# Why can the pion decay into two photons?

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

$$$$\pi^0 \rightarrow \gamma\gamma$$$$ 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

$$$$C_i = +1 \neq C_f = (-1)^3 = -1$$$$

and, for $$\pi^0 \rightarrow \gamma\gamma$$,

$$$$C_i = +1 = C_f = (-1)^2 = +1,$$$$ so regarding $$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?

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.