# Resonance propagator properties

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

## 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).

nrqed
Homework Helper
Gold Member

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

## 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?

Set ##p^2 = m^2## and then take the limit ##\epsilon \rightarrow 0##, what do you get? Now take ##p^2 \neq m^2## and then take the limit ##\epsilon \rightarrow 0##, what do you get? This is the first clue that it could be a Dirac delta. To prove it, define ##p^2-m^2 \equiv x## and integrate over x. Then look at what happens when ##\epsilon \rightarrow 0## when x =0 or not zero.

Set ##p^2 = m^2## and then take the limit ##\epsilon \rightarrow 0##, what do you get? Now take ##p^2 \neq m^2## and then take the limit ##\epsilon \rightarrow 0##, what do you get? This is the first clue that it could be a Dirac delta. To prove it, define ##p^2-m^2 \equiv x## and integrate over x. Then look at what happens when ##\epsilon \rightarrow 0## when x =0 or not zero.
Thank you for you reply. So I did it somehow, similar to what you suggested I assume, using Poisson kernel. Could you please tell me if this is ok? So letting ##x=p^2-x^2## we have: $$Im(M)=lim_{\epsilon \to 0}-\frac{-\epsilon}{x^2+\epsilon^2}=lim_{\epsilon \to 0}-\pi\int_{-\infty}^\infty e^{2\pi i \alpha x - |\alpha \epsilon|}d\alpha=-\pi\int_{-\infty}^\infty e^{2\pi i \alpha x}d\alpha=-\pi\delta(x)=-\pi\delta(p^2-x^2)$$ Is this formally correct? Also any suggestion about the second part? Thank you again!

nrqed
Homework Helper
Gold Member

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

## 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).
For the second part, you just have to use the same propagator for the two particles, but using simply a different mass in each case. $$\frac{i}{p^2-m^2+i\epsilon}$$
Now just write the first few diagrams and you can sum it up since it will simply be a geometric series.

For the second part, you just have to use the same propagator for the two particles, but using simply a different mass in each case. $$\frac{i}{p^2-m^2+i\epsilon}$$
Now just write the first few diagrams and you can sum it up since it will simply be a geometric series.
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? Thank you for help again!

For the second part, you just have to use the same propagator for the two particles, but using simply a different mass in each case. $$\frac{i}{p^2-m^2+i\epsilon}$$
Now just write the first few diagrams and you can sum it up since it will simply be a geometric series.
Any further suggestion on this?