One dimension The thermodynamic limit exists as long as the interaction decay is J_{ij} \sim |i - j|^{-\alpha} with α > 1. • In the case of
ferromagnetic interaction J_{ij} \sim |i - j|^{-\alpha} with 1 J_{ij} \sim |i - j|^{-2}, Fröhlich and Spencer proved that there is phase transition at small enough temperature (in contrast with the hierarchical case). • In the case of interaction J_{ij} \sim |i - j|^{-\alpha} with α > 2 (which includes the case of finite-range interactions), there is no phase transition at any positive temperature (i.e. finite β), since the
free energy is analytic in the thermodynamic parameters.
Ising's exact solution In the nearest neighbor case (with periodic or free boundary conditions) an exact solution is available. The Hamiltonian of the one-dimensional Ising model on a lattice of
L sites with free boundary conditions is H(\sigma) = -J \sum_{i=1,\ldots,L-1} \sigma_i \sigma_{i+1} - h \sum_i \sigma_i, where
J and
h can be any number, since in this simplified case
J is a constant representing the interaction strength between the nearest neighbors and
h is the constant external magnetic field applied to lattice sites. Then the
free energy is f(\beta, h) = -\lim_{L \to \infty} \frac{1}{\beta L} \ln Z(\beta) = -\frac{1}{\beta} \ln\left(e^{\beta J} \cosh \beta h + \sqrt{e^{2\beta J}(\sinh\beta h)^2 + e^{-2\beta J}}\right), and the spin-spin correlation (i.e. the covariance) is \langle\sigma_i \sigma_j\rangle - \langle\sigma_i\rangle \langle\sigma_j\rangle = C(\beta) e^{-c(\beta)|i - j|}, where
C(β) and
c(β) are positive functions for
T > 0. For
T → 0, though, the inverse correlation length
c(β) vanishes.
Proof The proof of this result is a simple computation. If
h = 0, it is very easy to obtain the free energy in the case of free boundary condition, i.e. when H(\sigma) = -J \left(\sigma_1 \sigma_2 + \cdots + \sigma_{L-1} \sigma_L\right). Then the model factorizes under the change of variables \sigma'_j = \sigma_j \sigma_{j-1}, \quad j \ge 2. This gives Z(\beta) = \sum_{\sigma_1,\ldots, \sigma_L} e^{\beta J \sigma_1 \sigma_2} e^{\beta J \sigma_2 \sigma_3} \cdots e^{\beta J \sigma_{L-1} \sigma_L} = 2 \prod_{j=2}^L \sum_{\sigma'_j} e^{\beta J\sigma'_j} = 2 \left[e^{\beta J} + e^{-\beta J}\right]^{L-1}. Therefore, the free energy is f(\beta, 0) = -\frac{1}{\beta} \ln\left[e^{\beta J} + e^{-\beta J}\right]. With the same change of variables \langle\sigma_j\sigma_{j+N}\rangle = \left[\frac{e^{\beta J} - e^{-\beta J}}{e^{\beta J} + e^{-\beta J}}\right]^N, hence it decays exponentially as soon as
T ≠ 0; but for
T = 0, i.e. in the limit β → ∞ there is no decay. If
h ≠ 0 we need the transfer matrix method. For the periodic boundary conditions case is the following. The partition function is Z(\beta) = \sum_{\sigma_1,\ldots,\sigma_L} e^{\beta h \sigma_1} e^{\beta J\sigma_1\sigma_2} e^{\beta h \sigma_2} e^{\beta J\sigma_2\sigma_3} \cdots e^{\beta h \sigma_L} e^{\beta J\sigma_L\sigma_1} = \sum_{\sigma_1,\ldots,\sigma_L} V_{\sigma_1,\sigma_2} V_{\sigma_2,\sigma_3} \cdots V_{\sigma_L,\sigma_1}. The coefficients V_{\sigma, \sigma'} can be seen as the entries of a matrix. There are different possible choices: a convenient one (because the matrix is symmetric) is V_{\sigma, \sigma'} = e^{\frac{\beta h}{2} \sigma} e^{\beta J\sigma\sigma'} e^{\frac{\beta h}{2} \sigma'} or V = \begin{bmatrix} e^{\beta(h+J)} & e^{-\beta J} \\ e^{-\beta J} & e^{-\beta(h-J)} \end{bmatrix}. In matrix formalism Z(\beta) = \operatorname{Tr} \left(V^L\right) = \lambda_1^L + \lambda_2^L = \lambda_1^L \left[1 + \left(\frac{\lambda_2}{\lambda_1}\right)^L\right], where λ1 is the highest eigenvalue of
V, while is the other eigenvalue: \lambda_1 = e^{\beta J} \cosh \beta h + \sqrt{e^{2\beta J} (\cosh \beta h)^2 -2 \sinh 2 \beta J}=e^{\beta J} \cosh \beta h + \sqrt{e^{2\beta J} (\sinh \beta h)^2 +e^{-2\beta J}}, and . This gives the formula of the free energy above. In the thermodynamics limit for the non-interaction case (J = 0), we got Z_N \to (\lambda_1)^N = (2\cosh \beta h)^N, as the answer for the open-boundary Ising model.
Comments The energy of the lowest state is −
JL, when all the spins are the same. For any other configuration, the extra energy is equal to 2
J times the number of sign changes that are encountered when scanning the configuration from left to right. If we designate the number of sign changes in a configuration as
k, the difference in energy from the lowest energy state is 2
k. Since the energy is additive in the number of flips, the probability
p of having a spin-flip at each position is independent. The ratio of the probability of finding a flip to the probability of not finding one is the Boltzmann factor: \frac{p}{1 - p} = e^{-2\beta J}. The problem is reduced to independent biased
coin tosses. This essentially completes the mathematical description. From the description in terms of independent tosses, the statistics of the model for long lines can be understood. The line splits into domains. Each domain is of average length exp(2β). The length of a domain is distributed exponentially, since there is a constant probability at any step of encountering a flip. The domains never become infinite, so a long system is never magnetized. Each step reduces the correlation between a spin and its neighbor by an amount proportional to
p, so the correlations fall off exponentially. \langle S_i S_j \rangle \propto e^{-p|i-j|}. The
partition function is the volume of configurations, each configuration weighted by its Boltzmann weight. Since each configuration is described by the sign-changes, the partition function factorizes: Z = \sum_{\text{configs}} e^{\sum_k S_k} = \prod_k (1 + p ) = (1 + p)^L. The logarithm divided by
L is the free energy density: \beta f = \log(1 + p) = \log\left(1 + \frac{e^{-2\beta J}}{1 + e^{-2\beta J}}\right), which is
analytic away from β = ∞. A sign of a
phase transition is a non-analytic free energy, so the one-dimensional model does not have a phase transition.
One-dimensional solution with transverse field To express the Ising Hamiltonian using a quantum mechanical description of spins, we replace the spin variables with their respective
Pauli matrices. However, depending on the direction of the magnetic field, we can create a transverse-field or longitudinal-field Hamiltonian. The
transverse-field Hamiltonian is given by H(\sigma) = -J \sum_{i=1,\ldots,L} \sigma_i^z \sigma_{i+1}^z - h \sum_i \sigma_i^x. The transverse-field model experiences a phase transition between an ordered and disordered regime at
J ~
h. This can be shown by a mapping of Pauli matrices \sigma_n^z = \prod_{i=1}^n T_i^x, \sigma_n^x = T_n^z T_{n+1}^z. Upon rewriting the Hamiltonian in terms of this change-of-basis matrices, we obtain H(\sigma) = -h \sum_{i=1,\ldots,L} T_i^z T_{i+1}^z - J \sum_i T_i^x. Since the roles of
h and
J are switched, the Hamiltonian undergoes a transition at
J =
h.
Renormalization When there is no external field, we can derive a functional equation that f(\beta, 0) = f(\beta) satisfies using renormalization. Specifically, let Z_N(\beta, J) be the partition function with N sites. Now we have:Z_N(\beta, J) = \sum_{\sigma} e^{K \sigma_2(\sigma_1 + \sigma_3)}e^{K \sigma_4(\sigma_3 + \sigma_5)}\cdotswhere K := \beta J. We sum over each of \sigma_2, \sigma_4, \cdots, to obtainZ_N(\beta, J) = \sum_{\sigma} (2\cosh(K(\sigma_1 + \sigma_3))) \cdot (2\cosh(K(\sigma_3 + \sigma_5))) \cdotsNow, since the cosh function is even, we can solve Ae^{K'\sigma_1\sigma_3} = 2\cosh(K(\sigma_1+\sigma_3)) as A = 2\sqrt{\cosh(2K)}, K' = \frac 12 \ln\cosh(2K). Now we have a self-similarity relation:\frac 1N \ln Z_N(K) = \frac 12 \ln\left(2\sqrt{\cosh(2K)}\right) + \frac 12 \frac{1}{N/2} \ln Z_{N/2}(K')Taking the limit, we obtainf(\beta) = \frac 12 \ln\left(2\sqrt{\cosh(2K)}\right) + \frac 12 f(\beta')where \beta' J = \frac 12 \ln\cosh(2\beta J). When \beta is small, we have f(\beta)\approx \ln 2, so we can numerically evaluate f(\beta) by iterating the functional equation until K is small.
Two dimensions In the ferromagnetic case there is a phase transition. At low temperature, the
Peierls argument proves positive magnetization for the nearest neighbor case and then, by the
Griffiths inequality, also when longer range interactions are added. Meanwhile, at high temperature, the
cluster expansion gives analyticity of the thermodynamic functions. In the nearest-neighbor case, the free energy was exactly computed by Onsager. The spin-spin correlation functions were computed by McCoy and Wu.
Onsager's exact solution obtained the following analytical expression for the free energy of the Ising model on the anisotropic square lattice when the magnetic field h=0 in the thermodynamic limit as a function of temperature and the horizontal and vertical interaction energies J_1 and J_2, respectively -\beta f = \ln 2 + \frac{1}{8\pi^2}\int_0^{2\pi}d\theta_1\int_0^{2\pi}d\theta_2 \ln[\cosh(2\beta J_1)\cosh(2\beta J_2) -\sinh(2\beta J_1)\cos(\theta_1)-\sinh(2\beta J_2)\cos(\theta_2)]. From this expression for the free energy, all thermodynamic functions of the model can be calculated by using an appropriate derivative. The 2D Ising model was the first model to exhibit a continuous phase transition at a positive temperature. It occurs at the temperature T_c which solves the equation \sinh\left(\frac{2J_1}{kT_c}\right)\sinh\left(\frac{2J_2}{kT_c}\right) = 1. In the isotropic case when the horizontal and vertical interaction energies are equal J_1=J_2=J, the critical temperature T_c occurs at the following point T_c = \frac{2J}{k\ln(1+\sqrt{2})} = (2.269185\cdots)\frac{J}{k} When the interaction energies J_1, J_2 are both negative, the Ising model becomes an antiferromagnet. Since the square lattice is bi-partite, it is invariant under this change when the magnetic field h=0, so the free energy and critical temperature are the same for the antiferromagnetic case. For the triangular lattice, which is not bi-partite, the ferromagnetic and antiferromagnetic Ising model behave notably differently. Specifically, around a triangle, it is impossible to make all 3 spin-pairs antiparallel, so the antiferromagnetic Ising model cannot reach the minimal energy state. This is an example of
geometric frustration.
Onsager's formula for spontaneous magnetization Onsager famously announced the following expression for the
spontaneous magnetization M of a two-dimensional Ising ferromagnet on the square lattice at two different conferences in 1948, though without proof''' Such a solution has not been found until now, although there is no proof that it may not exist. In three dimensions, the Ising model was shown to have a representation in terms of non-interacting fermionic strings by
Alexander Polyakov and
Vladimir Dotsenko. This construction has been carried on the lattice, and the
continuum limit, conjecturally describing the critical point, is unknown. In three as in two dimensions, Peierls' argument shows that there is a phase transition. This phase transition is rigorously known to be continuous (in the sense that correlation length diverges and the magnetization goes to zero), and is called the
critical point. It is believed that the critical point can be described by a renormalization group fixed point of the Wilson-Kadanoff renormalization group transformation. It is also believed that the phase transition can be described by a three-dimensional unitary conformal field theory, as evidenced by
Monte Carlo simulations, exact diagonalization results in quantum models, and quantum field theoretical arguments. Although it is an open problem to establish rigorously the renormalization group picture or the conformal field theory picture, theoretical physicists have used these two methods to compute the
critical exponents of the phase transition, which agree with the experiments and with the Monte Carlo simulations. This conformal field theory describing the three-dimensional Ising critical point is under active investigation using the method of the
conformal bootstrap. This method currently yields the most precise information about the structure of the critical theory (see
Ising critical exponents). In 2000,
Sorin Istrail of
Sandia National Laboratories proved that the spin glass Ising model on a
nonplanar lattice is
NP-complete. That is, assuming
P ≠
NP, the general spin glass Ising model is exactly solvable only in
planar cases, so solutions for dimensions higher than two are also intractable. Istrail's result only concerns the spin glass model with spatially varying couplings, and tells nothing about Ising's original ferromagnetic model with equal couplings.
Four dimensions and above In any dimension, the Ising model can be productively described by a locally varying
mean field. The field is defined as the average spin value over a large region, but not so large so as to include the entire system. The field still has slow variations from point to point, as the averaging volume moves. These fluctuations in the field are described by a continuum field theory in the infinite system limit. The accuracy of this approximation improves as the dimension becomes larger. A deeper understanding of how the Ising model behaves, going beyond mean-field approximations, can be achieved using
renormalization group methods. ==See also==