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

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

## Answers and Replies

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