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Time in QM vs. SR

  1. Jan 20, 2015 #1
    In QM, the energy operator is proportional to the time derivative

    E ~ d/dt

    So higher energy particles have higher frequencies, i.e. their wave functions change more often per time than when at rest

    But in SR, higher energy particles seem to exist in slow motion, appearing to age little

    How are these theories compatible and reconcilable?

    Do fast moving high energy particles change alot per unit time as per QM

    or freeze into slow motion as per SR ?
     
  2. jcsd
  3. Jan 20, 2015 #2

    Orodruin

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    First of all, QM is not a relatiistic theory so you should not expect it to reproduce SR results. The extension of QM is relativistic QM and quantum field theory.

    Second, your inference about SR is wrong. If you look at a highly relativistic particle in relativistic QM, it has a very high frequency compared to a particle at rest. If you want to make a classical analogy, you cannot forget about the particle momentum, which also will contribute to the phase, i.e., take into account that the particle is moving in space as well as in time.
     
  4. Jan 20, 2015 #3
    So you're saying that we look at the plane wave solutions

    Psi ~ exp( i( px - hwt )/h )
    ~ exp( i( hkx - hwt )/h )

    w/k --> c

    So if you were stationary, and the wave function flew by, you'd measure fast changing phase

    But comoving with the wave, you would see no change in phase
    The planes of constant phase would be speeding through space
    At the velocity of the particle

    So surfing those waves
    As it were
    One would see a frozen. Wave function
    As it were

    ?
     
  5. Jan 20, 2015 #4

    Orodruin

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    No. The classical QM (just as classical mechanics) ignores the main contribution to particle energy, its mass. Now this is fine in a classical theory since it is only an overall phase factor (or a constant energy shift if you will), but it is essential in relativistic QM in order to ensure Lorentz invariance.
     
  6. Jan 20, 2015 #5
  7. Jan 20, 2015 #6

    Orodruin

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    Yes, this does not contradict what I just said.
     
  8. Jan 20, 2015 #7

    PeterDonis

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    Would you? Try writing down a plane wave solution to the Klein-Gordon equation with the mass ##m## positive in a frame comoving with the wave. What do you get?
     
  9. Jan 20, 2015 #8
    From the link:
    It cannot be straightforwardly interpreted as a Schrödinger equation for a quantum state, because it is second order in time and because it does not admit a positive definite conserved probability density.

    What does the underlined mean exactly?
     
  10. Jan 20, 2015 #9

    PeterDonis

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    It means that the second time derivative appears instead of the first; ##\partial^2 \psi / \partial t^2## instead of ##\partial \psi / \partial t##.
     
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