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After a bipartite spin rotation, a leading Holstein–Primakoff transformation expansion gives $H_2=-J\Delta N{S}^2+JS{\sum }_k[2\Delta n_k+\cos k(a_k{a}_{-k}+a_k^{†}{a}_{-k}^{†})]$. The bosonic Bogoliubov diagonalization gives the displayed spectrum and vacuum energy $-J\Delta N{S}^2+JS{\sum }_k[\sqrt{{\Delta }^2-{\cos }^{2}k}-\Delta ]$. The anisotropy gap is $2JS\sqrt{{\Delta }^2-1}$. At $\Delta =1$ the finite-angle transformation is singular at its zero modes, and quantum depletion of Néel order is infrared divergent in one dimension.

For the XY model with equal ferromagnetic exchange $J$ in its plane and field term $-hS{\sum }_j{S}_j^x$, rotate the local quantization axis along the Classical spin. The Holstein–Primakoff transformation gives the number coefficient ${\alpha }_k=S(|h|+2J-J\cos k)$ and anomalous coefficient ${\Delta }_k=JS\cos k$ with one transverse-axis convention. Bosonic Bogoliubov diagonalization gives ${ϵ}_k=\sqrt{{\alpha }_k^2-{\Delta }_k^2}$ and field gap $S\sqrt{|h|(|h|+2J)}$. A nonzero field stabilizes every oscillator mode.

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 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.

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.

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.

The Holstein–Primakoff transformation acts on the physical number-state subspace $0\le n\le 2S$. Its operator ordering matters: $S^{+}=\sqrt{2S-\hat{n}}a$ lowers the number first. Thereforewhile $S^z|n⟩=(S-n)|n⟩$. The square roots vanish at the appropriate ends, so the spin operators preserve the Holstein–Primakoff occupation constraint. The bosonic Fock space outside this subspace is not part of the spin-$S$ representation.

These matrix elements give $[S^z,S^{\pm }]=\pm S^{\pm }$. The two products on $|n⟩$ are $(n+1)(2S-n)$ and $n(2S-n+1)$, respectively, so their difference is $2(S-n)$. ThusAlso $(S^z{)}^2+\frac{1}{2}(S^{+}{S}^{-}+S^{-}{S}^{+})=S(S+1)I$. The transformation therefore has both the correct algebra and the correct spin Casimir operator, not just its small-occupation approximation.

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.

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.

Use the Holstein–Primakoff transformation on the physical occupation number subspace $0\le n\le 2S$. With the canonical commutation relation $[a,a^{†}]=1$, the ordered square root givesBoth endpoints are respected: $S^{+}|0⟩=0$ and $S^{-}|2S⟩=0$. Consequently,Since these Fock states form a basis of the physical spin space, $[S^{+},S^{-}]=2S^z$ there. Similarly $[S^z,S^{\pm }]=\pm S^{\pm }$. The Holstein–Primakoff occupation constraint is essential: unrestricted bosonic occupation would not represent a spin-$S$ Hilbert space.

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]

Use the Holstein–Primakoff transformation only on the physical occupation sector $0\le n\le 2S$, with $n=a^{†}a$ and $S^{-}=a^{†}\sqrt{2S-n}$. Acting on number states givesConsequently $S^{+}{S}^{-}=(n+1)(2S-n)$ and $S^{-}{S}^{+}=n(2S-n+1)$, soThe endpoint factors keep the spin operators inside the physical sector. The square-root ordering matters; replacing it by an unrestricted linear oscillator would not give the exact finite-spin algebra.

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.

Quantize along the selected field direction by the right-handed relabeling ${\tilde{S}}^z=S^x$, ${\tilde{S}}^x=S^y$, ${\tilde{S}}^y=S^z$. Then the exact Holstein–Primakoff transformation in c gives $S_j^x=S-n_j$. Keep exchange terms of order $S^2$ and $S$, with $J$ and the printed $B$ held fixed. For a nearest-neighbor bond,On a periodic cubic lattice with $N$ sites and spacing $a$, there are $3N$ undirected bonds. Consequently,Use the unitary lattice Fourier transform $a_j=N^{-1/2}{\sum }_{\mathbf{k}}{e}^{i\mathbf{k}\cdot {\mathbf{r}}_j}{a}_{\mathbf{k}}$. Each positive lattice direction contributes $2(1-\cos k_{\alpha }a)$, so the Hamiltonian operator is already diagonal in momentum space:There are no anomalous $aa$ or $a^{†}{a}^{†}$ terms, so no Bogoliubov transformation or extra zero-point correction is needed. At long wavelengths,At $B=0$ this is the quadratic ferromagnetic Goldstone mode; for $B>0$ it is gapped by $BS$. Energies are written with $\hslash =1$; an excitation frequency is ${\omega }_{\mathbf{k}}={\epsilon }_{\mathbf{k}}/\hslash$. In this isotropic ferromagnet the fully aligned vacuum and its one-magnon dispersion are also exact; higher-magnon interactions are among the terms discarded from the linear spin-wave Hamiltonian operator.

Choose the positive-$x$ polarized ground state selected by $B>0$, or its symmetry-selected $B\to 0^{+}$ limit. It is exactly the boson vacuum in the Holstein–Primakoff transformation. Thus $n_j|\mathrm{GS}⟩=0$ and $S_j^x=S-n_j$ giveFor the transverse components, $S^y=({\tilde{S}}^{+}+{\tilde{S}}^{-})/2$ and $S^z=({\tilde{S}}^{+}-{\tilde{S}}^{-})/(2i)$. In the vacuum, the only nonzero quadratic ladder-operator expectation iswhich follows using the full ordered square-root operators, or directly from ${\tilde{S}}^{-}|S,S⟩=\sqrt{2S}|S,S-1⟩$. ThereforeAn independent exact spin argument starts from $S_j^x|S,S{⟩}_x=S|S,S{⟩}_x$ and $(S_j^x{)}^2+(S_j^y{)}^2+(S_j^z{)}^2=S(S+1)$. Rotational invariance about $x$ makes the two transverse second moments equal, so each is $[S(S+1)-S^2]/2=S/2$. This agrees exactly with the full boson representation.

The first moments describe the local magnetization. Subtracting their squares from the second moments gives spin component variances: zero longitudinal variance and transverse variance $S/2$. Transverse quantum fluctuations persist in a perfectly polarized state even though its magnon occupation is zero. Their relative scale decreases as $S^{-1/2}$ in the classical limit.

At precisely $B=0$, “the ground state” requires a choice within the ferromagnetic multiplet. For example, when $NS$ is an integer, a maximum-total-spin state with total $x$ projection zero is also a ground state but has $⟨S_j^x⟩=0$. The displayed local moments refer to the explicitly specified polarized member, not to every superposition or rotationally invariant mixture of zero-field ground states.

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.

The Holstein–Primakoff transformation maps the finite spin basis $|S,m⟩$ to bosonic Fock states $|n⟩$ with $n=S-m$. Thus $n=0$ is the fully polarized reference state, and creating a boson lowers $S^z$ by one. The physical subspace has the Holstein–Primakoff occupation constraint $0\le n\le 2S$, not the whole infinite oscillator Hilbert space.

The specified ordering of the square root givesThese are the exact spin ladder operator matrix elements; in particular, the lowering amplitude vanishes at $n=2S$. The underlying oscillator obeys its usual canonical commutation relation on the enlarged space, while the spin operators preserve the constrained physical subspace. Projecting the oscillator operators themselves to a finite space would change their commutator at its upper endpoint.

Expanding $\sqrt{2S-n}=\sqrt{2S}(1-n/(4S)+\cdots )$ converts an ordered spin problem to approximately free magnons, with interactions in higher orders of $1/S$. Two complications are the nonpolynomial, order-sensitive square roots, which generate interacting bosons, and loss of the occupation bound when a truncated expansion uses the full bosonic Fock space. Large depletion, high temperature, or soft modes near a phase transition can invalidate the assumption $⟨n⟩\ll S$ even when $S$ is large.