Four-spin Ising ring ($N=4,H=-J{\sum }_{i=1}^4{\sigma }_i{\sigma }_{i+1}$)

Take a cycle graph with four labeled graph vertices and four distinct edges $\{1,2\},\{2,3\},\{3,4\},\{4,1\}$. Set $h=0$, $J>0$, and impose periodic boundary conditions by writing ${\sigma }_5={\sigma }_1$. The square drawing represents a one-dimensional ring: it has no diagonal interactions and is not a two-dimensional lattice.

Call an edge between opposite Ising spins a domain wall. If $d$ is their number, the four bond products sum to $4-2d$, so the energy isEach domain wall costs $2J$ relative to an aligned bond. Returning around the ring must return to the starting Ising spin, so $d$ is even: equivalently, ${\prod }_{i=1}^4({\sigma }_i{\sigma }_{i+1})=1$. Choosing an even set of $d$ wall edges and then choosing ${\sigma }_1$ determines the entire pattern. The state degeneracy is therefore $g_d=2\left(\genfrac{}{}{0ex}{}{4}{d}\right)$, using a binomial coefficient.

The twelve middle microstates consist of eight patterns with one exceptional Ising spin and four patterns with two adjacent $+1$ Ising spins and two adjacent $-1$ Ising spins. The two alternating patterns have four domain walls. All $2+12+2=16$ microstates are shown below.

Writing $K=\beta J$, the partition function and the probability of each energy class areFor an original exact example, choose $K=(\ln 2)/4$. Then $e^{4K}=2$, $Z=17$, and $P(d=0)=4/17$, $P(d=2)=12/17$, $P(d=4)=1/17$. Each aligned microstate has probability $2/17$, each middle microstate $1/17$, and each alternating microstate $1/34$: multiply by state degeneracy before comparing whole classes. As $K\to 0$, all sixteen microstates become equally likely, but the class probabilities approach $1/8,3/4,1/8$. As $K\to \infty$, the two all-aligned microstates each have limiting probability $1/2$.

The Ising ring bond correlation is the neighboring two-point correlation function $c_1=⟨{\sigma }_i{\sigma }_{i+1}⟩$ in the four-spin Ising ring. All four edges are equivalent. At zero field each mean Ising spin is zero by spin inversion symmetry, so $c_1$ also equals the connected correlation function. Agreement gives bond product $+1$ and disagreement $-1$; thus $P({\sigma }_i={\sigma }_{i+1})=(1+c_1)/2$.

Differentiating the finite partition function with respect to $K=\beta J$ inserts the sum of four bond products. ThereforeHere $U$ is internal energy. At $K=(\ln 2)/4$, the class-weight calculation gives $c_1=(4-1)/17=3/17$ and $U=-12J/17$. The agreement probability for a selected bond is $10/17$, despite the exactly zero mean spin magnetization.

A second derivative gives the energy variance, and hence the heat capacity $C_H=dU/dT$ at fixed $J$ and $h=0$:To see the variance identity directly, a derivative of an expected value under weights $e^{KS}$ inserts $S-⟨S⟩$, where $S={\sum }_i{\sigma }_i{\sigma }_{i+1}=-H/J$. The factor $K^2$ follows from $dK/dT=-K/T$. At the same exact example, $\mathrm{Var}(H)/J^2=1216/289$. The heat capacity vanishes both as $K\to 0$ and as $K\to \infty$: at high temperature its dimensionless factor $K^2$ vanishes, while at low temperature excited microstates become exponentially rare.

Define the instantaneous spin magnetization $m=\frac{1}{4}{\sum }_i{\sigma }_i$. Global reversal sends $m$ to $-m$ without changing the zero-field energy. Pairing every microstate with its reversed partner proves $⟨m⟩=0$ at every positive temperature, even when almost all probability lies on the two fully aligned patterns. A vanishing mean does not imply weak alignment: the limiting probability distribution is half at $m=+1$ and half at $m=-1$.

The Ising ring magnetization fluctuations measure this distinction through variance. In the $d=0$ class, both patterns have $m^2=1$. Among the $d=2$ patterns, eight have $m^2=1/4$ and four have $m^2=0$; the two $d=4$ patterns also have $m^2=0$. ConsequentlyIt equals $6/17$ at $K=(\ln 2)/4$, approaches $1/4$ for four independent fair Ising spins at $K\to 0$, and approaches $1$ as $K\to \infty$. Unlike the energy variance, the spin magnetization variance remains large in the low-temperature symmetric mixture.

The same thermal fluctuation controls the field response. Allow $h$ to vary before setting it to zero; differentiating the Boltzmann distribution inserts $\beta {\sum }_i{\sigma }_i=4\beta m$, givingThis is a spin-normalized magnetic susceptibility with respect to the energy-valued parameter $h$; conversion to physical magnetization and magnetic field requires the magnetic moment and sample volume factors.

For finite $N$, $Z$ is a finite positive sum of exponential functions. Its free energy $-k_{B}T\log Z$ is a real analytic function for $T>0$, so the smooth heat capacity peak here is not a critical phase transition. Spontaneous magnetization concerns the order of limits: first a thermodynamic limit, then removal of a symmetry-breaking field. The four-spin Ising ring alone establishes neither that limit nor any two-dimensional Ising model result.^[ref]