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An antiferromagnetic Ising model on a closed one-dimensional ring, with Ising spins $S_j=\pm 1$ and the displayed periodic boundary conditions. Every opposite neighboring pair minimizes its bond energy. An even ring admits two completely alternating ground states; an odd ring must contain at least one parallel pair, an example of geometric frustration.

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.

The Ising model is a spin model with an Ising spin ${\sigma }_v\in \{-1,1\}$ at every graph vertex $v$. For a finite graph with vertex set $V$ and edge set $E$, a uniform coupling $J$ and field parameter $h$ give the HamiltonianEach edge is counted once. Here $T>0$ is temperature, $k_B$ is the Boltzmann constant, and $\beta$ is inverse temperature; $J$ and $h$ have units of energy. For magnetic applications, $h$ includes the magnetic moment multiplying the applied magnetic field. The canonical ensemble assigns the Boltzmann distributionThe partition function $Z$ makes the probabilities sum to one. Positive $J$ rewards equal neighboring Ising spins, modeling ferromagnetism; negative $J$ rewards opposite neighbors, giving an antiferromagnetic Ising model. Positive $h$ rewards $+1$, while negative $h$ rewards $-1$.^[ref]

A single local rule can organize a whole graph: rewarding agreement on each edge makes all-equal Ising spin patterns favorable even though no edge directly couples distant vertices. The four-spin Ising ring makes this visible with only sixteen microstates. It also separates three different questions: whether neighbors agree, whether the mean spin magnetization is nonzero, and how large its thermal fluctuations are.

A motivating experiment is the University of Colorado Boulder's Curie-temperature demonstration. An iron rod suspended from a twisting wire on a stand is initially held near a permanent magnet by ferromagnetism. A gas burner supplied from a propane cylinder heats the rod; the readout is its motion away from the magnet, followed by return as it cools. Repeated cooling supplies a reversible comparison on the same sample. The source explicitly calls for heating the rod while keeping the magnet cool, since heating the magnet can reduce its strength. The inference is that warming the sample weakens its magnetic attraction across the Curie temperature. This is a qualitative force readout, with no calibrated temperature or magnetization trace supplied: it motivates competition between alignment and thermal fluctuations, but does not identify $J$, measure individual Ising spins, or realize a four-spin Ising ring.^[ref]

Ernst Ising's 1925 paper examined a linear chain of two-orientation magnetic elements with neighboring interactions. In its calculation for a large chain, the zero-field mean magnetization vanishes and the model has no spontaneous magnetization. The inspected English translation states this immediately after equation (8) in §2. This historical result concerns the chain; it does not establish the behavior of a two-dimensional Ising model.^[ref]

Write $K=\beta J$, where $\beta =1/(k_{B}T)$ is the inverse temperature, $T$ is temperature and $k_B$ is the Boltzmann constant. In the canonical ensemble, each spin configuration has probability $e^{-\beta H}/Z_N$. The nearest-neighbor Boltzmann factors multiply, so the spin-chain transfer matrix, in the order $+1,-1$, isMultiplying its entries around the ring and summing over every Ising spin yields the partition function:The trace implements the periodic boundary conditions; replacing it by an endpoint sum would describe a different partition function. The normalized eigenvectors and corresponding eigenvalues areThe negative eigenvalue is essential for the antiferromagnetic Ising model. Introduce their ratioThe exact finite-$N$ partition function and Helmholtz free energy areFor finite $K>0$, $|t|<1$, so $Z_N>0$ for both even and odd $N$. The last term retains the finite-ring effect of geometric frustration.

To insert an Ising spin into this trace, use the diagonal matrix $\mathsf{S}=\mathrm{diag}(1,-1)$. In the eigenbasis of $\mathsf{T}$, it exchanges $|+⟩$ and $|-⟩$, and hence has vanishing diagonal entries. The one-point correlation function is thereforeThe same conclusion follows directly from spin inversion symmetry: reversing every Ising spin preserves the energy and the Boltzmann factor while reversing $S_1$.

For the two-point correlation function, the two insertions divide the ring into paths of $n$ and $N-n$ bonds. ThusIn particular, the exact finite-ring antiferromagnetic Ising correlation isBecause the one-point correlation functions vanish, this is also the connected correlation function. It gives $C_0=1$ and $C_n=C_{N-n}$, as required by $S_1^2=1$ and exchanging the two paths around the ring. At $K=0$, distinct Ising spins are independent and $C_n=0$ for $1\le n\le N-1$; $C_0=1$ is understood separately if a power $0^0$ occurs in the displayed expression.

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.

Let $\beta =(k_{B}T{)}^{-1}$, and sum over all $2^N$ configurations satisfying the periodic identification. The canonical partition function and finite-size free energy per site areThe plus sign multiplying $J$ in the energy means that $J>0$ favors opposite neighboring spins: this is an antiferromagnetic Ising model. Even $N$ permits an alternating configuration to close without frustration. The finite-size formula does not require taking a thermodynamic limit; that limit is taken only after evaluating $Z_N$.