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| -rw-r--r-- | frsb_kac-rice.tex | 94 |
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diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index 3d06e20..600c81d 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -3,9 +3,37 @@ \usepackage{fullpage,amsmath,amssymb,latexsym} \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 + for equilibrium, as it should. +\end{abstract} +\section{Introduction} + + +Although the Bray-Moore computation for the SK model was the first application of +some replica symmetry breaking scheme, it turned out that the problem has been open +ever since. + + + +to this date the program has been only complete for a subset of models + +here we present what we believe is the general scheme + + +\section{The model} + +Here we consider, for definiteness, the `toy' model introduced by M\'ezard and Parisi + + \section{Equilibrium} +Here we review the equilibrium solution. + \begin{equation} \beta F=\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-\frac12\log S_\infty \end{equation} @@ -112,8 +140,33 @@ $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. \right) \end{equation} + + + \section{Kac-Rice} +\subsection{The replicated Kac-Rice problem} + +\begin{eqnarray} +&=& \Pi_a \delta(Eq_a) \; \Pi_a \left| \det_a( )\right| \delta(E_a-E(s_a))\nonumber\\ +&\rightarrow& \overline{\Pi_a \delta(Eq_a)} \; \overline{ \Pi_a \left| \det_a( )\right|\delta(E_a-E(s_a))}\nonumber\\ +\end{eqnarray} + +the question of independence + +all saddles versus only minima + +The parameters: +\begin{eqnarray} +Q_{ab}&=&\nonumber\\ +R_{ab}&=&\nonumber\\ +D_{ab}&=& +\end{eqnarray} + + + + +\section{Replicated action} \begin{align*} \Sigma =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\left( @@ -138,6 +191,32 @@ The second equation implies \[ (R^2-DQ)^{-1}=Q^{-1}f'(Q) \] + +\section{Replica ansatz} + +\subsection{Motivation} + +One may write +\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} +\end{equation} + +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 +suggests that only the diagonal ${\bf Q}_{aa}$ depend on the $\theta$'s. This boils down +to putting: +\begin{eqnarray} +Q_{ab}&=& {\mbox{ a Parisi matrix}}\nonumber\\ +R_{ab}&=R \delta_{ab}&\nonumber\\ +D_{ab}&=& D \delta_{ab} +\end{eqnarray} +Not surprisingly, this ansatz closes, as we shall see. + +\subsection{Solution} + + Insert the diagonal ansatz $R=R_dI$, $D=D_dI$. Then \[ 0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_d(R_d^2I-D_dQ)^{-1} @@ -195,9 +274,11 @@ Finally, setting $0=\Sigma$ gives +\frac1{\hat\epsilon}\log\det(\hat\epsilon R_d^{-1} Q+I) \right) \] -which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\epsilon=y$. Therefore, a $(k-1)$-RSB ansatz in Kac-Rice will predict the correct ground state energy for a model whose equilibrium state at small temperatures is $k$-RSB. +which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\epsilon=y$. + +{\em Therefore, a $(k-1)$-RSB ansatz in Kac-Rice will predict the correct ground state energy for a model whose equilibrium state at small temperatures is $k$-RSB.} -\section{Full} +\subsection{Full} \begin{align*} \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I) @@ -293,4 +374,13 @@ All that changes is now \right) \] + +\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{Conclusion} + + \end{document} |