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According to quantum mechanics, existence of an electron at a place depends on the wavefunction which in turn gives us the probability of an electron being there. And for a few special places, like nodes in an atom, the probability of finding an electron diminishes to zero. But at every other possible point in the whole universe, there is some non-zero probability of finding an electron. This is what I know (correct me if I'm wrong).

Now this is my question. Let's suppose an electron wandered far away such that now it is more closer to the nucleus of another atom than it is to it's original atom's nucleus. Now how does the electron know that which atom did it originally belong to? And with so many number of electrons surrounding us, this process of intermixing of electrons from one atom to another should happen quite spontaneously. But as far as I know, we don't see electrons moving from one atom to another quite often. Yes, there are cases of ionic bond formation and conduction where there is apparent movement and transfer of electrons, but why not everywhere?

And the implications of electrons "getting lost" are drastic. Electronic configurations of atoms would no longer matter. Almost all the matter around us would get ionized, and also emit (hopefully) beautiful and colourful line spectra. But in reality, non of these fantasies exist. Why?

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  1. Quantum particles are indistinguishable. You cannot "label" the electrons. So, a state in which two electrons are exchanged is the same state as the original one.

  2. The probability of an electron to be found very far away from the nucleus is very low.

  3. In order to calculate the probability for an electron to "jump" from an atom to another atom you need to use the wavefunction for the entire system (with two or more atoms). Solving it you can find the probability for an electron transfer. If the two atoms are at a distance much larger than the typical chemical bonds, that probability will be very low.

In conclusion, the reason we do not see electrons "getting lost" is the very low probability of such an event to happen.

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  • $\begingroup$ Thanks for the answer!! But in your first point, you said that "Quantum particles are indistinguishable". But they can be distinguished from their spin. Also what if an electron having a positive spin ($+\frac{1}{2}$) enters an atom where all the quantum numbers corresponding to this wandering electron are already attributed to some other electron. Then won't it violate the Pauli's exclusion principle? $\endgroup$
    – user243267
    Nov 28, 2019 at 12:02
  • $\begingroup$ As I said, you need to use the wavefunction of the whole system (both atoms) in this case. There will be no eigenstate that violates the exclusion principle, so the probability of such an event to happen is 0. $\endgroup$
    – Andrei
    Nov 28, 2019 at 12:07
  • $\begingroup$ I think I will leave out the real mathematical treatment for another day, as I don't really know that much of quantum mechanics right now. I appreciate your help. Thank you again! $\endgroup$
    – user243267
    Nov 28, 2019 at 12:17
  • $\begingroup$ You do not need to know math to understand. If you consider a single atom, after doing the math you get a probability P1 to find the electron at the position X, 1Km away from the nucleus. This assumes that your atom is alone in the universe, that's why your system consists of a single atom. If you have two atoms, with the second one located at X you will have a different wavefunction, and, by solving it, you will get a different probability, P2 to find your electron inside the distant atom. The probability to find two electrons in the same state in this atom will be 0, in agreement with Pauli. $\endgroup$
    – Andrei
    Nov 28, 2019 at 12:37
  • $\begingroup$ I see!! Your comment clears my doubts which I asked you in the comments. Now I understand the fallacy in my claim of violatiin of the exclusion principle. Thanks. $\endgroup$
    – user243267
    Nov 28, 2019 at 13:22

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