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I am considering the following ferromagnetic Hamiltonian for the 2-d Ising Model, say with periodic boundary condition in the torus $\Lambda_n=\mathbb{T}^2_n := (\mathbb{Z}/ \mathbb{Z}_n)^2$:

$$ H_n(\sigma)= - \sum_{x\sim y}J \sigma_x \sigma_y- a_n\sum_{x}h \sigma_x $$ where $x\sim y$ denotes that $d_{\mathbb{T}^2_n}(x,y)=1$. Notice that $h_n$ is also changing according to $n$. We can also define the pressure $$ p(\beta,h)= \lim_{n \to \infty} \frac{1}{|\Lambda_n|} \log Z_{\Lambda_n,\beta,h\cdot a_n}. $$ where $Z_{\Lambda_n,\beta,h\cdot a_n}$ is the partition function for the volume $\Lambda_n$.

The Lee-Yang Circle Theorem guarantees that if I choose $a_n \equiv 1 \neq 0$, and $h \neq 0$, I have that my the pressure is analytic in $h$, and therefore I have no phase transition.

However, if one has a non-homogeneous magnetic field, one can still have phase transition (see BCCP14).

I would like to know if there there exists a "right order" for $a_n$ at which we could observe a phase in $h$. That is, can I choose ${a_n}$ such that there exists $h^* >0$ so that for all $h>h^*$, $p(\beta,h)$ is analytic in $h$, and for $0<h<h^*$, the pressure ceases to be differentiable?

If perhaps looking at the pressure is not the right type of phase transition, one could also think in terms of Gibbs measures.

Are there any articles in this direction? Comments and suggestions are appreciated.

EDIT: If this is this too difficult to explicit find in the Ising model, could one do that in the Curie-Weiss?

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Well, isn't it the case that any sequence $a_n$ that vanishes as $n \to\infty$ will give rise to the same pressure as the model with no field? Indeed, one can bound $$ e^{-\beta ha_nn^2} Z_{\Lambda_n,\beta,0} \leq Z_{\Lambda_n,\beta,ha_n} \leq e^{\beta ha_nn^2} Z_{\Lambda_n,\beta,0}, $$ so that $$ -\beta h a_n + \frac1{n^2}\log Z_{\Lambda_n,\beta,0} \leq \frac1{n^2}\log Z_{\Lambda_n,\beta,ha_n} \leq \beta h a_n + \frac1{n^2}\log Z_{\Lambda_n,\beta,0}, $$ which shows that $p(\beta,h) = p(\beta,0)$ whenever $\lim_{n\to\infty} a_n=0$.

On the other hand, if $\lim_{n\to\infty}a_n=a\neq 0$, then a similar argument shows that $p(\beta,h) = p(\beta,ah)$.

The situation does not seem to be much more interesting in terms of the Gibbs measure, I think, but I may be missing something.

Note that the problem becomes much more interesting when you use a square box with $-$ boundary condition, as in this case there is a nontrivial competition between the boundary condition and the magnetic field. In that case, the critical scale is $a_n = 1/n$ (as this makes the contribution to the energy due to the magnetic field and the one due to the boundary condition of the same order $O(n)$). For such a choice, there is a value $h_c\in (0,\infty)$ such that

  • for any $h<h_c$, the $-$ phase occupies the box (even though the magnetic field is positive);
  • for any $h>h_c$, the $+$ phase occupies the box, except for a microscopic layer of $-$ phase along the boundary (well, the situation is actually more interesting near the corners, but let's ignore that).

This question is discussed in detail in a paper by Schonmann and Shlosman from 1996. (Actually much more can be said nowadays about the behavior of the microscopic layer of $-$ phase when $h>h_c$, including the fact that its width is of order $n^{1/3}$ and its scaling limit is given by a Ferrari-Spohn diffusion; we are currently writing down a paper, with Dima Ioffe, Sébastien Ott and Senya Shlosman on this very topic.)

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    $\begingroup$ Sure. The pressure would also be unaffected. However, the behavior of typical configurations under the Gibbs measure is affected. $\endgroup$ – Yvan Velenik Oct 11 '19 at 10:56
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    $\begingroup$ Well, you do have the stochastic domination $\mu_{n,\beta,h/n} \succ \mu_{n-1,\beta,h/n}^-$. Therefore, since the latter converges to the $+$ state as soon as $a_n \gg 1/n$, the same should be true of the former. So, a vanishing field can indeed select the phase. Of course, if $a_n=o(n^{-2})$, then the magnetic field should have no impact and the limiting state should be the mixture $\tfrac12 \mu^+_{\beta,0} + \tfrac12 \mu^-_{\beta,0}$. It is not immediately clear to me (but I am a bit tired tonight) what happens when $n^{-2} \ll a_n \ll n^{-1}$... I'll think about it tomorrow if I have time. $\endgroup$ – Yvan Velenik Oct 15 '19 at 16:24
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    $\begingroup$ My guess (for tonight) is that the $+$ state is selected whenever $a_n\gg n^{-2}$, but I might very well be wrong about that. $\endgroup$ – Yvan Velenik Oct 15 '19 at 16:28
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    $\begingroup$ I thought a bit more and I confirm what I said yesterday. For any sequence $a_n\gg n^{-2}$, the $+$ phase will be selected (for $h>0$). The magnetization density $m_n=n^{-2}\sum_{i\in\Lambda_n} \sigma_i$ of the infinite-volume state will concentrate on $m^*(\beta)$ or $-m^*(\beta)$ and in particular, with high probability, will be outside the interval $[-\epsilon,\epsilon]$ for some $\epsilon(\beta)>0$ for all $\beta>\beta_c$. A simple argument based on flipping the signs of all spins shows that $\mu_{n,\beta,ha_n}(m_n<-\epsilon)/\mu_{n,\beta,ha_n}(m_n>\epsilon) = 0$. $\endgroup$ – Yvan Velenik Oct 16 '19 at 15:07
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    $\begingroup$ Presumably, for $a_n=O(n^{-2})$, the relative weights of the $+$ and $-$ components in the limiting state will change with the value of $h$ (but you keep a mixture of both states). I think that all this can be completely proved with a little effort, but I don't have time for that now. $\endgroup$ – Yvan Velenik Oct 16 '19 at 15:09

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