0
$\begingroup$

I am calculating the partition function of an Ising model with spin 1 ($\sigma_i \in \{-1,0,1\}$) for $n$ sites. The following Hamiltonian has been used:

$$H = -J \sum_{i=1}^{n} \sigma_i\sigma_{i+1},$$

assuming periodic boundary conditions, i.e. $\sigma_1 = \sigma_{n+1}$. From this I derived that $Z = \mathrm{tr}(P^n)$ where $P$ is the transfer matrix as follows:

\begin{equation} P = \begin{pmatrix} e^K & 1& e^{-K} \\ 1&1&1\\ e^{-K} & 1& e^K \end{pmatrix} \end{equation} in the basis $\{|1\rangle,|0\rangle,|-1\rangle\}$ where $K := -\beta J$.

I know that $Z \approx \lambda^N$ for the greatest eigenvalue of $P$, however I cannot find the eigenvalues. I ended up with this characteristic equation that I cannot solve:

$$(e^K-\lambda)^2(1-\lambda) - 2(e^{K}-\lambda) + 2e^{-K}(1-\lambda)=0.$$

Could someone clarify how to proceed further?

$\endgroup$
  • $\begingroup$ Have you tried using Mathematica to solve your cubic equation? $\endgroup$ – mike stone Sep 5 '19 at 15:41
  • $\begingroup$ No I am doing self-study on this topic and found this file: ocw.mit.edu/courses/physics/…. There it is stated that it should be done by symmetry considerations $\endgroup$ – Mathphys meister Sep 5 '19 at 15:48
  • 2
    $\begingroup$ Well one eigenvector is clearly $(1,0,-1)^T$ with eigenvalue $2\sinh K$ $\endgroup$ – mike stone Sep 5 '19 at 15:57
  • $\begingroup$ When the matrix is symmetric and has lots of "kinda the same" elements, it's often a good idea to try various vectors made up of $1$, $-1$, and $0$. @mikestone gave you one eigenvector in that style, and it's easy to find another one in that style right away. $\endgroup$ – Lagerbaer Sep 5 '19 at 16:39
  • $\begingroup$ I get some different terms in the char. equation. Why don't you have an $e^{-2K}$ term? I recommend you redo your equation without looking at your previous attempt. $\endgroup$ – Bill N Sep 5 '19 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.