From my understanding of quantum mechanics, when a wavefunction is observed, it collapses into a single state instantaneously (or at least in the length of a Planck time.) Is there a reason it has to take no time? Could it be briefly observed in a different state before settling into it's collapsed state?

# [Physics] Why are wavefunction collapses instantaneous

quantum mechanicsquantum-interpretationsquantum-measurementswavefunctionwavefunction-collapse

#### Related Solutions

A measurement is a type of entanglement. If you measure a photon at the opening of a slit in a double slit experiment then you can think of there being a spin state at this opening. If the photon passes through it the spin changes direction. So the wave function for the photon in the double slit experiment is $$ \psi(x)\rangle~=~C(e^{ikx}|1\rangle~+~e^{ik(x’)}|2\rangle) $$ where the x and x’ denote the different paths through the two slits. If you compute the modulus squared you get $$ \langle\psi|\psi\rangle~=~|C|^2(2~+~e^{ik(x’-x)}\langle 1|2\rangle~+~ e^{-ik(x'-x)}\langle 2|1\rangle) $$ The exponential terms give the interference between the superposed states $|1\rangle$ $|2\rangle$ which have a nonzero overlap $\langle 1|2\rangle~\ne~0$. We now consider the coupling of a spin to this $$ \psi(x)\rangle~\rightarrow~C(e^{ikx}|1\rangle|+\rangle~+~e^{ik(x’)}|2\rangle|-\rangle) $$ which serves as the detector. Since $\langle +|-\rangle~=~0$ if you compute $\langle\psi|\psi\rangle$ the interference term is gone. This gives a reason for why one can’t measure which slit the particle travels through. We have replaced a superposition of states type of nonlocality with an entanglement. Now in doing this we have not addressed the question of how one actually observes the spin. One might presume we entangle some other states and do so up some “chain.” This of course leads to the Schrodinger cat problem. A cat or a human being, our brains and the like are not described well as single quantum systems. In fact they are messy thermal systems with high entropy. This is one problem with the whole idea of quantum consciousness, which never got out of the starting gates as a serious physical problem.

Unless the wavefunction collapses to an eigenstate of the Hamiltonian, the subsequent time-evolution will produce a superposition.

The postulates clearly state that, if you measure the observable $\Lambda$ and obtain the outcome $\lambda$ (assumed non-degenerate for simplicity), then the state collapses to the eigenstate $\vert\psi_{\lambda}\rangle$ of $\hat \Lambda$, and the subsequent evolution is given by $$ \sum_{k}e^{-iE_k t/\hbar}\vert \Psi_{E_k}\rangle\langle \Psi_{E_k}\vert\psi_{\lambda}\rangle $$ where $\vert \Psi_{E_k}\rangle$ is an eigenstate of $H$ with eigenvalue $E_k$. Thus, unless $\langle \Psi_{E_k}\vert\psi_{\lambda}\rangle=\delta_{E_{k}\lambda}$, the system will revert to a superposition.

Edit: after the measurement the state $\vert \psi_{\lambda}\rangle$ functions as an initial state and its time development is obtained in the usual manner by expanding over a complete set of eigenstates of $H$ using $$ \hat 1=\sum_k\vert\Psi_{E_k}\rangle\langle \Psi_{E_k}\vert $$ so that $$ \vert\Psi(0)\rangle=\vert\psi_\lambda\rangle= \sum_k\vert\Psi_{E_k}\rangle\langle \Psi_{E_k}\vert\psi_\lambda\rangle $$ and evolving the $H$-eigenstates $$ \vert\Psi(t)\rangle= \sum_k\,e^{-iE_{k}t/\hbar }\vert\Psi_{E_k}\rangle\langle \Psi_{E_k}\vert\psi_\lambda\rangle\, . $$

## Best Answer

It’s important to remember that quantum mechanics is a tool that we use to describe the world — it is not the same as the world. For all that we love to talk about wavefunctions, it’s not clear at all that the wavefunction is a real thing. Questions like “can the wavefunction collapse instantaneously?” highlight this.

Let’s consider a specific example where nonrelativistic quantum mechanics and the Copenhagen probability interpretation of the wavefunction are both useful: a neutron interferometer. The fundamentally quantum-mechanical feature of any interferometer is that incident particles take two paths to the same destination, which you can demonstrate using interference effects. Neutron interferometers have the pedagogical advantage that, in every existing implementation, the number of neutrons in the interferometer at once is always zero or one.

A neutron in an interferometer has an intrinsic (strong-interaction) radius of about a femtometer. Its wavevector along its direction of motion is characterized by its de Broglie wavelength, typically a few angstroms. Perpendicular to its direction of motion, in a slice through the middle of the interferometer, the neutron’s probability distribution corresponds to a two-well potential: the neutron may be found in this arm or that arm of the interferometer, but not in the middle. The forbidden middle can be macroscopic. In a neutron interferometer it is centimeters. (In the LIGO optical interferometers, each laser’s path is several kilometers, and the probability of detecting a laser photon in the Louisiana swampland outside of the vacuum system is zero.)

You can establish this neutron wavefunction by solving the Schrödinger equation,

$$ \left( \frac{\hat p{}^2}{2m} + V(\vec x) \right)\psi(\vec x, t) = \hat E\psi(\vec x, t) $$

for a three-dimensional potential $V(\vec x)$. Your potential must correspond to the material which forms the interferometer (a lattice of silicon nuclei), plus any “sample” present in one of the arms, plus, ending somewhere upstream, a two-dimensional infinite well to explain that neutrons cannot originate from outside of the beamline. Solving this equation gives the probabilities that neutrons will be observed in each of the detectors downstream from the interferometer, and predicts how those probabilities change if the “sample” is adjusted. An interferometer experiment measures those probabilities by counting lots of neutrons at those downstream detectors, then adjusting the sample and measuring the new distributions.

When we say that “an observation collapses the wavefunction,” what we usually mean is the following: if we were to stick detectors

insidethe interferometer, where we have demonstrated that a single neutron follows both paths, we wouldneverobserve the same neutron being “classically detected” in both arms. Even in the case where the detection opportunities are spacelike-separated, so that relativity suggests they cannot influence each other, the probability of “detecting” the same neutron in both arms is zero. So the Copenhagen interpretation makes an ad-hoc adjustment to account for this by introducing the idea of collapse. There are some kinds of interactions which change the wavefunction “instantaneously,” using a mechanism to be worked out later.Pop-science books make a lot of hay out of instantaneous collapse because, in relativity, nothing is instantaneous. This is a little silly, because the Schrödinger equation is non-relativistic.

An early hint at a way out of this puzzle came from Mott, 1929 (see also): if the detection events are

alsoquantum-mechanical, the behavior of “classical” measurements becomes a question of correlated probabilities. There has been an enormous amount of research on the subject, especially in the last twenty years, using relativistic quantum mechanics, “weak measurements,” and sneaky business about entanglement. You currently have another answer which links to two papers from 2019 and 2020, and suggests that the mystery has mostly been wrung out of this historic puzzle.