# On the meaning of distinguishability, and wavefunctions for 3 electron atoms

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?

• All electrons are indistinguishable, which leads to the concept of a Slater determinant. – orthocresol Jan 11 '20 at 13:41
• Note that there is no $1p$ state. Above $1s$ is $2s$, though the point of your question is still clear. – Andrew Jan 11 '20 at 16:31