$\beta$. Thus, the linear space of the SSD eigenfunctions is equivalent to that of $\{\sin^{2n}(\tilde{x}) \}$ and $\{\cos(\tilde{x})\sin^{2n-1}(\tilde{x}) \}$. According to the standard formula of the trigonometric functions [@sinexpansion], moreover, the linear spaces spanned by $\{\sin^{2n}(\tilde{x}) \}$ and $\{\cos(\tilde{x})\sin^{2n-1}(\tilde{x}) \}$ are respectively equivalent to those by $\{\cos(2n\tilde{x})\}$ and $\{\sin(2n\tilde{x})\}$. Hence, the linear space of $\{ \pssd_n (x) \}^{N}_{n=0}$ can be spanned by the basis elements of $\{\cos(2n\tilde{x})\}^{N/2}_{n=0}$ and those of $\{\sin(2n\tilde{x})\}^{N/2}_{n=1}$, indicating that the single-particle basis of the SSD problem is identical to that of the plane waves.

Here, we should recall that the $\xi=0$ state in the $\beta=0$ sector is assumed to be empty of fermion, which implies that $u_0(x)(\propto \psi_0^{N}(x))$ is not included in $\{ \pssd_n (x) \}^{N}_{n=1}$. However, we can easily check $u_0(x)=\sum_{n=0}^{N/2} a_{0,2n} \sin^n(\tilde{x})\propto \cos(N\tilde{x})$ [@redundant]. Thus, the removal of $u_0(x)$ from the SSD basis results in the linear space $\{ \puni_n (x) \}^{N-1}_{n=0}$, which is identical to that of the plane waves below the Fermi level in the uniform system. Thus, we have obtained the desired relation Eq. (\[eq:theorem\])

If the total fermion number $N$ is odd, we should take eigenfunctions in the row labeled by $q=N$, which are antiperiodic in $x$. The proof proceeds 