In this post we will develop the transfer matrix method for solving the one dimensional Ising model. Hopefully the post will be pedagogical enough so that beginners can use it as a launching point into their investigations of more interesting systems.

We begin with a one-dimensional grid with $N$ vertices. The state of the $i^{th}$ vertex is denoted $\sigma_i\in \{1,-1\}$. We impose periodic boundary conditions so that $\sigma_{N+1} = \sigma_1$. The Hamiltonian can be written in a symmetric form:

The components of the transfer matrix are defined as

so that the transfer matrix reads

Some of you may now be thinking “why in the world did we decide to do that?” I apologize for the preemptive definition without any form of motivation, but bear with me. All we need to do is look at the partition function for this system to see the utility of the transfer matrix:

Voila. The partition function can be written as the trace of the transfer matrix exponentiated to the power of the number of vertices. Why is recasting the partition function in the language of linear algebra important here?  Well, because we can now use our arsenal of tricks to simplify the trace. If we can diagonalize the transfer matrix, $T = S^{-1}\Lambda S$, then, using the cyclic property of the trace, we can write:

where the $\lambda_i$ are the eigenvalues of the transfer matrix, ordered so that $\lambda_1 \ge \lambda_2$. The eigenvalues are the solutions to the characteristic equation, which for a $2\times 2$ matrix reads:

where $\text{Det}$ is the determinant. The two solutions are

Here’s a visualization of both of them. Clearly, the eigenvalue with the positive sign is greater than the other for all values of the coupling and the magnetic field (Can you prove this?). Hence it  plays the role of $\lambda_1=\lambda_+$. In the thermodynamic limit, we have then that:

I can’t stress how nice it is that we can derive an expression for the partition function here. For the majority of problems, calculating the partition function is not possible, so this example really is a very special case from which we can gain a lot of insight. We can instantly read off thermodynamic quantities by taking derivatives. In particular we find that the mean magnetization is:

Since this drops to 0 when the external field vanishes, we see that there is no spontaneous magnetization in the 1 dimensional Ising model, a fact originally discovered by Ising himself. Unfortunately, Ising then went on to assume that this means that there is no phase transition in the model in any number of dimensions, which is not true. You can read more about the history of how the critical temperature of the phase transition in 2-dimensions was discovered over the course of the three decades that followed Ising’s analysis of the one dimensional model here.