quantum mechanics - Orbital angular momentum of electrons

In a QM class, to study the hydrogen atom, we started by defining the Hamiltonian $H$ for a central potential, then made an orbital angular momentum operator appear as part of $H$, then down the line emerged spherical harmonics and probabilities of finding electrons in some regions around the nucleus. We never said that there was an electron that orbited the nucleus (as we would have done in classical mechanics, presumably) - that would be Bohr's model. There is no "trajectory" of the electron in the QM model, so I am not sure we could fall back on a classical interpretation and say that the electron actually carries an orbital angular momentum the way a planet does. Is there really for the electron a $\textbf{r}$ and a $\textbf{p}$ that we can observe and that we could use to calculate $\textbf{r}\times\textbf{p}$?

So, is the correct way of looking at the situation that there is a system that happens to be best modeled with some observables that happen to follow an angular momentum algebra, but not try to put too much classical meaning into those $J$'s ($\textbf{J}^2$, $J_x$, $J_y$, $J_z$, $J_+$, $J_-$)?

Answer

Angular momentum is that which is conserved under rotations. Equivalently, the angular momentum operators are the generators of rotations. This holds both classically and quantumly by (versions of) Noether's theorem.

Defining "angular momentum" as $\vec x \times \vec p$ classically and then showing that it is conserved is doing it the wrong way around from the Lagrangian and Hamiltonian perspective, and it is the Hamiltonian perspective that is the starting point for canonical quantization. So you should really start by looking at the generators of rotations $\mathfrak{so}(3)$.

Nevertheless, even quantumly, $L = x \times p$, as you may check by computing $[L_i,L_j]$ and observing that this operator also fulfills the commutation relations of $\mathfrak{so}(3)$.

This does not mean that there "is an $x$ and a $p$" (in the sense of eigenvalues) for every angular momentum eigenstate from which we could compute the angular momentum because the operators $x$ and $p$ do not commute with $L$, so we can have well-defined angular momentum of a state without having well-defined position or momentum.