How the express the position of a particle using the spherical harmonics?

42 Views Asked by At

I have a bit of a weird question, the spherical harmonics are defined as following for $m\ge0$:

$$ Y_{l,m}(\theta,\phi) = (-1)^m \sqrt{ \frac{(2l+1)(l-m)!}{4\pi (l+m)!} } P_{l,m}(\cos(\theta)) e^{im\phi} $$

Does anyone know if there's any operators able to "extract" the angles from this expression. For $\phi$, it should be something like this:

$$ \hat{\Phi} Y_{l,m}(\theta,\phi) = \left ( -i\frac{\partial}{\partial m} - \pi \right) Y_{l,m}(\theta,\phi) = \phi Y_{l,m}(\theta,\phi) $$

But for $\theta$, it's a bit more tricky and I don't know how to extract it from the Legendre polynomials. Any help would be appreciated. The Legendre polynomials are generated by the following expression:

$$ P_{l,m}(\cos(\theta)) = \frac{(-1)^l}{2^l \, l!} \sin^m(\theta) \frac{d^{l+m} \sin^{2l} (\theta) }{d\, \cos^{l+m} (\theta) } $$