Thursday, March 2, 2017

partition function - What are alternative ways to think about transfer matrix as used in Ising model?


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).



Answer



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.



No comments:

Post a Comment

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...