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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_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 $M = \begin{pmatrix}
		a & b \\
		b & c
	\end{pmatrix}$
	\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 $d = \flatfrac{\left(c-a\right)}{2}$
		\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}
	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}
	Therefore
	\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 $\flatfrac{\lambda}{2}$.

	We see that \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 $\lambda \rightarrow 0$, $\rho(\lambda) \rightarrow 0$.
	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.
	Therefore 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} \label{subsec:offdiag}
	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 $b$, $\sqrt{2}$.
	As we see, the numerator gets added in quadrature, $\sigma_d \propto \sqrt{2} \sigma_a$, denominator rescales by $\frac12$, becomes $\sqrt{2}$.
	This exactly parallels the derivation in \cref{subsec:offdiag}.
	In fact, you can pretty easily set up a homomorphism between sums and differences of variables with distributions of Gaussians of other means and standard deviations and distributions that sum in the ways implied by these expressions.\todo{What's the nice categorical view of this stuff? Morphisms of the category of probability spaces to some measurable space.}

	\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

\end{document}