# Why is the decay of a neutral rho meson into two neutral pions forbidden?

Why is the decay of a neutral rho meson into two neutral pions forbidden? (Other modes of decay are possible though.) Is it something with conservation of isospin symmetry or something else? Please explain in a bit more detail.

As far as I understand, due to conservation of angular momentum, the resulting system of neutral pions would need to have angular momentum 1, therefore, the identical neutral pions would be in an anti-symmetric state, which does not seem possible as they are bosons. Note that a neutral rho meson can decay into two neutral pions and a $\gamma$, although this decay is suppressed.

The charge conjugation symmetry is conserved in the strong interaction. In the above decay mode, neutral rho meson decay into two neutral Pion, neutral rho meson has -1 and neutral Pion has +1 for C such that decay violates C symmetry. Therefore it is forbidden.

• This answer is completely wrong, please vote it down so people do not get confused – Juanjo Nov 3 '18 at 15:31

If we look at isospin, $$\rho = |1,0\rangle$$ and $$\pi^0= |1,0\rangle$$.

Since SU(2) isospin is a really good symmetry in strong interactions, it must be conserved. Looking at the isospin of the final state:

$$$$|1,0\rangle \otimes |1,0\rangle = \sqrt{\frac{2}{3}} |2,0\rangle + 0 |1,0\rangle - \sqrt{\frac{1}{3}} |0,0\rangle$$$$

That is, there is no $$|1,0\rangle$$ component in the final state, and therefore the process is not allowed by SU(2) isospin symmetry.

Clebsch-Gordan isospin rule suggest decay of I=1,Iz=0 particle into two I=1,Iz=0 products has zero coefficients of Clebsch-Gordan [1x1].

• This doesn't look like a very clear explanation. You should expand on the details to make your answer somewhat understandable. – Anne O'Nyme Jun 13 '14 at 13:10