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The two point correlation function for the Ising model is defined as $\left[\langle\sigma_i\sigma_j\rangle -\langle\sigma_i\rangle\langle\sigma_j\rangle\right]$. Then the sum over $i$ $j$ of that function gives: \begin{equation} \sum_{ij}\left[\langle\sigma_i\sigma_j\rangle -\langle\sigma_i\rangle\langle\sigma_j\rangle\right] = \langle M^2\rangle - \langle M\rangle^2 \end{equation} Where $\sigma_i = \pm 1$ and $M =\sum_i \sigma_i$. The expression can be normalized by dividing it by $M^2 = N^2$ where $N$ is the number of spins of the system. My question is, I am assuming I can justify the long range correlations in the lattice by computing the sum of the correlation functions, is that correct? I should expect it to vanish when most of the two point correlation vanish, and to have peak when most two point correlation reach its maximum value.

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  • $\begingroup$ How did you get $\sum_{ij}\langle\sigma_i\sigma_j\rangle=\langle M^2\rangle$? $\endgroup$ – Iván Mauricio Burbano Jul 31 '20 at 22:35
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Let us be a bit more precise, so that the question can be answered accurately.

Let us thus assume that your system is composed of $N^d$ spins attached to the vertices of the box $\Lambda_N=\{1,\dots,N\}^d$. I am assuming that there is a positive magnetic field $h>0$ acting on the spins (this is technically useful to ensure that we are considering the proper state below, but the field will soon be set equal to $0$). Let me denote by $\langle\cdot\rangle_{N,\beta,h}$ the corresponding expectation.

Then, the susceptibility is given by $$ \chi_N(\beta,h) = \frac1{N^d}\sum_{i,j\in\Lambda_N} \Bigl[ \langle \sigma_i\sigma_j \rangle_{N,\beta,h} - \langle \sigma_i \rangle_{N,\beta,h} \langle \sigma_j \rangle_{N,\beta,h} \Bigr]. $$ We are really interested in the thermodynamic limit of this quantity, $$ \chi(\beta,h) = \lim_{N\to\infty} \chi_N(\beta,h) = \sum_{i\in\mathbb{Z}^d} \Bigl[ \langle \sigma_0\sigma_i \rangle_{\beta,h} - \langle \sigma_0 \rangle_{\beta,h} \langle \sigma_i \rangle_{\beta,h} \Bigr], $$ where I used the fact that the resulting infinite-volume state is translation invariant. We can now get rid of the magnetic field and define $$ \chi(\beta) = \lim_{h\downarrow 0} \chi(\beta,h) = \sum_{i\in\mathbb{Z}^d} \Bigl[ \langle \sigma_0\sigma_i \rangle_{\beta}^+ - \langle \sigma_0 \rangle_{\beta}^+ \langle \sigma_i \rangle_{\beta}^+ \Bigr], $$ where $\langle \cdot \rangle_\beta^+$ denotes expectation with respect to the $+$ state.

It is known, in any dimension $d$, that the truncated 2-point function decays exponentially with the distance when $\beta\neq\beta_{\rm c}$. Namely, for all $\beta\neq\beta_{\rm c}$, there exists $c=c(\beta,d)>0$ such that $$ 0\leq \langle \sigma_0\sigma_i \rangle_{\beta}^+ - \langle \sigma_0 \rangle_{\beta}^+ \langle \sigma_i \rangle_{\beta}^+ \leq e^{-c\|i\|}. $$ (In other words, the correlation length is finite at all non-critical temperatures.) This immediately implies that the susceptibility is finite away from the critical point: $$ \chi(\beta) < \infty\qquad\forall\beta\neq\beta_{\rm c}. $$ Moreover, it is also known that, in all dimensions $d\geq 2$, the 2-point function does not decay exponentially fast when $\beta=\beta_{\rm c}$ (much more precise information is available when $d=2$ and when $d$ is large enough).

Finally, it is known that the susceptibility diverges as $\beta\uparrow\beta_{\rm c}$: $$ \lim_{\beta\uparrow\beta_{\rm c}} \chi(\beta) = +\infty. $$

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  • $\begingroup$ @CMB : Does this answer your question? If not, what additional information do you need? $\endgroup$ – Yvan Velenik Aug 3 '20 at 9:36
  • $\begingroup$ Thank you very much for the answer. I was aware of that identification with susceptibility. I guess you prove my point when you state that the correlation function is semipositive defined. Being that the case, considering that the system has rotational symmetry , then the value of $\chi$ can be used to justify long range correlations in the system. I asked this because I'm doing a 2D simulation, and I wanted to know if I could justify this property without having to compute all the two point correlation functions step by step, as that'd cost a lot of computational power. $\endgroup$ – CMB Aug 5 '20 at 15:25
  • $\begingroup$ Yes, you can use the divergence of the susceptibility to locate the phase transition. Beware, however that if you use periodic boundary conditions, as is often the case in numerical simulations, then $\langle M\rangle = 0$ at all temperatures (by symmetry), so you have to be careful when evaluating the susceptibility below $T_{\rm c}$. $\endgroup$ – Yvan Velenik Aug 5 '20 at 16:25
  • $\begingroup$ To avoid that I define $M\equiv |M|$ so that $\chi = \langle M^2 \rangle - \langle |M| \rangle ^2$. $\endgroup$ – CMB Aug 5 '20 at 16:55
  • $\begingroup$ Which is fine, yes (if your system is not too small). $\endgroup$ – Yvan Velenik Aug 5 '20 at 17:32

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