# Sequential Stern-Gerlach devices - realizable experiment or teaching aid?

At least one textbook [1] uses sequential Stern-Gerlach devices to introduce to students that the components of angular momentum are incompatible observables. Viz., the $z$-up beam from a SG device with magnetic field in the $\hat z$ direction (an SG$z$ device) is passed through an SG$x$ device, and is found to split into two beams. Passing say, the $x$-up beam through an SG$z$ device, it too splits.

Of course, knowing quantum mechanics this is exactly what we expect.

But to someone who does not know quantum mechanics, is this convincing that there is no $\mid+x,+z\rangle$ state? I am not so sure it is if we consider it as a real experiment, with finite precision. We know that the beam entering the SG$x$ device has $S_z = \hbar/2$, we do not know anything about its $S_x$. We know that the beams leaving the SG$x$ device have $S_x = \pm \hbar/2$, respectively. By adding the second SG$z$ we wish to test if $S_x$ and $S_z$ can have definite values simultaneously, but there is then an assumption that the SG$x$ device does not disturb the value of $S_z$, or at least does so with a very small spread. But already in the classical picture the Stern-Gerlach device is not such a device.

In the $SG$z device the $\mathbf B$-field has a large homogeneous component $B_0\hat z$, such that the angular momentum around $\hat z$ is approximately conserved while the other components average to 0, and the force, on average, has only a $\hat z$ component [2]. But in the SG$x$ device the angular momentum precesses around $\hat x$, with a period that is quite short, $T = 10^{-9}$ s or less.

If the particle beam has a spread of velocities $v$ such that the spread in times-of-flight $t$ is not small compared to $T$, we should not expect the second beam to be $z$-polarized, even classically. The relation between the spreads is $\Delta t = t \Delta v /v$. In the original experiment [2] we can estimate $v$ and $t$ as being on the order of $10^2$ m/s and $10^{-4}$ s, requiring $\Delta v /v$ on the order $10^{-5}$. This seems entirely unreasonable for a thermal source, considering the finite width of the collimator and if nothing else the force component neglected initially seems liable to produce a spread of at least this order.

I tried to search the literature to see if the sequential experiment has actually been carried out, but could not find anything. I did find Ref. 3 that seems to talk about two-spinors, but I cannot access it.

References

1. Townsend, J.S. (2000). A Modern Approach to Quantum Mechanics. University Science Books
2. Stern, O. (1988). A way towards the experimental examination of spatial quantisation in a magnetic field. Zeitschrift für Physik D Atoms, Molecules and Clusters, 10(2), 114-116.
3. Darwin, C. G. (1927). The electron as a vector wave. Proceedings of the Royal Society of London A, 227-253.

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• The follow-up question is of course, if the method of sequential SG-devices was not available in the 20s, how and why were the Pauli matrices actually introduced? But perhaps that is a better fit for HSM. – Robin Ekman Aug 22 '15 at 0:44
• Also A. Peres: Quantum Theory: Concepts and Methods (Kluwer 1995) uses the same spin example. For examples with actual experiments (with quantum optics) check out U. Leonhardt: Measuring the Quantum State of Light (Cambridge U. Press 1997). Also W. M. de Muynck: Foundations of Quantum Mechanics, an Empiricist Approach (Kluwer 2002) may have examples inspired by actual experiments. – pglpm Mar 31 '18 at 12:33

In your first paragraph you describe a Stern-Gerlach device as one with a magnetic field in the $\hat z$ direction. And then later you talk about having a large homogeneous component of the magnetic field. I'm not sure you have an accurate physical model of a Stern-Gerlach device.

The Hamiltonian for a Stern-Gerlach has magnetic fields components combined with the Pauli matrices like $B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z$ well proportional to that. It is the quantum version of a magnetic moment in an external magnetic field and in this case the magnetic moment is proportional to the spin hence $\vec\mu\cdot\vec B$ is proportional to above.

The classical force comes from the gradient of this quantity. So it is inhomogeneous magnetic fields that you use to measure spin.

And while you want the field to only have $\hat z$ components to measure just the z component of spin you need the magnetic field to have a gradient (be inhomogeneous) to deflect the beam. And it is the direction that the field gets stronger that is as important as which direction it points. So it is in no way similar to just having a magnetic field that points some direction.

That said. It is rather straightforward to measure the spin z component twice in a row or three times in a row and you get the same answer each time as you got the first time. So it is in the nature of the outcome that it give those results again.

Same if you do two or three spin x component measurements. So it is in the nature of the outcome of the first experiment that the result be the kind of thing that gives those same results again and does so reliability.

These experiments are easy to do, so I don't really think that is what you are asking about.

Now if you measure z then x then z you do not always get the same result for the second z measurement as you got for the first z measurement. This has been done.

So we know for a certainty that the spin x "measurement" has changed the particle's state. Because it used to have a reliability under spin z measurements and then it no longer has that reliability.

I don't know what details you think need to be involved here, we definitely changed the particle when we measured a complementary (i.e. not equal) component.

• When you say that the SGz, SGx, SGz experiment has been done, please provide a reference. I do not doubt that it nowadays is easy to build the required setup. I do doubt that, using the --admittedly not very precise -- hypothesis the components of angular momentum can be well-defined simultaneously but are quantized when "measured", the setup has the precision required to state that the $z$-polarization of a beam exiting the second device is certain. Certainly this was not possible in the 20s. – Robin Ekman Aug 22 '15 at 1:28
• @RobinEkman I have trouble figuring out if I'm understanding your concern. If you think a Stern-Gerlach device exists then just buy three and rotate one of them when you set all three up. They aren't manufactured differently and they don't have to interfere if you have some room in the lab to set all three up. I just keep thinking I must be misreading what you are asking. – Timaeus Aug 22 '15 at 1:33
• I believe that while the quantum dynamics certainly predict the results Townsend describes, they don't distinguish very clearly between classical and quantum dynamics in the device, other than that there's some strange thing with discrete values going on. I believe this because in a classical model the experimental uncertainties should be large enough to produce the non-polarization that is inherent in the quantum model. If we had beams narrow enough in velocity space ($\Delta v/v \lessapprox 10^{-5}$) then the experiment would be convincing, but this seems very difficult to realize. – Robin Ekman Aug 22 '15 at 1:41
• @RobinEkman Again I see nothing unconvincing. And I never understand when someone thinks there is confusion about classical versus quantum dynamics. The dBB model clearly has a classical force and a quantum force and you can see that for Stern-Gerlach in, for instance, dx.doi.org/10.1119/1.4848217 the point is the previously existing reproducibility now longer exists and has been destroyed by the polarization that occurs in the device. – Timaeus Aug 22 '15 at 1:54
• Yes, in quantum mechanics we have that $\mid+z\rangle$ is a linear combination of $\mid\pm x\rangle$ and vice versa, so it is clear that an $x$-polarizer will destroy the $z$-polarization. But the classical dynamics also predict this for a beam with a spread of velocities, because the $x$-polarizer exerts a torque around $\hat x$ and there is a spread in time-of-flight. It seems to me that we need very large precision to distinguish between simple classical statistical uncertainty and non-commuting quantum observables. – Robin Ekman Aug 22 '15 at 15:20