I have been trying to work out a practice problem after reading about Transfer Matrices method for solving 1D Ising Model. Please, if you are able to, tell me whether the way I introduced transfer matrices and dealt with matrix multiplication is correct, I do not care much about whether I made any algebraic errors but rather if I got the concept of transfer matrices correctly.

Consider a modified 1D Ising Model with N sites, where N is odd, and magnetic field applied on every even site of strength $2\cdot h$. The interaction between neighbours has strength $J$. The Hamiltonian is given by:

$ H = -J \sum_{j=1}^{N-1} \sigma_j \cdot \sigma_{j+1} + 2\cdot h \sum_{j=1}^{\frac{N-1}{2}}\sigma_{2j}$

The objective is first to write a solution in a form:

$ \vec{v}^T \hat{T}^{\frac{N-1}{2}}\vec{u}$

Where $\vec{u}, \vec{v}$ are two dimensional vectors and $T$ is appropriately constructed transfer matrix.

My attempt:

Consider the energy when $N=3$:

$E = -J\sigma_1\sigma_2+h\sigma_2 -J\sigma_2\sigma_3 + h\sigma_2$

My idea was to write:

$E = E'(\sigma_1, \sigma_2) + E^*(\sigma_2,\sigma_3)$

Using this notation, we can write the partition function as:

$Z = \sum_{{\sigma_j}} e^{-\beta\cdot E'(\sigma_1, \sigma_2)}e^{-\beta\cdot E^*(\sigma_2, \sigma_3)}...e^{-\beta\cdot E^*(\sigma_{N-1}, \sigma_N)} $

Now the part where my mistake could be:

I introduce:

$T' = e^{-\beta\cdot E'(\sigma, \sigma')}$

$T^*= e^{-\beta\cdot E^*(\sigma, \sigma')}$

As $\sigma$s are independent this can be viewed as consecutive matrix multiplication of $T'$ and $T^*$. By introducing:

$ T(\sigma, \sigma') = \sum_{\sigma^*}T'(\sigma, \sigma^*)T^*(\sigma^*, \sigma')$

which can be computed to be:

$T = 2 \begin{bmatrix} cosh(2\beta(J-h)) & cosh(2\beta\cdot h) \\ cosh(2\beta\cdot h) & cosh(2\beta(J+h)) \end{bmatrix}$

Here the first matrix field corresponds to both spin up.

Using this matrix, I suppose that partition function could be written as:

$ Z = \sum_{\sigma_1}\sum_{\sigma_N} T^{\frac{N-1}{2}}(\sigma_1, \sigma_N)$

I understand this expression as sum of all the elements so I concluded that $\vec{u}^T=\vec{v}^T= (1,1)$.

The problem arises when I try to compute $Z$ when $h=0$. I get the following expression:

$Z = 2^{\frac{N+1}{2}} cosh^{\frac{N-1}{2}}(2\beta J)$

However, the well known expression should be:

$Z = 2(2 cosh(\beta J))^{N-1}$


1 Answer 1

  1. Yes, you understand the concept of the transfer matrix correctly.

  2. The way you get from $T$ to $Z$ is wrong. You can easily check this for $N=3$: In that case, $Z$ should be the sum of all entries of $T$ -- which it isn't, and if you check that sum instead, you will notice it is identical to the "well-known expression" you quote. For larger $N$, you will have to diagonalize $T$.

  • $\begingroup$ Thanks. This was indeed very stupid from me. I actually realised that what I did wrong is saying that cosh(0) = 0 where it should be 1. Probably lack of confidence when dealing with a completely new topic played its role here. Still, I believe that the sum of elements can be computed using: \begin{equation} \begin{bmatrix} 1 & 1 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = a+b+c+d \end{equation} $\endgroup$
    – Light
    Apr 12, 2020 at 22:13
  • $\begingroup$ Of course that is the sum of the elements - which is the correct solution for $N=3$. What you moreover claim is a general formula, and to get $(1,1)*T^K*(1,1)^T$ you need the eigenvalues and -vectors. --- BTW, I erased that comment, but: What I did is was to check the h=0 case along your calculation, and you should have done the same - set h=0 right from the beginning and follow your argument! $\endgroup$ Apr 12, 2020 at 22:32
  • $\begingroup$ Right! I easily got the desired result after correcting this error. Many thanks. $\endgroup$
    – Light
    Apr 12, 2020 at 22:42

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