T-2: Difference between revisions

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</ol>
</ol>
<br>
<br>
<!--=== Problem 3: the RS (Replica Symmetric) calculation===
<!-- Remember however that to get the free energy, we are interested in the limit <math>n \to 0</math>. A standard way to proceed is: (i) make an ansatz on the structure of the matrix Q, (ii) compute <math>\mathcal{A}[Q]</math> within this ansatz and expand <math>\mathcal{A}[Q]= n\mathcal{A}_0 + O(n^2)</math>, (iii) perform the saddle-point calculation on <math>\mathcal{A}_0</math>. -->
Let us consider the simplest possible ansatz for the structure of the matrix Q, that is the Replica Symmetric (RS) ansatz:
<center>
<math>
Q=\begin{pmatrix}
1 & q_0 &q_0 \cdots& q_0\\
q_0 & 1 &q_0 \cdots &q_0\\
&\cdots& &\\
q_0 & q_0 &q_0 \cdots &1
\end{pmatrix}
</math>
</center>
Under this assumption, there is a unique saddle point variable, that is <math>q_0</math>.
<ol>
<li>
Check that the inverse of the overlap matrix is
<center>
<math>
Q^{-1}=\begin{pmatrix}
\alpha & \beta &\beta \cdots& \beta\\
\beta & \alpha &\beta \cdots &\beta\\
&\cdots& &\\
\beta & \beta &\beta \cdots &\alpha
\end{pmatrix} 
\quad
\quad
\text{with}
\quad
\alpha= \frac{1}{1-q_0}
\quad
\text{and}
\quad
\beta=\frac{-1}{(1-q_0)[1+(n-1)q_0]}
</math>
</center>
Compute the saddle point equation for  <math>q_0</math> in the limit  <math>n \to 0</math>, and show that this equation admits always the solution <math>q_0= 0</math>: why is this called the <em>paramagnetic</em> solution?
</li>
</ol>
<br>
<ol start="2">
<li>
Compute the free energy corresponding to the solution <math>q_0= 0</math>, and show that it reproduces the annealed free energy. Do you have an interpretation for this?
</li>
</ol>
<br>
<ol start=“3">
<li>
Overlpa interpretation
</li>
</ol>
<br>
=== Problem 4: the 1-RSB (Replica Symmetry Broken) calculation===
In the previous problem, we have chosen a certain parametrization of the overlap matrix <math>Q</math>, which corresponds to assuming that typically all the copies of the systems fall into configurations that are at overlap  <math>q_0</math> with each others, no matter what is the pair of replicas considered. This assumption is however not the good one at low temperature. We now assume a different parametrisation, that corresponds to breaking the symmetry between replicas: in particular, we assume that typically the <math>n</math> replicas fall into configurations that are organized in <math>n/m</math> groups of size <math>m</math>; pairs of replicas in the same group are more strongly correlated and have overlap <math>q_1</math>, while pairs of replicas belonging to different groups have a smaller overlap <math>q_0<q_1</math>.  This corresponds to the following block structure for the overlap matrix:
<center>
<math>
Q=\begin{pmatrix}
1 & q_1 &q_1& q_0 & q_0 \cdots& q_0\\
q_1 & 1 &q_1& q_0 & q_0 \cdots& q_0\\
q_1 & q_1 &1& q_0 & q_0 \cdots& q_0\\
\cdots\\
\cdots\\
\cdots\\
q_0 & q_0 \cdots& q_0&1 & q_1 &q_1\\
q_0 & q_0 \cdots& q_0&q_1 & 1 &q_1\\
q_0 & q_0 \cdots& q_0&q_1 & q_1 &1\\
\end{pmatrix}
</math>
</center>
Here we have three parameters: <math>m, q_0, q_1</math> (in the formula above, <math>m=3</math>).
<ol>
<li>
Using that
<center>
<math>\log \det Q=n\frac{m-1}{m} \log (1-q_1)+ \frac{n-m}{m} \log [m(q_1-q_0) + 1-q_1]+ \log \left[nq_0+ m(q_1-q_0)+ 1-q_1 \right]</math>
</center>
show that the free energy now becomes:
<center>
<math>
f_{1RSB}= - \frac{1}{2 \beta} \left[ \frac{\beta^2}{2} \left(1+ (m-1)q_1^p - m q_0^p \right)+ \frac{m-1}{m} \log (1-q_1)+ \frac{1}{m} \log [m(q_1-q_0) + 1-q_1]+ \frac{q_0}{m(q_1-q_0)+ 1-q_1} \right]
</math>
</center>
Under which limit this reduces to the replica symmetric expression?
</li>
</ol>
<br>
<ol start="2">
<li>
Compute the saddle point equations with respect to the parameter <math> q_0, q_1 </math> and <math> m </math> are. Check that <math> q_0=0</math> is again a valid solution of these equations, and that for <math> q_0=0</math> the remaining equations reduce to:
<center>
<math>
(m-1) \left[ \frac{\beta^2}{q}p q_1^{p-1}-\frac{1}{m}\frac{1}{1-q_1}+ \frac{1}{m}\frac{1}{1+ (m-1)q_1} \right]=0, \quad \quad
\frac{\beta^2}{2} q_1^p + \frac{1}{m^2}\log \left( \frac{1-q_1}{1+ (m-1)q_1}\right)+ \frac{q_1}{m [1+(m-1)q_1]}=0
</math>
</center>
How does one recover the paramagnetic solution?
</li>
</ol>
<br>
<ol start="3">
<li>
We now look for a solution different from the paramagnetic one. To begin with, we set  <math> m=1 </math> to satisfy the first equation, and look for a solution of
<center>
<math>
\frac{\beta^2}{2} q_1^p + \log \left(1-q_1\right)+ q_1=0
</math>
</center>
Plot this function for <math> p=3</math> and different values of  <math> \beta</math>, and show that there is a critical temperature <math> T_c</math> where a solution <math> q_1 \neq 0</math> appears: what is the value of this temperature (determined numerically)?
</li>
</ol>
<br>-->

Revision as of 12:24, 19 December 2023

In this set of problems, we use the replica method to study the equilibrium properties of a prototypical mean-field toy model of glasses, the spherical -spin model.


The model: spherical p-spin

In the spherical -spin model the configurations that the system can take satisfy the spherical constraint , and the energy associated to each configuration is

where the coupling constants are independent random variables with Gaussian distribution with zero mean and variance and is an integer.

Quenched vs annealed, and the replica method

In TD1, we defined the quenched free energy density as the quantity controlling the scaling of the typical value of the partition function , which means:

The annealed free energy instead controls the scaling of the average value of . It is defined by

These formulas differ by the order in which the logarithm and the average over disorder are taken. Computing the average of the logarithm is in general a hard problem, which one can address by using a smart representation of the logarithm, that goes under the name of replica trick:

which can be easily shown to be true by Taylor expanding . Applying this to the average of the partition function, we see that

Therefore, to compute the quenched free-energy we need to compute the moments and then take the limit . The annealed one only requires to do the calculation with .

Problem 2.1: the annealed free energy

  1. Energy correlations. At variance with the REM, in the spherical -spin the energies at different configurations are correlated. Show that , where
    is the overlap between the two configurations. Why can we say that for this model converges with the REM discussed in the previous lecture?
  1. Energy contribution. Show that computing boils down to computing the average , which is a Gaussian integral. Compute this average. Hint: if X is a centered Gaussian variable with variance , then .
  1. Entropy contribution. The volume of a sphere of radius in dimension is given by . Use the large-N asymptotic of this to conclude the calculation of the annealed free energy:
    This result is only slightly different with respect to the free-energy of the REM in the high-temperature phase: can you identify the source of this difference?


Problem 2.2: the replica trick and the quenched free energy

  1. Step 1: average over the disorder. By using the same Gaussian integration discussed above, show that the -th moment of the partition function is

    Justify why averaging over the disorder induces a coupling between the replicas.


  1. Step 2: identify the order parameter. Using the identity , show that can be rewritten as an integral over variables only, as:

    In the derivation, you can use the fact that , where . The matrix is an order parameter: explain the analogy with the magnetisation in the mean-field solution of the Ising model.


  1. Step 3: the saddle point (RS). For large N, the integral can be computed with a saddle point approximation for general . The saddle point variables are the matrix elements with . Show that the saddle point equations read

    To solve these equations and get the free energy, one needs to make an assumption on the structure of the matrix .