9
$\begingroup$

I recently learned about how to find the partition function of Ising model using Transfer Matrix method. At my level of understanding things, it is a trick that happens to work! I would like to understand Transfer Matrices more deeply than that.

So I am sort of looking for things that would sound like "an equivalent formulation" or "an axiomatic treatment" or "an analogy with this technique in another field (say QFT)". I am not necessarily asking for something sophisticated. Even simple things or ideas that would motivate the ideas behind transfer matrices would be very useful! (Roughly I am looking for some kind of bigger context, where Transfer Matrices have a natural place).

$\endgroup$
7
+200
$\begingroup$

I don't think that there is a direct, natural physical interpretation (one can of course always cook up something ex post facto). There are however close relations with other topics. Here, I'll try to explain some close links with Markov chains.

I'll stick to the case of the one-dimensional Ising model, to keep things concrete, but it should be clear from the following that this holds much more generally.

Let $T$ be the transfer matrix of the one-dimensional Ising model, namely $$ T = \begin{pmatrix} e^{\beta + h} & e^{-\beta}\\ e^{-\beta} & e^{\beta-h} \end{pmatrix} = \bigl(T_{\sigma,\sigma'} \bigr)_{\sigma,\sigma'=\pm 1}\,, $$ with $T_{\sigma,\sigma'} = e^{\beta\sigma\sigma' + h(\sigma+\sigma')/2}$. (I use the mathematicians' convention of not multiplying $h$ by $\beta$, but this is of course irrelevant.)

Then one has, for example, for the partition in a system of length $N$ with boundary condition $\sigma$ (on the left-hand side) and $\sigma'$ (on the right-hand side): $$ \mathbf{Z}_N^{\sigma,\sigma'} = \bigl( T^N \bigr)_{\sigma,\sigma'}\,. $$ Let us denote by $\lambda_1>\lambda_2>0$ the two eigenvalues of $T$ and $\varphi^1,\varphi^2$ the corresponding eigenvectors, normalized so that $\|\varphi^1\|_2=\|\varphi^2\|_2=1$. All these quantities can easily be computed explicitly, but the resulting expressions are irrelevant for what I want to say.

Then, we can define a new matrix $\Pi=(\pi(\sigma,\sigma'))_{\sigma,\sigma'=\pm 1}$ with matrix elements $$ \pi(\sigma,\sigma') = \tfrac{\varphi^1_{\sigma'}}{\lambda_1\varphi^1_\sigma} T_{\sigma,\sigma'}\,. $$ (Note that, by the Perron-Frobenius theorem, all components of $\varphi^1$ are positive.) It is easy to check that $\Pi$ is the transition matrix of an irreducible, aperiodic Markov chain: for $\sigma=\pm 1$, $$ \sum_{\sigma'=\pm 1} \pi(\sigma,\sigma') = \tfrac{1}{\lambda_1\varphi^1_\sigma} \sum_{\sigma'=\pm 1} T_{\sigma,\sigma'} \varphi^1_{\sigma'} = \tfrac{1}{\lambda_1\varphi^1_\sigma} \bigl( T\varphi^1 \bigr)_{\sigma} = 1\,, $$ since, by definition, $T\varphi^1=\lambda_1\varphi^1$. Being irreducible, $\Pi$ possesses a unique invariant probability measure $\mu$: for $\sigma=\pm 1$, $$ \mu(\sigma) = (\varphi^1_\sigma)^2\,. $$ Indeed, $\mu(1)+\mu(-1) = \|\varphi^1\|_2^2=1$ and $$ \bigl( \mu\Pi \bigr)(\sigma') = \sum_{\sigma=\pm 1} \mu(\sigma)\,\pi(\sigma,\sigma') = \frac{1}{\lambda_1} \varphi^1_{\sigma'} \sum_{\sigma=\pm 1} \varphi^1_\sigma\,T_{\sigma,\sigma'} = (\varphi^1_{\sigma'})^2 = \mu(\sigma')\,, $$ since the matrix $T$ is symmetric. The measure $\mu$ describes the one-spin marginal of the infinite-volume Gibbs measure. Indeed, denoting the Gibbs measure on the interval $\{-N,\ldots,N\}$ with boundary condition $\sigma$ (one the left) and $\sigma'$ (on the right) by $\nu_N^{\sigma,\sigma'}$, the probability that the spin at $0$ takes the value $\sigma_0$ is given by $$ \nu_N^{\sigma,\sigma'}(\sigma_0) = \frac{\mathbf{Z}_N^{\sigma,\sigma_0}\mathbf{Z}_N^{\sigma_0,\sigma'}}{\mathbf{Z} _{2N}^{\sigma,\sigma'}} = \frac{\bigl(T^{N}\bigr)_{\sigma,\sigma_0} \bigl(T^{N}\bigr)_{\sigma_0,\sigma'}}{\bigl(T^{2N}\bigr)_{\sigma,\sigma'}}\,. $$ Now, for any $\sigma_1,\sigma_2=\pm 1$, $$ \bigl( T^N \bigr)_{\sigma_1,\sigma_2} = \lambda_1^N \frac{\varphi^1_{\sigma_1}}{\varphi^1_{\sigma_2}} \bigl( \Pi^N \bigr)_{\sigma_1,\sigma_2}\,, $$ which gives, after substitution in the above expression, $$ \nu_N^{\sigma,\sigma'}(\sigma_0) = \frac{\bigl(\Pi^{N}\bigr)_{\sigma,\sigma_0} \bigl(\Pi^{N}\bigr)_{\sigma_0,\sigma'}}{\bigl(\Pi^{2N}\bigr)_{\sigma,\sigma'}} \,. $$ Now, since the Markov chain is irreducible and aperiodic, $\lim_{N\to\infty} (\Pi^N)_{\sigma_1,\sigma_2} = \mu(\sigma_2)$ and we conclude that $$ \lim_{N\to\infty} \nu_N^{\sigma,\sigma'}(\sigma_0) = \frac{\mu(\sigma_0)\mu(\sigma')}{\mu(\sigma')} = \mu(\sigma_0)\,. $$ Similarly, the full infinite-volume Gibbs measure is given by the invariant path-measure of the Markov chain.

In more general situations, the transfer matrix might not be symmetric. This is not a problem, but it makes the definitions above slightly more complicated. Let me refer to this paper for an example in which these methods are used in a more complicated settings. I can't provide general references, as I don't know them; this is certainly part of the folklore now.

$\endgroup$
  • $\begingroup$ This is an excellent answer and taught me something I needed to know. I'd really like to know more about the general case, in which the transfer matrix might not be symmetric. $\endgroup$ – Nathaniel Jan 18 '16 at 7:44
  • $\begingroup$ @Nathaniel : thanks for your appreciation (and the unexpected bounty). Concerning the nonsymmetric case, I might either add a (short) second part to the answer, in which I point out the relevant changes, or simply point them out in the comments. What would be the best solution? $\endgroup$ – Yvan Velenik Jan 19 '16 at 10:50
  • $\begingroup$ Oh, either's fine! I can probably work through it myself, but if you have time to put some information in the answer it might be helpful. (I just figured out that if you use the convention that stochastic matrices left-multiply probability vectors then you have to use the left Perron-Frobenius eigenvector of $T$, which is what was holding me up.) $\endgroup$ – Nathaniel Jan 19 '16 at 11:03
  • $\begingroup$ @Nathaniel : yes, that's exactly the point. If you denote by $\varphi^{*,1}$ the left Perron-Frobenius eigenvector (associated, of course, to the same eigenvalue $\lambda_1$), then the only thing that changes is the expression for the invariant measure, which becomes $\mu( \sigma) = \varphi^{*,1}_\sigma \varphi^1_\sigma$. (Note also that you could define another Markov chain with transition probabilities $\pi^*(\sigma,\sigma') = \frac{\varphi^{*,1}_{\sigma'}}{\lambda_1\varphi^{*,1}_\sigma} T_{\sigma',\sigma}$; this corresponds to the time-reversed chain (which has the same invariant measure). $\endgroup$ – Yvan Velenik Jan 19 '16 at 12:26
  • $\begingroup$ For those interested in more information, in a more general framework, I'd recommend to have a look at Chapter 11 of this book. $\endgroup$ – Yvan Velenik Feb 6 '17 at 10:33

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.