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In quantum theory, when an electron is sent through a slit (or multiple slits),the electron is described using probability amplitudes and is said to be in a superposition of multiple quantum states. However, when the electron reaches a screen or a detector, the wavefunction of the electron is said to collapse and hence the electron localises itself in one place. Now, I understand that there is really no consensus on why this happens, but I would like understand the opposite process of the collapse of the wave function. If the electron reaching the screen is then resent into another slit, then I assume that the wavefunction "decollapses" and hence we cannot localise the electron anymore again right? How does this process of "decollapse" of the wavefunction work?

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    $\begingroup$ to simplify the process, just think of the wave function as a wave of probability, and hitting the screen forces an outcome $\endgroup$ Oct 14, 2019 at 15:56

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Collapse is an unfortunate terminology, imo. The wavefunction is a mathematical function , not a balloon

When you throw a dice and it comes up 6, has anything collapsed? It is but one event, that follows the probability distribution.

dice

Probabilities are the same in quantum and classical states.

The confusion arises by the people who want to think of the wavefunction $Ψ$ as real, instead of a mathematical model of a given potential and boundary conditions.

The quantum mechanical probability is given by $Ψ^*Ψ$ and here, as an example, is the electron probability distribution in a hydrogen atom:

hydr

If by some method the electron is scattered off, there will no longer be the same probability distribution ( $Ψ^*Ψ$ ) for the scattered electron , but a new one, dependent on the boundary conditions for the new state.

So "decollapsing" has no meaning. If the exact same conditions are set up, the original solution will hold, it is all about mathematics and solutions of equations. All the undisturbed hydrogen atoms will have the above probability distribution because they are described by the same wavefuntion.

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  • $\begingroup$ If it is all about mathematics so why people are trying to think this pure mathematics shows real physical world? If it is just mathmatics, but there would be no reason that it is the reality! Anything could happen in pure mathematical world but why it should describe the facts in our real world? You brought up dice example and I would recall the collapsing of dice probablity distribution as an event. You roll a dice and it shows for example 6. What happened to probability distribution? Still, it’s there but you had an event that pulled out a single number from probability distribution. $\endgroup$ Oct 14, 2019 at 17:23
  • $\begingroup$ @alone “All models are wrong, some are useful.” QM is useful. That doesn’t mean it’s a full representation of Truth, nor does it claim to be. It claims to predict the outcome of experiment, and it does. $\endgroup$
    – JPattarini
    Oct 14, 2019 at 17:39
  • $\begingroup$ @JPattarini I'm a computational guy, so that's our most famous motto: All models are wrong, some are useful. But I'm not sure it is applicable to QM or not. I mean it just bothers me if I want to think of QM as a mathematical tool that enables us to study some aspects of physical world and predict the outcome of experiments, but still we are not sure that is QM the very truth of our world? I know it's more philosophical question rather than physical one, but I don't know... $\endgroup$ Oct 14, 2019 at 17:43
  • $\begingroup$ I agree with this answer. However, I would qualify the statement Probabilities are the same in quantum and classical states.. The probabilities in a classical sense are due to limited knowledge of the system. In QM it's an inherent part of measurement outcomes even if you know everything about the system (the wavefunction). $\endgroup$ Oct 14, 2019 at 17:43
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    $\begingroup$ @daljit97 Throw a ball. The Newtonian mathematical model can give you an exact prediction of its trajectory given the boundary conditions:( mass , angle of throw, force of hand). Is it the mathematics that is real or the ball falling at an (x,y,z) predicted by the mathematics? It is a philosophical question.,not a physics one $\endgroup$
    – anna v
    Oct 15, 2019 at 4:13
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The wave function for a free particle experience dispersion (this means that components with different energies, travels at slightly different speeds). This causes a free particle, that is initially well localised, to become more and more delocalized. The wave function spreads as it moves around in space (propagates).

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The traditional interpretation of quantum mechanics was that during measurements the wave function of a particle switched instantaneously to one of the allowed 'eigenfunctions' associated with the property being measured. If you subsequently measured some other property with a different set of allowed eigenfunctions then the wave function would switch again to one of those. Depending on the nature of the property being measured, the 'before' state could be more or less widely distributed in space than the 'after' state, so the term 'collapse', with its implications of reduction or compression, is a misnomer. In that terminology, what you called 'decollapsing' is just another 'collapse' to a different wave function again.

This instantaneous jumping from one eigenfunction to another has been seen as problematic by many physicists, and has led to a range of efforts to explain it away.

That said, some physicists take the view that the wave-function is simply a mathematical representation of probabilities, and there's no problem with probabilities switching instantaneously- for example, if you lose your phone there is a range of places it might be, and you could associate probabilities with each of them, but as soon as you find it the range of probabilities 'collapses' to one.

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  • $\begingroup$ The problem is that if you have a distribution of probabilities, the probability of single point in this distribution is zero! Why?! cause: $\int_{a}^{a} P(x) dx = 0$ where $a \in [x_{1},x_{2}]$. But you could have: $\int_{x_{1}}^{x_{2}} P(x) dx \neq 0$. So, why if a single point has zero probability, but the integration of all these zeros is something non-zero?! $\endgroup$ Oct 14, 2019 at 18:50
  • $\begingroup$ @AloneProgrammer Your measuring device does not exist at a point, so there isn't an issue with that. $\endgroup$ Oct 14, 2019 at 20:52
  • $\begingroup$ I know but it's just jiggling me from philosophical point of view that why the probability of finding something in lower dimensions is always zero in a higher dimension space... For example the probability of finding points, lines, and planes are zero in a 3D space, because of that integral formula, but is that really true?! $\endgroup$ Oct 14, 2019 at 21:03

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