D.77 Harmonic oscillator revisited

This note rederives the harmonic oscillator solution, but in spherical coordinates. The reason to do so is to obtain energy eigenfunctions that are also eigenfunctions of square angular momentum and of angular momentum in the $z$-​direction. The derivation is very similar to the one for the hydrogen atom given in derivation {D.15}, so the discussion will mainly focus on the differences.

The solutions are again in the form $R(r)Y_l^m(\theta,\phi)$ with the $Y_l^m$ the spherical harmonics. However, the radial functions $R$ are different; the equation for them is now

\begin{displaymath}
- \frac{1}{R} \frac{{\rm d}}{{\rm d}r}\left(r^2\frac{{\rm ...
...c{1}{2}} m_{\rm e}\omega^2 r^4
= \frac{2m_e}{\hbar^2} r^2 E
\end{displaymath}

The difference from {D.15} is that a harmonic oscillator potential $\frac12m_e\omega^2r^2$ has replaced the Coulomb potential. A suitable rescaling is now $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\rho\sqrt{\hbar/m_{\rm e}\omega}$, which produces

\begin{displaymath}
- \frac{1}{R} \frac{{\rm d}}{{\rm d}\rho}\left(\rho^2\frac...
...\rho}\right)
+ l(l+1)
+ \rho^4
= \rho^2 \epsilon \qquad
\end{displaymath}

where $\epsilon$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E$$\raisebox{.5pt}{$/$}$${\textstyle\frac{1}{2}}\hbar\omega$ is the energy in half quanta.

Split off the expected asymptotic behavior for large $\rho$ by defining

\begin{displaymath}
R = e^{-\rho^2/2} f
\end{displaymath}

Then $f$ satisfies

\begin{displaymath}
\rho^2 f'' + 2 \rho f' - l(l+1)f = 2\rho^3 f' + (3-\epsilon)\rho^2 f
\end{displaymath}

Plug in a power series $f$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sum_pc_p\rho^p$, then the coefficients must satisfy:

\begin{displaymath}[p(p+1)-l(l+l)]c_p = [2(p-2) + 3 -\epsilon]c_{p-2}
\end{displaymath}

From that it is seen that the lowest power in the series is $p_{\rm {min}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l$, $p_{\rm {min}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-l-1$ not being acceptable. Also the series must terminate, or blow up will occur. That requires that $\epsilon$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2p_{\rm {max}}+3$. So the energy must be $(p_{\rm {max}}+\frac32)\hbar\omega$ with $p_{\rm {max}}$ an integer no smaller than $l$, so at least zero.

Therefore, numbering the energy levels from $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 like for the hydrogen level gives the energy levels as

\begin{displaymath}
E_n=(n+\frac12)\hbar\omega
\end{displaymath}

That are the same energy levels as derived in Cartesian coordinates, as they should be. However, the eigenfunctions are different. They are of the form

\begin{displaymath}
\psi_{nlm} = e^{-\rho^2/2} P_{nl}(\rho) Y_l^m(\theta,\phi)
\end{displaymath}

where $P_{nl}$ is some polynomial of degree $n-1$, whose lowest power of $\rho$ is $\rho^l$. The value of the azimuthal quantum number $l$ must run up to $n-1$ like for the hydrogen atom. However, in this case $l$ must be odd or even depending on whether $n-1$ is odd or even, or the power series will not terminate.

Note that for even $l$, the power series proceed in even powers of $r$. These eigenfunctions are said to have even parity: if you replace $r$ by $\vphantom0\raisebox{1.5pt}{$-$}$$r$, they are unchanged. Similarly, the eigenfunctions for odd $l$ expand in odd powers of $r$. They are said to have odd parity; if you replace $r$ by $\vphantom0\raisebox{1.5pt}{$-$}$$r$, they change sign.