# Question about torque in Stern Gerlach Experiment

1. Dec 5, 2011

### stevenb

I'm reading about the Stern Gerlach experiment in Sakurai's well-known book "Modern Quantum Mechanics". I remember studying this years ago, and somehow back then this question didn't enter my mind. However, I can't seem to figure out the answer.

I understand that a gradient in magnetic field can produce a force on the magnetic moment and split the beam of silver atoms into spin-up and spin-down. Sakurai describes how the beam is composed of randomly oriented atoms, and hence "there is no preferred direction for the orientation of the magnetic moment".

However, what confuses me is that I would expect the presence of the magnetic field to also put a torque on the magnetic moment. Why isn't there a torque that tries to orient the spin and give a bias to one of the two spin states?

2. Dec 6, 2011

### f.c.

well, would not the short answer be that there is a torque trying to re-orient the spin axis, but in order to actually flip the spin over, the angular momentum would need to change, and in a static external magnetic field that cannot happen, and the best that can be achieved is a precession, like a spinning top trying to tip over?

3. Dec 6, 2011

### Hans de Vries

The electron spin precesses due to the magnetic field. The spin four vector, the axial
current $j^\mu_\circlearrowleft$ changes in proper time $\tau$ just like the electric charge current density $j^\mu$

$$\frac{\partial j^\mu}{\partial \tau} ~~=~~ \frac{q}{mc}\,F^{\mu}_{~\nu}\,j^\nu$$
$$\frac{\partial j^\mu_\circlearrowleft}{\partial \tau} ~~=~~ \frac{q}{mc}\,F^{\mu}_{~\nu}\,j_{\circlearrowleft}^\nu$$

The latter is the Thomas-Bargmann-Michel-Telegdi eqation without the small magnetic
moment anomaly part (see for instance Jackson 11.164). Field tensor $F^{\mu}_{~\nu}$ couples the
E field with the boost generators K and the B field with the rotation generators J

$$F^{\mu}_{~\nu} ~~=~~ \Big(\,\mathsf{E}^i\,\hat{K}^i + \mathsf{B}^i\,\hat{J}^i\,\Big) \ =\ \left( \begin{array}{rrrr} ~\ 0\ \ & ~~\mathsf{E}_x & ~~\mathsf{E}_y & ~~\mathsf{E}_z \ \\ ~ \mathsf{E}_x & \ 0\ \ & ~~\mathsf{B}_z & - \mathsf{B}_y \ \\ ~ \mathsf{E}_y & - \mathsf{B}_z & \ 0\ \ & ~~\mathsf{B}_x \ \\ ~ \mathsf{E}_z & ~~\mathsf{B}_y & - \mathsf{B}_x & \ 0\ \ \ \end{array} \right)$$

The electric field boosts and the magnetic field rotates, both the charge-current density
as well as the spin-four vector. This happens already in a constant magnetic field while
the Stern Gerlach experiment requires a gradient in the magnetic field.

The precessing is caused because the phase change rates in time of the two spin states
differs due to the magnetic field. Therefor the phase between the two spin states changes
and it is this phase which determines the direction of the part of the axial current which is
orthogonal to the magnetic field.

Hans.

Last edited: Dec 6, 2011
4. Dec 6, 2011

### stevenb

Thank you both. It's always nice to get both the intuitive answer and the detailed mathematical answer. The former makes perfect sense to me, and the latter inspires me to break out Jackson and clean the rust out from between my ears.

5. Jan 11, 2012

### cygnet1

Another thing that I don't understand about the Stern-Gerlach experiment is that the results are usually attributed to the intrinsic spin of the electrons in the neutral silver atoms. Why couldn't the effect also be attributed to the nuclei? Why not to the orbital angular momentum of the electrons?

Last edited: Jan 11, 2012
6. Jan 11, 2012

### nonequilibrium

I think the spin magnetic moment of a nucleon is a factor 2000 smaller than that of the electron, so it is neglected in these cases.

Why it doesn't contribute, I cannot immediately say with confidence, but how you know that they aren't contributing you can deduce from seeing only two beams: two beams imply that there are only two m values and hence m is a half-integer. Orbital momentum doesn't give rise to such m values. Spin does.

EDIT: On second thought I think I also know why they don't contribute: it's performed with silver atoms, which, I think, only have one valence electron. All the other electrons are paired up (hence both their contributions of angular and spin momentum vanish), but the valence electron has a contributing spin (hence at least two beams). But since it is a sole valence electron, it will most likely be in the s state, hence l = 0 (hence no further splitting).

7. Jan 11, 2012

### cygnet1

Thank you. So I take it if you repeated the experiment with, say, iron atoms, you would expect to see more than two lines, correct?

8. Jan 11, 2012

### nonequilibrium

Well I don't know what you would expect with iron, because it has multiple valence electrons so I would suspect you would first have to do a summation of the separate angular momenta and then look what the possible z-projections are, which is not trivial (there might be an easier way to figure it out, but I don't know it), but your point is true: in general it need not be two lines. It's even possible to have no splitting at all, e.g. for noble gases (as all angular and spin momenta are cancelled).