Why is the decay of a neutral ρ meson to two neutral π's disallowed?

[Able]

Hi all. I'm looking at the decay of a neutral ρ meson to two neutral π's. I think it is disallowed but I can't figure out why. The $J^{P}$ of the ρ is $1^{-}$ and $0^{-}$ for the neutral π's. The formula $P(ρ^{0})=(-1)^{L}P(π^{0})P(π^{0})$ then says that the decay is allowed provided the π's are produced in an L=1 state. If we insist that total angular momentum is conserved, we initially start with J=1 for the ρ which will decay to two π's with S=0 so using J=L+S, L=1 for the pions. If this argument is correct then the decay is not disallowed by parity. What else could disallow the decay? A colour argument? If it helps solve the problem $ρ->π^{-}π^{+}$ is allowed.

 

[Vanadium 50]

What else is conserved in strong interactions?

 

[Meir Achuz]

Two identical pi0s cannot be in an antisymmetric L=1 state.