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=== 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.
* 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 ====
When the interaction energies <math>J_1</math>, <math>J_2</math> 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 <math>h=0</math>, 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]].
===== Transfer matrix =====
Start with an analogy with quantum mechanics. The Ising model on a long periodic lattice has a partition function
<math display="block">\sum_{\{S\}} \exp\biggl(\sum_{ij} S_{i,j} \left( S_{i,j+1} + S_{i+1,j} \right)\biggr).</math>
Think of the ''i'' direction as ''space'', and the ''j'' direction as ''time''. This is an independent sum over all the values that the spins can take at each time slice. This is a type of [[path integral formulation|path integral]], it is the sum over all spin histories.
A path integral can be rewritten as a Hamiltonian evolution. The Hamiltonian steps through time by performing a unitary rotation between time ''t'' and time ''t'' + Δ''t'':
<math display="block"> U = e^{i H \Delta t}</math>
The product of the U matrices, one after the other, is the total time evolution operator, which is the path integral we started with.
<math display="block"> U^N = (e^{i H \Delta t})^N = \int DX e^{iL}</math>
where ''N'' is the number of time slices. The sum over all paths is given by a product of matrices, each matrix element is the transition probability from one slice to the next.
Similarly, one can divide the sum over all partition function configurations into slices, where each slice is the one-dimensional configuration at time 1. This defines the [[Transfer-matrix method (statistical mechanics)|transfer matrix]]:
<math display="block">T_{C_1 C_2}.</math>
The configuration in each slice is a one-dimensional collection of spins. At each time slice, ''T'' has matrix elements between two configurations of spins, one in the immediate future and one in the immediate past. These two configurations are ''C''<sub>1</sub> and ''C''<sub>2</sub>, and they are all one-dimensional spin configurations. We can think of the vector space that ''T'' acts on as all complex linear combinations of these. Using quantum mechanical notation:
<math display="block">|A\rangle = \sum_S A(S) |S\rangle</math>
where each basis vector <math>|S\rangle</math> is a spin configuration of a one-dimensional Ising model.
Like the Hamiltonian, the transfer matrix acts on all linear combinations of states. The partition function is a matrix function of T, which is defined by the [[Trace (linear algebra)|sum]] over all histories which come back to the original configuration after ''N'' steps:
<math display="block">Z= \mathrm{tr}(T^N).</math>
Since this is a matrix equation, it can be evaluated in any basis. So if we can diagonalize the matrix ''T'', we can find ''Z''.
===== ''T'' in terms of Pauli matrices =====
The contribution to the partition function for each past/future pair of configurations on a slice is the sum of two terms. There is the number of spin flips in the past slice and there is the number of spin flips between the past and future slice. Define an operator on configurations which flips the spin at site i:
<math display="block">\sigma^x_i.</math>
In the usual Ising basis, acting on any linear combination of past configurations, it produces the same linear combination but with the spin at position i of each basis vector flipped.
Define a second operator which multiplies the basis vector by +1 and −1 according to the spin at position ''i'':
<math display="block">\sigma^z_i.</math>
''T'' can be written in terms of these:
<math display="block">\sum_i A \sigma^x_i + B \sigma^z_i \sigma^z_{i+1}</math>
where ''A'' and ''B'' are constants which are to be determined so as to reproduce the partition function. The interpretation is that the statistical configuration at this slice contributes according to both the number of spin flips in the slice, and whether or not the spin at position ''i'' has flipped.
===== Spin flip creation and annihilation operators =====
Just as in the one-dimensional case, we will shift attention from the spins to the spin-flips. The σ<sup>''z''</sup> term in ''T'' counts the number of spin flips, which we can write in terms of spin-flip creation and annihilation operators:
<math display="block"> \sum C \psi^\dagger_i \psi_i. \,</math>
The first term flips a spin, so depending on the basis state it either:
#moves a spin-flip one unit to the right
#moves a spin-flip one unit to the left
#produces two spin-flips on neighboring sites
#destroys two spin-flips on neighboring sites.
Writing this out in terms of creation and annihilation operators:
<math display="block"> \sigma^x_i = D {\psi^\dagger}_i \psi_{i+1} + D^* {\psi^\dagger}_i \psi_{i-1} + C\psi_i \psi_{i+1} + C^* {\psi^\dagger}_i {\psi^\dagger}_{i+1}.</math>
Ignore the constant coefficients, and focus attention on the form. They are all quadratic. Since the coefficients are constant, this means that the ''T'' matrix can be diagonalized by Fourier transforms.
Carrying out the diagonalization produces the Onsager free energy.
===== Onsager's formula for spontaneous magnetization =====
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