TITLE: How fermions arrange themseves when there are more allowed states than particles in the system?
QUESTION [0 upvotes]: I'm considering the problem of three fermions in a system where spin is considered and there are two possible orbital wavefunctions, $\phi_{1}$ and $\phi_{2}$. This amounts to a problem with three particles in a four-state system. In this case, we can have two fermions in the same orbital state with opposite spins and the remaining fermion in the remaining orbital state with one of two possible spins. There are four ways to do this:
$ \phi_{1+}\,\,\phi_{1-}\,\,\phi_{2+} \\  \phi_{1+}\,\,\phi_{1-}\,\,\phi_{2-} \\ \phi_{1+}\,\,\phi_{2+}\,\,\phi_{2-}\\ \phi_{1-}\,\,\phi_{2+}\,\,\phi_{2-}$
We can build a Slater determinant with each line and coordinates $r_1,\,r_2,$ and $r_3 $, obtaining then all four possible antisymmetrical wavefunctions. Now, if $E_{i}$ is the energy associated to the state $\phi_{i}$ and if $E_{1}<E_{2}$ how will the fermions distribute themselves?
If we're considering three fermions in a harmonic oscillator-like potential we say that the fermions will fill the ground state before occupying the levels with greater energy, so only the first two lines would be possible, but why? Why do they have to fill the ground state? Couldn't one of them be so energetic it jumps to the next level, leaving a hole in the ground state? Does it have to do with the probability of the wavefunctions generated by the first two lines?
Thank you.

REPLY [1 votes]: In general, any states of any energy can be filled - it just depends on the initial conditions.  We often consider situations where the lowest-energy states are filled because of statistical mechanics - this is the state of thermal equilibrium at zero temperature, which is often a realistic approximation.  But if the system isn't at thermal equilibrium at zero temperature, then other states are possible.
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