From 0ecfa0b5cc477a0f9eb62a5aae0e42aeb46bea52 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Sun, 5 Jun 2022 14:17:29 +0200
Subject: More work.

---
 frsb_kac-rice.tex | 135 +++++++++++++++++++++++++++++++++---------------------
 1 file changed, 82 insertions(+), 53 deletions(-)

(limited to 'frsb_kac-rice.tex')

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index df30c38..316ffda 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -169,13 +169,13 @@ complexity in the ground state are
 Here we review the equilibrium solution. \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical, Crisanti_2006_Spherical}
 
 \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
+  \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)
+  -\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
@@ -216,8 +216,8 @@ $\log S_\infty=1+\log2\pi$.
 
 \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)
+  -\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
@@ -230,8 +230,8 @@ $\log S_\infty=1+\log2\pi$.
 $q_0=0$
 \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)
+  -\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.\\
@@ -243,7 +243,7 @@ $q_0=0$
 $x_i=\tilde x_ix_k$, $x_k=y/\beta$, $q_k=1-z/\beta$
 \begin{align*}
   \beta F=
-  -\frac12\log S_\infty+
+  -\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
@@ -258,7 +258,7 @@ $x_i=\tilde x_ix_k$, $x_k=y/\beta$, $q_k=1-z/\beta$
 \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)
+  -\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.\\
@@ -270,7 +270,7 @@ $x_i=\tilde x_ix_k$, $x_k=y/\beta$, $q_k=1-z/\beta$
 \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)
+  \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}
 
@@ -370,45 +370,46 @@ The parameters:
 \end{align}
 
 \begin{equation}
-  S
-  =\mathcal D(\mu)+\hat\epsilon\epsilon+\lim_{n\to0}\frac1n\left(
-    \mu\sum_a^nR_{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})
-    \right]
-  +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix}
-  \right)
+  \begin{aligned}
+    S
+    =\mathcal D(\mu)+\hat\epsilon\epsilon+\lim_{n\to0}\frac1n\left(
+      \mu\sum_a^nR_{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})
+      \right] \right. \\ \left.
+    +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix}
+    \right)
+  \end{aligned}
 \end{equation}
 
 
-
 \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*}
-\[
-  0=\frac{\partial\Sigma}{\partial R_{ab}}
-  =-\mu\delta_{ab}+\hat\epsilon f'(Q_{ab})+R_{ab}f''(Q_{ab})+\sum_c(R^2-DQ)^{-1}_{ac}R_{cb}
-\]
-\[
-  0=\frac{\partial\Sigma}{\partial D_{ab}}
-  =\frac12 f'(Q_{ab})-\frac12\sum_c(R^2-DQ)^{-1}_{ac}Q_{cb}
-\]
+
+\begin{equation}
+  \begin{aligned}
+    S
+    =\mathcal D(\mu)+\hat\epsilon\epsilon+\lim_{n\to0}\frac1n\left(
+      \mu\sum_a^nR_{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})
+      \right] \right. \\ \left.
+    +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix}
+    \right)
+  \end{aligned}
+\end{equation}
+\begin{align}
+  0&=\frac{\partial S}{\partial R_{ab}}
+  =\mu\delta_{ab}+\hat\epsilon f'(Q_{ab})+R_{ab}f''(Q_{ab})+\sum_c(DQ+R^2)^{-1}_{ac}R_{cb} \\
+  0&=\frac{\partial S}{\partial D_{ab}}
+  =-\frac12 f'(Q_{ab})+\frac12\sum_c(DQ+R^2)^{-1}_{ac}Q_{cb}
+\end{align}
 The second equation implies
-\[
-  (R^2-DQ)^{-1}=Q^{-1}f'(Q)
-\]
+\begin{equation}
+  (DQ+R^2)^{-1}=Q^{-1}f'(Q)
+\end{equation}
 
 \section{Replica ansatz}
 
@@ -416,19 +417,28 @@ The second equation implies
 
 The reader who is happy with the ansatz may skip this section. 
  We may encode the original variables in a superspace variable:
-  \begin{equation}
-  \phi_i(1)= q_i(t) + \bar \theta a_i + a_i^\dag \theta + p_i \bar \theta \theta~,
-  \end{equation} 
+\begin{equation}
+  \phi_a(1)= s_a + \bar\eta_a\theta_1+\bar\theta_1\eta_a +  \hat s_a \bar \theta_1 \theta_1
+\end{equation}
 \begin{equation}
 \begin{aligned}
-{\bf Q}(1,2)&=\frac 1 N \sum_i \phi_i(1) \phi_i (2) =
-Q_{ab} + (\bar \theta_2 - \bar \theta_1)
-\theta_2  R_{ab}
-+ \bar \theta_1 \theta_1 R_{ab} + \bar\theta_1\theta_1 \bar \theta_2 \theta_2 D_{ab} \\ 
+  \mathbb Q_{a,b}(1,2)&=\frac 1 N \phi_a(1)\cdot\phi_b (2) =
+Q_{ab}  -i\left[\bar\theta_1\theta_1+\bar\theta_2\theta_2\right]  R_{ab}
+ +(\bar\theta_1\theta_2+\theta_1\bar\theta_2)F_{ab}
+ + \bar\theta_1\theta_1 \bar \theta_2 \theta_2 D_{ab} \\ 
 &+  \text{odd terms in the $\bar \theta,\theta$}~.
 \end{aligned}
 \label{Q12}
 \end{equation}
+\begin{equation}
+  \overline{\Sigma(\epsilon,\mu)}
+  =\hat\epsilon\epsilon\lim_{n\to0}\frac1n\left[
+    \mu\int d1\sum_a^n\mathbb Q_{aa}(1,1)
+    +\int d2\,d1\,\frac12\sum_{ab}^n(1+\hat\epsilon\bar\theta_1\theta_1)f(\mathbb Q_{ab}(1,2))(1+\hat\epsilon\bar\theta_2\theta_2)
+    +\frac12\operatorname{sdet}\mathbb Q
+  \right]
+\end{equation}
+
 Here $\theta_a$, $\bar \theta_a$ are Grassmann variables, and we denote the full set of coordinates 
 in a  compact form as
 $1=  \theta_1 \overline\theta_1$, $d1=   d\theta_1 d\overline\theta_1$, etc.
@@ -444,10 +454,29 @@ suggests that only the diagonal ${\bf Q}_{aa}$ depend on the $\theta$'s. This bo
 to putting:
 \begin{eqnarray}
 Q_{ab}&=& {\mbox{ a Parisi matrix}}\nonumber\\
-R_{ab}&=R \delta_{ab}&\nonumber\\
-D_{ab}&=& D \delta_{ab}
+R_{ab}&=R_d \delta_{ab}&\nonumber\\
+D_{ab}&=& D_d \delta_{ab}
 \end{eqnarray}
-Not surprisingly, this ansatz closes, as we shall see.
+Not surprisingly, this ansatz closes, as we shall see. That it closes under Hadamard products is simple.
+
+\begin{equation}
+  \begin{aligned}
+  \int d3\,\mathbb Q_1(1,3)\mathbb Q_2(3,2)
+  =\int d3\,(
+  Q_1  -i(\bar\theta_1\theta_1+\bar\theta_3\theta_3)  R_1
+   +(\bar\theta_1\theta_3+\theta_1\bar\theta_3)F_1
+ + \bar\theta_1\theta_1 \bar \theta_3 \theta_3 D_1
+ ) \\ (
+  Q_2  -i(\bar\theta_3\theta_3+\bar\theta_2\theta_2)  R_2
+   +(\bar\theta_3\theta_2+\theta_3\bar\theta_2)F_2
+ + \bar\theta_3\theta_3 \bar \theta_2 \theta_2 D_2
+ ) \\
+ =-i(Q_1R_2+R_1Q_2)
+ +Q_1D_2\bar\theta_2\theta_2+D_1Q_2\bar\theta_1\theta_1
+ -i\bar\theta_1\theta_1\bar\theta_2\theta_2R_1D_2
+ -i\bar\theta_1\theta_1\bar\theta_2\theta_2D_1R_2
+ \end{aligned}
+\end{equation}
 
 \subsection{Solution}
 
-- 
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