Comprehensive Study Notes

For the present, we confine ourselves to one-particle problems. In this section we consider the three-dimensional case of the problem solved in Section 2.2, the particle in a box.

There are many possible shapes for a three-dimensional box. The box we consider is a rectangular parallelepiped with edges of length $a, b$, and $c$. We choose our coordinate system so that one corner of the box lies at the origin and the box lies in the first octant of space (Fig. 3.2). Within the box, the potential energy is zero. Outside the box, it is infinite:

\(
\begin{align}
V(x, y, z) & =0 \text { in the region }\left{\begin{array}{l}
0<x<a \
0<y<b \
0<z<c
\end{array}\right. \tag{3.60}\
V & =\infty \quad \text { elsewhere }
\end{align}
\)

Since the probability for the particle to have infinite energy is zero, the wave function must be zero outside the box. Within the box, the potential-energy operator is zero and the Schrödinger equation (3.47) is

\(
\begin{equation}
-\frac{\hbar^{2}}{2 m}\left(\frac{\partial^{2} \psi}{\partial x^{2}}+\frac{\partial^{2} \psi}{\partial y^{2}}+\frac{\partial^{2} \psi}{\partial z^{2}}\right)=E \psi \tag{3.61}
\end{equation}
\)

To solve (3.61), we assume that the solution can be written as the product of a function of $x$ alone times a function of $y$ alone times a function of $z$ alone:

\(
\begin{equation}
\psi(x, y, z)=f(x) g(y) h(z) \tag{3.62}
\end{equation}
\)

It might be thought that this assumption throws away solutions that are not of the form (3.62). However, it can be shown that, if we can find solutions of the form (3.62) that satisfy the boundary conditions, then there are no other solutions of the Schrödinger equation that will satisfy the boundary conditions. (For a proof, see G. F. D. Duff and

FIGURE 3.2 Inside the box-shaped region, $V=0$.
D. Naylor, Differential Equations of Applied Mathematics, Wiley, 1966, pp. 257-258.) The method we are using to solve (3.62) is called separation of variables.

From (3.62), we find

\(
\begin{equation}
\frac{\partial^{2} \psi}{\partial x^{2}}=f^{\prime \prime}(x) g(y) h(z), \quad \frac{\partial^{2} \psi}{\partial y^{2}}=f(x) g^{\prime \prime}(y) h(z), \quad \frac{\partial^{2} \psi}{\partial z^{2}}=f(x) g(y) h^{\prime \prime}(z) \tag{3.63}
\end{equation}
\)

Substitution of (3.62) and (3.63) into (3.61) gives

\(
\begin{equation}
-\left(\hbar^{2} / 2 m\right) f^{\prime \prime} g h-\left(\hbar^{2} / 2 m\right) f g^{\prime \prime} h-\left(\hbar^{2} / 2 m\right) f g h^{\prime \prime}-E f g h=0 \tag{3.64}
\end{equation}
\)

Division of this equation by $f g h$ gives

\(
\begin{gather}
-\frac{\hbar^{2} f^{\prime \prime}}{2 m f}-\frac{\hbar^{2} g^{\prime \prime}}{2 m g}-\frac{\hbar^{2} h^{\prime \prime}}{2 m h}-E=0 \tag{3.65}\
-\frac{\hbar^{2} f^{\prime \prime}(x)}{2 m f(x)}=\frac{\hbar^{2} g^{\prime \prime}(y)}{2 m g(y)}+\frac{\hbar^{2} h^{\prime \prime}(z)}{2 m h(z)}+E \tag{3.66}
\end{gather}
\)

Let us define $E_{x}$ as equal to the left side of (3.66):

\(
\begin{equation}
E_{x} \equiv-\hbar^{2} f^{\prime \prime}(x) / 2 m f(x) \tag{3.67}
\end{equation}
\)

The definition (3.67) shows that $E{x}$ is independent of $y$ and $z$. Equation (3.66) shows that $E{x}$ equals $\hbar^{2} g^{\prime \prime}(y) / 2 m g(y)+\hbar^{2} h^{\prime \prime}(z) / 2 m h(z)+E$; therefore, $E{x}$ must be independent of $x$. Being independent of $x, y$, and $z$, the quantity $E{x}$ must be a constant.

Similar to (3.67), we define $E{y}$ and $E{z}$ by

\(
\begin{equation}
E{y} \equiv-\hbar^{2} g^{\prime \prime}(y) / 2 m g(y), \quad E{z} \equiv-\hbar^{2} h^{\prime \prime}(z) / 2 m h(z) \tag{3.68}
\end{equation}
\)

Since $x, y$, and $z$ occur symmetrically in (3.65), the same reasoning that showed $E{x}$ to be a constant shows that $E{y}$ and $E_{z}$ are constants. Substitution of the definitions (3.67) and (3.68) into (3.65) gives

\(
\begin{equation}
E{x}+E{y}+E_{z}=E \tag{3.69}
\end{equation}
\)

Equations (3.67) and (3.68) are

\(
\begin{gather}
\frac{d^{2} f(x)}{d x^{2}}+\frac{2 m}{\hbar^{2}} E{x} f(x)=0 \tag{3.70}\
\frac{d^{2} g(y)}{d y^{2}}+\frac{2 m}{\hbar^{2}} E{y} g(y)=0, \quad \frac{d^{2} h(z)}{d z^{2}}+\frac{2 m}{\hbar^{2}} E_{z} h(z)=0 \tag{3.71}
\end{gather}
\)

We have converted the partial differential equation in three variables into three ordinary differential equations. What are the boundary conditions on (3.70)? Since the wave function vanishes outside the box, continuity of $\psi$ requires that it vanish on the walls of the box. In particular, $\psi$ must be zero on the wall of the box lying in the $y z$ plane, where $x=0$, and it must be zero on the parallel wall of the box, where $x=a$. Therefore, $f(0)=0$ and $f(a)=0$.

Now compare Eq. (3.70) with the Schrödinger equation [Eq. (2.10)] for a particle in a one-dimensional box. The equations are the same in form, with $E_{x}$ in (3.70) corresponding to $E$ in (2.10). Are the boundary conditions the same? Yes, except that we have $x=a$ instead of $x=l$ as the second point where the independent variable vanishes. Thus we can use the work in Section 2.2 to write as the solution [see Eqs. (2.23) and (2.20)]

\(
\begin{aligned}
f(x) & =\left(\frac{2}{a}\right)^{1 / 2} \sin \left(\frac{n{x} \pi x}{a}\right) \
E{x} & =\frac{n{x}^{2} h^{2}}{8 m a^{2}}, \quad n{x}=1,2,3 \ldots
\end{aligned}
\)

The same reasoning applied to the $y$ and $z$ equations gives

\(
\begin{gathered}
g(y)=\left(\frac{2}{b}\right)^{1 / 2} \sin \left(\frac{n{y} \pi y}{b}\right), \quad h(z)=\left(\frac{2}{c}\right)^{1 / 2} \sin \left(\frac{n{z} \pi z}{c}\right) \
E{y}=\frac{n{y}^{2} h^{2}}{8 m b^{2}}, \quad n{y}=1,2,3 \ldots \quad \text { and } \quad E{z}=\frac{n{z}^{2} h^{2}}{8 m c^{2}}, \quad n{z}=1,2,3 \ldots
\end{gathered}
\)

From (3.69), the energy is

\(
\begin{equation}
E=\frac{h^{2}}{8 m}\left(\frac{n{x}^{2}}{a^{2}}+\frac{n{y}^{2}}{b^{2}}+\frac{n_{z}^{2}}{c^{2}}\right) \tag{3.72}
\end{equation}
\)

As with the particle in a one-dimensional box, the ground-state energy is greater than the classical-mechanical, lowest-energy value of zero.

From (3.62), the wave function inside the box is

\(
\begin{equation}
\psi(x, y, z)=\left(\frac{8}{a b c}\right)^{1 / 2} \sin \left(\frac{n{x} \pi x}{a}\right) \sin \left(\frac{n{y} \pi y}{b}\right) \sin \left(\frac{n_{z} \pi z}{c}\right) \tag{3.73}
\end{equation}
\)

The wave function has three quantum numbers, $n{x}, n{y}, n_{z}$. We can attribute this to the three-dimensional nature of the problem. The three quantum numbers vary independently of one another.

In a one-particle, one-dimensional problem such as the particle in a one-dimensional box, the nodes are where $\psi(x)=0$, and solving this equation for $x$, we get points where $\psi=0$. In a one-particle, three-dimensional problem, the nodes are where $\psi(x, y, z)=0$, and solving this equation for $z$, we get solutions of the form $z=f(x, y)$. Each such solution is the equation of a nodal surface in three-dimensional space. For example, for the stationary state with $n{x}=1, n{y}=1, n{z}=2$, the wave function $\psi$ in (3.73) is zero on the surface where $z=c / 2$; this is the equation of a plane that lies parallel to the top and bottom faces of the box and is midway between these faces. Similarly, for the state $n{x}=2$, $n{y}=1, n{z}=1$ the plane $x=a / 2$ is a nodal surface.

Since the $x, y$, and $z$ factors in the wave function are each independently normalized, the wave function is normalized:

\(
\int{-\infty}^{\infty} \int{-\infty}^{\infty} \int{-\infty}^{\infty}|\psi|^{2} d x d y d z=\int{0}^{a}|f(x)|^{2} d x \int{0}^{b}|g(y)|^{2} d y \int{0}^{c}|h(z)|^{2} d z=1
\)

where we used (Prob. 3.40)

\(
\begin{equation}
\iiint F(x) G(y) H(z) d x d y d z=\int F(x) d x \int G(y) d y \int H(z) d z \tag{3.74}
\end{equation}
\)

What are the dimensions of $\psi(x, y, z)$ in (3.73)? For a one-particle, three-dimensional system, $|\psi|^{2} d x d y d z$ is a probability, and probabilities are dimensionless. Since the dimensions of $d x d y d z$ are length ${ }^{3}, \psi(x, y, z)$ must have dimensions of length ${ }^{-3 / 2}$ to make $|\psi|^{2} d x d y d z$ dimensionless.

Suppose that $a=b=c$. We then have a cube. The energy levels are then

\(
\begin{equation}
E=\left(h^{2} / 8 m a^{2}\right)\left(n{x}^{2}+n{y}^{2}+n_{z}^{2}\right) \tag{3.75}
\end{equation}
\)

Let us tabulate some of the allowed energies of a particle confined to a cube with infinitely strong walls:

FIGURE 3.3 Energies of the lowest few states of a particle in a cubic box.

Note that states with different quantum numbers may have the same energy (Fig. 3.3). For example, the states $\psi{211}, \psi{121}$, and $\psi{112}$ (where the subscripts give the quantum numbers) all have the same energy. However, Eq. (3.73) shows that these three sets of quantum numbers give three different, independent wave functions and therefore do represent different states of the system. When two or more independent wave functions correspond to states with the same energy eigenvalue, the eigenvalue is said to be degenerate. The degree of degeneracy (or, simply, the degeneracy) of an energy level is the number of states that have that energy. Thus the second-lowest energy level of the particle in a cube is threefold degenerate. We got the degeneracy when we made the edges of the box equal. Degeneracy is usually related to the symmetry of the system. Note that the wave functions $\psi{211}, \psi{121}$, and $\psi{112}$ can be transformed into one another by rotating the cubic box. Usually, the bound-state energy levels in one-dimensional problems are nondegenerate.

In the statistical-mechanical evaluation of the molecular partition function of an ideal gas, the translational energy levels of each gas molecule are taken to be the levels of a particle in a three-dimensional rectangular box (the box is the container holding the gas); see Levine, Physical Chemistry, Sections 21.6 and 21.7.

In the free-electron theory of metals, the valence electrons of a nontransition metal are treated as noninteracting particles in a box, the sides of the box being the surfaces of the metal. This approximation, though crude, gives fairly good results for some properties of metals.