Does a Propagator Only Have an Imaginary Part On-Shell?

[Malamala]

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).

 

[nrqed]

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?

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.

 

[Malamala]

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 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).

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.

 

[Malamala]

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!

 

[Malamala]

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?