D.65 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

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

then that

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

$$\fbox{$\displaystyle Y_l^l = \sqrt{\frac{1}{4\pi} \fra... ...;\ldots\;(2l+1)}{2\;4\;6\;\ldots\;2l}} (x+{\rm i}y)^l $} %$$ (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 [24, pp. 42-47] for details. An equivalent derivation is given in addendum {A.15.5} based on quantum field theory.