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 statistics for iid

We denote

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

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

We need to understand two key relations:

• The first relation is exact:

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

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

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

.

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

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

Moreover, if we define $P^{<}(\epsilon )=\exp(-A(\epsilon ))$ we recover the famous Gumbel distribution:

$Q_{M}(\epsilon )\sim \exp \left(-e^{-A'(a_{M})(\epsilon -a_{M})}\right)$

Exercise L1-A: the Gaussian case

Specify these results to the Guassian case and find

• the typical value of the minimum

%

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

• The expression $A(\epsilon )={\frac {\epsilon ^{2}}{2\sigma ^{2}}}-{\frac {\sqrt {2\pi }}{\sigma }}\log |\epsilon |+\ldots$
• The expression of the Gumbel distribution for the Gaussian case

$Q_{M}(\epsilon )\sim \exp \left(-e^{-{\frac {\sqrt {2\log M}}{\sigma }}(\epsilon -a_{M})}\right)$

Density of states above the minimum

For a given disorder realization, we compute $d(x)$, the number of configurations above the minimum, but with an energy smaller than $E_{\min }+x$.

${\text{Prob}}(d(x)=k)=M{\binom {M-1}{k}}\int dE\;p(E)[P^{>}(E)-P^{>}(E+x)]^{k}P^{>}(E+x)^{M-k-1}$

Taking the average ${\overline {d(x)}}=\sum _{k}k{\text{Prob}}(d(x)=k)$, we derive

${\overline {d(x)}}=M(M-1)\int dE\;p(E)\left[P^{>}(E)-P^{>}(E+x)\right]P^{>}(E)^{M-2}$

The landscape

To characterize the energy landscape of the REM, we can determine the number ${\mathcal {N}}(E)dE$ of configurations having energy $E_{\alpha }\in [E,E+dE]$. The average of this number is given by

${\overline {{\mathcal {N}}(E)}}=e^{N\Sigma \left({\frac {E}{N}}\right)+o(N)},\quad \quad \Sigma (\epsilon )=\log 2-\epsilon ^{2},$

where $\Sigma (\epsilon )$ is the entropy of the model. A sketch of this function is in Fig. X. The point where the entropy vanishes, $\epsilon =-{\sqrt {\log 2}}$, is the energy density of the ground state, consistently with what we obtained with extreme values statistics. The entropy is maximal at $\epsilon =0$: the highest number of configurations have vanishing energy density.

• To derive the expression for ${\overline {{\mathcal {N}}(E)}}$, we can write ${\mathcal {N}}(E)dE=\sum _{\alpha =1}^{2^{N}}\chi _{\alpha }(E)dE$, where $\chi _{\alpha }(E)=1$ if $E_{\alpha }\in [E,E+dE]$ and $\chi _{\alpha }(E)=0$ otherwise. Use this together with $p(E)$ to obtain the entropy of the model.

• Why the point where the entropy vanishes is the ground state of the model? For $|\epsilon |>{\sqrt {\log 2}}$ the entropy is negative. This means that configurations with those energy are exponentially rare: the probability to find one is exponentially small in $N$. Do you have an idea of how to show this, using the expression for ${\overline {{\mathcal {N}}(E)}}$?

• For $|\epsilon |\leq {\sqrt {\log 2}}$ the quantity ${\mathcal {N}}(E)$ is self-averaging, meaning that its distribution concentrates around the average value ${\overline {\mathcal {N}}}(E)$ when $N\to \infty$. Show that this is the case by computing the second moment ${\overline {{\mathcal {N}}^{2}(E)}}dE$ and using the central limit theorem. Notice that this is no longer true in the region where the entropy is negative: this will be responsible of the fact that the partition function $Z$ is not self-averaging in the low-T phase, as we discuss below.

The partition function and the freezing transition

Let us compute the free energy $f$ of the REM. The partition function reads

$e^{-\beta Nf+o(N)}=Z=\sum _{\alpha =1}^{2^{N}}e^{-\beta E_{\alpha }}=\int dE\,{\mathcal {N}}(E)e^{-\beta E}$

Taking the average, we see that

${\overline {Z}}=\int dE\,{\overline {{\mathcal {N}}(E)}}e^{-\beta E}=\int _{-{\sqrt {\log 2}}}^{\sqrt {\log 2}}d\epsilon \,e^{N\left[\Sigma (\epsilon )-\beta \epsilon \right]+o(N)}$

In the limit of large $N$, this integral can be computed with the saddle point method, and one gets

${\overline {Z}}=e^{N\left[\Sigma (\epsilon ^{*})-\beta \epsilon ^{*}\right]+o(N)},\quad \quad \epsilon ^{*}={\text{argmax}}_{|\epsilon |\leq {\sqrt {\log 2}}}\left(\Sigma (\epsilon )-\beta \epsilon \right)$

Using the expression of the entropy, we see that the function is stationary at $\epsilon ^{*}=-1/2T$, which belongs to the domain of integration whenever $T\geq T_{c}=1/(2{\sqrt {\log 2}})$. This temperature identifies a transition point: for all values of $T<T_{c}$, the stationary point is outside the domain and thus $\epsilon ^{*}$ has to be chosen at the boundary of the domain, $\epsilon ^{*}=-{\sqrt {\log 2}}$.

The free energy becomes

$f=-{\frac {1}{\beta }}\lim _{N\to \infty }{\frac {\log {\overline {Z}}}{N}}={\begin{cases}\end{cases}}$

Bibliography

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