# Rho0 decays

1. Jul 2, 2012

### dingo_d

1. The problem statement, all variables and given/known data
I saw here, but since the thread is closed I need to ask my question in a new thread.

I have two decays: $\rho^0\to \pi^+\pi^-$ and $\rho^0\to \pi^0\pi^0$. The text says that both are allowed with respect to laws of conservation of energy and impulse. Why is the second one forbidden?

Both rho meson and pi meson, are a triplet in isospin space, so I don't see a problem there. But how are both allowed with respect to conservation of impulse, if the impulse of rho is 1, and of pions is 0 (impulse as in J=L+S)?

Is it just because the Clebsch-Gordan coefficient in the second decay is 0 and in the first it's $\frac{1}{\sqrt{2}}$?

So the coupling of two pi0 is forrbidden in the isospin coupling? Is that the answer?

2. Jul 3, 2012

### fzero

The final state is a two-particle state, so it is possible to have nonzero orbital angular momentum ($L\neq 0$). However, once you identify the possible orbital state, you must also consider the symmetry of the wavefunction under exchange of the $\pi$s.

This is not an issue, since that coupling exists. In fact, $\rho^0\to \pi^0\pi^0\gamma$ is an allowed decay channel.

3. Jul 3, 2012

Thank you :)