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4-momentum in relativistic QM

by pmb_phy
Tags: 4momentum, relativistic
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pmb_phy
#1
Jun30-08, 03:37 AM
P: 2,954
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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Hans de Vries
#2
Jun30-08, 03:57 AM
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P: 1,135
Quote Quote by pmb_phy View Post
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:

[tex]\psi(x)~=~e^{-iEt/\hbar + ipx/\hbar}[/tex]


Regards, Hans
Fredrik
#3
Jun30-08, 03:59 AM
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PF Gold
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P: 9,277
It's defined as the [itex]P^\mu[/itex] that appears in the translation operator [itex]e^{-iP^\mu a_\mu}[/itex], where [itex]a^\mu[/itex] 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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