Question on Two Decays of Rho Meson

[dingo_d]

Homework Statement

I saw here, but since the thread is closed I need to ask my question in a new thread.

I have two decays: [itex]\rho^0\to \pi^+\pi^-[/itex] and [itex]\rho^0\to \pi^0\pi^0[/itex]. 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 [itex]\frac{1}{\sqrt{2}}[/itex]?

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

 

[fzero]

Homework Statement

I saw here, but since the thread is closed I need to ask my question in a new thread.

I have two decays: [itex]\rho^0\to \pi^+\pi^-[/itex] and [itex]\rho^0\to \pi^0\pi^0[/itex]. 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 [itex]\frac{1}{\sqrt{2}}[/itex]?

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.

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

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

 

[dingo_d]

Thank you :)