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?

  • $\begingroup$ is it a one-dimensional potential well? $\endgroup$
    – hft
    Feb 24, 2015 at 22:01
  • $\begingroup$ do they have spin? $\endgroup$
    – hft
    Feb 24, 2015 at 22:02
  • $\begingroup$ Yes. Spin is 1/2. $\endgroup$
    – Artemisia
    Feb 24, 2015 at 22:10
  • $\begingroup$ it is a one-dimensional or two-dimensional well? $\endgroup$
    – hft
    Feb 24, 2015 at 22:11
  • $\begingroup$ I want to understand it for a 1D well first :) $\endgroup$
    – Artemisia
    Feb 24, 2015 at 22:14

1 Answer 1


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.


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