The corrected propagator is given by Δ′(q)=1q2+m2−Π∗(q2)−iϵ
(Π∗ is the sum of all irreducible one-particle amplitudes) I get that the residue of the original propagator around the pole q2=−m2 is 12πi∮around q2=−m2dq2q2+m2−iϵ=limq2→−m2q2+m2q2+m2=1 and that the corrected propagator must have the same residue 12πi∮Δ′(q)dq2=1 So how does the condition [dΠ∗(q2)dq2]q2=−m2=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