Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Action corresponding to photon emission

  1. Nov 7, 2014 #1
    Hi I am trying to write the probability of photon emission due to transition of electron in feynman's path integral formulation. I am stuck trying to figure out the action corresponding to the photon emission. Would anyone shed some light on this? Thanks
     
  2. jcsd
  3. Nov 7, 2014 #2
    If I'm not mistaken (and I could very well be, as I'm a dilettante dabbler), the corresponding 1st-order diagram would be an incoming electron with momentum p, and an outgoing electron with a momentum (p-k) and photon with momentum k, so the perturbation expansion term would have a destruction operator for the electron with momentum p, and the electron/photon creation operators with momenta (p-k)/k.

    However, if you're asking about the actual action S corresponding to that situation, as in the integral over a term in the Lagrangian, I don't think there is one; the "electron emitting a photon" and all of the other possibilities come from the perturbation expansion of the (exponential of?) the normal, (semi-)classical action, where the only term giving the interaction between the electron field and the Maxwell field is just the minimal-coupling term (##j_{\mu} A^{\mu}##?); everything else derives from that.

    But, again, caveat emptor.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Action corresponding to photon emission
  1. Photon emission (Replies: 3)

  2. Photon emission. (Replies: 5)

  3. Photon emission? (Replies: 2)

Loading...