Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2019 / Paper 2 / Section B / 21 / iv

For longitudinal motion in transverse subband $i$,For a completely free particle take $E_i=0$ and use the full wave vector. In a semiconductor the parabolic-band mass is the appropriate effective mass rather than necessarily the bare electron mass.

For a wire of length $L$, one spin state has $Ldk/(2\pi )$ states in one propagation direction. Including two degenerate spin states and using $dϵ/dk=\hslash v$ givesDivide by $L$ for density per unit length. The factor-of-two distinction is essential: the current from one reservoir uses only its incoming propagation branch. If spin degeneracy is lifted, count each resolved spin channel separately.

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.