Charge Conjugation to Analyze CPT Invariance

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Source: http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/cpt.html#:~:text=Charge%20conjugation(C)%3A%20reversing,like%20momentum%20and%20angular%20momentum.
In the image below, reaction (1) shows CP conservation whereas reaction (2) does not. Does this have to do with the fact that the charge conjugation of the RHS of the reaction 1 is $(-1)(-1)=1$ whereas the charge conjugation of the RHS of reaction 2 is $(-1)(-1)(-1)=-1$, which is not equivalent to the RHS Conservation of Charge C or $(1)(1)(1)=1$?
Is this how one would check for charge conjugation?

Lucas Baldo · Accepted Answer

Yes. More precisely, the neutral pion $pi_0$ is an eigenstate of the charge conjugation operator $mathcal{C}$, with eigenvaue $+1$. A photon is also an eigenstate of this operator but with eigenvalue $-1$.
The eigenvalue for a multiparticle state is given by the product of the eigenvalues of each individual particle (given that those particles are eigenstates themselves). So for an even number ($2n$) of photons, the C-parity of the system would be $(-1)^{2n}=+1$. Conversely, for an odd number ($2n+1$) of photons we have that the C-parity is $(-1)^{2n+1}=-1$.
To check whether Charge conjugation allows a certain transition you can then compare the C-parity of the initial an final states, as you have done. The transition is allowed by the Charge conjugation symmetry if and only if both eigenvalues are the same.

Charge Conjugation to Analyze CPT Invariance

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