## Parity in the \eta to two photon decay

Greetings, I'm curious about parity conservation in the decay$$\eta \rightarrow \gamma \gamma$$. The $$\eta$$ has odd parity, while the product of the two photon parities (each is odd) is even. Now, parity is conserved in the EM interactions, so there must be a factor of (-1) coming in from orbital angular momentum factors--but the two photon final state has no orbital angular momentum. What am I missing here?
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 IN SHORT : PARITY DOES NOT ADD UP LINEARLY... I mean : in order for the two-photon state to have J = 0 (conservation of J we must write it as a superposition of two states A and B. Each state has the two photons with anti-parallel spins and in state A the photonspin is aligned with the photon-momentum, in B photon spin is opposite wrt photon momentum. You can derive these states by using the Clebsch-Gordan coefficients. A photon has indeed parity -1 but since we are working with a superposition of TWO photonstates, the parity is relative. If you had just a single two-photon state then parity would be -1 * -1 but because of the superposition, it is the parity from state A with respect to state B that determins the actual parity of the entire wavefunction... regards marlon More to be found here

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 Quote by fliptomato Greetings, I'm curious about parity conservation in the decay$$\eta \rightarrow \gamma \gamma$$. The $$\eta$$ has odd parity, while the product of the two photon parities (each is odd) is even. Now, parity is conserved in the EM interactions, so there must be a factor of (-1) coming in from orbital angular momentum factors--but the two photon final state has no orbital angular momentum. What am I missing here?