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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

$$ \rho^0 \longrightarrow \pi^0 \pi^0 $$ The parity af $\rho^0$ and $\pi^{0}$ are $$ \begin{aligned} & P_{\rho^0}=(-1)(+1)(-1)^0=-1 \\ & P_{\pi^0}=(-1)(+1)(-1)^0=-1 \end{aligned} $$ Due to conservation of parity in strong interactions, we require $$ P_{\rho^0}=P_{\pi^0} \cdot P_{\pi^0} \cdot(-1)^2 $$ which means the orbital angular momentuen of the final state is $L=1$. Due to conservation of total angular momentum, the total angular momentum of the final state must be equal to the total angular momentum of the initial $\rho_0$ $$ J=J_{\rho^0}=1 $$ Since $\pi^{0}$ are bosons so we require the total wavefunction to be symmetric. $L=1$ means the spacial part is anti-symmetric. so the spin part has to be anti-symmetric, which is impossible for two spin-0 sion.

Answer 4

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].

Answer 5

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.