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Assume that we calculate $i\Sigma(\not{p})$ in Eq. (62.28) in Srednicki's book on QFT. If $p$ is the momentum of an external massive particle of mass $m$, then the equation $$\not{p}=-m$$ holds. In addition, the diagram depicting this $i\Sigma(\not{p})$ is going to be connected to whatever diagram depicts the interaction of the particle of momentum $p$. Thus, if $\mathcal{A}(p)$ is the matrix representing that interaction, then the amplitude for the diagram connecting the self-energy correction to that interaction will be given by $$i\mathcal{M}=\mathcal{A}(p) (-i)\frac{(-\not{p}+m)}{p^2+m^2}i\Sigma(\not{p})u(p).$$ This is because an internal line will connect the self-energy part of the diagram with the interaction part of the diagram.

My question is:

Isn't this infinite because of the internal line connecting the self-energy part of the diagram to the remaining diagram for the interaction being on-shell (i.e. $p^2=-m^2$ on the denominator)? I know that $i\Sigma(-m)=0$, but how is this connected to the fact that the contribution of the self-energy diagram, along with the counter-term diagram should altogether vanish? Even if it is zero, then aren't we multiplying zero with infinity?

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  • $\begingroup$ I've always taken this as a limit and just used L'Hospitals rule to justify why this is typically represented as $\frac{\partial \Sigma(\not p)}{\partial \not p}\Bigr|_{\not p \to m}$ $\endgroup$
    – Triatticus
    Commented Nov 14 at 15:57
  • $\begingroup$ So, you mean that in order for you to decide whether or not $i\mathcal{M}$ diverges due to $p^2+m^2=0$, you perform L'Hospital's rule to see if the quantity in the denominator $p^2+m^2$ goes to zero faster than $i\Sigma(\not{p})$? $\endgroup$
    – schris38
    Commented Nov 14 at 16:42

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I will attempt to answer my question for two reasons: first, to see if I have understood correctly the comment of @Triatticus and, second, to have a clear answer so that other people can see. So, please, if I misunderstood something, correct me in the comments.

From what I understand the answer must have something to do with the fact that I can Taylor expand $i\Sigma(\not{p})$ around $\not{p}=-m$. This gives $$ i\Sigma(\not{p})=i\Sigma(-m)+ i\frac{d\Sigma(\not{p})}{d\not{p}} (\not{p}+m)+ \frac{1}{2}i\frac{d^2\Sigma(\not{p})}{d\not{p}^2} (\not{p}+m)^2+... $$ So, substituting this expansion into $i\mathcal{M}$ yields $$ \mathcal{A}(p)(-i)\frac{(-\not{p}+m)}{p^2+m^2} \bigg\{i\Sigma(-m)+ i\frac{d\Sigma(\not{p})}{d\not{p}} (\not{p}+m)+ \frac{1}{2}i\frac{d^2\Sigma(\not{p})}{d\not{p}^2} (\not{p}+m)^2+...\bigg\}u(p) $$ I am now going to take the limit $\not{p}\rightarrow-m$, but before doing so, I will acknowledge that $i\Sigma(-m)=0$ and that $$i\frac{d\Sigma(\not{p})}{d\not{p}}\bigg|_{\not{p}=-m}=0$$ and, hence, now I have to take the limit $$\lim_{\not{p}\to-m} \mathcal{A}(p)(-i)\frac{(-\not{p}+m)}{p^2+m^2} \bigg\{\frac{1}{2}i\frac{d^2\Sigma(\not{p})}{d\not{p}^2} (\not{p}+m)^2+...\bigg\}u(p) $$ which is zero.

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