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# $$\S 14.2$$ Harmonic Oscillator in Three Dimensions

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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 harmonic oscillator in three dimensions is described by the potential

$$V(\vec{r}) =\frac{1}{2}m\omega_1^2 x^2 + \frac{1}{2}m\omega_1^2 y^2 + \frac{1}{2}m\omega_1^2 z^2 .$$

The Schr\"{o}dinger  equation for harmonic oscillator in three dimensions

$$- \frac{\hbar^2}{2m}\left( \PP[\psi(\vec{r})]{x} + \PP[\psi(\vec{r})]{y} + \PP[\psi(\vec{r})]{z}\right) + \left( \frac{1}{2}m\omega_1^2 x^2 + \frac{1}{2}m\omega_1^2 y^2 + \frac{1}{2}m\omega_1^2 z^2 \right) = E \psi(\vec{r}).\label{EQ05}$$

can be solved by separating the variables in cartesian coordinates. Substituting $\psi(\vec{r}) = X(x) Y(y) Z(z)$ gives separate harmonic oscillator equations for the three functions $X(x),Y(y),Z(z)$.

\begin{eqnarray} \left( - \frac{\hbar^2}{2m} \DD{x} + \frac{1}{2}m\omega_1^2 x^2\right) X(x) &=& E_1 X(x), \label{EQ06}\\ \left( - \frac{\hbar^2}{2m} \DD{y} + \frac{1}{2}m\omega_2^2 y^2\right)Y(y) &=& E_2 Y(y), \label{EQ07}\\ \left( - \frac{\hbar^2}{2m} \DD{z} + \frac{1}{2}m\omega_3^2 z^2\right) Z(z) &=& E_3 Z(z). \label{EQ08} \end{eqnarray}

where $E_1+E_2+E_3=E$. Using the known solutions of the harmonic oscillator problem, we get the energy as

$$E =\sum_{k=1}^3 \Big( n_k+ \frac{1}{2}\Big)\hbar \omega_k .$$

The corresponding wave function is given by

$$\psi(\vec{r})=(N_1N_2N_3) H_{n_1}(\alpha_1x) H_{n_2}(\alpha_2 y) H_{n_3}(\alpha_3 z)$$

where

$$N_k= \left(\frac{1}{2^{n_k}\sqrt{\pi} n_k!} \right), \qquad \alpha_k^2 =\frac{m\omega_k}{\hbar}, \qquad k=1,2,3.$$

and $n_1,n_2 n_3$ are nonnegative integers.

In the special case of {\it isotropic oscillator}, $\omega_1=\omega_2=\omega_3\equiv\omega$ and we have the energy eigenvalues given by

$$E = ( n+ 3/2) \hbar \omega,~~~ \text{ where } n=n_1+n_2+n_3$$

For a fixed value of $n$ several combinations of $n_1,n_2,n_3$ give the same energy and the energy levels are degenerate.

For a given $n$, $n_3$is determined in terms of $n_1$ and $n_2$ and $n_3=n-n_1-n_2$. The degeneracy of the $n^\text{th}$ excited state is therefore, given by

$$\sum _{n_1=0}^n \sum_{n_2=n-n_1}^n 1 = \sum_{n_1=0}^n n_1 = (n+1)(n+2)/2.$$

For the isotropic oscillator in three dimensions the potential is $\frac{1}{2}m \omega r^2$ and it depends on $r$ alone. This problem can therefore also be solved by separating the variables in spherical polar $r,\theta, \phi$ coordinates.

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