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# $$\S 14.1$$ Particle in Three Dimensional Box

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Problems in Higher Dimensions

$\newcommand{\DD}[2][]{\frac{d^2 #1}{d#2^2}}$

$\newcommand{\PP}[2][]{\frac{\partial^2 #1}{\partial#2^2}}$

A rigid rectangular box is represented by zero potential inside the box and infinite potential outside the box. Taking one of the corners of the box as the origin, a rigid box in three dimensions and of sides $a,b,c$ corresponds to the potential

$$V(x,y,z) = \begin{cases} 0, & \qquad \text{if } 0<x<a, \, 0<y<b,\, \text{and } 0<z,c\\ \infty \qquad \text{otherwise} \end{cases}. \label{EQ01}$$

Since the potential is infinity outside the box, the wave function must be taken to be zero. The boundary conditions to be satisfied are

\begin{eqnarray} \psi(0,y,z)=\psi(a,y,z)=0, &\quad& \text{for all } y,z, \label{EQ02}\\ \psi(x,0,z)=\psi(x,b,z)=0, &\quad& \text{for all } z,x,\label{EQ03}\\ \psi(x,y,0)=\psi(x,y,c)=0, &\quad& \text{for all } x,y. \label{EQ04} \end{eqnarray}

Recall that {\it there is no condition on the partial derivatives of the wave function when the potential jumps by an infinite amount across a boundary.} Inside  the box, the Schrodinger equation is  free particle equation. Thus, for $0<x<a, \, 0<y<b,\, 0<z<c$ we have
$$- \frac{\hbar^2}{2m}\left( \PP[\psi(\vec{r})]{x} + \PP[\psi(\vec{r})]{y} + \PP[\psi(\vec{r})]{z}\right) = E \psi(\vec{r}).\label{EQ05}$$
This solution is separable in several coordinate systems, %FIXME Which Systems but the geometry of the problem demands that we must use the Cartesian coordinates. Thus writing $\psi(\vec{r})$ as a product of three functions $X(x), Y(y)$ and $Z(z)$ leads to free particle, ordinary differential equation for each of the three functions.
\begin{eqnarray} -\frac{\hbar^2}{2m} \DD[X(x)]{x} &=& E_1 X(x), \label{EQ06}\\ -\frac{\hbar^2}{2m} \DD[Y(y)]{x} &=& E_2 Y(y), \label{EQ07}\\ -\frac{\hbar^2}{2m} \DD[X(z)]{z} &=& E_3 X(z). \label{EQ08} \end{eqnarray}
where $E_1+E_2+E_3=E$. The boundary condition \REF{EQ02} gives

$$X(0)Y(y)Z(z)= X(a)Y(y)Z(z) = 0, \forall y,z.\label{EQ09}$$

This condition can be satisfied by a nontrivial solution if and only if $X(0)=X(a)=0$. Similarly, $Y(0)=Y(b)=0$ and $Z(0)=Z(c)=0$. Thus the solution of the three dimensional box problem can be down in terms of solutions of the particle in  one dimensional box. A little thinking shows that we must have

\begin{eqnarray} X(x)= \sqrt{\frac{2}{a}} \sin k_1x, \qquad k_1 = \frac{n_1\hbar\pi}{a}, \qquad E_1=\frac{\hbar^2k_1^2}{2m} = \frac{\hbar^2n_1^2\pi^2}{2ma^2}\\ Y(y)= \sqrt{\frac{2}{b}} \sin k_2y, \qquad k_2 = \frac{n_2\hbar\pi}{b}, \qquad E_2=\frac{\hbar^2k_2^2}{2m} = \frac{\hbar^2n_2^2\pi^2}{2mb^2}\\ Z(z)= \sqrt{\frac{2}{c}} \sin k_3z, \qquad k_3 = \frac{n_3\hbar\pi}{c}, \qquad E_3=\frac{\hbar^2k_3^2}{2m} = \frac{\hbar^2n_3^2\pi^2}{2mc^2}\\ \end{eqnarray}

and the energy total wave function  $\psi(x,y,z)$ and $E$ are given by

$$E=\frac{\hbar^2}{2m}\Big(\frac{n_1^2}{a^2} +\frac{n_2^2}{b^2} + \frac{n_3^2}{c^2} \Big), \qquad \psi(x,y,z) = \sqrt{\frac{8}{abc}} \sin k_1x\,\sin k_2y \,\sin k_3z.$$

For a cubical box of side $L$ we have

$$E= \frac{\hbar^2}{2m L^2}(n_1^2+n_2^2+n_3^2).$$

The ground state corresponds to $n_1=n_2=n_3=1$ and is nondegenerate. For a cubical box The next energy level corresponds to the values $(n_1,n_2,n_3)=(2,1,1), (1,2,1), (1,1,2)$. and has energy equal to $4(\hbar^2\pi^2/2mL^2)$ and is threefold degenerate. For energy $E=14 E_0,\, (E_0=\hbar^2\pi^2/2mL^2)$, there are six permutations of 1,2,3 for $n_1,n_2$ and $n_3$, giving six wave functions for the which correspond to the same energy $E=14E_0$ .

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