$$(A^{(1n;\,\omega lm)}, A^{(1n';\,\omega' l'm')})
= 2\omega \langle \varphi^{(1n)}_{\omega l},
\varphi^{(1n')}_{\omega' l'}\rangle
\delta_{nn'}\delta_{ll'}\delta_{mm'}\,. \label{normal1}$$ Thus, by normalizing $\varphi^{(1n)}_{\omega l}(r)$ as $$\begin{aligned}
\varphi^{(1\rightarrow)}_{\omega l} & \approx &
 e^{i \omega r^*} + 
{\cal R}^{(1\rightarrow)}_{\omega l} e^{-i \omega r^*} 
\,\ \ \ (r^* \to -\infty)\,, \label{norm1}\\
\varphi^{(1\leftarrow)}_{\omega l} & \approx &
e^{-i\omega r^*} + {\cal R}^{(1\leftarrow)}_{\omega l} e^{i\omega r^*}
\,\ \ \ (r^* \to +\infty)\,, \label{norm2}\end{aligned}$$ we have $$(A^{(1n;\,\omega lm)}, A^{(1n';\,\omega' l'm')})
= 4\pi\omega \delta_{nn'}\delta_{ll'}\delta_{mm'}
\delta(\omega - \omega')\,.$$

Finally, we let $$R^{(2n)}_{\omega l}(r) \equiv r^{-(p-2)/2}\varphi^{(2n)}_{\omega l}(r) \,.$$ Then from Eq. (\[R\^2\]), we have $$\left( \omega^2 + \frac{d^2\ }{dr^{* 2}} - V_2(r^*) 
\right)\varphi^{(2n)}_{\omega l}
= 0\,, \label{varphi2}$$ where $$V_2 [r^*(r)]= f\frac{(l+1)(l+p-2)}{r^2} + \frac{(p-2)(p-4)}{4r^2}\frac{f}{h}
+ \frac{(p-2)}{4} \frac{f}{hr}\left(\frac{f'}{f} - \frac{h'}{h}\right)\,.$$ (We again note that $V_2 \to 0$ as $r\to r_+$ in Reissner-Nordström spacetime.) Notice that $V_1 = V_2$ if $p=2$. Hence, in the four-dimensional case $\varphi^{(1n)}_{\omega l}(r)$ and $\varphi^{(2n)}_{\omega l}(r)$ satisfy the same equation. These modes can easily be shown to be orthogonal to the modes previously discussed. By normalizing the functions $\varphi^{(2n)}_{\omega l}(r)$ in the same way as $\varphi^{(1n)}_{\omega l}(r)$ \[see Eqs. (\[norm1\])-(\[norm2\])\], we have $$(A^{(2n;\,\omega lm)}, A^{(2n';\,\omega' l'm')})
= 4\pi\omega \delta_{nn'}\delta_{ll'}\delta_{mm'}
\delta(\omega - \omega')\,.$$

In order to quantize the field $A_\mu$, we impose the equal-time 