= Solution
A <GaAs>–AlGaAs <semiconductor heterostructure> confines electrons normal to the interface in a <semiconductor quantum well>. Remote <modulation doping> supplies a high-mobility <two-dimensional electron gas>. Negative split gates deplete it laterally, leaving a narrow <quantum point contact> with discrete transverse subbands and continuous longitudinal momentum. A single occupied subband requires transverse excitation gaps larger than thermal and applied-bias energies.
Take $e>0$ and count \b[one spin-resolved channel]. In a length $L$, right-moving states have density $L\,dk/(2\pi)$; each contributes charge current $-ev_k/L$. Thus
$$
I^+=-\frac e{2\pi}\int_0^\infty v_k f_L\,dk,\qquad I^-_{\rm signed}=+\frac e{2\pi}\int_0^\infty v_k f_R\,dk.
$$
The second expression is the right reservoir's left-moving contribution expressed in the same left-to-right conventional-current sign. Since $v_k\,dk=dE/\hbar$,
$$
I=\frac eh\int_{E_0}^\infty[f(E;\mu,T)-f(E;\mu-eV,T)]\,dE,
\qquad\boxed{G=\frac{e^2}{h}f(E_0;\mu,T)\longrightarrow\frac{e^2}{h}.}
$$
The limit needs $k_BT,|eV|\ll\mu-E_0$ and <ballistic transport>. Velocity cancels the one-dimensional state density. Two degenerate spin channels give $2e^2/h$, not an extra factor hidden in the quoted single-channel result.
For elastic phase-coherent scattering, expand each incident flux-normalized mode in outgoing lead modes. Its transmitted fraction is $|(t_E)_{nm}|^2$; sum over all open outgoing $n$ and occupied incident $m$. Linearizing the reservoir occupations gives the <Landauer formula>
$$
\boxed{G=\frac{e^2}{h}\int dE\,[-f'(E)]\operatorname{tr}(t_E^\dagger t_E).}
$$
The source's $T_E^+$ denotes this amplitude matrix, not temperature or a probability matrix. Flux normalization is essential: ordinary wave-amplitude ratios need a velocity factor. Reservoirs equilibrate, contacts supply incoming Fermi distributions, and inelastic processes within the channel are neglected. Spin is included among channel labels. At $T\to0$, only transmissions at $\mu$ contribute; perfect channels each add one quantum. This is a two-terminal conductance including contact resistance, even with no internal backscattering.
Trace I has steps near integer $2e^2/h$ as less-negative gate voltage opens successive spin-degenerate subbands. At the printed parallel field $11\,\mathrm T$, <Zeeman splitting> separates the two spin thresholds, allowing approximately half-sized steps in the plotted units; the onset also shifts. Thermal broadening, imperfect transmission and unresolved thresholds round the plateaux. A parallel field primarily resolves spin in the ideal thin-interface model; finite layer thickness can also shift subband energies. The plot alone does not determine a unique effective $g$ factor or microscopic disorder model.
\Image[/past-exam-of-the-advanced-quantum-condensed-matter-physics-course-of-the-university-of-cambridge/2014/paper-1-aqc2014-channel-steps.png]
{title=Illustrative spin-resolved channel thresholds produce full or half-sized conductance steps}
{source=paper-1-aqc2014-channel-steps.py}
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