briefly summarize the SUSY quantum mechanics for the $1/\sin^2 $ potential problem. The eigenvalue problem of Eq. (\[SSDpotential\]) can be solved algebraically for non-negative integers $\beta=0,1,2,3 \cdots$. Let us start with defining intertwiners $$A_\beta = \frac{d}{dx} + W'_\beta(x),  
\quad  A^\dagger_\beta = -\frac{d}{dx} +W'_\beta(x)$$ with the superpotential $W_\beta(x) := - \beta \ln \left( \sin \frac{\pi}{L}x \right)$. Using these operators, the Hamiltonian $H_\beta$ in Eq. (\[SSDpotential\]) can be written as $$\begin{aligned}
H_\beta = A_\beta^\dagger A_\beta + \left(\frac{\pi}{L}\right)^2\beta^2.  
\label{eq:quad}\end{aligned}$$ According to SUSY quantum mechanics, it is well-known that $A^{}_\beta A^\dagger_\beta$ and $A^\dagger_\beta A^{}_\beta$ form the SUSY-partner Hamiltonians, and the shape invariance ensures the recursive relation $$A^{}_\beta A_\beta^\dagger = A^\dagger_{\beta+1} A_{\beta+1}^{} + \left(\frac{\pi}{L}\right)^2 [(\beta+1)^2- \beta^2]
=  H_{\beta+1} - \left(\frac{\pi}{L}\right)^2\beta^2.
\label{eq:recursion}$$ In addition, it is easily confirmed that $A_\beta \psi^0_\beta(x)=0$ with $\psi^0_\beta(x) := (\sin \frac{\pi}{L} x )^\beta$, which corresponds to the eigenfunction of $H_\beta$ with the lowest eigenvalue. Note that Eq. (\[eq:quad\]) implies that the eigenvalue of $H_\beta$ is bounded from below by $(\frac{\pi}{L})^2 \beta^2$. It follows from Eq. (\[eq:recursion\]) that, for a given non-negative integer $\beta$, the $\ell$th eigenstate of $H_\beta$ is constructed as $$\psi_\beta^\ell(x) = A^\dagger_\beta \cdots A^\dagger_{\beta+\ell-2}A^\dagger_{\beta+\ell-1}\psi^0_{\beta+\ell}(x).
\label{A_state}$$ The explicit form of the eigenfunction can be written as $$\psi_\beta^\ell(x) ={\cal N}^\ell_\beta\, C_\ell^\beta\left(\cos 