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

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

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

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

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

$$R = e^{-\rho^2/2} f$$

Then $f$ satisfies

$$\rho^2 f'' + 2 \rho f' - l(l+1)f = 2\rho^3 f' + (3-\epsilon)\rho^2 f$$

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

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

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

$$E_n=(n+\frac12)\hbar\omega$$

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

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

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.