LBan-1

Overview

This lesson is structured in three parts:

• Self-averaging and disorder in statistical systems

Disordered systems are characterized by a random energy landscape, however, in the thermodynamic limit, physical observables become deterministic. This property, known as self-averaging, does not always hold for the partition function which is the quantity that we can compute. When it holds the annealed average $\ln {\overline {Z}}$ and the quenched average ${\overline {\ln Z}}$ coincides otherwiese we have

${\overline {\ln Z}}<\ln {\overline {Z}}$

• The Random Energy Model

We study the Random Energy Model (REM) introduced by Bernard Derrida. In this model at each configuration is assigned an independent energy drawn from a Gaussian distribution of extensive variance. The model exhibits a freezing transition at a critical temperature​, below which the free energy becomes dominated by the lowest energy states.

• Extreme value statistics and saddle-point analysis

The results obtained from a saddle-point approximation can be recovered using the tools of extreme value statistics.

Part I

Random energy landascape

In a system with $N$ degrees of freedom, the number of configurations grows exponentially with $N$. For simplicity, consider Ising spins that take two values, $\sigma _{i}=\pm 1$, located on a lattice of size $L$ in $d$ dimensions. In this case, $N=L^{d}$ and the number of configurations is $M=2^{N}=e^{N\log 2}$.

In the presence of disorder, the energy associated with a given configuration becomes a random quantity. For instance, in the Edwards-Anderson model:

$E=-\sum _{\langle i,j\rangle }J_{ij}\sigma _{i}\sigma _{j},$

where the sum runs over nearest neighbors $\langle i,j\rangle$, and the couplings $J_{ij}$ are independent and identically distributed (i.i.d.) Gaussian random variables with zero mean and unit variance.

The energy of a given configuration is a random quantity because each system corresponds to a different realization of the disorder. In an experiment, this means that each of us has a different physical sample; in a numerical simulation, it means that each of us has generated a different set of couplings $J_{ij}$.

To illustrate this, consider a single configuration, for example the one where all spins are up. The energy of this configuration is given by the sum of all the couplings between neighboring spins:

$E[\sigma _{1}=1,\sigma _{2}=1,\ldots ]=-\sum _{\langle i,j\rangle }J_{ij}.$

Since the the couplings are random, the energy associated with this particular configuration is itself a Gaussian random variable, with zero mean and a variance proportional to the number of terms in the sum — that is, of order $N$. The same reasoning applies to each of the $M=2^{N}$ configurations. So, in a disordered system, the entire energy landscape is random and sample-dependent.

Self-averaging observables

A crucial question is whether the macroscopic properties measured on a given sample are themselves random or not. Our everyday experience suggests that they are not: materials like glass, ceramics, or bronze have well-defined, reproducible physical properties that can be reliably controlled for industrial applications.

From a more mathematical point of view, it means that the free energy $F_{N}(\beta )=Nf_{N}(\beta )$ and its derivatives (magnetization, specific heat, susceptibility, etc.), in the limit $N\to \infty$, these random quantities concentrates around a well defined value. These observables are called self-averaging. This means that,

$\lim _{N\to \infty }f_{N}(\beta )=\lim _{N\to \infty }f_{N}^{\text{typ}}(\beta )=\lim _{N\to \infty }{\overline {f_{N}(\beta )}}=f_{\infty }(\beta )$

Hence $f_{N}(\beta )$ becomes effectively deterministic and its sample-to sample fluctuations vanish in relative terms:

$\lim _{N\to \infty }{\frac {\overline {f_{N}^{2}(\beta )}}{{\overline {f_{N}(\beta )}}^{2}}}=1.$

The partition function

The partition function

$Z_{N}=\exp(-\beta Nf_{N}(\beta ))$

is itself a random variable in disordered systems. Analytical methods can capture the statistical properties of this variable. We can define to average over the disorder realizations:

• The annealed average corresponds to the calculation of the moments of the partition function. The annealed free energy is

$f^{\text{ann.}}=-{\frac {1}{\beta N}}\ln \,{\overline {Z_{N}}}$

• the quenched average corresponds to the average of the logarithm of the partition function, which is self-averaging for sure.

$f_{\infty }(\beta )\sim {\overline {f_{N}(\beta )}}=-{\overline {\ln Z_{N}(\beta )}}/(\beta N)$

Do these two averages coincide?

If the partition function is self-averaging in the thermodynamic limit, then

$\lim _{N\to \infty }Z_{N}(\beta )=\lim _{N\to \infty }Z_{N}^{\text{typ}}(\beta )=\lim _{N\to \infty }{\overline {Z_{N}(\beta )}}=e^{-\beta Nf_{\infty }(\beta )}$

As a consequence, the annealed and the quenched averages coincide.

If the partition function is not self-averaging, only typical partition function concentrates, but extremely rare configurations contribute disproportionately to its moments:

$\lim _{N\to \infty }Z_{N}^{\text{typ}}(\beta )=e^{-\beta Nf_{\infty }(\beta )}<\lim _{N\to \infty }{\overline {Z_{N}(\beta )}}=e^{-\beta Nf^{\text{ann.}}(\beta )}$

There are then two main strategies to determine the deterministic value of the observable :

• Compute directly the quenched average $f_{\infty }(\beta )=-{\overline {\ln Z_{N}(\beta )}}/(\beta N)$ using methods such as the replica trick and the Parisi solution.

• Determine the typical value $Z_{N}^{\text{typ}}(\beta )$ and evaluate $f_{\infty }(\beta )=-{\frac {1}{\beta N}}\ln Z_{N}^{\text{typ}}(\beta )$

Part II

Random Energy Model

The Random energy model (REM) neglects the correlations between the $M=2^{N}$ configurations. The energy associated to each configuration is an independent Gaussian variable with zero mean and variance $N$. The simplest solution of the model is with the microcanonical ensemble.

Microcanonical calculation

Step 1: Number of states .

Let ${\mathcal {N}}_{N}(E)dE$ the number of states of energy in the interval (E,E+dE). It is a random number and we use the representation

${\mathcal {N}}_{N}(E)dE\equiv \exp(S_{N}(E))=\sum _{\alpha =1}^{2^{N}}\chi _{\alpha }(E)dE\;$

with $\chi _{\alpha }(E)=1$ if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_\alpha \in [E, E+dE]} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi_\alpha(E)=0} otherwise. We can cumpute its average

${\overline {{\mathcal {N}}_{N}(E)}}=\sum _{\alpha =1}^{2^{N}}{\overline {\chi _{\alpha }(E)}}={\frac {2^{N}}{\sqrt {2\pi N}}}\exp \left(-{\frac {E^{2}}{2N}}\right)\sim \exp \left[N(\ln 2-\epsilon ^{2}/2)\right]$

Here Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon =E/N } is the energy density and the annealed entropy density in the thermodynamic limit is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s^{\text{ann.}}(\epsilon)=\ln 2 -\epsilon^2/2 }

Step 2: Self-averaging.

Let compute now the second moment

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{\mathcal{N}_N^2(E)} = \sum_{\alpha=1}^{2^N} \overline{\chi_\alpha} \left(\sum_{\beta\ne \alpha} \overline{\chi_\beta} \right) + \sum_{\alpha=1}^{2^N} \overline{\chi_\alpha^2} \sim \overline{ \mathcal{N}_N(E)} \left( \overline{\mathcal{N}_N(E)} - \exp\left(-\frac{E^2}{2 N}\right) \right) + \overline{\mathcal{N}_N(E)} }

We can then check the self averaging condition:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\overline{\mathcal{N}_N^2(E)}}{\overline{\mathcal{N}_N(E)}^2} \sim 1+ \frac{1}{\overline{\mathcal{N}_N(E)}}}

A critical energy density Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon^* = \sqrt{2 \ln 2}} separates a self-averaging regime for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\epsilon| < \epsilon^*} and a non self-averaging regime where for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\epsilon| > \epsilon^*} . In the first regime, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{\mathcal{N}_N(E)}} is exponentially large and its value is determinstic (average, typical, median are the same). In the secon regime, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{\mathcal{N}_N(E)}} is exponentially small but nonzero. The typical value instead is exactly zero, ${\mathcal {N}}_{N}^{\text{typ}}(E)=0$: for most disorder realizations, there are no configurations with energy below Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle - \epsilon^* N} and only a vanishingly small fraction of rare samples gives a positive contribution to the average. As a result, the quenched average on the entropy density is:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_\infty(\epsilon) = \begin{cases} \ln 2 - \dfrac{\epsilon^2}{2}, & \text{for } |\epsilon| < \epsilon^* \\ -\infty, & \text{for } |\epsilon| > \epsilon^* \end{cases} }

Back to canonical ensemble: the freezing transition

The annealed partition function is the average of the partition function over the disorder:

${\overline {Z_{N}(\beta )}}=\int _{-\infty }^{\infty }d\epsilon \;{\overline {{\mathcal {N}}_{N}(\epsilon )}}\,e^{-\beta N\epsilon }=\int _{-\infty }^{\infty }d\epsilon \;\exp \left[N(\ln 2-\epsilon ^{2}/2-\beta \epsilon )\right].$

Using the saddle point for large N we find $\epsilon _{saddle}=-\beta$ and thus

$f^{\text{ann.}}(\beta )=\ln 2/\beta +{\dfrac {\beta }{2}}$

The quenched partition function is obtained replacing the mean with the typical value:

$Z_{N}^{\text{typ.}}(\beta )=\int _{-\infty }^{\infty }d\epsilon \;{\mathcal {N}}_{N}^{\text{typ.}}(\epsilon )\,e^{-\beta N\epsilon }=\int _{-\epsilon ^{*}}^{\epsilon ^{*}}d\epsilon \;\exp \left[N(\ln 2-\epsilon ^{2}/2-\beta \epsilon )\right].$

Using the saddle point for large N we find a critical inverse temperature Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta_c = \epsilon^* = \sqrt{2 \ln 2}} separating two phases:

• For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta < \beta_c } , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_{saddle} =-\beta} and the annealed calculation works
• For $\beta >\beta _{c}$, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_{saddle} =-\beta_c} and the free energy freezes to a temperature independent value. As a result, the quenched average on the free energy density is:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_\infty(\beta) = \begin{cases} \ln 2/\beta + \dfrac{\beta}{2}, & \text{for } \beta < \beta_c \\ \sqrt{2 \ln 2}, & \text{for } \beta> \beta_c \end{cases} }

Part III

Detour: Extreme Value Statistics

Consider the REM spectrum of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M} energies Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_1, \dots, E_M} drawn from a distribution Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(E)} . It is useful to introduce the cumulative probability of finding an energy smaller than E

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(E) = \int_{-\infty}^E dx \, p(x)}

We also define:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{\min} = \min(E_1, \dots, E_M), \quad Q_M(E) \equiv \text{Prob}(E_{\min} > E) }

The statistical properties of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{\min}} are derived using two key relations:

• First relation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(E_{\min}^{\text{typ}}) = 1/M}

This is an estimation of the typical value of the minimum. It is a crucial relation that will be used frequently.

• Second relation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q_M(E) = (1-P(E))^M= e^{M \log(1 - P(E))} \sim \exp\left(-M P(E)\right) }

The first two steps are exact, but the resulting distribution depends on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M} and the precise form of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(E)} . In contrast, the last step is an approximation, valid for large Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M} and that allows one to express the random variable Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{\min}} in a scaling form: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{\min} = a_M + b_M z} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_M} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b_M} are deterministic and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M} -dependent, while Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z} is a random variable that is independent of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M} .

Back to REM

Thi chiediamo di dimostrare che per una distribuzione Gaussiana di media zero e varianza sigma^2 la cumulativa puo' essere scritta come Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(E) = \exp(A(E))} con

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A(E)= -\frac{E^2}{2 \sigma^2} - \log(\frac{\sqrt{2 \pi} |E|}{\sigma})+\ldots } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A'(E)= -\frac{E}{\sigma^2} +\ldots }

From the second relation we impose Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A(E_{\min}^{\text{typ}})=- \ln M} . For large Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M} we get:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{\min}^{\text{typ}} = -\sigma \sqrt{2 \log M} \left( 1- \frac{1}{4} \frac{\log (4 \pi \log M)}{\log M} + \ldots \right)}

We look for a scaling form of the random variable Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{\min} = a_M + b_M z} such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_M} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b_M} are deterministic and M dependent while Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z} is random and M independent.

In the Gaussian case it is better to A(a_M)=- \ln M a_M and expand Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A(E)} around this value:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q_M(E) \sim \exp\left(-M P(E)\right) \sim \exp\left(-M e^{A(a_M) +A'(a_M)\cdot (E-a_M)}\right) }

Hence setting

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b_M = \frac{1}{A'(a_M)}= \frac{ \sigma}{\sqrt{2 \log M}}}

In the REM the variance is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma^2_M = \frac{\log M}{\log 2} = N} . Then we have:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{\min} =a_M +b_M z = - \sqrt{2 \log 2}\; N +\frac{1}{2} \frac{ \log (4 \pi \log2 N)}{\sqrt{2 \log 2}} + \frac{z}{\sqrt{2 \log 2}}}

Key Observations:

• the ground state energy is self-averaging with an extensive deterministic part Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sim -\sqrt{2 \log 2} \cdot N = f_\infty(\beta>\beta_c) \cdot N } .
• Its fluctuations are very small (N independent) with a standard deviation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{2 \log 2} =\beta_c } .