T-II-3

Problem 3: the RS (Replica Symmetric) calculation

Let us consider the simplest possible ansatz for the structure of the matrix Q, that is the Replica Symmetric (RS) ansatz:

$Q={\begin{pmatrix}1&q_{0}&q_{0}\cdots &q_{0}\\q_{0}&1&q_{0}\cdots &q_{0}\\&\cdots &&\\q_{0}&q_{0}&q_{0}\cdots &1\end{pmatrix}}$

Under this assumption, there is a unique saddle point variable, that is $q_{0}$.

1. Check that the inverse of the overlap matrix is

   $Q^{-1}={\begin{pmatrix}\alpha &\beta &\beta \cdots &\beta \\\beta &\alpha &\beta \cdots &\beta \\&\cdots &&\\\beta &\beta &\beta \cdots &\alpha \end{pmatrix}}\quad \quad {\text{with}}\quad \alpha ={\frac {1}{1-q_{0}}}\quad {\text{and}}\quad \beta ={\frac {-1}{(1-q_{0})[1+(n-1)q_{0}]}}$

   Compute the saddle point equation for $q_{0}$ in the limit $n\to 0$, and show that this equation admits always the solution $q_{0}=0$: why is this called the paramagnetic solution?

2. Compute the free energy corresponding to the solution $q_{0}=0$, and show that it reproduces the annealed free energy. Do you have an interpretation for this?

1. Overlpa interpretation

Problem 4: the 1-RSB (Replica Symmetry Broken) calculation

In the previous problem, we have chosen a certain parametrization of the overlap matrix $Q$, which corresponds to assuming that typically all the copies of the systems fall into configurations that are at overlap $q_{0}$ 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 $n$ replicas fall into configurations that are organized in $n/m$ groups of size $m$; pairs of replicas in the same group are more strongly correlated and have overlap $q_{1}$, while pairs of replicas belonging to different groups have a smaller overlap $q_{0}<q_{1}$. This corresponds to the following block structure for the overlap matrix:

$Q={\begin{pmatrix}1&q_{1}&q_{1}&q_{0}&q_{0}\cdots &q_{0}\\q_{1}&1&q_{1}&q_{0}&q_{0}\cdots &q_{0}\\q_{1}&q_{1}&1&q_{0}&q_{0}\cdots &q_{0}\\\cdots \\\cdots \\\cdots \\q_{0}&q_{0}\cdots &q_{0}&1&q_{1}&q_{1}\\q_{0}&q_{0}\cdots &q_{0}&q_{1}&1&q_{1}\\q_{0}&q_{0}\cdots &q_{0}&q_{1}&q_{1}&1\\\end{pmatrix}}$

Here we have three parameters: $m,q_{0},q_{1}$ (in the formula above, $m=3$).

1. Using that

   $\log \det Q=n{\frac {m-1}{m}}\log(1-q_{1})+{\frac {n-m}{m}}\log[m(q_{1}-q_{0})+1-q_{1}]+\log \left[nq_{0}+m(q_{1}-q_{0})+1-q_{1}\right]$

   show that the free energy now becomes:

   $f_{1RSB}=-{\frac {1}{2\beta }}\left[{\frac {\beta ^{2}}{2}}\left(1+(m-1)q_{1}^{p}-mq_{0}^{p}\right)+{\frac {m-1}{m}}\log(1-q_{1})+{\frac {1}{m}}\log[m(q_{1}-q_{0})+1-q_{1}]+{\frac {q_{0}}{m(q_{1}-q_{0})+1-q_{1}}}\right]$

   Under which limit this reduces to the replica symmetric expression?

2. Compute the saddle point equations with respect to the parameter $q_{0},q_{1}$ and $m$ are. Check that $q_{0}=0$ is again a valid solution of these equations, and that for $q_{0}=0$ the remaining equations reduce to:

   $(m-1)\left[{\frac {\beta ^{2}}{q}}pq_{1}^{p-1}-{\frac {1}{m}}{\frac {1}{1-q_{1}}}+{\frac {1}{m}}{\frac {1}{1+(m-1)q_{1}}}\right]=0,\quad \quad {\frac {\beta ^{2}}{2}}q_{1}^{p}+{\frac {1}{m^{2}}}\log \left({\frac {1-q_{1}}{1+(m-1)q_{1}}}\right)+{\frac {q_{1}}{m[1+(m-1)q_{1}]}}=0$

   How does one recover the paramagnetic solution?

3. We now look for a solution different from the paramagnetic one. To begin with, we set $m=1$ to satisfy the first equation, and look for a solution of

   ${\frac {\beta ^{2}}{2}}q_{1}^{p}+\log \left(1-q_{1}\right)+q_{1}=0$

   Plot this function for $p=3$ and different values of $\beta$, and show that there is a critical temperature $T_{c}$ where a solution $q_{1}\neq 0$ appears: what is the value of this temperature (determined numerically)?