0
$\begingroup$

Question about quantum back action in hypothetical scenario:

We know that, at $t_0$, a certain kind of particle, with spin initially prepared to be “spin right” in the x basis, goes through a Stern-Gerlach magnet that affects spin in the z basis (as “spin up” or “spin down”).

Eventually, the particle’s z-basis spin will be measured by a detector. Hence, there is a 50/50 chance that the particle will end up being measured as spin up or spin down, after exiting the magnet and reaching the detector.

Well above the magnet (in the z-up direction and between the magnet and the detector) is a second, hypothetical, environment which experiences no internal forces causing its wave function, or density matrix, to change with time. It is also made to be completely isolated from all external influences, except for this particle, which may locally interact with it (pass through it). Assume if the particle passes through this second environment (after exiting the Stern-Gerlach magnet), the environment will experience a substantial back-action, but the environment will have no affect on the spin of the particle.

We know that, at $t_1 > t_0$ , the particle is measured to be spin down.

What happened with regard to the back-action of the second environment? Some different possibilities are:

1.)After exiting the magnet, the particles's wave function split into two packets, one going up and one going down. The "up" wave function interacted with the second environment. The second environment’s wave function, or density matrix, evolved according to the backaction caused by the particle, between $t_0$ and $t_1$. At $t_1$, when the particle was measured to be spin down, the evolution of the second environment was “undone”, so that, at $t_1$, the second environment is in the same state that it was in at $t_0$.

2.) The particle's spin wave function collapsed to spin down, upon exiting the magnet, so it never interacted with the second environment. Therefore, the second environment’s wave function, or density matrix, did not change between $t_0$ and $t_1$ (so, no undoing is necessary).

3.) Since there is no such thing as trajectory in QM, even a spin down particle might have gone through the second environment and so we might see back-action in the second environment even if the particle is measured to be spin down. But if the particle is measured to be spin down,then the z-momentum of the particle must have been down, so I do not see this possibility being good.

4) Other…

I suppose which possibility is correct may depend on one’s interpretation of quantum mechanics. What does your favorite interpretation say?

$\endgroup$
  • $\begingroup$ I am thinking that 1) could lead to some strange consequences for the second environment, if the particle's wave function splits (to up and a down wave functions) upon exiting the magnet and the up wave function becomes entangled with the second environment, but the particle is then measured as spin down. so am leading toward 2). $\endgroup$ – David Dec 15 '15 at 6:54
  • $\begingroup$ Actually, I think that this scenario might cause the joint system to go from a pure state into a mixed state of two wave functions that are not in superposition. One would be associated with spin up (and include entanglement of the particle with the second environment) and the other would be associated with spin down (and not include entanglement of the particle with the second environment). Each of those two wave functions could be rewritten as a density matrix and the sum of those density matrices would be the density matrix of the composite system of particle and second environment. $\endgroup$ – David Dec 16 '15 at 5:43
  • $\begingroup$ This above comment is similar to the double slit experiment, when different polarizing filters are places at the different slits. The result is the two emerging waves are in a mixed state (no no interference between them), because each wave is entangled to a different filter, and so must be independent of the other. $\endgroup$ – David Dec 16 '15 at 5:46
1
$\begingroup$

Until the particle spin is measured it is in a superposition of the up and down states. When it interacts with the second system the particle and second system become entangled and both exist as a superposition of the up + interaction and down + interaction states. When the particle reaches the detector the superposition collapses and both the particle and the second system acquire definite properties.

Well, in principle. In practice the second system may be complex enough to decohere. In that case the second system is effectively acting as a detector i.e. its interaction with the particle constitutes a measurement and collapses the particle wavefunction into one of the up and down states (or whatever property the second system is effectively measuring).

$\endgroup$
  • $\begingroup$ Thank you, John! Are you are saying that when the particle leaves the magnet, it's wave function packet (assumed to be compact) splits into two packets (one for up and one for down) that are in a superposition until the particle is measured? $\endgroup$ – David Dec 15 '15 at 7:13
  • $\begingroup$ @David: There's no splitting and while I'm not sure what you mean by assumed to be compact I doubt this is a safe assumption. The wavefunction of the particle is a superposition of all possible trajectories, and remains so until you interact with the particle and thereby measure it. $\endgroup$ – John Rennie Dec 15 '15 at 7:16
  • $\begingroup$ I mean when the particle exits the magnet, it is in a superposition of two spin states. Each spin state has a set of z-momenta entangled with it. Let's assume that the two sets of z-momenta do not overlap. If the position wave function of the particle was a "narrow gaussian", before it entered the magnet, then after exiting, having an spin up and a down (and their associated z-momenta) in superposition would mean having the position wave function consist of multiple position gaussians. Only a subset of those gaussians would enter the second environment. $\endgroup$ – David Dec 15 '15 at 7:29
  • $\begingroup$ @David: The wavefunction is an elementary object. You can't say only part of the wavefunction interacts with the second system. If any part of the wavefunction overlaps the second system then the whole wavefunction interacts with it. $\endgroup$ – John Rennie Dec 15 '15 at 7:33
  • $\begingroup$ Yes, but what confuses me is that the entanglement would only be mathematically attached to the up state, not the down state, so, when the measurement yields the down state, what happens to the entanglement? $\endgroup$ – David Dec 15 '15 at 7:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.