\documentclass[twocolumn]{article}
\usepackage{graphicx}
\begin{document}
\date{}
\title{Exercise Hydrogen}
\maketitle
\section{Gradient of Rosenbrock function}
\begin{enumerate}
\item Find the extremum of the
Rosenbrock function
\begin{equation}\label{rosen}
f(x,y) = (1-x)^2 + 100(y-x^2)^2 ,
\end{equation}
by finding the root of it's gradient (that is, the point where the
gradient is zero). You have to calculate the gradient analytically.

\item Print out the number of iterations it took the algorithm to converge.

\item Print out the points where the algorithm calculated the values of
the gradient.

\item (Optional) Make a contour plot of the Rosenbrock
function~(\ref{rosen}) together with the path the alogorithm used to
converge to the extremum.

\end{enumerate}

\section{Bound states of hydrogen atom with
shooting method for
boundary value problems}
\subsection{Introduction}
The s-wave radial Schrödinger equation for the Hydrogen atom reads
(in units "Bohr radius" and "Hartree"),

\begin{equation}
-\frac{1}{2}f'' -\frac{1}{r} f = \epsilon f \;,
\end{equation}

where $f(r)$ is the radial wave-function,
$\epsilon$ is the energy, and primes denote the derivative
over $r$.

The bound $s$-state wave-function satisfies this equation and the two
boundary conditions,
	\begin{eqnarray}
f(r \rightarrow 0) &=& r-r^2 \;, \mathrm{(prove this)} \\
f(r \rightarrow \infty) &=& 0 .
	\end{eqnarray}

These two boundary conditions can only be satisfied for certain
discrete (negative) values of the energy.

Since one cannot integrate numerically to $\infty$ one substitutes
$\infty$ with a reasonably large number, $r_{\mathrm{max}}$, such
that it is much larger than the typical size of the hydrogen atom but
still managable for the numerical inregrator
(say, $r_{\mathrm{max}} = 10$~Bohr radii),

	\begin{equation}
f(r_{\mathrm{max}})=0 .
	\end{equation}

Let $F_\epsilon(r)$ be the solution (to be found numericall
via {\tt gsl\_odeiv} to our differential equation with energy
$\epsilon$ and initial condition $F_\epsilon(r \rightarrow 0)=r-r^2$.

Generally, for a random negative $\epsilon$, this solution will not satisfy the
boundary condition at $r_{\mathrm{max}}$. It will only be satisfied when
$\epsilon$ is equal one of the bound state energies of the system.

Now define an auxiliary
function
	\begin{equation}
M(\epsilon) \equiv F_\epsilon(r_{\mathrm{max}}) \;.
	\end{equation}

The shooting method is then equivalent to finding the
(negative) root of the equation

	\begin{equation}
M(\epsilon) = 0 .
	\end{equation}

\subsection{Exercises}

\begin{enumerate}

\item Find the lowest root, $\epsilon_0$, of the equation
$M(\epsilon) = 0$ for, say, $r_{max}=8$. Plot
the resulting function and compare with the exact result (which is
$\epsilon_0=-\frac12$, $f_0(r)=r e^{-r}$ --
check this by inserting $\epsilon_0$ and $f_0(r)$
into the Schredinger equation above).

\item (Optional) Investigate the convergence of the solution towards
the exact result as function of $r_{max}$.

\item (Optional) Try also to use a more precise boundary condition for
bound states (which have $\epsilon < 0$),

	\begin{equation}
f(r \rightarrow \infty) = r e^{-kr} \;,
	\end{equation}
where $k=\sqrt{-2\epsilon}$.
This should allow you to use a smaller $r_{max}$.

\end{enumerate}
\begin{figure}
\includegraphics{../../examples/multiroot-H/plots.png}
\end{figure}

\end{document}
