\pssd_n (x) \}^{N}_{n=1}$ that is non-singular, i.e., the determinant of its transformation matrix is nonzero. For this purpose, however, the eigenfunctions in terms of the Gegenbauer polynomials (\[gegenbauer\]) are not so useful. In the following, we will consider a more direct treatment of $\psi_\beta^{N-\beta}(x)$. Let us rewrite the intertwiner $A^\dagger_\beta$ as $$A^\dagger_\beta  = -\sin^{-\beta} (\tilde{x}) \frac{d}{dx} \sin^\beta (\tilde{x}),$$ where we have introduced the variable $\tilde{x} := \frac{\pi}{L} x $ for simplicity. This leads us to the Rodrigues formula for the eigenfunctions Eq. (\[A\_state\]), $$\psi_\beta^{N-\beta}(x)= \sin^{-\beta +1 } \tilde{x}   \left( \frac{-1}{\sin \tilde{x}} \frac{d}{dx} \right)^{N-\beta} \sin^{2N-1} \tilde{x}. 
\label{eq:Rod}$$ This implies that the single-particle states can be described in terms of only trigonometric functions of ${\tilde x}$. Note that the explicit form depends on whether $N$ and $\beta$ are even or odd. Here, recall that $N$ should be even for the PBC. We can thus write $N= 2\theta $ with a positive integer $\theta$. For $\beta=2\nu$ ($\nu=0, 1, ..., \theta$), we get from Eq. (\[eq:Rod\]) the following expansion: $$u_\beta(x) :=(-1)^{\theta-\nu}\left(\frac{\pi}{L}\right)^{-2(\theta-\nu)} \frac{(2\theta+2\nu-1)!}{(4\theta-1)!}\psi^{2(\theta-\nu)}_{2 \nu}(x) 
= \sum_{n=\nu}^\theta  a_{2\nu,2n} \sin^{2n} (\tilde{x}),
\label{eq:expansion1}$$ where $$a_{2\nu,2n} = \prod_{k=n}^{\theta-1}\frac{(2k+1)(k+1)-\nu(2\nu-1)}{2(k - \theta)(k+\theta)}$$ with $a_{2\nu,2\theta} := 1$. For $\beta=2\nu-1$ ($\nu=1,2, ..., \theta$), we also obtain another linearly-independent solution of odd parity as 