# 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)?
• Eq (6) is just separation of variables in polar coordinates. I woudl readily become transparten, if you do the transformation to relative position and then to polar coordinates explicitly, rather than hiding them into $p_r$ and $L_i$. It is an excercise worth doing! Commented Mar 25, 2021 at 10:53
• @Vadim I was told that we can write the eigenfunction of an Hamiltonian $H$ in the factorized form only if the commutator between $H$ and the quantity you are trying to factorize is zero. Are you saying that this is not true and that we can always factorize? Commented Mar 25, 2021 at 10:56
• I am not saying that it is not true - it is a possible way to derive it, but clearly not the best for you. You will get a better insight by going through the calculations yourself. Btw, they are given in any good QM book (Schiff, Landau, etc.) Commented Mar 25, 2021 at 11:01
• @Vadim Maybe this topic needs a separate question. Commented Mar 25, 2021 at 11:17
• I have asked the more specific, related, question here. Commented Mar 25, 2021 at 11:50

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.

• Nice explanation, but I wonder if the OP is already familiar with the Heisenberg picture used in the first part. Furthermore, I think the angular momentum question could be answered more elegantly by using that the angular momentum operator is the infinitesimal generator of rotations and the Hamiltonian is rotation invariant. But frankly, I am to lazy at the moment to elaborate that. ;-) Commented Mar 25, 2021 at 17:27
– Fra
Commented Mar 25, 2021 at 17:58
• I am familiar with the Heisenberg picture. This is indeed a really well written answer. Commented Mar 26, 2021 at 10:56
• I am left with only one doubt: in the section Factorization of eigenvalues you say to project on the coordinate basis, this would leave us with $$\langle r, \theta, \phi| \frac{p_r}{2m}+\frac{L^2}{2mr^2}+U(r)|n,l,m\rangle=E\phi _{nlm}(r,\theta,\phi)$$ but you don't write that, you instead write: $$\langle r, \theta, \phi| \frac{p_r}{2m}+\frac{\hbar ^2 l(l+1)}{2mr^2}+U(r)|n,l,m\rangle=E\phi _{nlm}(r,\theta,\phi)$$ how did $L^2$ change into its eigenvalue? Commented Mar 26, 2021 at 11:05
• when you carry on the projection $L^2\rvert n,l,m\rangle = \hbar^2 l(l+1)\rvert n,l,m\rangle$
– Fra
Commented Mar 26, 2021 at 11:07

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.