Onsager solution of the 2D Ising model (zero field)

Exact infinite-volume free energy for the 2D nearest-neighbor Ising model on the square lattice and identification of the critical point.
Onsager solution of the 2D Ising model (zero field)

Context

The on the 2D square lattice is the canonical exactly solvable interacting lattice system exhibiting a genuine at positive temperature.

Consider the nearest-neighbor ferromagnet (see ) with Hamiltonian

H(σ)=Ji,jσiσj,σi{1,+1},J>0, H(\sigma) = -J\sum_{\langle i,j\rangle}\sigma_i\sigma_j , \qquad \sigma_i\in\{-1,+1\},\quad J>0,

in finite volume with partition function (see )

ZΛ(β)=σΛexp(βHΛ(σ)). Z_\Lambda(\beta)=\sum_{\sigma_\Lambda}\exp\bigl(-\beta H_\Lambda(\sigma)\bigr).

The (Helmholtz) free energy density is the thermodynamic limit (see )

f(β)=1βlimΛ1ΛlogZΛ(β), f(\beta)= -\frac{1}{\beta}\lim_{|\Lambda|\to\infty}\frac{1}{|\Lambda|}\log Z_\Lambda(\beta),

when the limit exists.

Theorem (Onsager, zero external field)

Let K=βJK=\beta J and define the modulus

k(K)=2sinh(2K)cosh2(2K). k(K)=\frac{2\sinh(2K)}{\cosh^2(2K)}.

For the 2D square-lattice Ising model at zero field, the infinite-volume free energy density exists and satisfies the exact formula

βf(β)=ln ⁣(2cosh(2K))+12π0πln ⁣(1+1k(K)2sin2θ2)dθ. -\beta f(\beta) ={} \ln\!\bigl(2\cosh(2K)\bigr) +\frac{1}{2\pi}\int_0^\pi \ln\!\left( \frac{1+\sqrt{1-k(K)^2\sin^2\theta}}{2} \right)\,d\theta.

Moreover, the nonanalyticity occurs precisely at

sinh ⁣(2Kc)=1Kc=12ln(1+2), \sinh\!\bigl(2K_c\bigr)=1 \qquad\Longleftrightarrow\qquad K_c=\frac12\ln(1+\sqrt2),

equivalently at Tc=J/(kBKc)T_c=J/(k_B K_c) (using ).

Key consequences

  • Second-order transition: the free energy is nonanalytic at TcT_c, yielding a continuous transition with a logarithmic divergence of the specific heat:

    CV(T)AlogTTcas TTc, C_V(T)\sim -A\log|T-T_c| \quad \text{as } T\to T_c,

    where CVC_V is the .

  • Spontaneous magnetization (Yang): below TcT_c the is strictly positive and has the closed form

    m(β)=(1sinh4(2K))1/8for K>Kc, m(\beta)=\bigl(1-\sinh^{-4}(2K)\bigr)^{1/8} \quad \text{for } K>K_c,

    while m(β)=0m(\beta)=0 for KKcK\le K_c.

  • Exact critical data: the solution provides a benchmark for ideas, , and .