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\begin{document}

\section{Scattering and Barrier Penetration in 1-D}

\subsection{Governing Equation}

We wish to examine the interaction of a free particle with a potential
in one dimension.  To do this we need the scaled, time-independent,
one-dimensional Schr\"{o}dinger equation for the wavefunction $\psi$:
\begin{equation}
     \frac{\partial^2 \psi}{\partial x^2} +
     2 (\epsilon - u) \psi = 0 ,
\label{scaled}
\end{equation}
where $\epsilon = mL^2 E / \hbar^2$, $u = mL^2U/ \hbar^2$, and $L$ is
the assumed $x$ scaling factor.  Notice that since the second
derivative of $\psi$ is the curvature of $\psi (x)$, this equation has
an oscillatory character if $\epsilon > u$ and an exponential
character if $\epsilon < u$.

The free wave solution to equation (\ref{scaled}), i. e., that solution
which applies where $u = 0$, is simply
\begin{equation}
     \psi = \psi_L \exp (-ik_f x) + \psi_R \exp (ik_f x) ,
\end{equation}
where the wavenumber $k_f $ is given by
\begin{equation}
     k_f  = (2 \epsilon)^{1/2} .
\end{equation}
The amplitude for the left-moving wave is $\psi_L$ and the amplitude for
the right-moving wave is $\psi_R$.

\subsection{Step Function Potential}

\begin{figure}[ht]
\psfig{figure=pstep.eps}
\caption{Defining sketch for the step function potential.}
\label{pstep}
\end{figure}

The simplest potential we can consider is a step function, as shown in
figure \ref{pstep}.  In this example the potential is zero for $x > 0$
and is a nonzero constant value, $u$, for $x < 0$.  The solution for
$x > 0$ consists of incident and reflected components, while for $x <
0$ a single transmitted component moving in the incident direction,
i. e., to the left, is found.  The transmitted component has a
nondimensional wavenumber equal to $k_t = [2(\epsilon - u)]^{1/2}$ for
$u < \epsilon$ and $k_t = i[2(u - \epsilon)]^{1/2}$ for $u >
\epsilon$.  The sign of the latter expression is chosen so that $\exp
(-i k_t x )$ decays exponentially to the left.  The wavefunction thus
has the form
\begin{equation}
     \psi = \left\{ \begin{array}{ll}
          I \exp (-i k_f x) + R \exp (i k_f x) & ,~x > 0 \\
          T \exp (-i k_t x) & ,~x < 0
     \end{array} \right. .
\end{equation}

Applying the connecting conditions at $x = 0$, namely continuity of
$\psi$ and $\partial \psi / \partial x$, we find relationships between
$I$, $R$, and $T$:
\begin{equation}
     I + R = T ,
\end{equation} 
\begin{equation}
     -I + R = -(k_t /k_f ) T .
\end{equation} 
The resulting solutions for the transmitted and reflected amplitudes
are
\begin{equation}
     \frac{T}{I} = \frac{2}{1 + k_t /k_f } ,~~~~~
     \frac{R}{I} = \frac{1 - k_t /k_f}{1 + k_t /k_f} .
\end{equation}

It is easy to show that the ratio of the minimum and maximum values of
$| \psi |$ for $x > 0$ is
\begin{equation}
     \eta \equiv \frac{| \psi |_{min}}{| \psi |_{max}} =
     \frac{1 - |R/I|}{1 + |R/I|} .
\end{equation}
Notice that there are three distinct cases, $u < 0$, $0 < u <
\epsilon$ and $u > \epsilon$:
\begin{equation}
     \eta = \left\{ \begin{array}{ll}
          k_f /k_t = (\epsilon /( \epsilon - u))^{1/2} & ,~u < 0 \\
          k_t /k_f = ((\epsilon - u)/ \epsilon)^{1/2} & ,~0 < u < \epsilon \\
          0 & ,~u > \epsilon
     \end{array} \right. .
\label{refl}
\end{equation}
For the first two cases there is partial reflection and a partial
standing wave for $x > 0$.  The magnitude of $\eta$ is inversely
related to the strength of the reflected wave.  As $u$ approaches
$+\epsilon$ or $-\infty$, the reflection becomes total, resulting in
$\eta = 0$.  For $u > \epsilon$ the region $x < 0$ is classically
forbidden to the particle, and only an evanescent, or exponentially
decaying wave exists for $x < 0$.

Quantum mechanics thus differs from classical mechanics in that there
is a small but finite probability of finding a particle in a region
that is classically forbidden on energy grounds.  This can be related
to the uncertainty principle: The part of the forbidden region
accessible to the particle is quite close to the $x = 0$ interface
between the two regions, so localizing the particle to within this
accessible region gives it an uncertainty in the momentum.  The energy
in turn is uncertain to the degree necessary to avoid a violation of
energy conservation.

\subsection{Arbitrary Potential}

For a potential $u(x)$ that is arbitrary within some distance $D$ of
$x = 0$, but is known to be zero (or very close to zero) outside this
region, equation (\ref{refl}) can be used to determine the
probabilities for reflection, $P_r$, and transmission, $P_t$, of an
incident particle as long as (\ref{refl}) is evaluated where $x > D$.
It is easy to show in this case that
\begin{equation}
     P_r = \left( \frac{1 - \eta}{1 + \eta} \right)^2
\end{equation}
and
\begin{equation}
     P_t = 1 - P_r .
\end{equation}

Equation (\ref{scaled}) can be solved numerically by rewriting it as
two coupled first order equations,
\begin{equation}
     \frac{\partial \psi}{\partial x} = \chi
\label{chi}
\end{equation}
and
\begin{equation}
     \frac{\partial \chi}{\partial x} = 2(u - \epsilon) \psi .
\label{psi}
\end{equation}
If we assume that particles are coming in from the right, then the
solution to the left of the potential has only a leftward-moving
component, as discussed above.  In the region where $u = 0$, the
spatial dependence of this solution is $\exp [-i(2\epsilon)^{1/2} x]$.
Equation (\ref{chi}) shows that $\chi = -i(2\epsilon)^{1/2} \psi$ in
this region.  Thus equations (\ref{chi}) and (\ref{psi}) can be
jointly solved by specifying initial values at the left end of the
domain and integrating numerically to the right in $x$.
  
The RLaB demonstration {\em scatter1\_demo.r} uses these equations to
compute $\psi$, $P_r$, and $P_t$ for a variety of potentials.

\subsection{Barrier Penetration}

\begin{figure}[ht]
\psfig{figure=barrier.eps}
\caption{Example of barrier penetration.  Upper plot shows $| \psi |^2$
for a particle entering from the right.  Lower plot shows the energy of
the particle (dashed line) and the potential energy function (filled
region).  This plot was made using the RLaB demonstration program
{\em scatter1\_demo.r}.}
\label{barrier}
\end{figure}

Since particles in quantum mechanics can potentially be found in
regions that are classically forbidden by energy considerations, this
raises the possibility of particles actually passing through narrow
forbidden regions.  Figure \ref{barrier} shows an example of such {\em
barrier penetration}, calculated using {\em scatter1\_demo.r}.  Notice
how the absolute square of the wavefunction decays exponentially to
the left in the barrier region where the potential energy is $1.5$
times the total energy.  As discussed above, the region to the right
of the barrier exhibits a standing wave, represented by the
oscillations in $| \psi |^2$.  The probability of the particle
penetrating this barrier is approximately $0.39$.

\subsection{Negative Potential Energies}

\begin{figure}[ht]
\psfig{figure=negpot.eps}
\caption{Passage of a wave through a region of negative potential
energy.  The total dimensionless energy is $0.5$.}
\label{negpot}
\end{figure}

Figure \ref{negpot} shows the passage of a wave through a region of
negative potential energy.  Even though the potential is not a barrier
in a classical sense, the wave is partially reflected.  Figure
\ref{edep} shows the reflection coefficient as a function of energy
for the potential shown in figure \ref{negpot}.  Particularly large
values of the reflection coefficient are seen for small total
energies.  This occurs because the ratio of the kinetic energies
inside and outside of the potential well is then large.  Note also
that the reflection coefficient oscillates with energy, with certain
energies resulting in zero reflection.  This is reminiscent of the
destructive interference that occurs between reflections from the
front and back surfaces of a thin film when light passes through it.

\begin{figure}
\psfig{figure=edep.eps}
\caption{Reflection coefficient as a function of energy for the
potential well shown in figure \ref{negpot}.}
\label{edep}
\end{figure}

\end{document}
