phase factor, where $\tilde\omega^2 = 
\omega^2-V_0[r^*(r_H)] \geq 0$ and ${\cal R}^{(0\rightarrow)}_{\omega l}$ and ${\cal R}^{(0\leftarrow)}_{\omega l}$ are constants (with $|{\cal R}^{(0\rightarrow)}_{\omega l}|
=
|{\cal R}^{(0\leftarrow)}_{\omega l}|$), we have $$\langle \varphi^{(0\leftarrow)}_{\omega l},
\varphi^{(0\leftarrow)}_{\omega' l}\rangle = 
\langle \varphi^{(0\rightarrow)}_{\omega l},
\varphi^{(0\rightarrow)}_{\omega' l}\rangle
= 2\pi \delta(\omega-\omega') \,.$$ Now, by using Eq. (\[explicit\]) we find $$\begin{aligned}
(A^{(0n;\,\omega lm)}, A^{(0n';\,\omega' l'm')})
 & = & - 2\omega \langle \varphi^{(0n)}_{\omega l},\varphi^{(0n')}_{\omega' l'}
 \rangle
\delta_{nn'}\delta_{ll'}\delta_{mm'}
\nonumber \\
& = & 
- 4\pi \omega \delta_{nn'}\delta_{ll'}\delta_{mm'}\delta(\omega -\omega')\,.\end{aligned}$$ where we have used $
\delta(\tilde\omega -\tilde\omega') 
= 
(\tilde\omega/\omega)\delta(\omega -\omega')\,.
$ Noting that $A^{(0n;\,\omega lm)}$ and $A^{(3n;\,\omega lm)}$ have the same $t$-components, we immediately obtain from Eqs. (\[explicit\]) and (\[A3A3\]) that $$(A^{(0n;\,\omega lm)}, A^{(3n';\,\omega' l'm')})
= - 4\pi \omega \delta_{nn'}\delta_{ll'}\delta_{mm'}\delta(\omega -\omega')\,.$$

Next let us examine the physical modes. By letting $$R^{(1n)}_{\omega l}(r) \equiv
\frac{\sqrt{l(l+p-1)}}{\omega}\, (fh)^{1/2} r^{-p/2-1} 
\varphi^{(1n)}_{\omega l}(r)\,,
\label{R1}$$ we find from Eq. (\[R\^1\]) $$\left( \omega^2 + \frac{d^2\ }{dr^{* 2}} - 
V_1(r^*) \right)\varphi^{(1n)}_{\omega l}
= 0\,,  \label{varphi1}$$ where $$V_1 [r^*(r)]= f\frac{l(l+p-1)}{r^2} + \frac{p(p-2)}{4r^2}\frac{f}{h}
- \frac{(p-2)}{4} \frac{f}{h r} \left(\frac{f'}{f} - \frac{h'}{h}\right)\,.$$ (We note that $V_1 \to 0$ as $r\to r_+$ for Reissner-Nordström spacetime unlike $V_0$.) We first note that the modes $A^{(1n;\,\omega lm)}$ are orthogonal to both the non-physical and pure-gauge modes. Next we note 