- #1

- 71

- 2

## Homework Statement

(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 $$\frac{i}{p^2-m^2+im\Gamma}$$

## Homework Equations

## The Attempt at a 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).