quantum mechanics - 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.

Answer

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 $\mathbf{\mu} = \frac{q\mathbf{L}}{2m}$.