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In a 2-electron atom at lowest energy, the $(1s)^2$ is occupied and the electronic wave-function must satisfy anti-symmetry requirements in the particle coordinates, as the spatial wave function is symmetric. How is the situation in a 3 electron system?

In a 3-electron atom (or in a nucleon with one excited quark) of lowest energy, say the $(1s)^2(1p)$ states are occupied, is the fermion in the $(1p)$ state distinguishable? I.e. must my wave-function still treat all 3 electrons indistinguishably, or are mixed symmetry wave-functions now physically viable?

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    $\begingroup$ All electrons are indistinguishable, which leads to the concept of a Slater determinant. $\endgroup$ – orthocresol Jan 11 '20 at 13:41
  • $\begingroup$ Note that there is no $1p$ state. Above $1s$ is $2s$, though the point of your question is still clear. $\endgroup$ – Andrew Jan 11 '20 at 16:31
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The only case in which electrons can be considered distinguishable is when they are so well separated that their wave functions do not overlap. Any electrons on the same atom or within the same molecule are considered indistinguishable.

In chemistry we like to talk about electrons being in specific orbitals, but the indisinguishable nature of them means that they are all in all orbitals simultaneously. This is why a mathematically rigorous description of a molecule using hybrid orbitals is just as valid as one using canonical orbitals. Neither is true in the sense of describing where each electron is, and both are true in the sense of describing the collective electron density.

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  • $\begingroup$ Thank you so much! $\endgroup$ – Craig Jan 11 '20 at 22:26

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