 Solution

Addition of two spin-one angular momenta gives $J=0,1,2$, with $M=-J,-J+1,\dots ,J$. To obtain the $J=1$ multiplet, its highest-weight state must have $M=1$ and be killed by $J_{+}=J_{+}^{(1)}+J_{+}^{(2)}$. On the two possible product states, the raising operator gives the same multiple of $|1;1⟩$, so their difference is required. Normalization and successive lowering giveFor example $J_{-}|1,1⟩=|1;-1⟩-|-1;1⟩=\sqrt{2}|1,0⟩$, and lowering once more gives the last expression. These states are all antisymmetric under interchanging the two constituents.

Identical spin-one particles are bosons, so the total state must be symmetric. Its $J=1$ spin factor is antisymmetric and must therefore multiply an antisymmetric spatial factor. Two particles in the same one-particle orbital have zero antisymmetric spatial state. The lowest allowed choice isFor distinguishable particles there is no exchange-symmetry constraint. Both can occupy the ground orbital, giving ${\psi }_1(r_A){\psi }_1(r_B)|1,M⟩$ and lowest energy $\boxed{2E_1}$, again with three spin states.