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What happens if we pass a particle through several Stern-Gerlach devices (say, oriented along x- and z-axes) without measuring the state of the system between them? My initial assumption has been that this is equivalent to measuring $\hat{s_z} \hat{s_x}$, but this operator is not hermitean, and hence doesn't correspond to a quantum mechanical observation.

Dmst
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    For a discussion of several SG devices, you should check out https://www.feynmanlectures.caltech.edu/III_05.html – curio Apr 27 '21 at 10:01
  • Related/possible duplicate: https://physics.stackexchange.com/q/455435/50583 – ACuriousMind Apr 27 '21 at 10:02
  • $\sigma_z\sigma_x=i\sigma_y$ But obviously the result of the experiment depends on the order of the SE devices - this is why straightforwardly it looks non-hermitian. – Roger V. Apr 27 '21 at 10:04
  • @Vadim could you please elaborate, what do you mean by "straightforwardly it looks non-hermitean"? Is it possible to interpret the reults of this experiment in terms of somehow measuring $i \sigma_y$? – Dmst Apr 27 '21 at 10:15
  • I think it can be interpreted in terms of $\sigma_y$. This is a real experiment, so the result is necessarily Hermitian. Also, you are probably not sending all that comes out from the first SG device into teh second one... Anyhow, Feynman really discusses this in details in his lectures, as recommended by @curio. – Roger V. Apr 27 '21 at 10:18
  • @Vadim when Feynman discusses sequential S-G, he filters out states after each device. It is also clear what quantum mechanical operator each S-G device corresponds to in this case. However, when you have a sequence of these devices, they act as a single black box with unclear correspondence to some specific quantum mechanical operator. – Dmst Apr 27 '21 at 10:35
  • If you don't measure after a device, then it is as if this device is not there. – Roger V. Apr 27 '21 at 11:34

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