# Why does interaction with a Stern-Gerlach machine, or any other laboratory apparatus, 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.

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.

• Just to clarify, what is the overlap $⟨\text{going right}|\text{going left}⟩$? If those two are orthogonal then there is never any interference anywhere to begin with. – Emilio Pisanty Sep 26 '16 at 6:48
• @EmilioPisanty Hey Emilio, the overlap is indeed zero, but I don't see where the $|\text{going left} \rangle$ state appears at all in the question. – knzhou Sep 26 '16 at 16:10
• – Bruce Greetham Oct 22 '16 at 15:25

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.)

• Thanks for the great answer! Just to make sure, the only features of the SG machine used here are that it's heavy, and that it moves like a single particle, right? Is there any way to state that latter thing in terms of something being in a coherent state? – knzhou Sep 26 '16 at 21:00
• I'm not sure I presume it moves like a single particle? Its momentum can be thought of as the momentum of the center of mass. Or is there another reason you say it's like a single particle? – Ruben Verresen Sep 26 '16 at 21:07
• My confusion is, you seem to be treating the SG apparatus as having only a single degree of freedom, the CM momentum. In that case, the result makes sense. But doesn't an SG apparatus have an enormous amount of microscopic degrees of freedom? – knzhou Oct 19 '16 at 18:29
• For example, why doesn't the electron push on a single atom in the SG machine instead? That atom also has a smeared momentum distribution. But since it's not very heavy, the impulse from the electron will be significant, and the overlap between the initial and final atom states will be small. – knzhou Oct 19 '16 at 18:30
• As a side-note, there is a well-studied phenomenon where the momentum of a photon is transferred to (and shared among) a large number of atoms in a complex system once rather than a single atom in that system: the Mössbauer effect. – Michael Seifert Dec 13 '16 at 21:07

I think the issue is resolved in the transactional picture (TI). In TI, you don't rely on a unitary-only 'decoherence' narrative. Rather, you have genuine collapse and that's what constitutes an actual measurement. That's also what establishes the classical level of phenomena in which objects of perception all have well-defined positions and momenta (defying the uncertainty relation). Note that in the above 'decoherence' approaches, one has to argue that the S-G device doesn't have a well-defined position; but of course it does. It is NOT in a superposition of positions. It's sitting right there with momentum=0 (relative to the lab) AND a well-defined position. According to TI, the reason it can do this (defy the uncertainty relation) is that the S-G is not a quantum system; it has entered the domain of classicality because its constituents are engaging in frequent collapses. This is a form of decoherence (a much stronger form that in the unitary-only theory.) That's why the S-G cannot enter into a coherent superposition with the electron state as presented in the question.

• Can you add some more explanation of what TI is, and what its axioms are? I've never heard of it before, and this sounds very different from the interpretations of QM that I know. – knzhou Oct 19 '16 at 20:29
• Sure, it was first proposed by John Cramer in 1986. I have 2 books out discussing TI. I've developed a relativistic version of it. For an introduction to the basic concepts you can see wordpress.com/post/transactionalinterpretation.org/372 – Ruth E Kastner Oct 20 '16 at 0:04
• Why do you say the issue is not resolved in the standard picture? It seems to me you want to say the SG machine has a fixed momentum and fixed position, but there is no justification for this. (And the common sense answer 'because you see it at a particular location' doesn't cut it: the accuracy to which we 'see' is well below the accuracy relevant for the discussion at hand.) – Ruben Verresen Oct 20 '16 at 1:56
• Thanks Ruben, to clarify: the issue is that if one allows the SG to be in a superposition, however microscopic, given unitary-only evolution that superposition can be reversibly amplified to an arbitrary macroscopic size--which we never see (basically the Schrodinger Cat situation). Appealing to environmentally-induced decoherence to eliminate superpositions depends on a circular argument in establishing a preferred basis for diagonalization of the density matrix for the system. I discuss this here: arxiv.org/abs/1406.4126 – Ruth E Kastner Oct 20 '16 at 19:20
• [cont] Curiously though, it seems to me your argument would actually give the wrong conclusion: if the SG machine were to have a fixed (and measurable!) momentum, then upon the neutron having gone through it, the SG machine would have picked up a corresponding momentum shift, which would exactly tell us whether the spin went up or down, which on its turn would imply that it is impossible to ever rejoin the neutron beams again in a quantum coherent fashion, inconsistent with experiment. – Ruben Verresen Oct 20 '16 at 20:23

There is no contradiction concerning the exchange of momentum if you take into account that it is after you check the electron trajectory that a measurement has been performed. At the level of the Stern-Gerlach interaction, all you have is entanglement.

Case 1: Deflection by a Stern-Gerlach followed by detection (measurement). Some momentum has been transferred from the electron to the apparatus.

Case 2: Deflection by a Stern-Gerlach followed by a second, upside-down, Stern-Gerlach (no measurement). There has been no momentum exchange, although there has been entanglement of electron and first apparatus, in a superposed state of two different exchanges of momentum, corresponding to the two spin states and associated trajectories.

In short: the interaction with the Stern-Gerlach is never a measurement by itself.

So why is the entanglement not destroying the interference? I guess the problem is the viability of semiclassical arguments here. If we take the Stern-Gerlach to be classical at the level of the first interaction, entanglement leads to decoherence. But if we do not, it is just part of the whole quantum system.

• I don't see where this answers my question. I'm concerned with case 2. Why does the entanglement between the electron and the apparatus not destroy interference? – knzhou Oct 17 '16 at 7:17
• I have edited my answer. – Stéphane Rollandin Oct 17 '16 at 7:37