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In the Ising Model with $+$ boundary condition in dimension $2$, but possibly one could ask about dimension $d$. Set $\Lambda_N:=[-N,N]^2$, let $\beta >0$ be the inverse of the temperatura and $h>0$ be the magnetic field. For $x,y \in \Lambda_N$, is it true that

$$ \langle \sigma_x \sigma_y \rangle^{+}_{\Lambda_N,\beta,h} \le \langle \sigma_x \rangle^{+}_{\Lambda_N,\beta,h}. $$

I am particularly interested in the case where $x$ and $y$ are nearest neighbors, but I suppose that this case is actually more difficult to prove.

Kernel
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1 Answers1

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No, the inequality $$ \langle\sigma_x\sigma_y\rangle_{\Lambda_N,\beta,h}^+ \leq \langle\sigma_x\rangle_{\Lambda_N,\beta,h}^+ \tag{1} $$ does not hold in general, in any dimension and for any pair $x\neq y$ in $\Lambda_N$. Indeed, assume that $\beta<\beta_c$. On the one hand, $$ \lim_{h\downarrow 0}\lim_{N\to\infty} \langle\sigma_x\rangle_{\Lambda_N,\beta,h}^+ = 0 , $$ so that the right-hand side of (1) can be made as small as you wish by taking $N$ large enough and $h$ small enough.

On the other hand, by the Griffiths inequalities, $$ \langle\sigma_x\sigma_y\rangle_{\Lambda_N,\beta,h}^+ \geq \langle\sigma_x\sigma_y\rangle_{\Lambda_N,\beta,0}^+ \geq (\tanh\beta)^{\|y-x\|_1} . $$ The last inequality is obtained by choosing a shortest path in $\Lambda_N$ connecting $x$ and $y$ and removing all coupling constants (that is, setting them to zero) not along this path; this yields the 2-point function for a one-dimensional Ising model, which is easily seen to equal the expression in the right-hand-side.

This shows that the left-hand side of (1) is bounded below away from zero, uniformly in $N$ and $h$. Combined with the observation above, we conclude that (1) does not hold in general.

Yvan Velenik
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  • Thank you very much for the answer! – Kernel Mar 06 '17 at 13:09
  • Do you think that is could be valid on the opposite case where $\beta$ is sufficiently large? – Kernel Mar 06 '17 at 14:10
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    Yes, it should be true for sufficiently large values of $\beta$. I am going to argue heuristically; a rigorous proof using cluster expansion should not be hard. Assume for simplicity that $h=0$ and that $x$ and $y$ are neighbors.

    First, it is not difficult to show, using the cluster expansion, that $\langle\sigma_x\rangle^+_{\Lambda_N,\beta,0} = 1 - 2e^{-8\beta} + o(e^{-8\beta})$; see Exercise 5.12 in this book.

     – Yvan Velenik Mar 06 '17 at 15:51
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    Now, observe that $\langle\sigma_x\sigma_y\rangle^+{\Lambda_N,\beta,0} = 1 - 2\mu{\Lambda_N,\beta,0}(C_{x,y})$, where $C_{x,y}$ is the event that there is a Peierls contour separating $x$ from $y$. It should not be too hard (although I haven't checked) to show that $\mu_{\Lambda_N,\beta,0}(C_{x,y}) = 2 e^{-8\beta} + o(e^{-8\beta})$, since there are two contours of minimal length: one surrounding $x$ and one surrounding $y$. – Yvan Velenik Mar 06 '17 at 15:51
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    This indicates that $\langle\sigma_x\rangle^+{\Lambda_N,\beta,0} - \langle\sigma_x\sigma_y\rangle^+{\Lambda_N,\beta,0} = 2e^{-8\beta} + o(e^{-8\beta})$, which will be positive for large enough $\beta$. The same type of computations (still using cluster expansion) can be done when $h>0$. – Yvan Velenik Mar 06 '17 at 15:51
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    Note that this requires $\beta\gg 1$. It will not apply for all $\beta>\beta_c$ and the inequality actually fails for $\beta$ slightly larger than $\beta_c$ by the same argument as in my answer (since the magnetization can be made as small as one wishes by taking $\beta$ close enough to $\beta_c$), while the 2-point function remains bounded away from $0$ (roughly by $\tanh\beta_c$). – Yvan Velenik Mar 06 '17 at 16:00
  • Again, thank you very much for your help! It was very useful! – Kernel Mar 06 '17 at 16:48