and $\cos(\theta_{\gamma,\vec{k},s}) << 1$. Therefore, for wavevectors near the band extrema one can make the approximation [@Efimkin13], $$F_{s,s'}(\vec{k},\vec{k}',\vec{q}) = e^{i(\tau\phi_{\vec{k}+\vec{q}} - \tau\phi_{\vec{k}} + \tau'\phi_{\vec{k}'} - \tau'\phi_{\vec{k}'-\vec{q}})/2} \label{eq:F2}$$ Exciton states are approximate eigenstates of the Hamiltonian $H_{o} + H_{eh}$. Assuming that the ground state of the semiconductor is $|\psi_{o}\rangle$, which consists of a filled valence band and an empty conduction band, an exciton state with in-plane momentum $\vec{Q}$ can be constructed from the ground state as follows [@Changjian14; @Efimkin13], $$|\psi_{s,s',\alpha}(\vec{Q})\rangle =   \frac{1}{\sqrt{A}}\sum_{\vec{k}} \psi_{\alpha,\vec{Q}}(\vec{k}) c_{\vec{k} + \frac{m_{e}}{m_{ex}}\vec{Q},s}^{\dagger} b_{\vec{k}-\frac{m_{h}}{m_{ex}}\vec{Q},s'}|\psi_{o} \rangle  \label{eq:exciton}$$ The exciton wavefunction is $\psi_{\alpha,\vec{Q}}(\vec{k})$. The electron and hole effective masses are $m_{e}$ and $m_{h}$, respectively. The exciton mass is $m_{ex} = m_{e} + m_{h}$, and the reduced electron-hole mass is $m_{r}$. If one writes the exciton wavefunction as, $$\psi_{\alpha,\vec{Q}}(\vec{k}) = \phi_{\alpha}(\vec{k}) e^{i(\tau\phi_{\vec{k}+ (m_{e}/m_{ex})\vec{Q}} + \tau'\phi_{\vec{k}'-(m_{h}/m_{ex})\vec{Q}})/2} \label{eq:psiphi}$$ then the exciton wavefunction $\phi_{\alpha}(\vec{k})$ satisfies the standard exciton eigenvalue equation [@Changjian14; @timothy], $$\begin{aligned}
& & \left[\bar{E}_{c}(\vec{k}) - \bar{E}_{v}(\vec{k}) \right]\phi_{\alpha}(\vec{k}) - \frac{1}{A} \sum_{\vec{q}} V(\vec{q}) \phi_{\alpha}(\vec{k}-\vec{q}) \nonumber \\
& & = E_{\alpha}(\vec{Q}) \phi_{\alpha}(\vec{k}) \label{eq:eigen}\end{aligned}$$ with an eigenvalue $E_{\alpha}(\vec{Q})$ given by, $$E_{\alpha}(\vec{Q}) = E_{g} - E_{\alpha} + \frac{\hbar^{2} Q^{2}}{2 m_{ex}}$$ where, $E_{\alpha}$ is the exciton binding energy. The energy $E_{\alpha}(\vec{Q})$ is measured with respect to the energy 