# Resonance propagator properties

(This is part of a problem from Schwarz book on QFT). 1. 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 $$iM=\frac{i}{p^2-m^2+i\epsilon}$$ 2. 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:

1. 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?
2. 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).

So this is what I have until now. For the normal propagator (no loop) we get: $$\frac{i}{p^2-M^2+i\epsilon}$$ M is the mass of $$\phi$$ and m is the mass of $$\psi$$. For the 1 loop correction, I get: $$(\frac{i}{p^2-M^2+i\epsilon})^2\int d^4k\frac{i}{k^2-m^2+i\epsilon}\frac{i}{(p-k)^2-m^2+i\epsilon}$$ For 2 loops $$(\frac{i}{p^2-M^2+i\epsilon})^3(\int d^4k\frac{i}{k^2-m^2+i\epsilon}\frac{i}{(p-k)^2-m^2+i\epsilon})^2$$ so the series will look in the end like this: $$\frac{i}{p^2-M^2+i\epsilon}\frac{1}{1-\frac{i}{p^2-M^2+i\epsilon}\int d^4k\frac{i}{k^2-m^2+i\epsilon}\frac{i}{(p-k)^2-m^2+i\epsilon}}$$ $$\frac{i}{p^2-M^2+i\epsilon-i\int d^4k\frac{i}{k^2-m^2+i\epsilon}\frac{i}{(p-k)^2-m^2+i\epsilon}}$$ I assume we can get rid of the first $$i\epsilon$$ term as we don't use it for any integral so we get $$\frac{i}{p^2-M^2+i\int d^4k\frac{1}{k^2-m^2+i\epsilon}\frac{1}{(p-k)^2-m^2+i\epsilon}}$$ and I think I need to show that this is equal to $$\frac{i}{p^2-M^2+i M \Gamma}$$ but I am not sure how. Also I am not sure what do they mean by "Show that if these give an imaginary number..." Do I need to assume that the integral is an imaginary number? And what then? Can someone tell me if this is correct and help me from here please?