$\mu = (\frac{\pi}{L})^2 N^2$, one can easily confirm that the states in the $q=N$ row with $\beta=0$ and $1$ are just located at $\xi=0$. Thus, the many-body ground state is two-fold degenerate, because either one of the two $\xi=0$ states is occupied by a fermion. In the following, we assume that $\beta=1$ state is occupied. Then, the single-particle states involved in the Slater determinant is written as $$\begin{aligned}
\puni_n (x) = \left\{
\begin{array}{cc}
\cos \frac{2n \pi}{L}x \quad & (n=0, 1, ..., \frac{N}{2}-1), \\
 & \\
\sin \frac{(2n-N+2)\pi}{L}x \quad & (n=\frac{N}{2}, \frac{N}{2}+1, ..., N-1),
\end{array}
\right.\end{aligned}$$ where $\cos \frac{N\pi}{L}x$ does not appear because we have assumed that $\beta=0$ state is empty. Note that, if we consider the case of odd $N$, the boundary conditions are antiperiodic, where we should take the wavefunctions only from $q=$odd sectors. In terms of $\puni_n (x)$, the $N$-fermion ground-state wavefunction is written as $$\Puni (x_1, ... , x_N) = \det_{1 \le i, j \le N} \, \left[ \puni_{i-1} (x_j) \right].
\label{SD1}$$

We next consider a many-body ground state of the SSD problem with the same setup, i.e., $N$ is even and $\mu= (\frac{\pi}{L})^2N^2$. As shown in the previous section, the single-particle eigenfunctions of the SSD problem 