The Sherrington-Kirkpatrick model §

Is a simple mean-field Disordered systems with gaussian coupings. The hamiltonian is

$${\mathcal{H}}^{SK}=\frac{1}{2\sqrt{N}}\sum _{(i,j)}{J}_{ij}{\sigma }_i{\sigma }_j{J}_{ij}\sim N(0,1)$$

with $J_{ij}$ i.i.d. We can add an external field $h$, this doesn’t complicates the study, since it factorazies.

Computing the annealed free energy §

Proposition The annealed free energy of the SK model is

$$f^A=-\frac{1}{\beta }\ln 2\cosh \beta h-\frac{\beta }{4}$$

Proof Since the couplings are i.i.d, the partizion function is the product of $\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ identical partion function.

$$Z=\sum _{\sigma }\prod _{1\le i<j\le N}\exp \left(\frac{\beta }{\sqrt{N}}{J}_{ij}{\sigma }_i{\sigma }_j\right)\exp \left(\beta h\sum _k{\sigma }_k\right)$$

we can compute the gaussian average

$$\mathbb{E}\exp \left(\frac{\beta }{\sqrt{N}}{J}_{ij}{\sigma }_i{\sigma }_j\right)$$

sice ${\sigma }_i{\sigma }_j=\pm 1$ $J_{ij}{\sigma }_i{\sigma }_j\sim N(0,1)$, from a simple calculation we find

$$=\exp \left(\frac{{\beta }^2}{2N}\right)$$

since we have $\left(\genfrac{}{}{0ex}{}{N}{2}\right)=N(N-1)/2$ terms in the product we get

$$\mathbb{E}Z=\sum _{\sigma }\exp \left[\frac{{\beta }^2}{2N}\cdot \frac{N(N-1)}{2}\right]\exp \left(\beta h\sum _k{\sigma }_k\right)$$

per sommare il termine dipendente dagli spin, lo fattorizziamo

$$\prod _{k}2\cosh (\beta h)=2^N{\cosh }^N(\beta h)$$

otteniamo:

$$Z_N=2^N{\cosh }^N(\beta h)\exp (\frac{{\beta }^2}{4}(N-1))$$

quindi l’energia libera nel TDL è

$$f^A=\underset{N\to \infty }{\lim }\frac{-1}{\beta N}\ln \mathbb{E}Z$$ $$=-\frac{1}{\beta }\ln 2\cosh (\beta h)-\frac{\beta }{4}$$

Remark se si calcola l’entropia si trova che è negativa per $\beta$ abbastanza grandi, questo non ha senso fisico, di conseguenza l’energia libera annealed non è quella corretta.

The square root normalization §

In the Modello di Curie-Weiss the normalization for the interacions $J$ was $\frac{1}{N}$, while here in the SK model we have $1/\sqrt{N}$.

The intuition: sum of $n$ gaussian variables is again a gaussian with standard deviation of $\sqrt{n}$.

Si può far vedere analiticamete usando il Isserlis-Wick theorem.

Let’s compute the average energy:

$$⟨\mathcal{H}⟩=\mathbb{E}\frac{{\sum }_{\sigma }\mathcal{H}(\sigma )e^{-\beta \mathcal{H}}}{Z}$$ $$=-\frac{1}{2\sqrt{N}}\sum _{i,j}\mathbb{E}{J}_{ij}\frac{{\sum }_{\sigma }{\sigma }_i{\sigma }_j{e}^{-\beta \mathcal{H}}}{Z}$$

we now use the theorem $\mathbb{E}[gF(g)]={\nu }^2\mathbb{E}[F^{\prime }(g)]$

$$=-\frac{1}{2\sqrt{N}}\sum _{i,j}\mathbb{E}{\partial }_{J_{ij}}\frac{{\sum }_{\sigma }{\sigma }_i{\sigma }_j{e}^{-\beta \mathcal{H}}}{Z}$$

let’s compute the partial deriviative

$$-\frac{\beta }{2\sqrt{N}}\frac{1}{Z^2}\left[\left(\sum _{\sigma }{\sigma }_i^2{\sigma }_j^2{e}^{-\beta H}\right)Z-\left(\sum _{{\sigma }^a}{\sigma }_i^a{\sigma }_j^a{e}^{-\beta H}\right)\left(\sum _{{\sigma }^b}{\sigma }_i^b{\sigma }_j^b{e}^{-\beta H}\right)\right]$$ $$=-\frac{\beta }{2\sqrt{N}}\left(1-\Omega ({\sigma }_i^a{\sigma }_j^a)\Omega ({\sigma }_i^b{\sigma }_j^b)\right)$$

dove con $\Omega$ abbiamo indicato la media rispetto alla misura di Boltzmann. Quindi alla fine

$$⟨\mathcal{H}⟩=-\frac{\beta }{4N}\sum _{i,j}1-\mathbb{E}[\Omega ({\sigma }_i^a{\sigma }_j^a)\Omega ({\sigma }_i^b{\sigma }_j^b)]$$ $$=-\frac{\beta N}{4}(1-⟨q_{ab}^2⟩)$$

che come voluto è di ordine $O(N)$.