Hamiltonian operator in spherical coordinates

Question

I'm studying the hydrogen atom from a quantum mechanics perspective, but I'm having troubles understanding a step.

Consider the stationary Schroedinger equation: $$\hat H \psi = E\psi$$

Let $M$ be the mass of the nucleus, and $m$ the mass of the electron. Then the reduced mass is $$\mu = \frac{mM}{m + M} \approx m$$

Hence Hamiltonian can be written as $$\hat H = -\frac{\hbar^2}{2m}\nabla^2 + P(\vec r) = -\frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\varepsilon_0 r}$$ where $\nabla^2$ is the Laplace operator and $Z$ is the atomic number, so it should be $1$. The next step is the one I fail to understand. Since the potential has spherical symmetry, we change to spherical coordinates. The Hamiltonian is then $$\hat H = -\frac{\hbar^2}{2m}\underbrace{\left(\frac{\partial^2}{\partial r^2} + \frac2r\frac{\partial}{\partial r}\right)}_{\nabla^2} + \underbrace{\frac{\hat L^2}{2mr^2}}_{\text{??}} - \frac{Ze^2}{4\pi\varepsilon_0 r}$$ where $\hat L$ is the angular momentum operator. This passage comes with no explanation whatsoever. Here is what I have trouble understanding:

1. How can I derive the new expression for $\nabla^2$ in spherical coordinates?
2. Why is the angular momentum appearing?

Answer 1

1. Deriving the expression of the Laplace operator in spherical coordinates: http://digitalcommons.uconn.edu/cgi/viewcontent.cgi?article=1034&context=chem_educ

2. The angular momentum operator comes from gathering the angular parts $ (\theta, \phi) $ of the Laplace Operator:
   $$ {1 \over r^2 \sin \theta} {\partial \over \partial \theta} \left(\sin \theta {\partial \over \partial \theta} \right) + {1 \over r^2 \sin^2 \theta} {\partial^2 \over \partial \phi^2}. $$

In addition if you assume spherical harmonic solutions, you find: $\hat{L}^2 = l (l+1) \hat{1} $ as the operator equality when acting on the spherical harmonics. (https://en.wikipedia.org/wiki/Particle_in_a_spherically_symmetric_potential).