Normalization of Spherical Harmonic Functions
Theorem
The standardized spherical harmonics are as follows.
$$ Y_{l}^{m}(\theta,\phi)=\sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P_{l}^{m}(\cos\theta)e^{im\phi} $$
$$ \nabla ^2 f = \frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial f}{\partial r} \right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left( \sin\theta \frac{\partial f}{\partial \theta} \right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 f}{\partial^2 \phi}=0 $$
$$ f(r,\theta,\phi)=R(r)\Theta (\theta)\Phi (\phi) $$
Description
In the Laplace equation on spherical coordinates, solutions for the polar angle $\theta$ and the azimuthal angle $\phi$ are referred to as spherical harmonics. $$ \Theta (\theta)\Phi (\phi)=Y_{l}^{m}(\theta,\phi)=e^{im\phi}P_{l}^{m}(\cos \theta) $$ In quantum mechanics, spherical harmonics are dealt with as wave functions and must be standardized. $$ \iiint |R(r)\Theta (\theta) \Phi (\phi)|^{2}r^{2}\sin \theta dr d \theta d\phi=\int_{0}^{\infty}|R(r)|^{2}r^{2}dr\int_{0}^{2\pi}\int_{0}^{\pi}|Y_{l}^{m}(\theta,\phi)|^{2}\sin\theta d\theta d\phi=1 $$ Let’s isolate the angular component that is a spherical harmonic and call the standardizing constant $C$. Then, $$ \begin{align*} &&& |C|^{2}\int_{0}^{2\pi}\int_{0}^{\pi}|Y_{l}^{m}(\theta,\phi)|^{2}\sin\theta d\theta d\phi=1 \\ \implies &&&|C|^{2}\int_{0}^{2\pi}|e^{im\phi}|^{2} d\phi\int_{0}^{\pi}|P_{l}^{m}(\cos \theta)|^{2}\sin\theta d\theta =1 \\ \implies &&&2\pi|C|^{2}\int_{0}^{\pi}|P_{l}^{m}(\cos \theta)|^{2}\sin\theta d\theta =1 \end{align*} $$ The integral over $\theta$ is due to the orthogonality of the associated Legendre polynomials and hence $\frac{2}{2l+1} \frac{(l+m)!}{(l-m)!}$, $$ C=\sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}} $$ Therefore, the standardized spherical harmonics are as follows. $$ Y_{l}^{m}(\theta,\phi)=\sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P_{l}^{m}(\cos\theta)e^{im\phi} $$ In quantum mechanics, it is usually assumed that spherical harmonics are standardized unless stated otherwise.