Resonance propagator properties

In summary, the first part shows that the imaginary part of a propagator is given by $$Im(M)=-\pi\delta(p^2-m^2)$$ when the propagator is in the form $$iM=\frac{i}{p^2-m^2+i\epsilon}$$. The second part discusses the effects of loops of particles on the propagator, showing that the imaginary parts can arise if certain kinematic conditions are met. The series of loop corrections to the propagator can be summed to obtain a new propagator with an imaginary mass term, $$\frac{i}{p^2-m^2+im\Gamma}$$. However, the specific details and calculations for this part are not fully understood at this time
  • #1
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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).
 
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  • #2
Malamala said:

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.
 
  • #3
nrqed said:
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!
 
  • #4
Malamala said:

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.
 
  • #5
nrqed said:
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!
 
  • #6
nrqed said:
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?
 

What is a resonance propagator?

A resonance propagator is a mathematical concept used in quantum field theory to describe the propagation of a particle in a resonant state. It represents the probability amplitude for a particle to transition between different energy levels.

What are the properties of a resonance propagator?

The properties of a resonance propagator include its poles, residues, and branch cuts. The poles represent the energy levels of the resonant state, the residues determine the strength of the resonance, and the branch cuts represent the energy values where the propagator is undefined.

How is a resonance propagator calculated?

A resonance propagator is calculated using Feynman diagrams, which are graphical representations of particle interactions. The propagator is obtained by summing over all possible Feynman diagrams that contribute to the process of interest.

What is the physical significance of a resonance propagator?

The physical significance of a resonance propagator lies in its ability to describe the behavior of resonant states in particle interactions. It is an essential tool for understanding and predicting the behavior of subatomic particles in high-energy physics experiments.

How is a resonance propagator used in practical applications?

A resonance propagator is used in practical applications such as particle accelerator experiments and in the development of new theories in particle physics. It also has applications in other fields such as condensed matter physics, where it is used to study the behavior of complex systems.

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