I happened to come across the following term while doing an excercise on perturbation theory

\begin{equation} H^2=J^2\sum_{<i,j>}\sum_{<k,l>}\sigma_i\ \sigma_j\ \sigma_k\ \sigma_l \end{equation}

Where $H$ is the Ising Hamiltonian, $J$ is the coupling constant, $\sigma_I=\pm1$ and $<i,j>$ means sum over neighbouring sites.

When computing the mean value $\langle H^2 \rangle$

\begin{equation} \langle H^2\rangle=J^2\frac{1}{Z}\sum_{\{\sigma\}}e^{-\beta H}\sum_{<i,j>}\sum_{<k,l>}\sigma_i\ \sigma_j\ \sigma_k\ \sigma_l \end{equation}

I started to wonder if every term in this sum is non-zero. for example, terms where $i=k, j=l$ give a contribution of $2J^2N$ because $\sigma_i \sigma_j \sigma_k \sigma_l=1$ and one is left out with a sum over pairs of sites. Using the fact that $\sum_{\sigma=-1}^1\sigma^2=2$ and $\sum_{\sigma=-1}^1\sigma=0$ is there any argument to determine if some combinations of $i,j,k,l$ don't contribute to the mean value?

Edit: I would be really interested in knowing, for example, how much this term is

\begin{equation} \langle H^2\rangle_{i=j}= J^2\frac{1}{Z}\sum_{\{\sigma\}}e^{-\beta H}\sum_{<i,j>}\sum_{<k,l>}\sigma_i\ \sigma_j\ \sigma_k\ \sigma_l\ \delta_{ik} \end{equation}

  • $\begingroup$ I don't understand the question. What do you mean by "this term is zero"? Each term is a function of the spin configuration, not a number... $\endgroup$ – Yvan Velenik Apr 4 '18 at 7:12
  • $\begingroup$ you are right! Sorry for the misunderstunding, I meant that the mean value is zero. Let me edit. $\endgroup$ – P. C. Spaniel Apr 4 '18 at 14:46
  • $\begingroup$ OK, then all the expectations $\langle \sigma_i\sigma_j\sigma_k\sigma_l\rangle$ are strictly positive (as long as $\beta>0$). This follows, for example, from the GKS correlation inequalities. The latter imply that $\langle\sigma_i\sigma_j\sigma_k\sigma_l\rangle \geq \langle\sigma_i\sigma_j\rangle \langle\sigma_k\sigma_l\rangle$. Now, for any $i,j$ neighbors, $\langle\sigma_i\sigma_j\rangle \geq \tanh(\beta J) > 0$ for all $\beta>0$. $\endgroup$ – Yvan Velenik Apr 4 '18 at 14:57
  • $\begingroup$ Thanks! I edited to add a small new question, if you have any clue on how to compute that particular mean value that would be great! thanks! $\endgroup$ – P. C. Spaniel Apr 5 '18 at 5:25
  • $\begingroup$ The term you are interested in will scale linearly with the number of spins in the system. The exact constant is probably impossible to determine, except in dimensions $1$ and $2$. Note, however, that $\langle H^2\rangle$ scales as the square of the number of spins, so the term you are interested in contributes negligibly to this expectation. $\endgroup$ – Yvan Velenik Apr 5 '18 at 5:49

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