The neutral pion is dominantly decaying to 2 photons via electromagnetic interaction [itex] \pi^0 \rightarrow 2 \gamma[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

However if one allows for P-violation in electromagnetic interactions, he will also get C violation (due to CPT theorem). In that case a decay of the form [itex] \pi^0 \rightarrow 3 \gamma[/itex] could be observable.

I read in "Discrete symmetries and CP Violation: From experiment to theory" by M.S.Sozzi that without the selection rule applied on the ##\pi^0## decay, then the rate of the three-gamma is reduced by a factor of order [itex]\mathcal{O}(\alpha)[/itex] (that means ~130 times less), since we have 3 photon vertices instead of 2. Would that imply that:

[itex]\frac{Br(\pi^0 \rightarrow 3 \gamma )}{Br(\pi^0 \rightarrow 2 \gamma)} \sim 10^{-2}[/itex]?

How can this be in agreement with the experimental result of [itex] <3~ 10^{-8}[/itex]? It's not and that's why we say that C-invariance is there...then why looking for the 3-photon decay?

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# C-violating pion0 decay

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