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diff --git a/2-point.tex b/2-point.tex index 1c11fc1..806b4dd 100644 --- a/2-point.tex +++ b/2-point.tex @@ -23,22 +23,6 @@ \usepackage{tikz} \usetikzlibrary{calc,fadings,decorations.pathreplacing,calligraphy} -\addbibresource{2-point.bib} - -\begin{document} - -\title{ - Arrangement of nearby minima and saddles in the mixed $p$-spin energy landscape -} - -\author{Jaron Kent-Dobias} -\affil{\textsc{DynSysMath}, Istituto Nazionale di Fisica Nucleare, Sezione di Roma} - -\maketitle -\begin{abstract} -\end{abstract} - - \newcommand\pgfmathsinandcos[3]{% \pgfmathsetmacro#1{sin(#3)}% \pgfmathsetmacro#2{cos(#3)}% @@ -87,20 +71,50 @@ \draw[current plane,dashed] (180-\angVis:1) arc (180-\angVis:\angVis:1); } -%% document-wide tikz options and styles - \tikzset{% >=latex, % option for nice arrows inner sep=0pt,% outer sep=2pt% } +\addbibresource{2-point.bib} + +\begin{document} + +\title{ + Arrangement of nearby minima and saddles in the mixed spherical energy landscapes +} + +\author{Jaron Kent-Dobias} +\affil{\textsc{DynSysMath}, Istituto Nazionale di Fisica Nucleare, Sezione di Roma} + +\maketitle +\begin{abstract} + The family of mixed spherical models was recently found to + violate long-held assumptions about mean-field glassy dynamics. In + particular, the threshold energy, where most stationary points are marginal + and which in the simpler pure models attracts long-time dynamics, seems to + loose significance. Here, we compute the typical distribution of stationary + points relative to each other in mixed models with a replica symmetric + complexity. We examine the stability of nearby points, accounting for the + presence of an isolated eigenvalue in their spectrum due to their proximity. + Despite finding rich structure not present in the pure models, we find + nothing that distinguishes the points that do attract the dynamics. Instead, + we find new geometric significance of the old threshold energy. +\end{abstract} + +\tableofcontents + +\section{Introduction} + +\cite{Biroli_1999_Dynamical} +\cite{Folena_2020_Rethinking, Folena_2021_Gradient} \cite{Ros_2020_Distribution, Ros_2019_Complex, Ros_2019_Complexity} \section{Model} -The mixed $p$-spin models are defined by the Hamiltonian +The mixed spherical models are defined by the Hamiltonian \begin{equation} \label{eq:hamiltonian} H(\mathbf s)=-\sum_p\frac1{p!}\sum_{i_1\cdots i_p}^NJ^{(p)}_{i_1\cdots i_p}s_{i_1}\cdots s_{i_p} \end{equation} @@ -138,6 +152,7 @@ The gradient and Hessian at a stationary point are then \end{align} where $\partial=\frac\partial{\partial\mathbf s}$ will always denote the derivative with respect to $\mathbf s$. +\section{Results} \section{Complexity} @@ -209,12 +224,11 @@ Therefore, all of the Lagrange multipliers are fixed identically to the stabilit \mathcal D(\mu) &=\int d\lambda\,\rho(\lambda+\mu)\log|\lambda| \\ &=\begin{cases} - \frac12+\log\left(\frac12\mu_\text m\right)+\frac\mu{\mu_\text m}\left(\frac\mu{\mu_\text m}-\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) - -\log\left(\frac{\mu}{\mu_\text m}-\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) & \mu>\mu_\text m \\ \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2} - & -\mu_\text m\leq\mu\leq\mu_\text m \\ - \frac12+\log\left(\frac12\mu_\text m\right)+\frac\mu{\mu_\text m}\left(\frac\mu{\mu_\text m}+\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) - -\log\left(\frac{\mu}{\mu_\text m}+\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) & \mu<-\mu_\text m + & \mu^2\leq\mu_\text m^2 \\ + \frac12+\log\left(\frac12\mu_\text m\right)+\frac{\mu^2}{\mu_\text m^2} + -\left|\frac{\mu}{\mu_\text m}\right|\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1} + -\log\left(\left|\frac{\mu}{\mu_\text m}\right|-\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) & \mu^2>\mu_\text m^2 \end{cases} \end{aligned} \end{equation} @@ -246,11 +260,11 @@ where we have introduced the linear operator \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) - + - \sum_a^n\delta(\mathbf t-\mathbf s_a)\left( + \right) + + + \sum_a^n\delta(\mathbf t-\mathbf s_a)\left( i\hat{\mathbf s}_a\cdot\partial_{\mathbf t}-\hat\beta_1 - \right) + \right) \end{equation} We have written the $H$-dependant terms in this strange form for the ease of taking the average over $H$: since it is Gaussian-correlated, it follows that \begin{equation} @@ -307,9 +321,9 @@ where &\quad+\frac1m\bigg\{ \frac12\sum_{ab}^m\left[ \hat\beta_1^2f(C^{00}_{ab})+(2\hat\beta_1R^{00}_{ab}-D^{00}_{ab})f'(C^{00}_{ab})+(R_{ab}^{00})^2f''(C_{ab}^{00}) - \right]+\frac12\log\det\begin{bmatrix}C^{00}&R^{00}\\R^{00}&D^{00}\end{bmatrix} - \bigg\} - \end{aligned} + \right]+\frac12\log\det\begin{bmatrix}C^{00}&R^{00}\\R^{00}&D^{00}\end{bmatrix} +\bigg\} +\end{aligned} \end{equation} is the action for the ordinary, one-point complexity, and remainder is given by \begin{equation} @@ -317,28 +331,28 @@ is the action for the ordinary, one-point complexity, and remainder is given by &\mathcal S(\mathcal Q_{00},\mathcal Q_{11},\mathcal Q_{01}) =\hat\beta_1E_1-r^{11}_\mathrm d\mu_1-\frac12\hat\mu_1(1-c^{11}_\mathrm d)+\mathcal D(\mu_1) \\ &\quad+\frac1n\sum_b^n\left\{-\frac12\hat\mu_{12}(q-C^{01}_{1b})+\sum_a^m\left[ - \hat\beta_0\hat\beta_1f(C^{01}_{ab})+(\hat\beta_0R^{01}_{ab}+\hat\beta_1R^{10}_{ab}-D^{01}_{ab})f'(C^{01}_{ab})+R^{01}_{ab}R^{10}_{ab}f''(C^{01}_{ab}) - \right]\right\} + \hat\beta_0\hat\beta_1f(C^{01}_{ab})+(\hat\beta_0R^{01}_{ab}+\hat\beta_1R^{10}_{ab}-D^{01}_{ab})f'(C^{01}_{ab})+R^{01}_{ab}R^{10}_{ab}f''(C^{01}_{ab}) + \right]\right\} \\ &\quad+\frac1n\bigg\{ \frac12\sum_{ab}^n\left[ \hat\beta_1^2f(C^{11}_{ab})+(2\hat\beta_1R^{11}_{ab}-D^{11}_{ab})f'(C^{11}_{ab})+(R^{11}_{ab})^2f''(C^{11}_{ab}) \right]\\ &\quad+\frac12\log\det\left( - \begin{bmatrix} - C^{11}&iR^{11}\\iR^{11}&D^{11} - \end{bmatrix}- - \begin{bmatrix} - C^{01}&iR^{01}\\iR^{10}&D^{01} - \end{bmatrix}^T - \begin{bmatrix} - C^{00}&iR^{00}\\iR^{00}&D^{00} - \end{bmatrix}^{-1} - \begin{bmatrix} - C^{01}&iR^{01}\\iR^{10}&D^{01} - \end{bmatrix} - \right) - \bigg\} + \begin{bmatrix} + C^{11}&iR^{11}\\iR^{11}&D^{11} + \end{bmatrix}- + \begin{bmatrix} + C^{01}&iR^{01}\\iR^{10}&D^{01} + \end{bmatrix}^T + \begin{bmatrix} + C^{00}&iR^{00}\\iR^{00}&D^{00} + \end{bmatrix}^{-1} + \begin{bmatrix} + C^{01}&iR^{01}\\iR^{10}&D^{01} + \end{bmatrix} + \right) + \bigg\} \end{aligned} \end{equation} Because of the structure of this problem in the twin limits of $m$ and $n$ to @@ -354,7 +368,7 @@ complexities, they have the following form at the saddle point: \begin{align} C^{01}= \begin{subarray}{l} - \hphantom{[}\begin{array}{ccc}\leftarrow&n&\rightarrow\end{array}\hphantom{\Bigg]}\\ + \hphantom{[}\begin{array}{ccc}\leftarrow&n&\rightarrow\end{array}\hphantom{\Bigg]}\\ \left[ \begin{array}{ccc} q&\cdots&q\\ @@ -365,7 +379,7 @@ complexities, they have the following form at the saddle point: \right]\begin{array}{c} \\\uparrow\\m-1\\\downarrow \end{array} - \end{subarray} +\end{subarray} && R^{01} =\begin{bmatrix} @@ -415,23 +429,23 @@ then the result is \begin{equation} \begin{aligned} & \begin{bmatrix} - C^{01}&iR^{01}\\iR^{10}&D^{01} - \end{bmatrix}^T - \begin{bmatrix} - C^{00}&iR^{00}\\iR^{00}&D^{00} - \end{bmatrix}^{-1} - \begin{bmatrix} - C^{01}&iR^{01}\\iR^{10}&D^{01} - \end{bmatrix} \\ - &\qquad=\begin{bmatrix} - q^2\tilde d_\mathrm d^{00}+2qr_{10}\tilde r^{00}_\mathrm d-r_{10}^2\tilde d^{00}_\mathrm d + C^{01}&iR^{01}\\iR^{10}&D^{01} + \end{bmatrix}^T + \begin{bmatrix} + C^{00}&iR^{00}\\iR^{00}&D^{00} + \end{bmatrix}^{-1} + \begin{bmatrix} + C^{01}&iR^{01}\\iR^{10}&D^{01} + \end{bmatrix} \\ + &\qquad=\begin{bmatrix} + q^2\tilde d_\mathrm d^{00}+2qr_{10}\tilde r^{00}_\mathrm d-r_{10}^2\tilde d^{00}_\mathrm d & i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right] \\ i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right] & d_{01}^2\tilde c^{00}_\mathrm d+2r_{01}d_{01}\tilde r^{00}_\mathrm d-r_{01}^2\tilde d^{00}_\mathrm d - \end{bmatrix} + \end{bmatrix} \end{aligned} \end{equation} where each block is a constant $n\times n$ matrix. @@ -451,7 +465,7 @@ point are well known, and take the values. &=\frac1{f'(1)} -\left( \frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} - \right)^2 + \right)^2 \end{align} $(r^{00}_\mathrm d)^2+d^{00}_\mathrm d=1/f'(1)$ @@ -477,13 +491,13 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case, \\ &+\left( (r^{01})^2-\frac{r^{11}_\mathrm d-r^{11}_0}{1-q^{11}_0}\left(2qr^{01}-\frac{(1-q^2)r^{11}_0-(q^{11}_0-q^2)r^{11}_\mathrm d}{1-q^{11}_0}\right) - \right)\big(f'(1)-f'(q_{22}^{(0)})\big) \\ + \right)\big(f'(1)-f'(q_{22}^{(0)})\big) \\ &+\frac{1-q^2}{1-q^{11}_0}+\frac{(r^{10}-qr^{00}_\mathrm d)^2}{1-q^{11}_0}f'(1) -\frac1{f'(1)}\frac{f'(1)^2-f'(q)^2}{f'(1)-f'(q^{11}_0)} +\frac{r^{11}_\mathrm d-r^{11}_0}{1-q^{11}_0}\big(r^{11}_\mathrm df'(1)-r^{11}_0f'(q^{11}_0)\big) \\ &+\log\left(\frac{1-q_{11}^0}{f'(1)-f'(q_{11}^0)}\right) - \Bigg\} + \Bigg\} \end{aligned} \end{equation} @@ -508,6 +522,75 @@ 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=-\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_0^{\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} + V_0^{\oldstylenums{1}\textsc{rsb}}=-\hat\beta_0f(q)-r^{11}_\mathrm df'(q)q-\frac12\left( + z(f(1)-f(q_0))+\frac{f'(1)}{y_1}-\frac{y_1(q^2-q_0)}{1+y_1z(1-q_0)}-(1+y_0z(1-q_0))\frac{f'(q)^2}{y_1f'(1)}+\frac1z\log\left(1+zy_1(1-q_0)\right) + \right) +\end{equation} +Though the saddle point in $y_1$ can technically 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} The two-point complexity depends on the spectrum at both stationary points @@ -1031,6 +1114,11 @@ assuming the last equation is satisfied. The trivial solution, which gives the b \] as expected. We need to first the nontrivial solutions with nonzero $X$, but because the coefficients are so nasty this will be a numeric problem... Specifically, we are good for $y$ where one of the eigenvalues of $B-yC$ is zero. +\section{Conclusion} + +The methods developed in this paper are straightforwardly (if not easily) +generalized to landscapes with replica symmetry broken complexities. + \paragraph{Acknowledgements} \paragraph{Funding information} |