enjoy $\varphi(0)=\varphi(L)=\varphi'(0)=\varphi'(L)=0$, which reflects the fact that they have wavepacket-like shapes and are disconnected at $x=0$ (mod $L$).

Equivalence of two Slater determinants {#sec: equivalence}
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In this section, we give a proof that the Slater determinant of the plane waves, corresponding to the ground state of the uniform system with PBC, is identical to the Slater determinant of $\psi^{q-\beta}_\beta (x)$, which are the eigenfunctions of the SSD problem. This can be thought of as a continuous version of the ground-state correspondence proved in a class of 1D lattice models reducible to free fermions [@Katsura2011; @Katsura2011B; @MaruyamaKH2011]. In the following, we focus on the case where the total fermion number $N$ is even. It is, however, very straightforward to extend the proof to the case of odd $N$.

Let us first consider the many-body ground state of $N$ (even) fermions on the uniform periodic ring with $\mu = (\frac{\pi}{L})^2 N^2$, which is the Slater determinant obtained by filling the plane-wave states up to $\xi=0$. It is easy to see from Table I that the plane-wave states with even $q$ can be obtained as linear combinations of eigenfunctions in the columns labeled by $\beta=0$ and $1$, which satisfy the PBC. Since 