I am reading the calculation of charge conductivity by Non-Equilibrium Green Function (NEGF) method in this following paper.

Van-Nam Do, Non-equilibirum Green function method: theory and application in simulation of nanometer electronic devices, Adv. Nat. Sci.: Nanosci. Nanotechnol. 5 (2014) 033001 (21pp)

According to it, the charge conductivity is computed by the following formula.

$$I=\frac{e_{0}}{\hbar}\int\frac{dE}{2\pi}T(E)[n_L(E)-n_R(E)]$$ Here, $T(E)$ is the transimission coefficient; $n_L(E)$ and $n_R(E)$ are the occupation number of carrier at the energy value $E$ on left and right leads. These occupation number is usually computed by the Fermi-Dirac distribution function.

If the left and right leads are the same materials; then, $n_L(E)$ and $n_R(E)$ should have the same value. This means $[n_L(E)-n_R(E)]$ should be zero and the total charge current would be zero as well.

Taking the Cu-benzene-Cu nanowire as example, there would be no charge current flowing through it, if the current conductivity is computed by this formula.

Obviously, this is not right.

Would anyone please tell me what is wrong with my understanding about this formula? Taking the Cu-benzene-Cu nanowire as example, would anyone please tell me how to compute $n_L(E)$ and $n_R(E)$ to make sure the final charge conductivity of the nanowire is not zero?

Thank you in advance.

  • $\begingroup$ Any chance that n_L and n_R are different because the wire is part of a larger electric circuit, so that more electrons are injected on one site, and drawn away on the other side? Also, usually "I" is used as abbreviation for current instead of conductivity $\endgroup$
    – Wouter
    Apr 27, 2023 at 10:29

1 Answer 1


Conductance vs. conductivity
Let me first point out that for nanodevices we do not speak of conductivity, but of conductance. Conductivity is a local property, which implies averaging over a physically small volume. Such an averaging is impossible/meaningless for nanodevices. Importantly, some basic phenomena, like conductance quantization, could not be discussed with such averaging.

Landauer-Büttiker formalism
The equation given in the OP is groudned in Landauer formalism (often referred to as Landauer-Büttikerr formalism, Büttiker having generalized the formalism to more than two terminals.) The formalism applies to non-interacting particles, so there is no much point in using Green's functions, although there are exist generalizations to interacting case (notably, the approach by Jauho, Meir and Wingreen, based on Keldysh Green's functions.)

Fermi distributions
To drive a current through a nanostructure we need to apply electric field. In case of nanostructures we usually think of applied the electric field as a difference in chemical potentials of the reservoirs feeding and collecting the electrons to/from the structure (this is what Landauer annd Büttiker have taught us). In other words, the distribution functions $n_{L,R}(E)$ are different not only due to the properties of the material, but because the materials held at different chemical potentials: $$ n_{L,R}=\frac{1}{e^{\beta(E-\mu_{L,R})}+1}, \text{ where the potential difference is } \mu_L-\mu_R=eV $$

I suggest several standard references:

The original articles by Landauer, Büttiker, Meir&Wingree, Jauho, Meir&WIngreen are also quite readable and highly recommended.

  • $\begingroup$ Thank you for the explanation and reference. I will refer to them. According to your explanation, I need to set up fermi level value for both left and right leads manually, which could come from the experiment, when I use this formula to compute the conductance. Am I correct? $\endgroup$
    – Kieran
    Apr 27, 2023 at 12:13
  • 1
    $\begingroup$ @Kieran The formula is for the current. If you set, e.g., $\mu_L = \mu + eV/2, \mu_R=\mu-eV/2$ and expand in $eV$ you will arrive to the conductance as defined by $G=I/V$ $\endgroup$
    – Roger V.
    Apr 27, 2023 at 12:17
  • $\begingroup$ Thank you for the explanation. Now, I understand. $\endgroup$
    – Kieran
    Apr 27, 2023 at 12:24

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