How is the 4-momentum 4-vector defined in relativistic QM?

pmb_phy
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I've been wondering about relativistic quantum mechanics. Elsewhere I'm addressing some comments about this branch of physics but I have never studied it. Is the 4-momentum 4-vector defined in the same way in relativsitic QM or is there a difference? I'm wondering if the time component of 4-momentum is defined in the same way in relativistic QM as in classical relativity. Thanks.

Pete
 
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pmb_phy said:
I've been wondering about relativistic quantum mechanics. Elsewhere I'm addressing some comments about this branch of physics but I have never studied it. Is the 4-momentum 4-vector defined in the same way in relativsitic QM or is there a difference? I'm wondering if the time component of 4-momentum is defined in the same way in relativistic QM as in classical relativity. Thanks.

Pete

Yes, generally the metric is (+---), although Weinberg uses (-+++) as in (flat) GR.

In QFT the 4-momentum is typically associated with the phase change rates in the
time and space components corresponding to the plane wave eigenfunctions:

\psi(x)~=~e^{-iEt/\hbar + ipx/\hbar}Regards, Hans
 
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It's defined as the P^\mu that appears in the translation operator e^{-iP^\mu a_\mu}, where a^\mu is the translation four-vector. This definition works in both relativistic and non-relativistic QM. (The best place to read about these things is chapter 2 of vol. 1 of Weinberg's QFT book).

In a relativistic quantum field theory, you can also construct the four-momentum operators expliclity from the Lagrangian, as the conserved quantities that Noether's theorem tells us must exist due to the invariance of the action under translations in space and time.
 
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