Square-lattice Ising model

In statistical mechanics, the two-dimensional square-lattice Ising model is a simple model of interacting magnetic spins. The model is notable for having nontrivial interactions, yet having an analytical solution. The model was solved by (Lars Onsager 1944) for the special case that the external magnetic field H = 0. An analytical solution for the general case for ${\displaystyle H\neq 0}$ has yet to be found.

Definition of the model

Consider the 2D Ising model on a square lattice ${\displaystyle \Lambda }$ with N sites, with periodic boundary conditions in both the horizontal and vertical directions, which effectively reduces the geometry of the model to a torus. In a general case, the horizontal coupling J is not equal to the coupling in the vertical direction, J*. With an equal number of rows and columns in the lattice, there will be N of each. In terms of

${\displaystyle K=\beta J}$
${\displaystyle L=\beta J^{*}}$

where ${\displaystyle \beta =1/(kT)}$ where T is absolute temperature and k is Boltzmann's constant, the partition function ${\displaystyle Z_{N}(K,L)}$ is given by

${\displaystyle Z_{N}(K,L)=\sum _{\{\sigma \}}\exp \left(K\sum _{\langle ij\rangle _{H}}\sigma _{i}\sigma _{j}+L\sum _{\langle ij\rangle _{V}}\sigma _{i}\sigma _{j}\right).}$

Critical temperature

The critical temperature ${\displaystyle T_{c}}$ can be obtained from the Kramers–Wannier duality relation. Denoting the free energy per site as ${\displaystyle F(K,L)}$, one has:

${\displaystyle \beta F\left(K^{*},L^{*}\right)=\beta F\left(K,L\right)+{\frac {1}{2}}\log \left[\sinh \left(2K\right)\sinh \left(2L\right)\right]}$

where

${\displaystyle \sinh \left(2K^{*}\right)\sinh \left(2L\right)=1}$
${\displaystyle \sinh \left(2L^{*}\right)\sinh \left(2K\right)=1}$

Assuming there is only one critical line in the (K,L) plane, the duality relation implies that this is given by:

${\displaystyle \sinh \left(2K\right)\sinh \left(2L\right)=1}$

For the isotropic case ${\displaystyle J=J^{*}}$, one finds the famous relation for the critical temperature ${\displaystyle T_{c}}$

${\displaystyle k_{B}T_{c}/J={\frac {2}{\ln(1+{\sqrt {2}})}}\approx 2.26918531421}$.

Dual lattice

Consider a configuration of spins ${\displaystyle \{\sigma \}}$ on the square lattice ${\displaystyle \Lambda }$. Let r and s denote the number of unlike neighbours in the vertical and horizontal directions respectively. Then the summand in ${\displaystyle Z_{N}}$ corresponding to ${\displaystyle \{\sigma \}}$ is given by

${\displaystyle e^{K(N-2s)+L(N-2r)}}$
Dual lattice

Construct a dual lattice ${\displaystyle \Lambda _{D}}$ as depicted in the diagram. For every configuration ${\displaystyle \{\sigma \}}$, a polygon is associated to the lattice by drawing a line on the edge of the dual lattice if the spins separated by the edge are unlike. Since by traversing a vertex of ${\displaystyle \Lambda }$ the spins need to change an even number of times so that one arrives at the starting point with the same charge, every vertex of the dual lattice is connected to an even number of lines in the configuration, defining a polygon.

Spin configuration on a dual lattice

This reduces the partition function to

${\displaystyle Z_{N}(K,L)=2e^{N(K+L)}\sum _{P\subset \Lambda _{D}}e^{-2Lr-2Ks}}$

summing over all polygons in the dual lattice, where r and s are the number of horizontal and vertical lines in the polygon, with the factor of 2 arising from the inversion of spin configuration.

Low-temperature expansion

At low temperatures, K, L approach infinity, so that as ${\displaystyle T\rightarrow 0,\ \ e^{-K},e^{-L}\rightarrow 0}$, so that

${\displaystyle Z_{N}(K,L)=2e^{N(K+L)}\sum _{P\subset \Lambda _{D}}e^{-2Lr-2Ks}}$

defines a low temperature expansion of ${\displaystyle Z_{N}(K,L)}$.

High-temperature expansion

Since ${\displaystyle \sigma \sigma '=\pm 1}$ one has

${\displaystyle e^{K\sigma \sigma '}=\cosh K+\sinh K(\sigma \sigma ')=\cosh K(1+\tanh K(\sigma \sigma ')).}$

Therefore

${\displaystyle Z_{N}(K,L)=(\cosh K\cosh L)^{N}\sum _{\{\sigma \}}\prod _{\langle ij\rangle _{H}}(1+v\sigma _{i}\sigma _{j})\prod _{\langle ij\rangle _{V}}(1+w\sigma _{i}\sigma _{j})}$

where ${\displaystyle v=\tanh K}$ and ${\displaystyle w=\tanh L}$. Since there are N horizontal and vertical edges, there are a total of ${\displaystyle 2^{2N}}$ terms in the expansion. Every term corresponds to a configuration of lines of the lattice, by associating a line connecting i and j if the term ${\displaystyle v\sigma _{i}\sigma _{j}}$ (or ${\displaystyle w\sigma _{i}\sigma _{j})}$ is chosen in the product. Summing over the configurations, using

${\displaystyle \sum _{\sigma _{i}=\pm 1}\sigma _{i}^{n}={\begin{cases}0&{\mbox{for }}n{\mbox{ odd}}\\2&{\mbox{for }}n{\mbox{ even}}\end{cases}}}$

shows that only configurations with an even number of lines at each vertex (polygons) will contribute to the partition function, giving

${\displaystyle Z_{N}(K,L)=2^{N}(\cosh K\cosh L)^{N}\sum _{P\subset \Lambda }v^{r}w^{s}}$

where the sum is over all polygons in the lattice. Since tanh K, tanh L ${\displaystyle \rightarrow 0}$ as ${\displaystyle T\rightarrow \infty }$, this gives the high temperature expansion of ${\displaystyle Z_{N}(K,L)}$.

The two expansions can be related using the Kramers-Wannier duality.

Exact solution

The free energy per site in the limit ${\displaystyle N\to \infty }$ is given as follows. Define the parameter k as:

${\displaystyle k={\frac {1}{\sinh \left(2K\right)\sinh \left(2L\right)}}}$

The Helmholtz free energy per site F can be expressed as:

${\displaystyle -\beta F={\frac {\log(2)}{2}}+{\frac {1}{2\pi }}\int _{0}^{\pi }\log \left[\cosh \left(2K\right)\cosh \left(2L\right)+{\frac {1}{k}}{\sqrt {1+k^{2}-2k\cos(2\theta )}}\right]d\theta }$

For the isotropic case ${\displaystyle J=J^{*}}$, from the above expression one finds for the internal energy per site:

${\displaystyle U=-J\coth(2\beta J)\left[1+{\frac {2}{\pi }}(2\tanh ^{2}(2\beta J)-1)\int _{0}^{\pi /2}{\frac {1}{\sqrt {1-4k(1+k)^{-2}\sin ^{2}(\theta )}}}d\theta \right]}$

and the spontaneous magnetization is, for ${\displaystyle T:

${\displaystyle M=\left[1-\sinh ^{-4}(2\beta J)\right]^{1/8}}$