I'm imagining a square lattice with Ising spins on the vertices and nearest-neighbor Ising interactions. The interaction on a given bond is ferromagnetic with probability $(1-p)$ and antiferromagnetic with probability $p$. I'm imagining the majority of bonds to be ferromagnetic, so $p<1/2$.

That is, I have $$H = -\sum_{\langle ij \rangle} c_{ij} \sigma_i \sigma_j$$ where $c_{ij}$ are i.i.d. random variables that are $1$ with probability $(1-p)$ and $-1$ with probability $p$.

Consider the correlation function between two spins at $\vec{r}$ and $\vec{r}'$, where $\mathbb{E}$ is the disorder average and $\langle \cdot \rangle$ is the thermal average: $\mathbb{E}[\langle \sigma_{\vec{r}} \sigma_{\vec{r}'} \rangle_{\beta}]$.

Does this correlation function decrease with $p$? That is, do we have

$$ \frac{\partial}{\partial p} \mathbb{E}[\langle \sigma_{\vec{r}} \sigma_{\vec{r}'} \rangle_{\beta}] \leq 0?$$

I'm most interested when $p$ and $\beta$ are such that the model is in the ferromagnetic phase. The rough picture I have in my mind is that when one is in the ferromagnetic phase, sprinkling in antiferromagnetic bonds will weaken the correlation function and even introduce some disorder realizations where the correlation function is negative. That is, I strongly suspect that $\frac{\partial}{\partial p} \mathbb{E}[\langle \sigma_{\vec{r}} \sigma_{\vec{r}'} \rangle_{\beta}] \leq 0$ is true in the ferromagnetic phase.

I also suspect that this kind of inequality is well-known and generalizable to other kinds of ferromagnetic models (not just nearest neighbor) with some dilute antiferromagnetic couplings and other kinds of multi-body correlation functions, and I would appreciate if someone can name it for me.


1 Answer 1


This is not a direct answer, but it may be useful.

Define, for any subset of sites $A$ $$ \sigma_A \equiv \prod_{i\in A} \sigma_i $$

For systems with only ferromagnetic couplings (ferromagnetic is defined as the couplings being positive as defined below, regardless of whether you preserve global spin-flip symmetry. For example, a magnetic field that prefers $\sigma_i = 1$ is allowed.) (note that there is no disorder averaging), $$ H = -\sum_{A} J_{A} \sigma_A ,\quad J_{A} \geq 0 $$ there are the Griffiths-Kelly-Sherman inequalities and its extension, the Ginibre inequality. One of its statements is under these conditions, $\langle \sigma_A\sigma_B\rangle -\langle \sigma_A\rangle \langle \sigma_B\rangle \geq 0 $. This is useful for us because if you consider \begin{align} \frac{\partial \langle \sigma_A \rangle}{\partial J_B} &= \frac{\partial}{\partial J_B}\Bigg(\frac{1}{\sum_{\sigma} e^{-H[\sigma]}} \sum_{\sigma}\sigma_A e^{-H[\sigma]}\Bigg) \\ &= \langle \sigma_A\sigma_B\rangle -\langle \sigma_A\rangle \langle \sigma_B\rangle \geq 0 \end{align} using the above theorem. It tells us that any correlation function is non-decreasing in any of the "ferromagnetic" couplings. In particular, if your $c_{ij}\geq 0$, you can show $$ \frac{\partial \langle \sigma_k \sigma_l \rangle}{\partial c_{ij}} \geq0 \quad \text{for all } i,j,k,l. $$ I am not sure if there are additional subtleties with spontaneous symmetry breaking, but I don't think so since this correlation function is invariant under the global symmetry.

  • 1
    $\begingroup$ +1 Thanks for the discussion. I agree that the GKS inequalities are useful for making rigorous the intuition about changing the strength of ferromagnetic couplings, and I've been trying to see whether I might be able to adapt them. However, since this system has both ferromagnetic and antiferromagnetic bonds, and because changing the disorder strength $p$ is only indirectly related to changing the couplings, I'm not sure whether they'll end up being useful. $\endgroup$
    – user196574
    Commented Nov 16, 2023 at 20:48
  • $\begingroup$ I completely agree. I haven’t seen any rigorous results when you have anti ferromagnetic couplings, but I am not an expert. $\endgroup$ Commented Nov 16, 2023 at 21:36

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