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The integral field equation. The Ampere-Maxwell law is $\nabla \times \mathbf{H}=\mathbf{J}+{\partial }_t\mathbf{D}$. Apply Stokes theorem to any fixed oriented surface bounded by the chosen contour:The added displacement current is required for compatibility with electric charge conservation. Taking a divergence and using $\nabla \cdot \mathbf{D}={\rho }_f$ gives $\nabla \cdot \mathbf{J}+{\partial }_t{\rho }_f=0$.

Field geometry. Choose the upper plate positive and let $z$ increase upward. In the long-plate, slow-discharge approximation, $\mathbf{E}=-(V/d)\hat{\mathbf{z}}$. Conventional wire current runs downward from the positive plate. The magnetic field intensity circulates clockwise as viewed from above. The Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{H}$ points radially toward the resistive wire.

Wire and displacement currents. The capacitance and discharge law areBecause $V$ decreases, ${\partial }_t\mathbf{D}$ points upward, opposing the downward wire current. Through a disk of radius $r$ between the plates its magnitude isThe effective downward current enclosed by the circular contour is therefore $I-I_D(r)$. The Ampere-Maxwell law givesThis is the magnitude; with the stated coordinates, $\mathbf{H}=-H\hat{\mathbf{\varphi }}$. The factor $1-r^2/a^2$ expresses cancellation between conduction and displacement currents. The approximation requires distances outside the thin wire and away from detailed electrode-edge fields.

Energy flux and power. Since $\hat{\mathbf{z}}\times \hat{\mathbf{\varphi }}=-\hat{\mathbf{r}}$,Its units are watts per square metre. Integrating over the cylinder's side area $2\pi rd$, the signed outward power is negative:There is no leading axial flux through the cylinder's ends in this ideal field geometry.

Where the energy goes. The Poynting flux in a centrally discharged parallel-plate capacitor carries field energy through the dielectric space into the resistor, where it becomes Joule heating at rate $V^2/R$. Inside radius $r$ the stored electric energy isHenceThe resistor receives the inward flux from outside that cylinder plus the field energy released inside it. As $r$ approaches the wire radius, essentially all its power arrives as radial electromagnetic flux. Over the whole capacitor, $-d(C{V}^2/2)/dt=V^2/R$. The Poynting theorem thus accounts for the complete discharge without treating energy as transported only along the conducting wire. High resistance makes the quasistatic electric-field approximation and neglect of changing magnetic energy consistent.

A transmission line must be treated as a distributed circuit when its propagation time is significant on the signal timescale. Let $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$ denote resistance, inductance, leakage conductance and capacitance per unit length. Applying Kirchhoff's voltage law and Kirchhoff's current law to a short length $dx$ gives the telegrapher's equationsFor a lossless line, $R^{\prime }=G^{\prime }=0$. Differentiating again gives wave equations withThe travelling solutions can be written $V=V_{+}(t-x/v)+V_{-}(t+x/v)$ and $I=[V_{+}(t-x/v)-V_{-}(t+x/v)]/Z_0$. The minus sign for the backward wave is essential to power transport.

At a load $Z_L$, continuity of voltage and current gives the voltage reflection coefficientA matched termination $Z_L=Z_0$ absorbs the incident signal without reflection. An open circuit has ${\Gamma }_L=+1$ and a short circuit has ${\Gamma }_L=-1$. Superposed incident and reflected waves produce standing waves. A sinusoidal lossless line of length $\ell$ transforms the termination intofor the $e^{i\omega t}$ convention. At a quarter wavelength, $Z_{\mathrm{in}}=Z_0^2/Z_L$, allowing impedance transformation. A coaxial line supports a transverse electromagnetic mode whose ideal $L^{\prime },C^{\prime }$ are calculated in A4. Its energy travels in the fields between conductors; the characteristic impedance is not the wire's DC resistance. Losses introduce attenuation and make the characteristic impedance and propagation constant complex.

Consider a long uniform length $\ell$ of a coaxial cylindrical capacitor, with permittivity ${ϵ}_0{ϵ}_r$ and inner-cylinder charge per unit length $\lambda$. A cylindrical Gaussian surface givesThe electric potential difference isThus its capacitance per unit length is $c({ϵ}_r)=2\pi {ϵ}_0{ϵ}_r/\ln (b/a)$.

In the partially filled coaxial capacitor, the vacuum section of length $L-x$ and the dielectric section of length $x$ share the same two conductors and hence the same electric potential difference. They are electrically in parallel, not in series. The radial electric field is tangent to the axial dielectric interface, consistent with the boundary condition in part (b). Adding their charges givesThis is the leading long-capacitor expression, neglecting end and interface-fringing corrections. The stated $x\gg b$ supports a long filled section; treating the other section as uniform also requires it to be well away from the end regions when those corrections matter.

Let increasing overlap $x$ mean that the dielectric moves farther into the capacitor. Its capacitance derivative isAt fixed electric potential difference $V$, the capacitor energy rises by $d{U}_{\mathrm{field}}=V^{2}dC/2$, but the voltage source supplies electrical work $VdQ=V^{2}dC$. The difference is available as mechanical work $F_{x}dx$:The dielectric insertion force at fixed voltage is thereforeFor the usual $ϵ>1$, it pulls the dielectric inward, increasing $x$; its magnitude is constant within the long-overlap approximation. In the original longitudinal diagram this is to the left. The units are newtons, since $V^{2}dC/dx$ is energy per length. Differentiating the field energy alone with a minus sign at fixed $V$ would miss the source contribution; the correct fixed-voltage capacitor energy balance is essential.

Use $Q=CV$ for the magnitude of charge on either plate. The fully filled capacitance is $C_f=2\pi {ϵ}_0ϵL/\ln (b/a)$, while the initial value is given in part (c). At the same fixed electric potential difference,For $ϵ>1$ and positive $V$, this is an increase in the positive plate's electric charge magnitude, supplied by the voltage source; the negative plate acquires an equal opposite change. The net free charge of the two-plate capacitor remains zero. Thus “charge on the capacitor” has its usual circuit meaning, not the algebraic sum over both plates.

The electric field is approximately radial in the long interior, but near an electrode end its lines spread into a fringing field of a capacitor. The electric field magnitude falls on moving outward through that end region. The dielectric's induced electric dipoles experience a net force toward the stronger field, pulling the material into the capacitor. For a weakly perturbing polarizable element this is expressed by $\mathbf{F}=\frac{1}{2}\alpha \nabla E^2$, with positive electric polarizability $\alpha$.

In the sketch the field lines run from the positive electrode to the negative one, curving outward near their ends. The material is attracted toward the interior where the field is stronger. A perfectly uniform interior field by itself would not give a longitudinal dipole-gradient force; the end region is essential to the local explanation. Neglecting its small contribution to the total capacitance does not justify omitting the finite insertion force that it transmits. The energy argument of part (d) gives that force without needing the detailed fringe solution.

For charge sharing between distant connected spheres, use the standard approximation that the spheres are sufficiently far apart for mutual electrostatic effects to be negligible, and that the wire adds negligible capacitance. The capacitance of an isolated conducting sphere is $C=4\pi {ϵ}_{0}r$. ThusThe initial charges are $Q_i=C_i{V}_i$. Connecting the spheres allows charge to flow until their potentials are equal. Charge conservation givesThe initial and final electrostatic energies are $U_i=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2$ and $U_f=\frac{1}{2}(C_1+C_2)V_f^2$. Their difference becomes Joule heating in the resistive wire:Numerically,The resistance determines how fast this happens, not the final energy difference. Sufficient damping suppresses sustained charge-current oscillations; the calculation also neglects energy radiated away. At finite sphere separation the capacitance matrix would be needed, and the radii and initial potentials alone would not specify an exact correction.

The quoted capacitance is that of the filled parallel-plate capacitor, $C_i=1.0\times 10^{-6}\mathrm{F}$. Disconnecting the source imposes charge conservation on the isolated plates:On removing the dielectric, the capacitance falls by its relative permittivity, so $C_f=C_i/5=2.0\times 10^{-7}\mathrm{F}$. The capacitor energy at fixed electric charge is $U=Q^2/(2C)$. ThereforeandThe increase is supplied by the external work needed to withdraw the dielectric against its attraction into the electric field, in the lossless quasistatic limit. The final electric potential difference is $Q/C_f=12500\mathrm{V}$, not the original charging value.

For sinusoidal steady state with convention $e^{i\omega t}$, complex impedance is the voltage phasor divided by the current phasor. Its modulus is the amplitude ratio and its argument is the voltage-current phase difference. Since $V_L=LdI/dt$,The original circuit is a series RLC circuit with output across the inductor. Its transfer function isFor the original capacitance, the outputs are $10.15\mathrm{mV}$ at $9\mathrm{MHz}$ and $4.65\mathrm{mV}$ at $10\mathrm{MHz}$, a ratio $0.458$. The current resonance is $f_0=1/(2\pi \sqrt{LC})=8.967\mathrm{MHz}$, so the circuit favors the first transmitter.

The resistor voltage is in phase with the current; the inductor voltage leads it by $\pi /2$ and the capacitor voltage lags by $\pi /2$. Large reactive voltages therefore nearly cancel in their phasor sum. A large inductor voltage is consistent with a small supply voltage: it is inductor voltage magnification, not power gain in a passive circuit.

Reducing the capacitance raises $f_0$ to $9.452\mathrm{MHz}$, between the two transmitters. Direct evaluation gives $7.02\mathrm{mV}$ and $7.22\mathrm{mV}$, respectively. Their similar detunings give approximately equal amplitudes; they are not exactly equal, partly because the numerator of $H$ also depends on frequency.

The two-tone output of a linear resonant circuit, for coherent monochromatic sources of equal input amplitude and arbitrary fixed input phases, isWith $A\simeq B=V_o$, this is approximatelyThus a $9.5\mathrm{MHz}$ rapid oscillation has a beat envelope repeating in magnitude at $1\mathrm{MHz}$. The signed envelope cosine has frequency $0.5\mathrm{MHz}$. Unequal amplitudes leave a small nonzero envelope minimum; independently drifting transmitter phases make the beating drift. No new frequency is generated by this ideal linear circuit.

Complex impedance. In sinusoidal steady state write physical voltage and electric current as real parts of $\tilde{V}{e}^{i\omega t}$ and $\tilde{I}{e}^{i\omega t}$. The electrical impedance $Z=\tilde{V}/\tilde{I}$ is a ratio of phasors; its magnitude gives the voltage-to-current amplitude ratio and its argument gives their phase difference. It is measured in ohms.

For an ideal capacitor, $q=Cv$ and $i=dq/dt=Cdv/dt$. Consequently $\tilde{I}=i\omega C\tilde{V}$, givingWith this time convention the voltage lags the electric current by $\pi /2$.

Bridge output. The original diagram has $R_1$ and $C_1$ in parallel in the upper-left arm, and $R_2$ and $C_2$ in series in the lower-left arm. The point labelled $C$ is connected to the upper supply node. Set the lower supply node to zero, so $V_C=V_i$. Assuming the measured output draws negligible electric current, the two voltage dividers giveTheir difference is thereforeThe reactive-arm electrical impedances are

Balance. Setting $V_{AB}=0$ and cross-multiplying gives $R_3{Z}_2=R_4{Z}_1$. For this Wien bridge,The right-hand side must be the real number $R_4/R_3$. Equating imaginary and real parts givesPositive capacitances require $R_4/R_3>R_2/R_1$. In the equal-component case $R_1=R_2=R$ and $C_1=C_2=C$, this reduces to

Band-pass output and phase. Preserve the requested voltage polarity: $V_{AC}=V_A-V_C$. Since $V_C=V_i$,Put $u=\omega RC>0$. The equal-component ratio is $Z_2/Z_1=2+i(u-u^{-1})$, so the Wien bridge band-pass response isAt finite positive frequency, its imaginary part vanishes exactly when $u-u^{-1}=0$. ThusThe magnitude $|H|=[9+(u-u^{-1}{)}^2{]}^{-1/2}$ is maximal there and tends to zero at both frequency extremes, as expected for a band-pass filter.

A convenient continuous, unwrapped wave phase isIt decreases from $-\pi /2$ as $\omega \to 0^{+}$, passes through $-\pi$ at ${\omega }_r$, and tends to $-3\pi /2$ as $\omega \to \infty$. A principal-value plot instead jumps from $-\pi$ to $+\pi$ at resonance; both represent the same phasor. Reversing the output to $V_{CA}$ would shift the phase by $\pi$ and make the resonant ratio $+1/3$, but that is a different output polarity from the one printed.

Let $C_T=C+C_1$ be the total parallel capacitance, since capacitors in a parallel circuit have additive capacitances. Put ${\omega }_{\ast }=2\pi (50\mathrm{Hz})$ into the lossy parallel coil resonance condition:The additional capacitance is thereforeThis is the added capacitance, rather than the new total, and the conversion from frequency to angular frequency is essential.

For a long solenoid, solenoid inductance obeys $L={\mu }_0{N}^{2}A/\ell$, where $N$ is the number of turns, $A$ the cross-sectional area and $\ell$ the axial winding length. The natural reading of half as many turns occupying half the axial length givesWith unchanged wire material and gauge, halving the wire length also halves its electrical resistance: $R^{\prime }=R/2$. With the total capacitance $C_T$ fixed, the new angular frequency satisfiesThusNeglecting the small resistive correction gives the estimate $\sqrt{2}$.

There is a documented alternative interpretation: if half the turns are distributed over the original axial length, then $L^{\prime }=L/4$ while $R^{\prime }=R/2$, givingThis corresponds to treating “length of the winding” as wire length while preserving axial geometry. The examiner's report explicitly accepted either inductance scaling, provided the resistance change was included. The first interpretation follows the statement that the axial winding length is halved; the second records the accepted ambiguity rather than silently choosing a different geometry.^[ref]

In the quasisteady diffusion approximation, the vapour-density disturbance outside a small stationary particle satisfies Laplace's equation, with a specified equilibrium value on its surface and a uniform far-field value. This has the same mathematical boundary-value problem as the electrostatic potential outside a conductor held at fixed potential. The integrated normal gradient is proportional to the conductor's geometrical capacitance.

Here $C$ is a length: a sphere of radius $r$ has $C=r$. The corresponding electrical capacitance in SI units would be $4\pi {ϵ}_{0}C$, not $C$ farads. Replacing the spherical diffusive conductance $4\pi r$ by $4\pi C$ accounts for the crystal's non-spherical geometry. Coupling the analogous heat-conduction problem to vapour diffusion supplies the ice-specific thermal and diffusional resistances.

The approximation assumes continuum diffusion, equilibration of the surface vapour concentration, an approximately isothermal particle, a slowly changing shape, weak perturbations and a far field sufficiently distant relative to the crystal size. It neglects ventilation by falling motion, interactions with neighbouring particles and facet-dependent attachment kinetics. Shape affects diffusive growth through capacitance, not merely through projected collecting area as in a collision calculation.

graphene is a two-dimensional honeycomb lattice with two sublattices. Nearest-neighbour hopping gives two bands $E_{\pm }(\mathbf{k})=\pm t|{\sum }_{j=1}^3{e}^{i\mathbf{k}\cdot {\mathbf{\delta }}_j}|$. They touch at two inequivalent valleys. Expanding near valley $\tau =\pm 1$ givesThe two-component spinor describes sublattice amplitudes, not real spin. This is a massless two-dimensional Dirac Hamiltonian with material velocity $v_F$ in place of $c$. The emergent band quasiparticles are relativistic in this mathematical sense; the electron's vacuum rest mass is not set to zero. Real spin and valley each contribute a degeneracy of two in the ideal low-energy model.

A single-walled carbon nanotube is a rolled graphene sheet with carbon nanotube chiral vector ${\mathbf{C}}_h=n{\mathbf{a}}_1+m{\mathbf{a}}_2$. Periodicity requires $\mathbf{k}\cdot {\mathbf{C}}_h=2\pi \ell$, selecting lines through graphene momentum space and producing one-dimensional transverse electronic modes. Near a valley,A quantization line through one of the Dirac valleys in graphene gives a metallic branch; ideal nanotube zone folding gives the criterion $n-m\equiv 0(\mathrm{mod}3)$. Otherwise the nearest line misses the valley and opens $E_g=2\hslash v_F|q_{\perp }|$. The transverse momentum gives a carbon-nanotube Dirac mass, with $m_{\mathrm{eff}}=\hslash |q_{\perp }|/v_F$. For a large-radius ideal semiconducting tube, $|q_{\perp }|\simeq 2/(3d)$ and $E_g\simeq 4\hslash v_F/(3d)$, showing the inverse-diameter gap. Curvature, strain, spin-orbit coupling and interactions can change the simple classification; nominally metallic non-armchair tubes can acquire small gaps. One-dimensional transverse electronic mode thresholds give Van Hove singularities, unlike graphene's low-energy density of states proportional to $|E|$.

A gate changes occupation and the Fermi level in both systems. Ideal monolayer graphene can be tuned through charge neutrality from holes to electrons, but an ordinary scalar gate does not open a bulk band gap. Including spin and valley,The gate relation includes geometric and quantum capacitances, so one should not equate gate voltage to Fermi energy without the capacitance model. A semiconducting carbon nanotube can instead be depleted when its Fermi level lies in a gap and turned on in electron or hole bands. Contact Schottky barriers can strongly influence the observed nanotube field effect. A truly gapless ideal metallic tube cannot be switched off simply by moving the Fermi level through its band crossing. These gate responses reflect band structure and contacts, not a change of wrapping indices.

Integer Hall transport and filling. A clean two-dimensional electron gas at low temperature in a strong perpendicular magnetic field forms Landau levels with orbital spacing $\hslash {\omega }_c$, where ${\omega }_c=e|B|/m^{\ast }$. Examples include semiconductor heterojunctions and inversion layers. On integer plateaus,with integer plateau index $\nu$. The density of states in one resolved Landau branch is $e|B|/h$ per unit area, so the actual filling factor isIt varies continuously with density and field. Within a disorder-broadened plateau, localized states can change occupation without changing the integer number of transporting edge branches; the filling need not equal the plateau integer at every point in that interval.

At a boundary, the confining potential bends the Landau energies through the Fermi energy, giving chiral edge states, interpreted semiclassically as skipping orbits. Opposite edges carry opposite propagation directions. An ideal spin-resolved transmitted edge channel contributes $e^2/h$; $\nu$ such channels give $I=\nu (e^2/h)(V_S-V_D)$. Chirality and suppressed backscattering between well-separated edges explain vanishing longitudinal voltage. Ideal reservoirs and appropriate probe equilibration are part of this edge-state transport model.

A gated antidot. A local repulsive gate potential depletes an island inside the Hall bar. Edge states circulate around its boundary as closed antidot orbits. When a localized orbit is weakly coupled to extended channels on opposite sides, resonance at the reservoir electrochemical potential permits backscattering and produces conductance dips on an otherwise quantized transmitting plateau. If the path instead adds a transmission route between otherwise disconnected channels, resonant peaks can occur: the sign depends on the device's tunnel connectivity and what conductance is measured. An isolated antidot too far from every conducting edge has negligible transport signature.

For the simplest opposite-edge backscattering picture with one active branch,where $f$ branches would otherwise transmit. A gate moves the resonance through the Fermi energy by changing both confinement and orbit area. Stronger depletion of the constrictions also changes $f$, producing broad conductance plateaus or pinch-off, with narrow resonances superposed where coupling is finite. Low temperature resolves individual resonances; thermal and tunnel broadening can merge them.

For a fixed-area single-electron orbit, a phase change of $2\pi$ corresponds toSweeping gate voltage at fixed field shifts enclosed area and level energy, giving oscillatory resonance tracks. This period is not automatically set by the lithographic gate area. Charging energy, self-consistent screening and spin branches modify spacings; in particular coupled spin branches in a charging model can generate two resonances per $h/e$ cycle without changing the electron flux quantum.^[ref]

No antidot dimensions, capacitances, lever arms or measured spectrum are supplied, so absolute periods and resonance voltages cannot be extracted. The sketch illustrates the mechanisms and regimes requested, rather than inventing a numerical device data set.

Improving the bare edge potential. A picture of noninteracting Landau levels simply intersecting a prescribed smooth confinement ignores the redistribution and screening of electron charge. A self-consistent electrostatic calculation combines gate and donor potentials, electron density and Coulomb interactions. Where a partially occupied Landau level has a large density response, electrons screen strongly and create a compressible strip with nearly pinned level energy. Regions with the chemical potential in a gap are incompressible and support much of the potential change. Consequently the screened edge alternates between compressible and incompressible strips, rather than preserving an arbitrary externally drawn slope. Finite temperature, disorder, exchange and edge reconstruction can require refinements beyond the simplest electrostatic model.^[ref]

Fractional states and composite fermions. Very clean two-dimensional systems show plateaus at selected fractional fillings, with small longitudinal resistance and a many-electron excitation gap. Electron-electron correlations are essential; partial filling of a noninteracting Landau level is not enough. The simplest composite fermion binds two statistical flux quanta to each electron. This is a representation of correlated electron motion, not a physical solenoid attached to a particle. With $B>0$, ${\Phi }_0=h/e$, and density $n_s$, the mean effective field isDefine a signed effective filling ${\nu }^{\ast }=n_s{\Phi }_0/B^{\ast }$. ThenFor ${\nu }^{\ast }=p>0$, filled composite-fermion levels produce $\nu =p/(2p+1)=1/3,2/5,3/7,\dots$. For reversed effective field write ${\nu }^{\ast }=-p$, giving $\nu =p/(2p-1)=1,2/3,3/5,\dots$; the nontrivial fractional upper sequence starts with $p=2$. If effective filling is defined as nonnegative using $|B^{\ast }|$, retain these as separate field-direction branches instead of using the signed expression blindly. At $\nu =1/2$ the mean $B^{\ast }$ vanishes and the effective filling diverges, so it is not an integer composite-fermion Hall plateau in this simplest construction. Attaching two flux quanta preserves fermionic statistics; the emergent integer-like gaps organize a prominent sequence of fractional electronic states.^[ref]

A gated quantum dot is an electron island confined in all spatial directions, weakly coupled by tunnel barriers to source and drain and tuned by a plunger gate. Small capacitance makes charge addition costly, while confinement resolves orbital levels. In the constant-interaction model,The charging contribution to successive addition potentials is $e^2/C_{\Sigma }$; the often used charging-energy scale $E_C=e^2/(2C_{\Sigma })$ differs by a factor of two. Orbital spacing and spin filling add further structure.

For weak coupling, $k_{B}T$ and lifetime broadening must be smaller than the relevant addition energies to see Coulomb blockade. Sequential tunnelling proceeds when an addition level is inside the source-drain bias window and the appropriate charge states are accessible. A gate sweep gives conductance peaks near charge degeneracies and low conductance between them. A bias-versus-gate map shows Coulomb diamonds, with excited-state lines when additional levels enter the window. Their slopes and sizes depend on source, drain and gate capacitances and the bias convention. Cotunnelling gives higher-order transport inside nominally blocked regions; stronger coupling and many-body effects can invalidate the simple sequential picture.

Coherent dot states as a qubit. A single-electron spin has basis $|0⟩=|\uparrow ⟩$, $|1⟩=|\downarrow ⟩$ in a magnetic field, split by $E_Z=|g^{\ast }|{\mu }_{B}B$. A qubit is a controllable coherent state $a|\uparrow ⟩+b|\downarrow ⟩$, not just a pair of classical current levels. Initialize through spin-selective loading or relaxation; resonant magnetic or spin-orbit-assisted electrical driving produces rotations. Adjacent occupied dots can couple through a gate-controlled exchange interaction. In the weak-tunnel symmetric Hubbard limit,Local spin rotations combined with a suitable exchange entangling operation provide universal control in the ideal spin-qubit proposal. Gate pulses must avoid unwanted orbital excitation and leakage, and complete coherent operations before dephasing destroys the information.^[ref]

For measurement, place the reservoir electrochemical potential between Zeeman levels so that only one spin state can tunnel out, or use Pauli spin blockade in a double dot. A nearby quantum point contact or single-electron transistor detects the resulting charge change. Such spin-to-charge conversion makes transport useful for reading a spin even when the stored information is not itself a net current.^[ref]

A double dot can alternatively encode a charge qubit in left/right occupation, with Hamiltonian $H=(\epsilon {\sigma }_z+2t_c{\sigma }_x)/2$. It allows electrical control and charge sensing, but couples strongly to charge noise. Spin qubits also suffer hyperfine, electrical and spin-orbit-related noise, as well as relaxation and readout errors. An observed Coulomb diamond alone establishes charging physics, not initialization, coherent control or a functioning qubit.

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.

An electrostatically defined quantum dot uses depletion gates to confine a small island laterally in a two-dimensional electron gas. Two narrow barriers allow weak tunnelling to source and drain; a plunger gate tunes the island's electrochemical levels, and additional gates can form a double quantum dot. Adjusting the gates resolves successive charge states and eventually reaches the single-electron regime. An electrical dot is defined by confinement and tunnel coupling, not simply by the drawn metal area.

In a constant interaction model,Here $Q_g$ is gate-induced charge and $C_{\Sigma }$ total island capacitance. Successive additions cost $E_{\mathrm{add}}=e^2/C_{\Sigma }+{ϵ}_{N+1}-{ϵ}_N$, with spin, exchange and correlations modifying the orbital term. Coulomb blockade appears when no allowed charge transition lies in the source–drain energy window. Gate sweeps give charge steps and transport peaks; finite-bias Coulomb diamonds constrain charging and orbital energies under a stated bias convention. Resolved single-electron operation requires thermal and tunnel broadening smaller than the relevant addition energies.

Quantum-dot charge sensing avoids relying on a measurable current through a nearly isolated dot. A nearby quantum point contact biased on a sensitive conductance slope, or a single-electron transistor, responds capacitively when one electron enters or leaves the dot. A time trace of sensor current can therefore distinguish $N$ from $N-1$ and resolve individual tunnelling events. The sensor measures charge; a spin-dependent step is needed to measure spin.

For one electron, Zeeman splitting gives a lower and a higher spin level separated by $E_Z=|g^{\ast }|{\mu }_B|B|$. In energy-selective spin readout, initialize an empty dot and pulse both levels below the reservoir chemical potential to load one electron. Then move to a readout configurationAn electron in the higher state can tunnel out, temporarily leaving the dot empty, after which a lower-state electron can refill it. An electron initially in the lower state remains trapped. The charge sensor records a pulse for the first outcome and no escape pulse for the second. This is spin-to-charge conversion; it measures the spin projection along the field through a state-dependent tunnelling probability. A primary single-shot experiment uses this conversion with a quantum-point-contact charge detector.^[ref]

The readout window must be long enough to detect escape but short compared with excited-spin relaxation:Detector bandwidth must also resolve the empty-state dwell time before refilling. Thermal activation, relaxation before escape, missed fast events and sensor backaction cause errors. Since the excited electron can leave and be replaced, this simple sequence does not preserve an arbitrary initial spin state.

For coupled dots, label charge occupation by $(n_L,n_R)$. Gate-controlled detuning can make $(1,1)$ and $(0,2)$ accessible. Two electrons in the lowest orbital of one dot must form a singlet state; a triplet state requires a different orbital or another higher-energy configuration. Choose a measurement window where the $(0,2)$ singlet is available but the $(0,2)$ triplet is not. Thenunder spin-conserving tunnelling. The second process is forbidden by spin symmetry and the Pauli exclusion principle, producing Pauli spin blockade. Charge sensing distinguishes transfer from a blocked $(1,1)$ state. This measures singlet versus triplet content; without a reference spin or additional control it is not a measurement of each electron's absolute spin direction separately.

The exchange interaction in a double quantum dot provides control of these states. In an effective spin description,For a symmetric weak-tunnelling two-site model with onsite charging cost $U$, $J\simeq 4t_c^2/U$. Gates change tunnel coupling $t_c$ and detuning, allowing preparation and evolution followed by a charge-conversion measurement. Spin mixing from local magnetic-field differences or spin-orbit effects can lift blockade, so the sensing interval must be chosen before unwanted relaxation obscures the initial state. Single-dot readout uses spin-dependent escape; double-dot readout can use spin-dependent charge transfer. In both cases the measurable signal is a resolved charge change.

Dopant atoms and carrier type. Substitutional semiconductor doping places a small fraction of foreign atoms on host lattice sites. In a tetrahedrally bonded group-IV semiconductor, a group-V substitution such as phosphorus in silicon has an extra valence electron. A shallow donor level just below the conduction edge can release that electron, leaving a positively charged donor. In an n-type semiconductor, mobile electrons are the majority carriers and the chemical potential shifts toward the conduction band.

A group-III substitution such as boron in silicon has one fewer valence electron. An acceptor level just above the valence edge captures an electron from the valence states, leaving a mobile electron hole and a negatively charged acceptor. In a p-type semiconductor, electron holes are the majority carriers and the chemical potential shifts toward the valence band. The material can remain electrically neutral in both cases: mobile carrier electric charge is compensated by fixed ionized impurity electric charge. Dopant freeze-out, compensation and degeneracy qualify the simplest one-carrier-per-dopant picture.

Making the junction. Choose p material at $x<0$ and n material at $x>0$. Electrons initially diffuse from n to p, and electron holes from p to n, recombining near the interface. Negative ionized acceptors are exposed on the p side and positive ionized donors on the n side. They produce an electric field directed from n to p. The field opposes further majority-carrier diffusion until drift and diffusion balance and the equilibrium electrochemical potential is uniform.

The depletion regions are depleted of mobile majority carriers, not of the fixed dopant charges that create the space charge. At equilibrium the conduction and valence edges bend together, while the chemical potential is flat. The specified $\Delta {\varphi }_0$ is a electric potential difference: the corresponding initial chemical-potential energy separation is $e\Delta {\varphi }_0$. This convention is required by the quoted units and width formulas. Write ${\epsilon }_s=ϵ{ϵ}_0$ for the absolute permittivity; the paper's $ϵ$ is relative permittivity.

Poisson equation and widths. Let the depleted intervals be $-w_p<x<0$ and $0<x<w_n$. Fully ionized number densities $N_A,N_D$ correspond to electric charge densities $-e{N}_A,+e{N}_D$, respectively. ThusThe factor $e$ converts a dopant concentration to a space-charge density. Requiring zero field in both neutral regions gives charge neutrality,With $\mathcal{E}=-{\varphi }^{\prime }$, direct integration gives the triangular fieldContinuity at the interface is the same neutrality condition. The field is negative throughout the depleted interval. Set $\varphi (-w_p)=0$. Its integral givesPotential continuity therefore requiresCombining it with neutrality yields the abrupt-junction depletion width and capacitance resultsThe more lightly doped side depletes farther, because equal and opposite space charges must be exposed.