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 order parameters: overlaps, and their meaning

In the lectures, we have introduced the Edwards-Anderson order parameter

$q_{EA}=\frac{1}{N}\sum _i\overline{⟨{\sigma }_i{⟩}^2}$

This quantity is a measure of ergodicity breaking: when ergodicity is broken, the system at equilibrium explores only a sub-part of the phase space; the Boltzmann measure clusters into pure states (labelled by $\alpha$) with Gibbs weight ${\omega }_{\alpha }$, meaning that one can re-write the thermal averages $⟨\cdot ⟩$ of any observable $A$ as

$⟨A⟩=\sum _{\alpha }{\omega }_{\alpha }⟨A{⟩}_{\alpha },{\omega }_{\alpha }=\frac{Z_{\alpha }}{Z},Z_{\alpha }=\underset{\overrightarrow{\sigma }\in \text{ state }\alpha }{\int }d\overrightarrow{\sigma }{e}^{-\beta E[\overrightarrow{\sigma }]}=⟨e^{-\beta E[\overrightarrow{\sigma }]}{⟩}_{\alpha }$

In the Ising model at low temperature there are two pure states, $\alpha =\pm 1$, that correspond to positive and negative magnetization. In a mean-field spin glass, there are more than two pure states. The quantity $q_{EA}$ measures the overlap between configurations belonging to the same pure state, that one expects to be the same for all states:

$q_{EA}=q_{\alpha \alpha }=\frac{1}{N}\sum _i⟨{\sigma }_i{⟩}_{\alpha }⟨{\sigma }_i{⟩}_{\alpha }.$

This non-zero in a spin-glass phase, but it is also different from zero in unfrustrated ferromagnets: in the low-T phase of an Ising model, it equals to $m^2$, where $m$ is the magnetization. One can generalize this and consider also the overlap between configurations in different pure states, and the overlap distribution:

$q_{\alpha \beta }=\frac{1}{N}\sum _i⟨{\sigma }_i{⟩}_{\alpha }⟨{\sigma }_i{⟩}_{\beta },P(q)=\sum _{\alpha ,\beta }{\omega }_{\alpha }{\omega }_{\beta }\delta (q-q_{\alpha \beta }).$

The disorder average of quantities can be computed within the replica formalism, and one finds:

$\bar{P}(q)=\underset{n\to 0}{\lim }\frac{2}{n(n-1)}\sum _{a>b}\delta (q-Q_{ab}^{SP}),q_{EA}=\max \left\{Q_{a\ne b}^{SP}\right\}$

where $Q_{ab}^{SP}$ are the saddle point values of the overlap matrix introduced in Problem 2.2. The solution of the saddle point equations for the overlap matrix $Q$ thus contain the information on whether spin glass order emerges, which corresponds to having a non-trivial distribution $\bar{P}(q)$. This distribution measures the probability that two copies of the system, equilibrating in the sae disordered environment, end up having overlap $q$. In the Ising case, a low temperature one has $q_{\alpha \alpha }=m^2$ and $q_{\alpha \ne \beta }=-m^2$, and thus $\bar{P}(q)$ has two peaks at $\pm m^2$.

Problem 3.1: the RS (Replica Symmetric) calculation

We go back to the saddle point equations for the spherical $p$-spin model derived in the previous problems. Let us consider the simplest possible ansatz for the structure of the matrix Q, that is the Replica Symmetric (RS) ansatz:

$Q=\left(\begin{array}{cccc}1& q_0& q_0\cdots & q_0\\ q_0& 1& q_0\cdots & q_0\\ & \cdots & & \\ q_0& q_0& q_0\cdots & 1\end{array}\right)$

Under this assumption, there is a unique saddle point variable, that is $q_0$. We denote with $q_0^{SP}$ its value at the saddle point.

1. Under this assumption, what is the overlap distribution $\bar{P}(q)$ and what is $q_{EA}$? In which sense the RS ansatz corresponds to assuming the existence of a unique pure state?

2. Check that the inverse of the overlap matrix is

   $Q^{-1}=\left(\begin{array}{cccc}\alpha & \beta & \beta \cdots & \beta \\ \beta & \alpha & \beta \cdots & \beta \\ & \cdots & & \\ \beta & \beta & \beta \cdots & \alpha \end{array}\right)\text{with}\alpha =\frac{1}{1-q_0}\text{and}\beta =\frac{-1}{(1-q_0)[1+(n-1)q_0]}$

   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?

3. 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?

Problem 3.2: 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:

$Q=\left(\begin{array}{cccccc}1& q_1& q_1& q_0& q_0\cdots & q_0\\ q_1& 1& q_1& q_0& q_0\cdots & q_0\\ q_1& q_1& 1& q_0& q_0\cdots & q_0\\ \cdots \\ \cdots \\ \cdots \\ q_0& q_0\cdots & q_0& 1& q_1& q_1\\ q_0& q_0\cdots & q_0& q_1& 1& q_1\\ q_0& q_0\cdots & q_0& q_1& q_1& 1\\ \end{array}\right)$

Here we have three parameters: $m,q_0,q_1$ (in the sketch above, $m=3$). We denote with $m^{SP},q_0^{SP},q_1^{SP}$ their values at the saddle point.

1. What is the overlap distribution in this case?

2. Using that

   $\log \det Q=n\frac{m-1}{m}\log (1-q_1)+\frac{n-m}{m}\log [m(q_1-q_0)+1-q_1]+\log [n{q}_0+m(q_1-q_0)+1-q_1]$

   show that the free energy now becomes:

   $f_{1RSB}=-\frac{1}{2\beta }[\frac{{\beta }^2}{2}(1+(m-1)q_1^p-m{q}_0^p)+\frac{m-1}{m}\log (1-q_1)+\frac{1}{m}\log [m(q_1-q_0)+1-q_1]+\frac{q_0}{m(q_1-q_0)+1-q_1}]$

   Under which limit this reduces to the replica symmetric expression?

3. 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:

   $(m-1)[\frac{{\beta }^2}{q}p{q}_1^{p-1}-\frac{1}{m}\frac{1}{1-q_1}+\frac{1}{m}\frac{1}{1+(m-1)q_1}]=0,\frac{{\beta }^2}{2}{q}_1^p+\frac{1}{m^2}\log \left(\frac{1-q_1}{1+(m-1)q_1}\right)+\frac{q_1}{m[1+(m-1)q_1]}=0$

   How does one recover the paramagnetic solution?

4. 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

   $\frac{{\beta }^2}{2}{q}_1^p+\log (1-q_1)+q_1=0$

   Plot this function for $p=3$ and different values of $\beta$, and show that there is a critical temperature $T_c$ where a solution $q_1\ne 0$ appears: what is the value of this temperature (determined numerically)?