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Decoherence theory predicts that any quantum particle coupled to any "large" environment should undergo decoherence and its wavefunction should collapse. This explains why measurement leads to wavepacket reduction.

However, in solids, liquids or gases, electrons within atoms don't reduce and stay as wavefunctions (orbits) somehow protected from the environment of the atoms.

This is surprising since the atoms are at room temperature, with a lot of things to interact with such as neighbouring atoms, light, thermal excitations, etc. So any idea why electrons seem 'protected' from a wavepacket reduction in atoms?

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Welcome to SE -- good question! Decoherence does not mean that there won't be a wavefunction anymore, it just means that if the electron becomes coupled to the surrounding environment, its state will be described by a probabilistic mixture of orbital wavefunctions rather than a (coherent) superposition thereof. The electron in an atom doesn't have some "non-quantum" state(s) it can collapse into -- collapse just means that it will end up in one of the orbital states.

As a simplified example, consider spin states of an electron (simpler than orbitals because there are only two of them). Let $|0\rangle$ and $|1\rangle$ be some (orthonormal) basis states for this system. Then if the electron is initially in the state $$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle),$$ a (coherent) superposition of the two basis states, after it has interacted with some noisy environment for a while we would expect its state to evolve towards a probabilistic mixture of the states $|0\rangle$ and $|1\rangle$ (assuming we are still representing in this basis), with probabilities 0.5 each unless there is some other factor to bias them. But the electron's spin cannot magically enter some other state that is not a linear combination of these; similarly, the orbital state remains an orbital state even when it decoheres.

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    $\begingroup$ "assuming we are still interested in this basis" How does what we are "interested" in affect the behavior of an electron? By "interested in", do you mean "subjecting the wavefunction to an operator that has these basis states as eigenstates"? $\endgroup$ – Acccumulation May 8 at 18:18
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    $\begingroup$ @A.JBeahv again, good questions. First of all, you cannot measure the position of particle exactly; the best you can do is measure whether it is inside some small region of space or not, and you can make that region arbitrarily small, but not a single point. If you do that and localize the electron within a very small region, then its momentum uncertainty will be be large, and so after the measurement it might not even be in a bound state any more! (Full disclosure: I'm not sure how you would actually do such a measurement.) See more in next comment. $\endgroup$ – Will May 9 at 1:15
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    $\begingroup$ @A.JBeahv But if it is still in a bound state, such a state can always be written as a superposition of orbital states, since these are a complete basis for the bound states. Again, if you localize the electron (via measurement) in some small region, there will be very high energy orbitals involved in such a superposition, such that there may be significant probability of dislodging the electron entirely, but the bound part of the wavefunction will still be a superposition of orbital states. This, though, brings us to Rococo's answer, which points out that the actual energies required for $\endgroup$ – Will May 9 at 1:22
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    $\begingroup$ @A.JBeahv localization are much higher than those typically found in every-day-scale noisy environments, so you probably wouldn't get such extreme localization due to environmental interactions in, say, a laboratory on Earth (unless you are creating high-energy noise on purpose!) The bottom line is: the type of effective "measurement" due to coupling with the environment that you describe is actually quite improbable (vanishingly improbable as the volume in which you are localizing the electron goes to 0.) Hope this helps! $\endgroup$ – Will May 9 at 1:27
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    $\begingroup$ Thanks again. That really helped. Just to be sure I got it right : if I compare it to the usual double slit exp with single electrons detection on the screen, here in the atom, there is nothing close to that process because not enough energy and coupling with a measurement device. And this mainly comes from the Heisenberg principle which keeps the electron delocalized over a typical angstrom-size area around the nucleus (by balancing momentum and position within the atom) with typical eV energies. Would you agree ? $\endgroup$ – A.J Beahv May 9 at 8:27
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I agree with Will's answer, but since there are multiple ways of looking at this here is another one: For an electron that is initially in its ground state to become spatially localized necessarily requires that some energy be added. For a hydrogen atom, the needed energy is at least 10 eV (to get to the second shell), and increasingly more than this to make an increasingly localized wavepacket. This requires high-energy photons, and there normally (at temperatures we find on Earth) aren't that many of those around, nor are there enough low-energy photons for multiple-photon transitions to be likely.

In a high-temperature environment in which there are lots of x-ray and gamma ray photons to drive these transitions, you probably would no longer have neutral hydrogen but instead a plasma. The electrons in this plasma might indeed be localized on a smaller scale than the hydrogen orbitals, depending on parameters such as the density.

This theme of needing higher energies to resolve smaller locations might sound familiar- it is just another manifestation of why we need huge accelerators like the LHC to directly probe the physics on very small length scales within a nucleon.

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