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A single gate-defined dot can trap one electron whose lower and upper Zeeman states define $|0⟩,|1⟩$. Initialize by energy-selective loading or relaxation, coherently rotate the spin with a resonant magnetic drive or electrically generated spin-dependent drive, and use neighbouring-dot exchange interaction for two-qubit operations. The dot supplies confinement; a full qubit also requires control, coherence and calibrated readout. A nearby quantum point contact or single-electron transistor detects charge capacitively, so spin-to-charge conversion is needed.

In energy-selective spin readout, set the reservoir chemical potential between the two spin-dependent removal energies. The upper state can tunnel out; the lower remains. Reloading can create a temporary empty-dot signal that a time-resolved charge detector records. Resolving this window requires sufficiently small electron-temperature broadening and stable level alignment. The escape should be faster than spin relaxation, but the empty interval must remain long enough for detection. Useful controls include known-state preparation, sensor-noise measurement, varied waiting time and readout pulses, and checks for missed events or thermally activated false escape.^[ref]

tunnel-rate-selective spin readout instead places both candidate removal energies on the allowed side of the reservoir and uses different escape rates ${\Gamma }_f\gg {\Gamma }_s$. After a window $\tau$ with ${\Gamma }_f^{-1}\ll \tau \ll {\Gamma }_s^{-1}$, the fast state has usually changed charge while the slow state has not. It does not require the reservoir Fermi edge to resolve the state splitting, although it still requires controlled charge occupation, substantial rate contrast and relaxation slower than the mapping. The contrast must have a physical origin. A demonstrated single-dot implementation uses two-electron singlet and triplet states: Pauli symmetry puts the triplet's second electron in a higher orbital with stronger lead coupling. Nearly degenerate states can then be distinguished by rate rather than by their energy difference.^[ref]

For a bare one-electron Zeeman pair in the same orbital with an unpolarized lead, different tunnel rates are not automatic. A spin-selective barrier/reservoir or another spin-dependent mechanism is needed. The two-electron singlet/triplet example is a different state encoding/readout arrangement, not evidence that the bare single-electron pair already has useful rate contrast.

For the optimal tunnel-rate readout time, assume no relaxation and equal prior probabilities. The fast-state missed-event probability is $e^{-{\Gamma }_f\tau }$ and the slow-state false-event probability is $1-e^{-{\Gamma }_s\tau }$. Their average is minimized atFinite detector bandwidth, preparation errors, state relaxation and unwanted reloading alter this optimum. energy selection is direct for a resolved single-electron spin but sensitive to the narrow alignment window; rate selection has a wider energy operating window but requires engineered rate contrast and a carefully timed charge measurement. Neither method is an ideal nondestructive spin measurement merely because the charge sensor itself is noninvasive.

A trajectory-resolving ballistic experiment. Use a high-mobility modulation-doped GaAs-AlGaAs heterostructure with a two-dimensional electron gas. Grow the layers by molecular-beam epitaxy, separate the donor layer from the conducting interface with an undoped spacer, and pattern an isolated mesa with ohmic reservoir contacts. Electron-beam lithography defines metal split gates; negative gate voltage depletes the underlying electron gas, leaving narrow adjustable injector and collector quantum point contacts along an edge. Keep the relevant path shorter than the momentum-relaxation mean free path, cool to suppress phonon scattering and thermal smearing, and use a small excitation to avoid heating.

Inject a low-frequency current through one contact to a distant drain and measure the collector's open-circuit voltage with a high-input-impedance amplifier. Sweep perpendicular magnetic field and gate settings. The cyclotron radius is $r_c=\hslash k_F/(e|B|)$, with $k_F=\sqrt{2\pi n_s}$ for a spin-degenerate single-valley gas. For contact separation $L$, semicircular or skipping trajectories focus into the collector whenThe appropriate field sign bends carriers toward the collector; the opposite sign suppresses the peaks. Density-dependent focusing positions, suppression with increased scattering and reproducible gate control distinguish directional ballistic propagation from an ordinary diffusive voltage. Quantum interference may add fine structure, but the basic focusing signal does not require phase coherence over every path. Split-gate injection and collection in a two-dimensional gas were directly demonstrated in the original focusing experiment.^[ref]

Multi-probe current bookkeeping. For one directed spin-resolved mode, velocity times one-dimensional density of states gives a particle flux $dE/h$. A small reservoir electrochemical-energy shift ${\mu }_j$ therefore supplies excess flux ${\mu }_j/h$. Let $N_i$ be the number of propagating modes at lead $i$, $R_{ii}$ their summed return probability, and $T_{ij}$ the total transmission from reservoir $j$ into lead $i$. Multiplying outgoing-minus-incoming electron flux by charge $-e$ gives the source's convention,The reservoirs equilibrate arriving carriers; an ideal voltage probe adjusts its chemical potential until its net current vanishes. Here ${\mu }_i$ is an energy, not a voltage. Choose the lead-voltage convention ${\mu }_i=-e{V}_i$ and positive conventional current injected into the junction. The working equation is thenConservation of scattering probability and gauge invariance require both column and row balance: every outgoing channel is accounted for and equal voltages give zero current. The counts $M,N$ below include every occupied internal channel; an extra factor of two must not be inserted if spin is already counted. If one instead counts orbital channels with unresolved spin degeneracy, double all transmission counts consistently.

Zero-field routing. The original diagram labels lead 1 at the left, 2 above, 3 at the right and 4 below. With columns denoting injectors and rows collectors, straight-through ballistic routing and the stipulated excess-wide-channel rule giveThere is no reflection in this ideal routing model. Set $V_4=0$, $I_2=I_3=0$, and $I_1=I=-I_4$. The two probe equations giveThusThe derivation assumes $N>M>0$. At $N=M$, the ideal network splits into two disconnected opposite-lead pairs and cannot carry the imposed 1-to-4 current at finite voltage; the divergence diagnoses failure of that limiting transport configuration, not an infinite dissipative resistance of each channel. Negative bend resistance is a nonlocal voltage sign, compatible with positive total dissipated power.

Positive field. At $+B_0$, left-deflected incoming carriers go $1\to 2$, $2\to 3$, $3\to 4$. Lead 4 supplies $M$ modes to narrow lead 1; the remaining $N-M$ enter wide lead 3. The balanced transmission matrix isTherefore $V_2=V_1$, $V_3=M{V}_1/N$ and $I=g_{0}M{V}_1$, givingNegative field. Reversing the chirality gives ${\mathsf{T}}_{-}={\mathsf{T}}_{+}^{\mathsf{T}}$, as expected from reciprocity for this spin-independent routing model. With $V_4=0$, the zero-current conditions become $V_3=V_4=0$ and $V_2=V_3=0$. ThusThis does not contradict reciprocity: the general relation interchanges the current and voltage lead pairs when reversing field; it does not require the same fixed-contact four-terminal resistance to be even in $B$.

Comparison with the measured trace. The PDF shows a negative region around zero field, approximately zero signal at sufficiently negative field, and a positive rising resistance near $+0.4\mathrm{T}$ and above. These signs match the three routing limits. Near zero field the depth is roughly a few tenths of a kilohm; the graph does not supply $M,N$, so there is no justified exact amplitude fit. Rounded corners, collimation and finite cyclotron radius continuously change which outgoing lead is reached between the ideal limits. Skipping trajectories and geometrical focusing can produce shoulders or resonances. At larger positive fields magnetic depopulation and spin resolution can change the mode counts, increasing the positive resistance or producing structure; disorder, incomplete equilibration and finite temperature round ideal features. These are candidate mechanisms, not uniquely established assignments of every small feature from one trace.

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]

Transverse confinement in a short clean channel gives subband dispersionsA gate varies the subband thresholds and channel width. If elastic backscattering is negligible and spin is degenerate, the two-terminal linear conductance isA fully open subband has $T_n\simeq 1$, producing conductance steps in units $2e^2/h$. Magnetic spin splitting can resolve $e^2/h$ steps. A finite two-terminal resistance remains in a perfect ballistic channel because reservoirs supply and absorb a finite number of transport modes; it does not prove momentum-randomizing scattering inside the constriction.

At finite source-drain bias the two reservoir distributions differ over an energy window rather than at one Fermi energy. Choose ${\mu }_S-{\mu }_D=e{V}_{\mathrm{sd}}$ and take positive current from source to drain. For spin-degenerate channels,At zero temperature with a rigid transmission spectrum, $I=(2e/h){\sum }_n{\int }_{{\mu }_D}^{{\mu }_S}{T}_n(E)dE$. For symmetric bias ${\mu }_{S,D}=E_F\pm e{V}_{\mathrm{sd}}/2$ and neglecting an explicit bias response of $T_n$,If the two window edges lie above different numbers of subbands, the differential conductance can lie at half-integer values in units $2e^2/h$. For example, two open subbands at one edge and one at the other give $3e^2/h$. This is a bias-window effect, not fractional charge or a fractional Hall state.

Mapping transconductance versus gate voltage and source-drain bias reveals boundaries where a reservoir chemical potential crosses a subband bottom, enabling energy-spacing spectroscopy under a specified lever arm and bias division. Finite temperature rounds thresholds over $k_{B}T$; reflection, disorder, series resistance and a saddle-shaped barrier reduce or smooth plateaus. Large bias can alter the electrostatic potential, heat carriers or produce inelastic scattering, so the rigid-transmission differential formula is not generally valid without its assumptions. One-dimensional interaction effects can add anomalies beyond this independent-channel model.

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.

The Clapeyron equation gives the slope of a two-phase phase coexistence curve. Let $s_l,s_v$ and $v_l,v_v$ be entropy and volume per molecule, or consistently per mole. Equality of the two chemical potentials and $d\mu =-sdT+vdP$ implyAt reversible coexistence the latent heat in the same amount convention is $\mathcal{L}=T(s_v-s_l)$. ThereforeThe relation assumes thermodynamic equilibrium between two bulk phases of a pure substance, neglecting curvature and other work terms. For vaporization both latent heat and volume change are positive, giving a positive vapor-pressure slope. If the vapor is approximately ideal and its volume dominates, the ideal-vapor Clausius-Clapeyron approximation gives $d\ln P/dT\simeq \mathcal{L}/(k_B{T}^2)$ with latent heat per molecule.

Experimentally, a sealed two-phase sample in a temperature-controlled bath supplies the coexistence pressure as temperature changes. Independent calorimetry and phase-density measurements supply $\mathcal{L}$ and $\Delta v$, allowing the pressure slope to test the equation. The formula connects those distinct measurements and is a practical route to latent heats.

Entropy is a thermodynamic state function. For a reversible heat transfer, $dS=\delta Q_{\mathrm{rev}}/T$; the Clausius inequality generalizes this to irreversible processes. The combined entropy of a system and its surroundings does not decrease. A subsystem can lose entropy by exporting heat or matter, so the second-law statement must specify the whole isolated combination. In a simple compressible system,which identifies temperature, pressure and chemical potential as derivatives of the energy representation of the state function. Thermodynamic equilibrium maximizes entropy under fixed total energy, volume and particle number. A heat reservoir instead leads to minimizing the Helmholtz free energy at fixed $T,V,N$, or the Gibbs free energy at fixed $T,P,N$.

Statistical mechanics explains the state count. In a microcanonical ensemble, equally probable compatible microstates give Boltzmann's entropy formula, $S=k_B\ln \Omega$. With general discrete probabilities, the Gibbs entropy isThe microcanonical formula follows by setting every nonzero probability to $1/\Omega$. For a canonical ensemble, $p_j=e^{-\beta E_j}/Z$, soThis turns microscopic energy levels and their populations into a measurable thermodynamic quantity. Classical phase-space counting needs a consistent microscopic normalization and indistinguishability convention; quantum state counting supplies that normalization. Independent systems have multiplicative state counts and therefore additive entropies.

For an ideal gas freely expanding from volume $V$ to $2V$ in an insulated enclosure, energy remains fixed while each particle gains twice the accessible position volume. The state count increases by $2^N$, giving $\boxed{\Delta S=N{k}_B\ln 2}$. The actual irreversible process transfers no heat. A reversible isothermal comparison path transfers heat $N{k}_{B}T\ln 2$ and computes the same entropy change because the initial and final thermodynamic equilibrium states are the same.

Calorimetry supplies another connection to experiment: at fixed pressure and within one phase, $S(T)-S(0)={\int }_0^T{C}_P(T^{\prime })d{T}^{\prime }/T^{\prime }$, with latent-heat contributions $\mathcal{L}/T$ at first-order transitions. The third law of thermodynamics fixes $S(0)=0$ for an ideal perfect crystal with a unique ground state. Degeneracy or frozen disorder can produce residual entropy. Entropy thus links heat-capacity measurements, phase thermodynamic equilibrium and the number of accessible microscopic configurations.

For a bulk liquid and vapor at common fixed temperature and pressure, transfer $d{N}_l$ molecules from vapor to liquid while holding their sum fixed. The Gibbs free energy change is $dG=({\mu }_l-{\mu }_v)d{N}_l$. Thermodynamic equilibrium with both phases present requires stationarity under either sign of transfer, hence $\boxed{{\mu }_l={\mu }_v}$. This is the bulk coexistence condition; an interface's curvature adds a surface-energy cost.

In the capillarity approximation, take an incompressible drop with constant liquid number density $n_l$, bulk chemical potential ${\mu }_l$ at the specified reservoir conditions, and size-independent surface tension $\Gamma$. The liquid bulk is extensive, so its Gibbs thermodynamic free energy is $N_l{\mu }_l$. Creating interface area reversibly at fixed temperature adds $\Gamma A$. Equivalently, along growth at fixed bulk parameters the stated differential integrates as $d{G}_d={\mu }_{l}d{N}_l+\Gamma dA$, with the no-drop reference set to zero. For a sphere,This approximation uses bulk properties for the interior and neglects the molecular interface thickness, curvature dependence of surface tension and higher compressibility corrections. It does not apply quantitatively to drops only a few molecules across.

The vapor contains $N-N_l$ molecules. Treat its chemical potential as approximately unchanged by forming the drop, as for a large surrounding reservoir or negligible vapor depletion. ThenThe radius is limited by the available molecules in a finite sample; the constant-reservoir expression is a local nucleation model, not a description of arbitrarily complete vapor depletion. The excess over the vapor reference has derivativeIf ${\mu }_l>{\mu }_v$, it increases for every $R>0$, leaving only the stable boundary minimum $R=0$. If ${\mu }_l<{\mu }_v$, $R=0$ is a metastable local minimum and the additional stationary radius isAt that radius $\Delta G^{\prime \prime }=-8\pi \Gamma <0$, so it is an unstable maximum. The classical nucleation theory barrier is $16\pi {\Gamma }^3/[3n_l^2({\mu }_v-{\mu }_l{)}^2]$.

A fluctuation below $R_c$ tends to shrink because reducing its surface cost lowers thermodynamic free energy. A supercritical drop tends to grow, because its bulk free-energy gain outweighs the added surface cost. Growth continues while material supply and supersaturation allow it, eventually changing the vapor state or leading to macroscopic liquid. Curved-interface pressure and chemical thermodynamic equilibrium can also be described through the Young–Laplace equation and the Kelvin equation.

Semiconductor doping introduces electrically active impurities. In gallium arsenide, silicon has one more valence electron than gallium but one fewer than arsenic. Silicon on a gallium site can therefore supply a shallow donor electron, while silicon on an arsenic site can accept an electron and leave a mobile hole. It is an amphoteric dopant, so its net effect depends on site occupation and growth conditions.^[ref]

For isotropic parabolic bands, shift the energy origin in the given free-electron density of states to the appropriate band edge and use its effective mass:and, for valence electron states,The stated normalization includes the two spin states and assumes one effective valley. In a nondegenerate semiconductor, the conduction occupation is approximately $e^{-(E-\mu )/(k_{B}T)}$ and the probability of a valence hole is approximately $e^{(E-\mu )/(k_{B}T)}$. Substituting $x=(E-{ϵ}_c)/(k_{B}T)$ in the first integral, and the corresponding hole energy in the second, giveswhere the effective density of states values are $N_c=2[m_e{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$ and $N_v=2[m_h{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. The factor $\sqrt{\pi }/2$ comes from the supplied Gaussian-weighted integral. Multiplying cancels the shared thermodynamic equilibrium chemical potential and proves the semiconductor mass-action law:For the given sample $N_D=2\times 10^{21}{\mathrm{m}}^{-3}$ and $N_A=10^{21}{\mathrm{m}}^{-3}$. With complete donor ionization and occupied acceptors, charge neutrality gives $n-p=N_D-N_A=10^{21}{\mathrm{m}}^{-3}$. At $293\mathrm{K}$ the given masses and gap giveSolving the compensated semiconductor equations, or using $n_i\ll N_D-N_A$, yieldsThe very small minority density is consistent with the wide gap and the strongly n-type balance. Also $n/N_c\simeq 2.6\times 10^{-3}\ll 1$ and $p/N_v\ll 1$, checking the nondegenerate approximation. The acceptor levels are occupied in this n-type regime; complete acceptor ionization in the compensation equation is appropriate to the model.

At low temperature dopant freeze-out reduces the conduction-electron population; a quantitative curve would require dopant binding energies and occupation factors. Once donors are ionized while intrinsic excitation is negligible, $n\simeq N_D-N_A$ gives an extrinsic plateau. At higher temperature intrinsic pairs eventually dominate, so $n\simeq p\simeq n_i(T)\propto T^{3/2}{e}^{-E_g/(2k_{B}T)}$ if the gap and masses are treated as constant. Real gap and mass changes modify that last expression.

In a conductor, mobile electrons redistribute in response to an electric potential, producing an induced charge density that opposes its source. Finite electronic compressibility means a finite density change requires a change of local kinetic Fermi energy. Screening is therefore not perfect at every microscopic distance. On scales comparable to atomic structure or the Fermi wavelength, a continuum local-density model itself fails.

Thomas-Fermi screening treats electrons as a locally uniform degenerate free electron gas, replaces the ionic background by a smooth neutralizing charge, and neglects nonlocal response, exchange and correlation corrections. The electrochemical chemical potential is uniform in equilibrium. Because an electron has charge $-e$, its electrostatic energy is $-eV$, givingAt zero potential $\mu =E_{F0}$. Linearizing at fixed $\mu$, with $|eV|\ll E_{F0}$, givesInsert this in Poisson's equation to obtain $({\nabla }^2-{\xi }^{-2})V=-{\rho }_{\mathrm{free}}/{ϵ}_0$, where the Thomas-Fermi screening length isAway from a point source, direct radial differentiation shows ${\nabla }^2(e^{-r/\xi }/r)={\xi }^{-2}{e}^{-r/\xi }/r$. The decaying solution is therefore $V=C{e}^{-r/\xi }/r$. Integrating Poisson's equation over a small sphere containing charge $Q$ fixes $C=Q/(4\pi {ϵ}_0)$. For a sufficiently large source, the linearized formula is only a far-field approximation, not a justified description of its singular core.

For monovalent sodium, the electron number density is its atomic number density,This gives $E_{F0}=3.15\mathrm{eV}$ andRoom temperature is far below the corresponding Fermi temperature, so degeneracy is well justified, and sodium's simple conduction band makes a free-electron estimate useful. However $k_F\xi =0.615$ and the mean atomic spacing is $n_0^{-1/3}=0.340\mathrm{nm}$. The potential varies on a scale shorter than $k_F^{-1}$ and the atomic spacing, so the slowly varying local-density and continuum assumptions are not quantitatively controlled at the screening length itself. A nonlocal treatment includes Friedel oscillations, absent from the pure exponential.

Nearly unchanged interlayer force constants in two-layer and twenty-layer films are consistent with very local screening and bonding forces: reducing the distant bulk need not strongly change the force between adjacent layers. They also indicate that metallic electronic response can survive in the thin film. The observation does not by itself validate the numerical Thomas-Fermi approximation, establish an identical full phonon spectrum, or rule out surface-induced electronic changes; it directly constrains the measured local force constant.

A positive electric potential difference between gate and substrate attracts electrons towards the insulating interface and repels the mobile electron holes of the p-type semiconductor substrate. The exposed, negatively charged acceptor impurities form a depletion region; the thin mobile electron layer at its surface is the semiconductor inversion layer. The conduction band and valence band bend downwards towards the positively biased gate, because the electron potential energy is $-e\psi$.

To specify the conventional MOS inversion threshold, let $N_A$ be the ionized acceptor density, $n_i$ the intrinsic carrier concentration, andIn the neutral bulk the intrinsic level lies $e{\varphi }_F$ above the semiconductor chemical potential. A positive surface potential ${\psi }_s$ lowers both band edges by $e{\psi }_s$, giving, in the nondegenerate approximation,At ${\psi }_s={\varphi }_F$, the surface is intrinsic: $n_s=p_s=n_i$. The usual onset of strong inversion instead meansThus the surface electron density has grown to the bulk majority-carrier density. The chemical potential need not actually enter the conduction band. The phrase “just forms” can also refer to the earlier weak-inversion crossover; the diagram uses the conventional strong-inversion threshold and labels it explicitly.

Within the depletion approximation the negative space charge is approximately $-e{N}_A$ over a widthwhere ${ϵ}_s$ is the absolute semiconductor permittivity. Its outer edge is where the bands become flat and mobile electron holes recover their bulk density. The electrical insulator blocks steady carrier transport while sustaining an electric field; the metal gate supplies positive surface charge balancing depletion and inversion charge. Under gate bias, the gate and substrate electrochemical potentials are different, so a single equilibrium Fermi level should not be drawn through the whole biased stack.

A solar cell uses the photovoltaic effect to turn photon energy into electrical work. In a semiconductor, above-gap photon absorption generates an electron in the conduction band and an electron hole in the valence band. A p-n junction, together with selective electrical contacts, collects the two carrier species at opposite terminals. Carrier diffusion and the built-in electric field help separate them before electron-hole recombination. Illumination maintains different quasi-Fermi levels for electrons and electron holes; their splitting, divided by $e$, sets the available terminal voltage in an ideal device. The equilibrium built-in potential alone cannot continuously deliver work.

For an ideal junction, the illuminated output current density isHere $J_{\mathrm{sc}}$ is the photogenerated short-circuit current density and $J_0$ the diode saturation current. The useful operating point maximizes $VJ$, rather than either current or voltage separately. With solar-cell fill factor $\mathrm{FF}=P_{\max }/(V_{\mathrm{oc}}{I}_{\mathrm{sc}})$,Series resistance, shunts, incomplete photon absorption and nonradiative recombination reduce the attainable efficiency.

The Shockley-Queisser limit removes these avoidable losses but retains the following physical constraints. There is one semiconductor junction with one band gap; each absorbed above-gap photon yields at most one collected electron-hole pair; carriers thermalize to the cell temperature; and the inverse optical process produces unavoidable radiative recombination. If ${\Phi }_{\odot }(E)$ is the incident photon flux per unit energy, ideal photon absorption givesThe same ability to absorb light requires the junction to emit it. At a forward voltage $V$, the emitted photon population has a chemical potential $eV$; in the ordinary nondegenerate limit the resulting recombination current is the exponential diode term above. Maximizing $VJ$ with this radiative $J_0$, and then choosing $E_g$, gives the limiting efficiency.

A small band gap collects many photons but gives a low voltage and large thermalization losses. A large band gap permits a larger voltage but transmits too much of the incident spectrum. Below-gap photons are lost to transmission, excess photon energy above the gap becomes heat, and radiative emission and its associated entropy loss keep $e{V}_{\mathrm{oc}}$ below $E_g$. The optimum single-junction conversion is roughly thirty percent under ordinary unconcentrated sunlight; the precise ceiling depends on spectrum, cell temperature and optical geometry. The original detailed-balance calculation gave about $30\%$ at $E_g\simeq 1.1\mathrm{eV}$ under its specified assumptions.^[ref] Multiple junctions and optical concentration change the assumptions and consequently change the bound.

Use two plane waves, $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$, as a basis near their Bragg degeneracy. The potential has Fourier coefficients $V_{\pm \mathbf{G}}$; its zero Fourier coefficient vanishes. All other plane waves are off resonance and are neglected in the leading nearly-free electron model. With ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$, the Schrödinger equation reduces to the Hermitian matrix problemSetting the characteristic polynomial equal to zero givesSolving this quadratic equation givesFor the real cosine potential its coefficient $V_G$ is real; the absolute-value form also allows an arbitrary choice of origin and corresponding Fourier phase.

At the Brillouin-zone boundary, $\mathbf{k}\cdot \mathbf{G}=G^2/2$, the free energies are equal. The two standing waves have different probability densities relative to the lattice potential and are split byRepeating this avoided crossing at the reciprocal lattice boundaries joins smooth branches into energy bands. In one dimension a nonzero Fourier coefficient opens a gap at the corresponding crossing. In higher dimensions a direct gap at one wavevector need not give a common forbidden energy interval at every wavevector: different bands may still overlap in energy. The present two-dimensional example remains metallic even though the crossings are locally gapped.

For the rectangular lattice, choose real-space primitive vectors $a\hat{\mathbf{x}}$ and $b\hat{\mathbf{y}}$. The reciprocal lattice vectors areDefine the half-widths of the first Brillouin zone by $K=\pi /a$ and $L=\pi /b$. Their values and the zone areA primitive reciprocal cell can be drawn with a corner at the origin; its centred translate is the first Brillouin zone here. Its area is $4{\pi }^2/(ab)=37.9600\mathrm{n}{\mathrm{m}}^{-2}$.

Including both spin states, counting the occupied free-electron states per unit area givesConsequentlyThe occupied disk has precisely the area of the first Brillouin zone, but their shapes differ: $K<k_F<L$. It protrudes through both vertical zone boundaries and does not reach the horizontal ones. Thus two electrons per cell do not imply a filled first band in the absence of a separating band gap.

Fold the protruding caps by $\pm {\mathbf{G}}_x$. The original circle and the translated circle intersect at $k_x=K$, $k_y=\pm \sqrt{k_F^2-K^2}$. The potential mixes these states and replaces each crossing by an avoided crossing. The lower branch reconnects into two open sheets running across successive zones in the $k_x$ direction; the upper branch forms a closed lens-shaped pocket about a vertical zone boundary. In a reduced-zone sketch its two halves sit at opposite edges, but periodic identification makes them one closed sheet. This is rectangular-lattice Fermi-surface reconstruction.

The largest pocket in this weak-potential construction is the lens-shaped intersection of two disks enclosed by the intersecting free circles. Each has radius $k_F$ and their centres are separated by $2K$. If $\theta =\arccos (K/k_F)$, two sectors minus their two triangles give the lens bound on a reconstructed Fermi pocket:Opening a gap raises the upper branch at fixed chemical potential and shrinks its pocket. The reported quantum oscillation area satisfiesIt is therefore consistent with this closed sheet. Onsager quantization gives $F=\hslash A_k/(2\pi e)\simeq 601\mathrm{T}$. An orbit on this closed contour can return to itself under the Lorentz force and occupy quantized Landau levels. The open sheets do not enclose a finite orbit area for a perpendicular field in this two-dimensional model. The observation is consistent with the reconstructed pocket, without uniquely determining a completely arbitrary band structure.

For a quantitative rough coupling estimate, use the supplied fixed-chemical-potential approximation and retain the two waves coupled by ${\mathbf{G}}_x$. WriteTheir upper eigenvalue isSetting it equal to $C{k}_F^2$ gives the pocket boundaryIts intersection with $k_y=0$ occurs at $q=\pm q_m$, whereThe two-wave estimate of a Bragg-gap coupling from a pocket area therefore usesAt $v=0$ this reproduces the lens bound on a reconstructed Fermi pocket; the pocket disappears at $v=C(k_F^2-K^2)=0.23786\mathrm{eV}$. Numerical integration gives $A(0.075\mathrm{eV})=6.22\mathrm{n}{\mathrm{m}}^{-2}$ and $A(0.100\mathrm{eV})=5.54\mathrm{n}{\mathrm{m}}^{-2}$. Matching the measured area gives $v\simeq 0.0931\mathrm{eV}$, so the justified rough result isThis estimate neglects remote plane-wave admixtures and other Fourier coefficients. It also holds the chemical potential fixed as requested; an exact fixed-density calculation would determine it self-consistently after reconstruction. The numerical root is a way to implement the approximate model, not a measurement of $V_G$ to four significant figures.