Consider a Stern-Gerlach machine that measures the $z$-component of the spin of an electron. Suppose our electron's initial state is an equal superposition of $$|\text{spin up}, \text{going right} \rangle, \quad |\text{spin down}, \text{going right} \rangle.$$ After going through the machine, the electron is deflected according to its spin, so we get $$|\text{spin up}, \text{going up-right} \rangle, \quad |\text{spin down}, \text{going down-right} \rangle.$$ In a first quantum mechanics course, we say the spin has been measured. After all, if you trace out the momentum degree of freedom, we no longer have a spin superposition. In simpler words, you can figure out the spin by which way the electron is going.

In a second course, sometimes you hear this isn't really a measurement: you can put the two beams through a second, upside-down Stern-Gerlach machine, to combine them into $$|\text{spin up}, \text{going right} \rangle, \quad |\text{spin down}, \text{going right} \rangle.$$ Now the original spin superposition is restored, just as coherent as before. This point of view is advanced in this lecture and the Feynman lectures.

Here's my problem with this argument. Why doesn't the interaction change the state of the Stern-Gerlach machine? I thought the two states would be $$|\text{spin up}, \text{going up-right}, \text{SG down} \rangle, \quad |\text{spin down}, \text{going down-right}, \text{SG up} \rangle.$$ That is, if the machine pushes the electrons up, it itself must be pushed down by momentum conservation. After recombining the beams, the final states are $$|\text{spin up}, \text{going right}, \text{SG down} \rangle, \quad |\text{spin down}, \text{going right}, \text{SG up} \rangle.$$ and the spins cannot interfere, because the Stern-Gerlach part of the state is different!

This is a special case of a general question: under what circumstances can interaction with a macroscopic system not cause decoherence? Intuitively, there is always a backreaction from the spin onto the system, which changes its state and destroys coherence, so it seems that every particle is always continuously being measured.

In the case of a magnetic field acting on a spin, like in NMR, there is a resolution: the system state is a coherent state, because it's a macroscopic magnetic field, and coherent states are barely changed by $a$ or $a^\dagger$. But I don't think this argument applies for the Stern-Gerlach machine.

Consider a Stern-Gerlach machine that measures the $z$-component of the spin of an electron. Suppose our electron's initial state is an equal superposition of $$|\text{spin up}, \text{going right} \rangle, \quad |\text{spin down}, \text{going right} \rangle.$$ After going through the machine, the electron is deflected according to its spin, so we get $$|\text{spin up}, \text{going up-right} \rangle, \quad |\text{spin down}, \text{going down-right} \rangle.$$ In a first quantum mechanics course, we say the spin has been measured. After all, if you trace out the momentum degree of freedom, we no longer have a spin superposition. In simpler words, you can figure out the spin by which way the electron is going.

In a second course, sometimes you hear this isn't really a measurement: you can put the two beams through a second, upside-down Stern-Gerlach machine, to combine them into $$|\text{spin up}, \text{going right} \rangle, \quad |\text{spin down}, \text{going right} \rangle.$$ Now the original spin superposition is restored, just as coherent as before. This point of view is advanced in this lecture and the Feynman lectures.

Here's my problem with this argument. Why doesn't the interaction change the state of the Stern-Gerlach machine? I thought the two states would be $$|\text{spin up}, \text{going up-right}, \text{SG down} \rangle, \quad |\text{spin down}, \text{going down-right}, \text{SG up} \rangle.$$ That is, if the machine pushes the electrons up, it itself must be pushed down by momentum conservation. After recombining the beams, the final states are $$|\text{spin up}, \text{going right}, \text{SG down} \rangle, \quad |\text{spin down}, \text{going right}, \text{SG up} \rangle.$$ and the spins cannot interfere, because the Stern-Gerlach part of the state is different! Upon tracing out the Stern-Gerlach machine, this is effectively a quantum measurement.

This is a special case of a general question: under what circumstances can interaction with a macroscopic piece of lab equipment not cause decoherence? Intuitively, there is always a backreaction from the spin onto the equipment, which changes its state and destroys coherence, so it seems that every particle is always continuously being measured.

In the case of a magnetic field acting on a spin, like in NMR, there is a resolution: the system state is a coherent state, because it's a macroscopic magnetic field, and coherent states are barely changed by $a$ or $a^\dagger$. But I'm not sure how to argue it for the Stern-Gerlach machine.