in eq.(\[toy1hamil1\]) is absorbed to $\hat{H}^>_0$.

Even in this case, however, one can derive a non-trivial fixed-energy amplitude under the assumption that $v$ is infinite bare coupling constant by adopting the coupling constant renormalization. To show this explicitly we re-express the modification term $\Delta \hat{G}^{\xi = 0}$ as folllows; $$\label{toy1dmodi}
\Delta \hat{G}^{\xi = 0}[a, b; E] = 
\lim_{\epsilon \rightarrow 0^+}
\frac{\hat{G}_F^{\xi = 0}[a, \epsilon; E] \hat{G}_F^{\xi = 0}[\epsilon, b; E]}
     {\frac{1}{v} - \hat{G}_F^{\xi = 0}[\epsilon, \epsilon; E]}.$$ Expanding the denominator and numerator separately one can conclude $$\label{toy1dmodi2}
\Delta \hat{G}^{\xi = 0}[a, b; E] =
\frac{2}{\sqrt{2E} - v^{ren}} e^{-\sqrt{2E}(|a| + |b|)}$$ where the renormalized coupling constant $v^{ren}$ is defined as $$\label{toy1rencop}
v^{ren} = \frac{1}{2 \epsilon^2} \left( 2 \epsilon - \frac{1}{v} \right).$$ It is easy to show that $v^{ren}$ has a same dimension with the bare coupling constant $v$. Following the philosophy of renormalization we regard $v^{ren}$ as a finite quantity. Combining Eq.(\[toy1dn\]) and Eq.(\[toy1dmodi2\]) we get finally $$\label{toy1dfinal}
\hat{G}^{\xi = 0}[a, b; E] = \frac{e^{-\sqrt{2E} |a - b|}}{\sqrt{2E}}
+ \frac{\frac{\sqrt{2E}}{v^{ren}} + 1}
       {\frac{\sqrt{2E}}{v^{ren}} - 1}
\frac{e^{-\sqrt{2E}(|a| + |b|)}}{\sqrt{2E}}$$ whose bound state energy is $B(\xi=0) = - (v^{ren})^2 / 2$.

In the next section we will apply the analysis in this toy 