# How to show that the interaction $p + \bar p \rightarrow \pi^0 + \pi^0$ is forbidden in the electromagnetic and strong interaction?

(I am at the introductory level of particle physics)

I know that the parity is conserved in strong and electromagnetic interaction, so I would like to show that the charge parity is violated.

I would like to show that the product of the charge conjugation parity of proton and its antiparticle is $$-1$$, which doesn't equal to the charge parity of $$\pi^0 + \pi^0$$

problem I have encountered:

by definition $$\hat C |p, \Psi\rangle = C_p |p,\Psi\rangle$$ and $$\hat C |p, \Psi\rangle = |\bar p,\Psi\rangle$$ then $$\hat C^2 |p, \Psi\rangle = C_p^2 |p,\Psi\rangle = |p,\Psi\rangle = C_\bar p |\bar p,\Psi\rangle$$

applying $$\bar C$$ once again

$$C_p|p,\Psi\rangle = C_\bar p |\bar p,\Psi\rangle \implies C_p = C_\bar p \implies C_pC_\bar p = 1$$ which doesn't lead to the result I wanted.

What went wrong for me?

Crystal Barrel has measured the branching ratios for antiproton proton annihilation into two neutral light mesons from about $$10^7$$ annihilations into 0-prong (Amsler, 1993b). These data have been collected by vetoing charged particles with the PWC’s and the internal layers of the JDC.The lowestγ-multiplicity was four (e.g.π0π0,π0η) and the highest nine (e.g.ηω, with η→3π0and ω→π0γ).