1
$\begingroup$

I'm looking at Lebedev quadrature for the integration of functions of a sphere, where it says:

[...] integrate exactly all polynomials up to a given order. On the unit sphere, this is equivalent to integrating all spherical harmonics up to the same order.

I would like to check this, so I need the exact value of $$ \int_{S^2} Y_l^m \,\text{d}S^2, $$ i.e., the integral of the spherical harmonic over the 2-sphere.

Are those values known explicitly?

$\endgroup$
1
  • 2
    $\begingroup$ $\int Y_\ell^m Y_{\ell'}^{m'} dS^2 = \delta_{\ell\ell'}\delta_{mm'}$ and $Y_0^0 = \frac{1}{\sqrt{4\pi}}$. $\endgroup$ Jul 31, 2017 at 12:03

2 Answers 2

2
$\begingroup$

The accepted answer was not easy for me to see and it took me a while to do it in a more step-by-step manner:

The spherical harmonics are orthonormal by definition:

$$\int_{\theta=0}^{\pi} \int_{\varphi=0}^{2 \pi} Y_{\ell}^{m} Y_{\ell^{\prime}}^{m^{\prime} *} d \Omega=\delta_{\ell \ell^{\prime}} \delta_{m m^{\prime}}$$

where $d \Omega=\sin (\theta) d \varphi d \theta$ and $\delta$ is the Kronecker delta and is 1 if the indices are the same and 0 otherwise. We can now set $m^\prime = \ell^\prime =0$. If you insert this into the definition of the spherical harmonic, $Y_{l^\prime}^{m^\prime}(\theta, \phi)=\sqrt{\frac{2 l^\prime+1}{4 \pi} \frac{(l^\prime-m^\prime) !}{(l^\prime+m^\prime) !}} P_{l^\prime}^{m^\prime}(\cos (\theta)) \exp (\mathrm{i} m^\prime \phi)$ you can see that it yields $1/\sqrt{4 \pi}$. We substitute this back into the equation above to obtain

$$ \int_{\theta=0}^{\pi} \int_{\varphi=0}^{2 \pi} Y_{\ell}^{m} 1/\sqrt{4 \pi} d \Omega=\delta_{\ell 0} \delta_{m 0} $$

and multiply by $\sqrt{4 \pi}$ to see that the result to your desired integral is

$$ \int_{S^{2}} Y_{l}^{m} \mathrm{~d} S^{2} = \sqrt{4 \pi} \delta_{\ell 0} \delta_{m 0} = \left\{\begin{array}{ll} \sqrt{4 \pi} & \text { if } l=0 \text { and } m=0 \\ 0 & \text { otherwise } \end{array}\right. $$

$\endgroup$
2
$\begingroup$

The representation $$ Y^m_l(\theta,\phi) = \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}} P_l^m(\cos(\theta)) \exp(\text{i}m\phi) $$ is instructive here. Clearly $Y^0_0(\theta,\phi)= (4\pi)^{-1/2}$, so the integral over the sphere is $\sqrt{4\pi}$. In all other cases, one can separate the integral into polar component ($\phi$) and zenithal component ($\theta$) and integrate separately. Since $$ \int_0^{2\pi} \exp(\text{i}m\phi)\,d\phi = 0,\quad \int_{0}^{\pi} P_l^m(\cos(\theta))\,d\theta = 0, $$ (the latter being a propery of the associated Legendre polynomials) one has $$ \int_{S^2} Y^m_l(\theta,\phi) = \begin{cases} \sqrt{4\pi} \quad&\text{if } l = 0 \text{ and } m = 0,\\ 0 \quad&\text{otherwise.} \end{cases} $$

$\endgroup$
1
  • 1
    $\begingroup$ Can someone explain why this is easy to see? $\endgroup$
    – A-P
    Oct 12, 2021 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.