Why are wavefunction collapses instantaneous?

Question

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?

Answer 1

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 inside the interferometer, where we have demonstrated that a single neutron follows both paths, we would never observe 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 also quantum-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.

Answer 2

A quantum system is always observed to be in a single state, but that doesn't mean that anything has "happened" to the wave function. It's just a point of view.

Replace Schrodinger's cat with a scientist observing the Geiger counter. To the "orginal" scientist, outside the box, the scientist in the box is in a superposition of two states: In one possibility, he has observed the atomic decay. In the other possibility he has not.

In both possible states, the internal scientist has observed the atom to be in one state or another, but nothing has really happened, according to the external scientist.

When the external scientist opens the box and asks "did the atom decay?", the situation will be resolved from his point of view... but he doesn't know if he's in a bigger box or not.

The notion of "collapse" really just constrains the ways in which the observer's state is entangled with the state of whatever he observed.

Answer 3

The collapse is not instantaneous. Rather, it proceeds over a finite time and by (reasonably, at least in principle) well understood dynamics. In this paper, they observe the trajectory of a quantum jump due to a measurement, by taking snapshots at different elapsed times.

Furthermore, a quantum jump from one state to another can be detected before it completes, and reversed in a deterministic fashion - as shown in this study where they achieved this by monitoring the population of a third state coupled to one of the other two. This technique may be useful in the future for quantum error correction.

Another question which might arise is - how surprising is this result? Does it revolutionise our understanding of quantum mechanics? The answer is no, as the top voted answer to this question explains very well and robustly.