81 lines
2.6 KiB
TeX
81 lines
2.6 KiB
TeX
\documentclass{article}
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% set up telugu
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\usepackage{fontspec}
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\newfontfamily\telugufont{Potti Sreeramulu}[Script = Telugu]
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\usepackage{polyglossia}
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\setdefaultlanguage{english}
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\setotherlanguage{telugu}
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%other packages
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\usepackage{amsmath}
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\usepackage{amssymb}
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\usepackage{physics}
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\usepackage{siunitx}
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\usepackage{todonotes}
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\usepackage[plain]{fancyref}
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\usepackage{luacode}
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\usepackage{titling}
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\usepackage{enumitem}
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% custom deepak packages
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\usepackage{luatrivially}
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\usepackage{subtitling}
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\begin{luacode*}
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math.randomseed(31415926)
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\end{luacode*}
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\title{Problem 1.6}
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\subtitle{Random matrix theory}
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\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}}
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% want empty date
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\predate{}
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\date{}
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\postdate{}
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% !TeX spellcheck = en_GB
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\begin{document}
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\maketitle
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Gaussian orthogonal ensemble: $N \times N$ matrix, all elements are random numbers with Gaussian distributions of mean zero and $\sigma = 1$.
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So each member from
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\begin{equation}
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\rho(x) = \frac{1}{\sqrt{2 \pi}} e^{-\flatfrac{x^2}{2}}
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\end{equation}
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Then add with its transpose to make symmetric.
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Let's look at $N = 2$, so $M = \begin{pmatrix}
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a & b \\
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b & c
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\end{pmatrix}$
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\begin{enumerate}[label=(\alph*), start=2]
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\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 $d = \flatfrac{\left(c-a\right)}{2}$
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\end{enumerate}
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\section{Solution} \label{sec:solution}
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\subsection{(b) Eigenvalue differences} \label{subsec:solb}
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\triv the eigenvalues $\nu_{\pm}$ are easy to find.
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Set up the characteristic equation.
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\begin{align}
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0 &= \left(a - \nu\right)\left(c - \nu\right) - b^2 \\
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&= (ac - b^2) - \left(a + c\right) \nu + \nu^2.
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\end{align}
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\thrf
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\begin{align}
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\nu_\pm &= \frac{\left(a + c \right) \pm \sqrt{\left(a + c\right)^2 - 4 \left(ac -b^2\right)}}{2} \\
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\nu_\pm &= \frac{\left(a + c \right) \pm \sqrt{\left(c - a\right)^2 + 4 b^2}}{2}
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\end{align}
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So the difference is $\sqrt{\left(c - a\right)^2 + 4 b^2}$.
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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 $\flatfrac{\lambda}{2}$.
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\triv \begin{align}
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\rho(\lambda) &= \int_{(d, b)} \rho(d, b) \dd{d} \dd{b} \\
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&\propto \int_{(d, b)} \rho(d, b) \lambda \dd{\lambda} \dd{\phi} \\
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&\propto 2 \pi \int_{(d, b)} \rho(d, b) \lambda \dd{\lambda}
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\end{align}
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The $\propto$ is because some factors of $2$.
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Important point is that as long as the probability $\rho_M$ is well-enough behaved, then as $\lambda \rightarrow 0$, $\rho(\lambda) \rightarrow 0$.
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\newpage
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\listoftodos
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\end{document}
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