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The moment an electron is observed (interaction), the electron takes on particle properties, but it is thought that it will recover its wave properties over time.

Even if an electron becomes an existence occupying a local area like a particle for a moment through some observation or interaction, shouldn't it be restored to an electron with wave properties over time?

Is there a function that describes how a collapsed wavefunction (particle or delta function?) is restored back into a (non-collapsed) wave?

D will
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    Required reading, or equivalent in your text or QM course. An electron wave function localized ("observed") by passing through a lead-foil pinhole evolves in the form provided in WP. Perhaps you can focus your question? – Cosmas Zachos Oct 20 '22 at 15:18
  • Thank you This is what I'm looking for. "Gaussian wave packets in quantum mechanics" Has this been proven to be consistent through experiments? – D will Oct 20 '22 at 15:36
  • It is a tractable paradigm *confirmed/vidnicated* throughout physics for virtually a century. "Confirmed" might be the wrong logical framing, here, however: Have you confirmed there are no purple dinosaurs on the dark side of the moon? – Cosmas Zachos Oct 20 '22 at 15:39
  • Can't we prove that by measuring the wavelength of the scattered(or spreading) wave at every unit of time? – D will Oct 20 '22 at 15:48
  • The momentum (hence wavelength) profile of that wave packet does not change over time. Consult your text. Tomonaga's QM does a good job on it. – Cosmas Zachos Oct 20 '22 at 16:03
  • width? Can't we just measure this width? This width eventually grows linearly in time, as ħt/(m√a), indicating wave-packet spreading.[6] For example, if an electron wave packet is initially localized in a region of atomic dimensions (i.e., 10−10 m) then the width of the packet doubles in about 10−16 s. Clearly, particle wave packets spread out very rapidly indeed (in free space):[7] For instance, after 1 ms, the width will have grown to about a kilometer. – D will Oct 20 '22 at 16:11
  • This is another question, altogether. The spreading of wave packets is accounted for in each and every beam experiment.... No discrepancies ever observed... – Cosmas Zachos Oct 20 '22 at 16:17
  • Individual electrons don't have wave properties. They have energy, momentum, angular momentum and charges (electric and leptonic). What can have wave properties is an ensemble of electrons. It's one of those "trivial" properties of quantum mechanics that we should teach in the first ten minutes but don't. Or, if we do, it doesn't register with most students, I guess. – FlatterMann Jun 17 '23 at 01:49

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Yep, it's called the Schrodinger equation! The Schrodinger equation states that if $H$ is the Hamiltonian, or total energy operator of a particle with wavefunction $\Psi$, then$$i\hbar\frac{\partial\Psi}{\partial t}=H\Psi.$$In other words, the Hamiltonian operator (which multiplies each basis state wavefunction by the corresponding energy of that wavefunction) also describes the time derivative of the wave function (how it changes). So if a particle is measured at $t=0$, then at that point its wavefunction suddenly becomes $1$ at a single point and $0$ at all other points. From there, the Schrodinger equation gives you the time derivative of the wavefunction, which shows how it will evolve and turn back into a delocalized probability density function again.

slithy_tove
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  • I would appreciate it if you could elaborate on this part in more detail. in the form of a function ( the Schrodinger equation gives you the time derivative of the wavefunction, which shows how it will evolve and turn back into a delocalized probability density function again.) – D will Oct 20 '22 at 15:18
  • In a point or delta function, is the wave-evolving form accurately expressed? – D will Oct 20 '22 at 15:20
  • The thing is, the Hamiltonian isn't a function - it's an operator (a function that acts of vectors to produce other vectors). In quantum mechanics, we express things like the particle's position as infinite-dimensional vectors. However, the Hamiltonian can be expressed in the general form $-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V(r)$, where $r$ is the position (this assumes motion in one dimension, for more dimensions you use the $\nabla$ operator instead of $\frac{\partial}{\partial x}$. Basically, the first term is the kinetic energy term and the second the potential term. – slithy_tove Oct 20 '22 at 15:33
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    https://physics.stackexchange.com/questions/334574/the-explicit-solution-of-the-time-dependent-schr%C3%B6dinger-equation-for-a-free-part – QCD_IS_GOOD Oct 20 '22 at 16:58