Spherical Harmonics

Section 15.4 Spherical Harmonics

We have found that normalized solutions of the \(\phi\) equation (6.5.18) satisfying periodic boundary conditions are

\begin{equation} \Phi(\phi) = \frac{1}{\sqrt{2\pi}} \> e^{im\phi} \qquad\qquad (m=0,\pm1,\pm2,...)\tag{15.4.1} \end{equation}

and normalized solutions of the \(\theta\) equation (6.5.17) which are regular at the poles are given by

\begin{equation} P(\cos\theta) = \sqrt{\frac{(2\ell+1)}{2} \frac{(\ell-\absm)!}{(\ell+\absm)!}} \>P_\ell^m(\cos\theta)\tag{15.4.2} \end{equation}

Combining these yields via multiplication (we assumed solutions of this type when we first did the separation of variables procedure), we obtain the spherical harmonics

\begin{equation} Y_\ell^m(\theta,\phi) = (-1)^{(m+|m|)/2} \sqrt{\frac{(2\ell+1)}{4\pi} \frac{(\ell-\absm)!}{(\ell+\absm)!}} \> P_\ell^m(\cos\theta) \> e^{im\phi}\tag{15.4.3} \end{equation}

where the somewhat peculiar choice of phase is conventional.

The spherical harmonics are orthonormal on the unit sphere:

\begin{equation} \int\limits_0^{2\pi} \int\limits_0^\pi \left(Y_{\ell_1}^{m_1}\right)^* Y_{\ell_2}^{m_2} \sin\theta\,d\theta\,d\phi = \delta_{\ell_1 \ell_2} \delta_{m_1 m_2}\tag{15.4.4} \end{equation}

since \(dz=\sin\theta\,d\theta\text{.}\) They are complete in the sense that any sufficiently smooth function \(f\) on the unit sphere can be expanded in a Laplace series as

\begin{equation} f(\theta,\phi) = \sum\limits_{\ell=0}^\infty \sum\limits_{m=-\ell}^\ell a_{\ell m} \, Y_\ell^m(\theta,\phi)\tag{15.4.5} \end{equation}

where

\begin{equation} a_{\ell m} = \int\limits_0^{2\pi} \int\limits_0^\pi \left(Y_\ell^m\right)^* f(\theta,\phi) \,\sin\theta\,d\theta\,d\phi\tag{15.4.6} \end{equation}

Example 3. Example.

Suppose you want a function of \((\theta,\phi)\) which satisfies

\begin{equation} f(\theta,\phi) = \begin{cases} \sin\theta \amp 0<\theta<\frac\pi2\\ 0 \amp \text{otherwise} \end{cases}\tag{15.4.7} \end{equation}

Then \(f\) takes the form (15.4.5), and the constants \(a_{\ell m}\) can be determined from (15.4.6), yielding

\begin{align} a_{\ell m} \amp= \int\limits_0^{2\pi} \int\limits_0^{\pi/2} \left(Y_\ell^m\right)^* \sin^2\!\theta\,d\theta\,d\phi\notag\\ \amp= N_{\ell m} \int\limits_0^{2\pi} e^{-im\phi} \,d\phi \int\limits_0^{\pi/2} P_\ell^m(\cos\theta)\,\sin^2\!\theta\,d\theta\tag{15.4.8} \end{align}

where

\begin{equation} N_{\ell m} = (-1)^{(m+|m|)/2} \sqrt{\frac{(2\ell+1)}{4\pi} \frac{(\ell-\absm)!}{(\ell+\absm)!}}\tag{15.4.9} \end{equation}

Thus,

\begin{equation*} a_{\ell m} = \begin{cases} 0 \amp (m\ne0)\\ \displaystyle\sqrt{(2\ell+1)\,\pi} \int\limits_0^{\pi/2} P_\ell(\cos\theta) \,\sin^2\!\theta \,d\theta \amp (m=0) \end{cases} \end{equation*}

For \(m=0\text{,}\) the integral is most easily computed with the substitution \(z=\cos\theta\text{;}\) the first few coefficients are:

\begin{align} a_{00} \amp= \frac\pi8 \qquad a_{10} = \frac12 \qquad a_{20} = -\frac{5\pi}{64}\notag\\ a_{30} \amp= -\frac7{12} \qquad a_{40} = -\frac{9\pi}{512} \qquad a_{50} = \frac{77}{240}\tag{15.4.10} \end{align}

(each of which should be multiplied by \(\sqrt{4\pi/(2\ell+1)}\)). As you can check by graphing, however, it requires at least twice this many terms to obtain a good approximation.

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