# Correlation length anisotropy in the 2D Ising model

In the Ising model, the two-spin correlation function is $$C(\vec{r}) = \langle \sigma_{\vec{r}_0+\vec{r}}\sigma_{\vec{r}_0}\rangle - \langle \sigma_{\vec{r}_0+\vec{r}}\rangle \langle \sigma_{\vec{r}_0} \rangle.$$ This quantity doesn't depend on $$\vec{r}_0$$ due to the translational invariance. When $$r = |\vec{r}|$$ is large compared to the lattice spacing, we expect the following approximate form $$C(\vec{r}) \sim \exp(-r/\xi),$$ where $$\xi$$ is the correlation length.

Different directions on the lattice are not equivalent. For example, in the Ising model on the square lattice, there are two directions, say vertical and horizontal, along which neighboring spins interact. I see no reasons to think that other directions are equivalent to these two. In the anisotropic Ising model, vertical and horizontal directions are also not equivalent.

Then the correlation length $$\xi$$ should depend on the direction of $$\vec{r}$$. Is the analytical form of this dependence known at least for the square lattice? The Ising model is probably the most studied model of statistical physics, but I was not able to find corresponding formulas. So any references would be appreciated.

P.S. I know that in the scaling limit the Ising model becomes isotropic. The question above is for systems far enough from the critical point.

The correlation length of the 2d Ising model has been computed explicitly. You can find the expression in the famous book by McCoy and Wu. Here's a plot of the inverse correlation length (i.e., $$1/\xi$$) at various temperatures, taken from this recent review paper:

This is only to show the directional dependence, as the radial scale is not the same for all pictures. The temperature decreases from left to right (you can see the isotropy appearing close to the critical temperature) from close to $$\infty$$ to close to the critical temperature. Below the critical temperature, the behavior is exactly the same, since self-duality of the model implies that, for any $$T, $$\xi(T) = \xi(T^*)/2$$ where the dual temperature $$T^*=T^*(T)$$ satisfies $$T^*>T_c$$.

• Thank you very much! Is this problem solved for the anisotropic Ising model on the triangular lattice?
– Gec
Commented Jul 28, 2019 at 15:18
• @Gec : Stephenson had a series of papers in the 1960s and early 1970s on correlations of the Ising model on the triangular lattice. The fourth one of the series might cover what you want: Ising‐Model Spin Correlations on the Triangular Lattice. IV. Anisotropic Ferromagnetic and Antiferromagnetic Lattices, J. Stephenson, Journal of Mathematical Physics 11, 420 (1970). Commented Jul 28, 2019 at 15:23
• (Note: I haven't read Stephenson's papers, so I might be mistaken. In any case, looking at which more recent papers cite those might lead you to what you need...) Commented Jul 28, 2019 at 15:26
• I am curious to know more about the relation $\xi(T) = \xi(T^*)/2$. Is there a simple explanation for this fact? I can create a new topic if it is inconvenient to respond in the comments.
– Gec
Commented Feb 12, 2023 at 20:28
• @Gec : this relation is a consequence of the duality enjoyed by the planar Ising model. Namely, the surface tension of the model at temperature $T<T_c$ is easily seen to be equal to the rate of exponential decay of the 2-point function at the dual temperature $T^*>T_c$ (essentially by a direct comparison of the low and high-temperature expansions of these two quantities). Commented Feb 13, 2023 at 7:59

You could study this problem near the fixed point (the right two images in Yvan's answer) by looking for the most relevant operator with the right symmetry charge.

For instance for a rectangular lattice we would be looking for spin 2 operators, a triangular lattice spin 3, and a square lattice spin 4.

Because these higher spin deformations in the Ising model are coming from descendant operators, one expects about an order of $$(T-T_c)$$ separation in the anisotropies between each case.

I don't know how to explain other interesting features though, like why the correlation length has a cusp at low temperatures. That's pretty cool!

• You only have a cusp in the limit $T\to 0$ or $T\to\infty$ (actually, the high- and low-temperature correlation lengths are proportional thanks to self-duality). At positive and finite temperatures, the correlation length is rigorously known to be analytic in the direction (in any dimension, actually). At very high and very low temperatures, you can understand the behavior of the correlation length using perturbative techniques (e.g., cluster expansion). Commented Jul 28, 2019 at 17:42