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Sometimes in Monte Carlo simulations we need to compute the correlation length, but this is a hard task without a formula. However, for instance, in an periodic cubic lattice of $L^3$ spins, some papers present a formula of the correlation length as follows:

$$ \xi = \frac{1}{2 \sin(\pi/L)} {\left (\frac{\chi}{F}-1 \right)}^{1/2} $$

where $\chi$ is the susceptibility, which is $\hat{G}(0)$ (i.e. the Fourier transformed two-point correlation function at $\vec{k} = (0,0,0)$), and $F$ is $\hat{G}(\vec{k}_{min})$, where $\vec{k}_{min} = (2\pi/L,0,0)$. My guess is that it comes from the following definition $$ {\xi}^2 = \frac{\sum_{\vec{r}} r^2 G(\vec{r})}{\sum_{\vec{r}}G(\vec{r})}, $$

but I can't get to the formula.

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To me this looks like a rearranged version of the Ornstein-Zernike form of the Fourier-transformed correlation function, which reads $$ \hat{G}(k) = \frac{\hat{G}(0)}{1+(k\xi)^2} $$ for any sufficiently small $k$. Obviously, this form applies to an isotropic system rather than a cubic lattice.

The assumptions behind that equation are discussed in this question and my answer to it; both the question and the answer give further pointers to the literature. Your equation reduces to this one, if we make the assumption that $k=k_\text{min}$ is small, so that $\sin\frac{1}{2}k_\text{min}\approx \frac{1}{2}k_\text{min}$. (I note that the lattice spacing is taken to be unity, making both $L$ and $k$ dimensionless).

I guess that the lattice structure of the problem is the reason your equation involves a $\sin$ function, while mine does not, but I don't know enough about the context to say for sure.

In any case, I hope that this sets you along the right lines.

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  • $\begingroup$ Hi LonelyProf. Thank you for your thoughts. The fomula was used in computing the spin-glass correlation length in the cubic lattice. Nevertheless, I have also seen the same formula for the two-dimensional Ising model. For instance, this is shown in Eq. (2.8) in Journal of Statistical Physics, Vol. 98, Nos. 3/4, 2000 by J. Salas et al. $\endgroup$ – O. Daniel Jun 6 at 1:45
  • $\begingroup$ ....the appearance of that $\sin(\pi/L)$ is still a mystery for me. $\endgroup$ – O. Daniel Jun 6 at 1:52
  • $\begingroup$ I see this formula in earlier papers e.g. Ballesteros et al J Phys A, 32, 1 (1999) and references therein. Clearly it is a standard one which the specialists should recognize; some of those people are active on this site for sure, so you should get a better answer from one of them before long, if those papers are not enough to track it down. $\endgroup$ – user197851 Jun 6 at 10:22

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