I am trying to prove that the operator $L^2 = -\partial_\theta^2 - \cot\theta\,\partial_\theta - \frac{1}{\sin^2\theta}\partial_\phi^2$ fulfills the following property:

For $y_{l,m} = L_{\,-}^{l-m}\left(e^{il\phi}\sin^l\theta\right),\; m = -l,-l+1,...,l$ and $L_\pm=e^{\pm i\phi}\left(\pm\partial_\theta + i\cot\theta\,\partial_\theta\right)$, $\langle f,g\rangle:=\int_0^\pi\int_0^{2\pi}\bar{f}\cdot g\sin^2\theta\,d\phi \,d\theta\;$ we have $$\langle L^2y_{l,m}, y_{l^\prime,m^\prime}\rangle = \langle y_{l,m},L^2 y_{l^\prime,m^\prime}\rangle.$$ The point of doing this is to show that the $y_{l,m}$ are an orthogonal set by using the eigenfunction property $L^2 y_{l,l} = l(l+1)y_{l,l}$. I already know that $L^2$ commutes with $L_{\,-}$(, so it suffices to show the above property for the $y_{l,l}=\sin^l\theta \,e^{il\phi}$). I have done it with explicit computation, but this is really tedious and I think there should be a more elegant way, using the property $L^2 = L_+ L_{\,-} + L_z^2 -L_z,\; L_z=-i\partial_\phi$ or the like. Any ideas?

EDIT: I just realized it does NOT suffice to show it for the $y_{l,l}$.



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