Ising Model

The goal of this section is to introduce the Ising model which is a simple model of systems with interparticle interactions and to compute its canonical partition function according to both the macroscopic approximation and the rigorous transfer matrix technique.

The 1-dimensional Ising model is described by a system of N spins arranged in a ring as represented in the following diagram:

The Hamiltonian of the system is

$$H = -\bar{\mu} B \sum_j S_j - J \sum_{jk} S_j S_k,$$ (285)

where $\bar{\mu}$ is the magnetic dipole moment, $B$ is an external magnetic field and $J$ is the coupling constant between spins. The sum of products $S_j S_k$ defines the interaction between spins, including only nearest neighbors. In the absence of an external magnetic field, the canonical partition function of the system is

$$Z=\sum_{S_1= -1}^1 \sum_{S_2= -1}^1 ... \sum_{S_N= - 1}^1 e^{\beta J S_1 S_2} e^{\beta J S_2 S_3} ... e^{\beta J S_N S_1},$$ (286)

The partition function, introduced by Eq. (286), is approximately equal to

$$Z \approx \sum_{b_1= \pm 1} ... \sum_{b_N= \pm 1} e^{\beta J \sum_{j=1}^N b_j} = [2 (\beta J) ]^N,$$ (287)

where we have replaced the products of interaction $S_k S_j$ by bonds $b_j = \pm 1$ and we have assumed that all bonds are linearly independent. Note, however, that such approximation consist in assuming that N is sufficiently large (i.e., $N >> 1$) as to neglect the energy of one bond relative to the total energy of the system, since only N-1 bonds are actually linearly independent.

In order to perform a rigorous calculation of the canonical partition function introduced by Eq. (286), we define the transfer function in the absence of an external magnetic field as follows,

$$T(S_i, S_{i+1}) \equiv (\beta J S_1 S_2).$$ (288)

Substituting Eq. (288) into Eq. (286) we obtain

$$Z = \sum_{S_1= -1}^1 \sum_{S_2= -1}^1 ... \sum_{S_N= - 1}^1 T(S_1, S_2) T(S_2, S_3) ... T(S_N, S_1).$$ (289)

This expression corresponds to the trace of a product of N identical $2 \times 2$ matrices. In order to show this we introduce the transfer matrix,

$${\bf T} \equiv \begin{pmatrix}T(1,1) & T(1,-1) \\ T(-1,1) & T(-1,-1)\\ \end{pmatrix}.$$ (290)

Note that the element (j,k) of ${\bf T}^2$ is

$${\bf T}^2(j,k) = \sum_{S_2= -1}^1 T(j, S_2) T(S_2, k),$$ (291)

and therefore

$$Z = \sum_{S_1 = -1}^1 {\bf T}^N(S_1, S_1) = \{ {\bf T}^N \}.$$ (292)

Thus the calculation of the canonical partition function for the 1-dimensional Ising model has been reduced to that of computing the trace of the $N$th power of the transfer matrix. Now, the trace of a matrix is the sum of its eigenvalues and the eigenvalues of ${\bf T}^N$ are $\lambda_{\pm}^N$, where $\lambda_{\pm}$ are the eigenvalues of ${\bf T}$ determined by the equation

$$\Bigg \vert \begin{matrix}e^{\beta J}-\lambda & e^{- \beta J} \\ e^{-\beta J} & e^{\beta J} - \lambda\\ \end{matrix} \Bigg \vert = 0,$$ (293)

with solutions

$$\lambda_{\pm} = e^{\beta J} \pm e^{- \beta J}.$$ (294)

Hence, the partition function is simply,

$$Z = \lambda_+^N +\lambda_-^N = 2^N ( ^N(\beta J)+ ^N(\beta J)).$$ (295)

Note that when N is sufficiently large, sinh$^N(\beta J) <<$   cosh$^N(\beta J)$ and Eq. (295) coincides with Eq. (287). In the presence of a magnetic field, however,

$$Z=\sum_{S_1= -1}^1 \sum_{S_2= -1}^1 ... \sum_{S_N= - 1}^1 e^{\bet......3 + \beta \mu B (S_2+S_3)/2} ... e^{\beta J S_N S_1 + \beta \mu B (S_N+S_1)/2},$$ (296)

Exercise: 1-dimensional Ising Model

Compute the canonical partition function introduced by Eq. (296) by implementing the transfer matrix approach.