(This is part c & d of problem 7.9 from Schwartz' book on QFT).
Show that a propagator only has an imaginary part if it goes on-shell. Explicitly, show that $$Im(M)=-\pi\delta(p^2-m^2)$$ when $$iM=\frac{i}{p^2-m^2+i\epsilon}.$$
Loops of particles can produce effective interactions which have imaginary parts. Suppose we have another particle $\psi$ and an interaction $\phi\psi\psi$ in the Lagrangian. Loops of $\phi$ will have imaginary parts if and only if $\psi$ is lighter than half of $\phi$, that is, if $\phi \to \psi \psi$ is allowed kinematically. Draw a series of loop corrections to the $\phi$ propagator. Show that if these give an imaginary number, you can sum the graphs to get for the propagator $$\frac{i}{p^2-m^2+im\Gamma}.$$
Solution:
So we have $$M=\frac{1}{p^2-m^2+i\epsilon} = \frac{p^2-m^2-i\epsilon}{(p^2-m^2)^2+\epsilon^2}$$ $$Im(M) = \frac{-\epsilon}{(p^2-m^2)^2+\epsilon^2}.$$ How do I get a $\pi$ and a delta function from this? And why would $p^2\neq m^2$ imply that this is zero?
I am not sure I understand this part. If we write corrections we would have first a straight line (simple propagator), then one loop, 2 loops and so on, so I assume we need a power series. But I am not sure how to proceed, what should I write for the propagators of the 2 particles? And I am not sure how does the imaginary part come into play (maybe if I do part 1 I can understand this better).