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diff --git a/2-point.tex b/2-point.tex index d39f12c..6d3c437 100644 --- a/2-point.tex +++ b/2-point.tex @@ -107,12 +107,61 @@ \section{Introduction} +Many systems exhibit ``glassiness,'' characterized by rapid slowing of dynamics +over a short parameter interval. These include actual (structural) glasses, +spin glasses, certain inference and optimization problems, and more +\cite{lots}. Glassiness is qualitatively understood to arise from structure of +an energy or cost landscape, whether due to the proliferation of metastable +states, to the raising of barriers which cause effective dynamic constraints. +However, in most models there is no quantitative correspondence between these +properties and the behavior. + +In perhaps the simplest mean-field model of glasses, such a correspondence +exists and is well understood. In the pure spherical models, the dynamic +transition corresponds precisely with the energy level at which all marginal +minima are concentrated. At that level, called the \emph{threshold energy} +$E_\mathrm{th}$, slices of the landscape at fixed energy undergo a percolation +transition. In fact, this threshold energy is significant in other ways: it +attracts the long-time dynamics after quenches in temperature to below the +dynamical transition from any starting temperature. All of this can be understood in terms of the landscape structure. \cite{Biroli_1999_Dynamical} -\cite{Folena_2020_Rethinking, Folena_2021_Gradient} -\cite{Ros_2020_Distribution, Ros_2019_Complex, Ros_2019_Complexity} +In slightly less simple models, the mixed spherical models, the story changes. +There are now a range of energies with exponentially many marginal minima. It +was believed that the energy level at which these marginal minima are the most +common type of stationary point would play the same role as the threshold +energy in the pure models. However, recent work has shown that this is +incorrect. Quenches from different starting temperatures above the dynamical +transition temperature result in dynamics that approach different energy +levels, and the purported threshold does not attract the long-time dynamics in +most cases \cite{Folena_2020_Rethinking, Folena_2021_Gradient}. + +This paper studies the two-point structure of stationary points in the mixed +spherical models, or their arrangement relative to each other, previously +studied only for the pure models \cite{Ros_2019_Complexity}. This gives various +kinds of information. When one point is a minimum, we see what other kinds of +minima are nearby, and what kind of saddle points (barriers) separate them. +When both points are saddles, we see the arrangement of barriers relative to +each other, perhaps learning something about the geometry of the basins of +attraction that they surround. + +In order to address the open problem of what attracts the long-time dynamics, +we focus on the neighborhoods of the marginal minima, to see if there is +anything interesting to differentiate sets of them from each other. Though we +find rich structure in this population, their properties pivot around the +debunked threshold energy, and the apparent attractors of long-time dynamics +are not distinguished by this measure. Therefore, with respect to the problem +of dynamics this paper merely deepens the outstanding problems. + +In \S\ref{sec:model}, we introduce the mixed spherical models and discuss their +properties. In \S\ref{sec:results}, we share the main results of the paper. In +\S\ref{sec:complexity} we detail the calculation of the two-point complexity, +and in \S\ref{sec:eigenvalue} and \S\ref{sec:franz-parisi} we do the same for +the properties of the isolated eigenvalue and for the zero-temperature +Franz--Parisi potential. \section{Model} +\label{sec:model} \cite{Crisanti_1992_The, Crisanti_1993_The} @@ -252,6 +301,7 @@ and a $3+8$ model tuned to maximize the ``interesting'' region of the dynamics r \end{figure} \section{Complexity} +\label{sec:complexity} We introduce the Kac--Rice \cite{Kac_1943_On, Rice_1944_Mathematical} measure \begin{equation} @@ -618,80 +668,9 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case, E_1=E_0+\frac12\frac{f'(1)(f'''(1)+f''(1))-f''(1)^2}{f(1)(f'(1)+f''(1))-f'(1)^2}(E_0-E_0^*)(1-q)^2+O\big((1-q)^3\big) \end{equation} -\section{Franz--Parisi potential} - -\cite{Franz_1995_Recipes} - -\begin{equation} \label{eq:franz-parisi.definition} - \beta V_\beta(q\mid E_0,\mu_0) - =-\frac1N\overline{\int\frac{d\nu_H(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu_H(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\, - \log\bigg(\int d\mathbf s\,\delta\big(\|\mathbf s\|^2-N\big)\,\delta(\pmb\sigma\cdot\mathbf s-Nq)\,e^{-\beta H(\mathbf s)}\bigg)} -\end{equation} -Both the denominator and the logarithm are treated using the replica trick, which yields -\begin{equation} - \beta V_\beta(q\mid E_0,\mu_0) - =-\frac1N\lim_{\substack{m\to0\\n\to0}}\frac\partial{\partial n}\overline{\int\left(\prod_{b=1}^md\nu_H(\pmb\sigma_b,\varsigma_b\mid E_0,\mu_0)\right)\left(\prod_{a=1}^nd\mathbf s_a\,\delta(\|\mathbf s_a\|^2-N)\,\delta(\pmb \sigma_1\cdot \mathbf s_a-Nq)\,e^{-\beta H(\mathbf s_a)}\right)} -\end{equation} -\begin{equation} - \mathcal O(\mathbf t) - =\sum_a^m\delta(\mathbf t-\pmb\sigma_a)\left( - i\hat{\pmb\sigma}_a\cdot\partial_{\mathbf t}-\hat\beta_0 - \right) - -\beta - \sum_a^n\delta(\mathbf t-\mathbf s_a) -\end{equation} -\begin{equation} - \beta V_\beta(q\mid E_0,\mu_0)=-\frac1N\lim_{\substack{m\to0\\n\to0}}\frac\partial{\partial n}\int d\mathcal Q_0\,d\mathcal Q_1\,e^{Nm\mathcal S_0(\mathcal Q_0)+Nn\mathcal S_\mathrm{FP}(\mathcal Q_1)} -\end{equation} - -\begin{equation} - n\mathcal S_{\mathrm{FP}} - =\frac12\beta^2\sum_{ab}^nf(Q_{ab}) - +\beta\sum_a^m\sum_b^n\left[ - \hat\beta_0f(C^{01}_{ab}) - +R^{10}_{ab}f'(C^{01}_{ab}) - \right] - +\frac12\log\det\left( - Q-\begin{bmatrix}C^{01}\\iR^{10}\end{bmatrix}^T\begin{bmatrix}C^{00}&iR^{00}\\iR^{00}&D^{00}\end{bmatrix}^{-1}\begin{bmatrix}C^{01}\\iR^{10}\end{bmatrix} - \right) -\end{equation} - -\begin{equation} - \begin{aligned} - \beta V_\beta&=\frac12\beta^2\big[f(1)-(1-x)f(q_1)-xf(q_0)\big] - +\beta\hat\beta_0f(q)+\beta r^{10}f'(q)-\frac{1-x}x\log(1-q_1) - +\frac1x\log(1-(1-x)q_1-xq_0) \\ - &+\frac{q_0-d^{00}_df'(1)q^2-2r^{00}_df'(1)r^{10}q+(r^{10})^2f'(1)}{ - 1-(1-x)q_1-xq_0 - } - \end{aligned} -\end{equation} -The saddle point for $r^{10}$ can be taken explicitly. After this, we take the -limit of $\beta\to\infty$. There are two possibilities. First, in the replica -symmetric case $x=1$, and in the limit of large $\beta$ $q_0$ will scale like -$q_0=1-(y_0\beta)^{-1}$. Inserting this, the limit is -\begin{equation} - V_\infty^{\textsc{rs}}=-\hat\beta_0 f(q)-r^{11}_\mathrm df'(q)q-\frac12\left(y_0(1-q^2)+\frac{f'(1)^2-f'(q)^2}{y_0f'(1)}\right) -\end{equation} -The saddle point in $y_0$ can now be taken, taking care to choose the solution for $y_0>0$. This gives -\begin{equation} - V_\infty^{\textsc{rs}}=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\sqrt{(1-q^2)\left(1-\frac{f'(q)^2}{f'(1)^2}\right)} -\end{equation} -The second case is when the inner statistical mechanics problem has replica symmetry breaking. Here, $q_0$ approaches a nontrivial limit, but $x=z\beta^{-1}$ approaches zero and $q_1=1-(y_1\beta)^{-1}$ approaches one. -\begin{equation} - \begin{aligned} - V_\infty^{\oldstylenums{1}\textsc{rsb}}(q\mid E_0,\mu_0) - &=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\frac12\bigg( - z(f(1)-f(q_0))+\frac{f'(1)}{y_1}-\frac{y_1(q^2-q_0)}{1+y_1z(1-q_0)} \\ - &\hspace{8pc}-(1+y_1z(1-q_0))\frac{f'(q)^2}{y_1f'(1)}+\frac1z\log\left(1+zy_1(1-q_0)\right) - \bigg) - \end{aligned} -\end{equation} -Though the saddle point in $y_1$ can be evaluated in this expression, it -delivers no insight. The final potential is found by taking the saddle over -$z$, $y_1$, and $q_0$. \section{Isolated eigenvalue} +\label{sec:eigenvalue} The two-point complexity depends on the spectrum at both stationary points through the determinant of their Hessians, but only on the bulk of the @@ -1232,7 +1211,82 @@ solution to make sense we must have $y^2>f''(1)$. In practice, there is at most \emph{one} $y$ which produces a zero eigenvalue of $B-yC$ and satisfies this inequality, so the solution seems to be unique. +\section{Franz--Parisi potential} +\label{sec:franz-parisi} + +\cite{Franz_1995_Recipes} + +\begin{equation} \label{eq:franz-parisi.definition} + \beta V_\beta(q\mid E_0,\mu_0) + =-\frac1N\overline{\int\frac{d\nu_H(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu_H(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\, + \log\bigg(\int d\mathbf s\,\delta\big(\|\mathbf s\|^2-N\big)\,\delta(\pmb\sigma\cdot\mathbf s-Nq)\,e^{-\beta H(\mathbf s)}\bigg)} +\end{equation} +Both the denominator and the logarithm are treated using the replica trick, which yields +\begin{equation} + \beta V_\beta(q\mid E_0,\mu_0) + =-\frac1N\lim_{\substack{m\to0\\n\to0}}\frac\partial{\partial n}\overline{\int\left(\prod_{b=1}^md\nu_H(\pmb\sigma_b,\varsigma_b\mid E_0,\mu_0)\right)\left(\prod_{a=1}^nd\mathbf s_a\,\delta(\|\mathbf s_a\|^2-N)\,\delta(\pmb \sigma_1\cdot \mathbf s_a-Nq)\,e^{-\beta H(\mathbf s_a)}\right)} +\end{equation} +\begin{equation} + \mathcal O(\mathbf t) + =\sum_a^m\delta(\mathbf t-\pmb\sigma_a)\left( + i\hat{\pmb\sigma}_a\cdot\partial_{\mathbf t}-\hat\beta_0 + \right) + -\beta + \sum_a^n\delta(\mathbf t-\mathbf s_a) +\end{equation} +\begin{equation} + \beta V_\beta(q\mid E_0,\mu_0)=-\frac1N\lim_{\substack{m\to0\\n\to0}}\frac\partial{\partial n}\int d\mathcal Q_0\,d\mathcal Q_1\,e^{Nm\mathcal S_0(\mathcal Q_0)+Nn\mathcal S_\mathrm{FP}(\mathcal Q_1)} +\end{equation} + +\begin{equation} + n\mathcal S_{\mathrm{FP}} + =\frac12\beta^2\sum_{ab}^nf(Q_{ab}) + +\beta\sum_a^m\sum_b^n\left[ + \hat\beta_0f(C^{01}_{ab}) + +R^{10}_{ab}f'(C^{01}_{ab}) + \right] + +\frac12\log\det\left( + Q-\begin{bmatrix}C^{01}\\iR^{10}\end{bmatrix}^T\begin{bmatrix}C^{00}&iR^{00}\\iR^{00}&D^{00}\end{bmatrix}^{-1}\begin{bmatrix}C^{01}\\iR^{10}\end{bmatrix} + \right) +\end{equation} + +\begin{equation} + \begin{aligned} + \beta V_\beta&=\frac12\beta^2\big[f(1)-(1-x)f(q_1)-xf(q_0)\big] + +\beta\hat\beta_0f(q)+\beta r^{10}f'(q)-\frac{1-x}x\log(1-q_1) + +\frac1x\log(1-(1-x)q_1-xq_0) \\ + &+\frac{q_0-d^{00}_df'(1)q^2-2r^{00}_df'(1)r^{10}q+(r^{10})^2f'(1)}{ + 1-(1-x)q_1-xq_0 + } + \end{aligned} +\end{equation} +The saddle point for $r^{10}$ can be taken explicitly. After this, we take the +limit of $\beta\to\infty$. There are two possibilities. First, in the replica +symmetric case $x=1$, and in the limit of large $\beta$ $q_0$ will scale like +$q_0=1-(y_0\beta)^{-1}$. Inserting this, the limit is +\begin{equation} + V_\infty^{\textsc{rs}}=-\hat\beta_0 f(q)-r^{11}_\mathrm df'(q)q-\frac12\left(y_0(1-q^2)+\frac{f'(1)^2-f'(q)^2}{y_0f'(1)}\right) +\end{equation} +The saddle point in $y_0$ can now be taken, taking care to choose the solution for $y_0>0$. This gives +\begin{equation} + V_\infty^{\textsc{rs}}=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\sqrt{(1-q^2)\left(1-\frac{f'(q)^2}{f'(1)^2}\right)} +\end{equation} +The second case is when the inner statistical mechanics problem has replica symmetry breaking. Here, $q_0$ approaches a nontrivial limit, but $x=z\beta^{-1}$ approaches zero and $q_1=1-(y_1\beta)^{-1}$ approaches one. +\begin{equation} + \begin{aligned} + V_\infty^{\oldstylenums{1}\textsc{rsb}}(q\mid E_0,\mu_0) + &=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\frac12\bigg( + z(f(1)-f(q_0))+\frac{f'(1)}{y_1}-\frac{y_1(q^2-q_0)}{1+y_1z(1-q_0)} \\ + &\hspace{8pc}-(1+y_1z(1-q_0))\frac{f'(q)^2}{y_1f'(1)}+\frac1z\log\left(1+zy_1(1-q_0)\right) + \bigg) + \end{aligned} +\end{equation} +Though the saddle point in $y_1$ can be evaluated in this expression, it +delivers no insight. The final potential is found by taking the saddle over +$z$, $y_1$, and $q_0$. + \section{Conclusion} +\label{sec:conclusion} The methods developed in this paper are straightforwardly (if not easily) generalized to landscapes with replica symmetry broken complexities. |