Fermions in a well 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? 
 A: 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.
