# Ising model on a cycle

The Ising model on $\mathbb{Z} / 2d\mathbb{Z}$ gives to the configuration $x=(x_0, \ldots, x_{2d-1}) \in \{-1,+1\}^{2d}$ a probability proportional to $\exp\\big(\beta \sum_i x_ix_{i+1} \\big)$. The Gibbs sampler with block updates is a Markov chain $X_k$ that evolves on the set of such configurations and updates the odd (resp. even) indices conditionally on the even (resp. odd) indices with probability a half.

It seems like a relatively straightforward application of the path coupling  approach (two configurations are neighbours if they agree on all odd or all even coordinates; distance between two neighbours is $1+H(x,y)/d$ where $H$ is the Hamming distance) shows that the mixing time of the Gibbs sampler stays bounded as the size $d$ of the system goes to infinity, which looks rather surprising. Any intuition behind that? If this is already written somewhere, any reference concerning this (or similar) result?

•  Chapter 14 of Markov Chains and Mixing Times by D. Levin, Y. Peres and E. Wilmer
• I guess you mean $\mathbb{Z}/2d\mathbb{Z}$ and $(x_0,\dots, x_{2d-1})\in \{-1, +1\}^{2d}$? Also $d$ is probably more naturally "size" than "dimension" here, since everything is really 1-dimensional. Anyway, are you sure you are asking the question you mean? Maybe I misunderstand, but for most sensible definitions the mixing time for, say $d=1$ (a system with only 2 sites) will not be the same as that for $d=1000$. Often in these kind of problems the mixing time is considered in an asymptotic regime as some parameter gets large - but that seems to be absent here. – James Martin Oct 19 '12 at 14:58
• Thank you for the comment, I have updated the notations. As you said, I really meant "size" instead of "dimension". And I should not have written that the mixing time does not depend on d; what I really meant is that the mixing time $\tau(d)$ seems to stay bounded as $d$ goes to infinity. – Alekk Oct 19 '12 at 15:26

Consider first why the mixing time might be large. If $\beta$ is very high, and we start with a configuration where half the circle is + and half -, it will take a fairly long time for the chain to converge to one of the extreme states. (Roughly $d^2$, as the interface will perform a random walk.)
However, if $\beta$ is fixed and $d$ is large, then at every step the process will create islands of the opposite sign, at distance of order $e^{4\beta}$, regardless of $d$. Notethat the stationary distribution also has a finite correlation length.
Finally, another way to see the bounded mixing time is by a coupling argument. The simplest local coupling, will create agreement with some density, and segments of agreement will grow at positive rate, so after roughly at most $e^{4\beta}$ steps any two starting configurations will couple. This bound can be improved.
• Thank, that's a great answer. I was trying the other day to see how long it would take to couple an all +1 configuration with an all -1 configuration, and one can see that the coupling time stays bounded wrt $d$. – Alekk Oct 21 '12 at 10:39