 Solution

For an infinite cubic potential well of side $L$, the one-particle energy eigenvalues are $E_{n_x{n}_y{n}_z}=E_0(n_x^2+n_y^2+n_z^2)$, where $E_0={\pi }^2{\hslash }^2/(2m{L}^2)$ and each quantum number is a positive integer. The lowest spatial orbital is $(1,1,1)$, with energy $3E_0$ and orbital degeneracy one. The next energy is $6E_0$, with three orbitals given by permutations of $(2,1,1)$.

Spin-zero identical particles are bosons. Both can occupy the lowest orbital, so the two-boson ground state is nondegenerate. The first excited total energy is $9E_0$: one particle occupies the lowest orbital and one an excited orbital. Symmetrizing the two-particle wavefunction gives one state for each of the three excited orbitals, so the first-excited degeneracy is three.

Spin-half identical particles are fermions, so the total state must be antisymmetric. Two particles in the same lowest orbital have a symmetric spatial wavefunction and must form the unique spin singlet state. Thus the fermion ground state is also nondegenerate. At the first excited energy, each choice of excited spatial orbital permits a symmetric spatial state with a singlet state, or an antisymmetric spatial state with a three-dimensional triplet state. ThereforeThe spin multiplicity is essential: counting only the three orbital choices would omit allowed triplet states. The result assumes ordinary spin-statistics, an unsplit spin degeneracy and the hard-wall cube model.