(PBC) for odd $N$ and anti-periodic boundary conditions (A-PBC) for even $N$, as the anti-symmetric unit function leads to $\Psi_F(...x_n+L...)=(-1)^{N-1}\Psi_F(...x_n...)$. Since the single-particle eigenfunctions, $\psi_n(x)$, form an orthogonal basis, the Slater determinant obeys these boundary conditions if $$\psi_n(L)=(-1)^{N-1} \psi_n(0).
  \label{eq:periodicBC}$$

In free space the single-particle solutions are given by $\psi_n(x) = \frac{1}{\sqrt{L}} e^{i k_n x}$, with $E_n=\frac{\hbar^2 k_n^2}{2m}$. For PBC (corresponding to an odd number of particles) we therefore have $k_n = 2 n \pi /L$ with $n=\lbrace 0,\pm 1,\pm 2,\pm 3 \dots\rbrace$. In this case the ground-state is non-degenerate, while all excited states $E_n,E_{-n}$ are two-fold degenerate. For A-PBC (corresponding to an even number of particles) all states are two-fold degenerate as $n=\lbrace \pm 1,\pm 2,\pm 3 \dots\rbrace$. The effect of the lattice is to introduce energy gaps into the single particle spectrum (see Fig. \[fig:SPspectrum\]), and while the free space degeneracy structure remains for a large range of lattice depths, this gap can break the degeneracy between the two states closest to the gap. For example, for an even number of sites this will be the case in the PBC spectrum, but not in the A-PBC spectrum. A more detailed discussion of this odd-even effect is given in 