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| -rw-r--r-- | 2-point.tex | 68 |
1 files changed, 5 insertions, 63 deletions
diff --git a/2-point.tex b/2-point.tex index 380d677..c5e404e 100644 --- a/2-point.tex +++ b/2-point.tex @@ -372,63 +372,12 @@ then the result is \end{equation} where each block is a constant $n\times n$ matrix. - -Define $\tilde C=C-\tilde c$, $\tilde R=R-\tilde r$, $\tilde D=D-\tilde d$. Then -\begin{align*} - &\Sigma_{12} - =\mathcal D(\mu_1)+\hat\beta_1E_1-\hat\mu_1 - +\hat\beta_0\hat\beta_1f(q)+(\hat\beta_0r_{01}+\hat\beta_1r_{10}-d_{01})f'(q)+r_{01}r_{10}f''(q) - \\& - +\lim_{n\to0}\frac1n\bigg\{ - \frac12\sum_{ab}\left[ - \hat\beta_1^2f(C_{ab})+(2\hat\beta_1R_{ab}-D_{ab})f'(C_{ab})+R_{ab}^2f''(C_{ab}) - \right] - \\& - +\hat\mu_1\operatorname{Tr}C-\mu_1\operatorname{Tr}R - +\frac12\log\det((C-\tilde c)(D-\tilde d)+(R-\tilde r)^2) - \bigg\} -\end{align*} -These equations for $D^*$ are the same as those for the unpinned case, or -\[ - 0=-\frac12f'(C)+\frac12((C-\tilde c)(D-\tilde d)+(R-\tilde r)^2)^{-1}(C-\tilde c) -\] -Solving, we get -\[ - D=\tilde d+f'(C)^{-1}-(C-\tilde c)^{-1}(R-\tilde r)^2 -\] -\begin{align*} - &\Sigma_{12} - =\mathcal D(\mu_1)+\hat\beta_1E_1-\frac12\hat\mu_1 - +\hat\beta_0\hat\beta_1f(q)+(\hat\beta_0r_{01}+\hat\beta_1r_{10}-d_{01})f'(q)+r_{01}r_{10}f''(q) - \\& - +\lim_{n\to0}\frac1n\bigg\{ - \frac12\sum_{ab}\left[ - \hat\beta_1^2f(C_{ab})+(2\hat\beta_1R_{ab}-(f'(C)^{-1}_{ab}-((C-\tilde c)^{-1}(R-\tilde r)^2)_{ab}-\tilde d))f'(C_{ab})+R_{ab}^2f''(C_{ab}) - \right] - \\& - +\frac12\hat\mu_1\operatorname{Tr}C-\mu_1\operatorname{Tr}R - +\frac12\log\det(C-\tilde c)-\frac12\log\det f'(C) - \bigg\} -\end{align*} - - - -\begin{align*} - 0&=C^*-f'(C)(C^*D^*+R^*R^*) \\ - 0&=\big[\hat\beta_1f'(C)-\mu_1I+R\odot f''(C)\big](C-\tilde c)+f'(C)(R-\tilde r) -\end{align*} - -\begin{align*} - 0&=-f'(q)+\frac12(C^*D^*+R^*R^*)^{-1}_{ij}\left( - C^*_{jk}\frac{D^*_{ki}}{d_{01}} - + - 2R^*_{jk}\frac{R^*_{ki}}{d_{01}} - \right) -\end{align*} - \section{Replica symmetric case} -With a replica-symmetric ansatz, these conditions are +We focus now on models whose equilibrium has at most one level of replica +symmetry breaking, which corresponds to a replica symmetric complexity. +For these models, the saddle point parameters for the reference stationary +point are well known, and take the values. \begin{align} \hat\beta_0 &=\frac{(\epsilon_0+\mu_0)f'(1)+\epsilon_0f''(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2}\\ @@ -441,7 +390,7 @@ With a replica-symmetric ansatz, these conditions are \right)^2 \end{align} -In the replica symmetric case, +Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case, the diagonals of the inverse block matrix from above are simple expressions: \begin{align} \tilde c_\mathrm d^{00}=\frac1{(r^{00}_\mathrm d)^2+d^{00}_\mathrm d} && \tilde r_\mathrm d^{00}=\frac{r^{00}_\mathrm d}{(r^{00}_\mathrm d)^2+d^{00}_\mathrm d} && @@ -458,13 +407,6 @@ In the replica symmetric case, \right\} \end{equation} -What about the average for the Hessian terms? - -\[ - \overline{ - |\det\operatorname{Hess}H(s_0)|\delta(\mu_0-\operatorname{Tr}\operatorname{Hess}H(s_0))|\det\operatorname{Hess}H(s_a)|\delta(\mu_1-\operatorname{Tr}\operatorname{Hess}H(s_a)) - } -\] \section{Isolated eigenvalue} |