From d95f3c5a3ddb04dd5d191715a467ab0db1cf60b3 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Fri, 19 May 2023 18:09:30 +0200
Subject: Computed the Franz-Parisi potential.

---
 2-point.tex | 214 ++++++++++++++++++++++++++++++++++++++++++------------------
 1 file changed, 151 insertions(+), 63 deletions(-)

(limited to '2-point.tex')

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}
-- 
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