ISEGORIA / MATH ENCYCLOPEDIA
Magnetism and the Ising model: order from alignment
Spins that only want to agree with their neighbours, and a temperature that decides whether they can. The two-dimensional Ising lattice with Onsager’s exact critical temperature and magnetisation, the one-dimensional chain that never orders and says exactly why, and the mean-field picture that predicts a transition everywhere and gets the numbers wrong in an instructive way.
Before you begin: Statistical mechanics and probability
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. The two-dimensional Ising model
Each site of a square lattice carries a spin ±1, and the energy is −J for every pair of agreeing neighbours. At high temperature the spins flip almost at random and the magnetisation m, the mean spin, is zero. Cool the lattice and at T꜀ = 2J / k ln(1 + √2) = 2.269 J/k the spins spontaneously choose a direction: below T꜀, m jumps away from zero with the exponent 1/8 that Onsager and Yang extracted from the exact solution. The lattice on the left is updated by the Metropolis rule, one random spin at a time, accepting a flip that costs energy ΔE with probability e^{−ΔE/kT}. Drag the temperature through T꜀ and watch domains form; the graph on the right compares the running average of |m| with the exact curve.
Worked example. At \(kT=2J\), \(\sinh(2J/kT)=\sinh1=1.175\), so \(m=(1-1.175^{-4})^{1/8}=(1-0.524)^{1/8}=0.911\): well below \(T_c\) the lattice is nearly saturated. At \(kT=2.2J\), \(m=0.785\); at \(kT=2.26J\), just under \(T_c\), \(m=0.613\), and the curve drops vertically to zero at \(T_c\) itself. In the disordered phase a flip that turns two agreeing neighbours into disagreeing ones costs \(\Delta E=8J\) at most, accepted with probability \(e^{-8J/kT}=0.20\) at \(kT=5J\) and \(0.0003\) at \(kT=J\).
Watch out. The lattice is 64 by 64 with periodic edges, and a finite lattice has no true phase transition: near \(T_c\) the whole lattice can flip its sign, which is why the graph shows \(|m|\) and why the measured points sit above the exact curve just above \(T_c\), where the correlation length exceeds the box. Metropolis dynamics also slows down critically near \(T_c\), so the running average there needs many sweeps to settle. A field \(h\) removes the transition entirely: \(m\) is then an analytic function of \(T\), and the exact \(m(T,h)\) of the two-dimensional model is not known.
Why does the Metropolis rule, which never computes the partition function, sample the Boltzmann distribution?
Because it satisfies detailed balance: for two configurations \(a\) and \(b\) differing by one spin, the rate \(a\to b\) is \(\min(1,e^{-(E_b-E_a)/kT})\) and the rate \(b\to a\) is \(\min(1,e^{-(E_a-E_b)/kT})\), and their ratio is exactly \(e^{-(E_b-E_a)/kT}=P(b)/P(a)\). A Markov chain whose transition rates stand in the ratio of the target probabilities, and which can reach every configuration, has that target as its unique stationary distribution. The normalising constant \(Z\) cancels in the ratio, which is why only energy differences ever appear.
2. The chain that never orders
In one dimension the same model has no transition at any temperature above zero, and the reason is a picture: it costs only 2J to insert a domain wall anywhere along a chain of N spins, and there are N places to put it, so the entropy k ln N wins over the energy 2J for every T > 0. The chain is solved exactly by the transfer matrix, whose largest eigenvalue gives the free energy and whose ratio of eigenvalues gives the correlation length ξ = −1/ln tanh(J/kT): correlations decay as tanh(J/kT)^r, and ξ diverges only as T → 0. Drag T and compare the measured correlation function with the exact one; the chain on the left shows the domain walls.
Worked example. At \(kT=J\), \(\tanh1=0.762\) and \(\xi=3.7\) spins: neighbours agree 76% more often than chance, but ten spins away the correlation is \(0.762^{10}=0.066\). At \(kT=0.2J\), \(\tanh5=0.99991\) and \(\xi=11{,}000\) spins, longer than the chain of 240 drawn here, which is why the cold chain looks ordered: it is one domain by accident, not by a phase transition. With a field the eigenvalues are \(\lambda_\pm=e^{J/kT}[\cosh(h/kT)\pm\sqrt{\sinh^2(h/kT)+e^{-4J/kT}}]\) and the magnetisation \(m=\sinh(h/kT)/\sqrt{\sinh^2(h/kT)+e^{-4J/kT}}\) is smooth in \(T\) and vanishes at \(h=0\).
Watch out. The measured correlation is an average over a chain of 240 spins and a few hundred sweeps, so it scatters around the exact curve, especially at large \(r\) where the signal is small; the exact curve is for the infinite chain. The domain-wall argument is Peierls’s in reverse: in two dimensions a wall enclosing a droplet has a length that grows with the droplet, so its energy beats the entropy at low temperature and order survives. The Mermin–Wagner theorem, which forbids continuous-symmetry order in two dimensions, does not apply to Ising spins because their symmetry is discrete.
Why does the transfer matrix turn a sum over 2ᴺ configurations into a product of two numbers?
The Boltzmann weight of a chain factorises into a product of factors, one per bond, each depending on two neighbouring spins. Summing over the interior spin between two bonds is exactly matrix multiplication of the two 2×2 matrices of bond weights, so the sum over all spins of a ring of \(N\) bonds is \(\operatorname{tr}V^N\). A symmetric matrix can be diagonalised, so this is \(\lambda_+^N+\lambda_-^N\), and for large \(N\) the larger eigenvalue dominates: the free energy per spin is \(-kT\ln\lambda_+\), and the ratio \(\lambda_-/\lambda_+<1\) sets the rate at which correlations decay, since inserting two spins into the trace is a matrix element of \(V^r\) between them.
3. Mean field: right idea, wrong numbers
Replace the neighbours of a spin by their average, and each spin feels an effective field zJm from its z neighbours plus the applied h. A single spin in a field has ⟨s⟩ = tanh((zJm + h)/kT), and requiring consistency, that the average spin is the m we assumed, gives the Curie–Weiss equation m = tanh((zJm + h)/kT). Below kT꜀ = zJ the line y = m crosses the tanh curve three times and the ordered solutions appear, with m ∝ (T꜀ − T)^{1/2}. For the square lattice z = 4, so mean field puts T꜀ at 4 J/k against Onsager’s 2.269, and its exponent 1/2 against the exact 1/8; for the chain it predicts a transition at 2 J/k where there is none. Drag T and watch the intersections; the right panel lays the three m(T) curves side by side.
Worked example. With \(z=4\) and \(kT=2J\), the self-consistent solution is \(m=\tanh(2m)\), i.e. \(m=0.957\); Onsager gives \(0.911\) at the same temperature. At \(kT=3J\) mean field still says \(m=0.776\) while the exact lattice is already disordered. Near the mean-field \(T_c\), expanding \(\tanh x\approx x-x^3/3\) gives \(m^2=3(1-T/T_c)\): at \(kT=3.9J\), \(m=\sqrt{3\times0.025}=0.27\).
Watch out. Mean field ignores fluctuations, and fluctuations are exactly what a phase transition is made of: they are what destroys order in one dimension and what lowers \(T_c\) in two. The approximation improves as the number of neighbours grows and becomes exact for infinite-range interactions or in more than four dimensions, where the fluctuation corrections are irrelevant in the renormalisation-group sense. The exponents \(\beta=1/2\), \(\gamma=1\) are therefore not wrong so much as the wrong universality class: the two-dimensional Ising values \(\beta=1/8\), \(\gamma=7/4\) are shared by every two-dimensional system with a two-state symmetry, whatever its lattice.
Why does the mean-field equation always produce a transition, even in one dimension where the exact answer has none?
Because it replaces the fluctuating neighbours by a fixed average field, it cannot see that in one dimension a single domain wall, costing a finite energy, breaks the alignment of everything beyond it. The self-consistency equation only asks whether a uniform \(m\) can sustain itself, and a uniform \(m\) with \(z\) neighbours always can once \(zJ/kT>1\), because the slope of \(\tanh(zJm/kT)\) at the origin exceeds 1. The question it never asks is whether that uniform state is stable against local excitations, and in one dimension it is not.