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An antiferromagnetic spin wave is a small collective variation of the ordered spin pattern in an antiferromagnet. In a bipartite Néel state, neighbouring spins point oppositely; the slowly varying Néel order parameter describes their staggered orientation. A long-wavelength disturbance changes that orientation smoothly rather than independently flipping each site.

For an isotropic ordered antiferromagnet in the usual long-wavelength regime, the low-energy dispersion relation is linear, $\omega \simeq c|\mathbf{k}|$, with spin-wave velocity $c$. This differs from the quadratic low-wavenumber ferromagnetic magnon in a conventional isotropic ferromagnet. Anisotropy or an applied magnetic field can alter the branches or introduce gaps. The starting ordered state also needs justification: strong quantum fluctuations, especially in low dimension, can invalidate a naive linear spin-wave approximation. A computed divergence is then a warning about the approximation, not a negative physical magnetization.

A connected isotropic spin-$S$ ferromagnet with positive exchanges has a fully polarized ground state and a maximal-total-spin multiplet. For $N$ sites the total spin is $NS$, with $2NS+1$ magnetic states generated by total-spin lowering. Global rotations span the same degenerate space. Slowly varying reorientations produce gapless ferromagnetic magnons; finite-volume degeneracy must be distinguished from a thermodynamic statement about spontaneous order.

A ferromagnetic magnon is a quantum of a collective spin wave about ferromagnetic order. In a uniform isotropic exchange ferromagnet, slowly varying misalignment costs less energy than rapid variation. Its long-wavelength energy is typically quadratic, $E(\mathbf{k})\simeq D{k}^2$, and its angular frequency is $E/\hslash$.

A one-magnon state in a fully polarized magnet reduces total spin projection by one quantum while spreading that change coherently across the sample. It is therefore different from assigning each site an independent classical tilt. The ferromagnetic spin chain gives an explicit cosine dispersion relation whose small-wavenumber limit is quadratic. An applied magnetic field or symmetry-breaking anisotropy can introduce a nonzero excitation gap, while interactions between magnons matter when their population is no longer dilute. The mode picture is most useful when the ordered reference and low-excitation approximation are justified.

A ferromagnetic spin chain is a one-dimensional array whose exchange interactions favor aligned neighbouring spins. For $J>0$, the nearest-neighbour Heisenberg model conventionhas fully aligned ground states. A collective small deviation from alignment produces a ferromagnetic magnon rather than a set of unrelated local spin flips.

For spin length $S$, lattice spacing $a$ and this exchange convention, the one-magnon energy is $E(k)=2JS[1-\cos (ka)]$, giving $E(k)\simeq JS{a}^2{k}^2$ at long wavelength. The gapless point reflects the freedom to rotate a uniform aligned state. This ground-state and excitation statement does not establish finite-temperature spontaneous order in an infinite one-dimensional short-range model. Finite length, anisotropy and coupling between chains can change what an experiment observes.

The edges of an itinerant spin-flip particle-hole continuum follow from both excitation energy and occupation constraints. Consider a zero-temperature three-dimensional parabolic-band model with majority states occupied for $|\mathbf{k}|\le k_F$ and an entirely empty minority band. For exchange splitting $2M$ and ${ϵ}_k={\hslash }^2{k}^2/(2m)$, a majority-to-minority flip carrying wavevector $\mathbf{q}$ hasFor $q>0$, the scalar product is bounded by $\pm k_{F}q$, giving continuum edgesThe extremes occur at opposite points of the occupied Fermi surface along $\mathbf{q}$. At $q=0$ the interval collapses to the single exchange energy $2M$.

The empty-minority assumption requires sufficient splitting; otherwise Pauli exclusion principle blocks some final states and the edges must be recomputed over the allowed region. These are independent particle-hole thresholds, not a calculation of the collective ferromagnetic magnon dispersion. A collective mode below the continuum can therefore have a different low-momentum energy scale.

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.

Spin reversal in exchange-split bands creates an electron-hole excitation with energy ${ϵ}_{k+q}-{ϵ}_k+2M$ for a majority-to-minority flip in the simple uniform model with $M>0$. Its allowed continuum depends on occupied initial and empty final states. At $q=0$ all such allowed transitions have exchange energy $2M$; finite momentum changes the thresholds. A collective ferromagnetic magnon can remain gapless below this continuum when spin rotation is exact. The continuum is not the same excitation as a longitudinal amplitude mode.

For physical spin ladder operators, $S_j^{\pm }=S_j^x\pm i{S}_j^y$. Operators at distinct sites commute. Expanding the two products cancels their cross terms and givesAdding the remaining component provesFor spin one-half, $S^{+}|\downarrow ⟩=\hslash |\uparrow ⟩$, $S^{-}|\uparrow ⟩=\hslash |\downarrow ⟩$, and raising an up spin or lowering a down spin gives zero. These formulas restore the physical $\hslash$ factors implicit in the dimensionless ladder hint.

Let $|n⟩$ abbreviate the state with the only down spin at site $n$, keeping $|{\psi }_0⟩$ for the fully up state. If $n$ lies on neither end of bond $(j,j+1)$, that bond contains two up spins. The ladder terms vanish, while $S_j^z{S}_{j+1}^z$ gives ${\hslash }^2/4$. ThusFor a down spin at $j$, the $z$ product is instead $-{\hslash }^2/4$, and $S_j^{+}{S}_{j+1}^{-}$ moves it to $j+1$ with amplitude ${\hslash }^2$. For a down spin at $j+1$, the other ladder product moves it back. ThereforeEvery bond in $|{\psi }_0⟩$ is aligned, so summing its $N$ bond energies givesThis is indeed a ground-state energy: a pair of spin-one-half particles has ${\mathbf{S}}_j\cdot {\mathbf{S}}_{j+1}$ eigenvalues ${\hslash }^2/4$ and $-3{\hslash }^2/4$, so each ferromagnetic bond is bounded below by $-J/4$, simultaneously attained by the fully aligned state. The isotropic Heisenberg model has a whole maximum-total-spin ground multiplet, not a unique all-up ground state.

In the single-excitation subspace, only the two bonds adjacent to the down spin differ from the aligned background. Their diagonal energy increase is $J$, and each neighboring hop has amplitude $-J/2$. Hence, with periodic site indices,This is an exact finite-dimensional hopping problem. Choose the active one-site translation convention $T|n⟩=|n+1⟩$. A periodic plane-wave coefficient pattern requires $e^{iq(n+N)a}=e^{iqna}$, soThere are $N$ independent such wavevectors, for example one set of $N$ consecutive integers $\ell$ in a Brillouin zone. The normalized one-magnon Bloch states on a ring are $|q⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{iqna}|n⟩$. Reindexing with the periodic coefficient condition gives $T|q⟩=e^{-iqa}|q⟩$ and hence $T^N|q⟩=|q⟩$.

There is a logical qualification to the condition stated in the paper. On an $N$-site ring $T^N=I$ already holds for every state. In particular, the finite sum written with an arbitrary real $q$ also returns to itself after $N$ translations; that condition alone does not quantize $q$. The displayed allowed values are the ones for which the ansatz is a single one-site translation eigenstate, equivalently a consistently periodic plane wave. This is the additional condition needed for the subsequent energy-eigenstate argument. Other $q$ values generally describe superpositions of the allowed momentum modes.

Apply the hopping Hamiltonian operator and shift the summation index using periodic boundary condition. The coefficients from $|n-1⟩$ and $|n+1⟩$ are respectively $e^{iqa}$ and $e^{-iqa}$ times the original coefficient. Thusand the nearest-neighbour ferromagnetic magnon dispersion isThe unnormalized state in the question has the same energy. Distinct allowed momenta are orthogonal, since $N^{-1}{\sum }_n{e}^{i(q-q^{\prime })na}={\delta }_{q{q}^{\prime }}$ over one set of representatives.

For large $N$ the sampled wavevectors become dense, and the requested sketch is the periodic curve $[E(q)-E_0]/J=1-\cos (qa)$ over $-2\pi <qa<2\pi$. It has a zero at $q=0$ and tends to the equivalent zero at either open endpoint; maxima $2J$ occur at $q=\pm \pi /a$. Near any equivalent zero,using $q$ measured from that zero. The low-momentum ferromagnetic magnon is quadratic, with group velocity ${\hslash }^{-1}dE/dq=(Ja/\hslash )\sin (qa)$. The $q=0$ state is proportional to the total spin lowering operator acting on $|{\psi }_0⟩$, so it belongs to the same ground multiplet and has exactly zero excitation energy. Extending the plot beyond one reciprocal period repeats states rather than adding new independent modes. The lattice spacing $a$ is the nearest-neighbor chord length specified in the paper; $Na$ is the translation length in this lattice description, not necessarily the continuous circumference of a circular arrangement.

Use dimensionless spin angular momentum operators, so their commutators have no additional $\hslash$. A fully polarized state $|F⟩={⨂}_n|S,S{⟩}_n$ is an exact ground state of the Heisenberg ferromagnet, with $E_0=-NJ{S}^2$. On each bond the largest eigenvalue of ${\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$ is $S^2$, obtained in the two-spin total-spin-$2S$ sector. Hence the bondwise lower bound on the Hamiltonian is attained by this state.

The ferromagnetic ground-state multiplet has total spin $NS$. Applying the global lowering operator $(S_{\mathrm{tot}}^{-}{)}^r$, $r=0,\dots ,2NS$, generates its $2NS+1$ magnetic states; their linear combinations are also ground states. Global spin rotations generate polarized ground states pointing in arbitrary directions and span the same multiplet. Uniform reorientation therefore costs no energy. Slowly varying orientations give gapless ferromagnetic magnons, with quadratic rather than linear long-wavelength dispersion. In the ordered thermodynamic ground state, the two broken spin generators are canonically paired because $⟨[S_{\mathrm{tot}}^x,S_{\mathrm{tot}}^y]⟩=i⟨S_{\mathrm{tot}}^z⟩\ne 0$, producing one type-B Goldstone boson. For a finite chain the zero-wavevector motion remains inside the exact ground multiplet; nonzero wavevectors have a finite-size spacing that vanishes as $N^{-2}$. This is a ground-state statement, not a claim of finite-temperature long-range order in one dimension.

Put $q=\delta \theta$, $p=\delta \varphi$, so $\theta =\pi /2-q$ and $\varphi =p$. The quadratic Lagrangian is obtained by takingThe two real fluctuations form a coordinate and its canonical momentum rather than two independent modes. Their Euler-Lagrange equations combine to giveand similarly for $p$. The ferromagnetic magnon therefore has the quadratic dispersion relation

Write $\theta =\pi /2-\vartheta$ and $\varphi =\phi$. To quadratic order,After dropping the total derivative $S{\phi }_t$, the quadratic Lagrangian density isIts Euler--Lagrange equations areFor modes proportional to $e^{i(kx-\omega t)}$, nontrivial amplitudes requireThusAt short wavelength the anisotropy is negligible and the quadratic dispersion is that of a conventional ferromagnetic magnon. At long wavelength, easy-plane anisotropy makes the out-of-plane fluctuation the conjugate density of the in-plane phase; eliminating it produces the linear Goldstone sound mode characteristic of a superfluid.

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 constant term in the spin sum cancels. Linearizing the saddle point equation givesThe paramagnetic auxiliary-field curvature isConsequently the Stoner criterion for the loss of local paramagnetic stability isThe printed $\nu (0)$ uses energy measured from the Fermi level. For a smooth density of states, $-n_F^{\prime }$ becomes a narrow unit-area peak at $\mu$; a singular density of states or appreciable temperature requires the integral form. The curvature establishes local instability, not the order or location of every possible thermodynamic transition.

With exact spin-rotation symmetry, itinerant ferromagnetism selects a magnetization direction. The two broken transverse generators form one quadratic type-B Goldstone boson, a long-wave ferromagnetic magnon with $E_q\sim D{q}^2$. Longitudinal amplitude motion is not symmetry-protected to be gapless. In a metallic ordered state, particle-hole excitations remain near each surviving exchange-split Fermi surface; these are distinct from the collective magnon. The itinerant spin-flip particle-hole continuum starts at exchange scale $2|M|$ at $q=0$ in this simple mean-field spectrum, while its finite-$q$ boundaries differ. Spin anisotropy, spin-orbit coupling or an external field can gap the transverse mode; none is present in the stated model.

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]