# How does the Hamiltonian operator contain the angular momentum operator?

The time-dependent Schrödinger equation is $$\hat H \Psi = i\hbar \partial_t \Psi$$ When solving this equation for the hydrogen atom (in position space) by separation of variables, one gets not only the the eigenvalues of the hamiltonian (i.e., possible energies), but also the quantum numbers $$\ell$$ and $$m$$, which are related to the eigenvalues of the angular momentum operators $$\hat L_z$$ and $$\hat L^2$$, though they were not used explicitly. Does $$\hat H$$ somehow contain those operators ? and if not where do $$\ell$$ and $$m$$ come from ?

Since the eigenstates of $$\hat L_z$$ and $$\hat L^2$$ are degenerate wrt those of $$\hat H$$, I thought they cannot be found so explicitly (except by a very happy coincidence!)

• I think this is a homework level question, as the answer is given in any good quantum mechanics textbook. Feb 23, 2021 at 9:30
• What is the homework ? Not every two operators $A$ and $B$ that commute $[A,B]=0$ are such that $A = cB + E$, but this Hamiltonian, may every other Hamiltonian, is such that $\hat H = c\hat L^2 + \cdots$, and you can't tell until $\hat H$ and $\hat L^2$ are represented in the position space or some other space Feb 23, 2021 at 9:32
• What you wrote above about operators doesn't make sense or at least requires more explanation. The conservation of angular momentum in atoms has to do with spherical symmetry - whether you describe it in classical or quantum way (remember Kepler laws?) Feb 23, 2021 at 9:38
• I restate here: if $A$ and $B$ are such that $[A,B] = 0$ it doesn't follow that $A = cB+\cdots$. But $\hat H$ is such that $\hat H = c\hat L^2+\cdots$, without $[\hat H,\hat L^2] = 0$. It depends on the dots, i.e. the potential $V(\mathbf r)$, whether $[\hat H,\hat L^2] = 0$ holds or not Feb 23, 2021 at 9:43
• @Vadim Just because something can be found in a text book does not mean it is a homework question. The site policy on homework questions says nothing about that. Feb 23, 2021 at 12:33

When operators commute there exists a basis of common eigenvectors. Since for the hydrogen atom $$\hat H$$ commutes with $$\hat{L^2}$$ as well as a component of $$\hat{\mathbf L}$$ (that we usually take to be $$\hat L_z$$) we know that there is a basis of common eigenstates to all three operators.

The topic of what is "in" the hamiltonian for the hydrogen atom:

$$\hat H = -\frac{\hbar^2}{2\mu}\nabla^2 - \frac{e^2}{4\pi\epsilon_0 r}$$

is a big topic.

It is invariant under rotations and can separated in spherical coordinates, with:

$$\nabla^2 = \frac 1 {r^2} \frac{\partial}{\partial r}\big(r^2\frac{\partial}{\partial r}\big) - \frac 1 {\hbar^2 r^2}\vec L^2$$

This leads to the $$l$$ quantum number and the S, P, D, ... shells solutions. Since $$[L^2, L_z]=0$$, the representation theory of SO(3) naturally leads to the integer $$m$$ quantum numbers with bound $$|m| \le l$$. Rotational invariance means the $$m$$ are degenerate for fixed $$l$$.

The hydrogen hamiltonian is also separable in parabolic coordinates because the hamiltonian commutes with the (classically conserved) Runge-Lenz vector:

$$\vec A = \frac 1 2(\vec p \times \vec L - \vec L \times \vec p) - \frac{\mu e^2 \vec r} r$$

This amounts to a hidden SO(4) symmetry and is the reason the $$l$$ are degenerate so that the energy only depends on $$n$$. (The classical counter part is that orbit energy does not depend on eccentricity, rather it only depends on the semi-major axis).

It's all in $$\hat H$$.

• I think the radial term of the Laplacian you've written isn't correct. Shouldn't it be $$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\right)?$$ Feb 22, 2021 at 16:12
• $$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\right) \iff \frac{1}{r}\frac{\partial^2}{\partial r^2}r$$ can you prove it ? Feb 23, 2021 at 9:47