\eta\to\pi^0\gamma decay (Why this decay is forbidden by J?)

[Alex_Rob]

Decay \eta\to\pi^0\gamma is forbidden by C parity. But why this decay is also forbidden by statistics (J)?

pi^0 spin is 0
photon spin is 1

But there are can be different situations with full orbital angular momentum of pi^0\gamma.
J=L+S and must be 0. Look like L is 0. But why? Why we can't have L=1 in this case? (unlike in \rho\to\pi^0\gamma decay)

 

[ChrisVer]

I would say that your first has spin 0, while the product has a spin 0 particle and a photon which is a spin 1 particle.

 

[ChrisVer]

If you put it in a J=1 state the total wavefunction is antisymmetric, and it has to be symmetric because of Bose statistics. [quoting from Vanadium 50 in the ref. link]

rho is a J=1 particle. It can't decay though to pi0 pi0. A similar question has be asked/answered here:
https://www.physicsforums.com/threads/why-cant-rho-o-decay-into-two-pi-o.278578/

 

[ChrisVer]

Again I am saying that the rho has J=1, so angular momentum conservation does not prohibit it to decay in a J=1 + J=0 particle.
The eta has J=0. And you try to make it decay into a J=1 and a J=0 particle? you can't compare these.
It's not only the CG coefficients, keep reading the conversation.

If you try to make the last configuration of pi0 and gamma to have a total angular momentum 0, then you have to set the orbital angular momentum of the two products to be L=1.
But then you are having an antisymmetric wavefunction describing your boson final states

 

[Vanadium 50]

I think this is much easier to see semiclassically: you are asking if a 0- --> 0- + radiation state is possible. It is not. A 0- state can have only monopole moments, and there are no magnetic monopoles, and the electric monopole is simply charge: zero in this case. Since no moment changes, there can be no radiation.