Ising model: How is $|langlesigmarangle|^{2}=lim _{r rightarrow infty} G^{(2)}(r)$?

A general statement.

Let us first state a general result, valid in any dimension. Let $f$ and $g$ be two local functions (that is, functions depending only on finitely many spins). Then $$ lim_{|i|toinfty} langle f cdot (gcirctheta_i) rangle^+ = langle f rangle^+ langle g rangle^+ , tag{1} $$ where I used $theta_i$ to denote the translation by $iinmathbb{Z}^d$ and $langlecdotrangle^+$ denotes the $+$ state (that is, the state obtained using $+$ boundary condition). Identity (1) is an easy consequence of the FKG inequality, see Exercise 3.15 in this book (note that its solution can be found in Appendix C).

Application to the 2-point function.

In particular, setting $f=g=sigma_0$, Identity (1) reduces to $$ lim_{|i|toinfty} langle sigma_0 sigma_i rangle^+ = bigl( langle sigma_0 rangle^+ bigr)^2. tag{2} $$ It can, in addition, be shown that $langle sigma_0 rangle^+=m^*$, where $m^*$ denotes the spontaneous magnetization density (defined as the limit, as $hdownarrow 0$, of the magnetization density in the presence of a magnetic field $h$); see Remark 3.30 in the book.

Extension to the free and periodic states

Note that (2) (but not Identity (1) above) also holds for the state $langlecdotrangle$ obtained using free or periodic boundary conditions. This follows from the decomposition $langlecdotrangle = frac12 langlecdotrangle^+ + frac12 langlecdotrangle^-$, combined with the identity $langlesigma_0sigma_irangle^+ = langlesigma_0sigma_irangle^-$: $$ lim_{|i|toinfty} langle sigma_0 sigma_i rangle = lim_{|i|toinfty} bigl( tfrac12 langlesigma_0 sigma_irangle^+ + tfrac12 langlesigma_0 sigma_irangle^-bigr) = lim_{|i|toinfty} langle sigma_0 sigma_i rangle^+ = bigl( langle sigma_0 rangle^+ bigr)^2. $$ Note that it is still $m^*=langle sigma_0 rangle^+$ that appears in the right-hand side (in fact, $langlesigma_0rangle = 0$ by symmetry).

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Additional remarks:

• Identity (1) is actually true for any extremal state in any model, but the proof is much more abstract (see Theorem 6.58 in the same book).
• The decomposition $langlecdotrangle = frac12 langlecdotrangle^+ + frac12 langlecdotrangle^-$ is not a trivial fact (it is, for instance, not always true for the Ising model on a tree). It follows from the fact that the state $langlecdotrangle$ is translation invariant and that all translation-invariant states are convex combinations of the states $langlecdotrangle^+$ and $langlecdotrangle^-$. Proofs can be found here or here.