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If the Hamiltonian commutes with total $S^z$, simultaneous spin rotations $e^{-i\varphi S_{\mathrm{tot}}^z}$ form a circle group symmetry. In a state of definite total $S^z$, the transverse one-point averages vanish because $S^x\pm i{S}^y$ change that eigenvalue. A chosen classical orientation or symmetry-breaking limit requires separate care when constructing a linear spin-wave approximation.

With unit lattice spacing and $\hslash =1$, a nearest-neighbour Heisenberg model has quadratic low-momentum ferromagnetic magnons and linear antiferromagnetic modes in the linear spin-wave approximation. The ferromagnetic polarized vacuum is an exact ground state, whereas the assumed antiferromagnetic ordered vacuum has quantum depletion of Néel order. In one dimension its logarithmic infrared divergence invalidates finite ordered magnetization and prevents the linear approximation from deciding the exact gap.

A spin-$S$ Hilbert space has dimension $2S+1$. Its Holstein–Primakoff transformation therefore uses only bosonic occupation numbers $0,1,\dots ,2S$. The square root annihilates the upper endpoint. A truncated linear spin-wave approximation formally enlarges this space; it is self-consistent only when boson depletion is small compared with $S$.

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.

For a nearest-neighbour Heisenberg antiferromagnet on a hypercubic lattice, the linear spin-wave approximation gives ${\gamma }_{\mathbf{k}}=d^{-1}{\sum }_j\cos (k_{j}a)$ and the displayed frequency. Near a gapless point its spin-wave velocity is $2JSa\sqrt{d}$. Its ground-state boson occupation is $[1/\sqrt{1-{\gamma }_{\mathbf{k}}^2}-1]/2$, producing quantum depletion of Néel order. Exact zero modes need an infrared regulator or separate global-rotation treatment.

The linear spin-wave approximation expands a Holstein–Primakoff transformation about a classical ordered state and retains the quadratic oscillator Hamiltonian. It gives the leading order-$S$ excitation energies above the order-$S^2$ classical energy. In a Heisenberg antiferromagnet, first make a bipartite spin rotation. The approximation also predicts a zero-point reduction of the order parameter; infrared-divergent quantum depletion of Néel order signals that the assumed ordered state is not a valid thermodynamic starting point.

For $H=-J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$ and unit spacing, the one-magnon energy is $2JS(1-\cos k)=4JS{\sin }^2(k/2)$. The exact one-magnon hopping matrix has diagonal $2JS$ and neighbour entries $-JS$. The same result follows from the linear spin-wave approximation. Its long-wavelength stiffness is $JS$, with energy proportional to $k^2$.

Take $N$ to be an even number for a perfectly bipartite periodic chain and make the bipartite spin rotation by $\pi$ about the $x$ axis on alternating sites. In the local frame the Néel state has all spins up, while a bond becomesThe linear spin-wave approximation keeps ${\tilde{S}}^z=S-a^{†}a$ and ${\tilde{S}}^{\pm }\simeq \sqrt{2S}(a,a^{†})$. ThereforeEvery site has two neighbours. Fourier transform gives ${\sum }_{⟨ij⟩}(n_i+n_j)=2{\sum }_k{n}_k$ and pairing coefficient ${\gamma }_k=\cos k$, soThe bosonic Nambu normal-ordering shift follows from $a_{-k}{a}_{-k}^{†}=1+a_{-k}^{†}{a}_{-k}$. It adds $NJS$ inside the matrix expression, which must be subtracted in its constant. HenceA periodic chain with an odd number of sites is frustrated at the boundary and lacks this exact two-sublattice reference; the bulk thermodynamic calculation uses the even-chain sequence.

For occupation much smaller than $2S$, expand the square root in the Holstein–Primakoff transformation. To leading order $S_n^{+}=\sqrt{2S}{a}_n+O(S^{-1/2})$, $S_n^{-}=\sqrt{2S}{a}_n^{†}+O(S^{-1/2})$ and $S_n^z=S-a_n^{†}{a}_n$. ConsequentlyPeriodic boundaries count each occupation in two bonds. The linear spin-wave approximation is thereforeThe displayed fluctuation Hamiltonian is bilinear and conserves boson number. The omitted terms are magnon interactions, quartic at the first omitted order. With occupation held of order one, they are smaller by $1/S$ than the order-$S$ quadratic terms. The polarized vacuum has no zero-point depletion in this ferromagnetic expansion.

Repeat the linear spin-wave approximation for bonds of lengths one and two. The polarized energy is $E_0=-N{S}^2(J_1+J_2)$ and each bond length contributes its own lattice Laplacian. ThusThe next-nearest-neighbour ferromagnetic magnon dispersion remains nonnegative for the stated positive exchanges and has its only zero at $k=0$ modulo $2\pi$ when $J_1>0$. At small $k$, ${\epsilon }_k\simeq S(J_1+4J_2)k^2$: longer-range ferromagnetic exchange increases the stiffness.

At this harmonic order, a bosonic occupation configuration has energy $E_0+{\sum }_k{\epsilon }_k{n}_k$. The one-magnon dispersion is also exact. General multi-magnon energies acquire the interactions omitted in the large-$S$ expansion; the formula does not claim an exact diagonalization of the full interacting many-spin Hamiltonian.

The linear spin-wave approximation expands about the rotated all-up state, with ${\hat{n}}_n=a_n^{†}{a}_n$:In the longitudinal product, $S_n^z{S}_{n+1}^z=S^2-S({\hat{n}}_n+{\hat{n}}_{n+1})+O(S^0)$. Each site belongs to two bonds, so the resulting quadratic Hamiltonian isThe omitted terms are of order $S^0$ at fixed small occupation number. The pair terms describe the quantum fluctuations that were missing from the classical Néel state.

Use dimensionless spin operators and nearest-neighbour exchange $J_H>0$ on a regular bipartite graph with N sites and coordination number of a lattice z. Count each bond once and define ${\gamma }_{\mathbf{k}}=z^{-1}{\sum }_{\delta }{e}^{i\mathbf{k}\cdot \delta }$, real for the inversion-symmetric examples considered here.

For the ferromagnetic Heisenberg model, $H_F=-J_H{\sum }_{⟨ij⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$. Expanding the Holstein–Primakoff transformation about all spins along +z givesup to interactions of higher order in boson density, with $E_{\mathrm{cl}}=-Nz{J}_H{S}^2/2$. The isotropic spin-wave vacuum energy comparison starts with this normal ordering. The linear spin-wave approximation then givesEach ferromagnetic magnon lowers total $S^z$ by one. The fully aligned state is an exact eigenstate and ground state; there is no harmonic vacuum depletion or added zero-point energy in this normal-ordered Hamiltonian. The global spin-rotation multiplet remains degenerate. On a hypercubic lattice of spacing a, $z=2d$ and ${\gamma }_{\mathbf{k}}=d^{-1}{\sum }_j\cos k_{j}a$, so $E_F\simeq J_{H}S{a}^2{k}^2$. The exact one-magnon sector agrees with this result even at finite S; interactions enter for multiple magnons.

For the Heisenberg antiferromagnet, take $H_A=+J_H{\sum }_{⟨ij⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$ and expand around a Néel state. On A, $S^z=S-a^{†}a$ and $S^{+}\simeq \sqrt{2S}a$; on B, $S^z=-S+b^{†}b$ and $S^{+}\simeq \sqrt{2S}{b}^{†}$. A bond becomesWith N/2 momenta in the magnetic Brillouin zone,The bosonic Bogoliubov transformation has $u_k^2-v_k^2=1$ and $\tanh 2{\theta }_k={\gamma }_k$. For example $a_k=u_k{\alpha }_k-v_k{\beta }_{-k}^{†}$ and $b_{-k}=u_k{\beta }_{-k}-v_k{\alpha }_k^{†}$. It gives two branches in this reduced zone,The reduced-zone counting of bipartite spin waves gives N oscillators in total, not 2N. The negative vacuum correction and quantum depletion of Néel order, $\delta S=(2/N){\sum }_{\mathrm{MBZ}}{v}_k^2$, distinguish the antiferromagnet from the ferromagnet. At nonzero temperature replace $v_k^2$ by $v_k^2+(u_k^2+v_k^2)n_B(E_k)$. Hypercubic small-momentum modes have $E_A\simeq 2J_{H}Sa\sqrt{d}|k|$, hence linear physical frequency $E_A/\hslash$.

Self-consistency requires depletion much smaller than S and respects the Holstein–Primakoff occupation constraint. In one dimension the antiferromagnetic vacuum depletion diverges logarithmically, invalidating ordered spin-wave theory; it does not alone prove an exact gap or its absence. At positive temperature, isotropic short-range models in d≤2 cannot sustain such magnetic order by the Mermin-Wagner theorem. In three dimensions the quadratic ferromagnetic modes give the low-temperature magnetization reduction proportional to $T^{3/2}$, whereas the antiferromagnetic thermal reduction scales as $T^2$. Anisotropy, fields, frustration and magnon interactions alter these conclusions. ^[ref]

Rotate alternate sites by $\pi$ about a transverse axis so all local reference spins point up. Choosing the transverse axes to give positive anomalous coupling, the linear spin-wave approximation yieldsUse lattice spacing one and $a_m=N^{-1/2}{\sum }_k{e}^{ikm}{a}_k$, with all $N$ momenta $k=2\pi j/N$, $j=-N/2,\dots ,N/2-1$, in $[-\pi ,\pi )$. Then ${\gamma }_k=\cos k$ andReordering $a_{-k}{a}_{-k}^{†}=a_{-k}^{†}{a}_{-k}+1$ gives the requested Nambu form with constant $-JNS(S+1)$. Its extra $JSN$ restores $-JN{S}^2$ before diagonalization. This full-zone convention has no additional sublattice or half-zone factor.

The Holstein–Primakoff transformation converts a spin algebra into bosonic operators while retaining the finite physical spin Hilbert space. With $[a_n,a_m^{†}]={\delta }_{nm}$, writeThe Holstein–Primakoff occupation constraint is $0\le n_n\le 2S$. Acting on $|n⟩$, the raising matrix element is $\sqrt{n(2S-n+1)}$ and the lowering element is $\sqrt{(n+1)(2S-n)}$. Their squared difference is $2(S-n)$, reproducing $[S^{+},S^{-}]=2S^z$. Operator ordering inside the square roots is essential. For low boson occupation relative to $S$, expand them to obtain $S^{+}\simeq \sqrt{2S}a$ and $S^{-}\simeq \sqrt{2S}{a}^{†}$. Keeping quadratic terms gives the linear spin-wave approximation; higher terms describe magnon interactions. Large $S$ helps control the expansion, but a low-dimensional infrared divergence can invalidate it even at large finite $S$.

For the ferromagnet choose $H_F=-J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, $J>0$, with unit lattice spacing and $\hslash =1$. Expand around the all-up state. A longitudinal bond contributes $-J{S}^2+JS(n_n+n_{n+1})$, and the transverse exchange contributes $-JS(a_n^{†}{a}_{n+1}+a_{n+1}^{†}{a}_n)$. After a Fourier transform,These are ferromagnetic magnons. The polarized vacuum is an exact ground state of the full ferromagnetic Hamiltonian: every bond is in its maximum total-spin sector, minimizing that bond energy. It therefore has no quantum zero-point reduction of its magnetization. The one-magnon dispersion is also exact, although interacting many-magnon states need not be free bosons. Rotating the polarization generates the maximum-total-spin ground-state multiplet. Selecting a definite direction in a thermodynamic symmetry-breaking description breaks $SU(2)$ to $U(1)$; the two transverse spin components form a canonical pair, producing a single quadratic type-B Goldstone boson. A small aligning field gaps it by the Zeeman energy.

At nonzero temperature the one-dimensional ferromagnet is different: its thermal magnon number contains $\int dk[e^{\beta JS{k}^2}-1{]}^{-1}$, with infrared behaviour $\int dkT/k^2$. The divergent magnetization reduction signals the absence of finite-temperature long-range order in an infinite short-range isotropic chain, consistent with the Mermin-Wagner theorem. It does not disallow the polarized zero-temperature ground states just constructed.^[ref]

For the antiferromagnet use $H_{AF}=+J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, still $J>0$, on an even bipartite chain. The classical reference state alternates up and down. Rotate one sublattice through $\pi$ so that both local quantization axes point along their reference spins. In unrotated physical components, to leading order $S_A^z=S-a^{†}a$, $S_B^z=-S+b^{†}b$, $S_A^{+}\simeq \sqrt{2S}a$ and $S_B^{+}\simeq \sqrt{2S}{b}^{†}$. Choosing Fourier phases using the actual site positions givesThe reduced Brillouin zone has $M=N/2$ momenta and length $\pi$. The pair creation and annihilation terms express the fact that an antiparallel reference product state is not an eigenstate. A bosonic Bogoliubov transformation with $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, $\tanh 2{\theta }_k=\cos k$ diagonalizes each two-mode block. With the sign convention $a_k=u_k{\alpha }_k-v_k{\beta }_{-k}^{†}$ and $b_{-k}=u_k{\beta }_{-k}-v_k{\alpha }_k^{†}$, the result isThus the harmonic antiferromagnetic modes have linear, rather than quadratic, long-wave dispersion. In an unfolded Brillouin-zone description the same result is represented by zeros at both zero and the antiferromagnetic wavevector; these are not extra independent degrees of freedom beyond the two-sublattice accounting.

The Bogoliubov vacuum has no $\alpha$ or $\beta$ excitations but contains original spin bosons, with $v_k^2=\frac{1}{2}(|\sin k{|}^{-1}-1)$. Its staggered magnetization is $m_s=S-M^{-1}{\sum }_{k\in \mathrm{RBZ}}{v}_k^2$. This quantum depletion of Néel order diverges logarithmically at small $k$ in one dimension. Linear spin-wave theory therefore does not establish a stable ordered antiferromagnetic ground state for this chain. Its nominal gapless dispersion is a result of expanding around an invalid ordered reference, not a general proof of the exact excitation spectrum. For the nearest-neighbour isotropic chain, integer-spin systems have a Haldane gap, whereas half-integer-spin systems are gapless; the spin-one-half chain in particular has fractional spin excitations rather than a simple dilute gas of ordered-state magnons. This distinction requires physics beyond the harmonic expansion, including the topological term in the continuum theory.^[ref]

For a local spin quantized along its positive $\tilde{z}$ axis, the exact Holstein–Primakoff transformation isThe bosonic operators obey $[a_j,a_k^{†}]={\delta }_{jk}$. The physical local subspace is $0\le n_j\le 2S$; the larger unconstrained boson Fock space is not itself the finite spin Hilbert space. The ordering of the square roots matters, since $n_j$ does not commute with $a_j$.

This represents spin lowering using a bosonic creation operator, making a fully polarized spin state the boson vacuum. On the physical subspace it reproduces the spin commutators and the Casimir ${\tilde{\mathbf{S}}}_j^2=S(S+1)$. Expanding the square roots for small occupation relative to $S$ organizes a controlled large-$S$ linear spin-wave approximation; it is not an assumption that a spin has an infinite number of physical levels.

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.

Expand around the $+z$ polarized ground state. With $n_j=a_j^{†}{a}_j$ fixed as $S\to \infty$, the Holstein–Primakoff transformation gives $S_j^{+}=\sqrt{2S}{a}_j+O(S^{-1/2})$, $S_j^{-}=\sqrt{2S}{a}_j^{†}+O(S^{-1/2})$ and $S_j^z=S-n_j$. The longitudinal exchange contributesUseThe transverse exchange then gives $-JS(a_{j+1}^{†}{a}_j+a_j^{†}{a}_{j+1})$, and the Dzyaloshinskii-Moriya interaction gives $-iDS(a_{j+1}^{†}{a}_j-a_j^{†}{a}_{j+1})$. Discarding the constant $-LJ(1+u)S^2$ and all $O(S^0)$ interactions yieldsHere periodic boundary conditions make each $n_j$ occur twice in the longitudinal sum. The approximation is the linear spin-wave approximation, controlled by small local magnon occupation compared with $S$.

Choose the unitary discrete Fourier transformThe finite-sum orthogonality identity gives $[b_k,b_q^{†}]={\delta }_{kq}$. It also gives ${\sum }_j{a}_{j+1}^{†}{a}_j={\sum }_k{e}^{-ik}{b}_k^{†}{b}_k$. Substituting $A,B$ from solution therefore gives the dispersion relationUsing the opposite sign in the Fourier transform reverses $k$ and the sine term together, without changing the spectrum.

For $u\gg 1$ every $ϵ(k)>0$, so the ground state of this quadratic Hamiltonian is the bosonic vacuum, $b_k|\mathrm{GS}⟩=0$ for every $k$. In the original spin variables it is the fully $+z$ polarized product state. Its energy is zero after the constant subtraction. Repeating the construction about the fully $-z$ polarized state describes the other easy-axis sector of the original Hamiltonian.

A one-magnon excitation $b_k^{†}|\mathrm{GS}⟩$ has energy $ϵ(k)$ above the reference ground state. A normalized many-magnon Fock state with occupations $n_k$ has excitation energy ${\sum }_k{n}_kϵ(k)$, within the dilute regime of the linear spin-wave approximation. The lowest excitation in this sector has the minimum allowed $ϵ(k)$.

For a linear spin-wave approximation, the ground state is empty of the diagonal quasiparticles but contains the original spin bosons. If $a_k=u_k{\alpha }_k-v_k{\alpha }_{-k}^{†}$, their occupation number is $v_k^2$. The staggered magnetization is reduced from $S$ to $S-N^{-1}{\sum }_k{v}_k^2$. For the one-dimensional nearest-neighbor Heisenberg antiferromagnet, $v_k^2=(|\sin k{|}^{-1}-1)/2$ and the integral diverges logarithmically. This invalidates finite ordered magnetization within the approximation; it is not itself a determination of the exact spectral gap.