Adds 1.6

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