Does Charmonium Decay into Two Photons Violate Parity Conservation?

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SUMMARY

The discussion centers on the decay of charmonium (c\bar c) in the state 11S0 into two photons (c\bar c → 2γ) and the implications for parity conservation. The initial parity of the charmonium state is calculated as -1, while the final parity of two photons is debated. The key insight is that the photons can have different polarization states, which affects the overall parity of the system. Specifically, if the photons are in a P-wave configuration, the parity considerations change, allowing for the decay to occur without violating parity conservation.

PREREQUISITES
  • Understanding of quantum mechanics and particle physics
  • Familiarity with charmonium states and their properties
  • Knowledge of photon polarization and its implications
  • Basic principles of parity conservation in electromagnetic interactions
NEXT STEPS
  • Study the properties of charmonium states, focusing on 11S0 and its decay modes
  • Research photon polarization states and their impact on particle decay processes
  • Learn about parity conservation and violations in quantum field theory
  • Investigate historical experiments related to parity measurements, such as those involving the pi zero meson
USEFUL FOR

Particle physicists, quantum mechanics students, and researchers interested in the properties of mesons and photon interactions will benefit from this discussion.

Derivator
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Hi,

one of my books says, that a charmonium c\bar c in the sate 1^1S_0 could decay into two phtons: c \bar c \rightarrow 2\gamma

But I think, conservation of parity is violated (not allowed, since we have electromagnetic interaction)

Parity of a charmonium state is (-1)^{L+1}, we have L=0, so before the decay we have a parity of -1.
Parity of a photon is -1, we have two photns, so the parity after the decay is +1

So, where have I made an error in reasoning?

--derivator
 
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The two photons can have mutually parallel polarization, and then the amplitude for their production has E.E, which is a scalar with [positive parity, or they have have mutually perpendicular polarization, corresponding to E.B, which is a psudoscalar with negative parity. In this case of psi decay, they have the perp polarization with negative parity.
This argument is not original with me. It was used by Yang over 50 years ago to propose a test to measure the parity of the pi zero meson. The experiment showed that the pi zero had negative parity.
 
Derivator said:
Hi,

one of my books says, that a charmonium c\bar c in the sate 1^1S_0 could decay into two phtons: c \bar c \rightarrow 2\gamma

But I think, conservation of parity is violated (not allowed, since we have electromagnetic interaction)

Parity of a charmonium state is (-1)^{L+1}, we have L=0, so before the decay we have a parity of -1.
Parity of a photon is -1, we have two photns, so the parity after the decay is +1

So, where have I made an error in reasoning?

--derivator

the photons could be in a P-wave, for example.
 

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