them as $\psi^{q-\beta}_{\beta} (x)$ ($\beta = 0, 1, \dots, q$) with the corresponding energy eigenvalues $\xi=-\left(\frac{\pi}{L}\right)^2\beta(\beta-1)$. Note that $\psi^0_q (x)$ can be thought of as the single-particle ground state of the SSD problem, while $\psi^{q}_{0}(x)$ and $\psi^{q-1}_{1} (x)$ are the plane-wave solutions with zero energy.

   B.C.   $q\backslash\beta$              0                         1             2                                                                       3                                                    4                                             5                         
  ------ -------------------- ------------------------- ------------------------- ----------------------------------------------------------------------- ---------------------------------------------------- --------------------------------------------- ------------------------- --
    AP             5           $\cos\frac{5 \pi x}{L}$   $\sin\frac{5 \pi x}{L}$  $\sin^2\frac{\pi x}{L}\cos\frac{\pi x}{L}(12\cos^2\frac{\pi x}{L}-5)$   $\sin^3\frac{\pi x}{L}(8\cos^2\frac{\pi x}{L}-1) $   $\sin^4\frac{\pi x}{L}\cos\frac{\pi x}{L} $   $\sin^5\frac{\pi x}{L}$   
    P             4            $\cos\frac{4 \pi x}{L}$   $\sin\frac{4 \pi x}{L}$  $\sin^2\frac{\pi x}{L}(6\cos^2\frac{\pi x}{L}-1) $                      $\sin^3\frac{\pi x}{L}\cos\frac{\pi x}{L} $          $\sin^4\frac{\pi x}{L} $                                                
    AP            3            $\cos\frac{3 \pi x}{L}$   $\sin\frac{3 \pi x}{L}$  $\sin^2\frac{\pi x}{L}\cos\frac{\pi x}{L} $                             $\sin^3\frac{\pi x}{L} $                                                                                                     
    P             2            $\cos\frac{2 \pi x}{L}$   $\sin\frac{2 \pi x}{L}$  $\sin^2\frac{\pi x}{L} $                                                                                                                                                                             
    AP            1             $\cos\frac{\pi x}{L}$     $\sin\frac{\pi x}{L}$                                                                                                                                                                                                        
    P             0                    const.                                                                                                                                                                                                                                          

In order to precisely see the correspondence between the solutions of the $1/\sin^2$-potential and the SSD problems, we further comment on the boundary conditions. In the literature, it is often assumed that $\beta \ge 1$ in Eq. (\[SSDpotential\]). Here, we have added the SUSY partners for $\beta=0$ and $1$. For both $\beta=0$ and 1, there is no potential term in $H_\beta$ and 