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?