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 $\ln Z$ coincides otherwiese we have

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

• 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:

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[\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 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 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 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_N(\beta)=N f_N(\beta)} and its derivatives (magnetization, specific heat, susceptibility, etc.), in the limit 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 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 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_N(\beta) } becomes effectively deterministic and its sample-to sample fluctuations vanish in relative terms:

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 \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

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^{\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.

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)\sim \overline{f_N(\beta)} = - \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:

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 \lim_{N \to \infty} Z_N^{\text{typ}}(\beta) = e^{-\beta N f_\infty(\beta)} < \lim_{N \to \infty} \overline{Z_N(\beta)} = e^{-\beta N f^{\text{ann.}}(\beta)} }

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

• Compute directly the quenched average using methods such as the replica trick and the Parisi solution.

• Determine the typical 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 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 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=2^N} configurations. The energy associated to each configuration is an independent Gaussian variable with zero mean and variance 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 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

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 \mathcal{N}_N(E)dE= \sum_{\alpha=1}^{2^N} \chi_\alpha(E) dE \;}

with 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)=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

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)}= \sum_{\alpha=1}^{2^N} \overline{\chi_\alpha(E)} = \frac{2^N}{\sqrt{2 \pi N}} \exp\left(-\frac{E^2}{2 N}\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 \mathcal{N}_N(E) \left( \mathcal{N}_N(E) - \exp\left(-\frac{E^2}{2 N}\right) \right) + \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, 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 \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} }

Step 3: partition function . 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 f_\infty(\beta) = \begin{cases} \ln 2 - \dfrac{\epsilon^2}{2}, & \text{for } |\epsilon| < \epsilon^* \\ -\infty, & \text{for } |\epsilon| > \epsilon^* \end{cases} }

Part III

Back to canonical ensemble

Step 3: partition function . As a result, the quenched average on the free energy density is:

$f_{\infty }(\beta )={\begin{cases}\ln 2/\beta +{\dfrac {\beta }{2}},&{\text{for }}\beta <\beta _{c}\\{\sqrt {2\ln 2}},&{\text{for }}\beta >\beta _{c}\end{cases}}$