Mean-Field Ising Model and Spontaneous Magnetization

Derivation

Mean-Field Ising Model and Spontaneous Magnetization

D-245 Home PU-302 Threads fields · chance · symmetry · matter Depends on The Partition Function and F = -kT ln Z, Energy Fluctuations and Heat Capacity

Statement

For the Ising model \(\mathcal{H} = -J\sum_{\langle ij\rangle} s_i s_j - h\sum_i s_i\) with \(s_i=\pm 1\) on a lattice of coordination number \(z\), the Weiss mean-field approximation replaces each neighbour spin by its thermal average \(m=\langle s\rangle\), yielding the self-consistency equation \(m=\tanh\!\big(\beta(zJm+h)\big)\). At \(h=0\) this admits a nonzero solution below the Curie temperature \(k_B T_c = zJ\), so the \(\mathbb{Z}_2\) symmetry \(s_i\to -s_i\) is spontaneously broken and the system acquires a spontaneous magnetization \(m(T)\).

Why it matters

This is the simplest microscopic model that shows a genuine continuous phase transition with a nonzero order parameter, and it captures the essential physics of ferromagnetism: cooperative alignment of spins winning over thermal disorder below a critical temperature. The same self-consistency structure reappears across physics — the BCS gap equation, the Curie–Weiss law of dielectrics, neural-network Hopfield models, and Landau theory of order parameters all descend from it.

Mean-field theory is also the first honest confrontation with spontaneous symmetry breaking: a Hamiltonian invariant under \(s_i\to -s_i\) produces states that are not, and the two degenerate ground states below \(T_c\) are the paradigm for everything from magnetization to the Higgs mechanism.

Assumptions

Nearest-neighbour, uniform ferromagnetic coupling \(J>0\).With frustration or antiferromagnetic bonds the uniform ansatz \(m\) fails and one needs sublattice or staggered order parameters; the single self-consistency equation is no longer sufficient. Every spin sees the same mean field (translational invariance, no boundaries).Near surfaces or defects the local field differs, so \(m\) becomes position dependent and one must solve coupled local equations \(m_i=\tanh(\beta\sum_j J_{ij}m_j)\). Fluctuations of neighbours about the mean are negligible: \((s_j-m)(s_k-m)\) is dropped.This is exact only as the number of neighbours \(z\to\infty\); in low dimension the neglected correlations dominate near \(T_c\) and every mean-field critical exponent comes out wrong. A single scalar order parameter with discrete \(\mathbb{Z}_2\) symmetry.For continuous symmetry (Heisenberg, XY) the Mermin–Wagner theorem forbids ordering in \(d\le 2\); the Ising discreteness is what lets order survive in \(d=2\).

Derivation

\[ \mathcal{H} = -J\sum_{\langle ij\rangle} s_i s_j - h\sum_i s_i \]

Start from the exact Ising Hamiltonian; the bilinear coupling \(s_i s_j\) is what makes the partition function intractable. A

\[ s_i s_j = \big[m+(s_i-m)\big]\big[m+(s_j-m)\big] = -m^2 + m(s_i+s_j) + (s_i-m)(s_j-m) \]

Write each spin as its mean \(m=\langle s\rangle\) plus a fluctuation \(\delta s_i = s_i-m\); this is an exact algebraic identity so far. A

\[ s_i s_j \approx -m^2 + m(s_i+s_j) \]

Drop the quadratic fluctuation term \((s_i-m)(s_j-m)\): the mean-field (Weiss) approximation. Legal when neighbour fluctuations are uncorrelated and small, exact as \(z\to\infty\). C

\[ \mathcal{H}_{\text{MF}} = \frac{1}{2}NzJ\,m^2 - (zJm+h)\sum_i s_i \]

Sum over the \(\tfrac12 Nz\) bonds: each spin has \(z\) neighbours, giving the constant \(+\tfrac12 NzJm^2\) and, for each site, an effective field \(h_{\text{eff}}=zJm+h\). B

\[ Z = e^{-\tfrac12\beta NzJ m^2}\left[\,2\cosh\!\big(\beta(zJm+h)\big)\right]^{N} \]

The Hamiltonian is now a sum of independent single-spin terms, so the partition function factorizes; each spin contributes \(\sum_{s=\pm1}e^{\beta h_{\text{eff}}s}=2\cosh(\beta h_{\text{eff}})\). B

\[ \langle s_i\rangle = \frac{1}{\beta}\frac{\partial \ln Z}{\partial h_{\text{eff},i}} = \tanh\!\big(\beta(zJm+h)\big) \]

The thermal average of a single spin in field \(h_{\text{eff}}\) is \(\tanh(\beta h_{\text{eff}})\), from \(\partial_x\ln(2\cosh x)=\tanh x\). B

\[ \boxed{\,m = \tanh\!\big(\beta(zJm+h)\big)\,} \]

Impose self-consistency: the average we assumed, \(m\), must equal the average the effective field produces, \(\langle s_i\rangle\). This closes the loop. A

\[ m = \tanh(\beta zJ\,m) \approx \beta zJ\,m - \frac{1}{3}(\beta zJ\,m)^3 \qquad (h=0) \]

Set \(h=0\) and expand \(\tanh x = x-\tfrac13 x^3+\dots\) for small \(m\) near the transition. The slope of the right side at \(m=0\) is \(\beta zJ\). B

\[ \left.\frac{d}{dm}\tanh(\beta zJ m)\right|_{m=0} = \beta zJ = 1 \;\Rightarrow\; k_B T_c = zJ \]

A nonzero root bifurcates from \(m=0\) exactly when the initial slope equals \(1\); above this the only solution is \(m=0\), below it a symmetric pair \(\pm m\) appears. This defines the Curie temperature. A

\[ 1 = \frac{T_c}{T} - \frac{1}{3}\Big(\frac{T_c}{T}\Big)^3 m^2 \;\Rightarrow\; m^2 = \frac{3\,(T_c/T-1)}{(T_c/T)^3} \xrightarrow{T\to T_c^-} 3\,\frac{T_c-T}{T_c} \]

Divide the cubic (step 8) by \(m\neq0\) and solve for \(m^2\), using \(\beta zJ = T_c/T\). Near \(T_c\) the prefactor \(\to1\). C

\[ m(T) \simeq \sqrt{3}\left(1-\frac{T}{T_c}\right)^{1/2},\qquad \beta_{\text{exp}}=\tfrac12 \]

Take the square root: the order parameter rises with a square-root law and mean-field critical exponent \(\beta_{\text{exp}}=1/2\). C

Result

\[ m = \tanh\!\big(\beta(zJm+h)\big),\qquad k_B T_c = zJ,\qquad m(T\to T_c^-)\simeq\sqrt{3}\Big(1-\tfrac{T}{T_c}\Big)^{1/2} \]

Reading. The magnetization per spin is determined self-consistently: the field each spin feels is set by the average alignment of its \(z\) neighbours. Above \(T_c\) thermal agitation wins and \(m=0\) is the only solution; below \(T_c\) two mirror-image ordered states \(\pm m\) appear and the system must pick one, breaking the up–down symmetry. The order parameter grows continuously from zero with a square-root law, so the transition is second order (continuous).

Units check. \(m\) and \(\tanh\) are dimensionless; the argument \(\beta(zJm+h)=(zJm+h)/k_BT\) is (energy)/(energy) and hence dimensionless. \(T_c = zJ/k_B\) has units \(\mathrm{J}/(\mathrm{J\,K^{-1}}) = \mathrm{K}\), a temperature, as required.

Limiting cases

\(T\to0\): \(\beta\to\infty\), so \(\tanh(\beta zJm)\to1\) and \(m\to1\) — every spin fully aligned, the saturated ferromagnet.
\(T\to T_c^-\): \(m\simeq\sqrt{3}\,(1-T/T_c)^{1/2}\to0\), continuous onset with infinite slope \(dm/dT\to-\infty\).
\(T>T_c,\;h\to0\): only \(m=0\); linearizing gives the Curie–Weiss susceptibility \(\chi = 1/\big(k_B(T-T_c)\big)\), diverging as \(T\to T_c^+\) with exponent \(\gamma=1\).
\(z\to\infty\) (or infinite-range coupling): fluctuations vanish and mean-field theory becomes exact.
\(h\neq0\): the symmetry is explicitly broken, \(m\neq0\) at all \(T\), and the sharp transition is rounded into a smooth crossover.

Breaks when

Low dimension near \(T_c\). In \(d=1\) there is no transition at all (\(T_c=0\)), yet mean field predicts \(k_BT_c=2J\); in \(d=2\) the true \(k_BT_c=2J/\ln(1+\sqrt2)\approx2.27J\) versus the mean-field \(4J\). The neglected fluctuations diverge and dominate.
Critical exponents. Mean field always gives \(\beta_{\text{exp}}=1/2,\ \gamma=1,\ \delta=3,\ \alpha=0\); the exact 2D Ising values are \(\beta_{\text{exp}}=1/8,\ \gamma=7/4,\ \delta=15\). Below the upper critical dimension \(d_c=4\) the exponents are wrong.
Strong short-range correlations / frustration. Spin glasses, antiferromagnets on frustrated lattices, and systems with competing interactions violate the uniform-\(m\) ansatz entirely.

Failure modes

Double-counting bonds. Writing the coupling energy as \(NzJm^2\) instead of \(\tfrac12 NzJm^2\); each bond is shared by two sites, so \(z\) neighbours give \(\tfrac12 Nz\) bonds.
Confusing \(z\) with dimension \(d\). The transition depends on \(z\) (6 for simple cubic, 4 for square, 8 for bcc), not directly on \(d\); using \(z=d\) gives nonsense.
Trusting \(T_c=zJ/k_B\) as exact. It systematically overestimates the true \(T_c\) because it ignores the fluctuations that disorder the system.
Choosing the wrong root. Below \(T_c\) the equation has three solutions \(0,\pm m\); \(m=0\) is a free-energy maximum (unstable) and must be discarded — check the second derivative of \(f(m)\).
Dropping the linear fluctuation term instead of the quadratic. The approximation deletes \((s_i-m)(s_j-m)\), not the \(m(s_i+s_j)\) term that supplies the effective field.
Applying the Landau square-root formula far from \(T_c\). \(m\simeq\sqrt3(1-T/T_c)^{1/2}\) is a small-\(m\) expansion; at \(T=0.5T_c\) it overshoots \(m>1\), which is impossible.

Discussion

The self-consistency equation is a feedback loop made quantitative: assume the neighbours have average alignment \(m\), compute the field they produce, work out the response of a single spin to that field, and demand the answer reproduce the assumption. Graphically, \(m\) is the intersection of the line \(y=m\) with the curve \(y=\tanh(\beta zJm)\). When the curve leaves the origin with slope \(<1\) (high \(T\)) the only crossing is at the origin; when the slope exceeds \(1\) (low \(T\)) two symmetric crossings split off — the geometric picture of a pitchfork bifurcation and of symmetry breaking.

The order parameter \(m\) is the prototype of Landau theory. Expanding the mean-field free energy per spin \(f(m)=\tfrac12 zJ m^2 - k_BT\ln\!\big(2\cosh\beta(zJm+h)\big)\) in powers of \(m\) gives \(f\approx f_0 + \tfrac12 a(T-T_c)m^2 + \tfrac14 b\,m^4\): a single well above \(T_c\), a double well below. The system rolls into one of the two minima, and which one it chooses is not fixed by the Hamiltonian — that is spontaneous symmetry breaking, and the flat direction connecting degenerate states foreshadows Goldstone modes when the symmetry is continuous.

The deeper lesson is that mean-field theory is a variational bound, not a truncation. The Bogoliubov–Gibbs inequality \(F\le F_0+\langle\mathcal{H}-\mathcal{H}_0\rangle_0\) with a trial single-spin Hamiltonian \(\mathcal{H}_0=-b_{\text{eff}}\sum_i s_i\) yields exactly the same self-consistency equation upon minimizing over \(b_{\text{eff}}\); the Weiss field is the optimal effective field. This also explains precisely why it fails: the Ginzburg criterion shows the neglected fluctuation energy stays subdominant only for \(d>d_c=4\), so below four dimensions the true critical behaviour belongs to a different universality class computable by the renormalization group, with the mean-field exponents recovered only at and above \(d_c\).

Common misconceptions. (i) A nonzero \(T_c\) in \(d=1\) is an artefact — mean field is qualitatively wrong there. (ii) \(T_c=zJ/k_B\) is an upper bound, not the real value. (iii) "The symmetry is broken by the field \(h\)" confuses explicit breaking (\(h\neq0\)) with spontaneous breaking (\(h\to0^+\) after \(N\to\infty\)); only the ordering of these limits produces a genuine phase transition.

Worked examples

Example 1 — Curie temperature of a simple-cubic ferromagnet.

\[ k_B T_c = zJ \quad\Rightarrow\quad T_c = \frac{zJ}{k_B} \]

Use the mean-field Curie temperature; a simple-cubic lattice has coordination number \(z=6\). A

\[ T_c = \frac{6\times(1.00\times10^{-21}\,\mathrm{J})}{1.381\times10^{-23}\,\mathrm{J\,K^{-1}}} \]

Insert \(z=6\), a representative exchange \(J=1.00\times10^{-21}\,\mathrm{J}\) (about \(6.2\,\mathrm{meV}\)), and \(k_B\). A

\[ T_c \approx 4.34\times10^{2}\ \mathrm{K} \approx 435\ \mathrm{K} \]

Reading. Mean field predicts ordering up to about \(435\,\mathrm{K}\). Because it neglects fluctuations, the true \(T_c\) of a real simple-cubic Ising magnet is lower, roughly \(0.75\,T_c^{\text{MF}}\approx330\,\mathrm{K}\).

Example 2 — Spontaneous magnetization at \(T=0.90\,T_c\).

\[ m = \tanh\!\Big(\frac{T_c}{T}\,m\Big) = \tanh(1.111\,m) \]

Set \(h=0\) and write \(\beta zJ = T_c/T = 1/0.90 = 1.111\). Solve by fixed-point iteration \(m_{n+1}=\tanh(1.111\,m_n)\). B

\[ 0.55 \to 0.543 \to 0.538 \to 0.535 \to \dots \to 0.531 \]

Iterate from a Landau seed \(m_0=\sqrt{3(1-0.9)}=0.548\); the map converges monotonically to the stable root. B

\[ m_{\text{Landau}} = \sqrt{3\Big(1-\tfrac{T}{T_c}\Big)} = \sqrt{0.30} = 0.548 \]

Compare with the near-\(T_c\) square-root formula; it overestimates by \(\sim3\%\) because \(T=0.9T_c\) is already outside the strictly-critical region. B

\[ m(0.90\,T_c) \approx 0.53 \]

Reading. Just \(10\%\) below \(T_c\) the magnet is already \(53\%\) saturated — the square-root law makes the order parameter rise steeply out of the transition. The Landau approximation (\(0.548\)) and the exact self-consistent solve (\(0.531\)) agree to a few percent here.

Problems

Lattice dependence of \(T_c\). For a fixed exchange \(J/k_B = 100\,\mathrm{K}\), compute the mean-field Curie temperature for the linear chain (\(z=2\)), square (\(z=4\)), simple cubic (\(z=6\)) and bcc (\(z=8\)) lattices. Comment on the \(d=1\) result.
Solution
Since \(T_c = zJ/k_B = z\times100\,\mathrm{K}\): \(z=2\Rightarrow200\,\mathrm{K}\); \(z=4\Rightarrow400\,\mathrm{K}\); \(z=6\Rightarrow600\,\mathrm{K}\); \(z=8\Rightarrow800\,\mathrm{K}\). The chain result is qualitatively wrong: the exact 1D Ising model has \(T_c=0\) (no finite-temperature order), because a single domain wall costs finite energy \(2J\) but gains extensive entropy, always destroying long-range order. Mean field ignores exactly these fluctuations.
Critical exponent. Starting from \(m=\tanh(\beta zJ m)\) at \(h=0\), expand to cubic order and derive \(m\propto(T_c-T)^{1/2}\), identifying the exponent \(\beta_{\text{exp}}\).
Solution
With \(t\equiv\beta zJ = T_c/T\), \(\tanh x\approx x-\tfrac13x^3\) gives \(m = tm-\tfrac13 t^3 m^3\). For \(m\neq0\) divide by \(m\): \(1=t-\tfrac13 t^3 m^2\), so \(m^2 = 3(t-1)/t^3\). As \(T\to T_c^-\), \(t\to1^+\) and \(t-1 = T_c/T-1=(T_c-T)/T\to(T_c-T)/T_c\), giving \(m^2\to 3(T_c-T)/T_c\) and \(m\simeq\sqrt3\,(1-T/T_c)^{1/2}\). Hence \(\beta_{\text{exp}}=\tfrac12\).
Curie–Weiss susceptibility. For \(T>T_c\) with a small field \(h\), linearize the self-consistency equation and show \(\chi\equiv(\partial m/\partial h)_{h\to0} = 1/\big(k_B(T-T_c)\big)\). What is the exponent \(\gamma\)?
Solution
For small arguments \(m\approx\beta(zJm+h)\). Solve: \(m(1-\beta zJ)=\beta h\Rightarrow m=\beta h/(1-\beta zJ)\). Then \(\chi=\beta/(1-\beta zJ)\). Multiply top and bottom by \(T\): \(\beta zJ = T_c/T\), so \(1-\beta zJ = (T-T_c)/T\) and \(\chi = (1/k_BT)\big/\big((T-T_c)/T\big) = 1/\big(k_B(T-T_c)\big)\). This is the Curie–Weiss law; the divergence \(\chi\sim(T-T_c)^{-\gamma}\) gives \(\gamma=1\).
Free-energy double well. The mean-field free energy per spin at \(h=0\) is \(f(m)=\tfrac12 zJm^2 - k_BT\ln\!\big(2\cosh(\beta zJm)\big)\). Expand to quartic order in \(m\) and show that the coefficient of \(m^2\) changes sign at \(T_c\), producing a double well below \(T_c\).
Solution
Let \(a\equiv\beta zJ m\). Use \(\ln\cosh a\approx\tfrac12 a^2-\tfrac1{12}a^4\). Then \(f\approx \tfrac12 zJm^2 - k_BT\ln2 - k_BT\big(\tfrac12\beta^2z^2J^2m^2-\tfrac1{12}\beta^4z^4J^4m^4\big)\). The quadratic coefficient is \(\tfrac12 zJ - \tfrac12 k_BT\beta^2 z^2J^2 = \tfrac12 zJ\big(1-\beta zJ\big) = \tfrac12 zJ\,(1-T_c/T)\). This is positive for \(T>T_c\) (single minimum at \(m=0\)) and negative for \(T<T_c\) (\(m=0\) becomes a maximum). The quartic coefficient \(+\tfrac1{12}k_BT\beta^4z^4J^4>0\) stabilizes two symmetric minima \(\pm m\) — the double well, hence spontaneous symmetry breaking.
Heat-capacity jump. Using \(u=-\tfrac12 zJ m^2\) per spin and the near-\(T_c\) result \(m^2\approx 3(T_c-T)/T_c\), find the discontinuity \(\Delta C\) in the specific heat per spin at \(T_c\).
Solution
Just below \(T_c\), \(u=-\tfrac12 zJ\cdot 3(T_c-T)/T_c = -\tfrac{3zJ}{2T_c}(T_c-T)\). Then \(C_{<}=du/dT = +\tfrac{3zJ}{2T_c}\). Above \(T_c\), \(m=0\) so \(u=0\) and \(C_{>}=0\) (in this leading approximation). Using \(zJ=k_BT_c\), \(C_{<}=\tfrac32 k_B\). Hence \(\Delta C = C_{<}-C_{>} = \tfrac32 k_B\) per spin — a finite jump (not a divergence), giving mean-field exponent \(\alpha=0\). The exact 2D Ising heat capacity instead diverges logarithmically, showing again that mean field misses the critical singularity.