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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 ($\textbf{J}^2$, $J_x$, $J_y$, $J_z$, $J_+$, $J_-$)?

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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.

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  • $\begingroup$ So, that's what I thought: there is some operator $L$ in QM with some algebra and some conservation law, and we are just extremely lucky that it turns out to be $r\times p$, although there is not really any $r$ or $p$ for the electron. $\endgroup$
    – Frank
    Commented May 1, 2015 at 14:27
  • $\begingroup$ Can we say that angular momentum somehow just models the rotation invariance of the system under consideration? $\endgroup$
    – Frank
    Commented May 1, 2015 at 14:29
  • $\begingroup$ @Frank: No, we aren't "lucky", it's a built-in feature! Canonical quantization turns the classical Poisson brackets into commutators, so the classical generators of any group will also be mapped to the quantum generators of the same group. The only thing that can go wrong is that the classical object is not Hermitian in its quantized form, but it's really no luck that the quantized $L$ looks like the classical $L$. (Caveat: There can be quantum operators that do not arise from canonical quantization, e.g. spin.) $\endgroup$
    – ACuriousMind ♦
    Commented May 1, 2015 at 14:32
  • $\begingroup$ Doesn't your caveat show that we are "lucky" in the sense that the more accurate QM picture could very well have been different - I mean there is some amount of arbitrariness in deciding how to carry out the canonical quantization procedure, and already spin doesn't fit :-) How about we tell the story another way: there is rotational symmetry, therefore there is a conserved quantity which we can call "$J$", that together with $J_+$ and $J_-$ follows an algebra typical of rotational symmetries? From that starting point, we should be able to express the symmetry in the Hamiltonian, $\endgroup$
    – Frank
    Commented May 1, 2015 at 16:13
  • $\begingroup$ then solve to find spherical harmonics, with a convenient choice of coordinates, e.g. $r, \theta, \phi$. No need to start from a classical analogy where the electron would have a $J$ because it orbits the nucleus? The only thing that bothers me in this story is that we choose a $z$ axis at some point, which breaks the spherical symmetry of the system. $\endgroup$
    – Frank
    Commented May 1, 2015 at 16:15

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