 Solution

Use the one-way density per unit length $g_{+}/L=2/(h|v|)$. The electron-flux magnitude in an interval is charge times velocity times line density:If $g$ is instead defined for a finite length $L$, the traversal frequency is $|v|/L$; the same cancellation follows. The printed $I=Nev$ therefore requires $N$ to be a line density, not an unscaled total particle number.

Subtract the opposite reservoir's incoming flux. With the stated electron electrochemical potentials, conventional positive current is opposite to electron-number flux, soAt low temperature and small bias, each fully transmitted occupied spin-degenerate subband contributes $2e^2/h$. Hence $\boxed{G=N_{b}2e^2/h}$. More generally, the Landauer formula weights each mode by its transmission probability; imperfect transmission removes exact integer plateaux.