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Antiferromagnetism is ordered compensation of magnetic moments. In the collinear two-sublattice model, ${\mathbf{M}}_A=-{\mathbf{M}}_B\ne 0$: net magnetization vanishes but staggered magnetization remains. Louis Néel developed this possibility.^[ref]

For an original local energy check, perturb the antiparallel pair in exchange interaction by writing $\theta =\pi -\delta$. ThenFor $J>0$, a small relative rotation costs positive energy. This verifies a pair's preference; it does not prove a bulk ordering temperature. A whole crystal must satisfy many bonds simultaneously, and thermal fluctuations and quantum effects matter. The Heisenberg antiferromagnet and antiferromagnetic Ising model specify different assumptions about those collective degrees of freedom.

Why is the distinction useful? A measurement of zero magnetization cannot decide between an ordered alternating arrangement, a disordered ensemble, and oppositely magnetized magnetic domains. A spatial probe such as magnetic neutron diffraction tests the pattern directly. The manganese monoxide neutron diffraction experiment is a particularly clear example: the order adds a new periodic scattering signal even though the net magnetic moment cancels.

Magnetic order is a reproducible spatial organization of microscopic magnetic moments. A net magnetization describes their sum; a staggered magnetization describes an alternating pattern. These are different order parameters. For example, an alternating row of equal up and down arrows has zero sum but a definite pattern, whereas randomly oriented arrows can have a small sum without that pattern.

A magnetic sublattice groups sites with a common ordered magnetic moment in a description of magnetic order. In the simplest two-sublattice description, assign magnetizations ${\mathbf{M}}_A$ and ${\mathbf{M}}_B$ per total specimen volume, so thatHere $\mathbf{L}$ is one normalization of the staggered magnetization; conventions may include a factor $1/2$. Using the same total volume for both contributions is essential. Equal magnetic moment per site does not imply equal sublattice contributions if the site populations differ.

In this six-site example, three A sites and three B sites give total magnetic moments $6m_0$, $0$, and $3m_0$, respectively, along the chosen axis. The antiferromagnetic case has $\mathbf{M}=0$ but $\mathbf{L}=6m_0\hat{\mathbf{z}}/V$. Cancellation of the sum leaves the alternating information intact.

Write $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant. For independent Ising spins, the one-site partition functions on the two sublattices are $z_B=2\cosh \beta (h+g)$ and $z_W=2\cosh \beta (h-g)$. Differentiating each logarithm with respect to its local field givesAt $h=0$, the two sublattices experience opposite fields, so spin reversal combined with sublattice exchange leaves the Hamiltonian invariant and gives $M_W=-M_B$.

For the antiferromagnetic Ising model, take $N$ sites, equal sublattice sizes and $q=2D$ neighbors per site. There are $Nq/2$ bonds counted once. The two-sublattice Ising mean-field approximation expands each bond asand drops only the product of fluctuations. Summing the remaining terms givesThe printed constant omits $N$; for a total Hamiltonian with unnormalized spin sums, this factor is necessary. Applying the independent-spin calculation now givesThe mean-field approximation is favored by large coordination number: averaging many neighbors reduces local relative fluctuations. The size of critical fluctuations still has to satisfy the Ginzburg criterion; high dimension is a reason for validity, not permission to discard every correlation in every regime.

Antisymmetric solutions and the transition. At $h=0$, putting $M_B=-M_W=M$ makes both equations reduce to $M=\tanh \beta (g+qJM)$. A zero-field continuous transition concerns $g=0$ as well: nonzero $g$ explicitly selects the staggered sign and rounds this critical bifurcation. Put $K=\beta qJ$. The graphical intersections of $M$ and $\tanh (KM)$ give only $M=0$ for $K\le 1$. For $K>1$ there are additionally two nonzero intersections $\pm M_{\ast }$: on $M>0$, $\tanh (KM)-M$ is strictly concave, initially increases, and is negative at $M=1$, so there is exactly one positive root. Spin reversal supplies the negative root. The origin becomes unstable in the staggered direction, whereas the nonzero roots have $K(1-M_{\ast }^2)<1$ and are stable. Expanding about the bifurcation givesThus $M_{-}=(M_B-M_W)/2$ turns on continuously, with $k_B{T}_c=qJ$ and order-parameter critical exponent $1/2$.

Uniform solutions. At $g=0$, imposing $M_B=M_W=M$ instead gives $M=\tanh \beta (h-qJM)$. The function $M-\tanh \beta (h-qJM)$ has derivative $1+\beta qJ{\mathrm{sech}}^2\beta (h-qJM)>0$, is negative at $M=-1$ and positive at $M=1$, and consequently has exactly one zero for every positive temperature. At $h=0$ this zero is $M=0$. Antiferromagnetic exchange opposes uniform polarization, so the uniform branch has no ferromagnetic ordering bifurcation. This does not remove the antisymmetric solutions below $T_c$, or exclude transitions of staggered order in the full problem at nonzero uniform field. In particular, uniqueness on the diagonal $M_B=M_W$ is not uniqueness of all solutions.

Free energy and stability. The appropriate off-shell two-sublattice Ising variational free energy is the energy minus temperature times the independent-spin entropy. DefineThere are $Nq/2$ bonds and $N/2$ spins of each kind, so the free-energy density per site isSince $s^{\prime }(m)=\mathrm{atanh}m$, stationarity givesand the analogous equation for $M_W$; these are exactly the two mean-field self-consistency equations. On a stationary solution, the identity $s(m)=m\mathrm{atanh}m-\ln (2\cosh (\mathrm{atanh}m))$ gives the equivalent on-shell expressionThe entropy form must be used to compare off-shell variations; treating this latter auxiliary expression as the variational potential would give spurious stability conclusions.

Using $s(m)=-\ln 2+m^2/2+m^4/12+O(m^6)$ and $M_B=M_{+}+M_{-}$, $M_W=M_{+}-M_{-}$ givesThis includes the requested quadratic expansion and first-order field terms. The positive quartic coefficient verifies a continuous phase transition in $M_{-}$ when the staggered quadratic coefficient changes sign at $k_{B}T=qJ$. The uniform quadratic coefficient remains positive.

Susceptibilities and a printed error. On a stable zero-field ordered branch let $M_B=-M_W=M$ and $a=\beta (1-M^2)$. Since the derivative of $\tanh x$ is $1-{\tanh }^{2}x$, differentiating the two equations with respect to the uniform field givesTheir sum and difference yieldThe difference equation is $(1-qJa){\chi }_{-}=0$; stability implies $qJa<1$ for $T<T_c$. Symmetry also explains the zero result: a uniform perturbation changes the two sublattice magnetizations equally to first order, and hence does not linearly change the staggered magnetization. This ${\chi }_{-}$ is a cross response to $h$, not the magnetic susceptibility to the staggered field $g$.

The displayed ${\chi }_{+}$ in the source instead contains $1+M^2$. It cannot follow from differentiating the stated equations. A decisive limiting check is $T\to 0$: $1-M^2$ vanishes exponentially and the actual uniform response tends to zero, whereas the printed expression tends to $1/(qJ)$. Above $T_c$, $M=0$ and the same linear system givesFor comparison, the actual staggered susceptibility is $\partial M_{-}/\partial g=a/(1-qJa)$, becoming $1/(k_{B}T-qJ)$ above $T_c$. Its divergence, rather than a divergence of the uniform response, detects antiferromagnetic criticality.

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.

For $h=g=0$, the expansion $I(m)=-\log 2+m^2/2+O(m^4)$ givesWith $m_{\pm }=(m_e\pm m_o)/2$, one has $m_e{m}_o=m_{+}^2-m_{-}^2$ and $m_e^2+m_o^2=2(m_{+}^2+m_{-}^2)$. HenceThe staggered magnetization $m_{-}$ becomes unstable atproducing the antiferromagnetic transition. The coefficient of the uniform magnetization $m_{+}$ remains positive for every $T>0$, so there is no $m_{+}$ transition.

For a linear spin-wave approximation, the ground state is empty of the diagonal quasiparticles but contains the original spin bosons. If $a_k=u_k{\alpha }_k-v_k{\alpha }_{-k}^{†}$, their occupation number is $v_k^2$. The staggered magnetization is reduced from $S$ to $S-N^{-1}{\sum }_k{v}_k^2$. For the one-dimensional nearest-neighbor Heisenberg antiferromagnet, $v_k^2=(|\sin k{|}^{-1}-1)/2$ and the integral diverges logarithmically. This invalidates finite ordered magnetization within the approximation; it is not itself a determination of the exact spectral gap.