Hamiltonian can be obtained by letting $g=15/8$, $c = 1/k \equiv R$, $v_1 = v_2 = 3 k /2$ and $z_0 = (e^{k r_c \pi} - 1) / k$. As we did in section 3, however, we will try to examine the fixed energy amplitude for arbitrary parameter as much as possible.

Next we impose $z$ is non-negative. This means we use the single copy of $AdS_5$ as a bulk spacetime. In this sense we have a same setting with that of $AdS$/CFT. In this setting Hamiltonian $\hat{H}_0$ in Eq.(\[rs1hamil1\]) becomes $$\label{rs1hamil2}
\hat{H}_0 = - \frac{1}{2} \partial_x^2 + \frac{g}{x^2}
\hspace{2.0cm}
(c \leq x \leq L)$$ where $L = c + z_0$.

Of course, the main problem is to compute the fixed-energy amplitude $\hat{G}_0[a, b; E]$ for $\hat{H}_0$ in Eq.(\[rs1hamil2\]). From $\hat{G}_0[a, b; E]$ it is simple to derive the fixed energy amplitude for $\hat{H}_{RS1}$ by applying the Schulman procedure twice; $$\label{rs1schul1}
\hat{G}_{RS1}[a, b; E] =
\hat{{\cal G}}_0[a, b; E] - 
\frac{\hat{{\cal G}}_0[a, L; E] \hat{{\cal G}}_0[L, b; E]}
     {\frac{1}{v_2} + \hat{{\cal G}}_0[L, L^-; E]}$$ where $$\label{rs1schul2}
\hat{{\cal G}}_0[a, b; E] =
\hat{G}_0[a, b;E] + 
\frac{\hat{G}_0[a, c; E] \hat{G}_0[c, b; E]}
     {\frac{1}{v_1} - \hat{G}_0[c^+, c; E]}.$$

The fixed-energy amplitude for $\hat{H}_0$ 