Timeline for Schrödinger evolution after a position measurement, without collapse
Current License: CC BY-SA 4.0
7 events
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| Mar 30, 2023 at 20:01 | comment | added | don't train ai on me | I'm trying to formulate more precisely in my mind what it would mean mathematically to follow your suggestion, to make the measurement at all times. Is this possible without solving the full Schrödinger equation? I'm very familiar with the density matrix formalism, I've heard of Master equations & used them in a very limited way. I wouldn't know which master equation to use for this purpose, though I can google that a bit. | |
| Mar 30, 2023 at 18:30 | comment | added | HTNW | But note that this model doesn't work for a macroscopic detector detector in continuous time. It is more apt for e.g. a quantum computer, where you can have qubits that "measure" each other (e.g. by CNOT). A macroscopic detector has "too many" states (variants of the "ready", "detected", etc. states due to thermal variations), and it cycles among them with time. This makes the "pure state" useless for calculation. You should look instead to use the mixed state formalism, where the detector is kept out of the quantum system. | |
| Mar 30, 2023 at 18:24 | comment | added | HTNW | @doublefelix It is in fact perfectly physical that a very thin detector has a very low chance of finding a particle distributed in space at any given instant. The "obvious" solution to both your points 1 and 2 is to "make the measurement" at all times. That is, the transition I've sketched in the answer is continuously happening; there is a (very strong?) coupling $\langle\text{detector trace at $x$}|\langle x|\hat H|x\rangle|\text{detector ready}\rangle\neq0.$ This is the "fundamentals" answer. | |
| Mar 30, 2023 at 17:04 | comment | added | don't train ai on me | For (2) above maybe I'm wrong, aka maybe the width of the w.f. is comparable to/smaller than the width of the detector. But I still in general feel like I'm not on solid ground when making such comparisons, especially because the initial width of the w.f. might not carry over to the moment of interaction, which messes things up. | |
| Mar 30, 2023 at 17:00 | comment | added | don't train ai on me | (btw, in my reading, it has been much more common for real position measurements not to have a fixed time $t$, albeit some of those show up too). The issue is then 1. When plugging in $\psi(t)$ after the measurement, which time $t$ should be chosen for the collapse? This is a rhetorical question, as it seems questionable to begin with. and 2. The volume of the detector may actually be tiny, and its relevance is more in its surface area. This seems to artificially put "nearly all" probability in the term you've written as $\psi_O$. But this seems like a nonphysical effect. | |
| Mar 30, 2023 at 16:57 | comment | added | don't train ai on me | Thanks for this nice answer. I went down this route myself as well, and I encountered some issues with this answer when considering the real position detector I'm thinking about. I've been using the pixel detector / semiconductor tracker, which are the innermost detectors (of position) at the LHC. If you'll allow I will give a short description. They are each basically a bunch of tiny, stationary pieces of silicon which the particles fly through. As such the measurement isn't done at a fixed time $t$, but on a fixed rectangular surface of one of these chips. | |
| Mar 29, 2023 at 5:39 | history | answered | HTNW | CC BY-SA 4.0 |