# Spin 1 Ising model

## Homework Statement:

Consider the 1-domensional Ising model given by the Hamiltonian

$$H = -J \sum_{i=1}^{N} s_{i} s_{i+1}$$

where $s_{i} = -1, 0, 1$

Calculate the partition function $Z$ and the phase transition temperature.

## Relevant Equations:

$$Z = Tr(T^{n})$$
I did the first part using the transfer matrix method:

$$Z = Tr(T^{N})$$

In this case, the transfer matrix will be

$$T(i,i') = \begin{pmatrix} e^{\beta J} & 1 & e^{-\beta J}\\ 1 &1 &1 \\ e^{-\beta J} & 1 & e^{\beta J} \end{pmatrix}$$

To get the trace of $T^N$, you find the eigenvalues of T:

$$\lambda_{1} = 2 sinh(\beta J)$$
$$\lambda_{2,3} = \frac{1}{2} (2cosh(\beta J) + 1 +- \sqrt{ ( 2cosh^2(\beta J) + 2sinh^2(\beta J) )^2 - cosh(\beta J) + e^{-2\beta J} +9})$$

With this, the partition function is:

$$Z = \lambda_{1}^{N}+\lambda_{2}^N + \lambda_{3}^N$$

Now, to find the transition temperature, I'd have to find the zeros of the partition function or the singular points of the Helmholtz free energy, but I'm not quite sure on how to proceed from here. Any ideas? Thanks in advance.

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hilbert2
Gold Member
If you can write a closed form expression for the partition function ##Z(T)## of a spin chain of ##N## spins, then the internal energy at temperature ##T## is, as described on this Wiki page:

##\displaystyle\langle E\rangle = -\frac{\partial\ln Z}{\partial\beta}##,

where ##\displaystyle\beta = \frac{1}{kT}## with ##k## the Boltzmann constant.

Then, when you know the internal energy as a function of temperature, you can divide it by ##N##, the number of spins, and take the limit ##N\rightarrow\infty## to find the internal energy per spin for the infinite chain of spins. If there is a phase transition at some temperature ##T_c##, then this internal energy per spin has a sudden "jump" at that temperature, or equivalently, the heat capacity ##C = d\langle E\rangle /dT## is infinite at ##T_c##.

In the case of finite but large ##N##, the heat capacity has a sharp peak at the approximate transition temperature, but not a real singular point.

Thales Castro
First of all, your Hamiltonian is wrong. It should be$$\mathcal{H}=J\sum_{i=1}^N \sigma_{i} \sigma_{i+1} - H\sum_{i=1}^N \sigma_i$$
where ##\sigma_i## and ##\sigma_{i+1} = \pm 1##. The transfer matrix is,$$\mathbf {T}=\begin{pmatrix} T(+,+) & T(+,-) \\ T(-,+) & T(-,-) \end{pmatrix}= \begin{pmatrix} exp(K+L) & exp(-K) \\ (exp(-K) & exp(K-L) \end{pmatrix}$$
where ##K=\beta J## and ##L=\beta H##. This should get you started. Also, the 1-d Ising model does't exhibit a phase transition because the spin correlations decay too rapidly as temperature goes to zero.

hilbert2
Gold Member
First of all, your Hamiltonian is wrong. It should be$$\mathcal{H}=J\sum_{i=1}^N \sigma_{i} \sigma_{i+1} - H\sum_{i=1}^N \sigma_i$$
where ##\sigma_i## and ##\sigma_{i+1} = \pm 1##. The transfer matrix is,$$\mathbf {T}=\begin{pmatrix} T(+,+) & T(+,-) \\ T(-,+) & T(-,-) \end{pmatrix}= \begin{pmatrix} exp(K+L) & exp(-K) \\ (exp(-K) & exp(K-L) \end{pmatrix}$$
where ##K=\beta J## and ##L=\beta H##. This should get you started. Also, the 1-d Ising model does't exhibit a phase transition because the spin correlations decay too rapidly as temperature goes to zero.
The second term in the Hamiltonian is only needed if there's an external magnetic field. I wasn't sure about whether the lack of phase transition in 1D is only in the spin-1/2 system, or for any total spin. Even in the 1D spin-half system the heat capacity has a peak at some temperature, but it doesn't become a singular 'spike' in the thermodynamic limit, like in real phase transition.

My apologies. I misinterpreted your question. I thought it was an homework assignment. Ising's idea was to model a 1-d ferromagnetic lattice. The purpose of the second term in the hamiltonian is to lift the degeneracy of the spin state of the valence electron. Without the second term in the hamiltonian the eigenvalue of the transfer matrix becomes zero. That means that there is no exchange of magnetic energy possible between adjacent sites in the lattice.
You propose that each lattice site have total spin = 1, but that implies that you have one spin up electron and one spin down electron per site (Pauli exclusion principle) . This would cancel the spin contribution to magnetic field per site and result in zero ferromagnetism.

hilbert2