The electromagnetic neutral pion decay is a three-point interaction: it decays into two virtual and charged Kaons or Protons, of which one then radiates a photon and then annihalates with the other to produce a second photon. (Obviously, a neutral particle cannot radiate photons directly) Questions: 1.) Let's go into the CMS of the neutral Pion. There, it neither has spin nor orbital momentum. The total angular momentum is therefore J=L+S=0 (where + denotes the spin addition). The Pion is a P=-1 eigenstate of parity and a C=+1 eigenstate of the charge conjugation operator. The C value of the two photons is C=(-1)(-1)(-1)^(l+s). Because of momentum conservation, it is L=0 and therefore l=0. Conservation of J then requires that the two photon spins are antiparallel, which requires S=0 and therefore s=0, where s is the spin-3-component. This shows that the two-Photon decay agrees with C-conservation. Using a similar argumentation, the one-Photon decay contradicts C-conservation. Let's have a look at parity conservation. The two photons have intrinsic parity -1 each, which "adds" up to +1. Orbital angular momentum must be zero according to classical conservation laws. So we have a violation of parity. Is that correct? 2.) Has that to do with the (a?) QFT anomaly, namely Ward-Identities as a substitution for classical conservation laws which get broken by the process of quantization? If yes, could someone please give an overview of what is actually happening here?