# 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.

1. 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:

1. 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.

• What is the source of the quote? Please don't just post random text copied from somewhere without saying what the source is. The Stern-Gerlach experiment was not actually done with electrons, and cannot be done with electrons. It was done with neutral silver atoms. – user4552 Apr 6 '18 at 0:51
• @BenCrowell That was intended as emphasis rather than a quote, sorry for the ambiguity. Does knzhou's answer carry over essentially verbatim (that is, $E=-\sum \mu_i \cdot \textbf{B}$ is the sum over the spins of the neutrons, protons and electrons in the atom) or are there interactions between the particles affecting the motion? – Meow Apr 6 '18 at 1:14
• No. Silver has a ground-state spin of 1/2, so you get 2 peaks. – user4552 Apr 6 '18 at 1:17
• The Stern-Gerlach experiment employed silver atoms, not electrons. An electron will move in a circular or helical fashion in a uniform magnetic field. In a non-uniform magnetic field there's also a "drift". See arxiv.org/abs/1504.07963 for some useful information. – John Duffield Apr 12 '18 at 12:33

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.
• Thank you very much! Could I ask for a couple of clarifications? Do you mean that $|\mu|$ is the same for all electrons, and for an individual electron the direction of $\mu$ can change with time, or that $|\mu|$ is the same for all electrons, and for each electron its direction is fixed too? Also, do you mean that the Dirac equation gives that $\textbf{F}=-q \textbf{S} \cdot \nabla \textbf{B} /m$, and is the derivation easy (I'm happy to try it myself, but it would be nice to know if it's doable)? – Meow Apr 5 '18 at 20:22
• @Meow You can change the direction of $\boldsymbol{\mu}$, by e.g. applying a magnetic field, just like in classical electromagnetism. Of course whenever you measure $\mu_z$ you will always get a result of $\pm |\boldsymbol{\mu}|$, just like for spins. – knzhou Apr 5 '18 at 20:25
• @Meow It's a bit of work but nothing impossible. It's easiest to derive the first equation I wrote; you basically write down the energy of a solution to the Dirac equation (by solving for $d\psi / dt$, since energy generates time translation) and look for the term proportional to $\mathbf{S} \cdot \mathbf{B}$. In a textbook it would be less than one page, but it's easy to get lost since the Dirac equation is a bit clunky to work with. – knzhou Apr 5 '18 at 20:28