as a geometric correction to $\chi_m$. For convenience, we fix the relation between $m$ and $\beta$ as $m = N-\beta$ so that the energy eigenvalues $\epsilon (m)$ of the lattice eigenfunctions are arranged in ascending order.

To verify Eq. (\[CLcomparison\]), we numerically diagonalize the lattice Hamiltonian (\[latticeSSD\]) for $L=101$ with $\mu=0$ and obtain the single-particle eigenfunctions $\chi_m (n)$. In this situation, we have $51$ energy levels below the zero energy: $m=0, \cdots, 50$, where the exact zero energy state corresponds to $m=50$. This can be seen in Fig. \[fig1\], where the eigenvalue spectrum is shown as a function of $m$. Note that, because of the finite-$L$ effect, the energy of the $m=49$ state slightly deviates from the zero energy to the low-energy side. Thus, the many-body ground state of the lattice SSD problem can be obtained by filling the levels $m=0, \cdots, 49$ with $N=50$ fermions.

![Eigenvalue spectrum of the lattice SSD model with $L=101$ and $\mu=0$. The energy eigenvalues $\epsilon (m)$ are arranged in ascending order. The Fermi level corresponds to $\epsilon (50)$, which is exactly at zero energy. The solid curve indicates $\epsilon (m) = -8.2 \times (m-49)(m-50)$ which is inferred from the spectrum of the continuous SSD 