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Feynman diagrams with odd number of vertices.

  1. Mar 10, 2014 #1
    If we scatter unpolarized electrons off each other and we calculate the amplitude for electrons to scatter into final states and one photon to be emitted do we get destructive interference because the photon can be emitted by either electron? Is one photon emission suppressed in favor of an even number of photons?

    Can we say anything interesting about amplitudes representing Feynman diagrams with odd number of vertices?

    Thanks for any help!
     
  2. jcsd
  3. Mar 10, 2014 #2
    I honestly don't think there is anything "special" about Feynman diagram with odd number of vertices in QED, but I could be wrong.
    In the case of a process like [itex]e^+e^-\to e^+e^-\gamma[/itex] you always have interference between different diagrams. The idea is that the photon could be emitted both by the positron or by the electron and both before or after the scattering. Since these diagrams produce all the same final state, by quantum mechanics they are indistinguishable processes and therefore they interfere.

    However, a similar thing happen to all orders, not necessarily in the case of just one photon or odd numbers of vertices.

    From my naive point of view there no big difference, unless some non-trivial reason comes out.
     
  4. Mar 11, 2014 #3
    Only when a diagram has odd number of photon vertices attached to a electron loop,in that case you can omit the diagram altogether.It is Furry's theorem.
     
  5. Mar 11, 2014 #4
    Does argument change much if the reaction were e- + e- ---> e- + e- + γ?

    I'm curious if one photon emitted when electrons scatter is suppressed because if interference?

    Thanks for your help!
     
  6. Mar 11, 2014 #5
    I think the same reasoning can be applied also to electron-electron scattering. Why do you think that it should be suppressed? Do you have any hint on that?
     
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