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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.

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.

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.

The vacuum of the diagonal quasiparticles contains original spin bosons: $⟨a_k^{†}{a}_k⟩=v_k^2$. Thus the quantum depletion of Néel order givesIn one dimension $\sqrt{1-{\gamma }_k^2}=|\sin k|$. Removing the global zero modes with an infrared cutoff, the continuum integral contains $\int dk/|k|$ near both gapless points and diverges logarithmically as the cutoff is removed. This is a breakdown of the assumed Néel order reference, not a physical infinitely negative magnetization. It indicates that the one-dimensional ordered spin-wave expansion cannot maintain a finite staggered moment; it does not determine the exact spectral gap or all properties of the spin chain for arbitrary $S$.

At zero temperature in $d$ dimensions the singular contribution scales as ${\int }_0{k}^{d-2}dk$. It is infrared finite for $d\ge 2$, permitting a finite quantum reduction and a self-consistent ordered spin-wave description in an appropriate regime. This differs from the positive-temperature contribution, where $n_B(\omega )\sim T/\omega$ yields ${\int }_0{k}^{d-3}dk$. Its divergence for $d\le 2$ agrees with the Mermin-Wagner theorem for short-range continuous-symmetry models. Finite-temperature nonordering and the zero-temperature one-dimensional depletion argument are separate statements.

Under the bipartite spin rotation, the staggered magnetization becomes the uniform transformed $S^z$. In the Bogoliubov transformation vacuum ${\alpha }_k|0_{\alpha }⟩=0$,The quantum depletion of Néel order therefore gives, with the zero modes regulated,Near each zero of $\sin k$, the integrand behaves as $1/(2|k-k_0|)$, producing a logarithmic divergence. With a finite-size cutoff of order $N^{-1}$, the depletion grows as ${\pi }^{-1}\log N+O(1)$. Thus the large-$S$ expansion about a state with finite Néel order is not self-consistent in the infinite one-dimensional chain. The divergent expression is not a negative physical magnetization; it signals breakdown of that ordered approximation. It does not determine whether the exact excitation spectrum is gapped.

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]

Two sublattices give two bosonic modes for each of N/2 wavevectors in a magnetic Brillouin zone. For a symmetric quadratic pairing block with number coefficient A and anomalous coefficient B, a bosonic Bogoliubov transformation gives two equal positive energies $E_k=\sqrt{A^2-B_k^2}$ and the displayed vacuum shift. There are N modes overall. The quantum depletion of Néel order per site is $(2/N){\sum }_{\mathrm{MBZ}}{v}_k^2$, where $v_k^2=(A/E_k-1)/2$. Equivalently one can use one branch over a full zone with the corresponding factors of one half. Mixing these conventions doubles the vacuum energy or depletion.