How do electrons actually move in the Stern Gerlach experiment? As I understand it, the Stern-Gerlach experiment discovered:
Fire an electron horizontally through a vertical magnetic field. Then, beyond the acceleration $e \textbf{v}\times \textbf{B}$ due to its charge, there is a motion upwards or downwards, whose magnitude depends only on $\textbf{B}$ and not on the particular electron or its initial speed.
However, so far I've not been able to find an actual description of how the electron moves. 


*

*What is the motion of an electron? That is, its motion is of the form $m\textbf{a}=e\textbf{v}\times \textbf{B}+ \square$ ; what precisely is $\square$ ?


I imagine $\square$ will look a bit like the motion of some sort of dipole moment. Given that the discreteness of this effect is a pretty non-classical, just knowing the form of $\square$ without a deeper reason would be pretty unsatisfying. So:


*Is the form of $\square$ deriveable from a more fundamental equation (Schrodinger, Dirac, ... ) for an electron?


It would be very nice to see an answer to 2 (and it's what I'm mostly interested in), especially because I've not yet seen a very concrete and interesting result come out of these fundamental equations.
 A: I'll answer your questions in order. The $\square$ term comes from the magnetic dipole moment of the electron. The energy of a magnetic dipole $\boldsymbol{\mu}$ in a magnetic field $\mathbf{B}$ is 
$$E = - \boldsymbol{\mu} \cdot \mathbf{B}$$
as one can compute using classical electromagnetism, for a current-loop model of a magnetic dipole. Differentiating, the force you wanted it just
$$\square = \mathbf{F} = \boldsymbol{\mu} \cdot \nabla \mathbf{B}$$
assuming that $\boldsymbol{\mu}$ is fixed. This is a really crucial assumption; if you don't make it you pick up extra terms comparable in size to the original one. In any case, this is how the Stern-Gerlach apparatus deflects electrons.
Now classically, if we imagine the electron as a little ball of charge, then it can have a magnetic moment because it spins, so we expect
$$\boldsymbol{\mu} \propto \mathbf{S}.$$
The quantum part is that $\boldsymbol{\mu}$ is fixed; the dipole moment is permanent and you can't change its magnitude like you could a real current loop. This is because it's proportional to the spin $\mathbf{S}$ which has the same characteristics, and those are fixed by the angular momentum commutation relations. Using the Dirac equation, which describes a spin $1/2$ particle, you can calculate
$$\boldsymbol{\mu} = \frac{q}{m} \mathbf{S}$$
which gives you the proportionality factor. But nothing else here requires the Dirac equation; the reasoning that $\boldsymbol{\mu} \propto \mathbf{S}$ is quite general. (Specifically, for an $SU(2)$ irrep there is a unique vector operator by Wigner-Eckart, so any two must be proportional.) Indeed the Stern-Gerlach experiment preceded the Dirac equation by several years.
