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Physical Review Letters 17, 1133 (1966), DOI 10.1103/PhysRevLett.17.1133. The abstract states the absence of ferromagnetic or antiferromagnetic order at nonzero temperature in finite-range one- or two-dimensional isotropic spin-$S$ Heisenberg models. This is a primary source for the low-dimensional limitation of mean-field magnetic ordering.

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.

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 two spin one-half systems, the dimensionless isotropic Heisenberg model coupling is $C=X\otimes X+Y\otimes Y+Z\otimes Z$, where $X,Y,Z$ are Pauli matrices. In the ordered computational basis $(|00⟩,|01⟩,|10⟩,|11⟩)$,where $S$ is the SWAP gate. This follows by acting on each basis state: the $XX+YY$ terms exchange the two opposite-spin states with coefficient two and annihilate the parallel-spin states. The symmetric three-dimensional sector has eigenvalue $1$ and the antisymmetric sector has eigenvalue $-3$. HenceThis identity explains how isotropic exchange interaction generates a SWAP gate at $a=\pi /2$ up to a global phase.

The Heisenberg model couples neighboring or more general pairs of spins through ${\mathbf{\sigma }}_i\cdot {\mathbf{\sigma }}_j$. For all-to-all equal coupling, the Hamiltonian can be written in terms of the square of the total spin.

Coauthor of the 1966 paper with N. D. Mermin on the absence of magnetic order in low-dimensional isotropic Heisenberg models. The primary publication identifies the author as H. Wagner. doi.org/10.1103/PhysRevLett.17.1133

The polarized ground state of an isotropic Heisenberg model with ferromagnetic exchange is an exact state. Its normal-ordered harmonic Hamiltonian has no pair-creation terms, no vacuum magnon occupation and no additional zero-point shift. An antiferromagnetic Néel state instead generates anomalous terms, a squeezed Bogoliubov transformation vacuum, a negative energy correction and reduced staggered magnetization. Neither statement licenses omitting normal-ordering constants from an arbitrary spin-wave model with anisotropy or a different reference state.

The Mermin-Wagner theorem excludes spontaneous continuous-symmetry order at positive temperature in important one- and two-dimensional systems with sufficiently short-range interactions. The original isotropic Heisenberg model result rules out both ferromagnetic and antiferromagnetic order under its assumptions.^[ref]

The physical obstruction is strong long-wavelength orientational fluctuation. In a two-dimensional Gaussian angle model, the fluctuation integral includes $\int d^{2}q/q^2$, which grows logarithmically with size. A locally stiff orientation need not remain aligned across an infinite sample. This leaves room for quasi-long-range order and a Berezinskii–Kosterlitz–Thouless transition, rather than forbidding every finite-temperature phenomenon in two dimensions. Discrete anisotropy, long-range interactions, interlayer coupling and finite size change the applicability of the ideal theorem.

Coauthor of the 1966 proof excluding finite-temperature magnetic order in finite-range one- and two-dimensional isotropic Heisenberg models. The primary publication identifies the author as N. D. Mermin. doi.org/10.1103/PhysRevLett.17.1133

For a periodic spin-one-half Heisenberg model $H=-(J/{\hslash }^2){\sum }_{n=1}^N{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, $J>0$, the fully aligned state has $E_0=-NJ/4$. In its single-excitation subspace, let $|n⟩$ denote a down spin at site $n$ in an otherwise up background. Direct action of each bond givesThe normalized translation eigenstates are $|q⟩=N^{-1/2}{\sum }_n{e}^{iqna}|n⟩$, with $q=2\pi \ell /(Na)$ modulo $2\pi /a$. Their energy eigenvalues are $E_0+J[1-\cos (qa)]$. This is an exact one-magnon calculation, not a many-magnon harmonic approximation. Near $q=0$ the energy above $E_0$ is $J{a}^2{q}^2/2$.

The wavenumber restriction comes from requiring a periodic plane-wave coefficient pattern, equivalently requiring this vector to be an eigenvector of a one-site translation with eigenvalue $e^{-iqa}$. The identity $T^N=I$ alone imposes no restriction on a general state: it already holds throughout the ring Hilbert space. An arbitrary finite sum with coefficients $e^{iqna}$ still returns after $N$ translations, but need not be a one-site translation eigenstate unless $e^{iqNa}=1$.

For a finite-range isotropic Heisenberg model, the Mermin-Wagner theorem excludes spontaneous magnetic order in one dimension at every nonzero temperature. Long-wavelength rotations cost very little energy: a smooth orientational fluctuation has $\delta F=({\rho }_s/2)\int dx({\partial }_x\theta {)}^2$, so its thermal variance contains $(k_{\mathrm{B}}T/{\rho }_s)\int dq/q^2$, which diverges in the infrared. In a Heisenberg ferromagnet, the same obstruction appears as a divergent thermal occupation of quadratic long-wave spin waves. This explains why a finite magnetization cannot be supported, although correlations can be strong over a long finite distance. The rigorous statement requires short-range interactions and the stated continuous symmetry; interchain coupling and anisotropy change the problem. Symmetry forcing the mean to zero in a finite chain alone would not prove absence of thermodynamic order. ^[ref]

For the Ising model, use dimensionless $s_i=2S_i^z=\pm 1$, $h=g{\mu }_{\mathrm{B}}B$, $K=\beta J/2$ and $b=\beta h/2$. Its pair Boltzmann factor isMultiplying N such factors around the periodic ring counts each bond once and each field term twice with weight one half. Summing the closed spin-index sequence therefore gives the transfer matrix for the one-dimensional Ising modelThe two eigenvalues areAt finite temperature every entry of P is positive, so the Perron–Frobenius theorem gives ${\lambda }_{+}>|{\lambda }_{-}|$. ConsequentlyThe exact finite-N correction to the extensive expression is $-k_{\mathrm{B}}T\ln [1+({\lambda }_{-}/{\lambda }_{+}{)}^N]$. No assumption that ${\lambda }_{-}$ is positive is needed.

Differentiate at fixed temperature. For $D=\sqrt{{\sinh }^{2}b+e^{-2\beta J}}$, ${\partial }_b\ln (\cosh b+D)=\sinh b/D$. The Ising-chain magnetization with spin-half normalization gives total magnetic momentThis is the thermodynamic-limit result; the moment per site is M/N. For every finite $T>0$, ${\lim }_{B\to 0}M/N=0$, even with the thermodynamic limit taken first. The zero-field magnetic susceptibility per site is $(g{\mu }_{\mathrm{B}}{)}^2{e}^{\beta J}/(4k_{\mathrm{B}}T)$, large but finite.

The Ising-chain domain-wall energy convention explains the result for $J>0$: changing a parallel bond into an antiparallel bond costs J in this spin-half normalization. Domain walls, occurring in pairs on a ring, acquire a nonzero density at any nonzero temperature. Indeed the zero-field correlation length is $\xi =-a/\ln [\tanh (\beta J/2)]\sim (a/2)e^{\beta J}$ at low temperature. The Ising symmetry is discrete, so this is a separate one-dimensional domain-wall mechanism rather than an application of the continuous-symmetry theorem. At zero temperature a ferromagnetic chain can select a fully polarized state; the $B\to 0$ and $T\to 0$ limits do not commute.

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.

Eu${}^{2+}$ has the half-filled $4f^7$ term with $L=0$. First-order orbital magnetism is therefore quenched already at the ion level, making the dominant moment approximately spin-only. Spin-orbit-induced magnetocrystalline anisotropy arises only through weaker admixtures and other small interactions; magnetic dipole interactions can also contribute. This explains why a cubic anisotropy term is allowed by symmetry but numerically weak.

An appropriate insulating local-moment model is a nearly isotropic Heisenberg model,with dominant effective ferromagnetic couplings in the sign convention $J_{ij}>0$. The Eu sublattice in rocksalt EuO is face-centered cubic; nearest- and further-neighbor effective exchanges can be retained as needed. Because the solid is not metallic, an itinerant-electron Stoner model or a generic free-carrier RKKY interaction mechanism is not the natural starting point. Exchange can still arise through virtual hopping and ligand-mediated processes in an insulator; it does not require mobile dc carriers.

For competing exchanges on a body-centred cubic antiferromagnet, use the Heisenberg model convention $H={\sum }_{⟨ij⟩}{J}_{ij}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$, with each bond counted once, dimensionless spins, and positive antiferromagnetic $J$. There are eight nearest and six next-nearest neighbours. Bond counting in the two classical collinear structures givesFor type II the eight $J_1$ contributions cancel, while all six $J_2$ bonds are antiparallel. Their crossing yields $c=2/3$. One can also exclude a lower spiral within the classical bilinear model: writing $u_i=\cos (q_{i}a/2)$ givesFor a favourable sign of the product, $u_x^2+u_y^2+u_z^2\ge 3|u_x{u}_y{u}_z{|}^{2/3}\ge 3|u_x{u}_y{u}_z|$. Hence when $J_2\ge 2J_1/3$ its minimum is at $u_x=u_y=u_z=0$, the type-II ordering vectors. When $J_2\le 2J_1/3$, compare with the type-I value and use $3-\sum u_i^2\le 3(1-|u_x{u}_y{u}_z|)$ to obtain the complementary bound. ThusThis is a classical/mean-field energy comparison; an exact quantum ground state is not generally a collinear product state.

For type I a magnetic cell is the simple cube of side $a$, with opposite moments at $0$ and $a(1,1,1)/2$. Its magnetic structure factor is proportional to $1-e^{i\pi (h+k+l)}$. Allowed magnetic reflections have integer $h,k,l$ with $h+k+l$ odd. The first two families are $(100)$ and $(111)$, with symmetry-equivalent members understood.

For type II choose magnetic primitive translations $a(1,1,0)$, $a(1,0,1)$ and $a(0,1,1)$, enclosing four spins in volume $2a^3$. Equivalently, a side-$2a$ cube is a convenient nonprimitive magnetic cell. On each original simple-cubic sublattice the moment at $a\mathbf{n}$ is proportional to $(-1{)}^{n_x+n_y+n_z}$. Accordingly,The translated body-centre basis gives $1+e^{i\pi (h+k+l)}=1\pm i$, so there is no further extinction. The first two families are $(1/2,1/2,1/2)$ and $(3/2,1/2,1/2)$. The neutron polarization factor may additionally suppress a reflection for a particular moment direction; the stated rules are the positional selection rules.

Each sublattice has number density $n_A=1/a^3$ and saturation magnetization $M_s=g{\mu }_{B}S/a^3$. Replacing neighbouring spins by their averages makes their exchange energy equivalent to a molecular magnetic field. Its minus sign expresses antiferromagnetic exchange, not an externally applied field. Since $⟨S_B⟩=M_B/(n_{A}g{\mu }_B)$,For $M_B=-M_A$, $B_{\mathrm{mf}}=({\lambda }_{AB}-{\lambda }_{AA})M_A$. Linearizing the Brillouin magnetization formula givesThe g-factor cancels because exchange is specified in energy units. A convention with $2J{\mathbf{S}}_i\cdot {\mathbf{S}}_j$ doubles these exchange energies and $T_N$.

Let $m=M_A/M_s$, $A=(S+1)/(3S)$ and let $b_B>0$ denote the cubic coefficient of the Brillouin magnetization formula. The argument is $y=(T_N/T)m/A$. Thus $m=Ay-b_B{y}^3$ impliesConsequently $\boxed{\beta =1/2}$ in mean-field theory. The positive branch falls from saturation at zero temperature to zero at $T_N$ and remains zero above it. An elastic magnetic neutron-reflection intensity scales as $I\propto M_A^2$; a log–log fit against $T_N-T$ has slope $2\beta$. Background subtraction and a suitable critical interval are needed. Fluctuations can change the measured exponent from its mean-field value; uniform bulk magnetization vanishes for a compensated antiferromagnet and is not itself this order parameter.

Using the supplied iron parameters givesAn internal Weiss molecular field depends on temperature through $M(T)$. Its low-temperature saturation estimate uses $M_s=ng{\mu }_{B}J\simeq 1.58\times 10^6\mathrm{A}{\mathrm{m}}^{-1}$:At and above $T_C$ the zero-applied-field magnetization vanishes in this model, so this is a saturation estimate, not the field at $T_C$.

Use the paper's stipulated face-centered cubic lattice, which has twelve nearest neighbours. The saturated one-ion alignment scale isDividing the effective field among twelve equivalent neighbours gives the per-bond exchange interaction scaleTo specify the convention, write the Heisenberg model as $\mathcal{H}=-\mathcal{J}{\sum }_{⟨ij⟩}{\mathbf{J}}_i\cdot {\mathbf{J}}_j$, with dimensionless angular-momentum operators and each bond counted once. Then $g{\mu }_B{B}_e=12\mathcal{J}⟨J_z⟩$, so $\mathcal{J}=3k_B{T}_C/[12J(J+1)]\simeq 11.2\mathrm{meV}$ here. A convention with $-2{\mathcal{J}}^{\prime }$ instead has ${\mathcal{J}}^{\prime }=5.62\mathrm{meV}$. The total saturated interaction energy per ion is $-12\mathcal{J}{J}^2/2\simeq -67.4\mathrm{meV}$; the factor of one half avoids counting each bond twice.

Use the ordered computational basis $(|00⟩,|01⟩,|10⟩,|11⟩)$, with $Z|0⟩=|0⟩$ and $Z|1⟩=-|1⟩$. Write $X,Y,Z$ for the Pauli matrices, $I$ for the identity operator, $J={\omega }_{12}$, $\Sigma ={\omega }_1+{\omega }_2$, and $\Delta ={\omega }_1-{\omega }_2$. As in the question, the Hamiltonian operators have angular frequency units, so the time-evolution operator is $e^{-iHt}$. The Heisenberg coupling matrix for two spins isConsequently the full Heisenberg model Hamiltonian operator and its proposed Ising coupling approximation are

For nondegenerate perturbation theory, choose $H_{\mathrm{I}}$ as the unperturbed Hamiltonian operator andas the perturbation. Only $|01⟩$ and $|10⟩$ are coupled, and their unperturbed energy levels differ by $\Delta$. All first-order energy corrections vanish. The first-order eigenvectors and second-order energy corrections areThe states $|00⟩,|11⟩$ are exact eigenvectors throughout. Hence the weak coupling condition for two NMR spins isA large common Zeeman energy alone does not ensure this condition. Crossings involving the uncoupled parallel-spin states do not invalidate the argument: the relevant denominator is the separation of the two states connected by $V$.

As an exact check, the middle $2\times 2$ matrix has eigenvaluesFor $\Delta \ne 0$, labeling by the limiting computational basis states reproduces the shifts above. The quantum transition probability from $|01⟩$ to $|10⟩$ isThus unwanted exchange has a uniformly small amplitude when $|J/\Delta |\ll 1$. If accumulated phases must also be approximated over a duration $t$, neglecting the second-order shifts additionally requires $J^{2}t/|\Delta |\ll 1$, or those shifts must be retained in an improved effective Hamiltonian operator.

The same result is the secular approximation for scalar-coupled spins. In the interaction picture of the Zeeman energy terms,These terms average rapidly on the coupling timescale $1/|J|$ when $|\Delta |\gg |J|$, while $JZ\otimes Z/2$ survives.

For fixed nuclei and electronic environments, the individual NMR angular frequencies scale with the static magnetic field $B_0$: schematically ${\omega }_i={\kappa }_i{B}_0$, where ${\kappa }_i$ includes the gyromagnetic ratio and magnetic shielding. The NMR scalar coupling $J$ in angular frequency units is approximately independent of $B_0$. Therefore a nonzero separation $|\Delta |=|{\kappa }_1-{\kappa }_2|B_0$ increases with magnetic field, improving the approximation. In particular, a fixed difference of chemical shifts gives a larger separation in hertz at higher magnetic field. If the two resonances remain exactly equal, increasing their common frequency does not suppress their exchange.

Set ${\omega }_1={\omega }_2=\omega$. Define the quantum rotating frame byThe isotropic Heisenberg model coupling commutes with the total $z$ spin angular momentum, so $[H_Z,C]=0$. The transformed Hamiltonian operator is thereforeHere removal of the common Zeeman energy is exact; the exchange is resonant and cannot be discarded using the weak coupling condition for two NMR spins.

Put $a=Jt$, $c=\cos a$, and $s=\sin a$. Exponentiating the parallel-spin entries and the middle $2\times 2$ matrix gives the time-evolution operatorFor example, at $a=\pi /2$ this equals the SWAP gate times $e^{-i\pi /4}$, which checks both the exchange angle and the overall phase.

The printed network reads from left to right as $U$, a Pauli Z gate on the upper spin only, $U$, and another upper-spin Pauli Z gate. LetIts total time-evolution operator, with the first operation on the right, is $W=PUPU$. For this Heisenberg exchange echo,The three paired Pauli operators $X\otimes X$, $Y\otimes Y$, $Z\otimes Z$ commute pairwise: for example both products $(X\otimes X)(Y\otimes Y)$ and $(Y\otimes Y)(X\otimes X)$ equal $-Z\otimes Z$. Thus $P{H}_{\mathrm{r}}P$ commutes with $H_{\mathrm{r}}$, andThe spin echo cancels the $XX+YY$ exchange terms and retains the Ising coupling. Equivalently, over the total free-evolution time $2t$ it produces the effective Hamiltonian operator $JZ\otimes Z/2$ exactly for ideal instantaneous gates. It is generally not the identity. A physical $\pi$ rotation about $z$ is $e^{-i\pi Z/2}=-iZ$; using two such pulses instead of the ideal $Z$ gates changes $W$ only by a physically irrelevant global phase of minus one.

The implementation requires frequency-selective spin control. Under the usual common-radio frequency NMR assumptions, equal resonance frequencies give no way to apply a pulse to the first spin while leaving the second untouched. For equally coupled spins in a spatially uniform control field, the transverse control Hamiltonian operator has the formBoth this control and the drift Hamiltonian operator commute with the SWAP gate $S$. Every reachable time-evolution operator consequently commutes with $S$. But $SPS=I\otimes Z\ne P$: the individual pulse required by the diagram fails this symmetry test. This is the exchange symmetry obstruction to individual spin control. Applying simultaneous $Z$ gates gives $Q=Z\otimes Z$, which commutes with $H_{\mathrm{r}}$ and instead yields $QUQU=U^2$; it does not refocus the exchange. A common change of the RF phase reference also supplies only a collective rotation.

This obstruction assumes the usual common-RF control with no extra handle distinguishing the spins. Equal resonance frequencies do not universally forbid individual addressing: a spatially selective field, unequal control couplings, or another symmetry-breaking interaction could alter controllability. Such resources are absent from the specified equivalent-spin NMR model.

The large effective field primarily represents exchange interactions, which arise from Coulomb interactions, electronic hopping and the Pauli exclusion principle. It is not usually the much smaller classical magnetic dipole field of neighbouring ions. Direct exchange involves overlapping magnetic-ion orbitals. Superexchange instead uses virtual charge transfer through a ligand while real charge transport remains gapped, allowing substantial magnetic order in an electrical insulator.

Use dimensionless spin operators ${\mathbf{S}}_i$ with eigenvalue $S=1/2$, and count each bond once. One consistent Heisenberg model convention isThen $J_{ij}<0$ favours a Heisenberg ferromagnet, while $J_{ij}>0$ favours a Heisenberg antiferromagnet. Replacing neighbouring spins by their mean values gives the effective fieldFor $z$ equivalent ferromagnetic neighbours with $J_{ij}=-J_F<0$ and uniform order, $M=n{g}_J{\mu }_B⟨S_z⟩$, soThe last equality is the same molecular-field result derived above, not an exact transition temperature for every lattice.

Microscopically, a half-filled orbital pair with virtual hopping $t_{ij}$ and a large on-site repulsion $U$ permits an energy lowering of the singlet but not the parallel-spin triplet, yielding antiferromagnetic $J\simeq 4|t_{ij}{|}^2/U$ in this convention. Ligand-mediated hopping is the analogous mechanism in many ionic insulators. Different orbital occupancies and nearly orthogonal ligand paths can suppress this antiferromagnetic term and permit ferromagnetic exchange to dominate. Thus the sign depends on electronic structure and bonding geometry, not simply whether the sample conducts.

For antiferromagnetic coupling on a bipartite lattice, the field on each sublattice is proportional to the opposite sublattice magnetization, with opposite alignment preferred. Two coupled self-consistency equations then permit $M_A=-M_B\ne 0$ and vanishing total magnetization: the order parameter is staggered. The single positive-$\lambda$ uniform equation describes ferromagnetism and must be extended to these sublattices to describe an antiferromagnet.

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]

Near zero the Langevin function has expansionFor positive $x$, $\mathcal{L}(x)$ rises toward one and its ratio $\mathcal{L}(x)/x$ decreases from $1/3$ to zero. Therefore a line $\tau x$ has only the zero intersection for $\tau \ge 1/3$, and also a nonzero positive intersection for $0<\tau <1/3$. In a fixed signed component there is a corresponding negative solution; in the isotropic vector problem the nonzero solutions have all spin-space orientations.

The mean-field transition temperature isAt the critical temperature the line is tangent at the origin. Just below it, $S\simeq \sqrt{(5/3)(1-T/T_c^{\mathrm{MF}})}$, the usual mean-field square-root onset. The ordered minimum is stable below this temperature while $S=0$ becomes unstable.

This is a mean-field prediction, not a genuine finite-temperature ordering transition for the ideal two-dimensional short-range isotropic model. Long-wavelength spin fluctuations invalidate the predicted nonzero thermodynamic magnetization at every $T>0$. The relevant obstruction is the Mermin-Wagner theorem for continuous spin symmetry; anisotropy, long-range interactions or a finite sample change the assumptions. Its original paper establishes the absence of finite-temperature magnetic ordering for one- and two-dimensional finite-range isotropic Heisenberg models.^[ref]