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Take the two-dimensional electron gas in the $xy$ plane, the Hall bar along $x$, and $\mathbf{B}=B\hat{\mathbf{z}}$ with $B>0$ perpendicular to it. An electron has charge $-e$. The confining electron potential energy $U(y)$ is approximately flat inside a wide strip and rises at either boundary; smooth quenched disorder adds hills and valleys. For a slowly varying edge, each spin-resolved Landau level approximately followsHere $s=\pm 1$, and $Y$ is the guiding centre. For $\mathbf{A}=(-By,0,0)$, $Y={\ell }_B^2{k}_x$ and the edge group velocity isThus electrons move along $+x$ at the upper edge and $-x$ at the lower edge. The Lorentz force gives counterclockwise bulk cyclotron motion in this convention. At a hard boundary its interrupted circles become directed skipping orbits; a smooth boundary gives the corresponding guiding-centre drift. Smooth bulk quenched disorder generally gives closed drift contours around localized hills or valleys, with extended trajectories appearing near a broadened level's critical energy. Realistic electrostatic screening can produce alternating compressible quantum Hall edge strips and incompressible quantum Hall edge strips, so the unscreened bending curves are an effective sketch rather than an exact self-consistent edge profile.

In the original diagram, contacts are ordered around the electron propagation direction: upper-left $1\to 2\to 3$, then lower-right $4\to 5\to 6$. Let $I_1=I$, $I_4=-I$ and the other leads be ideal equilibrating voltage probes. Define $R_H=(V_2-V_6)/I$ and $R_{xx}=(V_2-V_3)/I$. These choices fix the voltage sign; reversing the field reverses chirality and the Hall polarity. For spin-resolved plateau index $N$,At fixed sheet carrier density and increasing field, $N$ decreases, giving successively higher Hall plateaus; longitudinal peaks occur between them. Unresolved spin branches can suppress odd plateaus. The sketch uses resistance magnitudes, and every integer step is not promised in an arbitrary sample.

For Fermi-energy oscillations in a two-dimensional electron gas, first distinguish fixed density from fixed reservoir chemical potential. At negligible temperature the former is determined by state counting. Each resolved Landau level contains $eB/h$ states per unit area, soIgnoring spin and broadening for the counting illustration, if $N<\nu <N+1$ then $N$ levels are full and level $N$ is partly full: $\mu =E_N$. At fixed $n_s$, increasing $B$ increases both level energy and degeneracy. The partly filled level rises in energy until the lower levels can contain all electrons; $\mu$ then drops to the next lower level. The resulting sawtooth is periodic in $1/B$, with $\Delta (1/B)=e/(n_{s}h)$ per resolved branch. At an exactly integer filling in an ideal gap, any chemical potential within the gap gives the same zero-temperature particle number; the plotted discontinuity denotes its limiting values.

At fixed $B$ while $n_s$ increases, the ideal chemical potential stays at a partly occupied level and steps upward after each additional density $eB/h$. It therefore has a staircase superposed on the rising zero-field trend, not a repeated downward sawtooth. Landau-level broadening and finite temperature smooth both ideal graphs. With an externally pinned chemical potential, density changes instead; the fixed-density sawtooth cannot simultaneously be imposed. In a realistic disordered sample, the Fermi energy sweeps localized states between transport-carrying level centres. The following schematic omits spin for clarity and does not attempt an interaction-corrected density of states.

For $N$ transmitted spin-resolved quantum Hall edge states, the Landauer-Büttiker formula reduces, in that contact order, toZero probe currents imply $V_2=V_3=V_1$ and $V_5=V_6=V_4$. Hence $I=(N{e}^2/h)(V_1-V_4)$ and the boxed resistance results follow. The spin-resolved conductance quantum arises because the one-dimensional density of states times velocity is $1/h$ per unit energy; a reservoir imbalance supplies $e^2\Delta V/h$ per channel. Its value does not depend on edge slope or sample width.

On a plateau, opposite-moving edge channels are separated by a wide bulk with a mobility gap. Ordinary elastic scattering on one edge has no counterpropagating channel into which to reflect; closed bulk orbits do not connect the contacts. Changing $B$ or $n_s$ can alter the occupancy of localized states without changing the number of transmitting edge channels. This accounts for the finite plateau width, which an entirely clean set of sharp levels would not explain. Accuracy requires negligible backscattering between opposite edges, low enough current to avoid breakdown, equilibrating contacts and negligible probe loading. The finite-width plateau and the precision of its quantized value are related but distinct observations.

An antidot is a depleted potential hill, not a potential well. Its closed edge circulates oppositely to the outer boundary. In a constriction, tunnelling through an antidot orbit can transfer an electron from a forward edge into the returning edge: resonant antidot backscattering produces conductance dips. For $N$ otherwise open spin-resolved channels and one active resonance,At symmetric coupling a single resonance can remove $e^2/h$; asymmetry or thermal averaging reduces its depth. If the background constriction is closed and the resonant route instead supplies forward transmission, peaks occur. The lead connectivity must therefore accompany the conductance sketch. In the fixed-area single-particle limit, successive alignments change enclosed flux by ${\Phi }_0=h/e$, giving the Aharonov-Bohm antidot periodThe effective orbit area, not merely the etched island area, enters. Charging, screening and spin can modify this simple sequence; two coupled spin branches can yield a near-$h/(2e)$ pattern without changing the elementary flux quantum.^[ref]

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: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.

At zero magnetic field, one occupied spin-degenerate parabolic subband has constant density of states per area per energy, $D_0=m^{\ast }/(\pi {\hslash }^2)$ above its edge. A perpendicular field quantizes the in-plane orbital motion into Landau levels:Here $g^{\ast }$ 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 becomeswith Landau-level broadening in a real sample. The Landau filling factorcounts 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 $R_K/N=h/(N{e}^2)$ 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.

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:Use top-gate density tuning in the parallel-plate approximation. The geometric capacitance per unit area isThereforeThe 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 $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $e=1.602176634\times 10^{-19}\mathrm{C}$, and ${ϵ}_0\simeq 8.8541878188\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$.^[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 $1\to 2\to 3\to 4\to 1$. With $G_H=5e^2/h$,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 $I_1=G_H{V}_1$. Hencefor this circulation. Reversing the field reverses the contact order and gives $R_{13,24}=-h/(5e^2)$ with the same voltage ordering. 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,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 givesThis 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.^[ref]

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 resistanceResonant 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.