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authorJaron Kent-Dobias <jaron@kent-dobias.com>2020-12-10 12:21:16 +0100
committerJaron Kent-Dobias <jaron@kent-dobias.com>2020-12-10 12:21:16 +0100
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Added some equation labels and shorted some lines.
Diffstat (limited to 'bezout.tex')
-rw-r--r--bezout.tex11
1 files changed, 7 insertions, 4 deletions
diff --git a/bezout.tex b/bezout.tex
index 67544c4..f4ebf7e 100644
--- a/bezout.tex
+++ b/bezout.tex
@@ -278,7 +278,7 @@ Green function.
Introducing replicas to bring the partition function to
the numerator of the Green function gives
\begin{widetext}
- \begin{equation}
+ \begin{equation} \label{eq:green.replicas}
G(\sigma)=\frac1N\lim_{n\to0}\int d\zeta\,d\zeta^*\,(\zeta_i^{(0)})^*\zeta_i^{(0)}
\exp\left\{
\frac12\sum_\alpha^n\left[(\zeta_i^{(\alpha)})^*\zeta_i^{(\alpha)}\sigma
@@ -295,7 +295,7 @@ the numerator of the Green function gives
vectors. Taking the replica-symmetric ansatz leaves all off-diagonal elements
and vectors zero, and $\alpha_{\alpha\beta}=\alpha_0\delta_{\alpha\beta}$,
$\chi_{\alpha\beta}=\chi_0\delta_{\alpha\beta}$. The result is
- \begin{equation}
+ \begin{equation}\label{eq:green.saddle}
\overline G(\sigma)=\lim_{n\to0}\int d\alpha_0\,d\chi_0\,d\chi_0^*\,\alpha_0
\exp\left\{nN\left[
1+\frac{p(p-1)}{16}a^{p-2}\alpha_0^2-\frac{\alpha_0\sigma}2+\frac12\log(\alpha_0^2-|\chi_0|^2)
@@ -309,8 +309,11 @@ smallest value of $\mathop{\mathrm{Re}}\alpha_0$ appears gives the correct
solution. A detailed analysis of the saddle point integration is needed to
understand why this is so. Given such $\alpha_0$, the density of singular
values follows from the jump across the cut, or
-\begin{equation}
- \rho(\sigma)=\frac1{i\pi}\left(\lim_{\mathop{\mathrm{Im}}\sigma\to0^+}\overline G(\sigma)-\lim_{\mathop{\mathrm{Im}}\sigma\to0^-}\overline G(\sigma)\right)
+\begin{equation} \label{eq:spectral.density}
+ \rho(\sigma)=\frac1{i\pi}\left(
+ \lim_{\mathop{\mathrm{Im}}\sigma\to0^+}\overline G(\sigma)
+ -\lim_{\mathop{\mathrm{Im}}\sigma\to0^-}\overline G(\sigma)
+ \right)
\end{equation}
\textcolor{red}{\textbf{Missing a factor of two? Please check...}}