Pi 0 Decay: Chances of e+ e- Gamma vs e+ e-

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Discussion Overview

The discussion centers around the decay modes of the pi zero meson, specifically comparing the likelihood of its decay into an electron-positron pair with a photon (e+ e- gamma) versus decay into just an electron-positron pair (e+ e-). The scope includes theoretical considerations from particle physics and quantum field theory.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants note that the decay to a single electron-positron pair is suppressed because the leading order contribution, involving the exchange of a single photon, evaluates to zero.
  • Others suggest that the decays must occur via an intermediate state, and the most significant contributing intermediate state for the electron-positron decay is not allowed based on coupling considerations.
  • One participant mentions helicity suppression as a factor, explaining that a pseudoscalar decaying to leptons is suppressed by the mass of the lepton, which requires a spin flip. The presence of a photon allows for a decay without this spin flip, thus mitigating the mass suppression.
  • A request for resources on the topic is made, indicating a desire for further reading.
  • It is suggested that a complete understanding of the decay processes would require an introductory course in quantum field theory, typically taken in the later years of university study.

Areas of Agreement / Disagreement

Participants express differing views on the mechanisms behind the decay processes, with no consensus reached on the primary reasons for the differences in decay likelihood.

Contextual Notes

The discussion involves complex concepts from quantum field theory and particle physics, which may not be fully accessible without a background in these areas. Some assumptions and dependencies on specific theoretical frameworks are present but not fully elaborated.

Manel
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Hello everyone,
I found in the PDG booklet that pi 0 decay into e+ e- gamma is more likely to happen than to e+ e- only ..is there a reason that favorises the first decay?
 
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The decay to a single electron-positron pair is suppressed because what would be the leading order contribution, the diagram with the exchange of one single photon, evaluates to zero.
 
Orodruin said:
the diagram with the exchange of one single photon, evaluates to zero.
I didn't understand what you mean here. Would you please clarify. Thanks a lot
 
The decays must happen via an intermediate state. The intermediate state which would be contributing the most to the electron-positron decay based on the couplings is not allowed.
 
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Any resource detailing that ? So grateful
 
Complete understanding of this would require an introductory course in quantum field theory, usually given at universities somewhere in year 3-4.
 
Thanks a lot
 
also helicity suppression probably plays a role. a pseudoscalar decaying to to leptons will be suppressed by the mass of the lepton, due to conservation of angular momentum requiring a spin flip on one lepton. adding the photon will allow the lepton to not have to flip, thus trading the mass suppression for alphaEM.
 

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