Time in QM vs. SR

  • #1
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Main Question or Discussion Point

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 ?
 

Answers and Replies

  • #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.
 
  • #3
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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

?
 
  • #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
Orodruin
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Yes, this does not contradict what I just said.
 
  • #7
PeterDonis
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comoving with the wave, you would see no change in phase
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?
 
  • #8
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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?
 
  • #9
PeterDonis
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What does the underlined mean exactly?
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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