The Ising Model

6.1 Definition

The Ising model is the simplest model of a phase transition. On a lattice of $N$ sites, each site $i$ has a spin variable $s_i \in \{+1, -1\}$. The Hamiltonian is

$H = -J\sum_{\langle i,j\rangle} s_i s_j - h\sum_i s_i$

Where $J$ is the coupling constant, $\langle i,j\rangle$ denotes nearest-neighbour pairs, and $h$ is an external magnetic field.

• $J > 0$: ferromagnetic (spins prefer to align).
• $J < 0$: antiferromagnetic (spins prefer to anti-align).

6.2 Exact Solution in One Dimension

Theorem 6.1. The 1D Ising model with $h = 0$ has no phase transition at any finite temperature.

Proof (Transfer matrix method). Consider a chain of $N$ spins with periodic boundary conditions ($s_{N+1} = s_1$). The partition function is:

$Z = \sum_{\{s_i\}} \prod_{i=1}^{N} e^{\beta J s_i s_{i+1}}$

Define the transfer matrix $\mathbf{T}$ with elements $T_{s_i, s_{i+1}} = e^{\beta J s_i s_{i+1}}$:

$\mathbf{T} = \begin{pmatrix} e^{\beta J} & e^{-\beta J} \\ e^{-\beta J} & e^{\beta J} \end{pmatrix}$

The partition function is $Z = \mathrm{Tr}(\mathbf{T}^N) = \lambda_+^N + \lambda_-^N$ where $\lambda_\pm$ are the eigenvalues of $\mathbf{T}$:

$\lambda_\pm = e^{\beta J} \pm e^{-\beta J}$

In the thermodynamic limit ($N \to \infty$), $Z \approx \lambda_+^N$ and the free energy per spin is:

$f = -k_BT \ln(e^{\beta J} + e^{-\beta J}) = -k_BT \ln(2\cosh\beta J)$

The magnetisation $m = -\partial f/\partial h|_{h=0} = 0$ for all $T > 0$Confirming no spontaneous magnetisation and hence no phase transition. $\blacksquare$

6.3 Mean Field Theory

Theorem 6.2 (Mean field approximation). In mean field theory, each spin feels an effective field due to its neighbours. Replacing $s_j$ by its average $\langle s_j \rangle = m$ in the Hamiltonian:

$H_{\mathrm{MF} = -Jz\, m\sum_i s_i - h\sum_i s_i}$

Where $z$ is the coordination number (number of nearest neighbours). Each spin behaves as if in an effective field $h_{\mathrm{eff} = h + Jz\,m}$.

The self-consistency equation (mean field equation) is:

$m = \tanh\!\left(\frac{\beta(h + Jz\,m)}{k_B}\right) = \tanh\!\left(\frac{h + Jz\,m}{k_BT}\right)$

For $h = 0$: $m = \tanh(Jz\,m / k_BT)$.

Critical temperature. Expanding $\tanh x \approx x - x^3/3$ for small $x$:

$m = \frac{Jz\,m}{k_BT} - \frac{1}{3}\left(\frac{Jz\,m}{k_BT}\right)^3$

For $m \neq 0$Dividing by $m$:

$1 = \frac{Jz}{k_BT_c} - \frac{1}{3}\left(\frac{Jz}{k_BT_c}\right)^3$

At $T = T_c$: $T_c = Jz/k_B$.

6.4 Critical Exponents

Near the critical point, thermodynamic quantities follow power laws:

$m \sim (T_c - T)^{1/\beta}, \quad \chi \sim |T - T_c|^{-\gamma}, \quad C \sim |T - T_c|^{-\alpha}$

Mean field theory predicts:

$\beta = \frac{1}{2}, \quad \gamma = 1, \quad \alpha = 0\ \text{(jump discontinuity)}$

These are the classical critical exponents. They are independent of the spatial dimension $d$ and the lattice structure --- a deficiency of mean field theory. Exact results and renormalisation group calculations give dimension-dependent exponents that agree with experiment.

6.5 Worked Example: Mean Field Theory for the 2D Square Lattice

Problem. For the 2D Ising model on a square lattice ($z = 4$), find $T_c$ in mean field theory and compare with the exact result $k_BT_c / J = 2/\ln(1 + \sqrt{2}) \approx 2.269$.

6.6 Worked Example: Susceptibility above $T_c$

Problem. Calculate the magnetic susceptibility $\chi = \partial m/\partial h|_{h=0}$ above $T_c$ in mean field theory.