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I have an equation given in the following form:

$$D_{l,m} = \int_{\mathbf{\Omega}}^{} \mathrm{d}\Omega \ Y_{l,m}(\mathbf{\hat{s}})\int_{\mathbf{\Omega'}}^{} \mathrm{d}\Omega' \ K(\mathbf{\hat{s}} \cdot \mathbf{\hat{s}'}) \ w(\mathbf{\hat{s}'}) \tag{1}$$

where the vector symbol $\mathbf{\hat{s}}$ denotes a pair of variables $\theta$, $\phi$ to be, respectively, the polar and azimuthal angles specifying a point on the unit sphere with reference to a coordinate system at its center. Similarly, the vector symbol $\mathbf{\hat{s}'}$ denotes a pair of variables $\theta'$, $\phi'$ and $\mathrm{d}\Omega=\sin\theta \ \mathrm{d}\theta \ \mathrm{d}\phi$.

By expanding $w(\mathbf{\hat{s}'})$ using spherical harmonics we get

$$\ w(\mathbf{\hat{s}'}) = \sum_{l'=0}^{\infty} \sum_{m'=-l'}^{l} W_{l',m'} \ Y_{l',m'}(\mathbf{\hat{s}'}) \tag{2}$$

where $ W_{l',m'}$ are the coeffecients.

Inserting $(2)$ in $(1)$ we get

$$D_{l,m} = \sum_{l'=0}^{\infty} \sum_{m'=-l'}^{l} W_{l',m'} \int_{\mathbf{\Omega}}^{} \mathrm{d}\Omega \ Y_{l,m}(\mathbf{\hat{s}})\int_{\mathbf{\Omega'}}^{} \mathrm{d}\Omega' \ Y_{l',m'}(\mathbf{\hat{s}'}) \ K(\mathbf{\hat{s}} \cdot \mathbf{\hat{s}'}) \tag{3}$$

The kernel $K(\mathbf{\hat{s}} \cdot \mathbf{\hat{s}'})$ is defined as

$$ K(\mathbf{\hat{s}} \cdot \mathbf{\hat{s}'}) = k(\mathbf{\hat{s}} \cdot \mathbf{\hat{s}'}) \ \delta(\mathbf{\hat{s}} \cdot \mathbf{\hat{s}'}-a) \tag{4}$$

where $k$ is some function of $ \mathbf{\hat{s}}$ and $ \mathbf{\hat{s}'}$.

Inserting $(4)$ in ($3$) gives

$$D_{l,m} = \sum_{l'=0}^{\infty} \sum_{m'=-l'}^{l} W_{l',m'} \int_{\mathbf{\Omega}}^{} \mathrm{d}\Omega \ Y_{l,m}(\mathbf{\hat{s}}) \int_{\mathbf{\Omega'}}^{} \mathrm{d}\Omega' \ Y_{l',m'}(\mathbf{\hat{s}'}) \ k(\mathbf{\hat{s}} \cdot \mathbf{\hat{s}'}) \ \delta(\mathbf{\hat{s}} \cdot \mathbf{\hat{s}'}-a) \tag{5}$$

My question is, how can I use the orthogonality of spherical harmonics or any other form/ relation of $Y_{l,m}(\mathbf{\hat{s}})$ and $Y_{l',m'}(\mathbf{\hat{s}'})$ to get rid of the integrals?. I know I can use addition theorem for spherical harmonics as:

$$P_l''(\mathbf{\hat{s}} \cdot \mathbf{\hat{s}'}) = \frac{4\pi}{2l''+1} \sum_{m''=-l''}^{l''} Y_{l'',m''}(\mathbf{\hat{s}}) \ Y_{l'',m''}(\mathbf{\hat{s}'}) \tag{6}$$

but how can I use some kind of a sifting property of delta functions to remove the integrals in equation $(5)$?

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I won't give the full formulae (all those indices!) but I believe the answer is simple enough to explain without them. Your strategy is correct, and you already gave the answer: it is the orthogonality of the spherical harmonics (not the sifting property of the delta function in the definition of $K$) which enables you to do the angular integrals.

  1. Expand $K(\hat{\mathbf{s}}\cdot\hat{\mathbf{s}}')$ in Legendre polynomials $$K(\hat{\mathbf{s}}\cdot\hat{\mathbf{s}}')=\sum_{\ell''} K_{\ell''}P_{\ell''}(\hat{\mathbf{s}}\cdot\hat{\mathbf{s}}').$$ This gives (you may want to double check this) $$K_{\ell''}=\frac{2\ell''+1}{2} P_{\ell''}(a)k(a).$$
  2. Use the spherical harmonic addition theorem for each term, giving you $$K(\hat{\mathbf{s}}\cdot\hat{\mathbf{s}}')=\sum_{\ell''}\sum_{m''} K_{\ell''}\left(\frac{4\pi}{2\ell''+1}\right) Y_{\ell''m''}^*(\hat{\mathbf{s}})\,Y_{\ell''m''}(\hat{\mathbf{s}}').$$ Be careful, one of the $Y$'s should be complex-conjugated (at least, in my books that's the case).
  3. Having expanded $w(\hat{\mathbf{s}}')$ in spherical harmonics, there are no other angular variables left to worry about, so the angular integrations are separate, and of the forms $$\int d\Omega \, Y_{\ell m}(\hat{\mathbf{s}}) Y_{\ell'' m''}^*(\hat{\mathbf{s}}) \qquad \int d\Omega' \, Y_{\ell' m'}(\hat{\mathbf{s}}') Y_{\ell'' m''}(\hat{\mathbf{s}}') $$ both of which give you Kronecker deltas in the various indices. Again, you need to take care with the complex conjugation which will dictate some signs of the indices appearing in the Kronecker deltas. Things may look simpler if your expansion of $w(\hat{\mathbf{s}}')$ is written in terms of the conjugates of $Y_{\ell'm'}$, assuming $w$ is real, so both orthogonality relations turn out to have one $Y$ and one $Y^*$ each.
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  • $\begingroup$ Great, I got it now. Thank you. $\endgroup$ – dykes Aug 26 '18 at 19:42

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