Revision as of 17:02, 19 November 2023

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 $1/2$ .

Edwards Anderson order parameter

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, $M=2^{N}$ nameley independent random variables $(x_{1},...,x_{M})$ drawn from the Gaussian distribution $p(x)$ .

It is useful to use the following notations:

$P^{<}(x)=\int _{-\infty }^{x}dx'p(x')\sim {\frac {\sigma }{{\sqrt {2\pi }}|x|}}e^{-{\frac {x^{2}}{2\sigma ^{2}}}}\;$ for $x\to -\infty$ . It represents the probability to draw a number smalle than x
$P^{>}(x)=\int _{x}^{+\infty }dx'p(x')=1-P^{<}(x)$ . It represents the probabilitty to draw a number larger than x.

Extreme value stattics for Gaussian variables

We denote

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

Our goal is to compute the cumulative distribution $Q_{M}(y)={\text{Prob}}(y_{M}>$ for large M in the case of i.i.d .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})=1/M

From the second relation you can get that, in the Gaussian case:

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 =2 \sigma \sqrt{\ln M}-\frac{1}{2}\sqrt{\ln(\lnM)} +O(1) }

Close to 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 } , where P_<(y) sim 1/M we can use the limt $\lim _{M\to \int }(1+{\frac {k}{M}})^{M}=\exp(k)$ and re-write the 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 Q_M(y) \sim \exp\left(-M P^>(y)\right) We want now to give a natural definition for the number <math>a_N} 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_N} .

Consider $P^{>}({\tilde {y}})={\frac {1}{2}}$ . If you draw N independent exponential variables, how many variables (in average) will be greater than 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 \tilde y} ? Repeat the same exercise 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 \tilde \tilde y} such that $P^{>}({\tilde {\tilde {y}}})={\frac {2}{3}}$

In the large N limit, the 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 \pi(z)} becomes 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} independent.

Show that in this limit its cumulative takes the from

\Pi (z)=e^{-e^{-z}}

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 \Pi(z)= e^{-e^{-z}}}

This is the cumulative distribution of the famous Gumbel distribution.

Let us remark that the precise definition 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 a_N} and $b_{N}$ fix the mean and the variance of the rescaled 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 \pi(z)} At variance with the central limit case the mean will be different from zero and the variance different from one.

Compute the mean, the variance and the asymptotic behavior of the Gumbel distribution. Draw the distribution. Explain why 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=0} is a special point

Generic case: Universality of the Gumbel distribution

The Gumbel distribution is the limit distribution of the maxima of a large class of function. We can say that the Gumbel distribution plays, for extreme statistics, the same role of the Gaussian distribution for the central limit theorem.

By contrast the behavior 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 a_N} 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_N} as a function 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 N} strongly depend on the particular distributions 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(x)} . We discuss here a family of distribution characterized by a fast decay 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 x}

p(x)\sim ce^{-x^{\alpha }}

where $\alpha >0$ The key point is to be able to determine 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(x)} 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 P^>(x)=\exp(-A(x))}

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 p(x) = e^{- x}} shows 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(x)=x}

Otherwise 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(x)} should be determined asymptotically 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 x}

Show 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(x)=x^\alpha +(\alpha-1) \log x+...}
Show that in general 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(a_N)= \log N+...} and compute 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_N} as a function 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 \alpha } 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 N } .
Show that the maximum distribution take the 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 \lim _{N\to \infty } Q_N(y)=\left( y= a_N+ \frac{z}{A'(a_N)} \right)}

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 z } Gumbel distributed

Identify 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_N} and discuss its behavior as a function 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 \alpha }

Number

Bibliography

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

@@ Line 52: / Line 52: @@
 === 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 one associated to the <math>2^N</math> configurations.
+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 <math>M=2^N</math> configurations.
+Consider the  <math>M=2^N</math>  energies,  <math>M=2^N</math>  nameley independent random variables <math>(x_1,...,x_M)</math> drawn from the Gaussian distribution <math>p(x)</math>.
+It is useful to  use the following notations:
+* <math>P^<(x)=\int_{-\infty}^x dx' p(x')  \sim \frac{\sigma}{\sqrt{2 \pi}|x|}e^{-\frac{x^2}{2 \sigma^2}} \; </math> for  <math>x \to -\infty</math>. It  represents the probability to draw a number smalle than ''x''
+* <math> P^>(x)=\int_x^{+\infty} dx' p(x') = 1- P^<(x) </math>. It represents the probabilitty to draw a number larger than ''x''.
 ====  Extreme value stattics for Gaussian variables ====
-We consider ''N'' independent random variables <math>(x_1,...,x_N)</math> drawn from the Gaussian distribution <math>p(x)</math>.
 We denote
-<center><math>y_N=\min(x_1,...,x_N)</math></center>
+<center><math>y_M=\min(x_1,...,x_M)</math></center>
+Our goal is to compute the cumulative distribution  <math>Q_M(y)=\text{Prob}(y_M> </math> for large ''M'' in the case of i.i.d .variables.
+We need to understand two key relations:
+* the first relation is exact:
+<center><math>Q_M(y) = \left(P^<(y)\right)^M </math> </center>
+* the second relation identifies the typical value of the minimum, namely <math> a_M  </math>:
+<center><math>P_<(a_M) = 1/M </math> </center>
+From the second relation you can get that, in the Gaussian case:
+<center><math>a_M =2 \sigma \sqrt{\ln M}-\frac{1}{2}\sqrt{\ln(\lnM)} +O(1) </math> </center>
-It is useful to  use the following notations for the cumulative distribution <math>P^<(x)=\int_{-\infty}^x dx' p(x')  \sim </math> which represents the probabilitty to draw a number smalle than ''x'' and  <math> P^>(x)=\int_x^{+\infty} dx' p(x') = 1- P^<(x) </math> which represents the probabilitty to draw a number larger tthan ''x''.
+Close to <math> a_M  </math>, where P_<(y) sim 1/M we can use the limt  <math>\lim_{M\to \int} (1+\frac{k}{M})^M =\exp(k)</math> and re-write the first relation:
+  <center><math>Q_M(y) \sim \exp\left(-M  P^>(y)\right)
-Let us denote by <math>q_N(y)</math> the distribution of <math>y_N</math> and by <math>Q_N(y)=\text{Prob}(y_N>y)</math> its  cumulative distribution.
-* Write <math>Q_N(y)</math> in terms of <math>P^>(y) </math>. (Help: Start to write this relation for <math>N=2,3,...</math>).
-This is the fundamental relation of Extreme statistics and we analyze its consequences in the large ''N'' limit where, analogously to the central limit theorem, extremes statistics  display universal features.
-* In particular shows that in the  large ''N'' limit  we can write
-  <center><math>Q_N(y) \sim \exp\left(-N  P^>(y)\right) </math></center>
-Consider now the case <math>\lambda=1</math>
-* Write  <math>P^>(x)</math> and <math>P^<(x)</math>. (Remember that <math>x</math>  is a positive number.)
-* Write <math>Q_N(y)</math> and <math>q_N(y)</math>.
-* Plot <math>q_N(y)</math> for different values of ''N''.
@@ Line 78: / Line 85: @@
 <math>P^>( \tilde \tilde y)=\frac 23</math>
-* Justify that <math>a_N</math> can be estimated from
-<center><math>P^>(a_N)=\frac 1N</math></center>
-* Compute  <math>a_N</math> for the exponential distribution and justify that
-<center><math>Q_N\left(y=a_N+z\right)</math></center>
 In the large ''N'' limit, the distribution  <math>\pi(z)</math> becomes <math>N</math> independent.