No r dependence in L operator?

[Demystifier]

I find it surprising that the r dependency vanishes. How can we explain this physically ?

A simple heuristic argument is to use the Bohr quantization condition of "old" quantum mechanics
$$L=n\hbar$$
which clearly does not depend on $r$. For a quick "derivation" of this condition see the second box in
http://hyperphysics.phy-astr.gsu.edu/hbase/Bohr.html

 

[Karolus]

I try to answer, let's see ... The mistake I think you try to find an analogy with the case "classic." But unfortunately does not go very far, in the classical case it makes sense to speak of a physical object rolling. In MQ, for example, we associate an angular momentum to the electron, which also has a "radius", but can not conceive an object as "rotating". For example, the photon has a spin equal to 1, but how do you represent a photon "rotating" on itself? In addition, the electron orbital angular momentum would make sense if it were a solid macroscpic object that rotates on itself, and it is not the quantum case. Any analogy with the classical case vanishes, as vanishes r. The same "radius" r is not a concept that has an analogy with the classical case. (Positions and spatial coordinates are "operators"). So the only thing consistent in your hand is the electron wave function, and the only thing you can do is calculate the eigenstates of L applied to $\psi$ . The rest of the work it does mathematics..

 

[Karolus]

we see it in a more concrete way a little bit maybe you can somehow get closer to a "picture" classic. Indeed on "large atoms" for example with more electrons, you would expect a "big angular momentum" as a fact (in a quantum) occurs.
Let us take the simplest case of the hydrogen atom in a first approximation, only Coulomb potential without further correction, relativistic spin-orbit etc.
The eigenfunctions of the Hamiltonian are of the type: $\psi_ {nlm} (\ r, \Theta, \Phi) = \ R_ {nl} (\ r) \ Y ^ {m} _ {l} (\Theta, \Phi)$
These are eigenfunctions of the total angular momentum $\hat {\mathbf L}^2$ eigenvalues $\hbar ^2l (l + 1)$
Where n, l and m are respectively the main quantum numbers, angular momentum and angular orientation with respect to an axis.
The condition on the number l is : $0 \leq l \leq n-1$
So, effectively increasing the quantum number n, the quantum number can take on progressively higher values with "more" angular momentum as a mechanic.
Remember that the shape of the wave function is spherical only in the case n = 1, but with n higher this shape is complicated, assuming shapes or ellipsoids with lobes arranged on the axes, for which it is impossible to speak of a "radius", although in the spherical case we can speak of "radius" only in a probabilistic sense