T-II-3

Goal of these problems: In this set of problems, we compute the free energy of the spherical $p$-spin model in the Replica Symmetric (RS) framework, and in the 1-step Replica Symmetry Broken (1-RSB) framework.

Key concepts:

The meaning of the overlaps

In the lectures, we have introduced the order parameter

${\frac {1}{N}}\sum _{i}{\overline {\langle \sigma _{i}\rangle ^{2}}}$

Problem 3.1: 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 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} , and show that it reproduces the annealed free energy. Do you have an interpretation for this?

1. Overlpa interpretation

Problem 3.2: the 1-RSB (Replica Symmetry Broken) calculation

In the previous problem, we have chosen a certain parametrization of 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} , which corresponds to assuming that typically all the copies of the systems fall into configurations that are at 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} 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 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/m} groups of size 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} ; 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:

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

   Under which limit this reduces to the replica symmetric expression?

2. Compute the saddle point equations with respect to the parameter 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 } 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 $q_{0}=0$ is again a valid solution of these equations, and that 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 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 $m=1$ to satisfy the first equation, and look for a solution 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 \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 $\beta$, and show that there is a critical temperature 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 T_c} where a solution $q_{1}\neq 0$ appears: what is the value of this temperature (determined numerically)?