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\title{Problem 1.6}
\subtitle{Random matrix theory}
\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}}
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	\maketitle
	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}}
	\end{equation}
	Then add with its transpose to make symmetric.
	Let's look at $N = 2$, so $M = \begin{pmatrix}
		a & b \\
		b & c
	\end{pmatrix}$
	\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 $d = \flatfrac{\left(c-a\right)}{2}$
	\end{enumerate}
	\section{Solution} \label{sec:solution}
	\subsection{(b) Eigenvalue differences} \label{subsec:solb}
	\triv the eigenvalues $\nu_{\pm}$ are easy to find.
	Set up the characteristic equation.
	\begin{align}
		0 &= \left(a - \nu\right)\left(c - \nu\right) - b^2 \\
		&= (ac - b^2) - \left(a + c\right) \nu + \nu^2.
	\end{align}
	\thrf
	\begin{align}
		\nu_\pm &= \frac{\left(a + c \right) \pm \sqrt{\left(a + c\right)^2 - 4 \left(ac -b^2\right)}}{2} \\
		\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} \\
		&\propto 2 \pi \int_{(d, b)} \rho(d, b) \lambda \dd{\lambda}
	\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 $\lambda \rightarrow 0$, $\rho(\lambda) \rightarrow 0$.
	\newpage
	\listoftodos

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