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Consider the Ferromagnetic Ising Model ($J>0$) on the lattice $\mathbb{Z}^2$ with the Hamiltonian with boundary condition $\omega\in\{-1,1\}$ formally given by $$ H^{\omega}_{\Lambda}(\sigma)=-J\sum_{<i,j>}{\sigma_i\sigma_j} - \sum_{i\in\Lambda} {h_i\sigma_i}, $$ where the first sum is over all unordered pairs of first neighbors in $\Lambda\cup\partial \Lambda$.

Suppose that $h_i=h>0$ if $i\in\Gamma\subset\mathbb{Z}^d$ and $h_i=0$ for $i\in \mathbb{Z}^2\setminus \Gamma$. If $\Gamma$ is a finite set then this model has transition.

I expect that if $\Gamma$ is a subset of $\mathbb{Z}^2$ very sparse, for instance $\Gamma=\mathbb{P}\times\mathbb{P}$, where $\mathbb{P}$ is the set of prime numbers this model also has phase transition. So my question is, there exist an explicit example where $\Gamma$ is an infinite set and this model has phase transition?

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This question is related to the random-field problem in the Ising model for which there exists extensive literature starting from 70ies. Will try to lookup references. –  Slaviks Oct 5 '11 at 6:05
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up vote 5 down vote accepted

It seems pretty clear that if you take a very diluted subset of, say, the horizontal line through $0$, then you'll be able to make a Peierls argument.

For example, put $h=+\infty$ (worst possible case, amounting to fixing the corresponding spins to $+1$) at all vertices with coordinates (10^k,0), with $k\geq 1$. Then, when removing a contour surrounding a given site $i$, i.e. flipping all spins inside the contour, except for the frozen ones, we gain an energy proportional to the length of the contour, and only lose an energy at worst proportional to the number of frozen spins surrounded by the contour. The latter term is always much smaller than the former one (at least if $i$ is taken far enough from the frozen spins).

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Thanks for the answer Velenik, this is a nice example. –  Leandro Oct 5 '11 at 22:14
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