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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.

An irreducible spin representation is a nonzero Hilbert space with no proper nonzero closed invariant subspace under all three angular momentum operators. Equivalently, the associated unitary representation of SU(2) cannot be decomposed into two nonzero orthogonal invariant pieces. The irreducible possibilities are labelled by a spin $j\in \{0,\frac{1}{2},1,\frac{3}{2},\dots \}$ and have dimension $2j+1$.

Choose an orthonormal basis $|j,m_s⟩$, with $m_s=-j,-j+1,\dots ,j$. The angular momentum commutation relations imply $[{\mathbf{S}}^2,S_i]=0$, $[S_3,S_{\pm }]=\pm \hslash S_{\pm }$, $[S_{+},S_{-}]=2\hslash S_3$, and $S_{+}^{†}=S_{-}$. The complete action isIn particular, the spin raising operator annihilates $|j,j⟩$ and the spin lowering operator annihilates $|j,-j⟩$. The coefficient follows by taking the squared norm and using $S_{\mp }{S}_{\pm }={\mathbf{S}}^2-S_3^2\mp \hslash S_3$. Thus the ladder operators connect every member of the basis, while terminating at its two endpoints.

Make a bipartite spin rotation: rotate every odd site's spin through $\pi$ about the $x$ axis, using $U={\prod }_{n\mathrm{odd}}{e}^{-i\pi S_n^x}$. This unitary conjugation preserves the spin commutation relations. At an odd site it sends $(S^x,S^y,S^z)$ to $(S^x,-S^y,-S^z)$ and exchanges the spin raising operator and spin lowering operator.

Each bond joins one rotated and one unrotated site. In the transformed operators,ThusThe selected Néel state becomes an all-up reference state, so a single Holstein–Primakoff transformation convention works on both sublattices.

The spin raising operator and spin lowering operator areThe angular momentum commutation relations implyso $S_{\pm }|s,\sigma ⟩$ is proportional to $|s,\sigma \pm 1⟩$. Sinceits squared norm isChoosing the conventional positive phase gives

Using the spin ladder operators,and thereforeIn the ordered product basis of the tensor product of quantum systemsthe matrix representation isThe two parallel-spin states are already eigenvectors. Diagonalizing the central block gives the four energy eigenvaluesThis is the spectrum of the Two-spin Heisenberg Hamiltonian in an opposing longitudinal field.

The asserted ground-state ordering requires the antiferromagnetic case $\alpha >0$. As $B\to 0$,The corresponding state at $E_{-}$ is the unique spin-one-half singlet statewhereas the three states at the first excited energy form the spin-one-half triplet state.

Indeed, for the total angular momentum operatorone has at $B=0$The Hamiltonian then has rotational symmetry, so energy depends only on the total-spin sector. The singlet has total spin $s=0$ and multiplicity one; the triplet has $s=1$ and multiplicity $2s+1=3$. This rotational symmetry explains why the triplet energy eigenspace has dimension three.