# Non-equilibrium green method based on the Hamiltonian with non-orthogonal basis

I am thinking about one question. If the Hamiltonian matrix is based on the non-orthogonal basis, how to compute the charge conductance with the non-equilibrium green function (NEGF) method.

Suppose the Hamiltonian of the system is denoted by $$H$$ and the overlap matrix is denoted by $$S$$. They are written below. $$\left(\varepsilon_{F}+i\right)\times S-H=\varepsilon^{'}\times S-H=\begin{pmatrix} \varepsilon^{'}\times S_{ll}-H_{ll}&\varepsilon^{'}\times S_{ls}-H_{ls}&0\\ \varepsilon^{'}\times S_{sl}-H_{sl}&\varepsilon^{'}\times S_{ss}-H_{ss}&\varepsilon^{'}\times S_{sr}-H_{sr}\\0&\varepsilon^{'}\times S_{rs}-H_{rs}&\varepsilon^{'}\times S_{rr}-H_{rr} \end{pmatrix}$$ $$=\begin{pmatrix}H^{'}_{ll}&H^{'}_{ls}&0\\H^{'}_{sl}&H^{'}_{ss}&H^{'}_{sr}\\0&H^{'}_{rs}&H^{'}_{rr}\end{pmatrix}$$

Here, $$\varepsilon_{F}$$ is the Fermi energy level of the central scattering zone and $$i$$ is the imaginary number.

$$G_{ll}=\left[H^{'}_{ll}\right]^{-1}$$ and $$G_{rr}=\left[H^{'}_{r}\right]^{-1}$$

$$\Gamma_{left}=H^{'}_{sl}\times G_{ll}\times H^{'}_{ls}$$ and $$\Gamma_{right}=H^{'}_{sr}\times G_{ll}\times H^{'}_{rs}$$

Self energy for the left and right leads and the green function for the scattering zone are written below. $$\Sigma_{left}=i \times \left[\Gamma_{left}-\left(\Gamma_{left}\right)^{+}\right]$$ , $$\Sigma_{right}=i \times \left[\Gamma_{right}-\left(\Gamma_{right}\right)^{+}\right]$$, $$G_{ss}^{R}=\left[H^{'}_{ss}-\Gamma_{left}-\Gamma_{right}\right]^{-1}$$ and $$G_{ss}^{A}=\left(G_{ss}^{R}\right)^{+}$$

Then, the charge conductance is written below.

$$G_{k}=\frac{2e^{2}}{h}Tr\left[\Sigma_{left}G_{ss}^{A}\Sigma_{right}G_{ss}\right]$$

This conductance is only for one single k point in the reciprocal space. The total charge coneuctance is the integral of this single-k-point-conductance over the whole first Brillouin zone.

$$G_{total}=\frac{e}{h}\int G_{k}\times \left[f_{(\varepsilon_{F}-\mu_{left})}-f_{(\varepsilon_{F}-\mu_{right})}\right]d_{k}$$

where,$$f_{(\varepsilon_{F}-\mu_{left})}$$ and $$f_{(\varepsilon_{F}-\mu_{right})}$$ are Fermi-Dirac distribution function for left and right leads with the chemical potential $$\mu_{left}$$ and $$\mu_{right}$$ for the left and right leads.