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In what sense do within-H-atom electrons "move" at ~1/137c?

  1. Aug 6, 2015 #1
    Hello,

    1.
    I read here and here that in the hydrogen atom, electrons move at approximately ~1/137c. In the first link they speak of "zipping around the nucleus", presumably figuratively, because it is often stressed that QM has superseded the earlier model of electrons flying around. Instead we are to think of a probability distribution of both location and velocity - right?

    But is one to think of the electron as being that probability distribution? If yes, then what does that ~1/137c apply to? It seems like the distribution itself should not be thought of as moving - at least not in an atom unaffected by exterior influences. If no - if the electron is something to which the distribution applies, but is not the distribution itself - then how is it not true that it "zips around" in the classical sense?

    2.
    Here someone writes (emphasis added) "The ground state has momentum zero, so the electron doesn't move at all in any classical sense. Excited states have a non-zero angular momentum, but you shouldn't think of this as a point like object spinning around the atom." How does that relate to the nonzero ~1/137c velocity mentioned? I was assuming that the latter velocity refers to an isolated atom, with non-excited electrons. But here it is said that momentum, and hence velocity (for the electron does have mass), is zero. Or does that refer to the (nill) momentum of the probability distribution?

    Thanks in advance.
     
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  3. Aug 6, 2015 #2

    bhobba

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    I had a look at the first link. They were speaking about the Bohr model which has been well and truly superseded. Forget about it - its of zero value.

    Thanks
    Bill
     
  4. Aug 6, 2015 #3

    DrDu

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    That's not true. Only the expectation value of the momentum vanishes which is quite trivial as the vector can point in any direction. But there is still a distribution of possible p values.
    The average kinetic energy in a H atom is ##T=\langle p^2/2m\rangle ## and equals 13.6 eV. So ##\sqrt{2T/m}## would be the average velocity of the electron. You can check whether this is of the order of c/137 (there are different estimates of the mean velocity which differ by a factor of the order one, so you shouldn't expect to obtain exactly c/137 from this calculation.)

    PS: After correcting a small error, this seems to be the correct average to use to get the 137.
     
    Last edited: Aug 6, 2015
  5. Aug 6, 2015 #4
    The fine-structure constant is 1/137. Is there a connection between that and the mean velocity of the lowest energy electron in the H atom being c/137?
     
  6. Aug 6, 2015 #5

    DrDu

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    Yes, the fine structure constant showed first up in the analysis of the hydrogen spectrum.
     
  7. Aug 6, 2015 #6

    Avodyne

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    The expectation value of the kinetic energy of the hydrogen atom (in its ground state) is one Rydberg, with a numerical value of 13.6 eV. This can be expressed as ##{\rm Rydberg}=\frac12 \alpha^2 m c^2##, where ##m## is the electron mass (actually the reduced electron mass), and ##\alpha = e^2/4\pi\epsilon_0 \hbar c## is the fine structure constant. Since the expectation value of the kinetic energy is ##\frac12 m\langle v^2\rangle##, we see that ##\langle v^2\rangle^{1/2}=\alpha c##.
     
    Last edited: Aug 6, 2015
  8. Aug 6, 2015 #7
    Thanks all for taking the time to reply. However, I don't think my main question has been answered yet. It's not so much about the value of the fine-structure constant, it's about the meaning of velocity in this situation.

    Bill, I realize that the Bohr model is outdated. But the question is this: given that we have to drop classical intuitions along with the Bohr model, then when the speed of (within-atom) electrons is mentioned, should I interpret that entirely independently of classical intuitions about velocity? Because the electrons do not 'whizz around'. So what is it they do at the velocities mentioned?
     
  9. Aug 6, 2015 #8

    Avodyne

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    There is a probability distribution ##\rho(\vec v)## for the velocity. This means that if we have a hydrogen atom in its ground state, and we measure the velocity of the electron, the probability that the measured velocity is within ##d^3 v## of a particular velocity ##\vec v## is ##\rho(\vec v)d^3 v##. There is an explicit formula for ##\rho(\vec v)##, which is
    [tex]\rho(\vec v)={(8/\pi^2)\alpha^5 c^5\over\bigl(\vec v^2+\alpha^2 c^2\bigr)^4}[/tex]
    What words you want to attach to this set of facts is up to you.
     
    Last edited: Aug 6, 2015
  10. Aug 6, 2015 #9

    bhobba

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    In QM the electrons in the hydrogen atom have no velocity - the Bohr model is incorrect. In fact since they are not observed you cant say they are doing anything.

    What others have been talking about is average velocity which is understood via Ehrenfests Theorem:
    http://www.physics.drexel.edu/~bob/PHYS517/Ehrenfest.pdf

    Thanks
    Bill
     
    Last edited: Aug 6, 2015
  11. Aug 7, 2015 #10

    DrDu

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    But velocity is a well defined operator. So it makes perfect sense to talk about the distribution and mean of the velocity.
     
  12. Aug 7, 2015 #11
    Thanks all, that's helpful. I also belatedly found a related thread where this seems to make sense to me:
    Combined with some info on 'Fermi velocity', I get this as a summary-to-myself: given QM, there's no thinking in terms of velocity of a only-particle-like individual electron within the atom. However because the electron does evidently not drop to the nucleus, it must have kinetic energy, hence velocity, which is born out by relevant equations. But this is a property of the probabilistic particle/wave phenomenon we call 'electron'. And since it's probabilistic, 'the' electron really has a velocity distribution, and a value like 1/137c for the electron in the H atom, is the average of that distribution.

    Is that right, or still way off the mark? Pardon me if I'm looking for a too qualitative understanding. I realize this has its limits.
    Is this practically feasible? Maybe it's routine - I've no idea. Would that then again mean the average velocity of that electron?
     
  13. Aug 7, 2015 #12

    bhobba

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    Of course its a well defined operator. And I agree that it has a well defined mean and statistical distribution. What it doesn't have is an actual velocity until it observed to have it.

    Thanks
    Bill
     
  14. Aug 7, 2015 #13

    Avodyne

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    Just to be precise, it's the "root mean square" average. The average of the velocity itself is zero (because all directions for the velocity are equally likely), but the average of the velocity-squared is ##(\alpha c)^2##. In equations,
    [tex]\int d^3v\,\rho(\vec v)\vec v = 0, \qquad \int d^3v\,\rho(\vec v)\vec v^2 = (\alpha c)^2.[/tex]
    It is not practically feasible to measure the velocity of an electron in an atom, AFAIK. The discussion here has been in terms of an idealized measurement theory that assumes that any observable (represented by a hermitian operator) can be measured.
     
  15. Aug 9, 2015 #14
    Thanks Avodyne, and everyone - much clearer now :)
     
  16. Aug 9, 2015 #15
  17. Aug 9, 2015 #16
    That video is awfully vague in its description. I'm trying to find some background on it.
    I found this so far: http://phys.org/news/2008-02-electron.html

    I wonder why it appears to me that the orbital levels of electrons in atoms is due to the electrons and not the nucleus itself? Does it appear that way or is that an artifact of the imaging technique...
     
    Last edited: Aug 9, 2015
  18. Aug 10, 2015 #17
    I dunno if we can measure the velocities of electrons in atoms in practice. But in principle we can imagine ways to do it. For example, say we have a hydrogen atom: an electron orbiting a proton. Now we very very quickly remove the proton, leaving just the electron undisturbed in whatever state it was in previously. Since there is no proton anymore the electron is now free, and will fly off with some velocity. We can measure this velocity by letting the electron hit a wall and seeing how much energy is deposited (i.e. by measuring the kinetic energy of the now-free electron). If we repeat this experiment we will not get the same electron velocity every time; rather we will get a distribution of velocities which will agree with the formula given above in this thread in post #8.
     
  19. Aug 11, 2015 #18
    All these misunderstandings arise only because we obstinately want to believe that in QM there still are things like pointlike particles. As long we will continue to get stuck in this classical aristotelian worldview and will not hammer out of our heads once and forever that there is no such thing, we will permanently remain confused by words. Simply accept what QM is telling us. There are no particles with a position or momentum in the classical sense, but eventually only forces and events taking place in something we naively perceive as "space" and "time". The rest is only an anthropomorphic construction of human mind.
     
  20. Aug 11, 2015 #19
    Agree with much of what has been said. Hope this addition helps... There is no true velocity, as that would only be the case for restricted "motions", e.g. an orbit. Just thinking in terms of orbits, of course, is inherently non-quantum-mechanical. As said, the energy, however, (particularly its variations) DO related to the fine structure constant. The variations in the orbits (with the s, p , d, f.. early atomic levels being sharp, principal , diffuse, or fine , see wiki https://en.wikipedia.org/wiki/Electron_configuration .. So, the variations of energies in the orbital configureations, and particularly their transitions can relate to these quantum terms...
    but velocity, i do not think so.
     
  21. Aug 14, 2015 #20
    Such a good answer, with a bonus philosophical comment. Very satisfying.
     
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