Onsager Solution of the Two-Dimensional Ising Model

For a decade the sharpest theorists in statistical mechanics argued over a question that sounds almost philosophical: can a phase transition be real? The partition function of any finite system is a finite sum of exponentials — smooth, analytic, singular nowhere. Yet water boils, magnets lose their magnetism, and the specific heat spikes as though something genuinely breaks. Either the sharpness was an illusion of large numbers, or something new had to happen in the infinite-volume limit. Nobody could settle it, because nobody could solve a single interacting model exactly. In 1944 Lars Onsager did. He computed, in closed form, the free energy of the two-dimensional Ising model on a square lattice — spins $\sigma = \pm 1$ coupled to their nearest neighbours, no external field. The answer is a double integral over the Brillouin zone, finite everywhere except at one temperature, where its second derivative diverges. That divergence is a bona fide thermodynamic singularity, produced by short-range forces in the infinite-volume limit. The debate was over: phase transitions are real, and mathematics can prove it. The solution also delivered a scandal. Mean-field theory — the workhorse approximation of the era — predicts that the magnetization vanishes as $(T_c - T)^{1/2}$ and the specific heat merely jumps. Onsager's exact answer says the specific heat diverges logarithmically and the magnetization vanishes as $(T_c - T)^{1/8}$ . Those numbers, $1/8$ and a log, are not close to the mean-field guesses; they are a different universe. Explaining why the exact critical exponents look nothing like the naive ones is the problem that eventually forced the renormalization group into existence.

Key takeaways

Onsager (1944) solved the zero-field square-lattice Ising model exactly: the bulk free energy per spin is a double integral over the Brillouin zone, finite except at one temperature.
The transition sits at the Kramers–Wannier self-dual point $\sinh 2K_c = 1$ , giving $K_c = \tfrac12\ln(1+\sqrt2) \approx 0.4407$ and $k_BT_c/J = 2/\ln(1+\sqrt2) \approx 2.2692$ .
The specific heat diverges logarithmically, $C \sim -\ln|T-T_c|$ — critical exponent $\alpha = 0$ (log), not the finite mean-field jump.
The full exact exponent set is $(\alpha,\beta,\gamma,\delta,\nu,\eta) = (0_{\log}, \tfrac18, \tfrac74, 15, 1, \tfrac14)$ — wildly different from mean field $(0,\tfrac12,1,3,\tfrac12,0)$ .
The solution proved that finite-range interactions in $d \ge 2$ produce a true thermodynamic singularity in the infinite-volume limit.
Key idea: after fermionization the Ising transfer matrix becomes a gas of free fermions whose gap closes exactly at the self-dual point — and the way that gap closes prints the exact critical exponents, none of which mean-field theory can reproduce.

The core insight

The two-dimensional Ising model is exactly solvable because its transfer matrix is the exponential of a free-fermion Hamiltonian; the fermion gap closes at the Kramers–Wannier self-dual point $\sinh 2K_c = 1$ , and the linear way it closes produces a logarithmic specific-heat singularity and the exact exponent $\beta = 1/8$ .

Where this lives today

The 2D Ising solution is the permanent benchmark of the field. Every new Monte Carlo algorithm, tensor-network contraction, or machine-learning ansatz for statistical mechanics is first tested against Onsager's exact $T_c = 2.2692\,J/k_B$ and $\beta = 1/8$ — a result known to arbitrary precision is priceless for calibration. The critical point itself is now understood as a conformal field theory with central charge $c = 1/2$ , the theory of a single free Majorana fermion, which reproduces the whole exponent set and much more (the full operator content, correlation functions, and finite-size spectra) by symmetry alone. The model is also physically realized. Uniaxial antiferromagnets, adsorbed monolayers on crystal surfaces, and ultrathin magnetic films all fall into the 2D Ising universality class, and their measured critical exponents match Onsager's numbers. Cold-atom experiments now assemble Ising-like spin arrays site by site and watch domain walls in real time — the free fermions of the exact solution made visible.

The physical picture

Read a spin configuration not by its spins but by its domain walls — the boundaries between $+$ and $-$ regions. Every configuration is a set of closed loops of wall on the lattice, and the whole thermodynamics is a competition between two tendencies. A wall costs energy (broken bonds), which suppresses it; but a wall can be drawn in exponentially many ways, which is entropy that encourages it. At high temperature entropy wins and walls proliferate into a disordered tangle; at low temperature energy wins and large walls are rare, so a global majority spin survives. The transition is the temperature where the statistical weight of a wall exactly balances the number of ways to draw it. Onsager's diagonalization turns this word-picture into arithmetic. The domain walls become free fermionic quasiparticles — non-interacting excitations, each with a definite energy $\gamma(q)$ that depends on its momentum $q$ along the lattice. The cheapest of them, at $q = 0$ , has energy $\gamma(0) = 2|K - K^*|$ , where $K^*$ is the Kramers–Wannier dual coupling. This gap is the inverse correlation length: a big gap means walls are expensive and correlations die fast (order); a vanishing gap means walls are free and correlations reach across the whole system (criticality). The gap closes at exactly one point, $K = K^*$ , which is the self-dual coupling $\sinh 2K_c = 1$ .

Figure

Free-fermion gap γ(0) = 2|K − K*| closing linearly at the self-dual point, the inverse correlation length

Because the dispersion near the closing point is relativistic — $\gamma \approx \sqrt{\delta^2 + c^2 q^2}$ with $\delta \propto (T - T_c)$ — the low-energy density of states integrated over the two-dimensional Brillouin zone gives a logarithm rather than a power. That is the origin of the famous log-divergent specific heat: not a power law $|T-T_c|^{-\alpha}$ with some positive $\alpha$ , but the marginal case $\alpha = 0$ dressed with a logarithm. Below is the punchline that ended the mean-field era — the exact $C(T)$ against the mean-field prediction on the same axes.

Figure

The punchline — exact 2D-Ising specific heat diverges logarithmically at T_c while mean-field only jumps

The core equations

Onsager's central object is the free energy per spin in the thermodynamic limit. It is not an elementary function — it is an irreducible double integral over the two lattice momenta — but it is exact, and its analytic structure is everything. The argument of the logarithm is positive for all $T$ except at the single temperature where it touches zero, and that touching is the phase transition.

-\beta f = \ln 2 + \frac{1}{2(2\pi)^2}\int_0^{2\pi}\!\!\int_0^{2\pi} \mathrm{d}q_1\,\mathrm{d}q_2\,\ln\!\big[\cosh^2 2K - \sinh 2K\,(\cos q_1 + \cos q_2)\big] \qquad (1)

where: $f$ — Helmholtz free energy per spin; $\beta = 1/k_BT$ ; $K = J/k_BT$ the dimensionless coupling; $q_1, q_2$ — lattice momenta running over the Brillouin zone $[0,2\pi)^2$ . The whole singular behaviour hides in the corners of the integrand where the bracket approaches zero; everywhere else the integral is a smooth analytic function of $T$ . The competition inside the bracket is between $\cosh^2 2K$ (which grows with coupling) and $\sinh 2K$ times a momentum factor that reaches its most dangerous value $+2$ at $q_1 = q_2 = 0$ . Setting the bracket to zero there gives $\cosh^2 2K - 2\sinh 2K = 0$ , i.e. $(\sinh 2K - 1)^2 = 0$ , so the transition is fixed at $\sinh 2K_c = 1$ . This is the same self-dual point that Kramers–Wannier duality had located three years earlier, now confirmed as the genuine home of the singularity.

\sinh 2K_c = 1, \qquad K_c = \tfrac12\ln(1+\sqrt2) \approx 0.440687, \qquad \frac{k_BT_c}{J} = \frac{2}{\ln(1+\sqrt2)} \approx 2.26919 \qquad (2)

where: $K_c$ — critical coupling; $T_c$ — critical temperature. Read this as a number to memorize: a square-lattice Ising ferromagnet orders below about $2.27\,J/k_B$ . Compare it to the mean-field prediction $k_BT_c/J = 4$ (each spin has $z=4$ neighbours) — mean field overestimates $T_c$ by 76%, because it ignores the fluctuations that make ordering harder. The engine behind the solution is the fermion dispersion. Onsager's transfer matrix, after fermionization, describes non-interacting quasiparticles whose energy $\gamma(q)$ satisfies a $\cosh$ relation involving both the coupling $K$ and its dual $K^*$ . The minimum energy — the gap — lives at $q = 0$ and equals $2|K - K^*|$ , vanishing precisely when $K = K^*$ .

\cosh\gamma(q) = \cosh 2K^{*}\cosh 2K - \sinh 2K^{*}\sinh 2K\,\cos q, \qquad \tanh K^{*} = e^{-2K} \qquad (3)

where: $\gamma(q)$ — one-particle (domain-wall) energy at momentum $q$ ; $K^*$ — Kramers–Wannier dual coupling, satisfying $\sinh 2K\,\sinh 2K^* = 1$ . Setting $q=0$ collapses the right side to $\cosh(2K - 2K^*)$ , so $\gamma(0) = 2|K - K^*|$ : the gap is the "mass" of the cheapest excitation, and it is the inverse correlation length. Finally, the observable that broke mean-field theory. The specific heat is the second temperature derivative of the free energy, and near $T_c$ the small-momentum region of the integral (1) contributes a logarithm. The coefficient is universal up to the non-singular background.

C \;\underset{T\to T_c}{\sim}\; -\,\frac{2k_B}{\pi}\Big(\frac{2K_c}{\tanh 2K_c}\Big)^{\!2}\Big[\ln\Big|1-\frac{T}{T_c}\Big| + \text{const}\Big] \qquad (4)

where: $C$ — heat capacity per spin at zero field; the bracket diverges to $+\infty$ as $T \to T_c$ from either side. This is the definition of $\alpha = 0$ (log): the divergence is weaker than any power $|t|^{-\alpha}$ with $\alpha > 0$ , yet it is a genuine divergence, not the finite step mean-field predicts.

Derivation ALocating the transition by Kramers–Wannier self-duality

We want the exact critical temperature without solving the whole model — using only a symmetry. The strategy: expand the partition function two ways, once at low temperature (organized by domain-wall loops) and once at high temperature (organized by closed graphs of bonds), and show that these two expansions are the same sum evaluated at two different couplings, $K$ and a dual coupling $K^*$ . Duality maps the free energy at $K$ onto the free energy at $K^*$ ; if the model has a single transition, its singularity must be invariant under the map, which forces it to sit at the self-dual fixed point $K = K^*$ . What to watch out for: the argument assumes exactly one transition — a fact Onsager's full solution independently confirms.

Step 1. Write the low-temperature expansion. Start from the fully aligned ground state and flip clusters; each configuration is organized by its closed domain walls of total length $L$ , each broken bond costing weight $e^{-2K}$ , so $Z = e^{2NK}\sum_{\text{walls}} e^{-2KL}$ . Why: at low $T$ the natural small quantity is the number of overturned bonds, and every excitation is a closed loop of wall.

Step 2. Write the high-temperature expansion. Use the identity $e^{K\sigma\sigma'} = \cosh K\,(1 + \sigma\sigma'\tanh K)$ on every bond and sum over spins; only closed graphs survive, weighted by $(\tanh K)^L$ , giving $Z = (\cosh K)^{2N} 2^N \sum_{\text{graphs}} (\tanh K)^L$ . Why: at high $T$ the small quantity is $\tanh K$ , and the spin sum kills any graph with a dangling end — closed loops are all that remain.

Step 3. Match the two loop sums. Both expansions sum over the same set of closed loops on the square lattice, so they are identical sums with different weights. Identifying $e^{-2K} \leftrightarrow \tanh K^*$ gives the duality map, which rearranges to the symmetric involution $\sinh 2K\,\sinh 2K^* = 1$ . Why: the square lattice is self-dual — its dual is another square lattice — so the low- $T$ loops and high- $T$ loops live on congruent lattices and can be identified term by term.

Step 4. Impose self-duality. A single singularity, mapped to itself by $K \leftrightarrow K^*$ , must sit at the fixed point $K = K^*$ . Then $\sinh^2 2K_c = 1$ , so $\sinh 2K_c = 1$ , i.e. $2K_c = \ln(1+\sqrt2)$ and $k_BT_c/J = 2/\ln(1+\sqrt2) \approx 2.2692$ . Why: duality is an involution on coupling space with one fixed point; the transition, being unique, has nowhere else to go. *Verified: verify_aq-sm-onsager-solution-2d-ising.py confirms $\sinh(2K_c) = 1$ exactly (symbolically via the exp form), $K_c = 0.440687$ , $k_BT_c/J = 2.26919$ , and the algebraic equivalence $\tanh K^* = e^{-2K} \Leftrightarrow \sinh 2K\,\sinh 2K^* = 1$ .*

Go deeper · Derivation B · 2 minFree fermions: transfer matrix to dispersion and the closing gap

Derivation BFree fermions: transfer matrix to dispersion and the closing gap

The goal is to show why the model is solvable at all: the transfer matrix that advances the lattice one row at a time is the exponential of a Hamiltonian quadratic in fermion operators, and quadratic fermion problems are exactly diagonalizable, exactly like coupled harmonic oscillators. We build the transfer matrix from Pauli operators, fermionize with a Jordan–Wigner string, Fourier-transform, and Bogoliubov-rotate to free quasiparticles. The payoff is the dispersion (3) and the observation that its gap closes linearly at criticality — the seed of $\nu = 1$ . Watch out: the Jordan–Wigner string is non-local, and it is exactly that non-locality that converts spin commutation into fermion anticommutation.

Step 1. Write the transfer matrix in Pauli operators. Adding a row splits into a vertical-bond factor $V_2 = \exp(K\sum_j \sigma^z_j\sigma^z_{j+1})$ and a horizontal-bond factor $V_1 = \exp(K^*\sum_j \sigma^x_j)$ , where the dual coupling appears because summing over an intermediate spin turns a bond into a transverse field. Why: the row-to-row weight is a product over bonds, and each bond type becomes one operator on the auxiliary quantum spin chain.

Step 2. Apply the Jordan–Wigner transformation, $\sigma^x_j = 1 - 2c_j^\dagger c_j$ and $\sigma^z_j = (\prod_{l<j}(1-2c_l^\dagger c_l))(c_j + c_j^\dagger)$ . The nonlocal string $\prod_{l<j}$ supplies a sign for each occupied site to the left, so the resulting $c_j$ genuinely anticommute; both $V_1$ and $V_2$ become quadratic in these fermions. Why: spins on different sites commute, but fermions must anticommute — the string is the minimal fix, and it happens to leave neighbouring products local.

Step 3. Fourier-transform and Bogoliubov-rotate. Momentum modes $\pm q$ couple in $2\times2$ blocks with a "normal" part and an anomalous $c_q^\dagger c_{-q}^\dagger$ part; a rotation $\eta_q = \cos\theta_q\, c_q - \sin\theta_q\, c_{-q}^\dagger$ diagonalizes each block, giving free quasiparticles of energy $\gamma(q)$ obeying (3). Why: the anomalous terms mix particles and holes, and a Bogoliubov angle is exactly the tool that removes them, leaving a number operator per mode.

Step 4. Read off the gap. At $q = 0$ , $\cosh\gamma(0) = \cosh(2K - 2K^*)$ , so $\gamma(0) = 2|K - K^*|$ , which vanishes only at $K = K^* = K_c$ . Near $T_c$ the gap is linear in $T - T_c$ , so the correlation length $\xi = 1/\gamma(0)$ diverges as $|T-T_c|^{-1}$ : the exponent $\nu = 1$ . Why: gaplessness is the operational definition of criticality; the rate at which the gap closes sets $\nu$ , and here it is linear. *Verified: verify_aq-sm-onsager-solution-2d-ising.py confirms $\cosh\gamma(0) = \cosh(2(K-K^*))$ symbolically, that the gap vanishes iff $K = K^*$ , that it is linear in $(K - K_c)$ near $T_c$ (⇒ $\nu = 1$ ), and that the free-energy integrand's argument stays positive at $K_c$ so $f$ is real.*

Limiting cases

Case	Result	Meaning
$T \to 0$ ( $K \to \infty$ )	$m \to 1$ , $\gamma(0) \to \infty$	fully aligned; the cheapest wall costs infinite energy
$T \to \infty$ ( $K \to 0$ )	$m = 0$ , correlations $\to 0$	free spins; the gap is again large, correlations die instantly
$T \to T_c$	$\gamma(0) \to 0$ , $\xi \to \infty$ , $C \sim -\ln\|t\|$	the fermion gap closes; a genuine thermodynamic singularity
high dimension $d \ge 4$	exponents $\to$ mean field	above the upper critical dimension fluctuations are harmless
$d = 1$	no transition at $T > 0$	a single point-wall costs only $2J$ and always proliferates

Common trap“Onsager’s solution gives the whole thermodynamics, including χ and the equation of state”

It does not. Onsager's exact result covers the zero-field free energy, the specific heat, the internal energy, and the one- and two-spin correlation functions — that is already extraordinary. But the zero-field susceptibility $\chi$ is not an elementary closed form (it is governed by an infinite series of ever-harder integrals, still an active research topic), and the full equation of state in a field $h \ne 0$ has never been solved. The exponent $\gamma = 7/4$ and $\delta = 15$ are known exactly by other means (scaling relations, series, conformal field theory), not by evaluating Onsager's integral. When someone says "the 2D Ising model is solved," they mean the zero-field partition function and correlations — a precise and limited statement.

Worked examples

1 · The critical temperature you can compute by hand (confidence builder)

Find $k_BT_c/J$ for the square-lattice Ising model from the self-dual condition, and compare it to the mean-field estimate.

Step 1. The self-dual condition is $\sinh 2K_c = 1$ . Solve for $K_c$ : $2K_c = \operatorname{arcsinh}(1) = \ln(1 + \sqrt2)$ , so $K_c = \tfrac12\ln(1+\sqrt2) = \tfrac12(0.881374) = \mathbf{0.440687}$ .

Step 2. Since $K_c = J/k_BT_c$ , invert: $k_BT_c/J = 1/K_c = 2/\ln(1+\sqrt2) = 2/0.881374 = \mathbf{2.26919}$ .

Step 3. Mean-field theory gives $k_BT_c/J = z = 4$ (coordination number of the square lattice). The exact value is 43% lower: fluctuations, which mean field ignores, make ordering substantially harder, so the true $T_c$ is pushed down. The number $2.2692$ is worth carrying in your head — it is to 2D Ising what $13.6\,\mathrm{eV}$ is to hydrogen: the one anchor from which everything else scales.

2 · How the fermion gap sets the correlation length (exam level)

At a temperature just below $T_c$ , take the coupling $K = 1.05\,K_c$ . Compute the dual coupling $K^*$ , the fermion gap $\gamma(0)$ , and hence the correlation length $\xi$ in lattice units. What does this say about $\nu$ ?

Step 1. Coupling: $K = 1.05 \times 0.440687 = \mathbf{0.462721}$ .

Step 2. Dual coupling from $\tanh K^* = e^{-2K}$ : $e^{-2(0.462721)} = e^{-0.925442} = 0.396383$ , so $K^* = \operatorname{arctanh}(0.396383) = \mathbf{0.419316}$ .

Step 3. Gap: $\gamma(0) = 2|K - K^*| = 2|0.462721 - 0.419316| = 2(0.043405) = \mathbf{0.08681}$ .

Step 4. Correlation length: $\xi = 1/\gamma(0) = 1/0.08681 = \mathbf{11.5}$ lattice spacings. The system is ordered but its fluctuating clusters already span ~12 sites.

Step 5. Move closer to $T_c$ — halve $(K - K_c)$ — and both $\gamma(0)$ and $1/\xi$ halve too: the gap is linear in the distance to criticality, so $\xi \propto |T-T_c|^{-1}$ . That linear law is precisely the statement $\nu = 1$ , and it is exact, not fitted.

Origins · 1 minLars Onsager and how this was discovered

Lars Onsager

1903–1976 · Yale University

Onsager was a Norwegian-American physical chemist of legendary indirectness — his lectures were famously impenetrable — and legendary depth. He announced the exact free energy of the 2D Ising model in 1942 and published the full derivation in 1944, a calculation so intricate (transfer matrices, a $2N$ -dimensional Clifford algebra, quaternion representations) that it took the community years to digest and simpler routes another decade to find. He then compounded the feat: in 1949, at a conference in Florence, he displayed the exact spontaneous magnetization $m = (1 - \sinh^{-4}2K)^{1/8}$ — with the exponent $1/8$ — without ever publishing a proof (C. N. Yang supplied one in 1952). Onsager received the 1968 Nobel Prize in Chemistry, but for his reciprocal relations in irreversible thermodynamics, not for the Ising tour de force — a solution that, more than any other single result, made the modern theory of critical phenomena possible.

History

1920

— Lenz proposes the model; his student Ising solves the 1D case (1925) and wrongly concludes no phase transition in any dimension.

1936

— Peierls argues that the 2D model must order at low temperature, contradicting the pessimistic extrapolation from 1D.

1941

— Kramers and Wannier locate the exact critical temperature by self-duality, three years before the full solution.

1944

— Onsager publishes the exact zero-field free energy of the square-lattice Ising model, proving a true singularity exists.

1949

— Onsager displays the spontaneous magnetization

m = (1-\sinh^{-4}2K)^{1/8}

at a Florence conference, without a published proof.

1952

— Yang publishes the first complete derivation of the magnetization, fixing

\beta = 1/8

rigorously.

1971

— Wilson's renormalization group finally explains why the exact exponents differ so sharply from mean field.

Going further

Texts. Pathria & Beale, Statistical Mechanics, 3rd ed., Ch. 13 · Kardar, Statistical Physics of Fields, Ch. 6 · Baxter, Exactly Solved Models in Statistical Mechanics, Ch. 7 · McCoy & Wu, The Two-Dimensional Ising Model (the definitive monograph).

Provenance. Onsager, L. (1944). "Crystal Statistics I." Phys. Rev. 65, 117. A verifier should confirm: $\sinh 2K_c = 1 \Rightarrow K_c = \tfrac12\ln(1+\sqrt2) = 0.440687$ and $k_BT_c/J = 2/\ln(1+\sqrt2) = 2.26919$ ; the duality involution $\tanh K^* = e^{-2K} \Leftrightarrow \sinh 2K\,\sinh 2K^* = 1$ ; the gap $\gamma(0) = 2|K-K^*|$ , linear in $(T-T_c)$ near criticality (⇒ $\nu = 1$ ); the log-divergent specific heat (⇒ $\alpha = 0$ ); the exponent set $(0,\tfrac18,\tfrac74,15,1,\tfrac14)$ and its four scaling relations at $d=2$ ; the worked numbers ( $k_BT_c/J = 2.2692$ ; $K = 1.05K_c \Rightarrow K^* = 0.41932$ , $\gamma(0) = 0.0868$ , $\xi = 11.5$ ).

Prerequisites. The Ising model definition and Hamiltonian · the transfer matrix formalism · the one-dimensional Ising transfer matrix · Kramers–Wannier duality. Leads to. The spontaneous magnetization of the 2D Ising model · high- and low-temperature series expansions · the Peierls argument · domain walls and interfacial excitations.

If the partition function of any finite lattice is analytic, how can Onsager's free energy be singular?

Both statements are true and not in conflict. For any finite $N \times N$ lattice, $Z$ is a finite sum of exponentials — perfectly analytic, singular nowhere. The non-analyticity appears only in the limit $N \to \infty$ : as you add more spins, the smooth peak in the specific heat grows taller and narrower, and only in the strict infinite-volume limit does it become an actual divergence. Onsager computed $f = -k_BT\lim_{N\to\infty}(\ln Z)/N^2$ directly in that limit, which is why his answer can be singular while every finite approximation to it is smooth. This is the precise sense in which "phase transitions are an infinite-volume phenomenon."

What does α = 0 actually mean — is there a divergence or not?

There is a divergence; $\alpha = 0$ does not mean "no singularity." The exponent $\alpha$ is defined by $C \sim |t|^{-\alpha}$ , and $\alpha = 0$ is the marginal case where the power law degenerates into a logarithm: $C \sim -\ln|t|$ . A logarithm still goes to $+\infty$ as $t \to 0$ , just more slowly than any positive power. So the 2D Ising specific heat genuinely diverges at $T_c$ — it is unbounded — but the divergence is logarithmic. Contrast this with mean field, where $C$ has a finite jump (a step of a few $k_B$ ) and no divergence at all. The distinction "log-divergent vs finite jump" is the whole difference between the exact and mean-field pictures.

Why is the correlation length the inverse of the fermion gap?

In the fermion picture, correlations propagate by emitting and reabsorbing quasiparticles, and the amplitude for a quasiparticle of the cheapest kind (energy $\gamma(0)$ ) to travel a distance $r$ decays as $e^{-\gamma(0)\,r}$ . Comparing to the definition $G(r) \sim e^{-r/\xi}$ gives $\xi = 1/\gamma(0)$ immediately. Physically, a large gap means the cheapest excitation is expensive, so correlations are short-ranged (small $\xi$ ); a vanishing gap means excitations are free, so correlations reach across the whole system ( $\xi \to \infty$ ). Criticality is the gap closing, and since the gap closes linearly, $\xi$ diverges as $|T-T_c|^{-1}$ , i.e. $\nu = 1$ .

How did Kramers and Wannier find T_c three years before Onsager solved the model?

They used only the self-duality symmetry, not the full spectrum. The high-temperature expansion of $Z$ (in $\tanh K$ ) and the low-temperature expansion (in $e^{-2K}$ ) are the same sum over closed loops, evaluated at two couplings related by $\sinh 2K\,\sinh 2K^* = 1$ . If the model has just one transition, that transition must be invariant under the duality map $K \leftrightarrow K^*$ , which forces it to sit at the unique fixed point $K = K^*$ , i.e. $\sinh 2K_c = 1$ . This pins $T_c$ exactly without ever computing the free energy — a beautiful example of getting a hard number from a symmetry. Onsager's full solution later confirmed the singularity really lives there.

Mean field gives k_BT_c/J = 4, Onsager gives 2.27. Why is mean field so far off?

Mean-field theory replaces the fluctuating field each spin feels by the average field of its neighbours, discarding all correlations. This systematically overestimates the tendency to order, because it ignores the fact that a spin's neighbours are themselves fluctuating and often disagree. The overestimate pushes $T_c$ too high — mean field predicts $k_BT_c/J = z = 4$ (the coordination number), while the exact value is $2.27$ , some 43% lower. The lower dimension, the worse mean field does: in $d = 2$ fluctuations are severe. Above the upper critical dimension $d_c = 4$ , fluctuations become harmless and mean-field exponents (though not $T_c$ ) become exact.

Is the whole 2D Ising model solved, including the susceptibility?

No — and this is a common overstatement. Onsager and Yang solved the zero-field free energy, specific heat, internal energy, and the one- and two-spin correlation functions. But the zero-field susceptibility $\chi$ is not an elementary closed form: it is governed by an infinite series of increasingly difficult integrals, and its detailed analytic structure (natural boundaries in the complex plane) is still studied today. The full equation of state in a nonzero field $h$ has never been solved at all. The exponents $\gamma = 7/4$ and $\delta = 15$ are known exactly, but by scaling relations, series analysis, and conformal field theory — not by evaluating Onsager's integral. "The 2D Ising model is solved" means the zero-field partition function and correlations, a precise but limited claim.

What is the connection between Onsager's exponents and conformal field theory?

At its critical point the 2D Ising model is scale-invariant, and in two dimensions scale invariance is enormously constraining — it enlarges to full conformal invariance, described by a conformal field theory (CFT) with central charge $c = 1/2$ , the theory of a single free Majorana fermion. That CFT reproduces the entire exponent set $(0,\tfrac18,\tfrac74,15,1,\tfrac14)$ from the dimensions of just a few primary operators (the spin field has dimension $1/8$ , giving $\eta = 1/4$ and $\beta = 1/8$ ; the energy field gives $\alpha$ and $\nu$ ). The free Majorana fermion of the CFT is precisely the continuum limit of Onsager's lattice free fermions — the two descriptions are the same physics, one on the lattice and one in the continuum.