Why is ρ⁰ → π⁻π⁺ allowed but ρ⁰ → π⁰π⁰ forbidden by isospin conservation?

[Dilatino]

The decay

$$\rho^0 \rightarrow \pi^- \pi^+$$

occurs with a probability of $100\%$, whereas the decay

$$\rho^0 \rightarrow \pi^0 \pi^0$$

does not occur in nature, due to isosphin conservation.
I don't understand this.
Looking at the Isospin and its third component $¦I,I_3\rangle$ we have in the first decay

$$¦1,0\rangle \rightarrow ¦1,-1\rangle \otimes¦1,-1\rangle$$

and for the second decay

$$¦1,0\rangle \rightarrow ¦1,0\rangle \otimes¦1,0\rangle$$

There are a few things that confuse me to the effect, that I don't understand why the first decay is allowed (by isospin conservation) whereas the second is not:

1. To obtain the isospin of the final two-meson state, does one not have to add the isospin of the two mesons, such that it would be $2 \neq 1$ which is not the same as the isospino of the initial meson, such that both decays should be disallowed?

2. Adding the third components of the isospin for the two final-state mesons gives 0 which is the same as the third component of the isospin of the inital meson.

 

[Orodruin]

1. To obtain the isospin of the final two-meson state, does one not have to add the isospin of the two mesons, such that it would be 2≠12 \neq 1 which is not the same as the isospino of the initial meson, such that both decays should be disallowed?

No, you have to construct the decomposition of the final state into isospin irreps. The product of two isospin 1 states result in one quintuplet (isospin 2), one triplet (isospin 1) and one singlet (isospin 0). The part contributing to the decay is that with the same isospin as the in state, ie, the triplet.

Adding the third components of the isospin for the two final-state mesons gives 0 which is the same as the third component of the isospin of the inital meson.

Yes, but it is not part of the triplet and therefore not contributing.