quantum mechanics - Why can interaction with a macroscopic apparatus, such as a Stern-Gerlach machine, sometimes not cause a measurement?

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.

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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.

Answer

It's a very good question, since indeed if the original Stern-Gerlach machine had a well-defined momentum, then you are right that there could be no coherence upon rejoining the beams! The rule of thumb for decoherence: a superposition is destroyed/decohered when information has leaked out. In this setting that would mean that if by measuring, say, the momentum of the Stern-Gerlach machine you could figure out whether the spin had curved upwards or downwards, then the quantum superposition between up and down would have been destroyed.

Let's be more exact, as it then will become clear why in practice we can preserve the quantum coherence in this kind of set-up.

Let us for simplicity suppose that the first Stern-Gerlach machine simply imparts a momentum $\pm k$ to the spin, with the sign depending on the spins orientation. By momentum conservation, the Stern-Gerlach machine gets the opposite momentum, i.e. (using that $\hat x$ generates translation in momentum space)

$$\left( |\uparrow \rangle + |\downarrow \rangle \right) \otimes |SG_1\rangle \to \left( e^{- i k \hat x} |\uparrow \rangle \otimes e^{ i k \hat x} |SG_1\rangle \right) + \left( e^{i k \hat x} |\downarrow \rangle \otimes e^{- i k \hat x} |SG_1\rangle \right)$$ Let us now attach the second (upside-down) Stern-Gerlach machine, with the final state $$\to \left( |\uparrow \rangle \otimes e^{ i k \hat x} |SG_1\rangle \otimes e^{-i k \hat x} |SG_2\rangle \right) + \left( |\downarrow \rangle \otimes e^{- i k \hat x} |SG_1\rangle \otimes e^{ i k \hat x} |SG_2\rangle \right)$$

For a clearer presentation, let me now drop the second SG machine (afterwards you can substitute it back in since nothing really changes). So we now ask the question: does the final state $\boxed{ \left( |\uparrow \rangle \otimes e^{ i k \hat x} |SG_1\rangle \right) + \left( |\downarrow \rangle \otimes e^{- i k \hat x} |SG_1\rangle \right) }$ still have quantum coherence between the up and down spins?

Let us decompose

$$e^{ -i k \hat x} |SG_1\rangle = \alpha \; e^{i k \hat x} |SG_1\rangle + |\beta \rangle$$ where by definition the two components on the right-hand side are orthogonal, i.e. $\langle SG_1 | e^{ -2 i k \hat x} | SG_1 \rangle = \alpha$. Then $|\alpha|^2$ is the probability we have preserved the quantum coherence! Indeed, the final state can be rewritten as $$\boxed{ \alpha \left( |\uparrow \rangle +| \downarrow \rangle \right) \otimes e^{ i k \hat x} |SG_1\rangle + |\uparrow\rangle \otimes | \gamma \rangle + |\downarrow \rangle \otimes |\beta\rangle }$$ where $\langle \gamma | \beta \rangle = 0$. In other words, tracing out over the Stern-Gerlach machine, we get a density matrix for our spin-system: $\boxed{\hat \rho = |\alpha|^2 \hat \rho_\textrm{coherent} + (1-|\alpha|^2) \hat \rho_\textrm{decohered}}$.

So you see that in principle you are right: the quantum coherence is completely destroyed if the overlap between the SG machines with different momenta is exactly zero, i.e. $\alpha = 0$. But that would only be the case if our SG has a perfectly well-defined momentum to begin with. Of course that is completely unphysical, since that would mean our Stern-Gerlach machine would be smeared out over the universe. Analogously, suppose our SG machine had a perfectly well-defined position, then the momentum-translation is merely a phase factor, and $|\alpha|=1$ so in this case there is zero information loss! But of course this is equally unphysical, as it would mean our SG machine has completely random momentum to begin with. But now we can begin to see why in practice there is no decoherence due to the momentum transfer: in practice we can think of the momentum of the SG machine as being described by some mean value and a Gaussian curve, and whilst it is true that the momentum transfer of the spin slightly shifts this mean value, there will still be a large overlap with the original distribution, and so $|\alpha| \approx 1$. So there is strictly speaking some decoherence, but it is negligible. (This is mostly due to the macroscopic nature of the SG machine. If it were much smaller, than the momentum of the spin would have a much greater relative effect.)