The decay [tex] \rho^0 \rightarrow \pi^- \pi^+ [/tex] occurs with a probability of [itex]100\%[/itex], whereas the decay [tex] \rho^0 \rightarrow \pi^0 \pi^0 [/tex] does not occur in nature, due to isosphin conservation. I dont understand this. Looking at the Isospin and its third component [itex]¦I,I_3\rangle[/itex] we have in the first decay [tex] ¦1,0\rangle \rightarrow ¦1,-1\rangle \otimes¦1,-1\rangle [/tex] and for the second decay [tex] ¦1,0\rangle \rightarrow ¦1,0\rangle \otimes¦1,0\rangle [/tex] There are a few things that confuse me to the effect, that I dont 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 [itex]2 \neq 1[/itex] 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.