I have two identical fermions in an infinite potential well. They are non-interacting. How should I show that the first excited state is four-fold degenerate? Is the wavefunction just the superposition of the wavefunction of each fermion?
In the ground state, both electrons are in the state with the lowest value of "n". E.g., in the case of an infinite potential well, the lowest quantum number is n=1. In this case, both electrons have n=1, but one electron is spin up and one electron is spin down, because of exclusion.
The first excited state is one in which one of the electrons has n=1 (lowest single particle level) and one of the electrons has n=2 (first excited single particle level). In this case, either spin is okay for either electron.
So there are four states: (n=1,up; n=2,up), (n=1,up; n=2, down), (n=1,down; n=2, up), (n=1,down; n=2,down).
And, No, the wave function is not just the superposition for each one. The wave function is a Slater Determinant of the single-particle wavefunctions.
For example, in the case of the ground state, the spatial part of the wavefunction is symmetric $$ \sin(x_1\pi/L)\sin(x_2\pi/L) $$ and the spin part of the wavefunction is anti-symmetric $$ |\uparrow\downarrow>-|\downarrow\uparrow>\;. $$
You can work out the four excited states similarly.