Timeline for Why can interaction with a macroscopic apparatus, such as a Stern-Gerlach machine, sometimes not cause a measurement?
Current License: CC BY-SA 3.0
13 events
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| Dec 13, 2016 at 21:07 | comment | added | Michael Seifert | 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. | |
| Oct 22, 2016 at 6:40 | history | bounty ended | knzhou | ||
| Oct 22, 2016 at 6:40 | vote | accept | knzhou | ||
| Oct 19, 2016 at 18:52 | comment | added | Ruben Verresen | But the main point is that I agree that in by only explicitly taking into account the COM momentum, I presume that our machine moves like a single particle. And I would also argue that this is the case for any realistic SG machine. But it need not be /in principle/ true, and if you could devise such a freak SG machine, then yes the superposition would have decohered, as I tried to argue in my comment above. | |
| Oct 19, 2016 at 18:49 | comment | added | Ruben Verresen | Photons are only discrete when you measure them. But okay I agree that in principle the transfer won't be 100% homogeneous (when is such a statement every strictly true), but I mean to say, to any accuracy relevant here. | |
| Oct 19, 2016 at 18:46 | comment | added | knzhou | I'm not entirely sure the transferred momentum is evenly distributed. After all, isn't it transferred via photons, which are discrete? | |
| Oct 19, 2016 at 18:45 | comment | added | Ruben Verresen | [Cont.] Suppose you had a freak SG machine where all momentum would be transferred to one atom, then you could attach a dial to this atom which would move if the atom suddenly jumps up, so this macroscopic dial would effectively measure through which path our particle goes (not an electron btw, that is charged) and so it's consistent that this would decohere/collapse the superposition. But it also follows from the above reasoning, as your intuition suggested, since then the momentum change is huge and $\alpha \approx 0$. | |
| Oct 19, 2016 at 18:44 | comment | added | Ruben Verresen | I agree with you, in the freak case where the transferred momentum is not distributed throughout the SG machine but instead to a single atom, our superposition would have decohered! The thing is that that is not how the momentum is transferred: the SG machine is a big magnet made out of many small magnets, and each microscopic spin in this SG machine will get about an equal amount of momentum imparted to it, making the change imperceptible. [Cont.] | |
| Oct 19, 2016 at 18:30 | comment | added | knzhou | 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. | |
| Oct 19, 2016 at 18:29 | comment | added | knzhou | 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? | |
| Sep 26, 2016 at 21:07 | comment | added | Ruben Verresen | 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? | |
| Sep 26, 2016 at 21:00 | comment | added | knzhou | 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? | |
| Sep 26, 2016 at 19:47 | history | answered | Ruben Verresen | CC BY-SA 3.0 |