If $\rm p\bar p$ (proton-antiproton) annihilation at rest proceeds via the $S$ state ($L=0$), why is it that the reaction: $\rm p\bar p\to \pi^0 +\pi^0$ is forbidden as strong interaction (i.e. parity conserved).
The initial state has odd parity as $L=0$, the final state must have the same: $L=$ odd for the final state. How do I determine the value of $L$ for pions? (Do I use the fact that total angular momentum $J$ is conserved and the known values of spin $S$?)