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## Main Question or Discussion Point

I'm reading the QFT textbook by Weinberg. In volume one chapter 10 page 451, at the lower part of the page he says,

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)

[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.[itex]\Pi^*_{\mu\nu}(q)[/itex] receives contributions only from one-photon-irreducible graphs, it is expected not to have any pole at [itex]q^2=0[/itex]

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)