# 1D Ising Model with magnetic field on even sites: Transfer Matrices

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:

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

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 $$T$$.
• 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! Apr 12, 2020 at 22:32