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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.

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]

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]

Let $[a,a^{†}]=1$, $n=a^{†}a$, and take $S^{+}=(S^{-}{)}^{†}=\sqrt{2S-n}a$. The physical Holstein–Primakoff occupation constraint limits the Hilbert space, which contains only $n=0,1,\dots ,2S$. Acting in this number basis givesThe upper and lower endpoints are annihilated as required. ConsequentlyThe change in $n$ also gives $[S^z,S^{\pm }]=\pm S^{\pm }$. These matrix elements prove the spin algebra on the full physical spin-$S$ space. Extending the oscillator space to arbitrary occupation would not give a physical finite spin: the occupation constraint and operator ordering in the square root matter. bosons on distinct sites have $[a_m,a_n^{†}]={\delta }_{mn}$ and all other site commutators zero.

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.