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| -rw-r--r-- | bezout.tex | 33 |
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@@ -195,7 +195,6 @@ and their conjugates. The result is, to leading order in $N$, \end{equation} where \begin{widetext} - \textcolor{red}{\textbf{[appendix?? I'm putting too much right now so as to trim later...]}} \begin{equation} f=2+\frac12\log\det\frac12\begin{bmatrix} 1 & a & d & b \\ @@ -232,9 +231,6 @@ shift, or $\rho(\lambda)=\rho_0(\lambda+p\epsilon)$. The Hessian of \partial_i\partial_jH_0 =\frac{p(p-1)}{p!}\sum_{k_1\cdots k_{p-2}}^NJ_{ijk_1\cdots k_{p-2}}z_{k_1}\cdots z_{k_{p-2}}, \end{equation} - -{\color{red} \bf here I would explain the question of the det and also of the appearance of the gap, would draw a picture of ellipse etc, and would send the reader to an appendix for most of this part of the calculation} - which makes its ensemble that of Gaussian complex symmetric matrices. Given its variances $\overline{|\partial_i\partial_j H_0|^2}=p(p-1)a^{p-2}/2N$ and $\overline{(\partial_i\partial_j H_0)^2}=p(p-1)\kappa/2N$, $\rho_0(\lambda)$ is constant inside the ellipse @@ -276,14 +272,31 @@ elements of $J$ are standard complex normal, this corresponds to a complex Wishart distribution. For $\kappa\neq0$ the problem changes, and to our knowledge a closed form is not known. We have worked out an implicit form for this spectrum using the saddle point of a replica symmetric calculation for the -Green function. {\color{red} the calculation is standard, we outline it in appendix xx} The result is +Green function. Introducing replicas to bring the partition function to +the numerator gives \begin{widetext} \begin{equation} - G(\sigma)=\lim_{n\to0}\int d\alpha\,d\chi\,d\chi^*\frac\alpha2 - \exp nN\left\{ - 1+\frac{p(p-1)}{16}a^{p-2}\alpha^2-\frac{\alpha\sigma}2+\frac12\log(\alpha^2-|\chi|^2) - +\frac p4\mathop{\mathrm{Re}}\left(\frac{(p-1)}8\kappa^*\chi^2-\epsilon^*\chi\right) - \right\} + 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 + -\Re\left(\zeta_i^{(\alpha)}\zeta_j^{(\alpha)}\partial_i\partial_jH\right) + \right] + \right\} + \end{equation} + with sums taken over repeated latin indices. + The average can then be made over $J$ and Hubbard--Stratonovich used to change + variables to replica matrices + $N\alpha_{\alpha\beta}=(\zeta^{(\alpha)})^*\cdot\zeta^{(\beta)}$ and + $N\chi_{\alpha\beta}=\zeta^{(\alpha)}\cdot\zeta^{(\beta)}$ and a series of replica + 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} + \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) + +\frac p4\mathop{\mathrm{Re}}\left(\frac{(p-1)}8\kappa^*\chi_0^2-\epsilon^*\chi_0\right) + \right]\right\} \end{equation} \end{widetext} The argument of the exponential has several saddles, but the one with the smallest value of $\mathop{\mathrm{Re}}\alpha$ gives the correct solution \textcolor{red}{\textbf{we have checked this, but a detailed analysis of the saddle-point integration is still needed to justify it.}}. |