# Resonance propagator properties

(This is part c & d of problem 7.9 from Schwartz' 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?

• Your integral represents polarization operator in 1-loop approximation and it obtains imaginary part for specific momenta of scalar particle. – Artem Alexandrov Mar 7 at 17:08

To start, we would like to calculate polarization operator (=fermion bubble) in scalar Yukawa theory, $$g\bar{\psi}\phi\psi$$. The expression for polaziation operator is $$\Pi(k^2)=(-1)(-g)^2\int_p\frac{(\gamma\cdot p+m)(\gamma(p+k)+m))}{[(p^2-m^2)-i\epsilon][(p+k)^2-m^2+i\epsilon]}$$ Then, you can show that it is possible to sum the series with bubbles (it is just geometric series). The key point is to understand when $$\Pi(k^2)$$ obtains imaginary part. You can show it in a variety of ways (I can write down the calculation). The main thesis is

Polarization operator has imaganiry part if $$k^2>4m^2$$, where $$m$$ is fermion mass in the theory and $$k$$ is scalar 4-momentum

I assume that if you perform accurate calculation of $$\Pi(k^2)$$ and then consider renormalized propagator, you should find something like (in one bubble approximation) $$\frac{i}{k^2-M^2-\Pi(k^2)}$$ and if $$\Pi(k^2)$$ has imagianary part you exactly obtain your formula.

Indeed, one of the possible ways to calculate $$\Pi(k^2)$$ is to cut diagram and consider decay of scalar particle. Decay width will be $$\Gamma=\frac{g^2}{8\pi M}k^2\sqrt{1-\frac{4m^2}{k^2}}$$ and it relates to imaginary part as $$\text{Im}\,\Pi(k^2)=-M\Gamma$$. To see it, you can just calculate $$\text{Im}\,\Pi(k^2)$$ using another way and compare results.

If my answer is unclear or you need to see all the derivations, I can provide derivations and try to clarify my statements.

Suggested strategy is:

1. Consider 1-loop polarization operator and show how it obtains imaginary part
2. Then consider renormalized scalar propagator
3. Understand how to sum all 1PI (=1-particle irreducible) diagrams