D.66 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

$$\fbox{$\displaystyle \L_x = \frac{\hbar}{{\rm i}} \lef... ...\phi}{\sin\theta}\frac{\partial}{\partial\phi} \right) $}$$ (D.39)

then that

$$\fbox{$\displaystyle L^+ = \hbar e^{{\rm i}\phi} \left... ...heta}{\sin\theta}\frac{\partial}{\partial\phi} \right) $}$$ (D.40)

Note that the spherical harmonics are of the form $Y^m_l=e^{{\rm i}{m}\phi}\Theta^m_l(\theta)$, 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 $\L ^+Y^l_l=0$. 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.41)

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.13.5} based on quantum field theory.