Is QM Phase Inherently Relativistic?

[LarryS]

The phase of a pure momentum state for a single particle traveling freely is

(p.r – Et) / h-bar.

This also happens to be the Minkowski space-time dot product of the 4-momentum and the space-time 4-vector (relativistic action).

Is that just a coincidence or is phase inherently relativistic?

 

[tom.stoer]

Good question.

If you use a relativistic wave equation this phase somehow survives in the non-relativistic limit. But I guess that is not really the explanation you are expecting.

 

[matonski]

In The Feynman Lectures Vol 3 Ch 7, Feynman begins by postulating that the wave function for a particle at rest is constant in space and varies as $e^{-iE t/ \hbar}$ in time. He then uses the relativistic dot product to derive the wave function for a moving particle.

In other words, a particle at rest can be thought of as being in a plane wave that is headed purely in the time direction. Its wavefronts are planes of constant time. However, with respect to a moving frame, those surfaces of constant time will now be "tilted" in spacetime, leading to the spatial variation of $e^{ipx/\hbar}$.

 

[LarryS]

In The Feynman Lectures Vol 3 Ch 7, Feynman begins by postulating that the wave function for a particle at rest is constant in space and varies as $e^{-iE t/ \hbar}$ in time. He then uses the relativistic dot product to derive the wave function for a moving particle.

In other words, a particle at rest can be thought of as being in a plane wave that is headed purely in the time direction. Its wavefronts are planes of constant time. However, with respect to a moving frame, those surfaces of constant time will now be "tilted" in spacetime, leading to the spatial variation of $e^{ipx/\hbar}$.

Interesting. I will have to read up on that.

 

[Meir Achuz]

The phase of a pure momentum state for a single particle traveling freely is

(p.r – Et) / h-bar.

This also happens to be the Minkowski space-time dot product of the 4-momentum and the space-time 4-vector (relativistic action).

Is that just a coincidence or is phase inherently relativistic?

In non-relativistic QM, E=p^2/2m.

 

[LarryS]

In non-relativistic QM, E=p^2/2m.

Yes, and that expression normally occurs in operator form on left side of the Schrödinger equation. But the right side of that equation and also the definition of the momentum and energy operators is based on the expression (p.r - Et)/h-bar .