From 3b8d6014a60030fd38e2fd1a7a416d84ae731555 Mon Sep 17 00:00:00 2001 From: Deepak Mallubhotla Date: Mon, 14 Feb 2022 19:32:46 -0600 Subject: [PATCH] Adds 1.6 --- tex/1.6.tex | 49 ++++++++++++++++++++++++++++++++++++++++++++++--- 1 file changed, 46 insertions(+), 3 deletions(-) diff --git a/tex/1.6.tex b/tex/1.6.tex index 9c0a547..ac11876 100644 --- a/tex/1.6.tex +++ b/tex/1.6.tex @@ -40,7 +40,7 @@ Gaussian orthogonal ensemble: $N \times N$ matrix, all elements are random numbers with Gaussian distributions of mean zero and $\sigma = 1$. So each member from \begin{equation} - \rho(x) = \frac{1}{\sqrt{2 \pi}} e^{-\flatfrac{x^2}{2}} + \rho_g(x) = \frac{1}{\sqrt{2 \pi}} e^{-\flatfrac{x^2}{2}} \end{equation} Then add with its transpose to make symmetric. Let's look at $N = 2$, so $M = \begin{pmatrix} @@ -49,6 +49,8 @@ \end{pmatrix}$ \begin{enumerate}[label=(\alph*), start=2] \item Show that the eigenvalue difference for $M$ is $\lambda = \sqrt{\left(c - a\right)^2 + 4 b^2} = 2 \sqrt{d^2 + b^2}$, where $d = \flatfrac{\left(c-a\right)}{2}$ + \item Calculate analytically the standard deviation of a diagonal and off-diagonal element of the GOE matrix. + Calculate the standard deviation of $d = \flatfrac{\left(c - a\right)}{2}$ of $N = 2$ ensemble, and show it equals standard deviation of $b$. \end{enumerate} \section{Solution} \label{sec:solution} \subsection{(b) Eigenvalue differences} \label{subsec:solb} @@ -64,9 +66,9 @@ \nu_\pm &= \frac{\left(a + c \right) \pm \sqrt{\left(c - a\right)^2 + 4 b^2}}{2} \end{align} So the difference is $\sqrt{\left(c - a\right)^2 + 4 b^2}$. - + We have thus $\flatfrac{\lambda^2}{4} = d^2 + b^2$, so the region of between fixed $\lambda$ and $\lambda + \Delta$ is an annulus, around the circle with radius $\flatfrac{\lambda}{2}$. - + \triv \begin{align} \rho(\lambda) &= \int_{(d, b)} \rho(d, b) \dd{d} \dd{b} \\ &\propto \int_{(d, b)} \rho(d, b) \lambda \dd{\lambda} \dd{\phi} \\ @@ -74,6 +76,47 @@ \end{align} The $\propto$ is because some factors of $2$. Important point is that as long as the probability $\rho_M$ is well-enough behaved, then as $\lambda \rightarrow 0$, $\rho(\lambda) \rightarrow 0$. + Basically because the phase space is shaped like that. + + \subsection{(c) Standard Deviations} + + \subsubsection{diagonal elements} + For the diagonal elements, say $a$, we have a double of a Gaussian variable. + \thrf for instance $\rho(a) = \rho_g(\flatfrac{a}{2})$. + \begin{align} + \rho(a) &\propto \frac{1}{\sqrt{2 \pi}} e^{-\flatfrac{a^2}{8}}, + \end{align} + which means that $\sigma_a = 2$. + + \subsubsection{off-diagonal} + For off diagonal elements, say $b$, we have two independent Gaussians being summed. + \begin{align} + \rho(b) &= \int_{-\infty}^\infty \dd{x} \rho_g(x) \rho_g(b - x) \\ + \rho(b) &= \int_{-\infty}^\infty \dd{x} \frac{1}{\sqrt{2 \pi}} e^{-\flatfrac{x^2}{2}} \frac{1}{\sqrt{2 \pi}} e^{-\flatfrac{(b -x)^2}{2}} \\ + \rho(b) &= \frac{1}{2 \pi} \int_{-\infty}^\infty \dd{x} e^{-\flatfrac{x^2}{2}} e^{-\flatfrac{(b -x)^2}{2}} \\ + \rho(b) &= \frac{1}{2 \pi} \int_{-\infty}^\infty \dd{x} e^{-\flatfrac{\left(x^2 + \left(b - x\right)^2\right)}{2}} \\ + \rho(b) &= \frac{1}{2 \pi} \int_{-\infty}^\infty \dd{x} e^{-\flatfrac{\left(2 x^2 + b^2 - 2 b x\right)}{2}} \\ + \rho(b) &= \frac{1}{2 \pi} e^{- \flatfrac{b^2}{4}} \sqrt{\pi} \\ + \rho(b) &= \frac{1}{\sqrt{2 \pi} \sqrt{2}} e^{- \flatfrac{b^2}{2 \sqrt{2}^2}}, + \end{align} + which means $\sigma_b = \sqrt{2}$. + + \subsubsection{standard deviation of difference} + We want to show that the standard deviation of $d = \flatfrac{\left(c - a\right)}{2}$ is equal to the standard deviation of $b$, $\sqrt{2}$. + \triv the numerator gets added in quadrature, $\sigma_d \propto \sqrt{2} \sigma_a$, denominator rescales by $\frac12$, becomes $\sqrt{2}$. + + \subsection{(d) Find the formula for probability distirbution of $\lambda$} + + The two boys $d$ and $b$ are independent Gaussians, with standard deviations $\sqrt{2}$. + Remember that $\lambda^2 = 4 d^2 + 4 b^2$, and that $\dd{d} \dd{b} = \frac12 \lambda \dd{\lambda} \dd{\phi}$. + \begin{align} + \rho_\lambda &= \int_M \dd{d} \dd{b} \rho_M \delta(\lambda^2 - 4b^2 - 4d^2) \\ + \rho_\lambda &= \pi \int_M \dd{\lambda} \lambda \rho_M \delta(\lambda^2 - 4b^2 - 4d^2) \\ + \rho_\lambda &= 2 \pi \frac{1}{\sqrt{4 \pi}} \frac{1}{\sqrt{4 \pi}} \int_M \dd{\lambda} e^{-\flatfrac{d^2}{4}} e^{-\flatfrac{b^2}{4}} \delta(\lambda^2 - 4b^2 - 4d^2) \\ + \rho_\lambda &= 2 \pi \frac{1}{\sqrt{4 \pi}} \frac{1}{\sqrt{4 \pi}} \frac{1}{4} \int_M \dd{\lambda} \lambda e^{-\flatfrac{\lambda^2}{16}} \delta(\lambda^2 - 4b^2 - 4d^2) + \end{align} + And our integrand is as expected. + \newpage \listoftodos