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(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?

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In this paper , at page 9:

4.1 Annihilation into two neutral mesons

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).

So the reaction is seen experimentally, table at page 54 . The possible diagrams discussed are also strong interaction ones, as discussed in paragraph 4.2, illustrated in fig. 9

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