Resonances on propagators

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  • Thread starter 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
Orodruin
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It is an intermediate bound electron-positron state. Not an intermediate virtual photon.
 
  • #3
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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?
 
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Orodruin
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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.
 
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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?
 
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Orodruin
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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
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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?
 
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Orodruin
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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
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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?
 

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