Spins in a Stern-Gerlach Experiment

Most recent answer: 11/15/2013

In the Stern–Gerlach experiment, the spin of a particle assumes either spin up or spin down states once it is in the presence of the magnetic field. If a particle assumes the spin state which is anti-aligned with the magnetic field, then why doesn't it spontaneously transition to the spin state of lower energy, parallel to the magnetic field. It seems to me, that since the particle is not in a potential well, it should behave this way. Thanks
- Jeff (age 23)
Idaho Falls,ID,USA

Hi Jeff,

If you put a spin in a magnetic field, it precesses about the field: . Spins only align with a field if there is some more complicated interaction. If you put a macroscopic sample in a magnetic field, most free spins will align with the field because all the spins can gain and lose energy via interactions with themselves or the medium.

If the field and the spin are aligned or anti-aligned, then they are eigenstates, which evolve in time simply by gaining phase*. 

So, if you do the Stern-Gerlach experiment carefully, with no perturbing fields acting on each spin as it passes through, then the spins won't transition to their lower-energy state, aligned with the field. I'm not sure how difficult this is in practice.


David Schmid

*This isn't quite true. Classically, an anti-aligned spin is not a stable state, so you might suspect that it also wouldn't be stable quantum mechanically. As it turns out, the anti-aligned spin state is an eigenstate of a Hamiltonian that treats the EM field classically, and thus would be stable. But it is not an eigenstate of the full Hamiltonian, in which the EM field is quantized too. If you solve this full problem exactly, you find a rate of spontaneous emission to the ground state+photon. This spontaneous emission from the anti-aligned to the aligned state does not happen in reverse, so it provides a nice analog to the classical notion of "unstable." However, if you make sure that the timescale of this spontaneous emission is long, then the analysis I gave above is accurate.

(published on 11/15/2013)