The neutral pion belongs to the pseudoscalar meson octet, so it has, in the ground state ($L=0$):
\begin{align}
P_{\pi^0}&=-1 \\
C_{\pi^0}&=+1.
\end{align}
And the photon has
\begin{align}
P_\gamma = -1 \\
C_\gamma=-1.
\end{align}
Therefore, since electromagnetic interactions conserve parity and charge conjugation, why does the process
\begin{equation}
\pi^0 \rightarrow \gamma\gamma
\end{equation}
occur? Doesn't it violate parity?
In the example I have seen in class, $C$ conservation is used to explain why the $\pi^0$ cannot decay into three photons, since for $\pi^0 \rightarrow \gamma\gamma\gamma$ we have
\begin{equation}
C_i = +1 \neq C_f = (-1)^3 = -1
\end{equation}
and, for $\pi^0 \rightarrow \gamma\gamma$,
\begin{equation}
C_i = +1 = C_f = (-1)^2 = +1,
\end{equation}
so regarding $C$ conservation it should be allowed. But, considering P conservation,
\begin{align}
\pi^0 \rightarrow \gamma\gamma \qquad \Rightarrow \qquad P_i = (-1)^{L}\times \underbrace{(-1)}_{\text{intrinsic parity}} = -1 \neq P_f = (-1)^2 = +1
\end{align}
so it would be forbidden for $L=0$. And, with the same argument, the decay into three photons would be allowed.
What am I missing?
Best Answer
The photons have intrinsic spin (or, better, helicity) one, so the pair can have odd orbital angular momentum, still conserving total angular momentum (which has to be zero, as the pion is spinless).
Specifically, the spins of the two photon can combine to give total spin $S=1$. This, conmbined with an angular momentum $L=1$, has a $J=0$ component which permits the pion to decay into two photons. You can check from the Clebsch-Gordan table that the final two photon wavefunction is symmetric under particle permutation, as required by Bose statistics.