correspond to the free particle case without any constraint at the origin.

Following Schulman procedure one can calculate the fixed-energy amplitude $\hat{G}^{\xi}$ for $\hat{H}$ from $\hat{G}_F^{\xi}$ as follows; $$\label{toy1modi}
\Delta \hat{G}^{\xi}[a, b; E] \equiv
\hat{G}^{\xi}[a, b; E] - \hat{G}_F^{\xi}[a, b; E] = 
\frac{4 \xi^2}{\frac{\sqrt{2E}}{v} - 2 \xi}
\frac{e^{-\sqrt{2E}(|a| + |b|)}}{\sqrt{2E}}.$$ At $\xi = 1$ and $\xi=1/2$ the fixed-energy amplitudes are simply reduced to $$\begin{aligned}
\label{toy1nrs}
\hat{G}^{\xi=1}[a, b; E]&=& \frac{e^{-\sqrt{2E} |a -b|}}{\sqrt{2E}} +
\frac{\frac{\sqrt{2E}}{v} + 2}{\frac{\sqrt{2E}}{v} - 2}
\frac{e^{-\sqrt{2E} (|a| + |b|)}}{\sqrt{2E}}    \\   \nonumber
\hat{G}^{\xi=\frac{1}{2}}[a, b; E]&=& \frac{e^{-\sqrt{2E} |a -b|}}{\sqrt{2E}} +
\frac{e^{-\sqrt{2E} (|a| + |b|)}}{\sqrt{2E} \left(\frac{\sqrt{2E}}{v} - 1\right)
                                   }\end{aligned}$$ and the corresponding bound state energies $B(\xi)$ arising due to the $\delta$-function potential are $$\begin{aligned}
\label{toy1bound1}
B(\xi = 1)&=& -2 v^2   \\    \nonumber
B(\xi = \frac{1}{2})&=& = - \frac{v^2}{2}.\end{aligned}$$

Finally, let us consider $\xi = 0$ case. In this case Eq.(\[toy1modi\]) shows that the modification term $\Delta \hat{G}^{\xi = 0}$ vanishes. This means the $\delta$-function potential in Eq.(\[toy1hamil1\]) does not play any important role. In fact this is obvious if we consider the fact that at $\xi = 0$ the Hamiltonian $\hat{H}^>_0$ describes the free particle system plus $\lim_{\alpha \rightarrow \infty} \alpha \delta(x)$ which makes the half-line constraint. Thus, the $\delta$-function potential 