= Solution
A high-mobility <modulation-doped GaAs-AlGaAs electron gas> is grown by <molecular beam epitaxy> as a clean layered <semiconductor heterostructure>. An undoped <gallium arsenide> region is followed by an undoped <aluminium gallium arsenide> spacer and a donor-doped barrier region, commonly with a protective cap. The closely matched crystal structures allow a high-quality <heterojunction>. The spacer supplies a crucial separation: the mobile <electrons> reside away from the ionized donors, reducing impurity scattering compared with putting the donors directly in the conducting layer.
The larger <conduction band> energy on the <aluminium gallium arsenide> side permits donor electrons to transfer into <gallium arsenide>. The resulting charge separation creates <band bending>; together with the <band offset>, it forms a roughly triangular <quantum well> at the interface. The normal-direction wavefunction is confined, while motion parallel to the interface remains approximately free:
$$
E_j(k_x,k_y)=E_j+\frac{\hbar^2(k_x^2+k_y^2)}{2m^*}.
$$
When only the lowest confinement subband is occupied, this is a <two-dimensional electron gas>. It is the quantized normal motion, not a physically zero layer thickness, that makes the low-energy system two-dimensional.
\Image[/past-exam-of-the-physics-of-nanoelectronic-systems-course-of-the-university-of-cambridge/2016/paper-1-heterostructure.png]
{title=Remote donors supply a clean interface channel with quantized normal motion}
At zero <magnetic field>, one occupied spin-degenerate parabolic subband has constant <density of states> per <area> per <energy>, $D_0=m^*/(\pi\hbar^2)$ above its edge. A perpendicular field quantizes the in-plane orbital motion into <Landau levels>:
$$
E_{N,\sigma}=E_0+\hbar\omega_c\left(N+\frac12\right)
+\frac{\sigma g^*\mu_BB}{2},\qquad
\omega_c=\frac{e|B|}{m^*},\quad \sigma=\pm1.
$$
Here $g^*$ is the material's effective spin factor and $e$ is the positive <elementary charge>; no free-electron $g=2$ assumption is made for this semiconductor. Each resolved spin branch contains $e|B|/h$ states per unit <area>. The ideal <density of states> becomes
$$
D(E)=\frac{e|B|}{h}\sum_{N,\sigma}\delta(E-E_{N,\sigma}),
$$
with <Landau-level broadening> in a real sample. The <Landau filling factor>
$$
\boxed{\nu=\frac{n_sh}{e|B|}}
$$
counts the carrier density in units of one spin-resolved level's degeneracy. Two unresolved spin branches have twice that degeneracy; resolved branches must not each be counted twice.
Disorder localizes states in the broadened level tails, while extended states occur near level centres. As $B$ varies, occupation of localized states can change without changing the number of current-carrying edge channels. This produces a finite-width <quantum Hall plateau>, rather than quantization only at isolated fields. In a <mobility gap>, <longitudinal resistivity> approaches zero and the Hall magnitude is $R_K/N=h/(Ne^2)$ for $N$ occupied transmitting branches. When the <chemical potential> passes through extended states, longitudinal dissipation peaks and the Hall value changes to another plateau. A spectral <density of states> can remain nonzero where transport is localized. Very strong disorder can destroy the resolved quantum Hall regime; the statement is not that any amount of disorder improves it.
\Image[/past-exam-of-the-physics-of-nanoelectronic-systems-course-of-the-university-of-cambridge/2016/paper-1-hall-response.png]
{title=Broadened <Landau levels> and an illustrative integer Hall staircase with longitudinal peaks}
The original plot is schematic, not fitted data. At fixed sheet density, increasing $|B|$ decreases <Landau filling factor>, so the Hall magnitude climbs through $R_K/N$ plateaus while longitudinal peaks separate the plateaus. The signed Hall voltage depends on carrier and probe conventions.
For the gate calculation, the target sheet density is fixed by the desired <Landau filling factor>:
$$
n_{\rm target}=\frac{3e(3\,\mathrm T)}h
=2.17619\times10^{15}\,\mathrm{m^{-2}},
\qquad \Delta n_s=-1.30381\times10^{15}\,\mathrm{m^{-2}}.
$$
Use <top-gate density tuning> in the parallel-plate approximation. The geometric <capacitance> per unit <area> is
$$
C_A=\frac{\epsilon_0\epsilon_r}{d}=1.06250\times10^{-3}\,\mathrm{F\,m^{-2}},
\qquad \Delta V_g=\frac{e\Delta n_s}{C_A}.
$$
Therefore
$$
\boxed{\Delta V_g=-0.1966\,\mathrm V\simeq-0.20\,\mathrm V.}
$$
The gate must become more negative relative to the <electron gas>, reducing its density. The voltage is the change from the gate setting associated with the stated initial density; if that setting is defined as zero, it is the required gate voltage. Work-function offsets, trapped charge and <quantum capacitance> are neglected. An absolute voltage referenced to some other uncalibrated potential cannot be inferred from geometry alone. Numerical values use exact SI $h=6.62607015\times10^{-34}\,\mathrm{J\,s}$ and $e=1.602176634\times10^{-19}\,\mathrm C$, and $\epsilon_0\simeq8.8541878188\times10^{-12}\,\mathrm{F\,m^{-1}}$.<Fundamental Physical Constants — Complete Listing by National Institute of Standards and Technology and others>[2022 CODATA listing: Planck constant, elementary charge and vacuum electric permittivity]{ref}
For the subsequently specified five-channel plateau, the <chemical potential> between the fifth and sixth resolved branches means $N=5$, not ten. The physical contact order in the PDF is $1,2,3,4$ along one side of the mesa. Use ideal equilibrating <Ohmic contacts> and the <Landauer-Büttiker formula>, rather than inferring a longitudinal measurement merely from the contacts being drawn on one edge.
Choose one field polarity so the chiral order is $1\to2\to3\to4\to1$. With $G_H=5e^2/h$,
$$
I_i=G_H(V_i-V_{i-1}).
$$
Contacts 2 and 4 are voltage probes, so $I_2=I_4=0$. With $V_3=0$, this gives $V_2=V_1$, $V_4=V_3=0$, and $I_1=G_HV_1$. Hence
$$
\boxed{R_{13,24}\equiv\frac{V_2-V_4}{I_1}
=\frac h{5e^2}=5.163\,\mathrm{k\Omega}}
$$
for this circulation. Reversing the field reverses the contact order and gives $R_{13,24}=-h/(5e^2)$ with the same voltage ordering. \b[The resistance magnitude is $5.16\,\mathrm{k\Omega}$; it is not zero.] Contact 2 samples the outgoing source potential and contact 4 the outgoing drain potential for the chosen chirality.
The <antidot> is a depleted island supporting closed <quantum Hall edge states>. Tunnelling between those states and opposite propagating edges can produce resonant backscattering and resistance peaks as $B$ varies. One electronic <Aharonov-Bohm phase> winds by $2\pi$ when flux through a fixed effective orbit <area> $S$ changes by $h/e$. Using the supplied area as that effective area,
$$
S=(0.2\,\mathrm{\mu m})^2=4.0\times10^{-14}\,\mathrm{m^2},\qquad
\boxed{\Delta B_{h/e}=\frac h{eS}=0.1034\,\mathrm T.}
$$
With the two spin-split branches of the lowest orbital level occupied, the common <Coulomb-dominated antidot oscillation> model has two charge rearrangements per added $h/e$ flux. Both branches contribute to charge buildup; screening and their capacitive coupling can make tunnelling through the outer edge occur twice during the cycle. It then gives
$$
\boxed{\Delta B_{\rm two\ spin}\simeq\frac h{2eS}=51.7\,\mathrm{mT}.}
$$
This is two electron-tunnelling resonances per flux cycle, not evidence that the mobile particles have charge $2e$. The primary charging model specifically addresses this two-spin regime.<Coulomb blockade of tunnelling through compressible rings formed around an antidot by M. Kataoka and others>[Primary abstract: two spins of the lowest Landau level and two outer-spin resonances per h/e period]{ref}
\Image[/past-exam-of-the-physics-of-nanoelectronic-systems-course-of-the-university-of-cambridge/2016/paper-1-edge-and-antidot.png]
{title=Chiral reservoir potentials and the distinction between flux period and two-spin charging resonances}
The sketch shows resistance peaks in a weak-backscattering configuration, with arbitrary height and linewidth because tunnel couplings and <temperature> are unspecified. A two-spin count alone does not force two independent noninteracting resonance series to be exactly half a flux period apart: their relative positions also depend on spin splitting and confinement. Thus $51.7\,\mathrm{mT}$ describes the stated charging-model interpretation; $103.4\,\mathrm{mT}$ is the single-electron fixed-area flux period. The actual enclosed edge-orbit area can differ from the geometrically depleted area.
For bulk filling two and total forward transmission $\mathcal T$ in spin-resolved units, ideal chiral probes give the longitudinal four-terminal resistance
$$
R_{14,23}=\frac h{e^2}\left(\frac1{\mathcal T}-\frac12\right).
$$
Resonant loss of transmission therefore gives peaks. Their size, background and detailed shape cannot be calculated from the supplied area and bulk filling alone. If the gates instead make a tunnelling path that enhances forward transmission, the resonant contrast can be reversed; the plotted peak convention is explicitly the backscattering one.
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