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Why does this condition ensure that the residue of the propagator is 1?

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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.

This post imported from StackExchange Physics at 2014-09-16 10:47 (UCT), posted by SE-user 0celo7
asked Sep 16, 2014 in Theoretical Physics by 0celo7 (50 points) [ no revision ]
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.

This post imported from StackExchange Physics at 2014-09-16 10:47 (UCT), posted by SE-user TwoBs
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?

This post imported from StackExchange Physics at 2014-09-16 10:47 (UCT), posted by SE-user 0celo7
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.

This post imported from StackExchange Physics at 2014-09-16 10:47 (UCT), posted by SE-user TwoBs
Yeah thanks. It's essentially a case of L'Hospital's rule.

This post imported from StackExchange Physics at 2014-09-16 10:47 (UCT), posted by SE-user 0celo7

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