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The answer to this question is probably negative, but let me ask it anyway.

If I have an expression of the form

$$\tag{1}I =\int\!\mathrm{d}\Omega\, Y_1^0Y_l^{*m}(\theta,\phi) \times f(\cos\theta), $$

where $f(\cos\theta)$ is some function of $\theta$.

My question is: can I set $l=1, m=0$ in equation (1) and write the $Y$'s in explicit form i.e. cosines and sines and a mix thereof, and perhaps be able to do the integral?

Without $f$ I would have

$$\tag{2}I =\int\!\mathrm{d}\Omega\, Y_1^0Y_l^{*m}(\theta,\phi)=\delta_{l1}\delta_{m0}.$$


See e.g. this mathworld page for conventions and definitions.

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    $\begingroup$ The expression is only nonzero for $m = 0 $ but you can't assume $l=1$ without more information about $f$ $\endgroup$
    – Spencer
    Jun 11 '14 at 9:46
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    $\begingroup$ Quickest explanation is that the $\phi$ only shows up in the spherical harmonics as a multiple of $e^{im\phi}$. When you do the $\phi$ integration the result is zero since $m$ is an integer. $\endgroup$
    – Spencer
    Jun 11 '14 at 9:54
  • $\begingroup$ Obviously this only works since nothing else in the integral depends on $\phi$. $\endgroup$
    – Spencer
    Jun 11 '14 at 9:54
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    $\begingroup$ @Spencer and Love I fixed the comments to address the typos. Hope all's clear now. $\endgroup$ Jun 11 '14 at 9:58
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While as Spencer commented, you cannot in general assume $\ell = 1$, you can do the next best thing. As he noted, necessarily we have to set $m = 0$, as the only $\phi$ dependence in the spherical harmonics comes from the $e^{im\phi}$ factor and since your function $f(\cos\theta)$ is independent of $\phi$, for all $m \neq 0$ the integral vanishes.

With $m = 0$ you have $$ Y^0_1(\theta,\phi) = P_1(\cos\theta) =\cos\theta $$ and $$ Y^0_\ell(\theta,\phi) = P_\ell(\cos\theta) $$ where $P_\ell$ are the Legendre Polynomials. In particular, your integral can be written as $$ I_\ell = 2\pi \int_0^\pi \sin\theta \mathrm{d}\theta~ \cos\theta P_\ell(\cos\theta) f(\cos\theta) = 2\pi \int_{-1}^1 x P_\ell(x) f(x) ~\mathrm{d}x. \tag{1}$$

Now you can make use of Bonnet's recursion formula for the Legendre Polynomials $$ (n+1) P_{n+1}(x) = (2n+1) x P_n(x) - n P_{n-1}(x) $$ which we can rewrite as $$ x P_{\ell}(x) = \frac{\ell+1}{2\ell+1} P_{\ell+1}(x) + \frac{\ell}{2\ell+1} P_{\ell-1}(x) \tag{2}.$$

Inserting (2) into (1) we have $$ I_{\ell} = 2\pi \left( \frac{\ell+1}{2\ell+1} F_{\ell+1} + \frac{\ell}{2\ell+1} F_{\ell-1}\right) $$ where the coefficients $F_\ell$ are the projection of $f$ onto the basis given by the Legendre polynomials, that is $$ F_{\ell} = \int_{-1}^1 P_\ell(x)f(x) ~\mathrm{d}x; \qquad f(x) = \sum_{\ell = 0}^{\infty} F_{\ell} P_{\ell}(x).$$

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  • $\begingroup$ Just for the record, your conventions differ from mine. $\endgroup$ Jun 11 '14 at 12:18
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    $\begingroup$ That is known to happen on occasions. :-) Thankfully, I think, the normalisation factors are not too hard to insert. $\endgroup$ Jun 11 '14 at 13:42

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