 Solution

Use $\hslash =1$ throughout. Let $N_a=a^{†}a$, $N_b=b^{†}b$ and $N=N_a+N_b$. Operators belonging to different boson species commute. The proposed Schwinger boson representation givesConsequently the first of the spin commutation relations isThe number operator commutators $[N_a,a^{†}]=a^{†}$, $[N_a,a]=-a$ and their $b$ counterparts giveThus all the spin commutation relations hold. Also $[N,S^{\pm }]=[N,S^z]=0$: each fixed-$N$ sector is invariant. To identify its spin, calculate the Casimir operator:The irreducible spin representation with prescribed $s$ therefore requires $N=2s$, and its normalized Fock basis correspondence isConversely, $s=(n_a+n_b)/2$ and $m=(n_a-n_b)/2$. The nonnegative integer occupation numbers give precisely $m=-s,-s+1,\dots ,s$, and there are $2s+1$ such states. As a check, the spin raising operator acts aswhich is the usual spin ladder operator coefficient $\sqrt{s(s+1)-m(m+1)}$.

For a lattice of fixed spins, the local spin constraint in the Schwinger boson representation is essential:The full two-species bosonic Fock space also contains other local irreducible spin representations; fixing only the total boson number does not remove them.