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\begin{center}
\Large \bf
Numerieke methoden in de natuurwetenschappen\\
\large 
\end{center}
\vspace{0.4cm}
\noindent 
{\bf Ex.~1: Bound states in a finite square well}\\

Consider a square potential well of finite height V$_0$
and width $2a$, as in figure ~\ref{fig:put}. 


\begin{figure}[hbt]
 \centering
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\put(6001,-6661){\small 0}
\put(3901,-6661){\large $-a$}
\put(7801,-6661){\large $a$}
\put(6226,-1411){\large $V(x)$}
\put(6301,-2761){\large $V_{0}$}
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 \caption[1]{Finite potential well}
 \label{fig:put}

\end{figure}

The Schr\"odinger equation:
\[
\frac{d^2\Psi}{dx^2}-\frac{2m}{\hbar^2}\left[V(x)-E\right]\Psi=0
\]

with $V(x)$ given by:
\begin{eqnarray*}
V(x)&=&V_0 \qquad x<-a \mbox{~~~~~~~~~~~~~~~~~~~~~region $I$~~~}\\
V(x)&=&0 \qquad -a<x<a \mbox{~~~~~~~~~~~~~~~~region $II$~~}\\
V(x)&=&V_0 \qquad x>a \mbox{~~~~~~~~~~~~~~~~~~~~~~~~region $III$~}
\end{eqnarray*}

has general solutions for the bound states at energy less than $V_0$
in the three regions given by:
\begin{eqnarray*}
\Psi_{I}(x)&=&A~\exp^{\beta x}+B~\exp^{-\beta x} \qquad x<-a\\
\Psi_{II}(x)&=&C~\sin\alpha x+D~\cos\alpha x \qquad -a<x<a\\
\Psi_{III}(x)&=&E~\exp^{\beta x}+F~\exp^{-\beta x} \qquad x>a\\[1ex]
\alpha&=&\sqrt{\frac{2mE}{\hbar^2}}\\
\beta&=&\sqrt{2m(V_0-E)/\hbar^2}
\end{eqnarray*}

The boundary conditions require continuity of the wavefunction and of 
its derivative at the boundary, whence   
\\[1ex]
Even states: $B=E=0, C=0, D\neq0, A=F \qquad \alpha \tan \alpha a = \beta$\\
Odd states: $B=E=0 , C\neq0, D=0, A=-F \qquad \alpha \cot \alpha a = -\beta$\\

Therefore eigenvalues of the Schr\"odinger equation are found by solving
numerically:
\begin{center}
\begin{tabular}{ccc}
$\alpha tg(\alpha a)-\beta$=0 &for even states\\
$\alpha ctg(\alpha a)+\beta$=0 &for odd  states
\end{tabular} 
\end{center}

A more convenient form is:

\begin{center}
\begin{tabular}{ccc}
$\alpha sin(\alpha a)-\beta cos(\alpha a)$=0 &for even states\\
$\alpha cos(\alpha a)+\beta sin(\alpha a)$=0 &for odd  states
\end{tabular} 
\end{center}


Use $2a$=10~\AA, $m=1~m_e$. 
It is useful to use $\hbar^2=7.6199682 m_e$~eV~\AA$^2$   

\begin{enumerate}
\item The repetitive nature of the tangent suggests that there might be 
several solutions, one per each cycle of the function. Establish the 
intervals in which roots exist. Remember that, in one dimension,  a square well
has always at least one bound state at $E\neq 0$.
\item Find all bound states with precision 10$^{-4}$~eV
by use of the bisection, Newton and secant
methods for several values of $V_0$ between $V_0$=0.2~eV and $V_0$=5~eV. 
\item 
Make a plot of the 
eigenvalues as a function of $V_0$, comparing also to the limiting case of an 
infinite well: 

\begin{equation}
E_n= {\hbar^2\over 2m_e}({n\pi\over 2a})^2. 
\end{equation}
Do not forget to label the axis giving the used units and for which values of 
the parameters  (e.g which value of $a$) the calculation is done. 
The idea is that, even without the text of the exercise, by reading the 
report one can understand and reproduce the calculation. 
Comment briefly the results.
If you use gnuplot, you could plot the eigenvalues with linespoints so that you can trace back for which values of $V_0$ the calculations has been performed.
\item Facultative: Make a plot of the eigenvalues as a function of 
$a$ for $V_0$=3 eV.
%calculate the wavefunction corresponding to the first bound 
%state for two values of $V_0$ and comment the results.
%\item Could you find out which is the value of $V_0$ which gives the first 3
%bound states at energy $E_1$=0.26230 $E_2$=1.03633  $E_3$= 2.27006 eV?
\end{enumerate}
\footnote{ More details on this problem and on computational strategy can be 
found in: P.L. De Vries {\it A first course in computational Physics},
J. Wiley \& sons, N.Y. 1994}
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