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The corrected propagator is given by $$\Delta'(q)=\frac{1}{q^2+m^2-\Pi^*(q^2)-i\epsilon}$$ ($\Pi^*$ is the sum of all irreducible one-particle amplitudes) I get that the residue of the original propagator around the pole $q^2=-m^2$ is $$\frac{1}{2\pi i}\oint_{\text{around }q^2=-m^2} \frac{dq^2}{q^2+m^2-i\epsilon}=\lim_{q^2\rightarrow -m^2}\frac{q^2+m^2}{q^2+m^2}=1$$ and that the corrected propagator must have the same residue $$\frac{1}{2\pi i}\oint \Delta'(q)dq^2=1$$ So how does the condition $$\left[\frac{d\Pi^*(q^2)}{dq^2}\right]_{q^2=-m^2}=0$$ ensure the second integral above?

EDIT: Devouring complex analysis literature. Have already edited some things that weren't quite right. For anyone interested, I'm using Weinberg Vol 1 and this in section 10.3, ~p. 430.

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    $\begingroup$ You know that the denominator vanishes at the pole mass. Just Taylor expand the denominator around that point, and then you get your condition for the residues by taking the limit as in the unperturbed propagator. $\endgroup$ – TwoBs Sep 16 '14 at 1:19
  • $\begingroup$ Does this also imply that all higher derivatives of $\Pi^*$ vanish at the mass shell as well? The first term in the denominator's Taylor series is obviously 0. The second term is $[1-d\Pi^*/dq^2]_{-m^2}(q^2+m^2)$. Integrating this obviously leads to the desired condition, but how can I be sure that the series terminates after the first-order $q^2$ derivative? $\endgroup$ – Ryan Unger Sep 16 '14 at 1:44
  • $\begingroup$ The higher order terms in the expansions vanish faster than the $q^2+m^2$ term, and so they do not contribute to the limit. Just take the limit. $\endgroup$ – TwoBs Sep 16 '14 at 2:16
  • $\begingroup$ Yeah thanks. It's essentially a case of L'Hospital's rule. $\endgroup$ – Ryan Unger Sep 16 '14 at 2:24
  • $\begingroup$ I will note this was already answered elsewhere: reddit.com/r/Physics/comments/2gigu8/… (and by the way 0celo7, the fact that nearly the same text is posted here and there may constitute a violation of our plagiarism policy... I'll be looking into this) $\endgroup$ – David Z Sep 16 '14 at 6:13
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The original propagator has a pole at $q^2=-m^2$, the mass shell. For $m=\sqrt{-q^2}$ be the true mass of the particle, we have $\Pi^*(-m^2)=0$ and require that the residue of the modified propagator is unity around $q^2=-m^2$. Recall that for a meromorphic function $f(z)$ we have $$\oint f(z)dz=2\pi i\sum_k\operatorname{Res}_{z_k}(f)$$ Thus $$\oint \Delta'(q^2)dq^2=2\pi i\operatorname{Res}_{-m^2}(\Delta')$$ Note that the pole in $\Delta'$ is simple. Thus we have $$\operatorname{Res}_{-m^2}(\Delta')=\lim_{q^2\rightarrow-m^2}(q^2+m^2)\Delta'(q^2)=1$$ Inserting the definition of $\Delta'$, we get $$\lim_{q^2\rightarrow-m^2}\frac{(q^2+m^2)}{q^2+m^2-\Pi^*-i\epsilon}=1$$ which is $0/0$ on the left, i.e. indeterminate. Using L'Hopital's rule, we find $$\left.\frac{d\Pi^*}{dq^2}\right|_{-m^2}=0$$ as was to be shown.

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