& 0 & \frac{1}{2}\xi i \\

 0 & -\frac{\sqrt{3}}{2}\xi i & 0 & -\frac{1}{2}\xi i & 0 & -\frac{1}{2}\xi & \frac{1}{2}\xi i & 0 & 0 & 0 \\

 -\frac{\sqrt{3}}{2}\xi i & 0 & -\frac{1}{2}\xi i & 0 & \frac{1}{2}\xi & 0 & 0 & -\frac{1}{2}\xi i & 0 & 0
  \end{matrix} \right)
 }
\label{eq:ham_soc}.
\nonumber\end{aligned}$$ The inversion symmetry breaking field terms are given by $$\begin{aligned}
 \mathcal{H}_\mathrm{E}+\mathcal{V}_\mathrm{sf}=  
 {
 \left( \begin{matrix}
 \delta_2 & 0 & 0 & -2i\gamma_2 \sin k_x & -2i\gamma_2 \sin k_y \\

 0 & \delta_3 & 0 & -2i\gamma_3 \sin k_x & 2i\gamma_3 \sin k_y \\

 0 & 0 & \delta_1 & 2i\gamma_1 \sin k_y & 2i\gamma_1 \sin k_x \\

 2i\gamma_2 \sin k_x & 2i\gamma_3 \sin k_x & -2i\gamma_1 \sin k_y & 0 & 0 \\

 2i\gamma_2 \sin k_y & -2i\gamma_3 \sin k_y & -2i\gamma_1 \sin k_x & 0 & 0 \\
 \end{matrix} \right),
 }
\label{eq:ham_efield}
\nonumber\end{aligned}$$ where $$\begin{aligned}
 &\delta_1&=\varepsilon(d_{xy})-\varepsilon(d_{xz/yz}) \\
 &\delta_2&=\varepsilon(d_{z^2})-\varepsilon(d_{xz/yz}) \\
 &\delta_3&=\varepsilon(d_{x^2-y^2})-\varepsilon(d_{xz/yz})\\
 &\gamma_1&=\langle d_{xy}| \mathcal{H}_\mathrm{E} |d_{xz} \rangle_{\hat{y}} 
 =\langle d_{xy}| \mathcal{H}_\mathrm{E} |d_{yz} \rangle_{\hat{x}} \\ 
 &\gamma_2&=\langle d_{xz}| \mathcal{H}_\mathrm{E} |d_{z^2} \rangle_{\hat{x}} 
 =\langle d_{yz}| \mathcal{H}_\mathrm{E} |d_{z^2} \rangle_{\hat{y}} \\
 &\gamma_3&=\langle d_{x^2-y^2}| \mathcal{H}_\mathrm{E} |d_{yz} \rangle_{\hat{y}} 
 =\langle d_{xz}| \mathcal{H}_\mathrm{E} |d_{x^2-y^2} \rangle_{\hat{x}} .\end{aligned}$$

The Hamiltonian near the $\Gamma$ point can be written as $$\begin{aligned}
 