# How to prove spherical harmonics are orthogonal

A lot of texts and derivations eg here simply say:

"The Spherical Harmonics are orthonormal, so:

$$\int{ Y_l^m Y_{l'}^{m'} } = \delta_{ll'}\delta_{mm'}$$

And if you try any (l,m) pair you will find this always works out.

But how do you prove they are orthonormal for every $l$ and $m$? Where do you start? What principles do you use?

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## 1 Answer

Maybe not really an answer but you may get the idea nontheless: this is true more or less by construction. You get the spherical harmonics (as an example) as eigenfunctions of the angular part of the Laplace Operator, that is, they satisfy $$\Delta_{S^2} Y_{lm}(\vartheta,\phi) = \lambda Y_{lm} (\vartheta,\phi)$$ (Actually it turns out that this implies $\lambda = -l(l+1)$ with integer $l$) If you have such eigenfunctions for different eigenvalues it is a matter of linear algebra to show they are orthogonal, by looking at $$\int_{S^2}\langle \nabla_{S^2}Y_{lm}, \nabla_{S^2}Y_{l'm'}\rangle d\mu_{S^2}= -\int_{S^2}\langle Y_{lm}, \Delta_{S^2}Y_{l'm'}\rangle d\mu_{S^2}$$ This implies that the functions are orthogonal if $l\neq l'$, since otherwise you could derive $l(l+1) = l'(l'+1)$ from this. For fixed $l$ it turns out that you may solve the equation by a separation approach which leads to an ODE which is known to be solvable by orthogonal polynomials by ODE theory.

You can also write down the $Y_{ml}$ quite explicitly, see e.g. the german wikipedia page on "Kugelflächenfunktionen" http://de.wikipedia.org/wiki/Kugelfl%C3%A4chenfunktionen. If you look at these more closely and and do have some ODE background you may notice that the $\vartheta$ part are well known orthogonal polynomials in $\cos(\vartheta)$ (Legendre Polynomials), while the $\phi$ part is more or less just $e^{im\phi}$ which is known to a system of orthogonal functions. To then actually prove orthogonality is still a bit of work, but it kind of shows you the direction you should take.

To make them actually orthonormal you have to norm them, of course, that's where the complicated looking factors come from.

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Neat answer. I was not aware it was by construction! So there isn't a way to write an explicit proof, starting with $Y_l^m$. –  bobobobo Jun 4 '12 at 14:49
Sure there is. That's what I wanted to say with the 'You can also write down...' part. It's just a bit tedious. –  user20266 Jun 4 '12 at 14:50