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

Now, because $\Pi^*_{\mu\nu}(q)$ 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.

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?

Update: I think what Weinberg had in mind was Luboš Motl's answer, that why he's so brief. In addition Peskin & Schroeder used the same reasoning in page 245:

...the only obvious source of such a pole would be a single-massless-particle intermediate state, which cannot occur in any 1PI diagram

However P&S also put a footnote immediately after:

One can prove that there is no such pole, but the proof is nontrivial. Schwinger has shown that, in two spacetime dimensions, the singularity in $\Pi$ due to a pair of massless fermion is a pole rather than a cut; this is a famous counterexample to our argument. There is no such problem in four dimensions.

Thus my original question stands justified. I'd be grateful if one can give a reference that elaborates P&S's footnote. Of course explanations by any SE user himself/herself are even more welcomed.

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1PI graphs don't contain singularities at $q^2=0$ because those only arise from propagators that carry the external photon momentum $q$. The external ones are omitted (as factors), as you said, and if the graphs had a single propagator with the momentum $q$, it could be cut to two pieces by cutting this propagator and this is by definition a diagram that is not 1PI.

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    $\begingroup$ How about the internal lines? How do we argue after a bunch of momentum integrations of the internal lines, there still won't be a pole at $q^2=0$? $\endgroup$
    – Jia Yiyang
    Commented May 14, 2013 at 9:51
  • $\begingroup$ Hi, try to study section 1.4 here, mathematik.hu-berlin.de/~maphy/SkriptII11.pdf , Cutkosky rules - or elsewhere. It's a much stronger result than what you want. If there were a $1/q^2$ singularity, there would also be the imaginary part behaving as a delta-function, and those can be fully calculated via the rules. Those rules give you all the singularities and 1PI diagrams need to be cut by cutting several propagators at once and their singularity structure therefore depends on more momenta, as exactly dictated by the rules. $\endgroup$
    – Luboš Motl
    Commented May 14, 2013 at 10:03
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    $\begingroup$ I just read a few materials on cutting rules including the link you gave. I'm not sure how to apply it to my problem. Firstly the cutting rules only give the imaginary parts, but maybe the real parts can have poles? Secondly, in the link you gave there's a sentence on page 5:"Before we start, let us briefly note that even though everything that Cutkosky says is there, is actually there, but then again, there might be more that Cutkosky did not discuss" So maybe it doesn't cover all the singularities? $\endgroup$
    – Jia Yiyang
    Commented May 15, 2013 at 8:39

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