I've been delving into Bell test experiments and woefully most sources fall into either dense physics papers, or very low level simplifications with lots of handwavium.

One thing in particular I'd like to understand a bit better is the Stern-Gerlach device. The original experiment used a rectangular or fan-shaped beam, which results in a pattern on the detector kind of like a Gaussian shape, with a mirror image, like this:


Accounting for blurring/uncertainty and such, if you trace the local maxima you end up with this cat-eye shape.

Now as I understand it, spin magnetic moment $\vec{\mu}_S$ is a vector in 3-space, and firing something with net SMM (i.e. silver atom) through the SG device "measures" $\vec{\mu}_S$ along the up-down axis of the device, which we call $z$. Atoms which deflect maximally up or down are said to have spin aligned to that axis. But what happens to particles that don't have $z$-aligned spin? In other words, if the aperture were beam-shaped, how would the pattern on the screen appear, and why?

My intuition is that it would be ring-shaped or elliptical, since particles with non-perfectly-aligned spin would have to deflect less (along the SG device's major axis). But intuition and QM rarely mix.

Edit: additional question: Does the SG device alter the spin moment of the particle (much like torque on a gyroscope causes procession)?

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    $\begingroup$ Why do you think the silver atom beam was spin aligned to begin with? Or, in a spin measurement along an axis, what are the possible measurement values you can attain? $\endgroup$ – Jon Custer Oct 5 '16 at 15:46
  • $\begingroup$ The beam won't be spin aligned, as far as I understand. It will be a mixture of particles with spin oriented any which way, I believe. Some of those will have no deflection left or right, but full deflection up/down, and we would say after the fact it was aligned with $z$, correct? $\endgroup$ – DeusXMachina Oct 5 '16 at 17:41
  • $\begingroup$ If you measure the spin of an atom on some random axis, what possible values will you measure? $\endgroup$ – Jon Custer Oct 5 '16 at 18:07

$\newcommand{\ket}[1]{\left|#1\right>}$ As you say yourself “intuition and QM rarely mix”, and it specifically is the source of your misconception. If the magnetic moment $\vec{μ_S}$ were indeed a classical well defined 3D vector interacting classically with a magnetic field gradient.

But the magnetic moment of silver atoms is quantified, and is a spin $½$, and therefore any measurement of this quantity can only take two values, $±μ_S$. The probability to obtain each value would depend on the original state.

More quantitatively, let’s look at the quantum state corresponding to the classical vector $\vec{μ_S}$ oriented along the direction $(θ,φ)$ : \begin{align} \vec{μ_S}&=μ_S\begin{bmatrix}\sin θ \cosφ\\\sin θ\sin φ \\\cos θ\end{bmatrix}\\ \ket{\vec{μ_S}}&=\cos\tfrac{θ}{2}\ket{↑}+e^{iφ}\sin\tfrac{θ}{2}\ket{↓} \end{align} where $↑$ (spin-up) corresponds to $θ=0$ and $↓$ (spin-down) to $θ=π$. Each direction in 3D space indeed corresponds to a “different” quantum state, but two states are orthogonal (in the quantum sense i.e. in the Hilbert space) iff they correspond to diametrically opposed direction in the 3D space ($θ' = π-θ$ and $φ'=φ ±π$).

Suppose you prepare a polarized pencil beam in a well defined direction $(θ,φ)$, and that the magnetic gradient of your Stern—Gerlach apparatus is along the $z$ direction, the beam splits along the $z$ direction, with a fraction $\left(\cos\tfrac{θ}{2}\right)^2=\tfrac{1+\cosθ}2$ of the atoms going up and a fraction $\left(\sin\tfrac{θ}{2}\right)^2=\tfrac{1-\cosθ}2$ of the atoms going down (and $φ$ has no influence).

Therefore, if the beam is polarized in a direction deviating from $z$ ($θ∉\{0,π\}$), the atoms split into two different spots. If the beam is totally unpolarized, you average over all angels and find a 50:50 split among the two spots.

If the gradient is not strictly along a constant direction, the computation becomes tricky, and we can have some fancy interference effects.

edit to answer your “edit” question: As a measuring apparatus, the Stern—Gerlach apparatus project the spin in the $\ket{↑},\ket{↓}$ basis, so if the spin is not aligned with the $z$-axis, it forces it to be aligned. It actually does so by coupling the spin (along $z$) to the physical position of the atom, which is itself projected by interaction with the environment or at the latest with the measuring plate.

  • $\begingroup$ Didn't you read my answer? Never been done. Can't be done. You can't build an apparatus with a mangnetic gradient along the z (up-and-down) direction without also having just as much gradient along the side-t0-side direction. You cannot split a pencil beam into two dots with your Stern-Gerlach machine. $\endgroup$ – Marty Green Oct 6 '16 at 15:58
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    $\begingroup$ @DeusXMachina: Exactly $\endgroup$ – Frédéric Grosshans Oct 7 '16 at 7:46
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    $\begingroup$ @MartyGreen: Yes there is also a transverse magnetic field gradient, but no it does not prevent the experiment to be done. I’ve written more details on your blog. $\endgroup$ – Frédéric Grosshans Oct 7 '16 at 12:00
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    $\begingroup$ @DeusXMachina: If the beam is polarised, one can express its state as a superposition of the 2 possible states $\uparrow$ and $\downarrow$ along a given direction. The direction itself is arbitrary (and the superposition coefficients depend on it), but it is convenient to chose the direction of the measurement, that is the one of the Bell field. When the beam is unpolarised, the situation is a bit different, since the beam is a classical mixture of all possible moment directions, each of it corresponding to the above superposition... $\endgroup$ – Frédéric Grosshans Oct 9 '16 at 15:23
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    $\begingroup$ ... The latter situation happens to be equivalent to a classical mixture of both possible states in any based. If you know some linear algebra, this equivalence is related to the invariance of the identity matrix under basis change $\endgroup$ – Frédéric Grosshans Oct 9 '16 at 15:26

From a practical point of view the reason for using of a broad flat beam may be as simple as getting easy alignment and decent rate while not blurring the signal with a significant z-direction dispersion.

No need to over-complicate things.

  • $\begingroup$ Furthermore the available collimation methods for atomic beams in 1922 were much cruder than the one available now. «What?? No transverse laser cooling??» says the modern experimentalist who chose to time-travel to the golden-era of quantum mechanics discoveries. $\endgroup$ – Frédéric Grosshans Oct 7 '16 at 13:38

But what happens to particles that don't have z-aligned spin?

The shape doesn't change, it's the density of measurements that forms the shape that changes.

Does the SG device alter the spin moment of the particle?

Someone can correct me if i'm wrong but i don't think it does.


This is a topic which is the subject of huge misconceptions even going so far as Feynman. There is no machine which splits a ray (a pencil-shaped beam) of silver atoms into two paths. The reason is obvious. You cannot build a magnetic field whose strength varies in the up-down direction without at the same time having it vary just as much in the x-y direction.

The unpolarized ray of silver atoms is actually spread out into a ring. The polarized ray does something even more interesting. I wrote about this on my blog a couple of years ago: The Quantization of Spin Revisited

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Oct 8 '16 at 11:45
  • $\begingroup$ There is a machine (a calcite crystal) that splits a ray of photons into two paths. Why should silver atoms be any different? $\endgroup$ – Peter Shor Oct 8 '16 at 21:52
  • $\begingroup$ That's a good point, @Peter, but unfortunately you're joining in late to a very good discussion that has been deleted by the moderators. $\endgroup$ – Marty Green Oct 9 '16 at 3:49
  • $\begingroup$ @PeterShor: The point of Marty Green (as one can see on the chat) was not against quantization itself, but on magnetic field gradient. The usual simplified analysis given is incompatible with Maxwell's equation, but Marty Green's analysis is wrong for other reasons (it overlooks the constant B) $\endgroup$ – Frédéric Grosshans Oct 10 '16 at 7:18

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