is given by $$\pssd_{\beta} (x) = \psi^{N-\beta}_{\beta} (x) \quad (\beta=0,1,2, ..., N),$$ where $\psi^{N-n}_{n} (x)$ is defined in Eq. (\[A\_state\]). The many-body ground state is then described by the Slater determinant of $\psi_\beta^{N-\beta} (x)$ ($1\le \beta \le N$), which span the subspace of $\{ \pssd_n (x) \}^N_{n=0}$. Here, we again assume that not $\beta=0$, but $\beta=1$ state with $\xi=0$ is occupied by a fermion. Then, the $N$-fermion ground-state wavefunction is given by $$\Pssd (x_1, ..., x_N) = \det_{1 \le i,j \le N}\, \left[ 
\pssd_{i} (x_j) \right].
\label{SD2}$$

At first sight, one might think that the Slater determinants Eqs. (\[SD1\]) and (\[SD2\]) are not related to each other because $\{ \puni_n (x) \}$ and $\{ \pssd_n (x) \}$ look completely different. Nevertheless, we will show that the following relation $$\Pssd (x_1, ..., x_N) = C_N\, \Puni (x_1, ..., x_N)
\label{eq:theorem}$$ holds for any $N$ (even), where $C_N$ is a constant independent of $x_1, ..., x_N$. This means that the many-body ground states of the uniform and SSD systems are identical to each other. The above relation can be proved by noting that the two Slater determinants become equivalent if there exists a linear transformation between $\{ \puni_n (x) \}^{N-1}_{n=0}$ and $\{ 