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?(adsbygoogle = window.adsbygoogle || []).push({});

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# Photon propagator

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