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.


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.

Would anyone please confirm whether my understanding is correct or not? If not, would anyone please help me correct it?

Thank you in advance.

  • $\begingroup$ Would anyone please give me some suggestions on my enquiry or recommend any reference paper, which discusses the calculation with non-orthogonal basis Hamiltonian through the NEGF method? Thank you in advance. $\endgroup$
    – Kieran
    May 1, 2023 at 9:39


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