Antiferromagnetic Ising model ($J>0$)

With the convention $H=J{\sum }_{⟨ij⟩}{\sigma }_i{\sigma }_j-h{\sum }_i{\sigma }_i$, positive $J$ favors opposite neighboring Ising spins. On a bipartite graph, the Néel order parameter measures opposite sublattice magnetizations; it is distinct from the uniform magnetization. Changing the signs of spins on one sublattice maps the zero-field model to the ferromagnetic Ising model.

Take $2M$ base Ising spins on a ring and add one apex Ising spin over every other base bond. Give all base and apex bonds the antiferromagnetic energy $Js{s}^{\prime }$ with $J>0$. There are $3M$ sites, and adjacent triangles are connected by a single undecorated bridge rather than sharing a corner. With $K=J/(k_{B}T)$, summing the apex spins yields alternating spin-chain transfer matricesTheir common symmetric eigenvector gives the leading cell eigenvalue $3+e^{-4K}+e^{-2K}+3e^{2K}$. A minimum-energy triangle has one frustrated bond and energy $-J$; its connecting bridge can have energy $-J$. Requiring the bridge to be antiparallel leaves the counting transfer matrix $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ on successive first base spins. Its eigenvalues $3,1$ give ground-state degeneracy $g_M=3^M+1$ and ground-state energy $-2JM$. The residual entropy per site is $k_B\ln 3/3$.

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.

An odd periodic nearest-neighbour Ising model with antiferromagnetic coupling has unavoidable geometrical frustration. For $k$ parallel bonds, $H=-NJ+2Jk$, while ${\prod }_j{S}_j{S}_{j+1}=1$ forces $N-k$ to be even. Thus odd $N$ requires at least one parallel bond. A ground state has exactly one, with $N$ possible locations and two global spin inversion symmetry choices, proving the displayed ground-state degeneracy and ground-state energy. The zero-temperature canonical ensemble distributes that defect uniformly. A directed arc of $n$ bonds contains it with probability $n/N$, giving two-point correlation function $(-1{)}^n(1-2n/N)$. The finite antiferromagnetic Ising-ring correlation gives the corresponding positive-temperature result.

The exact zero-field two-point correlation function $C_n=⟨S_1{S}_{n+1}⟩$ in the finite canonical ensemble of an antiferromagnetic Ising ring. The spin-chain transfer matrix has eigenvalues ${\lambda }_{+}=2\cosh (\beta J)$ and ${\lambda }_{-}=-2\sinh (\beta J)$. In their eigenbasis, an Ising spin insertion exchanges the two eigenvectors. The trace with two insertions is therefore ${\lambda }_{-}^n{\lambda }_{+}^{N-n}+{\lambda }_{+}^n{\lambda }_{-}^{N-n}$, while the partition function is ${\lambda }_{+}^N+{\lambda }_{-}^N$, proving the formula. Its two terms retain both paths around the ring. It has $C_0=1$ and $C_n=C_{N-n}$. At fixed $n$ and finite positive inverse temperature its thermodynamic limit is $(-\tanh \beta J{)}^n$.

At fixed odd $N$, the zero-temperature canonical ensemble equally weights the $2N$ ground states of an antiferromagnetic Ising ring. The single frustrated bond in an antiferromagnetic Ising ring lies on a chosen $n$-bond path with probability $n/N$. The Ising spin product is $(-1{)}^n$ if the path excludes that bond and $-(-1{)}^n$ if it includes it, proving the displayed two-point correlation function. The loss of staggered alignment upon averaging persists at zero temperature and is caused by geometric frustration, rather than a positive density of thermal defects. The expression remains symmetric under $n\mapsto N-n$ for odd $N$.

A neighboring parallel pair in an antiferromagnetic Ising ring. It costs $2J$ relative to a satisfied antiparallel pair. If there are $m$ such pairs, the energy is $-JN+2Jm$, and the periodic boundary conditions force $N-m$ to be even because ${\prod }_j{S}_j{S}_{j+1}=1$. Thus an odd ring has $2N$ ground states, obtained by locating one such pair and choosing one overall Ising spin orientation. The even ring has two ground states with no such pair. This definition uses the local preferred ordering; a parallel pair is a defect in staggered order rather than a wall between uniformly magnetized domains.

On a regular bipartite graph with coordination number $q$, replace each bond product by ${\sigma }_B{\sigma }_W\simeq M_W{\sigma }_B+M_B{\sigma }_W-M_B{M}_W$. For $N$ sites there are $Nq/2$ bonds, so the constant in the effective Hamiltonian is $-NqJ{M}_B{M}_W/2$. The mean-field self-consistency equations are $M_B=\tanh \beta (h+g-qJ{M}_W)$ and $M_W=\tanh \beta (h-g-qJ{M}_B)$. At zero fields the antisymmetric branch has a continuous mean-field Ising critical bifurcation at $k_B{T}_c=qJ$.

On the stable zero-field branch $M_B=-M_W=M$, put $a=\beta (1-M^2)$. Differentiating the two-sublattice Ising mean-field approximation with respect to the uniform field gives ${\chi }_B=a(1-qJ{\chi }_W)$ and ${\chi }_W=a(1-qJ{\chi }_B)$. Thus ${\chi }_{+}=a/(1+qJa)$ and $\partial M_{-}/\partial h=0$. The true staggered magnetic susceptibility, $\partial M_{-}/\partial g=a/(1-qJa)$, is a different response and diverges at the continuous transition. The factor $1-M^2$ is essential: the uniform response vanishes when the spins saturate at zero temperature.

Here $s(m)=\frac{1+m}{2}\ln \frac{1+m}{2}+\frac{1-m}{2}\ln \frac{1-m}{2}$. This free-energy density is the energy minus temperature times entropy of independent sublattice Ising spins. Since $s^{\prime }(m)=\mathrm{atanh}m$, stationarity yields the two-sublattice Ising mean-field approximation. Writing $M_{\pm }=(M_B\pm M_W)/2$, the quadratic coefficients are $(k_{B}T+qJ)/2$ for $M_{+}^2$ and $(k_{B}T-qJ)/2$ for $M_{-}^2$. The positive quartic coefficient $k_{B}T/12$ in the staggered direction stabilizes the continuous transition. Unlike a log-cosh auxiliary potential, this variational expression can be minimized away from self-consistency.