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 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_0<q_1} . This corresponds to the following block structure for the overlap matrix:

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=\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, 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=3} ).

1. Using 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 \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:

   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_{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] }

   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 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 } are. Check 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 q_0=0} is again a valid solution of these equations, and that for $q_{0}=0$ the remaining equations reduce 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 (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 }

   How does one recover the paramagnetic solution?

3. We now look for a solution different from the paramagnetic one. To begin with, we set 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=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 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=3} and different values 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 \beta} , and show that there is a critical temperature $T_{c}$ where a solution 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_1 \neq 0} appears: what is the value of this temperature (determined numerically)?