Many questions have been asked on this site about the Stern-Gerlach experiment, but as far as I can tell this one hasn't. Does the following classical explanation of the SG experiment work?

Model electrons as a finite-sized hollow sphere of rotating charge $e$. Immediately after entering the SG device, the dipole moment of the electron quickly reorients itself to align with the magnetic field of the device. This is what I would expect to happen if a small bar magnet were placed in a region of high magnetic field gradient. Suppose this happens before the electron has traveled 1% of the distance of the device. Then, for the remaining 99% of the distance in the SG device, the electron is completely oriented either "spin-up" or "spin-down," so the binary "all-or-nothing" measurement is naturally predicted by this classical picture.

This would also work to explain sending the beam through multiple differently oriented SG devices, since the previous spin orientation of the electron is completely altered (very quickly) each time the electron enters a new device.


I don't think I was clear enough about the overall point of the question, since a few people have now brought up the fact that the spherical electron model has issues. I'm aware of that, but not really concerned with the specific electron structure model, so much as why some classical model wouldn't work to describe the binary output beam behavior (which knzhou gave a very nice answer to).

Perhaps a better hypothetical classical model would be: A classical point-particle with intrinsic angular momentum / magnetic dipole moment, the correct gyromagnetic ratio (supposing this could be a tunable classical parameter for point-particles), and some "braking mechanism" that allows it to quickly align with a magnetic field and stay aligned.

I think this question matters because the SG experiment is often used as a pedagogical example of quantum mechanics in introductory courses. When I first learned QM, I remember being confused about why this example was supposed to be so convincing, since it seemed that there could exist classical explanations of it.

  • 2
    $\begingroup$ Your model needs some kind of friction, otherwise a classical magnet would settle its direction in an external field, but just precess. $\endgroup$ Commented May 16, 2019 at 20:56
  • $\begingroup$ Models of electrons as spinning shells of charge have been tried. They don't work. $\endgroup$
    – G. Smith
    Commented May 16, 2019 at 20:58
  • $\begingroup$ @Sebastian Riese How about back-radiation emitted by the accelerated dipole? $\endgroup$
    – WillG
    Commented May 16, 2019 at 20:58
  • $\begingroup$ @WillG That can easily be checked by estimated the radiated power. $\endgroup$ Commented May 16, 2019 at 21:01
  • $\begingroup$ "Model electrons as a finite-sized hollow sphere of rotating charge $đť‘’$." This model has the wrong gyromagnetic moment, and the wrong quantization of angular momentum. $\endgroup$ Commented May 17, 2019 at 15:51

5 Answers 5


It's a decent question, as many people probably thought the same when the Stern-Gerlach experiment was first released. But there are many hurdles if you try to explain it classically. To get you started:

  • For spin $1$ particles, you get three spots and not two.
  • There is no mechanism given to dissipate the energy, to make it stay.
  • It is not simple to make a classical model where both spin up and spin down are energetically favored. In a straightforward model, if spin up has the highest energy, spin down has the lowest.
  • Microscopic classical models of spin generically get the spin wrong by a factor of $2$ and, given experimental data available at the time of the Stern-Gerlach experiment, require parts of the electron to be moving faster than light.
  • If you pass electrons through a vertical SG apparatus and select the spin up ones, then pass it through a horizontal SG apparatus, recombine the beams, and pass the result through a vertical SG apparatus, all of them will be spin up. This makes no sense in a model where a horizontal SG apparatus just rotates the spin to horizontal.

This is also ignoring the issue that permanent magnetic moments can't even exist in classical mechanics.

  • 2
    $\begingroup$ First point is pretty interesting—has SG been performed on higher-spin particles? $\endgroup$
    – WillG
    Commented May 16, 2019 at 21:34
  • 1
    $\begingroup$ I think the recombining has never been performed experimentally. There is also the 'humpty dumpty' paper doi.org/10.1007/BF01909939 which suggests this is not possible in reality. $\endgroup$
    – lalala
    Commented May 17, 2019 at 16:35
  • $\begingroup$ @unzhou, you write, "There is no mechanism given to dissipate the energy, to make it stay." What are you referring to? Please expand on what you is being made to stay, and where is it being made to stay? $\endgroup$ Commented Dec 6, 2021 at 14:33

First, in the Stern-Gerlach experiment, a beam of silver atoms was used, and not a beam of electrons. Secondly, the interpretation of data at that time (1922) based on the classical equations of motion. It is assumed that the silver atom has a magnetic moment on which the force acts

$$\vec {F}=\nabla (\vec {\mu}.\vec {B})……(1)$$

The role of quantum mechanics is reduced to the statement that the magnetic moment in equation (1) is quantized in proportion to the angular momentum or spin. In this form, the theory of the scattering of atoms in a non-uniform magnetic field was used to solve practical problems related to the determination of the magnetic moment of atoms using the Stern-Gerlach method down to the present time. As it is known, Stern and Gerlach obtained the following figure of silver atoms (on the left without a magnetic field, on the right in a magnetic field) Fig1 The question is, can we get the right figure using equation (1) with $\mu =\pm \mu_B$? I simulated the scattering of silver atoms in a magnetic field with a gradient as in the Stern-Gerlach experiment. All parameters of the gap and velocity of atoms were as in experiment. The result is shown in Fig. 2, where a) magnetic field lines and gap projection (red rectangle); b)magnetic field gradient (solid line) and experimental data (points) from Stern and Gerlach; c) atomic scattering pattern in a magnetic field; d) the same without a magnetic field. Comparing the scattering figures in Figs. 1 and 2, we find that Bohr’s hypothesis about quantizing the magnetic moment with projection $\pm\mu _B$ to the direction of the magnetic field is confirmed. Fig2

A similar result can be obtained without this hypothesis, simply using the Schrödinger–Pauli equation: $$i\hbar \frac {\partial \psi}{\partial t}=-\frac {\hbar ^2}{2m}\nabla^2\psi+\mu _B(\vec {B}.\vec {\sigma })\psi $$ $\vec {\sigma }=(\sigma _1,\sigma _2, \sigma _3)$ there is a Pauli matrix. Figure 3 shows the scatter pattern in the quantum model. Fig3

Update 1. Data shown above have been taken from several papers and books including

W . Gerlach, O . Stern . Der experimentelle Nachweis des magnetischen Moments des Silberatoms // Zeitschrift fur Physik, 1922, Volume 8, Issue 1, pp 110 - 111.

W. Gerlach,O. Stern. Der experimentelle Nachweis der Richtungsquantelung im Magnetfeld, Zeitschrift fur Physik 9 (1922) 349-352. The code for Mathematica shown below can be used to plot Figures as follows

Clear[p, \[Theta], \[Phi], q]
ge = .5; g = 0; L0 = .035; Wy = 
 0.0008/L0; Wz = .00003/L0; nz0 = 6; eps1 = 0; eps = 
 Pi/10; mB = -9.274078*10^(-24);
mAg = 107.868*1.6605655*10^(-27); kB = 1.380662*10^(-23); Tov = 
 1300 + 273.15;
v0 = Sqrt[
     mAg];(* data taken from W.Gerlach,O.Stern.Der experimentelle Nachweis des magnetischen Moments des Silberatoms//Zeitschrift fur 
Physik,1922,Volume 8,Issue 1,pp 110-111*)

t1 = 1; t0 = L0/v0;

h = 1.0545887*10^(-34);
\[Omega] = mB/h;

y0 = Wz;
z0 = Wy;

f[t_] := (1 - Sign[t - t1])/2;
(*Magnetic field data taken from W.Gerlach,O.Stern. Der experimentelle Nachweis der Richtungsquantelung im Magnetfeld, Zeitschrift fur Physik 9 (1922) 349-352 *)
Bsh = {{.00015, 23.6}, {.00020, 17.3}, {.0003, 13.5}, {.0004, 11.2}};
By[y_, z_, y0_, 
   z0_] := -(z + z0) (1/((y - y0)^2 + (z + z0)^2) - 
     1/((y + y0)^2 + (z + z0)^2));
Bz[y_, z_, y0_, 
   z0_] := (y - y0)/((y - y0)^2 + (z + z0)^2) - (y + 
      y0)/((y + y0)^2 + (z + z0)^2);
Byy[y_, z_, y0_, 
   z0_] := -(z + z0) (-((2 (y - y0))/((y - y0)^2 + (z + z0)^2)^2) + (
     2 (y + y0))/((y + y0)^2 + (z + z0)^2)^2);
Byz[y_, z_, y0_, z0_] := -1/((y - y0)^2 + (z + z0)^2) + 
   1/((y + y0)^2 + (z + z0)^2) - (z + 
      z0) (-((2 (z + z0))/((y - y0)^2 + (z + z0)^2)^2) + (
      2 (z + z0))/((y + y0)^2 + (z + z0)^2)^2);
Bzy[y_, z_, y0_, 
   z0_] := -((2 (y - y0)^2)/((y - y0)^2 + (z + z0)^2)^2) + 
   1/((y - y0)^2 + (z + z0)^2) + (
   2 (y + y0)^2)/((y + y0)^2 + (z + z0)^2)^2 - 
   1/((y + y0)^2 + (z + z0)^2);
Bzz[y_, z_, y0_, 
   z0_] := -((2 (y - y0) (z + z0))/((y - y0)^2 + (z + z0)^2)^2) + (
   2 (y + y0) (z + z0))/((y + y0)^2 + (z + z0)^2)^2;

Visualization of magnetic field and gradient

Show[{StreamDensityPlot[{Bz[y, z, y0*L0, z0*L0], 
    By[y, z, y0*L0, z0*L0]}, {z, -.0005, .0005}, {y, -0.0005, .0005}, 
   FrameLabel -> {"z", "y"}], 
  Graphics[{Opacity[.6], Red, 
    Rectangle[{-z0*L0 + 21*Wz*L0 - Wz*L0/2, -Wy*L0/2}, {-z0*L0 + 
       21*Wz*L0 + Wz*L0/2, Wy*L0/2}]}]}]


   23.6*Bzz[0, z, y0*L0, z0*L0]/
     Bzz[0, .00015, y0*L0, z0*L0], {z, .00015, .0005}, 
   AxesLabel -> {"z, m", "dBz/dz,T/cm"}], ListPlot[Bsh]}]

Figure 5

Visualization of silver deposition

k = ge*mB*L0/(mAg*v0^2); k1 = k*23.6*100/Bzz[0, .00015/L0, y0, z0];
eq = {Mx'[t] == 
    f[t]*g*(Bz[y[t], z[t], y0, z0] My[t] - 
       By[y[t], z[t], y0, z0] Mz[t]), 
   My'[t] == f[t]*g*(-Bz[y[t], z[t], y0, z0] Mx[t]), 
   Mz'[t] == f[t]*g*(By[y[t], z[t], y0, z0] Mx[t]), 
   Mx[0] == Cos[\[Theta]], My[0] == Sin[\[Theta]]*Cos[\[Phi]], 
   Mz[0] == Sin[\[Theta]]*Sin[\[Phi]], x''[t] == 0, 
   y''[t] == (1 + .0*(y[t]/y0)^2)*f[t]*
     k1*(My[t]*Byy[y[t], z[t], y0, z0] + 
       Mz[t]*Byz[y[t], z[t], y0, z0]), 
   z''[t] == 
    f[t]*k1*(My[t]*Bzy[y[t], z[t], y0, z0] + 
       Mz[t]*Bzz[y[t], z[t], y0, z0]), x'[0] == 1, x[0] == 0, 
   y'[0] == 0, y[0] == p, z'[0] == 0, z[0] == -z0 + 21*Wz + q};
pfuny = ParametricNDSolveValue[eq, 
  y[1], {t, 0, 1}, {p, \[Theta], \[Phi], q}]
pfunz = ParametricNDSolveValue[eq, 
  z[1], {t, 0, 1}, {p, \[Theta], \[Phi], q}] 
      pfunz[p, Pi/2 + eps1*j, Pi/2 + eps*j, q]*L0, {j, -nz0, 
       nz0}]/(2*nz0 + 1), 
    Sum[pfuny[p, Pi/2 + eps1*j, Pi/2 + eps*j, q]*L0, {j, -nz0, 
       nz0}]/(2*nz0 + 1)}, { q, -Wz/2, Wz/2}, {p, -Wy/2, Wy/2}, 
   PlotRange -> All, Mesh -> All], 
      pfunz[p, Pi/2 + eps1*j, -Pi/2 + eps*j, q]*L0, {j, -nz0, 
       nz0}]/(2*nz0 + 1), 
    Sum[pfuny[p, Pi/2 + eps1*j, -Pi/2 + eps*j, q]*L0, {j, -nz0, 
       nz0}]/(2*nz0 + 1)}, { q, -Wz/2, Wz/2}, {p, -Wy/2, Wy/2}, 
   PlotRange -> All, Mesh -> None]}]

Figure 6

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    $\begingroup$ i came to this thread late but still have the same question as knzhou If I drop a bunch of little magnets(magnetized ball bearings) through a much stronger magnetic field that is oriented horizontally, I get the same result as stern gerlach. Two little piles at the bottom (and a bunch stuck to the ends of the magnetic field generator poles). Clearly I am not doing a 'quantum' experiment. I do notice that angular alignment happens a lot faster than lateral displacement(ball bearings have nail polish dots). If I use a weaker magnetic field, the piles are sloppier (a more continuous distribution) $\endgroup$ Commented Jul 26, 2019 at 17:47
  • 1
    $\begingroup$ effectively, i have 'simulated' 2 quantum states by allowing classical forces to operate over time on a fixed uniform population. How is this different than using silver ? So if the same result can be obtained macroscopically with classical physics, why do we start saying it proves quantum spin? $\endgroup$ Commented Jul 26, 2019 at 17:52
  • $\begingroup$ @aquagremlin I solved this problem by assuming that classical particles with a magnetic moment oriented randomly create a similar scattering pattern. But I could not get a similar picture without the Bohr’s hypothesis about quantizing the magnetic moment with projection $\pm \mu _B$. $\endgroup$ Commented Jul 26, 2019 at 18:07
  • $\begingroup$ @AlexTrounev may I ask you, what software did you use to make the simulation ? whether it is a physics software or sth like Mathematica $\endgroup$
    – Physor
    Commented Sep 13, 2020 at 20:14
  • 1
    $\begingroup$ @user5419 Please, see update to my answer with code for Mathematica. $\endgroup$ Commented Jun 6, 2022 at 9:01

Modeling the electron as a finite-sized, spinning, charged spherical shell doesn’t work. Among a variety of problems, this model predicts the wrong value for the electron’s magnetic moment.

Consider a spherical shell of mass $m$, charge $e$, and radius $a$, spinning at angular velocity $\omega$. Its moment of inertia is


and its magnetic moment (in Gaussian units) is


(Source: http://www.physics.princeton.edu/~mcdonald/examples/rotatingshell.pdf)

The angular momentum is


and we know that for an electron this is $\hbar/2$.

We can then express the magnetic moment as


But the magnetic moment is measured to be slightly greater than twice this value.

  • $\begingroup$ Fair, but my point isn't so much about the specific model as it is about a classical explanation of the SG experiment, specifically the binary (not continuous) output beam separation. So if you prefer, assume classical point particles with a little extra physics thrown in to allow for intrinsic angular momentum and an arbitrary parameter for gyromagnetic ratio. These could easily be features of a classical theory, in the sense of not relying on quantum states, wavefunction collapse upon measurement, etc. $\endgroup$
    – WillG
    Commented May 16, 2019 at 21:16
  • $\begingroup$ This site is about mainstream physics. Rejecting quantum mechanics is about as far out of the mainstream as you can get. $\endgroup$
    – G. Smith
    Commented May 16, 2019 at 21:24
  • $\begingroup$ Asking about why alternative explanations don't work ≠ rejecting mainstream physics. $\endgroup$
    – WillG
    Commented May 16, 2019 at 21:26
  • $\begingroup$ Does that mean all questions about Bohmian mechanics should not be allowed on this site either? Bizarre, since there is a tag for it. $\endgroup$
    – WillG
    Commented May 16, 2019 at 21:27
  • $\begingroup$ My (extremely limited) understanding of Bohmian mechanics is that it is simply an interpretation of quantum mechanics, making exactly the same predictions as standard QM. So I think you are making a false comparison. $\endgroup$
    – G. Smith
    Commented May 16, 2019 at 21:31

There has been a recent demonstration of a Stern-Gerlach atom interferometer with high accuracy control of magnetic gradient on an atom chip (https://arxiv.org/pdf/1801.02708.pdf). Any attempt to picture the magnetic moment as a classical dipole pointing in a certain direction will fail to explain the interference pattern observed (see Figure 2 in the paper).


There are semiclassical equations of motion that fully explain the Stern-Gerlach experiment, see e. g. Section 5 of Gat, Lein & Teufel, Annales Henri Poincaré 15, 1967 (2014). It is important, though, that in this approach it is not the individual trajectories that matter, but averages of an ensemble of initial spins (which are obtained from a generalized Wigner transform of the quantum spin state). And this is how quantum mechanics comes into play, because only states from quantum mechanics give you a consistent interpretation.


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