7
$\begingroup$

If we measure position of a quantum particle, we force its wavefunction to collapse into a wavefunction whose probability density is given by a Dirac delta function (all the probability density of position is "squeezed" into one point in space).

Immediately after the measurement the wavefunction starts to delocalise (spread over the space), but if we measure the position very quickly again, its outcome (new position) shouldn't be far from the previous position. Moreover, by the new measurement we "localise" the wave-function again into a new Dirac delta function.

So, it looks to me that in the limit of continuous position measurements (extremely frequent measurements) the positions of the particle should form a continuous but stochastic trajectory that is probably described by a stochastic differential equation. Is it true? And, if it is the case, how does this equation look? What is its name?

ADDED

Is Balavkin equation a correct way to go? It looks like it describe a case of continuous measurement and, in particular, a case of continuous measurement of position is considered? Does this equation address some special (unrealistic) case? Is it based on some (unrealistic) assumptions?

Improve this question
$\endgroup$
15
  • 5
    $\begingroup$ Have you heard of the Quantum Zeno Paradox? You might find it interesting. $\endgroup$
    – Philip
    Commented Sep 4, 2020 at 10:22
  • $\begingroup$ Hi Roman. Unfortunately this collapse into a dirac delta business is not true. Why this is not mentioned in books is an oddity.. I guess to make things easier to teach. However the actual state-after-measurement is not really agreed upon. Eventually one would have to take a projection of some kind to account for the fact that consecutive msmts give the same result, but it seems that there is no such projection that can be seen as realistic except as an approximation. However this has not stopped people from working in the area you describe and as Chiral Anomaly says there are many references $\endgroup$
    – don't train ai on me
    Commented Sep 8, 2020 at 13:50
  • $\begingroup$ @doublefelix could you clarify please? It is true that the $\delta$ is unrealistic but POVMs are the commonly accepted approach to “smearing out” the position to account for the finite-resolution/non-$\delta$ outcome of the measurement. Of course continuous spectrum is always tricky... $\endgroup$
    – ZeroTheHero
    Commented Sep 8, 2020 at 15:41
  • $\begingroup$ A few issues. 1. A POVM is a measure and so it describes the probabilities of an outcome. But what is the new state of the system and/or apparatus after such a measurement? That is different than asking what the probabilities of outcomes are. 2. Supposing you did define a well-defined state-after-measurement alongside POVM, how do you know that your POVM is an accurate model of nature? There is in general more than one way to construct a POVM for a system, and none of those ways have been confirmed in experiment to accurately model the msmt process for a real apparatus. $\endgroup$
    – don't train ai on me
    Commented Sep 9, 2020 at 12:07
  • $\begingroup$ One common POVM (but not the only POVM) just chops off the position space wave function beyond a certain point. For this POVM you can straightforwardly guess a state after msmt (just renormalize psi). Yet the cutoff introduces some artifacts in the fourier transform which would influence momentum results. This has not been seen in experiment. $\endgroup$
    – don't train ai on me
    Commented Sep 9, 2020 at 12:10

1 Answer 1

Reset to default
5
+100
$\begingroup$

  1. The position measurement can't have infinite resolution, because then further time-evolution would be undefined. (A delta-function "wavefunction" doesn't belong to the Hilbert space.) So the resolution must be finite.

  2. For a single measurement event, we could approximate the effect by projecting the wavefunction into a spatial cell whose size represents the resolution of the measurement. Which cell? Whichever one we observe to be the result of the measurement. But if you really want to model a continuous position measurement (with finite resolution, of course), then this isn't a good approximation. It leads to the quantum Zeno effect, but in an artificial way: it's an artifact of the projection-approximation, which isn't how real position measurements work.

  3. To do better, we need to use a quantum model that includes more than just the particle of interest. We need to use a model that also includes the measurement equipment (at least). We can make that more manageable by taking a partial trace over the rest of the system, resulting in a master equation, such as a Lindblad equation, for the particle's density matrix. The density matrix evolves smoothly in time, but it doesn't remain pure: the interaction with the rest of the system causes the particle to become entangled with the rest of the system, so the entropy of the reduced density matrix increases.

  4. A stochastic trajectory emerges from the master equation if we occasionally apply the projection rule to account for where we actually observe the particle to be (with finite resolution), but to avoid artifacts, we should only do this occasionally, as explained above. We can do better by not taking a partial trace over the rest of the system, and letting the rest of the system include something like a physical periodically-updated digital readout of the particle's location. Then we can apply the projection rule to an observable associated with the periodic digital readout device, and you'll get a natural stochastic trajectory for the particle, without the artifacts. (Intuitively: the projection rule works great as long as you apply it far enough "downstream" from the process of interest, so that any mathematical "sharp edges" from the projection get smoothed out by the long chain of intervening physical interactions.)

For more information about point 3, this review paper looks pretty good:

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Thank you for the answer. Is there a way to connect your explanation with Belavkin equation ( en.wikipedia.org/wiki/Belavkin_equation). It seems to be relevant since it also describes a continuous observation of position (according to the linked wikipedia page). How is it connected to your approach? $\endgroup$
    – Roman
    Commented Sep 10, 2020 at 9:29
  • 1
    $\begingroup$ I have also specified my question by extending the original question. Thanks again for your explanations. $\endgroup$
    – Roman
    Commented Sep 10, 2020 at 9:32
  • $\begingroup$ @Roman That's a great question. I'm not familiar with the Belavkin equation, so I can't answer that part without lots of study, but here's what I would look for: consider a more complete model (one that includes measurement equipment and recording devices in the quantum system) that uses unitary time-evolution together with periodic applications of the projection rule to the recording devices (not to the particle). Maybe some kind of period $\to 0$ limit gives something like the Belavkin equation for the particle itself. I don't know if it does, but that's what I would look for. $\endgroup$
    – Chiral Anomaly
    Commented Sep 11, 2020 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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