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Having a particle entering the apparatus with spin state $|+\rangle$, for which $\hat S_x|+\rangle=+\frac\hbar 2|+\rangle$, I have a question about how to express the spin state when it comes out. I mean: will the spin state be an eigenstate of the spin $z$ component? or would it be a superposition of the $z$ component's eigenstate ?

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A Stern-Gerlach apparatus oriented along the $z$ axis acts as a measurement on the basis of the $z$ component of the particle's spin. What that means is that the particle will always come out in an eigenstate of $\hat S_z$. You don't know which one, of course, as that is decided probabilistically.

Even if you don't actually observe the outcome of the measurement (in which case the system would be in the corresponding eigenstate), it is not accurate to say that the particle comes out in a superposition of $z$ eigenstates. While what comes out is a probabilistic mixture, it is a mixed state in which the coherence between the two contributions has been lost.

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  • $\begingroup$ Do you have a reference for this (mixed state)? What is the source for the decoherence? I am pretty sure you can recombine beams from the Stern-Gerlach apparatus and observe interference. physics.mq.edu.au/~jcresser/Phys301/Chapters/Chapter6.pdf See page 66. $\endgroup$ – lionelbrits Nov 10 '13 at 23:36
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    $\begingroup$ You can do that but then you can't measure the outcome. You can even have them interact with some other system to give each component a different phase. However, there you are in a unitary evolution regime and not in a measurement regime. Usually, saying "a Stern-Gerlach apparatus" means a measurement device, but you can also use them unitarily. In that case, though, the state that comes out of the magnet is an entangled state, in which the spin degree of freedom is entangled with the position. The reduced density matrix is a mixed state until you recombine the beams. $\endgroup$ – Emilio Pisanty Nov 10 '13 at 23:50
  • $\begingroup$ Ok, the reduced density matrix is mixed, I'll give you that, but that is not the same as a mixed state altogether. Since you said "even if you don't observe the outcome", I took it to mean no measurement. Because you can take the position-spin entangled state, and pass it through a time-reversed z-oriented apparatus, and recombine the beams. Passing through an x-oriented apparatus will find them again in $|+\rangle_x$. $\endgroup$ – lionelbrits Nov 11 '13 at 0:03
  • $\begingroup$ Yes, you can get all sorts of technical with this. You can even extend this formalism to get all many-worldsian and say that the coherence is never lost, and instead you're entangled with the observer. The answer, however, was directed at the tone and perceived level of the OP. $\endgroup$ – Emilio Pisanty Nov 11 '13 at 0:06
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    $\begingroup$ Hmm... I think it would better serve OP to say that the particle comes out in a superposition of spin-position eigenstates, i.e., $|\uparrow\rangle\otimes|overhere\rangle + |\downarrow\rangle \otimes |overthere\rangle$. I don't think this is "all sorts of technical" any more than the double slit experiment is. $\endgroup$ – lionelbrits Nov 11 '13 at 0:20
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It's hard to believe that a correspondent as knowledgeable (and as highly rated) as Emilio Pisanty could show such a fundamental misunderstanding of quantum mechanics as he does in his answer to this question. If it were possible to send a single silver atom through an inhomogenous magnetic field of the type alluded to in this question (that is what we understand by the phrase "a Stern-Gerlach apparatus"), then if it enters the apparatus with a "sideways" spin, then it leaves with a net "sideways" spin (in the xy plane). There is no torque exerted by the magnets on the atom which is capable of flipping the spin into the z orientation, either z-up or z-down.

By the way, the notion that there is a machine which you can send a silver atom through and have it come out either spin-up or spin-down...well, no such machine exists. That's not what the Stern-Gerlach experiment did. I explain more about these things in this blogpost: The Quantization of Spin Revisited

DISCLAIMER: I am a recognized crackpot in this site whose answers are routinely and massively downvoted by people who know much more than me.

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  • $\begingroup$ A quantum spin isn't just a classical rigid rod. In general, for a quantum system, measurements can change the expectation values of quantities that would otherwise be conserved. So the expectation value of the spin changes even without any external torque. $\endgroup$ – Dominic Else Sep 24 '16 at 3:18
  • $\begingroup$ Whether or not I buy into the whole measurement paradigm, my real problem with Emilio's argument is that he has the angular momentum changing even WITHOUT a "measurement"...that is, detecting the location of the particle and inferring from that its angular momentum. Unless you think that passing the particle through an inhomogenous magnetic field is itself a "measurement". Which it isn't. $\endgroup$ – Marty Green Sep 24 '16 at 3:23
  • $\begingroup$ Well, fair enough. In quantum mechanics we are free to declare that a measurement has happened whenever the system has become sufficiently entangled with its environment that the coherence is unrecoverable in practice. I imagine that that is the scenario he had in mind. $\endgroup$ – Dominic Else Sep 24 '16 at 3:32
  • $\begingroup$ I have no idea what any of that means. $\endgroup$ – Marty Green Sep 24 '16 at 3:51
  • $\begingroup$ In that case, I would recommend reading about decoherence as it's the central idea behind the modern understanding of what measurement means in quantum mechanics. en.wikipedia.org/wiki/Quantum_decoherence $\endgroup$ – Dominic Else Sep 24 '16 at 4:29

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