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4-momentum in relativistic QM

  1. Jun 30, 2008 #1
    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.

  2. jcsd
  3. Jun 30, 2008 #2

    Hans de Vries

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    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
    Last edited: Jun 30, 2008
  4. Jun 30, 2008 #3


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    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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