My understanding of the Stern–Gerlach experiment is that neutral (0 total charge) particles are sent through a non-homogeneous magnetic field, with the expectation that the field will push that particle's path higher or lower on a detector because of the collective spin of that particle. While the detector can detect particles in a two-dimensional surface, the results of the experiment are that particles appear in only two localized areas directly above the path of the particle stream, or directly below the path of the particle stream - spin up or spin down. The conclusions from these measurements are that the particle, when measured, will always have about the same magnitude.

Why does this not follow from classical mechanical theory related to magnetism? If you shot a magnet through a similar apparatus, I would expect the magnet to be rotated to align with the magnetic field in some way which, at high enough field strengths relative to the mass of the magnet, would cause us to measure basically the same magnitude as if the magnet entered the apparatus pre-aligned with the field.

How is my explanation incorrect?


The magnet has a finite moment of inertia. What would happen when the magnet with "wrong" orientation enters Stern-Gerlach apparatus? Of course, the magnetic field will exert torque on it. The magnet starts rotating. After it comes to the equilibrium orientation, i.e. is oriented along the field, the torque is zero, but angular velocity is at maximum, and the magnet overshoots — just as in motion of usual oscillator.

If you find average magnetic moment over all the motion time, i.e. multiple periods of oscillation, you'll find that it has smaller magnitude than actual magnetic moment of the magnet. This means that net displacement in the direction of field is smaller. Now if there're lots of such identical magnets with random initial orientations, they all will have random average magnetic moment, and thus their displacement will form a continuum instead of just two points.

  • $\begingroup$ Ah, i see, that makes sense. $\endgroup$ – B T Jan 24 '15 at 7:02
  • $\begingroup$ You wrote that the magnet starts oscillating, reaches equilibrium and overshoots etc. But i don't think that the flight time in the real stern Gerlach experiment (using silver atoms, for example), is enough to change the orientation of the spin by making it oscillate. If this was true then the experiment would change the orientation of the incoming atoms in complicated way and we would not be able to say that the atoms that are deflected in the same direction all have the same spin. My guess is that the moment of inertia is too large for the flight path to deviate the spin by large amount. $\endgroup$ – Prem kumar Aug 16 '17 at 9:16
  • $\begingroup$ @Raja I don't know what you expect to be changed WRT the classical expectation vs actual outcome of the experiment. If the moment of inertia is too large, then classically we'd still expect one elongated spot — because the spins then don't noticeably align with the field. $\endgroup$ – Ruslan Aug 16 '17 at 9:19
  • $\begingroup$ @ruslan yes, i agree. Actually my concern was not directly related to the OP's question that you were addressing in your answer. In many textbooks, they say that if an atom comes out as spin up from one stern Gerlach apparatus, then if we let it enter another stern Gerlach apparatus with the same orientation, then the atom will show the same deviation even in the second apparatus. But your discussion of oscillating magnets made we wonder how it is possible if the first apparatus makes the atom oscillate the way you describe. The solution i think lies in assuming a large moment of inertia. $\endgroup$ – Prem kumar Aug 16 '17 at 9:55
  • $\begingroup$ @Raja I think you misunderstood what I mean by oscillation. I don't mean translational oscillation — I mean rotational oscillation, i.e. the field applies some torque to the magnet, then the magnet starts rotation, at some moment of time aligns with the field, and then rotationally overshoots this orientation. Average translational motion will still be in one particular direction, not back and forth in the direction of the field. So you don't have to assume large moment of inertia. $\endgroup$ – Ruslan Aug 16 '17 at 9:58

Suppose you shot a large number of small classical magnetic dipoles with magnetic moment $\vec{\mu}$ through the field. Imagine the dipoles to be small enough that they could be treated as the particles of an ideal gas, and they are "boiled" out of some source into the magnetic field.

We would then expect each of the particle's components of velocity to be randomly distributed according to the Maxwell velocity distribution - because that is the classical result for an ideal gas. So the alignment of their magnetic moments would also start out random.

The dipoles would experience both a force and a torque from the magnetic field. The torque would cause them to rotate, and the force would, as you said, tend to line them up with the magnetic field, until they are in a minimum energy state. This alignment would take some time however, and, since they started out with random velocity and random orientation of their magnetic moment vector to the field, their final velocities once aligned would show some variation.

The key, though, that makes the motion of the particles vary AFTER they are aligned, is the nonuniform magnetic field. Suppose the field is in the z direction, and varies with z.

The particles are in a minimum potential energy state once aligned, with potential energy

$E=-\vec{\mu} \cdot\vec{B} = -\mu B $

But the magnetic field $B(z)$ varies with z, so the dipole still experiences a force

$\dfrac{\partial E}{\partial z} = F(z) = \mu\dfrac {\partial B(z)}{\partial z}$

So the classical dipoles, with randomly distributed magnetic moment orientations and velocities at start, would drift in varying directions, hit various positions on the detector.

But if the magnetic dipoles were somehow constrained to be on only two possible initial directions, you would expect to see a concentration of hits on two locations of the detector, and nothing anywhere else. They would start out with only two orientations with respect to the field and end up being deflected into only two concentrations on the detector. They'd have only two end states of "lining up" with the magnetic field, and then drift apart due to the nonuniform magnetic field.

So the Stern Gerlach experiment is evidence that the magnetic moment of electrons in atoms, and thus electron spin, is quantized, because the results resemble the second case above, not the first. The initial direction of the magnetic moment of the electron is limited by quantization of spin.

  • 1
    $\begingroup$ My question is around why classical mechanics would make one "expect a uniform distribution of hits on the detector", which your answer doesn't make clear. You didn't address my assertion of the magnets rotating to align with the magnetic field at all. $\endgroup$ – B T Jan 24 '15 at 2:37
  • 1
    $\begingroup$ Ah I see some of what you wrote here: physics.mq.edu.au/~jcresser/Phys301/Chapters/Chapter6.pdf . Thanks for the additional info, but I'm still a little confused as to why my explanation of a magnet aligning to the detection device is not correct. Since you seem to be saying that both atoms and magnets experience both force and torque, that isn't a reason their behavior should be different. $\endgroup$ – B T Jan 24 '15 at 3:38
  • 1
    $\begingroup$ It's not the fact that a dipole orients to (more precisely, precesses about) a magnetic field that differentiates the classical and quantum case. It's the fact that in the classical case there are infinitely many possible orientations. In the quantum case, there are only a finite number of possible orientations (with electrons, only two). That' $\endgroup$ – paisanco Jan 24 '15 at 5:00
  • 1
    $\begingroup$ Hmm.. this isn't helping me understand. I'm approaching this from the opposite direction you are. I want to know why classical mechanics doesn't predict the detection of particles in two localities, and you are explaining to me why QM does predict that. They are very different questions. $\endgroup$ – B T Jan 24 '15 at 5:13
  • 1
    $\begingroup$ Can someone show with a calculation how the classical magnet will behave? I haven't been able to find any "proof" that a classical magnet would not eventually align, just the assumption that it won't and that it will continue to oscilate. thanks $\endgroup$ – andy Jan 25 '16 at 14:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.