If electrons aren't revolving around the nucleus, why do atoms have orbital magnetic moment? In most introductory textbooks, the explanation of orbital magnetic moment is based on Bohr's model and electrons orbiting around the nuclues, which can be modeled as a current loop. For example, here.
But I've never seen an explanation without Bohr's model, using Schrödinger's equation. Would this be possible, or do we need some experimental hypothesis? In particular, I'd prefer avoid charge density orbitals-based explanations, if possible.
 A: Let's consider coupling a charged particle to a magnetic field in quantum mechanics.  Assume a uniform magnetic field for simplicity. The prescription for coupling to an EM field is the substitution $\mathbf{p} \rightarrow \mathbf{p} - q\mathbf{A}$. The Hamiltonian is then
\begin{equation}
H = \frac{\left(\mathbf{p} - q\mathbf{A}\right)^2}{2m} + V
\end{equation}
Or, expanding,
\begin{equation}
H = \frac{1}{2m}\left[p^2 +q^2A^2 - q\left(\mathbf{p}\cdot\mathbf{A} +\mathbf{A}\cdot\mathbf{p}\right)\right] + V
\end{equation}
If we work in the Coulomb gauge, $\nabla\cdot\mathbf{A} =0$, then $\mathbf{p}\cdot\mathbf{A} = \mathbf{A}\cdot\mathbf{p}$ and
\begin{equation}
H = \frac{1}{2m}\left[p^2 +q^2A^2 - 2q\mathbf{A}\cdot\mathbf{p}\right] + V
\end{equation}
We still have some gauge freedom here, so let's choose explicitly $\mathbf{A} = \frac{1}{2}\mathbf{B} \times \mathbf{r}$ so that $\mathbf{A} \cdot \mathbf{p} = \frac{1}{2}\left(\mathbf{B} \times \mathbf{r}\right) \cdot \mathbf{p} = \frac{1}{2} \left(\mathbf{r} \times \mathbf{p}\right) \cdot \mathbf{B} = \frac{\mathbf{L}\cdot\mathbf{B}}{2}$ by the cyclic symmetry of the triple product.
The full Hamiltonian is
\begin{equation}
H = \frac{1}{2m}\left[p^2 +q^2A^2\right] - \frac{q}{2m}\mathbf{L}\cdot\mathbf{B} + V
\end{equation}
and by analogy to the classical energy of interaction between a magnetic dipole and magnetic field, we define $\mathbf{\mu} = \frac{q\mathbf{L}}{2m}$.
A: Have you had a 3rd or 4th year QM course yet?  The simple answer is that the angular momentum is in the wave-function.  For spherically symmetric (3-D) potentials the solution to the Schrodinger equation is in terms of spherical harmonics (at least the radial part of the solution.)  For any of the solutions you can act on it with the angular momentum operator and find the angular momentum of that state. (Since you are a physics student some of that should make sense.)  (Way back when in undergrad we used McGervey's "Introduction to Modern Physics".. it's still on my shelf and he does a pretty good job of laying all that out.)    
Adding some to avoid excess comments,
"What's a magnetic moment?" Well for me a magnetic moment is the magnetic analogy of a dipole moment.  (dipole is two charges separated by a distance.)  A magnetic moment is a loop of current with some area.   
A: At the quantum mechanical level, past the Bohr model, to every observable there corresponds an operator. It is convenient to use the semiclassical definition of the magnetic moment, and the Bohr model, but an operator format does exist.

That then makes the total magnetic dipole moment operator of a single proton equal to 


The link goes on to explore many forms of magnetic moment at the quantized level..
This operator acting on the  wave function which gives the orbitals, where the probability of finding an electron is described can be used to extract the magnetic moment of a state. Thus it  is only the correspondence of "observable" to "operator" that is important. Not any hypothetical motion of an electron around the nucleus.
This is true for all observables at the quantum mechanical level. 
