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I saw a discussion on this on Reddit so I thought I'd ask here to get a more authoritative answer.

Say you have two atoms, separated so that the electron wavefunctions have negligible overlap. Through some process, each atom loses an electron. Each of these electrons then travels towards the other atom, keeping their distance at all times so that the wavefunctions still have negligible overlap. Finally, the electrons are recaptured by the atoms.

Can we say that the electrons switched places? I can think of a couple of answers:

  1. The "common sense" approach: Yes, clearly the electrons moved while spatially separated, and each ended up on a different atom than where it started.

  2. No, electrons are indistinguishable particles. The final state differs from the initial state by a sign, which is unobservable.

  3. The question is somehow meaningless; this seems similar to option 2 but maybe in some sense you can't even ask the question.

I would think the correct answer is option 2, but I don't know how to refute option 1. Which is it?

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    $\begingroup$ Well, how do you know which electron is which to start with? If for example the two have opposite spins, you can measure that property to determine which is which. If they are identical, not only can you not tell, but I'd argue it's even irrelevant and you stray into philosophy. $\endgroup$
    – Asher
    Jun 13, 2017 at 16:01

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I would refute argument 1 for the following reason:

You are assuming the electrons have a definite trajectory: Each of these electrons then travels towards the other atom, keeping their distance at all times. In basic quantum mechanics, there is an invariably a carryover from classical mechanics, in that a definite trajectory is assumed, but actually there is no definite notion of motion, (sorry:) according to quantum field theory.

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We want to distinguish the process of extracting the two electrons and moving them to the other atoms from extracting the electrons and putting them back to the same atoms. Suppose that we have a system that can do either of the two, it is controlled by qubit, if it is in the state $\left|0\right>$ then there is no switch, while in the state $\left|1\right>$ the electrons are swapped. We can then start with a qubit in the state $\left|0\right>$, apply the Hadamard transform $U$ that acts on the qubit according to:

$$ \begin{split} U\left|0\right> &= \frac{1}{\sqrt{2}}\left[\left|0\right> + \left|1\right>\right]\\ U\left|1\right> &= \frac{1}{\sqrt{2}}\left[\left|0\right> - \left|1\right>\right] \end{split} $$

which yields the qubit in the state $\frac{1}{\sqrt{2}}\left[\left|0\right> + \left|1\right>\right]$, and then let the qubit control the process of extracting the electrons and putting them back. Since there is only a sign change when the electrons are interchanged, the qubit state factors out of the state of the qubit-atom system, but it is now in the state $\frac{1}{\sqrt{2}}\left[\left|0\right> - \left|1\right>\right]$, if we then apply the Hadamard transform $U$ to the qubit, we find that it's now in the state $\left|1\right>$.

If the electrons would always be put back to the same atoms regardless of the qubit state, then the final qubit state would become $\frac{1}{\sqrt{2}}U\left[\left|0\right> + \left|1\right>\right] = \left|0\right>$.

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