 Solution

Direct formation in $e^{+}{e}^{-}$ through one virtual photon selects the vector meson quantum numbers $J^{PC}=1^{--}$. The nonzero $K_S^0{K}_L^0$ mode and absence of $K_S^0{K}_S^0$ are consistent with this assignment: the two spin-zero kaons require odd orbital angular momentum for negative parity, whereas identical $K_S^0$ bosons cannot have an odd relative wave. This argument neglects the tiny CP violation in neutral kaons.

Assuming the $\rho \pi$ decays conserve isospin, their equal rates indicate $I=0$. The Clebsch-Gordan coefficients giveAn $I=1,I_3=0$ state lacks ${\rho }^0{\pi }^0$, while $I=2,I_3=0$ has squared coefficients $1:4:1$ rather than $1:1:1$. The branching fractions alone determine neither $J$ nor these relative phases; the single-photon formation and approximate strong interaction isospin conservation are essential additional inputs.