What does Schwarz's QFT theorem say about poles in Green's functions?

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In summary: I am a bit confused. On page 474 of Schwarz's book, just below the formula I just mentioned, it says: "positronium (an ##e^+e^-## bound state) would appear as a pole in a Green's function corresponding to ##e^+e^-## scattering". I assumed this means that positronium can be created in electron-positron scattering. But from what you told me, it can't. So what does this quote actually mean?The quote means that in electron-positron scattering, there is a potential for the production of positronium. However, this is only a potential and does not always happen.
  • #1
Malamala
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Hello! In Schwarz's QFT, Chapter 24.3 there is a theorem stating that Green's functions have poles when on-shell intermediate particles can be produced. I am not sure I understand how this works. If we have $$e^+e^- \to \gamma^* \to \mu^+\mu^-$$ we can have a positronium as an intermediate state. So when calculating the S matrix for this scattering we should have a term of the form $$\frac{1}{p^2-m_{positronium}^2}$$ but the only fraction appearing in this reaction is the photon propagator which is (in a renormalized theory) of the form $$\frac{1}{p^2(1+\Pi(p^2))}$$ which doesn't necessary have a pole at the positronium mass. I assume that I am missing something. Can someone explain this to me please? Thank you!
 
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  • #2
It is an intermediate bound electron-positron state. Not an intermediate virtual photon.
 
  • #3
Orodruin said:
It is an intermediate bound electron-positron state. Not an intermediate virtual photon.
Yes, but shouldn't it still appear in the S-matrix? The Green function is part of S-matrix and only on-shell external particles get amputated (so their poles disappear). Why don't we get a term of the form ##\frac{1}{p^2-m_{positron}^2}## in the S-matrix coming from the green function?
 
  • #4
You will not see the bound state propagator if you just stick to leading order Feynman diagrams.

The reaction you quoted will also never produce positronium on resonance. The positronium energy is lower than the CoM energy.
 
  • #5
Orodruin said:
You will not see the bound state propagator if you just stick to leading order Feynman diagrams.

The reaction you quoted will also never produce positronium on resonance. The positronium energy is lower than the CoM energy.
Oh ok, so what is the actual reaction to produce muon-antimuon and have a positron in between?
 
  • #6
Malamala said:
Oh ok, so what is the actual reaction to produce muon-antimuon and have a positron in between?
That cannot happen either. That would break Lorentz invariance. You seem to be confusing positronium with positrons.
 
  • #7
Orodruin said:
That cannot happen either. That would break Lorentz invariance. You seem to be confusing positronium with positrons.
Oh sorry, I miss-typed that. So my question is if you collide an electron and a positron and you produce positronium and then the positronium decays to something else (maybe electron-positron again), what are the intermediate steps? Do we have any intermediate photon at a point? Or in this case the positronium would act like the propagator instead of the photon?
 
  • #8
You seem to be missing the fact that the positronium mass is smaller than ##2m_e## since it is a bound state. It cannot decay to electron-positron and you cannot (only) produce it from electron-positron collisions.
 
  • #9
Orodruin said:
You seem to be missing the fact that the positronium mass is smaller than ##2m_e## since it is a bound state. It cannot decay to electron-positron and you cannot (only) produce it from electron-positron collisions.
I am a bit confused. On page 474 of Schwarz's book, just below the formula I just mentioned, it says: "positronium (an ##e^+e^-## bound state) would appear as a pole in a Green's function corresponding to ##e^+e^-## scattering". I assumed this means that positronium can be created in electron-positron scattering. But from what you told me, it can't. So what does this quote actually mean?
 

What are resonances on propagators?

Resonances on propagators refer to the phenomenon where the propagator, which is a mathematical calculation of the probability amplitude of a particle's motion, has a peak at a certain energy value. This peak is known as a resonance, and it indicates a high probability for the particle to exist at that energy level.

How do resonances on propagators affect particle physics experiments?

Resonances on propagators play a crucial role in particle physics experiments as they provide information about the properties of particles, such as their mass and decay width. By studying the shape and position of these resonances, scientists can better understand the fundamental forces and interactions between particles.

What causes resonances on propagators?

Resonances on propagators are caused by the interaction between particles and the force carriers of the fundamental forces. When the energy of the particles matches the energy of the force carriers, a resonance is produced in the propagator, indicating a strong interaction between the particles.

Can resonances on propagators be observed directly?

No, resonances on propagators cannot be observed directly as they are mathematical calculations based on experimental data. However, their effects can be observed through the behavior of particles in experiments, such as the shape and position of particle tracks in particle detectors.

Are resonances on propagators only found in high-energy particle collisions?

No, resonances on propagators can also be found in low-energy collisions or in stable particles. However, they are more commonly observed in high-energy particle collisions as the energy of the particles is high enough to produce these resonances.

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