# What does the * mean in spherical harmonics?

In Jackson's book about classical electrodynamics, this formula comes up: $$q_{lm} = \int \mathrm d^3 x' \, Y^*_{lm}\left(\theta', \phi'\right) r'^l \rho\left(\vec x'\right)$$

What does that $^*$ mean?

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what page? I have my copy right here –  Dylan Sabulsky Nov 4 '12 at 22:41
In my German one it is 170. It is equation (4.3). –  queueoverflow Nov 4 '12 at 22:43
–  Qmechanic Nov 4 '12 at 22:58

The superscript $*$ is a common notation for complex conjugate. Going back to check, (3.53) in the blue English edition states $$Y_{l,m} = \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P^m_l(\cos\theta)e^{im\phi}$$ which is followed by (3.54) $$Y_{l,-m}(\theta,\phi) = (-1)^m Y^*_{l,m}(\theta,\phi),$$ making is clear that it has to be complex conjugation.

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Unfortunately, I do not have this sense of “clear” yet. But this should help understanding what is going on. –  queueoverflow Nov 5 '12 at 12:32
To check (3.54) explicitly, if you replace $m$ with $-m$ in the equation for $Y_{l,m}$ and substitute (3.51) for $P^{-m}_l$, you will get something that looks like $Y_{l,m}$, but with an extra factor of $(-1)^m$ and $e^{im\phi}$ replaced by $e^{-im\phi}$. Except for $(-1)^m$, this is just the complex conjugate of $Y_{l,m}$, as $e^{im\phi}$ is the only complex factor. –  Stan Liou Nov 5 '12 at 16:36
$\exp(im\phi)$ makes sense. Thanks! –  queueoverflow Nov 5 '12 at 16:54

This is one way to denote the complex conjugate.

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The star * is probably what you think it is; the complex conjugate. Think about it like in quantum mechanics, $\langle \psi | \psi\rangle$ = $\int \psi^{*}\psi dx$. The spherical harmonics are complex.