 Four-spin Ising ring

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 $1/2$.