Rearranged and updated FRSB section.

1 files changed, 39 insertions, 43 deletions

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 52e65bd..a285c97 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -591,6 +591,44 @@ complexity in the ground state are
   d_d=\hat\beta r_d
 \end{align}
 
+\subsection{Full RSB}
+
+This reasoning applies equally well to FRSB systems.  Using standard
+manipulations (Appendix B), one finds also a continuous version of the
+supersymmetric complexity
+\begin{equation} \label{eq:functional.action}
+    \Sigma(E,\mu)
+    =\mathcal D(\mu)
+    +
+      \hat\beta E-\mu r_d
+      +\frac12\hat\beta r_df'(1)+\frac12r_d^2f''(1)+\frac12\log r_d^2
+      +\frac12\int_0^1dq\,\left(
+        \hat\beta^2f''(q)\chi(q)+\frac1{\chi(q)+r_d/\hat\beta}
+      \right)
+\end{equation}
+where $\chi(q)=\int_1^qdq'\int_0^{q'}dq''\,P(q)$, as in the equilibrium case.
+Though the supersymmetric solution leads to a nice tractable expression, it
+turn out to be useful only at one point of interest: the ground state. Indeed,
+we know from the equilibrium that in the ground state $\chi$ is continuous in
+the whole range of $q$. Therefore, the saddle solution found by extremizing
+\begin{equation}
+  0=\frac{\delta\Sigma}{\delta\chi(q)}=\frac12\hat\beta^2f''(q)-\frac12\frac1{(\chi(q)+r_d/\hat\beta)^2}
+\end{equation}
+given by
+\begin{equation}
+  \chi(q)=\frac1{\hat\beta}\left(f''(q)^{-1/2}-r_d\right)
+\end{equation}
+is correct. This is only correct if it satisfies the boundary condition
+$\chi(1)=0$, which requires $r_d=f''(1)^{-1/2}$. This in turn implies
+$\mu=\frac1{r_d}+f''(1)r_d=\sqrt{4f''(1)}=\mu_m$. Therefore, the FRSB ground state
+is exactly marginal! It is straightforward to check that these conditions are
+indeed a saddle of the complexity.
+
+This has several implications. First, other than the ground state, there are
+\emph{no} energies at which minima are most numerous; saddles always dominante.
+As we will see, stable minima are numerous at energies above the ground state,
+but these vanish at the ground state.
+
 \section{Examples}
 
 \subsection{1RSB complexity}
@@ -675,21 +713,7 @@ E\rangle_2$.
 \end{figure}
 
 \subsection{Full RSB complexity}
-Using standard manipulations (Appendix B), one finds also a continuous version
-\begin{equation} \label{eq:functional.action}
-  \begin{aligned}
-    \Sigma(E,\mu)
-    =\mathcal D(\mu)
-    +
-      \hat\beta E-\mu R_d
-      +\frac12(2\hat\beta R_d-D_d)f'(1)+\frac12R_d^2f''(1)+\frac12\log R_d^2
-      \\
-      +\frac12\int_0^1dq\,\left(
-        \hat\beta^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d}
-      \right)
-  \end{aligned}
-\end{equation}
-where $\chi(q)$ is defined with respect to $Q$ exactly as in the equilibrium case.
+
 
 \begin{figure}
   \centering
@@ -738,34 +762,6 @@ where $\chi(q)$ is defined with respect to $Q$ exactly as in the equilibrium cas
   } \label{fig:2rsb.comparison}
 \end{figure}
 
-In the case where any FRSB is present, one must work with the functional form
-of the complexity \eqref{eq:functional.action}, which must be extremized with
-respect to $\chi$ under the conditions that $\chi$ is concave, monotonically
-decreasing, and $\chi(1)=0$, $\chi'(1)=-1$. The annealed case is found by
-taking $\chi(q)=1-q$, which satisfies all of these conditions. $k$-RSB is
-produced by breaking $\chi$ into $k+1$ piecewise linear segments.
-
-Forget for the moment these tricky requirements. The function would then be
-extremized by satisfying
-\begin{equation}
-  0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\beta^2f''(q)-\frac12\frac1{(\lambda(q)+R_d^2/D_d)^2}
-\end{equation}
-which implies the solution
-\begin{equation}
-  \lambda^*(q)=\frac1{\hat\beta}f''(q)^{-1/2}-\frac{R_d^2}{D_d}
-\end{equation}
-If $f''(q)^{-1/2}$ is not concave anywhere, there is little use of this
-solution. However, if it is concave everywhere it may constitute a portion of
-the full solution.
-
-We suppose that solutions are given by
-\begin{equation}
-  \lambda(q)=\begin{cases}
-    \lambda^*(q) & q<q_\textrm{max} \\
-    1-q          & q\geq q_\textrm{max}
-  \end{cases}
-\end{equation}
-Continuity requires that $1-q_\textrm{max}=\lambda^*(q_\textrm{max})$.
 
 Here a picture of $\chi$ vs $C$ or $X$ vs $C$ showing limits $q_{max}$, $x_{max}$
 for different energies and typical vs minima.