# C-violating pion0 decay

1. Jun 13, 2015

### ChrisVer

The neutral pion is dominantly decaying to 2 photons via electromagnetic interaction $\pi^0 \rightarrow 2 \gamma$.
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 $\pi^0 \rightarrow 3 \gamma$ 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 $\mathcal{O}(\alpha)$ (that means ~130 times less), since we have 3 photon vertices instead of 2. Would that imply that:
$\frac{Br(\pi^0 \rightarrow 3 \gamma )}{Br(\pi^0 \rightarrow 2 \gamma)} \sim 10^{-2}$?
How can this be in agreement with the experimental result of $<3~ 10^{-8}$? It's not and that's why we say that C-invariance is there...then why looking for the 3-photon decay?

2. Jun 13, 2015

### Staff: Mentor

To check if the electromagnetic interaction is really C-conserving. Smaller C-violations lead to smaller branching fractions.

3. Jun 13, 2015

Staff Emeritus
This.

4. Jun 13, 2015

Staff Emeritus
Good reason: to see if the EM interaction is really C-conserving.
Better reason: to see if there is some new interaction that interferes with electromagnetism, thus making it experimentally accessible.

5. Jun 13, 2015

### ChrisVer

Yup but if we find some C-violation it won't be from EM interactions since EM interactions cannot explain the 3photon decay. It would have to be something else.
So I would go with the "better reason"...which atm I cannot challenge

6. Jun 13, 2015

### Staff: Mentor

It can, if it has a small C-violating term.
Other C-violating interactions are possible as well, of course (and given the precision experiments done with electromagnetism, I guess that would be a more likely explanation).