L-1

Spin glass Transition

Experiments

Parlare dei campioni di rame dopati con il magnesio, marino o no: trovare due figure una di suscettivita e una di calore specifico, prova della transizione termodinamica.

Edwards Anderson model

We consider for simplicity the Ising version of this model.

Ising spins takes two values $\sigma =\pm 1$ and live on a lattice of $N$ sitees $i=1,2,\ldots ,N$. The enregy is writteen as a sum between the nearest neighbours <i,j>:

$E=-\sum _{<i,j>}J_{ij}\sigma _{i}\sigma _{j}$

Edwards and Anderson proposed to study this model for couplings $J$ that are i.i.d. random variables with zero mean. We set $\pi (J)$ the coupling distribution indicate the avergage over the couplings called disorder average, with an overline:

${\bar {J}}\equiv \int dJ\,J\,\pi (J)=0$

It is crucial to assume ${\bar {J}}=0$, otherwise the model displays ferro/antiferro order. We sill discuss two distributions:

• Gaussian couplings: $\pi (J)=\exp \left(-J^{2}/2\right)/{\sqrt {2\pi }}$
• Coin toss couplings, $J=\pm 1$, selected with probability $1/2$.

The SK model

Sherrington and Kirkpatrik considered the fully connected version of the model with Gaussian couplings:

$E=-\sum _{i,j}{\frac {J_{ij}}{2{\sqrt {N}}}}\sigma _{i}\sigma _{j}$

At the inverse temperature $\beta$, the partion function of the model is

$Z=\sum _{\alpha =1}^{2^{N}}z_{\alpha },\quad {\text{with}}\;z_{\alpha }=e^{-\beta E_{\alpha }}$

Here $E_{\alpha }$ is the energy associated to the configuration $\alpha$. This model presents a thermodynamic transition at $\beta _{c}=??$.

Random energy model

The solution of the SK is difficult. To make progress we first study the radnom energy model (REM) introduced by B. Derrida.

Derivation of the model

The REM neglects the correlations between the $2^{N}$ configurations and assumes the $E_{\alpha }$ as iid variables.

• Show that the energy distribution is

$p(E_{\alpha })={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}e^{-{\frac {E_{\alpha }^{2}}{2\sigma ^{2}}}}$

and determine $\sigma ^{2}$

The Solution: Part 1

We provide different solutions of the Random Energy Model (REM). The first one focus on the statistics of the smallest energies among the ones associated to the $M=2^{N}$ configurations.

Consider the $M=2^{N}$ energies: $(E_{1},...,E_{M})$. They are i.i.d. variables, drawn from the Gaussian distribution $p(E)$. It is useful to use the following notations:

• $P^{<}(E)=\int _{-\infty }^{E}dxp(x)\sim {\frac {\sigma }{{\sqrt {2\pi }}|E|}}e^{-{\frac {E^{2}}{2\sigma ^{2}}}}\;$ for $x\to -\infty$. It represents the probability to find an energy smaller than E.
• $P^{>}(E)=\int _{E}^{+\infty }dxp(x)=1-P^{<}(E)$. It represents the probability to dfind an energy larger than E.

Extreme value stattics for Gaussian variables

We denote

$y_{M}=\min(E_{1},...,E_{M})$

Our goal is to compute the cumulative distribution $Q_{M}(y)\equiv {\text{Prob}}(y_{M}>y)$ for large M and iid variables.

We need to understand two key relations:

• The first relation is exact:

$Q_{M}(y)=\left(P^{>}(y)\right)^{M}$

• The second relation identifies the typical value of the minimum, namely $a_{M}$:

$P^{<}(a_{M})={\frac {1}{M}}$

. Hence in the Gaussian case we get:

$a_{M}=2\sigma {\sqrt {\log M}}-{\frac {1}{2}}{\sqrt {\log(\log M)}}+O(1)$

Close to $a_{M}$, we expect $P^{<}(y)\approx 1/M$. Hence, from the limit $\lim _{M\to \infty }(1-{\frac {k}{M}})^{M}=\exp(-k)$ we re-write the first relation:

$Q_{M}(y)\sim \exp \left(-MP^{<}(y)\right)$

Moreover, we define $P^{<}(y)=\exp(-A(y))$. For Gaussian variables we have $A(y)={\frac {y^{2}}{2\sigma ^{2}}}-{\frac {\sqrt {2\pi }}{\sigma }}\log |y|+\ldots$. It is then useful to write

Bibliography

• Theory of spin glasses, S. F. Edwards and P. W. Anderson, J. Phys. F: Met. Phys. 5 965, 1975