\omega[/itex]. However, if you look into Zitterbewegung you will find that the actual angular frequency predicted for a fermion is
    so that [itex]E=\frac{1}{2}\hbar\omega[/itex]. So does this factor of two come from the half-integer spin of the electron? Is there some general formula [itex]E=L\omega[/itex] and L is just quantized into units of [itex]\hbar[/itex]?? If so how does that explain the hydrogen atom where L=0 for all those states?

    Related to this: I came across this site (which I won't link to because it's complete bogus) which made the claim that [itex]\hbar[/itex] is just a conversion constant between angular frequencies and energies, similar to the speed of light (i.e. it has no real meaning, we're just using the wrong units). They made the claim that [itex]E=\hbar\omega[/itex] always holds, which I just showed is wrong. However they did seem to have a good point, since Planck's constant acts and is treated very similarly to the speed of light, in that its frequently set to 1 and frequencies and energies are usually interchangeable by some factor of it. I suppose it could be like Boltzmann's constant, which although it isn't a simple conversion factor between temperature and energy it is still a completely redundant and meaningless artifact of our unit system (temperature is energy per degree of freedom).

    Also, I found that any classical cylindrically symmetric system will obey:
    [tex]L\omega = E - \dfrac{m^2}{E}[/tex]
    Where E is the total energy and m is the energy of the system when L=0. So the quantum description of particles matches the classical prediction for a massless rotating system (e.g. photon quantization and electron Zitterbewegung). The hydrogen atom is still a big mystery to me though
    Last edited: May 14, 2014
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  3. May 14, 2014 #2


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    Something is wrong here; [itex]E=\hbar\omega[/itex] is always true.
  4. May 14, 2014 #3


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    Don't blame it on Planck's constant! Your examples are taken from different theories, and relate to different frequencies. It follows from dimensionality arguments alone that E ~ ħω in all the cases, but don't be surprised to see a numerical factor as well.

    Ok, this is not Schrodinger QM, this is semi-classical. The factor of 2 comes from two places: kinetic energy = 1/2 mv2, and the Virial theorem.

    This example, on the other hand, is from relativistic QM, and applies to fermions and bosons alike. But it is not the deBroglie wavelength (which is nonrelativistic), in relativistic QM it is the Compton wavelength that matters, ħ/mc, the scale at which single-particle theory begins to fail. The corresponding frequency mc2/ħ is the frequency of the wavefunction ψ ~ exp(iωt) of a particle at rest, with energy E = mc2. Zitterbewegung results from interference between a positive frequency wave and a negative frequency one, that is, 2ω. So there's the factor of 2 in this case.

    If it's bogus, there's no point in mentioning what it says.
  5. May 14, 2014 #4
    Michael, I too am seeking a clarification of this. If you take the reciprocal 1/h it seems to act as a coupling constant in some way. It appears in the fine structure constant for example and in gravity equations. If this is indeed the case what is being coupled to what?
  6. May 14, 2014 #5
    So you're saying EVERY system with a frequency [itex]\omega[/itex] will have an energy equal to [itex]E=\hbar\omega[/itex]??? That makes no sense
  7. May 14, 2014 #6
    Yes I understand the dimensionality argument, but if Planck's constant shows up in completely independent branches of physics then it's really hard to claim that it is "redundant". My point was that even though it appears all over, it is usually in association with an angular frequency with some physical significance.
    Yes of course, I derived the equation I know literally where the factor of 2 comes from! But this is a non-relativistic system so the classical kinetic energy formula should hold. Surely you're not telling me that if I switched to a purely QM (non-relativistic still) framework this factor of two would dissapear?
    Again, I know mathematically where the factor of 2 comes from but there are interpretations of Zitterbewegung that make that frequency a physically meaningful quantity (unlike the compton wavelength which is just derived from dimensional arguments). So my confusion is what the meaning behind the factor of 2 in the energy/frequency equation is.

    Of course, I could just make up some frequency and claim it adds a factor to the equation and ask why (I suspect this may be what you're accusing me of). However I'm only talking about frequencies that correspond to the movement of some physical system. This is QM, so what is physical and what isn't is always open to interpretation, but I would say Zitterbewegung is the only meaningful frequency intrinsic to an electron
    I only brought it up because I thought that their conclusion was interesting, even if their "derivation" was bogus
  8. May 14, 2014 #7


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    The Schrodinger wavefunction goes like ψ ~ exp(iEt/ħ), so in that sense E = ħω always holds in QM. So I think the conclusion we have to draw is that the semiclassical ω you calculate by setting ω = v/r is not meaningful in a quantum system.
  9. May 14, 2014 #8


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    If it is quantum, it applies; if it is classical, then no.
  10. May 14, 2014 #9
    What you're claiming is just a tautology... Sure, if you define angular frequency as [itex]E/\hbar[/itex] then of course it will hold everywere. However, the quantity [itex]E/\hbar[/itex] does not always have physical significance...

    I mean, in the case of an electron if you just make the transformation [itex]\omega\rightarrow 2\omega[/itex] then yea, the relation will hold. But [itex]\omega[/itex] is the frequency of the Zitterbewegung
  11. May 14, 2014 #10


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    We don't seem to be discussing the same topic. I was not discussing mathematical tautologies.

    You seem to have some question about the meaning of Planck's constant, but apparently I am not understanding your question.
  12. May 14, 2014 #11
    Ok, I still think the semi-classical approach works fine here, but I'll concede that example for now. Let's just focus on the Zitterbewegung of a fermion. I derived the following relation that holds for any relativistic, cylindrically symmetric system:
    [tex]\vec{L} \bullet \vec{\omega} = E - \dfrac{m^2}{E}[/tex]
    In QM r and p are operators and I would need to rederive this to take into account their non-zero commutator, but I suspect the results would be similar enough. This formula, plus the fact that the electron's spin is [itex]\hbar/2[/itex] along any axis would suggest the relation [itex]\frac{1}{2}\hbar\omega = mc^2[/itex] for a fermion which is consistent with Zitterbewegung... It's also consistent with the photon, which as far as I know is the only particle where we can actually measure it's intrinsic frequency.

    Also, this example is purely relativistic so the Schrodinger equation doesn't apply here
  13. May 14, 2014 #12
    Sorry, I'm pretty exhausted and I'm probably not being very clear. My overall interest is in Planck's constant, but this question is focusing more on the angular frequency of quantum systems. That is, what is the angular frequency corresponding to some physical "movement". Clearly in QM this is entirely up to interpretation, but what I was trying to say is that simply defining [itex]E=\hbar\omega[/itex] is circular since I'm trying to define frequency without Planck's constant. Thinking about it though, maybe this just isn't possible since the only way we have to make a frequency measurement is to measure the energy....

    That being said, Zitterbewegung is a theoretically motivated angular frequency (which can never be experimentally measured) that has some kind of meaning. If you treat that as the frequency of an electron though, you end up with a factor of 1/2.

    *The larger problem I'm investigating is the elimination of "fundamental" constants. Boltzmann's constant and the speed of light can easily be eliminated as they are just manifestations of flawed unit systems (time IS distance, temperature IS energy per degree of freedom). The difference between these two cases though, is that the speed of light is a simply linear relationship connecting time to space. Boltzmann's constant on the other hand, doesn't directly relate energy to temperature, and the exact relation depends on the system. I'm trying to establish:
    1) Is there some physical justification for eliminating Planck's constant? i.e. is there some intuitive reason why energy and frequency are simply two forms of the same thing
    2) What is the general relation between frequency and energy? Is it like the speed of light where they are identical after a change of units? Or is it like Boltzmann's constant where frequency isnt quite the same as energy, and the details of the system need to be considered (e.g. is the general formula relating energy to frequency [itex]E=J\omega[/itex] or [itex]E=\hbar\omega[/itex])

    *edit* on another tangent, for a Kerr-Newmann black hole the relation
    [tex]M = \dfrac{J}{a} = J\omega[/tex]
    holds, which is very reminiscent of the QM formula. The close relationship between the Kerr-Newmann metric and the Dirac equation has been shown in a number of papers, so I would be tempted to view this as evidence for [itex]E=J\omega[/itex] being the "general" formula
    Last edited: May 14, 2014
  14. May 14, 2014 #13


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    We can measure the frequency of light and sound - though this is often done indirectly by measuring wave lengths and computing frequency=speed/wavelength.

    For light we have the photon; for sound we have the phonon - both satisfy the Planck relation.
    There are also plasmons, etc.

    But you have to quantize the wave before you can apply the Planck relation.

    "Quantum Physics (Berkeley Physics Course, Volume 4)" by Wichmann discusses many such equivalences in some detail; there is even a convenient table. Your college library should have a copy. Not sure if any of it was intuitive - but it did follow from the physics of the quantum systems.
  15. May 14, 2014 #14
    Can you name any experiment where when you break it down to the quantum level it isn't simply measuring energy? Even if you could, there is definitely no way to measure the frequency of a single photon except to assume the planck relation and measure its energy. You might be able to do this by switching from the classical to quantum realms. For example, you generate some electromagnetic wave by moving some charge. You can predict the frequency of the light purely by classical means, and then measure the energy of the photons produced. But this seems like a loophole that would boil down to just two agreeing energy measurements in the end...

    However, this is all rather irrelevant since photons are unfortunately spin 1! So nobody is debating the planck relation for photons, the issue is how to relate frequencies to energies of other particles/systems. For electronic energy levels in an atom I used semi-classical methods (which may or may not be valid, I need to reevaluate after some sleep) to derive an alternate formula. Also the Dirac equation, relativistic mechanics, and black hole physics suggest a modified planck relation [itex]E=s\hbar\omega[/itex] where s is the "spin".

    To be honest at this point I'm pretty confused as to what exactly this angular frequency corresponds to in all of these cases. Clearly in the non-quantum ones it corresponds to rotating objects. For the hydrogen atom there is no orbital angular momentum in the states considered, so I don't even know what it would mean to talk about an angular frequency! Then when we move to fundamental particles its even more confusing, because as far as I can tell frequency is a purely mathematical object that can't be directly measured (how would you measure the frequency of an electron to determine if it has that factor of 2 or not??)

    Thanks for the reference, I will check that out
  16. May 15, 2014 #15
    It's probably important to remember that ##\omega## can represent any periodic fluctuation in the wave packet associated with the particle. In early QM, ##\omega## represented the fundamental harmonic in an idealized monochromatic photon. Stating ##\omega##n accommodates all of the harmonics or discrete Fourier components of the quantized wave.

    The factor of 2 shifts ##\omega## up an octave while an unstated factor of 1 will still allow you to obtain all ##\omega##n values that the factor of 2 allows. In other words, the factor 2 case is less general.

    The difference between ##\omega## and ##2\omega## can be determined experimentally by interferometry. But of course, you'd need to know which of the those 2 possibilities applies to your reference wave.
  17. May 15, 2014 #16
    On a similar topic, does anyone have any rational why planck's constant appears in the uncertainty principle?? I know all the math behind it, but qualitatively I'm stumped. So far I've been able to reduce every appearence to hbar to some form of the planck relation, except this one. If energy and frequency are the same thing related by Planck's constant, how does that lead to the uncertainty principle??
  18. May 15, 2014 #17


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    Your question seems equivalent to the question why the Schrödinger equation and the canonical commutation relation involve the same hbar. Or speaking from a symmetry point of view, why the generators of translations in time and space involve the same constant.

    Sakurai comments on this in his section about the time evolution operator. He basically argues that hbar wouldn't disappear in the classical limit if you had two different constants. I guess you can justify it more directly if you use relativistic reasoning.
  19. May 15, 2014 #18
    Yes, that's basically my source of confusion. Why would the same constant appear in (what at least appears to be) two completely unrelated equations? That classical limit argument is interesting, but the hbar->0 limit is fundamentally flawed and doesn't reproduce classical physics in all situations. The best example is in thermodynamics, where hbar->0 has some drastic effects observable at classical scales
  20. May 15, 2014 #19


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    You seem to accept that hbar has to be the same for different components of position and momentum. Momentum is the generator of translations in space and it's components are connected by Galilei transformations. Relativistically, the Lorentz transformation connects space and time. So I think we should expect a similar argument.

    As usual, things may get a bit complicated because there's no well-defined relativistic single particle theory.

    Taking hbar to zero is not the only way to get a classical limit. If Δx is small compared to the scale of changes in the potential V(x), Ehrenfest's theorem yields the classical equations of motion. If the hbars weren't equal, they would occur in these equations.
  21. May 15, 2014 #20
    Do you have a reference for this? I want to read more about it

    *edit* nevermind I found the comment in Sakurai, now I just have to work out the details myself
  22. May 15, 2014 #21
    Plancks constant represents the minimum amount of action. Consider a system with energy E1 at t1 and E2 at t2. The change in action action would be E2t2-E1t1 or
    (E1 + ΔE)(t1 + Δt) - E1t1 = ΔEt1 + ΔEΔt + E1Δt
    and because we are always free to set E1=0 and t1=0 this reduces to
    ΔEΔt = h
    Although how this relates to the uncertainty principle I have absolutely no idea!
  23. May 15, 2014 #22
    Ok you are right. I took the 5 different places hbar appears in classical QM and assumed they all had different values. The x/p commutator one needs the same value as the translator operator, the time evolution operator needs the same one as the schrodinger equation, and the shrodinger equation one needs to be the same as the x/p commutator in order to be consistent with classical physics.

    The one I left out is the planck relation. It's easy enough to show that if you take some constant energy wave-function with a characteristic frequency, and apply the time evolution operator you arrive at the conclusion that those two constants must be equal as well.

    However, this is all classical QM which can't describe an electron. Granted, I think this rules out the hydrogen atom situation, but I think in a relativistic context the planck relation for an electron would still have that factor of 2. In the above I showed how to prove the equality between the hbar in the planck relation and the hbar anywhere else. However, to do so I made the assumption that a 2π "rotation" of the time evolution operator would return it to its original value. For fermions though, you need a 4π spatial rotation to arrive at the original wave function. It still seems to me that once you make this relativistic and space=time that 4π could sneak in and add that factor of 2
  24. May 16, 2014 #23
    I thought a bit more about this. Perhaps it is because any measurement will involve some action, so it is because we cannot disturb the system in an arbitrarily small way. Any thoughts?
  25. May 16, 2014 #24
    There is a fairly recent but growing study of the relationships between

    1. A degree of commutativity (or non-commutativity) and non-locality
    2. A degree of commutativity (or non-commutativity) and angular dependence of fields
    3. A degree of commutativity (or non-commutativity) and ##\hbar##
    4. How #1 through #3 affect the form of a wave

    Those studies tend to be from mathematicians rather than physicists. Some example papers:

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