Understand Eta Particle Decay into 3 Pions

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SUMMARY

The discussion centers on the decay of the \(\eta\)-particle into three pions, highlighting that this decay is prohibited by the strong interaction due to symmetry constraints. According to Povh et al.'s 'Particles and Nuclei', three pions cannot couple to zero isospin, which accounts for the long lifetime of the \(\eta\)-particle. The analysis reveals that the decay violates conservation of G-parity, a critical factor in strong interactions, leading to the conclusion that the decay is not feasible.

PREREQUISITES
  • Understanding of isospin symmetry in particle physics
  • Familiarity with G-parity conservation laws
  • Knowledge of non-reducible representations in quantum mechanics
  • Basic principles of particle decay mechanisms
NEXT STEPS
  • Study the implications of G-parity in particle interactions
  • Explore the role of isospin symmetry in decay processes
  • Investigate the lifetime and decay modes of the \(\eta\)-particle
  • Learn about the mathematical framework of non-reducible representations in quantum field theory
USEFUL FOR

Particle physicists, students studying quantum mechanics, and researchers focusing on meson decay processes will benefit from this discussion.

eendavid
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1. The statement, all variables and given/known data
I am studying the decay of the \eta-particle. Povh et Al, 'Particles and nuclei' say that a decay into 3 pions is not possible via the strong interaction. "For reasons of symmetry 3 pions (isospin equals 1) can not couple to zero isospin." This explains the long lifetime of the eta-particle. I do not understand this.


Homework Equations


Denote by D^{(j)} the non-reducible representations of a rotation in isospin space, then D^{(1)}\otimes D^{(1)}=D^{(2)}\oplusD^{(1)}\oplusD^{(0)}


3. The problem
As
D^{(1)}\otimes D^{(1)}\otimes D^{(1)}= ...\oplus D^{(1)}\otimes D^{(1)}\oplus..., their are D^{(0)} components in the 3 pion system. Am I making a misinterpretation somewhere?
 
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For the purpose of archiving, here's the answer: the decay violates conservation of G-parity, which is conserved in the strong interaction.
 

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