\documentclass[fleqn]{article} \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} $\log S_\infty=1+\log2\pi$. \begin{align*} \beta F= -\frac12\log S_\infty+ \frac12\lim_{n\to0}\frac1n\left(\beta^2nf(1)+\beta^2\sum_{i=0}^kn(x_i-x_{i+1})f(q_i) +\log\left[ \frac{ 1+\sum_{i=0}^k(x_i-x_{i+1})q_i }{ 1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0 } \right]\right.\\ +\frac n{x_1}\log\left[ 1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0 \right]\\ \left.+\sum_{j=1}^kn(x_{j+1}^{-1}-x_j^{-1})\log\left[ 1+\sum_{i=j+1}^k(x_i-x_{i+1})q_i-x_{j+1}q_j \right] \right) \end{align*} \begin{align*} \lim_{n\to0}\frac1n \log\left[ \frac{ 1+\sum_{i=0}^k(x_i-x_{i+1})q_i }{ 1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0 } \right] &= \lim_{n\to0}\frac1n \log\left[ \frac{ 1+\sum_{i=0}^k(x_i-x_{i+1})q_i }{ 1+\sum_{i=0}^k(x_i-x_{i+1})q_i-nq_0 } \right] \\ &=q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1} \end{align*} \begin{align*} \beta F= -\frac12\log S_\infty+ \frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i) +q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1}\right. \\ +\frac1{x_1}\log\left[ 1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i-x_1q_0 \right]\\ \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[ 1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j \right] \right) \end{align*} $q_0=0$ \begin{align*} \beta F= -\frac12\log S_\infty+ \frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i) +\frac1{x_1}\log\left[ 1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i \right]\right.\\ \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[ 1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j \right] \right) \end{align*} $x_i=\tilde x_ix_k$, $x_k=y/\beta$, $q_k=1-z/\beta$ \begin{align*} \beta F= -\frac12\log S_\infty+ \frac12\left(\beta^2f(1)+\beta^2(y\beta^{-1}-1)f(1-z\beta^{-1})+y\beta\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)\right. \\ +\frac\beta{\tilde x_1 y}\log\left[ y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta \right]\\ +\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[ y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta-y\tilde x_{j+1}q_j \right]\\ \left.-\frac\beta{\tilde x_1 y}\log\beta-\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\beta+(1-\beta y^{-1})\log\left[ z/\beta \right] \right) \end{align*} \begin{align*} \lim_{\beta\to\infty}F= \frac12\left(yf(1)+zf'(1)+y\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i) +\frac1{\tilde x_1 y}\log\left[ y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z \right]\right.\\ \left.+\sum_{j=1}^{k-1}\frac1 y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[ y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-y\tilde x_{j+1}q_j \right] -\frac1y\log z \right) \end{align*} $F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. $\tilde x_0=0$ and $\tilde x_k=1$. \begin{equation} \label{eq:ground.state.free.energy} \lim_{\beta\to\infty}F=\lim_{n\to0}\frac1n\frac12\left(nzf'(1)+y\sum_{ab}f(\tilde Q_{ab})+\frac1y\log\det(yz^{-1}\tilde Q+I) \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( \sum_a\mu(F_{aa}-R_{aa}) +\frac12\sum_{ab}\left[ \hat\epsilon^2f(Q_{ab})+2\hat\epsilon R_{ab}f'(Q_{ab}) +D_{ab}f'(Q_{ab})+R_{ab}^2f''(Q_{ab})-F_{ab}^2f''(Q_{ab}) \right]\right.\\\left. +\frac12\log\det\begin{bmatrix}Q&-iR\\-iR&-D\end{bmatrix} -\log\det F \right) \end{align*} \[ 0=\frac{\partial\Sigma}{\partial R_{ab}} =-\mu\delta_{ab}+\hat\epsilon f'(Q_{ab})+R_{ab}f''(Q_{ab})+\sum_c(R^2-DQ)^{-1}_{ac}R_{cb} \] \[ 0=\frac{\partial\Sigma}{\partial D_{ab}} =\frac12 f'(Q_{ab})-\frac12\sum_c(R^2-DQ)^{-1}_{ac}Q_{cb} \] 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 \[ 0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_d(R_d^2I-D_dQ)^{-1} =(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_dQ^{-1}f'(Q) \] and \[ Q^{-1}f'(Q)=(I+D_df'(Q))/R_d^2 \] Substituting the second into the first, we have \[ 0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+\frac1{R_d}(I+D_df'(Q)) \] \[ 0=(R_df''(1)-\mu+R_d^{-1})I+(\hat\epsilon+D_d/R_d)f'(Q) \] The only way for this equation to be satisfied off the diagonal for nontrivial $Q$ is for $D_d=-R_d\hat\epsilon$. We therefore have \begin{align*} \Sigma =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\left( n\mu(F_d-R_d)+\frac12n\left[ \hat\epsilon R_df'(1)+R_d^2f''(1)-F_d^2f''(1) \right] +\frac12\sum_{ab} \hat\epsilon^2f(Q_{ab}) ]\right.\\\left. +\frac12\log\det(\hat\epsilon R_d^{-1} Q+I) +n\log R_d -n\log F_d \right) \end{align*} Taking the saddle with respect to $\mu$ and $F_d$ yields \[ F_d=R_d \] \[ \mu=R_d^{-1}(1+R_d^2f''(1)) \] and gives \begin{align*} \Sigma =\epsilon\hat\epsilon+\hat\epsilon R_d f'(1)+\frac12D_df'(1)+\lim_{n\to0}\frac1n\frac12\left( \hat\epsilon^2\sum_{ab} f(Q_{ab}) +\log\det(-D_dR_d^{-2} Q+I) \right) \end{align*} Finally, setting $0=\Sigma$ gives \[ \epsilon =\lim_{n\to0}\frac1n\frac12\left(nR_df'(1)+\hat\epsilon\sum_{ab} f(Q_{ab}) +\frac1{\hat\epsilon}\log\det(\hat\epsilon R_d^{-1} Q+I) \right) \] which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\epsilon=y$. {\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.} \subsection{Full} \begin{align*} \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I) =x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\int_{q_0^+}^{q_{k-1}}dq\,\mu(q)\log\left[\hat\epsilon R_d^{-1}\lambda(q)+1\right] \end{align*} where \[ \mu(q)=\frac{\partial x^{-1}(q)}{\partial q} \] Integrating by parts, \begin{align*} \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I) &=x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\left[x^{-1}(q)\log[\hat\epsilon R_d^{-1}\lambda(q)+1]\right]_{q=q_0^+}^{q=q_{k-1}}-\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac{\lambda'(q)}{x(q)}\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1}\\ &=\log[\hat\epsilon R_d^{-1}\lambda(q_{k-1})+1]+\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1} \end{align*} \begin{align*} \Sigma =-\epsilon\hat\epsilon+ \frac12\hat\epsilon R_df'(1) +\frac12\int_0^1dq\,\left[ \hat\epsilon^2\lambda(q)f''(q) +\frac1{\lambda(q)+R_d/\hat\epsilon} \right] \end{align*} for $\lambda$ concave, monotonic, $\lambda(1)=0$, and $\lambda'(1)=-1$ \[ 0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\epsilon^2f''(q)-\frac12\frac1{(\lambda(q)+R_d/\hat\epsilon)^2} \] \[ \lambda^*(q)=\frac1{\hat\epsilon}\left[f''(q)^{-1/2}-R_d\right] \] We suppose that solutions are given by \begin{equation} \lambda(q)=\begin{cases} \lambda^*(q) & q