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?


4 Answers 4


NOTE: this answer ignores initial angular momentum along the magnetic moment, so it isn't directly applicable to the silver atoms used in the actual Stern-Gerlach experiment. See the answer by Michael Seifert that takes angular momentum into account.

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, 2015 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
    Aug 16, 2017 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, 2017 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
    Aug 16, 2017 at 9:55
  • 2
    $\begingroup$ I don't think this is correct — it neglects that the electron already has an angular momentum aligned with the magnetic moment. The net result is then gyroscopic precession rather than oscillation, and the component of $\vec{\mu}$ along the magnetic field axis is expected to stay constant. See my answer below. $\endgroup$ Mar 19, 2021 at 20:14

The key realization is that the particles have an angular momentum $\vec{L}$ aligned with their magnetic moment $\vec{\mu}$. Specifically, $\vec{\mu} = \gamma \vec{L}$, where $\gamma$ is the gyromagnetic ratio.

So if such a particle enters a magnetic field, and we try to treat it classically, we would expect it to experience a torque, which will change its angular momentum: $$ \vec{\tau} = \frac{d\vec{L}}{dt} = \vec{\mu} \times \vec{B} = \gamma \vec{L} \times \vec{B} = -\vec{\Omega} \times \vec{L}, $$ where $\vec{\Omega} = \gamma \vec{B}$.

This can be recognized as the equation for gyroscopic precession. This implies that classically, the angular momentum (and therefore the magnetic moment of the particle) will not "flip" over to point along $\vec{B}$. Instead, it will precess about the axis of $\vec{B}$; and importantly, the component of $\vec{L}$ along $\vec{B}$ will remain constant at all times.

So if we fired a bunch of classical particles into a Stern-Gerlach device with a magnetic field along the $z$-direction, we would expect them to be sorted according to their $L_z$ component. Classically, this quantity is entirely continuous, and so we would expect a continuous spread of impacts along the screen. Of course, this is not what we actually observe.

  • $\begingroup$ Shouldn't we expect similar precession of the spin state in any evolution implied by Schroedinger's equation with external magnetic field as well? Something else than Schr. equation evolution is making the spins align/antialign irreversibly. Maybe it's like with the Rabi oscillations (when we have isolated non-degenerate energy eigenstates) vs. irreversible absorption (when we have continuum of states). $\endgroup$ Feb 27, 2023 at 1:34
  • $\begingroup$ @JánLalinský: It's not hard to show that the quantum expectation values $\langle S_x \rangle$, $\langle S_y \rangle$, and $\langle S_z \rangle$ obey their classical equations of motion, as one would expect. But the point is that in neither case (quantum or classical) does a strong magnetic field force the spins to completely align with the field, as the OP postulated. And once you realize that, the fact that there are only two possible $L_z$ values becomes very hard to explain in the classical case. $\endgroup$ Feb 27, 2023 at 3:27
  • 1
    $\begingroup$ I agree. But since the same is the case in the quantum case described by SE, it seems that fact is hard to explain even in quantum theory. $\endgroup$ Feb 27, 2023 at 4:06

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, 2015 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, 2015 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, 2015 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, 2015 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, 2016 at 14:05

Years have passed since the original post but this article by De Raedt et al. from last year by might be helpful. (The paper is about neutrons, not silver atoms.)

Classical, Quantum and Event-by-Event Simulation of a Stern–Gerlach Experiment with Neutrons

Three different ways of modelling a Stern-Gerlach experiment are implemented with the authors' software. The QM and classical models differ in results, with the QM model matching experimental results. The interesting bit is their third model named "Event by event". This is, in essence the same as the classical model but with the minimal necessary 'artificial' modification required such that the results match the results of the QM model.

The paper provides an answer to your question because it pinpoints how the classical model fails and approximately when and where in the trajectory of the particle some hocus pocus has to be applied to the classical model to achieve results that match experimental data.

The hocus pocus is this: The event by event model assumes that the axis of spin will instantly assume one of two orientations as the particle enters a magnetic field of a strength higher than a certain threshold. A random number generator is used to decide that orientation. Other than that, this event by event model is purely classical in nature.

  • $\begingroup$ This seems like an amazing paper and post. I have been revisiting this question for years wondering how strong do the usual arguments hold, and I've been tempted to create a simulation myself. I appreciate this. $\endgroup$ Feb 26, 2023 at 19:51
  • $\begingroup$ That is very interesting, and was basically the hypothesis I started with when asking the original question. What this says to me is that classical analysis of a magnet will approach the quantum behavior as the magnet aligns more quickly to the magnetic field. I might guess that as the "magnet" gets smaller, it will align more quickly. Others have said that in classical analysis, the magnet will simply oscilate back and forth, however there will realistically always be energy loss from that oscilation, whether its an electron or a bar magnet. Eventually that loss results in alignment of poles $\endgroup$
    – B T
    Mar 1, 2023 at 19:01
  • 1
    $\begingroup$ Below is another article that might be useful. Personally, the maths is beyond me and it is a preprint that hasn't been peer reviewed. However, the author claims to have devised a 'phenomenologcal' mathematical model of the original S-G experiment from a classical point of view. The impressive bit is that his modelled experimental results don't look like the ideal 'two spots' that QM predicts but a lot like the messier experimental results including the asymmetry. $\endgroup$
    – jon_m
    Mar 3, 2023 at 19:19
  • 1
    $\begingroup$ Rashkovskiy, S. Phenomenological Theory of the Stern-Gerlach Experiment. Preprints 2022, 2022100478. doi.org/10.20944/preprints202210.0478.v1. $\endgroup$
    – jon_m
    Mar 3, 2023 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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