$m_r + E^1_r(p_{rr'}) \ge 0$ holds. On the other hand, since $p_{rr'}^2 = E^1_{r}(r')^2 -m_r^2$, $E^1_{r}(r') - m_r \ge 0$ must be satisfied in order that $p_{rr'}$ exists. Similarly, since $\varepsilon_{r'{\bf p}}^{-} \le 0$ for the valence band, $m_{r'} - E^2_{r'}(p_{rr'}) \le 0$ holds. On the other hand, since $p_{rr'}^2 = E^2_{r'}(r)^2 - m_{r'}^2$, $E^2_{r'}(r) + m_{r'} \ge 0$ must be satisfied in order that $p_{rr'}$ exists.

Finally, we get a formula of the layered conductivity for $j=1,\cdots,N$, $$\begin{aligned}
 \sigma^N_j(\omega) = g_s g_v \frac{e^2}{\hbar}
 \sum_{r,r'} \frac{m_r m_{r'} + E^1_{r}(r')E^2_{r'}(r)}{\hbar \omega |E^2_{r'}(r) + E^1_{r}(r')|}
 \frac{\sin^2(jk_r)}{N+1}
 \frac{\sin^2(jk_{r'})}{N+1}
 \Theta(E^1_{r}(r')-m_r)\Theta(E^2_{r'}(r)+m_{r'}).
 \label{eq:sigma_j}\end{aligned}$$ For the general $N$, we evaluate Eq. (\[eq:sigma\_j\]) numerically. For a small $N$, we can evaluate Eq. (\[eq:sigma\_j\]) analytically. Particularly for the monolayer $N=1$, this formula reproduces the result $$\begin{aligned}
 \sigma^1_1 = g_s g_v \frac{e^2}{16\hbar}=\pi \alpha \epsilon_0 c,\end{aligned}$$ which is $\omega$-independent. For the bilayer $N=2$, $\sigma^2_1(\omega)$ and $\sigma^2_2(\omega)$ are the same because $\sigma^N_{j}(\omega) = \sigma^N_{N+1-j}(\omega)$ holds, and $$\begin{aligned}
 & \sigma^2_1(\omega) = \sigma^2_2(\omega) = g_s g_v \frac{e^2}{16\hbar} \times \nn \\
 & \frac{1}{4}\left\{ 
 \left[
 1 + \frac{\gamma_1}{\hbar \omega+\gamma_1} \right] + 
 2 \left[
 1+ \left( \frac{\gamma_1}{\hbar \omega}\right)^2 \right]
 \Theta(\hbar \omega-\gamma_1)
 + \left[
 1- \frac{\gamma_1}{\hbar \omega-\gamma_1} \right]
 \Theta(\hbar \omega-2\gamma_1)
 \right\}.\end{aligned}$$ We note that $\sigma^2_1(0)$ is exactly the 