advanced32 min studyLesson 37 in Physics

The Ising model and mean-field theory

Derive self-consistent magnetization, test stability, and understand why replacing neighbors by an average misses important fluctuations.

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What you’ll understand

  • Construct a nearest-neighbor ferromagnetic Ising Hamiltonian with consistent bond counting.
  • Derive and linearize the mean-field self-consistency equation.
  • Separate approximate mean-field ordering from finite-size symmetry and low-dimensional exact behavior.

A magnet can order collectively even when each microscopic moment has only two possible values. The Ising model isolates that possibility by rewarding neighboring alignment. Its exact state count grows exponentially with lattice size, so we introduce a controlled question rather than pretend to solve every lattice: what does replacing fluctuating neighbors by their average predict, and where does that replacement fail? Mean-field theory gives a useful self-consistency equation, but its limitations are part of the calculation.

Interactions create correlated configurations

Place variables si=±1s_i=\pm1 on a lattice and assign ferromagnetic coupling J>0J>0 to nearest-neighbor pairs. Count each unordered bond once. A field term −h∑isi-h\sum_i s_i biases the global orientation. On a regular bulk lattice with coordination number zz, there are approximately Nz/2Nz/2 bonds; boundaries modify that count for a finite open sample. Energy favors alignment, while the exponentially many disordered configurations can win entropically at high temperature.

Unlike independent spins, the partition function no longer factorizes into one-site factors because neighboring spins appear together in the energy. Dimensionality, coupling range, and boundary conditions matter. A finite zero-field system with spin-flip symmetric boundaries has exactly zero mean total magnetization when averaged over the complete equilibrium ensemble, even if most configurations have a large positive or negative magnetization. A symmetry-broken thermodynamic phase requires a specified limiting or conditioning procedure rather than discarding the opposite sign by assertion.

H=−J∑⟨ij⟩sisj−h∑isi,si∈{−1,+1},J>0H=-J\sum_{\langle ij\rangle}s_is_j-h\sum_i s_i,\qquad s_i\in\{-1,+1\},\qquad J>0
Each nearest-neighbor bond appears once. The field h is an energy, while the dimensionless mean magnetization per spin is m=⟨si⟩ in a uniform phase.↔ Scroll wide equations horizontally.

Replace a neighborhood by its average

Write a spin as its mean plus a fluctuation, and neglect products of the neighboring fluctuations. Each spin then sees an effective field h+zJmh+zJm. The one-spin thermal mean in that field must equal the mean mm used to construct it. This produces a self-consistency equation, not an externally prescribed magnetization. The decoupling also produces a constant correction that prevents double-counting interaction energy when constructing the mean-field free energy.

At zero field, expand the hyperbolic tangent near zero magnetization. The disordered solution changes its mean-field stability when βzJ=1\beta zJ=1, giving TcMF=zJ/kBT_c^{\mathrm{MF}}=zJ/k_B. Above that value, linearizing in a small field yields susceptibility per spin 1/(kBT−zJ)1/(k_BT-zJ). The divergence belongs to the infinite-system approximation. A numerical iteration of the self-consistency map can locate solutions, but its iteration index is not physical time and its convergence should not be interpreted as a measured magnetic relaxation rate.

m=tanh⁡ ⁣[β(zJm+h)],TcMF=zJkB,∂m∂h∣h=0,T>TcMF=1kBT−zJm=\tanh\!\left[\beta(zJm+h)\right],\qquad T_c^{\mathrm{MF}}=\frac{zJ}{k_B},\qquad \left.\frac{\partial m}{\partial h}\right|_{h=0,T>T_c^{\mathrm{MF}}}=\frac1{k_BT-zJ}
These are mean-field predictions for a uniform nearest-neighbor ferromagnet. They are not exact finite-dimensional Ising identities.↔ Scroll wide equations horizontally.

A useful approximation with visible failure modes

Below the mean-field transition, two oppositely magnetized solutions appear at zero field. Expanding the mean-field free energy near the transition gives a positive quartic term and predicts a square-root onset of spontaneous magnetization. Choosing a phase means first taking a thermodynamic limit with a small selecting field, then allowing that field to vanish. Reversing those limits restores the exact zero finite-system mean. Absolute magnetization and long-distance correlations can still reveal ordering tendencies in finite calculations.

Mean field discards spatial correlations, which become especially important near criticality and in low dimensions. A one-dimensional short-range Ising chain has no positive-temperature ordered phase, despite the nonzero transition temperature predicted by substituting z=2z=2 into the mean-field formula. Two- and three-dimensional critical exponents also differ from the simplest mean-field values. Large coordination, long-range couplings with suitable scaling, or dimensions above an appropriate upper critical dimension can improve its usefulness, but none of those conditions should be assumed for an arbitrary lattice.

A high-temperature response on a square lattice

Use the mean-field approximation for a square lattice with z=4z=4, JJ as the energy unit, and kBT=5Jk_BT=5J. Estimate magnetization per spin for h=0.05Jh=0.05J and inspect the local slope of the zero-field iteration map.

  1. The predicted transition scale is kBTcMF=4Jk_BT_c^{\mathrm{MF}}=4J, so 5J5J is on the disordered side of this approximate transition.

  2. The linear susceptibility per spin is 1/(5J−4J)=1/J1/(5J-4J)=1/J. Therefore m≈(1/J)(0.05J)=0.05m\approx(1/J)(0.05J)=0.05. Substitution gives tanh⁡[(4×0.05+0.05)/5]=tanh⁡(0.05)≈0.0499584\tanh[(4\times0.05+0.05)/5]=\tanh(0.05)\approx0.0499584, close to the estimate.

  3. At zero field and zero magnetization, the iteration-map slope is βzJ=4/5=0.8\beta zJ=4/5=0.8. Small perturbations shrink under this particular fixed-point iteration, although this numerical contraction is not the physical spin dynamics.

The result

Self-consistency predicts a strongly enhanced response compared with isolated spins. Every quoted transition or susceptibility remains a mean-field estimate.

Explore the model

Interactive lab

Interactive lab · network

Inspect one nonlinear mean-field-shaped map

The lab evaluates the single sigmoid σ(wx−0.5). Using tanh(y) = 2σ(2y)−1, compare a bounded sigmoid response with the shape needed for a mean-field spin average. Vary input x and weight w to inspect saturation and gain.

Inspect one nonlinear mean-field-shaped mapOne inspectable sigmoid unit, σ(wx − 0.5). Playback traces the computation graph. There is no training, attention mechanism or hidden deep network. Current readings: Product · wx: 1.2; Logit · wx − 0.5: 0.7; Sigmoid output: 0.6682. All parameters are adjustable with the labeled controls below.One input. One weight. One sigmoid.x1.20× w1.20+ b0.70σ0.668output0.668w = 1.00b = −0.50σ(z) = 1 / (1 + exp(−z))
Computation graphSignal trace
Product · wx
1.2
Logit · wx − 0.5
0.7
Sigmoid output
0.6682
t = 0 s

One inspectable sigmoid unit, σ(wx − 0.5). Playback traces the computation graph. There is no training, attention mechanism or hidden deep network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe fixed bias and sigmoid output are not the Ising self-consistency equation. This is only an analogy for a nonlinear response; the lab does not solve lattice spins, iterate the Ising map, compute Tc, or sample magnetic domains.

Check your understanding

Why is the exact equilibrium mean magnetization zero for a finite zero-field Ising system with spin-flip symmetric boundaries?

Try it yourself

At zero field and T=TcMF/2T=T_c^{\mathrm{MF}}/2, the mean-field equation becomes m=tanh⁡(2m)m=\tanh(2m). Verify that zero is a solution, test its local iteration stability, and check the approximate positive solution m=0.9575m=0.9575.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Differentiate the right-hand side at zero, then substitute the proposed nonzero value.

Compare with the solution

Zero solves the equation, but the derivative there is two, so simple fixed-point iteration moves away from a small nonzero perturbation. At m=0.9575m=0.9575, tanh⁡(1.915)≈0.957503\tanh(1.915)\approx0.957503, confirming the nonzero solution to the quoted precision. The negative solution follows by symmetry. These branches describe mean-field phase selection; they do not overturn the exact zero signed mean of a complete finite symmetric ensemble.

Take it with you

  • Ising interactions correlate spins and prevent the independent-spin partition factorization.
  • Mean field replaces neighbor fluctuations with a self-consistent uniform average.
  • Finite-size symmetry, dimensionality, and critical correlations delimit its predictions.

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