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.
 A: 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:
\begin{equation}
|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
\end{equation}
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.
A: 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. 
A: 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]. 
A: 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.
