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Electrons do not follow fixed orbits around an atom's nucleus but exist within "clouds of probability" (orbitals), where there is a high chance of finding them. As one extends the search for electrons farther from the nucleus, the probability diminishes, though it never reaches zero. Consequently, there is a non-zero probability that an atom's electron could be located "on the other side of the Universe."

The question is: if an electron is everywhere in space and not confined in fixed orbits, how does it maintain an association with a specific nucleus and thus form an atom (or a molecule!), considering that there may be further atoms bonded with it and conversely more indistinguishable electrons? Is it a mere convention?

Michael Seifert
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Marco Fabbri
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    Could you clarify exactly what "...is a mere convention?". – Albertus Magnus Feb 27 '24 at 14:49
  • i was referring to the concept of "atom and his electrons" – Marco Fabbri Feb 27 '24 at 17:23
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    how does it maintain an association with a specific nucleus Because the wavefunction decreases exponentially away from that nucleus. – Ghoster Feb 27 '24 at 18:23
  • For practical purposes we usually neglect very low probabilities, since they are not likely to manifest themselves experimentally during our lifetime (or the lifetime of the Universe). Related: Why do atomic charges balance?, Is limited computational capacity a fundamental obstacle? – Roger V. Mar 06 '24 at 12:40
  • In this context, "very low probabilities" are as low as the probability of you tunneling through your chair right now and staying in one piece: it is not zero, but one should stick around for several hundred thousand lifetimes of our universe to observe it. For all practical purposes, the probability to observe an electron bound to a nucleus on another side of the universe as it is still in that bound state is zero. – MsTais Mar 14 '24 at 19:03
  • "how does it maintain an association with a specific nucleus" You mean in real life, or in the model where you are solving for a wavefunction based on fixed classical nucleus at the origin (by convention)? In real life an electron does not "maintain an association" since there is no mechanism for a point particle to "maintain" a record of anything. In real life, say in a solid, an electron will not necessarily be associated with a single nucleus. OTOH, in the usual first discussion of the hydrogen atom, say, the nuclear is fixed and classical, it is just part of the problem setup. – hft Mar 14 '24 at 20:04

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So in Quantum mechanics an electron being in a bound state in an Atom does not just mean, that we don’t know exactly where it is and have a probability distribution describing at what point it actually might be. That would be the case, if we would know, that the wave function was a delta peak in position space and we just had incomplete information about, where exactly that delta peak is centered. In a bound state the electron is in fact delocalized around the center of your binding potential and exponentially decaying at large distances to the center. Collapsing to specific position is only happening, when measuring the position. Since the wave function of the electron in an atomic bound state is clearly centered around the nucleus, there is a definite association of that electron with the nucleus. Actually the fact, that this electron does occupy this specific bound state (including definite spin state), is the only way you can distinguish it from other electrons as they are indistinguishable otherwise.

Now, if you do some measurement, you might cause the electron wave function to collapse into small spacial area far away from the nucleus. If as mentioned in your question the system also does contain a lot of other stuff apart from the electron and the nucleus, which cause stronger interaction at the new position of the electron, the association with the previous nucleus would indeed be lost. In fact that is, how I imagine a tunneling transition to work. But it is key here that you changed the state of the electron by the measurement and only after the measurement the binding is lost.

For completeness I would like to mention that this whole picture of first looking at the electron nucleus interaction, then a measurement and then the interaction of the electron with new particles closer to its new positions is of course just a model. As for exact time development you would have to solve the entire many particle system without neglecting any pair interactions at any time. From my current understanding this whole hand wavy fuzz about “state → measurement → new state” probably is an artifact of the inability to solve a many particle system exactly, though I am somewhat unsure on this take. I mainly write this clarification to avoid getting asked what exactly I mean by measurement in this context.

Also if you are interested in a more statistical take on something somewhat related to your question you might be interested in reading this https://physics.stackexchange.com/a/774294/325089 .

Zaph
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    Folks have addressed the concerns about state$\rightarrow$measurment$\rightarrow$new state, and found them not to be of any real importance, i.e. the model is not very hand wavy at all. Here is a link on quantum measurement processes: https://qquest.lbl.gov/~carney/qtm-notes-full.pdf – Albertus Magnus Mar 06 '24 at 14:11
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    It is perhaps worth noting that in molecules, the electrons are shared (not neccesarily equally) by the multiple nuclei involved. Such electrons live in bonding states and are expressed by the method Linear Combinations of Atomic Orbitals (LCAO), in organic chemistry, for example, carbon forms bonds which are hybridizations between s and p orbitals, i.e. the $sp^3$ orbital which has a tetrahedral geometry and admits four single bonds. – Albertus Magnus Mar 06 '24 at 14:18