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Photon propagator

  1. May 5, 2009 #1
    Since the 4-vector A couples to the conserved current j^\mu = \psi-bar gamma^\mu \psi in QED, k_\mu Fourier-transform(j^\mu) = 0. The Fourier-transform of j is a mess. Is there an easy way to see why terms containing photon momentum k_\mu drop out of the photon propagator in practical QED calculations?
  2. jcsd
  3. May 5, 2009 #2
    You can find discussion of this point in section 8.5 in Weinberg's "The quantum theory of fields" vol. 1. The idea is the following:

    The true interaction Hamiltonian in QED is

    [tex] V(t) =-\int d^3x j(x,t)A(x,t) -\frac{1}{2}\int d^3x d^3y \frac{j^0(x,t)j^0(y,t)}{4 \pi|x-y|} [/tex]

    and the true photon propagator is

    [tex] D_{\mu \nu}(p) = \frac{-i}{(2 \pi)^4} \frac{P_{\mu \nu}(q)}{q^2} [/tex]

    However, results for the S-matrix would not change if you simplify both the interaction and the propagator. In the interaction operator you drop the (current)x(current) term

    [tex] V'(t) =-\int d^3x j(x,t)A(x,t) [/tex]

    and in the photon propagator you replace the momentum-dependent function [tex] P_{\mu \nu}(q)[/tex] simply by [tex] g_{\mu \nu}[/tex]

    [tex] D'_{\mu \nu}(p) = \frac{-i}{(2 \pi)^4} \frac{g_{\mu \nu}}{q^2} [/tex]

    Weinberg does not prove that this trick works at all orders, but apparently it works.
    Last edited: May 6, 2009
  4. May 6, 2009 #3
    Thanks for the pointer to Weinberg's book.

    Weinberg's discussion also seems to merely make the claim without offering any proof;
    he refers to Feynman's 1949 paper (section 8). Feynman has an argument to show
    that the divergence of the amplitude of a process that emits a photon (real/virtual) vanishes,
    implying that the dot product of $k$ and the Fourier transform of the amplitude vanishes. His
    argument appears plausible (even if a bit sketchy) and doesn't appear to invoke current conservation to show that the divergence vanishes. If Feynman's argument implies that the $k^\mu k^\nu$ term can be ignored in the photon propagator, it's not obvious as to why/how. The $k^\mu k^\nu$ term does not appear in the propagator in Feynman gauge, but that is not the same as disregarding the term in other gauges. Perhaps there is some simple way to see that the Fourier-transform of the current occurs dotted to the $k_\mu$ term in all Feynman diagrams containing QED vertices?
  5. May 6, 2009 #4
    It's also discussed in Zee chapter II.7
    As usual, you need external legs to be on shell.
  6. May 6, 2009 #5
    I think what you have is a special case of the Ward-Takahashi identity, which is most easily proven in the path integral formalism. If I recall, the identity says that classical equations of motion hold true inside the vacuum expectation value of time-ordered products.
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