# Completeness relation of spherical harmonics

In spherical coordinates, the resolution of the identity can be written as $$1=\int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta\, d\theta\, |\theta,\phi\rangle\langle\theta,\phi| \equiv \int d\Omega |\Omega\rangle\langle \Omega|,$$ where $$|\Omega\rangle = |\theta,\phi\rangle$$. For spherical harmonics $$Y_{lm}(\Omega)$$, we then have $$\delta_{l'l}\delta_{m'm} = \int d\Omega\, Y_{l'm'}^\ast(\Omega)\, Y_{lm}(\Omega).$$ The resolution of the identity in the angular momentum basis is given by $$1=\sum_{l=0}^{\infty}\sum_{m=-l}^l |l,m\rangle\langle l,m|,$$ so that $$\langle \Omega \mid\Omega'\rangle = \sum_{l=0}^{\infty}\sum_{m=-l}^l \langle\Omega\mid l,m\rangle\langle l,m\mid \Omega'\rangle\iff \delta(\Omega-\Omega')=\sum_{l=0}^{\infty}\sum_{m=-l}^l Y_{lm}(\Omega) Y_{lm}(\Omega').$$

Now, the term $$\delta(\Omega-\Omega')$$ is often rewritten as $$\frac1{\sin\theta}\delta(\theta-\theta')\delta(\phi-\phi')$$. How does one find this expression?

For $$\delta^{(2)}(\Omega-\Omega')$$ to behave like a delta function, we should get $$1$$ when we integrate it over the surface of the unit sphere. In other words, we should have that
$$\begin{equation} 1=\int {\rm} d^2 \Omega \delta^{(2)}(\Omega-\Omega') = \int {\rm d}\theta {\rm} d \phi \sin \theta\delta^{(2)}(\Omega-\Omega') \end{equation}$$ You can see this will work out if we take $$\begin{equation} \delta^{(2)}(\Omega-\Omega')=\frac{1}{\sin\theta} \delta(\theta-\theta')\delta(\phi-\phi') \end{equation}$$ but will not work out if we do not include the $$(\sin\theta)^{-1}$$ factor. Indeed if we do not include this factor, then we will get $$\sin\theta'$$ instead of $$1$$.
More generally, in $$D$$ spacetime dimensions, one should write the $$D$$-dimensional Dirac delta function as $$\begin{equation} \frac{1}{\sqrt{|g|}}\delta^{(D)}(x) \end{equation}$$ In spherical coordinates on the unit sphere, $$\sqrt{|g|}=\sin \theta$$. This is another argument to explain the factor of $$(\sin\theta)^{-1}$$.