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?


1 Answer 1


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 $(\sin\theta)^{-1}$.


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