# Parity in the \eta to two photon decay

1. Feb 4, 2006

### 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?

2. Feb 7, 2006

### marlon

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

3. Feb 8, 2006

### Meir Achuz

In \eta-->2\gamma, there is orbital angular momentum l=1, which gives the factor (-1)^l. The angular momentum addition for the photon spins must be S=1+1=1, and then l+S=1+1=0.
The eta decay is the same as the pi^0 decay. The parity of the pi^0, and of the eta^0, were determined by the relative plane polarizations of the two photons, which can be found from the spin addition 1+1=1.
This was first done by Yang.