From 3d52e9942d3e796cfc9f844fd9d73c4f5df72b23 Mon Sep 17 00:00:00 2001 From: "kurchan.jorge" Date: Sat, 4 Jun 2022 16:00:18 +0000 Subject: Update on Overleaf. --- frsb_kac-rice.tex | 100 ++++++++++++++++++++++++++++++++++++++++++++++++------ 1 file changed, 90 insertions(+), 10 deletions(-) (limited to 'frsb_kac-rice.tex') diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index 600c81d..2be5afc 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -1,13 +1,24 @@ \documentclass[fleqn]{article} \usepackage{fullpage,amsmath,amssymb,latexsym} +\binoppenalty=10000 +\relpenalty=10000 + + +\usepackage{float} + +% Fix \cal and \mathcal characters look (so it's not the same as \mathscr) +\DeclareSymbolFont{usualmathcal}{OMS}{cmsy}{m}{n} +\DeclareSymbolFontAlphabet{\mathcal}{usualmathcal} + + \begin{document} \title{Full solution of the Kac-Rice problem for mean-field models} \maketitle \begin{abstract} We derive the general solution for the computation of saddle points - of complex mean-field landscapes. The solution incorporates Parisi's solution + of mean-field complex landscapes. The solution incorporates Parisi's solution for equilibrium, as it should. \end{abstract} \section{Introduction} @@ -24,6 +35,49 @@ to this date the program has been only complete for a subset of models here we present what we believe is the general scheme +\subsection{What to expect?} + +In order to try to visualize what one should expect, consider two pure p-spin models, with +\begin{equation} + H = H_1 + H_2=\alpha_1 \sum_{ijk} J^1_{ijk} s_i s_j s_k + + \alpha_2 \sum_{ijk} J^2_{ijk} \bar s_i \bar s_j \bar s_k +\epsilon \sum_i s_i \bar s_i +\end{equation} +The complexity of the first and second systems in terms of $H_1$ and of $H_2$ +have, in the absence of coupling, the same dependence, but are stretched to one another: +\begin{equation} + \Sigma_1(H_1)= \Sigma_o(H_1/\alpha_1) \qquad ; \qquad \Sigma_2(H_2)= \Sigma_o(H_2/\alpha_2) +\end{equation} +Each system has a ground state energy $E_{gs}^{1,2}$, a threshold energy $E_{thres}^{1,2}$ (a well-defined notion, since we are considering pure p-spins), abd the corresponding limit values $X^{1,2}_{gs}=\left. \frac{d \Sigma_1}{dE_{1,2}}\right|_{E^{gs}_{12}}$ +and $X^{1,2}_{thres}=\left. \frac{d \Sigma_1}{dE_{1,2}}\right|_{E^{thres}_{12}}$ +Considering the cartesian product of both systems, we have, in terms of the total energy +$H=H_1+H_2$ three regimes: +\begin{itemize} +\item {\bf Unfrozen}: +\begin{eqnarray} +& & X_1 \equiv \frac{d \Sigma_1}{dE_1}= X_2 \equiv \frac{d \Sigma_2}{dE_2} + \end{eqnarray} +\item {\bf Semi-frozen} +As we go down in energy, one of the systems (say, the first) reaches its ground state, +At lower temperatures, the first system is thus frozen, while the second is not, +so that $X_1=X_1^{gs}> X_2$. The lowest energy is such that both systems are frozen. +\item {\bf Semi-threshold } As we go up from the unfrozen upwards in energy, +the second system reaches its threshold $X_2^{thres}$. At higher energies minima are extremely rare, +so the second system remains stuck at its threshold for higher energies. + +\item{\bf Both systems reach their thresholds} There essentially no more minima above that. +\end{itemize} +Consider now two combined vectors $({\bf s},{\bf \hat s})$ and $({\bf s}',{\bf \hat s}')$ +chosen at the same energies.\\ + +$\bullet$ Their normalized overlap is close to one when both subsystems are frozen, +close to a half in the semifrozen phase, and zero at all higher energies.\\ + +$\bullet$ In phases where one or both systems are stuck in their thresholds (and only in those), the +minima are exponentially subdominant with respect to saddles. + +$\bullet$ {\bf note that the same reasoning leads us to the conclusion that +minima of two total energies such that one of the systems is frozen have nonzero overlaps} + \section{The model} Here we consider, for definiteness, the `toy' model introduced by M\'ezard and Parisi @@ -145,7 +199,7 @@ $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. \section{Kac-Rice} -\subsection{The replicated Kac-Rice problem} +\subsection{The replicated problem} \begin{eqnarray} &=& \Pi_a \delta(Eq_a) \; \Pi_a \left| \det_a( )\right| \delta(E_a-E(s_a))\nonumber\\ @@ -196,15 +250,32 @@ The second equation implies \subsection{Motivation} -One may write +The reader who is happy with the ansatz may skip this section. + We may encode the original variables in a superspace variable: + \begin{equation} + \phi_i(1)= q_i(t) + \bar \theta a_i + a_i^\dag \theta + p_i \bar \theta \theta~, + \end{equation} \begin{equation} - {\bf Q}_{ab}(\bar \theta \theta, \bar \theta ' \theta')= -Q_{ab} + \bar \theta ... R_{ab} +\bar \theta \theta \bar \theta ' \theta' D_{ab} +\begin{aligned} +{\bf Q}(1,2)&=\frac 1 N \sum_i \phi_i(1) \phi_i (2) = +Q_{ab} + (\bar \theta_2 - \bar \theta_1) +\theta_2 R_{ab} ++ \bar \theta_1 \theta_1 R_{ab} + \bar\theta_1\theta_1 \bar \theta_2 \theta_2 D_{ab} \\ +&+ \text{odd terms in the $\bar \theta,\theta$}~. +\end{aligned} +\label{Q12} \end{equation} +Here $\theta_a$, $\bar \theta_a$ are Grassmann variables, and we denote the full set of coordinates +in a compact form as +$1= \theta_1 \overline\theta_1$, $d1= d\theta_1 d\overline\theta_1$, etc. +The odd and even fermion numbers decouple, so we can neglect all odd terms in $\theta,\bar{\theta}$. + + + The variables $\bar \theta \theta$ and $\bar \theta ' \theta'$ play the role of `times' in a superspace treatment. We have a long experience of -making an ansatz for replicated quantum problems. The analogy strongly +making an ansatz for replicated quantum problems, which naturally involve a (Matsubara) time. The analogy strongly suggests that only the diagonal ${\bf Q}_{aa}$ depend on the $\theta$'s. This boils down to putting: \begin{eqnarray} @@ -375,11 +446,20 @@ All that changes is now \] -\section{Overview of the landscape} -For the full model minima are always dominated exponentially by saddles, whose -index density goes smoothly down to zero with energy density. (I hope) -The meaning of the parameter $Q_{ab}$ is ???????? +\section{Ultrametricity rediscovered} + +Three states chosen at the same energy share some common information if there is some `frozen' element common to all. Suppose we choose randomly +these states but restrict to those whose overlaps +take values $Q_{12}$ and $Q_{13}$. Unlike an equilibrium situation, where the Gibbs measure allows us to find such pairs (in a FRSB case) the cost in probability of this in the present case will be exponential. +Once conditioned this way, we compute $Q_{23}= \min(Q_{12},Q_{13})$ + + + +{\tiny wild guess: +Consider two states chosen at different energies, corresponding to maximal values $q^{max}_1$, $q^{max}_2$. Their mutual overlap +should be (I am guessing this) $Q_{12}=\min(q^{max}_1, q^{max}_2)$ +} \section{Conclusion} -- cgit v1.2.3-70-g09d2