analogy with lattice systems, the Hamiltonian for the system with a generic deformation is defined as $${\hat {\cal H}}_{\rm cont} = \int^L_0 \, dx\, {\hat c}^\dagger (x)\,
                f(x) \left( -\frac{d^2}{dx^2} - \mu \right) {\hat c}(x) 
\label{Hamiltonian}$$ where $\mu$ is the “chemical potential" introduced for later convenience, and $f(x)$ encodes the spatially varying couplings. Here we set $\hbar=2m=1$. If $f(x)=1$, the systems is nothing but the uniform ring with non-relativistic free fermions. For the case of SSD, $f(x)$ is defined as $$f(x) := \sin^2 \left(\frac{\pi }{ L} x\right).$$ Thus, the energy scale of the system smoothly approaches zero as $x \to 0$ and $x \to L$, implying that the ring is effectively disconnected at $x=0$ (mod $L$). In this sense, the boundary conditions of the SSD problem can be interpreted as open ones.

A general $N$-particle eigenstate of ${\hat {\cal H}}_{\rm cont}$ is of the form: $$|\Phi \rangle = \int^L_0 dx_1 \cdots \int^L_0 dx_N\, 
{\hat c}^\dagger (x_1) \cdots {\hat c}^\dagger (x_N) |0\rangle 
\Phi (x_1, ..., x_N)\, ,$$ where $|0\rangle$ is the state with no fermions. Since the Hamiltonian is quadratic in ${\hat c}$ and ${\hat c}^\dagger$, the $N$-body wavefunction in the coordinate representation $\Phi (x_1, ..., x_N)$ can be constructed as 