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Why doesn't one-photon-irreducible function have any pole at q^2=0?

  1. May 16, 2013 #1
    I'm reading the QFT textbook by Weinberg. In volume one chapter 10 page 451, at the lower part of the page he says,
    [itex]\Pi^*_{\mu\nu}(q)[/itex] is the sum of all one-photon-irreducible graphs, with the two external photon propagators omitted, and q being the external photon momentum.

    Weinberg states it within one sentence as if it's self-explanatory, but I cannot understand why it is true. Is there something simple I missed?

    Cross-posted: why doesn't one-photon-irreducible function have any pole at q^2=0(stackexchange)
  2. jcsd
  3. May 16, 2013 #2


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    I guess the idea is the following: Free photon propagators have a pole at q=0 but in irreducible graphs, there are always integrals over the momentum of the photon propagators, so there is no pole left from the photon propagators.
  4. May 17, 2013 #3
    I think the same, but this is still a bit speculative. After all, it's a infinite sum of almost arbitrarily complicated diagrams, and it won't be surprising if nasty stuff happens.
    Weinberg uses this to argue photon mass is protected during renormalization, i.e. radiative corrections don't give photon a mass, so this is something important that I really wish to understand. Or is there some other way to show this?
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