From 3d73beb37993ac2167b0f903ae4f78a32d2500d0 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Tue, 12 Jul 2022 14:55:30 +0200
Subject: Rewrote some of the concrete example section.

---
 frsb_kac-rice.tex | 69 ++++++++++++++++++++++++++++++++-----------------------
 1 file changed, 40 insertions(+), 29 deletions(-)

(limited to 'frsb_kac-rice.tex')

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 0da2e88..741cc80 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -1287,40 +1287,51 @@ exponentially many stationary points.
 
 One can construct a schematic 2RSB model from two 1RSB models.
 Consider two independent pure models with $p_1$ and $p_2$ couplings,
-respectively, with energies $H_{p_1}({\mathbf s})$ and $H_{p_2}({\mathbf
-\sigma})$ of sizes  $N$, and couple them weakly with $\varepsilon \; {\mathbf
-\sigma} \cdot {\mathbf s}$. The landscape of the pure models is much simpler
-than that of the mixed because, in these models, fixing the stability $\mu$ is
-equivalent to fixing the energy: $\mu=pE$. This implies that at each energy
-level there is only one type of stationary point. Therefore, for the pure
-models our formulas for the complexity and its Legendre transforms are
-functions of one variable only, $E$, and each instance of $\mu^*$ inside mus be
-replaced with $pE$.
+respectively, with energies $H_{p_1}({\mathbf s})$ and
+$H_{p_2}({\boldsymbol\sigma})$ of sizes  $N$, and couple them weakly with
+$\varepsilon \; {\boldsymbol \sigma} \cdot {\mathbf s}$. The landscape of the
+pure models is much simpler than that of the mixed because, in these models,
+fixing the stability $\mu$ is equivalent to fixing the energy: $\mu=pE$. This
+implies that at each energy level there is only one type of stationary point.
+Therefore, for the pure models our formulas for the complexity and its Legendre
+transforms are functions of one variable only, $E$, and each instance of
+$\mu^*$ inside must be replaced with $pE$.
 
 In the joint model, we wish to fix the total energy, not the energies of the
 individual two models. Therefore, we insert a $\delta$-function containing
 $(E_1+E_2)-E$ and integrate over $E_1$ and $E_2$. This results in a joint
 complexity (and Legendre transform)
-\begin{eqnarray}
-    e^{N\Sigma(E)}&=&\int dE_1\, dE_2\, d\lambda \, e^{N[ \Sigma_1(E_1) + \Sigma_2(E_2) + O(\varepsilon) -\lambda N [(E_1+E_2)-E]}\nonumber \\
-    e^{NG(\hat \beta)}&=&\int dE\, dE_1\, dE_2\, d\lambda\, e^{N[-\hat \beta E+ \Sigma_1(E_1) + \Sigma_2(E_2) + O(\varepsilon) -\lambda N [(E_1+E_2)-E]}
-\end{eqnarray}
-The maximum is given by $\Sigma_1'=\Sigma_2'=\hat \beta$, provided it occurs in the phase
-in which both $\Sigma_1$ and $\Sigma_2$ are non-zero. The two systems are `thermalized',
-and it is easy to see that, because many points contribute, the overlap between two 
-global configurations $$\frac1 {2N}({\mathbf s^1},{\mathbf \sigma^1})\cdot ({\mathbf s^2},{\mathbf \sigma^2})=\frac1 {2N}[ {\mathbf s^1}\cdot {\mathbf s^2}+ {\mathbf \sigma^1}\cdot {\mathbf \sigma^2}] =0$$ This is the `annealed' phase of a Kac-Rice calculation.
-
-Now start going down in energy, or up in $\hat \beta$: there will be a point $e_c$, $\hat \beta_c$
-at which one of the subsystems freezes at its lower energy density, say it is system one,
-while system two is not yet frozen. At an even higher $\hat \beta=\hat \beta_f$, both systems are frozen. 
-For $\hat \beta_f > \hat \beta> \hat \beta_c$ one system is unfrozen, while the other is,
-because of coupling, frozen at inverse temperature $\hat \beta_c$.
-The overlap between two solutions  is:
-$$\frac1 {2N}({\mathbf s^1},{\mathbf \sigma^1})\cdot ({\mathbf s^2},{\mathbf \sigma^2}) =\frac1 {2N}[ {\mathbf s^1}\cdot {\mathbf s^2}+ {\mathbf \sigma^1}\cdot {\mathbf \sigma^2}] = \frac1 {2N} {\mathbf s^1}\cdot {\mathbf s^2}   $$
-The distribution of this overlap is the one of a frozen spin-glass at temperature $\hat \beta$,  a $1RSB$ system like the Random Energy Model. The value of $x$
-corresponding to it depends on $\hat \beta$, starting at $x=1$ at $\hat \beta_c$ and decreasing wuth increasing $\hat \beta$. 
-Globally, the joint Kac-Rice system is  $1RSB$, but note that the global overlap between different states is at most $1/2$. 
-At $\hat \beta>\hat \beta_f$ there is a further transition.
+\begin{align}
+    e^{N\Sigma(E)}&=\int dE_1\, dE_2\, d\lambda \,\exp\left\{N\left[\Sigma_1(E_1) + \Sigma_2(E_2) + O(\varepsilon) -\lambda\Big((E_1+E_2)-E\Big)\right]\right\} \\
+    e^{NG(\hat \beta)}&=\int dE\, dE_1\, dE_2\, d\lambda\, \exp\left\{N\left[-\hat\beta E+\Sigma_1(E_1) + \Sigma_2(E_2) + O(\varepsilon) -\lambda\Big((E_1+E_2)-E\Big)\right]\right\}
+\end{align}
+The saddle point is given by $\Sigma_1'(E_1)=\Sigma_2'(E_2)=\hat\beta$, provided that
+both $\Sigma_1(E_1)$ and $\Sigma_2(E_2)$ are non-zero. In this situation, two systems are `thermalized',
+and, because many points contribute, the overlap between two global
+configurations is zero:
+\begin{equation}
+  \frac1{2N}({\mathbf s^1},{\boldsymbol\sigma^1})\cdot ({\mathbf s^2},{\boldsymbol\sigma^2})=\frac1 {2N}[ {\mathbf s^1}\cdot {\mathbf s^2}+ {\boldsymbol\sigma^1}\cdot {\boldsymbol\sigma^2}] =0
+\end{equation}
+This is the `annealed' phase of a Kac-Rice calculation.
+
+Now start going down in energy, or up in $\hat \beta$: there will be a point
+$E_c$ or $\hat \beta_c$ at which one of the subsystems (say it is system one)
+freezes at its lowest energy density, while system two is not yet frozen. At
+this point, $\Sigma_1(E_1)=0$ and $E_1$ is the ground state energy. At an even
+higher value $\hat \beta=\hat \beta_f$, both systems will become frozen in
+their ground states.  For $\hat \beta_f > \hat \beta> \hat \beta_c$ one system
+is unfrozen, while the other is, because of coupling, frozen at inverse
+temperature $\hat \beta_c$.  The overlap between two solutions  is:
+\begin{equation}
+  \frac1 {2N}({\mathbf s^1},{\boldsymbol\sigma^1})\cdot ({\mathbf s^2},{\boldsymbol\sigma^2}) =\frac1 {2N}[ {\mathbf s^1}\cdot {\mathbf s^2}+ {\boldsymbol\sigma^1}\cdot {\boldsymbol\sigma^2}] = \frac1 {2N} {\mathbf s^1}\cdot {\mathbf s^2}\neq0
+\end{equation}
+The distribution of this overlap is one-half the overlap distribution of a
+frozen spin-glass at temperature $\hat \beta$, a 1RSB system like the Random
+Energy Model. The value of $x$ corresponding to it depends on $\hat \beta$,
+starting at $x=1$ at $\hat \beta_c$ and decreasing with increasing $\hat
+\beta$.  Globally, the joint complexity of the system is  1RSB, but note that the
+global overlap between different states is at most $1/2$.  At $\hat \beta>\hat
+\beta_f$ there is a further transition.
 
 \subsection{\textit{R} and \textit{D}: response functions}
 
-- 
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