Furry's theorem and soft bremsstrahlung

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SUMMARY

Furry's theorem states that diagrams with an odd number of photons vanish due to charge conjugation symmetry, where the photon field transforms as A_\mu → -A_\mu. However, the "soft bremsstrahlung" diagram, which involves e+e- scattering with an additional photon emission, does not vanish because it is not connected to the same internal fermion loop. This distinction is crucial for understanding infrared divergences in Quantum Electrodynamics (QED).

PREREQUISITES
  • Understanding of Furry's theorem in Quantum Field Theory
  • Knowledge of charge conjugation symmetry
  • Familiarity with Quantum Electrodynamics (QED)
  • Basic concepts of particle scattering and diagrams
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  • Research the implications of Furry's theorem on particle interactions
  • Study the role of charge conjugation symmetry in quantum field theories
  • Explore infrared divergences in Quantum Electrodynamics (QED)
  • Learn about soft bremsstrahlung processes in particle physics
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Physicists, particularly those specializing in Quantum Field Theory and Quantum Electrodynamics, as well as students and researchers interested in particle scattering processes and their underlying principles.

Einj
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Hi everyone, I have a doubt which was actually raised by a recent discussion on this forum.
We know that Furry's theorem says that diagrams with an odd number of photon vanish. Roughly the reason behind that is the charge conjugation symmetry, since under such operation the photon field behaves like A_\mu\to -A_\mu.

However, I have seen many times the "soft bremsstrahlung" diagram, i.e. e+e- scattering with the emission of an additional photon. This diagram is usually used to deal with infrared divergences in QED. Why doesn't it vanish?

Thank you very much.
 
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As was said in that thread, Furry's Theorem applies only when an odd number of photons is connected to the same internal fermion loop.
 
Let's not discuss this in two places.
 

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