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Reconciling DeBroglie and Dirac?

  1. Sep 16, 2012 #1
    So in 1927 Davisson and Germer showed that electrons shot at a crystal do indeed have a DeBroglie wavelength inversely proportional to their momentum (h/p)? That would mean that their wavelength is a function of their velocity, the voltage used to accelerate them, etc. But I seem to remember that Dirac showed that electrons always move at c, and the macroscopic velocity we measure is just the average of all those back and forth trips over some finite period of time.

    Are these two ideas compatible?

    Math Questions: I am interested in doing my own calculations, so what velocity do the electron wavelets have after being scattered by the atoms? Do they interact with only the outermost electrons of the atoms, or also some of the inner ones, or even the nucleus?
  2. jcsd
  3. Sep 16, 2012 #2


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    That doesn't sound like anything I've ever read. Maybe someone who knows more about the early history of QM than I do will recognize what this refers to.

    Maybe you're thinking of Brian Greene and his "everything moves through spacetime at speed c" meme?
  4. Sep 16, 2012 #3
    Nope, not Brian Greene. I'll look back through my Dirac books to see if I can find it again.

  5. Sep 16, 2012 #4


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    I'm sure all of the "great minds" came up with some ideas that didn't pan out. :wink:
  6. Sep 16, 2012 #5

    Doc Al

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    Are you talking about the infamous "Zitterbewegung"?
  7. Sep 16, 2012 #6
    OK, here it is: P. A. M. Dirac, “The Principles of Quantum Mechanics”, Fourth Edition 1957-58, Clarendon Press, Oxford. (The first edition was in 1930.) Chapter XI – Relativistic Theory of the Electron, Section 69 – The motion of a free electron, p. 261 and following.

    “It is of interest to consider the motion of a free electron in the Heisenberg picture corresponding to the above theory and to study the Heisenberg equations of motion. These equations of motion can be integrated exactly, as was first done by Schrodinger. (…)
    As Hamiltonian we must take the expression which we get as equal to when we put the operator on ψ in (10) equal to zero, i.e.

    We see at once that the momentum commutes with H and is thus a constant of the motion. Furthermore, the -component of the velocity is

    This result is rather surprising, as it means an altogether different relation between velocity and momentum from what one has in classical mechanics. (…) …we can conclude that a measurement of a component of the velocity of a free electron is certain to lead to the result ±c (italics Dirac’s). This conclusion is easily seen to hold also when there is a field present.

    Since electrons are observed in practice to have velocities considerably less than that of light, it would seem that we have here a contradiction with experiment. The contradiction is not real, though, since the theoretical velocity in the above conclusion is the velocity at one instant of time while observed velocities are always average velocities through appreciable time intervals. We shall find upon further examination of the equations of motion that the velocity is not at all constant, but oscillates rapidly about a mean value which agrees with the observed value.

    It may easily be verified that a measurement of a component of the velocity must lead to the result ±c in a relativistic theory, simply from an elementary application of the principle of uncertainty…”

    Well I see that the equations from Word don't paste over here...
  8. Sep 16, 2012 #7


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    As Doc Al says, this is the well-known Zitterbewegung effect. It's a rapid fluctuation in d<x>/dt, and only occurs when both positive and negative frequency solutions are present.

    You don't want d<x>/dt anyway, it fluctuates about the group velocity. The diffraction pattern relates to the phase velocity, and that in turn relates to p.
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