\boldsymbol{a}_{2,1} =a_{\mathrm{cc}}\frac{1}{2}\left(\begin{array}{c} \pm\sqrt{3}\\ 3 \end{array}\right)

\boldsymbol{b}_{2,1} =\left(\begin{array}{c} \pm\frac{2\sqrt{3}}{3a_{\mathrm{cc}}}\pi\\ \frac{2}{3a_{\mathrm{cc}}}\pi \end{array}\right)

\boldsymbol{\delta}_{1}=\left(\begin{array}{c} 0\\ -1 \end{array}\right) \\ \boldsymbol{\delta}_{2}=\frac{1}{2}\left(\begin{array}{c} \sqrt{3}\\ 1 \end{array}\right)\\ \boldsymbol{\delta}_{3}=\frac{1}{2}\left(\begin{array}{c} -\sqrt{3}\\ 1 \end{array}\right) \\

d_{1}=t_{0}\mathrm{e}^{-\beta\left(\frac{|\left(\boldsymbol{\varepsilon}+1\right)\boldsymbol{\delta}_{1}|}{a_{0}}-1\right)} \\ d_{2}=t_{0}\mathrm{e}^{-\beta\left(\frac{|\left(\boldsymbol{\varepsilon}+1\right)\boldsymbol{\delta}_{2}|}{a_{0}}-1\right)} \\ d_{3}=t_{0}\mathrm{e}^{-\beta\left(\frac{|\left(\boldsymbol{\varepsilon}+1\right)\boldsymbol{\delta}_{3}|}{a_{0}}-1\right)}

\boldsymbol{\varepsilon}=\varepsilon\left(\begin{array}{cc} \mathrm{cos}^{2}\theta-\rho\mathrm{sin^{2}\theta} & \left(1+\rho\right)\mathrm{cos}\theta\mathrm{sin}\theta\\ \left(1+\rho\right)\mathrm{cos}\theta\mathrm{sin}\theta & \mathrm{sin^{2}\theta}-\rho\mathrm{cos}^{2}\theta \end{array}\right)

t_0 = 2.7\, \mathrm{eV}

\beta \approx 3

\left|\alpha,\underline{m}\right\rangle =\left|\alpha\right\rangle \left|\underline{m}\right\rangle ,\ \left|A\right\rangle =\left(\begin{array}{c} 1\\ 0 \end{array}\right),\ \left|B\right\rangle =\left(\begin{array}{c} 0\\ 1 \end{array}\right)

\hat{H}_\mathrm{Gr}=\sum_{\underline{m}} d_1\left|B,\underline{m}\right\rangle \left\langle A,\underline{m}\right|+ d_2\left|B,\underline{m}\right\rangle \left\langle A,\underline{m}+\boldsymbol{a}_1\right| + d_3\left|B,\underline{m}\right\rangle \left\langle A,\underline{m}+\boldsymbol{a}_2\right|+\text{h.c.}

\boldsymbol{R}_{\underline{m}}=m_{1}\boldsymbol{a}_{1}+m_{2}\boldsymbol{a}_{2}=\underline{m}\cdot\underline{\boldsymbol{a}}

\hat{H}_\mathrm{KeK} =\Delta\sum_{\underline{M}} d_{2} \left|B,\underline{M}\cdot \underline{\boldsymbol{A}}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right| \\ + d_{3}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right| \\ +d_{1}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right|\\ +d_{1}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right| \\ +d_{2}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}+\boldsymbol{a}_{1}\right|\\ +d_{3}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}+\boldsymbol{a}_{1}\right| +\text{h.c.}

\mathbf{A}_{1}=2\mathbf{a}_{1}-\mathbf{a}_{2}\qquad\mathbf{A}_{2}=-\mathbf{a}_{1}+2\mathbf{a}_{2}.

\hat{H}_\mathrm{SOI}=\frac{im}{\sqrt{3}}\sum_{\underline{M}}\left(d_{4}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}\right\rangle \left\langle B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right|+\right. \\ d_{6}\left|B,\underline{M} \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right\rangle \left\langle B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right|+ \\ d_{5}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right\rangle \left\langle B,\underline{M}\cdot \underline{\boldsymbol{A}}\right|+\\ d_{5}\left|A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}+\boldsymbol{a}_{2}\right|+ \\ d_{4}\left|A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}+\boldsymbol{a}_{2}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right|+ \\ \left.d_{6}\left|A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right|\right)+\text{h.c.}.

\mathbf{A}_{1}=2\mathbf{a}_{1}-\mathbf{a}_{2}\qquad\mathbf{A}_{2}=-\mathbf{a}_{1}+2\mathbf{a}_{2}.

\hat{H}=\hat{H}_\mathrm{Gr}+\hat{H}_\mathrm{Kek}+\hat{H}_\mathrm{SOI}

\hat{H}_\mathrm{Gr}=\sum_{\underline{m}} d_1\left|B,\underline{m}\right\rangle \left\langle A,\underline{m}\right| \\ + d_2\left|B,\underline{m}\right\rangle \left\langle A,\underline{m}+\boldsymbol{a}_1\right| \\ + d_3\left|B,\underline{m}\right\rangle \left\langle A,\underline{m}+\boldsymbol{a}_2\right|+\text{h.c.}

\hat{H}_\mathrm{KeK} =\Delta\sum_{\underline{M}} d_{2} \left|B,\underline{M}\cdot \underline{\boldsymbol{A}}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right| \\ + d_{3}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right| \\ +d_{1}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right|\\ +d_{1}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right| \\ +d_{2}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}+\boldsymbol{a}_{1}\right|\\ +d_{3}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}+\boldsymbol{a}_{1}\right| +\text{h.c.}

\hat{H}_\mathrm{SOI}=\frac{im}{\sqrt{3}}\sum_{\underline{M}}\left(d_{4}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}\right\rangle \left\langle B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right|+\right. \\ d_{6}\left|B,\underline{M} \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right\rangle \left\langle B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right|+ \\ d_{5}\left|B,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right\rangle \left\langle B,\underline{M}\cdot \underline{\boldsymbol{A}}\right|+\\ d_{5}\left|A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}+\boldsymbol{a}_{2}\right|+ \\ d_{4}\left|A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}+\boldsymbol{a}_{2}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right|+ \\ \left.d_{6}\left|A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{2}\right\rangle \left\langle A,\underline{M}\cdot \underline{\boldsymbol{A}}+\boldsymbol{a}_{1}\right|\right)+\text{h.c.}.

\left|B,\underline{m}\right\rangle =\frac{1}{\sqrt{N}}\sum_{\boldsymbol{k}}\mathrm{e}^{i\boldsymbol{k}\boldsymbol{R_{\underline{m}}}}\left|B,\boldsymbol{k}\right\rangle, \\ \left|A,\underline{m}\right\rangle =\frac{1}{\sqrt{N}}\sum_{\boldsymbol{k}}\mathrm{e}^{i\boldsymbol{k}\left(\boldsymbol{R_{\underline{m}}}+\boldsymbol{\delta}_{1}\right)}\left|A,\boldsymbol{k}\right\rangle

\boldsymbol{K}_{\pm}=\frac{2\boldsymbol{b}_{1}+\boldsymbol{b}_{2}}{3},\frac{2\boldsymbol{b}_{2}+\boldsymbol{b}_{1}}{3}\\ \mathbf{G}=\boldsymbol{K}_{+}-\boldsymbol{K}_{-}=\frac{\mathbf{b}_{1}-\mathbf{b}_{2}}{3}

\sum_{\boldsymbol{k},\boldsymbol{k}'}\left|B,\boldsymbol{k}\right\rangle \left\langle A,\boldsymbol{k}'\right|\frac{1}{3}\left(\delta_{\boldsymbol{k}\boldsymbol{k}'}+\delta_{\boldsymbol{k}\boldsymbol{k}'+\boldsymbol{G}}+\delta_{\boldsymbol{k}\boldsymbol{k}'-\boldsymbol{G}}\right)

\hat{H}(\boldsymbol{k})=\underline{\Psi}_{\boldsymbol{k}}^{\dagger}\left(\begin{matrix} \boldsymbol{\Omega} & \boldsymbol{\Gamma} \\ \boldsymbol{\Gamma}^\dagger & -\boldsymbol{\Omega}^\dagger \\ \end{matrix} \right)\underline{\Psi}_{\boldsymbol{k}}

\underline{\Psi}_{\boldsymbol{k}}=\left(\begin{array}{cccccc} \left|B,\boldsymbol{k}\right\rangle & \left|B,\boldsymbol{k}+\boldsymbol{G}\right\rangle & \left|B,\boldsymbol{k}-\boldsymbol{G}\right\rangle & \left|A,\boldsymbol{k}\right\rangle & \left|A,\boldsymbol{k}+\boldsymbol{G}\right\rangle & \left|A,\boldsymbol{k}-\boldsymbol{G}\right\rangle \end{array}\right)

\Psi_\boldsymbol{k}=\left(-\left| A, \boldsymbol{k}+ \boldsymbol{G} \right. \rangle \left|B,\boldsymbol{k}-G\right.\rangle \left|B,\boldsymbol{k}+G\right.\rangle \left|A,\boldsymbol{k}+G\right.\rangle \right)

\hat{H}\left(\boldsymbol{k}\right)=-\frac{3}{2}t\left(1+\frac{2\Delta}{3}\right)\left[\boldsymbol{\sigma}\boldsymbol{A}\otimes\tau_{z}+\boldsymbol{\sigma}\left(\boldsymbol{1}+\left(1-\beta\right)\boldsymbol{\varepsilon}\right)\boldsymbol{k}\otimes\tau_{0}\right]-\\[0.5em] \frac{t\Delta}{2}\left[\left(\beta\mathrm{Sp}\left(\boldsymbol{\varepsilon}\right)-2\right)\sigma_{z}\otimes\tau_{x}+(\boldsymbol{A}\times\boldsymbol{k})_{z}\sigma_{0}\otimes\tau_{y}+\boldsymbol{A}\boldsymbol{k}\sigma_{0}\otimes\tau_{x}\right]+\\[0.5em] m\left[\boldsymbol{A}\boldsymbol{k}\sigma_{z}\otimes\tau_{z}-\left(\frac{1}{2}\beta\mathrm{Tr}\left(\varepsilon\right)-1\right)\sigma_{z}\otimes\tau_{0}\right]-\\[0.5em] m\left[\left(\begin{array}{cc} (\boldsymbol{M}\boldsymbol{k})_{x}-i(\boldsymbol{M}\boldsymbol{k})_{y}\\[0.5em] (\boldsymbol{M}^{*}\boldsymbol{k})_{x}+i(\boldsymbol{M}^{*}\boldsymbol{k})_{y} \end{array}\right)+\left[\left(1+\boldsymbol{\varepsilon}\right)\boldsymbol{k}\right]\boldsymbol{\sigma}\right]\otimes\tau_{x},

\boldsymbol{A}=\frac{\beta}{2a_{0}}\varepsilon\left(1+\rho\right)\left(\begin{matrix} \cos^2\vartheta\\ -\sin^2\vartheta \end{matrix}\right)

\Psi_\boldsymbol{k}=\left(-\left| A, \boldsymbol{k}+ \boldsymbol{G} \right. \rangle \left|B,\boldsymbol{k}-G\right.\rangle \left|B,\boldsymbol{k}+G\right.\rangle \left|A,\boldsymbol{k}+G\right.\rangle \right)