# Isotropic Harmonic Oscillator

We now solve the isotropic harmonic oscillator using the formalism that we have just developed. While it is possible to solve it in Cartesian coordinates, we gain additional insight by solving it in spherical coordinates, and it is easier to determine the degeneracy of each energy level.

The radial part of the Schrödinger equation for a particle of mass $M\!$ in an isotropic harmonic oscillator potential $V(r)=\frac{1}{2}M\omega^{2}r^2$ is given by:

$-\frac{\hbar^2}{2M}\frac{d^2u_{nl}}{dr^2}+\left(\frac{\hbar^2}{2M}\frac{l(l+1)}{r^2} + \frac{1}{2}M\omega^{2}r^2\right)u_{nl}=Eu_{nl}.$

Let us begin by looking at the solutions $u_{nl}\!$ in the limits of small and large $r.\!$

As $r\rightarrow 0\!$, the equation reduces to

$-\frac{\hbar^2}{2M}\frac{d^2u_{nl}}{dr^2}+\frac{\hbar^2}{2M}\frac{l(l+1)}{r^2}u_{nl}=Eu_{nl}.$

The only solution of this equation that does not diverge as $r\rightarrow 0$ is $u_{nl}(r)\simeq r^{l+1}.$

In the limit as $r\rightarrow \infty,$ on the other hand, the equation becomes

$-\frac{\hbar^2}{2M}\frac{d^2u_{nl}}{dr^2}+\frac{1}{2}M\omega^{2}r^2u_{nl}=Eu_{nl}$

whose solution is given by $u_{nl}(r)\simeq e^{-M\omega r^2/2\hbar}.$

We may now assume that the general solution to the equation is given by

$u_{nl}(r)=r^{l+1}e^{-M\omega r^2/2\hbar}f_{nl}(r).$

Substituting this expression into the original equation, we obtain

$\frac{d^2f_{nl}}{dr^2}+2\left(\frac{l+1}{r}-\frac{M\omega}{\hbar}r\right)\frac{df_{nl}}{dr}+\left[\frac{2ME}{\hbar^2}-(2l+3)\frac{M\omega}{\hbar}\right]f_{nl}=0.$

We now use a series solution for this equation:

$f_{nl}(r)=\sum_{n=0}^{\infty}a_{n}r^n= a_{0}+a_{1}r+a_{2}r^2+a_{3}r^3+\ldots +a_{n}r^n+\ldots$

Substituting this solution into the reduced form of the equation, we obtain

$\sum_{n=0}^{\infty} \left[n(n-1)a_{n}r^{n-2}+2 \left( \frac{l+1}{r}- \frac{M\omega}{\hbar}r\right) na_nr^{n-1} + \left[\frac{2ME}{\hbar^2} - (2l+3)\frac{M\omega}{\hbar}\right] a_n r^n\right]=0,$

which reduces to

$\frac{2(l+1)}{r}a_1+\sum_{n=0}^{\infty}\left[(n+2)(n+2l+3)a_{n+2}+\left(-\frac{2M\omega}{\hbar}n+\frac{2ME}{\hbar^2}-(2l+3)\frac{M\omega}{\hbar}\right)a_n\right]r^n=0.$

For this equation to hold, the coefficients of each of the powers of $r\!$ must vanish seperately. Doing this for the positive powers of $r\!$ yields the following recursion relation:

$(n+2)(n+2l+3)a_{n+2}=\left[-\frac{2ME}{\hbar^2}+(2n+2l+3)\frac{M\omega}{\hbar}\right]a_n$

In addition, we have an $r^{-1}\!$ term; for it to vanish, we must set $a_1=0.\!$ This, combined with the above recursion relation, means that the function $f_{nl}(r)\!$ contains only even powers of $r.\!$ In other words,

$f_{nl}(r)=\sum_{n=0,2,4,\ldots}^{\infty}a_{n}r^{n}=\sum_{n'=0}^{\infty}a_{n'}r^{n'}.$

By a similar argument as the one that we employed for the one-dimensional harmonic oscillator, we find that, unless the series for $f_{nl}(r)\!$ terminates, the resulting full wave function will diverge as $r\rightarrow\infty.$ Because the series must only contain even powers of $r,\!$ the resulting quantization condition on the energy is

$\frac{2M}{\hbar^2}E_{n'l}-\frac{M\omega}{\hbar}(4n'+2l+3)=0,\,n'=0,1,2,3,\ldots,$

or

$E_{nl}=\left(n+\frac{3}{2}\right)\hbar\omega,\,n=0,1,2,3,\ldots,$

where $n=2n'+l.\!$

The degeneracy corresponding to the $n^{\text{th}}\!$ level may be found to be $\tfrac{1}{2}(n+1)(n+2).$ We see that energy levels with even $n\!$ correspond to even values of $l,\!$ while those with odd $n\!$ have odd values of $l.\!$

The total wave function of the isotropic harmonic oscillator is thus given by

$\psi_{nlm}(r,\theta,\phi )=r^{l+1}e^{-M\omega r^2/2\hbar}f_{nl}(r)Y_{lm}(\theta,\phi)=R_{nl}(r)Y_{lm}(\theta ,\phi ).$

One may show that, in fact, $f_{nl}(r)\!$ is an associated Laguerre polynomial in $\frac{M\omega}{\hbar}r^2.$