Current and conductance from the Landauer formula

Question

The Landauer formula for a one dimensional quantum system (potential step scattering) can be written as
$$
I(V)=frac{2e}{h}int_{-infty}^infty dE T(E) (f_S(E) - f_D(E)),
$$
where $T(E)$ is the transmission probability and $f_i(E)$ is the Fermi function of source $S$ or drain $D$. In Cuevas it is claimed that if the temperature is zero (Fermi functions are potential steps) and if low voltages is assumed, the expression reduces to
$$
I = GV,
$$
where the conductance is given by $G=(2e^2/h)T$.
What is the low voltages assumption? In other words, if I assume low voltages, along with zero temperature, what is left to compute in the integral?

Massimo Ortolano · Answer

In general, the relationship $I(V)$ for arbitrary voltages is nonlinear in the voltage difference $V$, and the assumption of low voltage allows one to write the linear approximation $I approx GV$ by expanding at first order the difference of the Fermi-Dirac functions.
At zero temperature, the relationship $I(V)$ is still nonlinear unless $T(E)$ is a constant independent of $E$.

Current and conductance from the Landauer formula

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