Quantum Central Force Problem and Angular Momentum I am currently studying the quantum mechanics of the hydrogen atom. We have a proton and an electron orbiting around, so the Hamiltonian is:
$$H=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}+U(|\vec{r}_2-\vec{r}_1|) \tag{1}.$$
It's a classic instance of the two body problem; so we can separate the Hamiltonian into the center of mass Hamiltonian and the relative Hamiltonian in the following way:
$$H=\frac{p^2_{CM}}{2M}+\frac{p^2_{rel}}{2\mu}+U(r)=H_{CM}+H_{rel} \tag{2}$$
where $r$ is the relative coordinate: $r=|\vec{r}_2-\vec{r}_1|$. So now, since the Hamiltonian is separable, we can write the eigenfunctions of it in the form:
$$\psi=\psi_{CM}\psi_{rel}. \tag{3}$$
Perfect. $\psi _{CM}$ it's easy to find. We are now left with the task of finding the eigenfunctions of $H_{rel}$. $H_{rel}$ clearly represents the motion of a particle in a central force field, we are dealing with the quantum analog to the classic central force problem. So with a little but of work we can rewrite $H_{rel}$ in the following way:
$$H_{rel}=\frac{p_{r}^2}{2\mu}+\frac{L^2}{2\mu r^2}+U(r) \tag{4}$$
where $p_r$ stands for:
$$p_r := -i\hbar\left(\frac{1}{r}+\frac{\partial}{\partial r}\right) \tag{5}.$$
Ok, now comes the problem: In my lecture notes is stated that we can now write the eigenfunctions of (4) in the following factorized form:
$$\psi(r , \theta , \phi) = \mathcal{R} _{\mathcal{E},l}(r)Y_{l,m}(\theta,\phi). \tag{6}$$
where $Y_{lm}$ are the eigenfunctions of $L_z,L^2$. From what I understand the reasoning behind this consists of three steps:

*

*We notice that, by construction, $[H_{rel},L_i]=0 \ \ ; \ \ i=1,2,3$

*The fact that the previous commutator is zero implies that the angular momentum is a conserved quantity under a system governed by the Hamiltonian $H_{rel}$

*The conservation of angular momentum implies then that we are allowed to write the wave function in the factorized form (6)

However I have problems understanding this; specifically I have the following doubts:

*

*Why is $[H_{rel},L_i]=0$ true by construction? I would have check this by actually computing the commutator; is there a better way?

*Why does the fact that the commutator is zero imply that the angular momentum is conserved? How do we prove this statement?

*And mainly: why does the fact that the angular momentum is conserved imply that we can write the wave function in the factorized form (6)?

 A: Conservation laws
I would start from the most basic question. Any operator evolves in time as:
$$O(t) = e^{\frac{i}{\hbar} Ht} O(0) e^{-\frac{i}{\hbar} Ht}$$
(this is a consequence of the Hamiltonian being the generator of time translations)
where $U(t) = e^{-\frac{i}{\hbar} Ht}$ is the time evolution operator. It follows straightforwardly that when we derive this equation with respect to time, we get:
$$\frac{d}{dt}O(t) = -\frac{i}{\hbar}\left[O(t),H\right]$$
therefore, if:
$$\left[O(t),H\right] = 0\rightarrow \frac{d}{dt}O(t) = 0$$
and so you can say that $O$ is conserved, i.e. it doesn't change with respect to time evolution.
Conservation of angular momentum
The conservation of angular momentum can be understood in terms of invariance under rotations (see edit below). While this explanation is quite elegant, it requires alot of prior knowledge, a more direct approach can be taken by considering the explicit definition of the  orbital angular momentum $L$:
$$L_i = (x\times p)_i =\sum_{jk}\epsilon_{ijk}x_j p_k =-i\hbar\sum_{jk}\epsilon_{ijk}x_j\frac{\partial}{\partial x_k} $$
The Hamiltonian for a central potential is written as:
$$H = \frac{p^2}{2\mu}+U(r) = \frac{p^2_r}{2\mu}+\frac{L^2}{2\mu r^2}+U(r)$$
where $r = \sqrt{x^2+y^2+z^2}$.
Now, remembering that $[x_i,p_j] = i\hbar \delta_{ij}$ we have:
\begin{align}
[L_i,p^2] &= \sum_l [L_i,p_l]p_l+\sum_l p_l[L_i,p_l]\\
&=\sum_{ljk}\epsilon_{ijk}([x_j,p_l]p_kp_l+p_lp_k[x_i,p_l])\\
&=i\hbar \sum_{jk}\epsilon_{ijk}(p_jp_k+p_kp_j) = 0
\end{align}
since $\epsilon_{ijk}$ is antisymmetric.
Now, we use the know result $[p,f(x)]=-i\hbar\frac{\partial}{\partial x}f(x)$ to obtain:
\begin{align}
[L_i,U(r)] = \sum_{jk}\epsilon_{ijk}x_j[p_k,U(r)]  =-i\hbar\sum_{jk}\epsilon_{ijk}x_j \frac{\partial r}{\partial x_k} \frac{\partial}{\partial r}U(r)
\end{align}
now:
$$\frac{\partial r}{\partial x_k} = \frac{x_k}{r} $$
therefore:
\begin{align}
[L_i,U(r)] = \sum_{jk}\epsilon_{jk}x_j[p_k,U(r)]  =-i\hbar\frac{1}{r}\frac{\partial}{\partial r}U(r)\sum_{jk}\epsilon_{ijk}x_j x_k = 0
\end{align}
notice that $x_j x_k= \frac{1}{2}\left([x_j,x_k]+\{x_j,x_k\}\right) =\frac{1}{2}\{x_j,x_k\}$ since the coordinates commute among themeslves.
We conclude that: $$[L_i,H] = 0$$ and that: $$[L^2,H] = 0,\qquad [L_i,L^2] = 0$$
so the angular momentum is a conserved quantity. (As was specified in a comment this follows trivially from the rotational invariance of the Hamiltonian)
Factorization of eigenvalues
Commutatitvity implies that there exists a common basis of eigenvalues between $H$, $L_i$ and $L^2$ (commonly we choose $i=3$, i.e. $L_z$), we call these $\rvert n,l,m\rangle$:
\begin{align} 
&H \rvert n,l,m\rangle= E_n \rvert n,l,m\rangle\\
&L_z \rvert n,l,m\rangle =\hbar m \rvert n,l,m\rangle\\
&L^2 \rvert n,l,m\rangle= \hbar^2 l(l+1)\rvert n,l,m\rangle
\end{align}
we define the wave functions in spherical coordinates:
\begin{align} 
&x = r \sin\theta\cos\phi\\
&y =r \sin\theta\sin\phi\\
&z = r\cos\theta  
\end{align}
as:
$$\psi_{nlm} = \langle r,\theta,\phi \rvert n,l,m\rangle=\psi_{nlm}(r,\theta,\phi)$$
now we project the first eigen-equation on the coordinate basis:
$$H\psi_{nlm}(r,\theta,\phi) =E_n\psi_{nlm}(r,\theta,\phi) = (\frac{p^2_r}{2\mu}+\frac{\hbar^2 l(l+1)}{2\mu r^2}+U(r))\psi_{nlm}(r,\theta,\phi)$$
now its obvious, this equation does not depend anymore on $\theta$ or $\phi$ and therefore  the solution must simplify as:
$$\psi_{nlm}(r,\theta,\phi) = u_{nl}(r)Y_{lm}(\theta,\phi)$$
where $Y_{lm}(\theta,\phi)$ are the common eigenvalues of $L^2$ and $L_z$ just do the same on the eigen-equations of $L_i$ and $L^2$ and you'll see that:
\begin{align} 
&L_z \psi_{nlm}(r,\theta,\phi) =\hbar m \psi_{nlm}(r,\theta,\phi)\\
&L^2 \psi_{nlm}(r,\theta,\phi)= \hbar^2 l(l+1)\psi_{nlm}(r,\theta,\phi)
\end{align}
where in the coordinate basis:
\begin{align} 
&L_z = -i\hbar\frac{\partial}{\partial\phi}\\
&L^2 = -\hbar^2\left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right)
\end{align}
and therefore variables separate in these equations too (only $Y_{lm}(\theta,\phi)$ remains).
We conclude that:
$$E_n u_{nl}(r) = (\frac{p^2_r}{2\mu}+\frac{\hbar^2 l(l+1)}{2\mu r^2}+U(r))u_{nl}(r)$$
Rotational invariance
Consider now  a generic rotation of the coordinate axis representated by a matrix $R_{ij}$ such that:
$$x'_i = \sum_{j}R_{ij} x_j$$
and
$$p'_i = \sum_{j}R_{ij} p_j$$
it is clear that the square of any vector is left invariant under rotations since a rotation by definition cannot change its length:
$$x'^2 = \sum_i\sum_{j}R_{ij} x_j \sum_{k}R_{ik} x_k = x^2$$
then we find the condition:
$$\sum_{j} R_{ij}R_{jk} = \delta_{ik} $$
namely:
$$R^T R=1$$
where $R^T$ is the transpose matrix.
A rotation $R$ induces on the Hilbert space a unitary transformation $U(R)$ such that the coordinate operators tranform as:
$$U(R)^{\dagger}\hat{x}_i U(R) =\sum_j R_{ij}\hat{x}_j$$
this implies that our Hamiltonian is rotationally invariant:
$$U^{\dagger}(R)\,\hat{H}\, U(R) =\hat{H}$$
since it is only a function of $r$ and $p^2$.
Consider now a rotation by an infitesimal  angle:
$R_{ij} = \delta_{ij}+\omega_{ij}+O(\omega^2)$
then it can be proven that:
$$U(1+\omega) = 1+ \frac{i}{2\hbar}\sum_{ij} \omega_{ij}L_{ij}$$
namely, the angular momentum is the generator of rotations. If we substitute this infinitsimal transformation into the tranformation law for the Hamiltonian we find:
$$[H,L] = 0$$
so the invariance of Hamiltonian under roations causes the angular momentum to be conserved.
A: The serial nature of the skewed statements and braided questions makes it very hard to come up with a decent answer without teaching the subject properly as mainstream texts do: Sakurai, Landau-Lifshits, Merzbacher, Messiah...

We notice that, by construction, $[H_{rel},L_i]=0 \ \ ; \ \ i=1,2,3$

"Construction" is unfortunate. "Inspection" is better. It is trivial to show that r is rotationally invariant, because $r^2=\vec x \cdot \vec x$ is, trivial to show in standard introductions of angular momentum.

The fact that the previous commutator is zero implies that the angular momentum is a conserved quantity under a system governed by the Hamiltonian $H_{rel}$

Noether's theorem: every continuous symmetry (exponentiation of your commutator) corresponds to a conserved quantity (charge). All good texts and courses cover it, and so is Wikipedia.

The conservation of angular momentum implies then that we are allowed to write the wave function in the factorized form (6)

"Implies we are allowed" is a sadistic way of saying "leads to". As detailed in another question, the three conserved $\vec L$s can be put together into the operator $L^2$ in the hamiltonian, which does not act on radial coordinates, so you see the corresponding p.d.e. separates its variables into functions of r and functions of θ,φ, so not quite complete separation.

Why is $[H_{rel},L_i]=0$ true by construction? I would have check this by actually computing the commutator; is there a better way?

There is an efficient straightforward way, as indicated above.
See WP.

Why does the fact that the commutator is zero imply that the angular momentum is conserved? How do we prove this statement?

Noether's theorem, see above.

And mainly: why does the fact that the angular momentum is conserved imply that we can write the wave function in the factorized form (6)?

Hinted above. See this answer.  The conservation of angular momentum is useful for the separation of variables, but even in classical mechanics proving superintegrability is nontrivial.
