I understand that the dynamical principle behind a Stern-Gerlach experiment is the coupling of a magnetic dipole moment to a (non-homogeneous) magnetic field. I also understand that when an electric charge is spinning (more generally: when an electric charge has angular momentum), it also has a magnetic dipole moment, proportional to the angular momentum.

My question is: doesn't the "orbital" angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ (with $\mathbf{r}$ and $\mathbf{p}$ given relative to the laboratory's frame) also have its own magnetic moment? In that case, when the particle is going through the magnetic field, under what conditions may I describe the dynamics (the interaction hamiltonian) via $\mathbf{S} \cdot \mathbf{B}$ instead of $\mathbf{J} \cdot \mathbf{B}$, with $\mathbf{J} = \mathbf{L} + \mathbf{S}$?


1 Answer 1


Simple answer is that the Stern-Gerlach experiment does couple to the total angular momentum. The original experiment was done with silver atoms, which have an electron in their outermost orbital with $L=0$ so that there would be no interaction with the magnetic field other than that of the spin.

Edit: The electron configuration for silver is $[\text{Kr}]4d^{10}5s^1$ by the way, the $5s$ electron is the one I'm talking about.


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