D.76 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

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

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

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

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 $\vphantom{0}\raisebox{1.5pt}{$-$}$$r$, they are un­changed. Sim­i­larly, the eigen­func­tions for odd $l$ ex­pand in odd pow­ers of $r$. They are said to have odd par­ity; if you re­place $r$ by $\vphantom{0}\raisebox{1.5pt}{$-$}$$r$, they change sign.