PlanetPhysics/Particle in a Potential

As a simple example of the calculation of eigenvalues and eigenfunctions, we shall consider the [[../CosmologicalConstant2/|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 [[../CosmologicalConstant2/|energy]] on distance, Figure 1:

$V(x)=\{\begin{array}{cccc}{V}_0& for& -\mathrm{\infty }<x<0& (regionI)\\ 0& for& 0<x<l& (regionII\\ V_0& for& l<x<\mathrm{\infty }& (regionIII)\end{array}$

\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

${\psi }^{\prime \prime }=\frac{d^2\psi }{d{x}^2}$

and

$k^2=\frac{2m_0}{{\hslash }^2}E=\frac{p^2}{{\hslash }^2}>0$

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

${\psi }^{\prime \prime }+\frac{2m_0}{{\hslash }^2}(E-V_0)\psi =0$

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 [[../Bijective/|type]]). It is given by Eq. (3), the value of $k$ being

$k=\frac{1}{\hslash }\sqrt{2m_0(E-V_0)}$

No restrictions need to be imposed on the [[../CosmologicalConstant/|wave]] [[../Bijective/|functions]] at infinity. Therefore, the energy $E$ can assume any value in a continous [[../CohomologyTheoryOnCWComplexes/|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/|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

$k^2=\frac{1}{{\hslash }^2}2m_0(V_0-E)=\frac{|p{|}^2}{{\hslash }^2}>0$

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

${\psi }_I=A_I{e}^{kx}=A{e}^{-k|x|}$

${\psi }_{III}=B_{III}{e}^{-kx}=B{e}^{-k(x-l)}$

where, for the sake of simplicity, we have made

$B_{III}=B{e}^{kl}$

By joining the solutions at the [[../GenericityInOpenSystems/|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 $k^2=\frac{2m_0}{{\hslash }^2}E$, we obtain the following equation for the energy spectrum (the eigenvalues):

$E_n=\frac{{\pi }^2{\hslash }^2{n}^2}{2m_0{l}^2}$

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

$\underset{0}{\overset{l}{\int }}{\psi }_n^{2}dx=A_n^2\underset{0}{\overset{l}{\int }}si{n}^2\left(\pi n\frac{x}{l}\right)dx=\frac{l}{2}{A}_n^2=1$

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 [[../Formula/|theorem]] of eigenfunctions, the eigenfunctions (7) of the Schr\"odinger equation satisfy the orthogonality condition

$\underset{0}{\overset{l}{\int }}{\psi }_n^{*}{\psi }_{n}dx=0for{n}^{\prime }\ne n$

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:

$E_1=\frac{{\pi }^2{\hslash }^2}{2m_0{l}^2}{\psi }_1=\sqrt{\frac{2}{l}}\sin \left(\frac{\pi x}{l}\right)$

$E_2=4E_1{\psi }_2=\sqrt{\frac{2}{l}}\sin \left(2\pi \frac{x}{l}\right)$

$E_3=9E_1{\psi }_3=\sqrt{\frac{2}{l}}\sin \left(3\pi \frac{x}{l}\right)$

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.

• derivative of the Public domain work of [Sokolov].