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A bipartite spin rotation rotates the spin on one sublattice by $\pi$, converting an alternating classical Néel state to an all-up reference state. Rotation about $x$ sends $S^z\mapsto -S^z$ and $S^{\pm }\mapsto S^{\mp }$ on that sublattice. For an isotropic nearest-neighbor Heisenberg antiferromagnet, a bond becomes $-S_i^z{S}_j^z+(S_i^{+}{S}_j^{+}+S_i^{-}{S}_j^{-})/2$. The transformation is unitary and preserves the spin commutation relations; it changes the reference frame, not the model's spectrum.

The Holstein–Primakoff transformation represents a spin-$S$ operator algebra with a bosonic creation operator and annihilation operator:The order of the square root and oscillator operator matters. On the physical Fock states $|n⟩$, $0\le n\le 2S$, the raising and lowering matrix elements are $\sqrt{n(2S-n+1)}$ and $\sqrt{(n+1)(2S-n)}$. Their squared difference gives $[S^{+},S^{-}]|n⟩=2(S-n)|n⟩$, proving the spin commutation relations. Expanding the square root yields the linear spin-wave approximation and its interaction corrections.

The Holstein–Primakoff transformation realizes a spin algebra using bosonic creation operators and bosonic annihilation operators. A boson counts a unit of spin deviation from the local positive quantization axis: ${\tilde{S}}^z=S-n$ with $n=a^{†}a$. The square root in ${\tilde{S}}^{+}$ supplies the correct spin-ladder matrix elements. In particular,Their commutator is $2(S-n)$ on every physical number state, reproducing the Spin commutation relations. Although the oscillator Fock space is infinite, the physical subspace is onlyThe order of the square root and oscillator operator matters in obtaining these coefficients.

When the spin deviations are dilute, expand the square root in $n/(2S)$. The leading transverse operators are linear in the bosonic creation operators and bosonic annihilation operators and turn exchange interactions into a quadratic Hamiltonian, which can be diagonalized. Higher powers describe interactions between spin waves. The method therefore organizes corrections by inverse powers of $S$ about an ordered classical configuration. Large $S$ alone is insufficient: the average depletion must remain small compared with $S$. Possible problems include the occupation constraint after truncation, noncollinear order requiring site-dependent axis rotations, zero modes, and infrared fluctuations that destroy the assumed dilute expansion.

Choose the local positive $\tilde{z}$ axis along the classical spin. For nonzero field, a proper rotation isIt preserves the Spin commutation relations and turns the field term into $-H_{f}S{\sum }_j{\tilde{S}}_j^z$. At zero field choose the classical direction $+x$, so $s=1$. Expanding about the original $z$ axis without this rotation would put the reference state away from the classical minimum and would not give a controlled small-deviation expansion.

To the required order,The longitudinal exchange gives $-J{S}^2+JS(n_j+n_{j+1})+O(S^0)$ on each bond. The transverse exchange givesFinally the field contributes its classical constant and $H_{f}S{n}_j$. Each site belongs to two bonds, so after removing the order-$S^2$ constant,These are the order-$S$ terms; the discarded products of number operators and the next square-root terms are order $S^0$ for occupations held fixed as $S\to \infty$. A different transverse-axis phase can reverse the sign of $B$ without changing the spectrum, but the sign above follows from the stated rotation and oscillator convention.

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.

The Holstein–Primakoff transformation acts on the physical Fock states $|n⟩$, $0\le n\le 2S$, with the specified operator ordering. Its matrix elements areThe unavailable endpoint states have zero coefficient. These formulas directly give $[S^z,S^{\pm }]|n⟩=\pm S^{\pm }|n⟩$. MoreoverThus the spin commutation relations hold in units $\hslash =1$. The Holstein–Primakoff occupation constraint is essential: the spin Hilbert space has dimension $2S+1$, so unrestricted boson Fock space is not itself this finite-spin representation. The square root and oscillator operators must retain their order for the displayed matrix elements and endpoint conditions to agree.

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 commutation relations give $[S_{\mathrm{tot}}^z,S_n^{\pm }]=\pm S_n^{\pm }$. Using the product rule for a commutator,and the conjugate term also vanishes. The longitudinal field commutes with every $S_n^z$, so $[H,S_{\mathrm{tot}}^z]=0$. The Hamiltonian is invariant under the global spin rotation $U(\theta )=e^{-i\theta S_{\mathrm{tot}}^z}$: its continuous symmetry group is $U(1)$. Simultaneous energy and total-magnetization eigenstates can be chosen, and evolution preserves each fixed-magnetization sector. Local magnetization can still flow between sites; generally $[H,S_n^z]\ne 0$. In the fermion representation the conserved number will be $N_f=N/2-S_{\mathrm{tot}}^z$.

The Heisenberg equation of motion and the spin commutation relations give, componentwise,Thus $\boxed{d\mathbf{S}/dt=\gamma \mathbf{S}\times \mathbf{B}}$. For constant magnetic field this is Larmor precession: the component along $\mathbf{B}$ is conserved and the transverse components rotate. In particular, for $\mathbf{B}=B\hat{\mathbf{z}}$,The signed angular velocity is $-\gamma \mathbf{B}$; the positive precession rate is $|\gamma B|$.

There is a normalization mismatch in the printed definitions: ${\sigma }^x\pm i{\sigma }^y$ have nonzero entries of magnitude two, whereas a single fermionic creation or annihilation operator has magnitude one. To obtain the intended Jordan–Wigner transformation, use normalized spin ladders $s_j^{\pm }=({\sigma }_j^x\pm i{\sigma }_j^y)/2$. The given fermion expressions represent these $s_j^{\pm }$. The two-site Jordan-Wigner normalization and signs then fix every factor unambiguously.

Put $n_j=f_j^{†}{f}_j$ and $P_1=e^{i\pi n_1}=1-2n_1$. The canonical anticommutation relations give $P_1^2=I$, $P_1{f}_1=-f_1{P}_1$, and $[P_1,f_2]=[P_1,f_2^{†}]=0$. HenceandThese are the spin commutation relations, with distinct sites commuting. If one retains the unscaled symbols ${\sigma }_j^{\pm }=2s_j^{\pm }$, the first and last displayed commutators instead equal $-4{\sigma }_j^z$; the off-site commutators remain zero. Setting the unscaled ${\sigma }_j^{-}$ equal to $f_j$ would contradict the original Pauli matrices.

The Holstein–Primakoff transformation replaces each finite-dimensional spin algebra by a bosonic creation operator and annihilation operator, while restricting the local occupation number to $0\le n_j\le 2S$. The number $n_j=a_j^{†}{a}_j$ counts deviations from the fully polarized reference quantum state. Its zero-boson Fock vacuum has ${\tilde{S}}^z=S$.

The representation is exact on this physical subspace: the ordered square root gives ${\tilde{S}}^{+}|n⟩=\sqrt{n(2S-n+1)}|n-1⟩$ and ${\tilde{S}}^{-}|n⟩=\sqrt{(n+1)(2S-n)}|n+1⟩$, reproducing the spin commutation relations. For $n_j\ll S$,so the leading Hamiltonian is quadratic in bosons and describes spin waves.

The square root generates boson interactions beyond leading order and its operator ordering matters. Truncation also enlarges the Hilbert space to unphysical occupations unless the low-density approximation remains self-consistent. Large fluctuations, particularly near a gap closing or in low dimensions, can invalidate the linear spin-wave approximation even when its algebraic manipulation is straightforward.

Choose the positive-$x$ reference ground state and use the cyclic rotationThis is a right-handed permutation, so it preserves the spin commutation relations. It puts the ordered direction along the Holstein–Primakoff transformation's quantization axis. To the required order,The longitudinal exchange contributes $-L{J}^x{S}^2+2S{J}^x{\sum }_j{n}_j+O(S^0)$. The transverse terms contributeDropping the constant and all terms of order $S^0$ or smaller gives the stated linear spin-wave approximation, withThe anomalous pair terms are present because the two transverse couplings differ. A different right-handed choice of transverse axes can change the sign of $B$ by a boson phase convention, while leaving the dispersion relation unchanged.

The Holstein–Primakoff transformation realizes a spin algebra using bosonic creation operators and bosonic annihilation operators. A boson counts a unit of spin deviation from the local positive quantization axis: ${\tilde{S}}^z=S-n$ with $n=a^{†}a$. The square root in ${\tilde{S}}^{+}$ supplies the correct spin-ladder matrix elements. In particular,Their commutator is $2(S-n)$ on every physical number state, reproducing the Spin commutation relations. Although the oscillator Fock space is infinite, the physical subspace is onlyThe order of the square root and oscillator operator matters in obtaining these coefficients.

When the spin deviations are dilute, expand the square root in $n/(2S)$. The leading transverse operators are linear in the bosonic creation operators and bosonic annihilation operators and turn exchange interactions into a quadratic Hamiltonian, which can be diagonalized. Higher powers describe interactions between spin waves. The method therefore organizes corrections by inverse powers of $S$ about an ordered classical configuration. Large $S$ alone is insufficient: the average depletion must remain small compared with $S$. Possible problems include the occupation constraint after truncation, noncollinear order requiring site-dependent axis rotations, zero modes, and infrared fluctuations that destroy the assumed dilute expansion.

Choose the local positive $\tilde{z}$ axis along the classical spin. For nonzero field, a proper rotation isIt preserves the Spin commutation relations and turns the field term into $-H_{f}S{\sum }_j{\tilde{S}}_j^z$. At zero field choose the classical direction $+x$, so $s=1$. Expanding about the original $z$ axis without this rotation would put the reference state away from the classical minimum and would not give a controlled small-deviation expansion.

To the required order,The longitudinal exchange gives $-J{S}^2+JS(n_j+n_{j+1})+O(S^0)$ on each bond. The transverse exchange givesFinally the field contributes its classical constant and $H_{f}S{n}_j$. Each site belongs to two bonds, so after removing the order-$S^2$ constant,These are the order-$S$ terms; the discarded products of number operators and the next square-root terms are order $S^0$ for occupations held fixed as $S\to \infty$. A different transverse-axis phase can reverse the sign of $B$ without changing the spectrum, but the sign above follows from the stated rotation and oscillator convention.

The continuous internal symmetry is a common rotation around $z$:Both the transverse exchange and the Dzyaloshinskii-Moriya interaction contain one raising and one lowering spin operator, so the phases cancel. Equivalently, $[H,{\sum }_j{S}_j^z]=0$ and the continuous group is U(1).

The discrete internal symmetry is time reversal, $\mathcal{T}{\mathbf{S}}_j{\mathcal{T}}^{-1}=-{\mathbf{S}}_j$, together with $\mathcal{T}i{\mathcal{T}}^{-1}=-i$. Each term in $H$ contains two spin operators, so it is invariant. The operation is antiunitary; treating the reversal of all three spin components as a unitary rotation would violate the spin commutation relations. It reverses the magnetization and interchanges the two easy-axis ordered orientations. A convenient equivalent representative, composed with a $z$ rotation, is $R_z(\pi )\mathcal{T}$, which reverses $S^z$ while preserving $S^x,S^y$ as operators.

The statement about one continuous and one discrete internal symmetry describes generic parameters. At $D=0$, additional unitary $\pi$ rotations about transverse axes are allowed; at $D=u=0$, the continuous internal symmetry enlarges to full spin rotation symmetry. Spatial translations of the uniform chain are separate symmetries of the sites.