Why can we separate the radial and angular dependence for the eigenstates of this Hamiltonian? In other words, how can we be sure that all of the solutions must be separable? The Hamiltonian is the one given for a single particle in a central potential: (the reduced form for the simple 2-body  Hamiltonian)
$$
\hat{H}=\frac{\hat{\mathbf{p}}^{2}}{2 \mu}+V(|\hat{\mathbf{r}}|)
$$
From text:
One of the most evident features of the position-space representations (9.117), (9.127), and (9.128) of the angular momentum operators is that they depend only on the angles $\theta$ and $\phi$, not at all on the magnitude $r$ of the position vector. Rotating a position eigenstate changes its direction but not its length. Thus we can isolate the angular dependence and determine $\langle\theta, \phi \mid l, m\rangle$, the amplitude for a state of definite angular momentum to be at the angles $\theta$ and $\phi$. These amplitudes, which are functions of the angles, are called the spherical harmonics and denoted by
$$
\langle\theta, \phi \mid l, m\rangle=Y_{l, m}(\theta, \phi)
$$
Expressed in terms of these amplitudes, the energy eigenfunctions of the Hamiltonian (9.57) are given by
$$
\langle r, \theta, \phi \mid E, l, m\rangle=R(r) Y_{l, m}(\theta, \phi)
$$
I cannot quite understand the argument here. How is the fact "Rotating a position eigenstate changes its direction but not its length." and the fact that the generator of these rotations are angular momentum operators imply that the energy eigenstates must be separable?
I can include further details if necessary.
 A: We know the eigenstates can be separable because we can decompose the equation into two independent components - the angular and the radial one. They are independent because rotation doesn't affect the radial part. In that sense, it is not different than saying that if we can decompose the equation into two independent components - one only depending on $x$ and the other on the $y$ coordinate, than there exist a set of solutions which is separable in $x$ and $y$. That is because translations in $x$ do not affect the $y$ coordinate and vice-versa.
Mathematicaly, we can see it in the following way: we have an equation $H\Psi(\vec{r}) = E\psi(\vec{r})$ where $H$ is some differential operator, and we find out that we can decompose it into
$$ E\Psi(\vec{r}) = H_{\rm ang}\Psi(\vec{r}) + H_{\rm radial}\Psi(\vec{r}) $$
where $H_{\rm ang}$ only affects the angles and $H_{\rm radial}$ only affects the radial part. Then we can use an ansatz of separable solution $\Psi(\vec{r}) = f(r)\Theta(\theta, \varphi)$ and get that the equation now reads
$$ E f(r) \Theta(\theta, \varphi) = f(r) H_{\rm ang}\Theta(\theta, \varphi) + \Theta(\theta, \varphi) H_{\rm radial}f(r)$$ we can divide by $\Psi = f\Theta$ and get
$$ E = \frac{H_{\rm ang}\Theta(\theta, \varphi)}{\Theta(\theta, \varphi)} + \frac{H_{\rm radial}f(r)}{f(r)}$$
the left hand side is constant, so if we change $\theta, \varphi$ only the first term on the right-hand-side can change, but it should still maintain the equation, so it must be constant as well, and likewise if we change only $r$. So both the terms on the right-hand-side are constant, and we can solve them separately. The fact that we can change one of the parts without affecting the other one is crucial here. Without that, we wouldn't be able to get to the conclusion that each of the is constant.
Also note, that this means that there is a set of solution of separable form. There might be other sets of solution (assuming that there is a degeneracy and the set of solutions is not unique) which will not be separable along these lines. An example of that you can get when examining the $2d$ harmonic oscillator:
$$ H = \frac{p_x^2 + p_y^2}{2m} + \frac{1}{2}m\omega^2(x^2+y^2)$$
you can separate it on $x$ and $y$ and get one set of solutions, or go to radial coordinates and separate is on $r$ and $\varphi$ and get another set of solutions. Both of these sets are valid, and related to one another by a change of basis transformation.
