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Consider the following physical situation. Two position detectors are located next to each other, each one carries only YES/NO information on whether the particle hit that detector. A particle, let's say in a Gaussian state, is headed toward the two detectors. The width of the wavefunction of the incident particle is such that it would overlap with both detectors upon reaching the detection surface.

My question is, what would be (heuristically) the post-measurement state of this system, disregarding wavefunction collapse? For spin measurements it is common to write something like

$$(\alpha |\uparrow\rangle + \beta |\downarrow\rangle)\otimes |ready\rangle \to \alpha |\uparrow\rangle |\uparrow \text{detected} \rangle + \beta |\downarrow \rangle |\downarrow \text{detected} \rangle $$

where the first state in each tensor product represents the measured particle, and the second represents the state of the detector/environment. But for a continuous variable like position the answer is not so clear to me.

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  • $\begingroup$ Before the measurement alpha and beta satisfies $\alpha^2+\beta^2=1$ and after that one of them take the value zero (the one who is not detected) and the other the value 1 (maybe up to a fase). That's the difference between before and after states. $\endgroup$ Mar 29, 2023 at 2:09
  • $\begingroup$ @GabrielPalau That is if there is collapse. My question concerns what happens if we just have Schrödinger evolution and no collapse. $\endgroup$ Mar 29, 2023 at 2:20
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    $\begingroup$ How can a measurement be made without wavefunction collapse? $\endgroup$
    – Aiden
    Mar 29, 2023 at 3:40
  • $\begingroup$ Yes/no information is insufficient to tell the state of the system after the measurement. We don't know how much of the energy of the system was absorbed by the detectors. We also don't know what the aperture function of these detectors looks like and what the quantum efficiency is. While this sounds like a trivial quantum mechanics textbook question, in reality it's a non-trivial scattering problem. I am sure somebody can give you a full solution with a complicated integral and somebody else can give you a trivial false solution. I can't do the first and I won't do the second. $\endgroup$ Mar 29, 2023 at 5:53
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    $\begingroup$ @Aiden "Collapse" is a very unfortunate terminology. What happens physically is that the quantum system transfers some of its energy in form of a quantum to the detector. In the most simple scenario (atomic spectroscopy) the quantum of energy is the total energy that the system (atom) had and the process can't repeat itself because there is no energy left to be detected. In case of a position measurement of a free particle the detector may absorb the particle or it may "bounce", which alters the energy in the system and then we need to specify the scattering function of the detector. $\endgroup$ Mar 29, 2023 at 6:02

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$\newcommand\ket[1]{{|{#1}\rangle}}$The idea for position is the same as for spin. For simplicity, say your particle is moving in 1D and your detectors fire at some fixed time. Say the left detector covers some space region $[l_1,l_2]$ and the right detector covers $[r_1,r_2],$ and these are disjoint. The state (of particle + detectors) right before the measurement may be written $$\ket\psi\ket{\text{ready}_L}\ket{\text{ready}_R},$$ where the first factor is the state of the particle, given by a wavefunction $\psi(x)$ ($\ket\psi=\int_{-\infty}^\infty\psi(x)\ket x\,dx$) and the latter two factors describe the states of the detectors. The state right after measurement is $$\ket{\psi_L}\ket{\text{yes}_L}\ket{\text{no}_R}+\ket{\psi_R}\ket{\text{no}_L}\ket{\text{yes}_R}+\ket{\psi_O}\ket{\text{no}_L}\ket{\text{no}_R},$$

where the states $\ket{\psi_L},\ket{\psi_R},\ket{\psi_O}$ are the "collapsed" wavefunctions resulting from restricting $\psi$ to the appropriate regions. $$\psi_L(x)=\begin{cases}\psi(x)&\text{if $x\in [l_1,l_2]$}\\0&\text{else}\end{cases},\quad\psi_R(x)=\begin{cases}\psi(x)&\text{if $x\in [r_1,r_2]$}\\0&\text{else}\end{cases},\quad\psi_O(x)=\begin{cases}\psi(x)&\text{if $x\notin [l_1,l_2]\cup[r_1,r_2]$}\\0&\text{else}\end{cases}$$ (where again $\ket{\psi_i}=\int_{-\infty}^\infty\psi_i(x)\ket x\,dx$). Note that the overall wavefunction of particle + detectors hasn't collapsed. But the un-collapsed total wavefunction does (and must) encode all the ways the particle wavefunction could have collapsed.

I hope you can see how this relates to the formalism for a spin detector. You write the initial state of the particle as a superposition of (simultaneous) eigenstates of your detector(s)'s observable(s). For each possible outcome on the detector(s), you get a term ("branch") in the wavefunction where the detector(s) shows that result, and it's tensored to the part of the original particle wavefunction that is "consistent" with that result.

Putting the previous paragraph in math terms, you should consider one detector measuring some observable $O$ on a Hilbert space $\mathcal H,$ with possible outcomes (eigenvalues) $\lambda_1,\ldots,\lambda_n.$ (The two detectors in your position experiment can be considered as one detector measuring an observable with eigenvalues $\lambda_1=0$ for "not in left or right region", $\lambda_2=1$ for "in left region", $\lambda_3=2$ for "in right region". In excruciating detail, $O=\int_{l_1}^{l_2}\ket x\langle x|\,dx+\int_{r_1}^{r_2}2\ket x\langle x|\,dx.$) The detector should also have states $\ket{\lambda_1},\ldots,\ket{\lambda_n}$ for when it detects the given result, along with the special initial state $\ket{\text{ready}}.$ When the detector is activated, it sends any state $\ket\psi\otimes\ket{\text{ready}}$ (where $\ket\psi\in\mathcal H$) to the state $$\sum_{i=1}^n[P_i\ket\psi\otimes\ket{\lambda_i}],$$ where $P_i$ is the projection operator onto the eigenspace of $O$ associated to eigenvalue $\lambda_i.$ (In our example, $P_1=\int_{\mathbb R\backslash([l_1,l_2]\cup[r_1,r_2])}\ket x\langle x|\,dx,P_2=\int_{l_1}^{l_2}\ket x\langle x|\,dx,P_3=\int_{r_1}^{r_2}\ket x\langle x|\,dx,$ and these operators "implement" the wavefunction collapse: $\ket{\psi_L}=P_2\ket\psi,\ket{\psi_R}=P_3\ket\psi,\ket{\psi_O}=P_1\ket\psi.$)

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  • $\begingroup$ 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. $\endgroup$ Mar 30, 2023 at 16:57
  • $\begingroup$ (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. $\endgroup$ Mar 30, 2023 at 17:00
  • $\begingroup$ 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. $\endgroup$ Mar 30, 2023 at 17:04
  • $\begingroup$ @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. $\endgroup$
    – HTNW
    Mar 30, 2023 at 18:24
  • $\begingroup$ 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. $\endgroup$
    – HTNW
    Mar 30, 2023 at 18:30

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