From 0ecfa0b5cc477a0f9eb62a5aae0e42aeb46bea52 Mon Sep 17 00:00:00 2001 From: Jaron Kent-Dobias Date: Sun, 5 Jun 2022 14:17:29 +0200 Subject: More work. --- frsb_kac-rice.tex | 135 +++++++++++++++++++++++++++++++++--------------------- 1 file changed, 82 insertions(+), 53 deletions(-) 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} -- cgit v1.2.3-70-g09d2