D.64 Spher­i­cal har­mon­ics by lad­der op­er­a­tors

One ap­pli­ca­tion of lad­der op­er­a­tors is to find the spher­i­cal har­mon­ics, which as noted in chap­ter 4.2.3 is not an easy prob­lem. To do it with lad­der op­er­a­tors, show that

$$\fbox{$\displaystyle \L_x = \frac{\hbar}{{\rm i}} \left( ... ...sin\phi}{\sin\theta}\frac{\partial}{\partial\phi} \right) $}$$ (D.41)

then that

$$\fbox{$\displaystyle L^+ = \hbar e^{{\rm i}\phi} \left( ... ...s\theta}{\sin\theta}\frac{\partial}{\partial\phi} \right) $}$$ (D.42)

Note that the spher­i­cal har­mon­ics are of the form $Y^m_l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^{{\rm i}{m}\phi}\Theta^m_l(\theta)$, so

$$L^+ Y^m_l = \hbar e^{{\rm i}(m+1)\phi} \sin^m\theta \frac{{\rm d}(\Theta^m_l/\sin^m\theta)}{{\rm d}\theta}$$

$$L^- Y^m_l = - \hbar e^{{\rm i}(m-1)\phi} \frac{1}{\sin^m\theta} \frac{{\rm d}(\Theta^m_l\sin^m\theta)}{{\rm d}\theta}$$

Find the $Y_l^l$ har­monic from $\L ^+Y^l_l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That gives

$$\fbox{$\displaystyle Y_l^l = \sqrt{\frac{1}{4\pi} \frac{(... ...;5\;\ldots\;(2l+1)}{2\;4\;6\;\ldots\;2l}} (x+{\rm i}y)^l $} %$$ (D.43)

Now ap­ply $\L ^-$ to find the rest of the lad­der.

In­ter­est­ingly enough, the so­lu­tion of the one-di­men­sion­al har­monic os­cil­la­tor prob­lem can also be found us­ing lad­der op­er­a­tors. It turns out that, in the no­ta­tion of that prob­lem,

$$H^+ = -{\rm i}{\widehat p}+ m\omega{\widehat x} \quad H^- = {\rm i}{\widehat p}+ m\omega{\widehat x}$$

are com­mu­ta­tor eigen­op­er­a­tors of the har­monic os­cil­la­tor Hamil­ton­ian, with eigen­val­ues $\pm\hbar\omega$. So, you can play the same games of con­struct­ing lad­ders. Eas­ier, re­ally, since there is no equiv­a­lent to square an­gu­lar mo­men­tum to worry about in that prob­lem: there is only one lad­der. See [25, pp. 42-47] for de­tails. An equiv­a­lent de­riva­tion is given in ad­den­dum {A.15.5} based on quan­tum field the­ory.