# PlanetPhysics/Particle in a Potential

As a simple example of the calculation of eigenvalues and eigenfunctions, we shall consider the motion of a particle in a potential well. Since the chief interest of this problem is simply that it provides an illustration of methods used in the solution of this example, we may assume a very simple dependence of the potential energy on distance, Figure 1:
   $V(x)=\left\{{\begin{matrix}{llll}V_{0}&for&-\infty <x<0&(region\,I)\\0&for&0<x<l&(region\,II\\V_{0}&for&l<x<\infty &(region\,III)\\\end{matrix}}\right.$
\begin{figure}[h] \centering \includegraphics{Figure1.eps}
\centerline{Figure 1: The motion of a particle in a potential well} (1) \end{figure}
In the potential well (region II), where $E>V=0$, the Schr\"odinger equation takes the form
   $\psi _{II}^{\prime \prime }-k^{2}\psi _{II}=0$
where
$\psi ^{\prime \prime }={\frac {d^{2}\psi }{dx^{2}}}$
and
$k^{2}={\frac {2m_{0}}{\hbar ^{2}}}E={\frac {p^{2}}{\hbar ^{2}}}>0$
We note that the case $E<0$ has no physical meaning in this problem. Since the general solution of Eq. (2) is oscillatory, we have
   $\psi _{II}=B_{II}\cos(kx)+A_{II}sin(kx)$
In regions I and III, the Schr\"odinger equation has the form
$\psi ^{\prime \prime }+{\frac {2m_{0}}{\hbar ^{2}}}(E-V_{0})\psi =0$
Here two cases must be distinguised. In the first case $(E>V_{0})$, the solution for these regions is also oscillatory in character (an equation of the elliptic type). It is given by Eq. (3), the value of $k$ being
   $k={\frac {1}{\hbar }}{\sqrt {2m_{0}(E-V_{0})}}$
No restrictions need to be imposed on the wave functions at infinity. Therefore, the energy $E$ can assume any value in a continous spectrum of energies. It is better, however, not to investigate the case of a contiuous spectrum on the basis of this example, but rather on the basis of the motion of a free particle. The potential well only adds to the mathematical difficulties of the problem, without changing the general character of the solution.
In the second case, namely, the case of a potential barrier $(E<V_{0})$, the solution of Eq. (3) is exponential in character (an equation of the hyperbolic type). The general solution can be written in the form
$\psi _{I,III}=A_{I,III}e^{kx}+B_{I,III}e^{-kx}$
where
$k^{2}={\frac {1}{\hbar ^{2}}}2m_{0}(V_{0}-E)={\frac {|p|^{2}}{\hbar ^{2}}}>0$
If the energy can assume any value without restriction, the wave function inside the potential barrier $(0<E<V_{0})$ will contain both an exponentially increasing part and an exponentally decreasing part (see Fig. 2). Therefore, we must choose only those values of $E$ for which exponentially increasing solutinos do not exist inside the potential barrier.
\begin{figure}[h] \centering \includegraphics{Figure2.eps}
\centerline{Figure 2: Wave function for a given value of $E$.} (2) \end{figure}
Accordingly, we require that the coefficient $B_{I}=0$ in region I$(x<0)$, and the coefficient $A_{III}=0$ in region III$(x>l)$. We then have
$\psi _{I}=A_{I}e^{kx}=Ae^{-k|x|}$
   $\psi _{III}=B_{III}e^{-kx}=Be^{-k(x-l)}$
where, for the sake of simplicity, we have made
$B_{III}=Be^{kl}$
By joining the solutions at the boundary of regions I and II $(x=0)$, and also at the boundary of regions IIand III $(x=l)$, and making use of the requirement that the exponentially increasing solution vanish, we obtain the equation for the eigenvalues of the energy $E$.
We shall now further simplify our problem by requiring that $V_{0}$, together with $k$, go to infinity (see Fig.3). It is apparent from Eq. (5) that $\psi _{I}=\psi _{III}=0$, and therefore the boundary conditions for the solution (3) inside the potential well (region II) take the form
$\psi _{II}=0$
for $x=0$, and
$\psi _{II}=0$
for $x=l$.
Applying these two equations to the general solution (3) in region II, we find that $B_{II}=0$, and the eigenvalue are described by the equation
$sin(kl)=0$
from which
$kl=\pi n$
where $n=1,2,3,4,....$ We exclude the value $n=0$ from further considerations, since the wave function in this case is identically equal to zero. It is not necessary to consider separately the negative values of $n$, since the wave functions for negative $n$ are equal to the wave functions for positive $n$, taken with the opposite sign.
\begin{figure}[h] \centering \includegraphics{Figure3.eps}
\centerline{Figure 3: Particle in a infinite potential well.} (3) \end{figure}
Since $k^{2}={\frac {2m_{0}}{\hbar ^{2}}}E$, we obtain the following equation for the energy spectrum (the eigenvalues):
$E_{n}={\frac {\pi ^{2}\hbar ^{2}n^{2}}{2m_{0}l^{2}}}$
The wave functions corresponding to these values of energy (eigenfunctions) are
   $\psi _{n}=A_{I}sin\left(\pi n{\frac {x}{l}}\right)$
The coefficient $A_{n}$ can be found from the normalization condition
$\int _{0}^{l}\psi _{n}^{2}dx=A_{n}^{2}\int _{0}^{l}sin^{2}\left(\pi n{\frac {x}{l}}\right)dx={\frac {l}{2}}A_{n}^{2}=1$
which gives
$A_{n}={\sqrt {\frac {2}{l}}}$
Substituting the expression for $A_{n}$ into Eq. (6), we finally obtain
   $\psi _{n}={\sqrt {\frac {2}{l}}}\sin \left(\pi n{\frac {x}{l}}\right)$
Accoring to the general theorem of eigenfunctions, the eigenfunctions (7) of the Schr\"odinger equation satisfy the orthogonality condition
$\int _{0}^{l}\psi _{n}^{*}\psi _{n}dx=0\,\,\,\,\,for\,\,\,\,\,\,n^{\prime }\neq n$
as can be readily seen by performing the direct integration after substituting Eq. (7) for $\psi _{n}$.
We shall now write down a few specific eigenvalues $E_{n}$ and eigenfunctions $\psi _{n}$, shown in Fig. 3:
   $E_{1}={\frac {\pi ^{2}\hbar ^{2}}{2m_{0}l^{2}}}\,\,\,\,\,\,\psi _{1}={\sqrt {\frac {2}{l}}}\sin \left({\frac {\pi x}{l}}\right)$
   $E_{2}=4E_{1}\,\,\,\,\,\,\psi _{2}={\sqrt {\frac {2}{l}}}\sin \left(2\pi {\frac {x}{l}}\right)$
   $E_{3}=9E_{1}\,\,\,\,\,\,\psi _{3}={\sqrt {\frac {2}{l}}}\sin \left(3\pi {\frac {x}{l}}\right)$
These solutions are very similar to the familiar standing-wave solutions for a vibrating string with fixed ends. The case $n=1$ (8) corresponds to the fundamental mode, the case $n=2$ (9), to the first harmonic, etc.