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

$$ \psi^{\prime \prime}_{II} - k^2 \psi_{II} = 0 $$

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

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

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

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

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

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

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} = 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 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 $$k^2 = \frac{2 m_0}{\hbar^2}E$$, we obtain the following equation for the energy spectrum (the eigenvalues):

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

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

$$ \int_0^l \psi_n^* \psi_n dx = 0 \,\,\,\,\, for \,\,\,\,\,\, n^{\prime} \ne  n $$

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:

$$ E_1 =\frac{\pi^2 \hbar^2}{2 m_0 l^2} \,\,\,\,\,\, \psi_1 = \sqrt{\frac{2}{l}} \sin \left(\frac{\pi x}{l} \right) $$

$$ E_2 =4E_1 \,\,\,\,\,\, \psi_2 = \sqrt{\frac{2}{l}} \sin \left(2\pi\frac{x}{l} \right) $$

$$ E_3 = 9 E_1 \,\,\,\,\,\, \psi_3 = \sqrt{\frac{2}{l}} \sin \left( 3 \pi \frac{x}{l} \right) $$

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].