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The giant magnetoresistance mechanism is spin-dependent conduction through the two ferromagnets. Exchange splitting gives majority and minority $d$ bands different density of states values at the Fermi energy. Scattering into those bands, together with interface scattering, produces different resistances for the two spin channels. The relevant quantity is the spin-dependent transport/scattering rate, not simply a count of all available states.

For parallel layer magnetizations, one global spin channel encounters relatively weak scattering in both Co and NiFe, providing a low-resistance path. For antiparallel magnetizations, each global spin channel encounters one more strongly scattering ferromagnet, so both paths are comparatively resistive. The channels conduct in parallel while each crosses the layers in series. This is current-perpendicular-to-plane magnetoresistance, because the current crosses the interfaces.^[ref]

In the simplest two-current resistor model, with equal magnetic layers having channel resistances $a,b$ and spacer resistance $r$ per channel,Real Co/NiFe layers need not have equal parameters, but the same-sign spin asymmetries normally give $R_{AP}>R_P$. Their different magnetic coercivities allow the two configurations to be selected separately.

The nuclear Fermi gas model treats protons and neutrons as two species of spin one-half fermions moving approximately independently in a confining nuclear mean field. At a temperature small compared with their Fermi temperature, the Pauli exclusion principle fills the lowest available momentum states. It does not allow all nucleons to sit at zero momentum.

For species $i=p,n$, two spin states occupy each momentum-space cell $(2\pi \hslash {)}^3/V$. Counting a filled sphere giveswhere $n_i=N_i/V$. The Fermi energy is measured from the mean-field well bottom; it need not equal the energy needed to remove a nucleon into vacuum. Occupied-state averaging givesEven at zero temperature there is kinetic energy and degeneracy pressure. As an explicitly illustrative input, total density $n=0.16{\mathrm{fm}}^{-3}$ and $N=Z$ give $p_{F}c\simeq 263\mathrm{MeV}$ and $E_F\simeq 37\mathrm{MeV}$. These numbers illustrate the formula; they are not additional data supplied by the paper.

Separate proton and neutron seas explain a kinetic contribution to nuclear symmetry energy. At fixed $n$, put $\delta =(N-Z)/A$ and define $E_F$ at $n_p=n_n=n/2$. ThenIncreasing one species' population raises its Fermi surface and costs energy, favoring similar populations before electrostatic and interaction terms are included.

pauli blocking suppresses collisions deep in a filled sea because scattering needs unoccupied final states. An energetic external probe can knock a nucleon into an empty state and reveal its initial Fermi motion, making the model useful for broad inclusive scattering distributions. A finite surface can be approximated using a position-dependent nuclear mean field.

A free gas cannot produce nuclear binding or saturation: those require an attractive interaction and its density dependence. The smooth gas picture also omits discrete nuclear shells, nuclear spin-orbit splitting, nuclear pairing, short-distance correlations and deformation. It cannot assign all ground-state spins from a momentum sphere. Its useful prediction is the scale and occupancy of nucleon motion, with a surrounding mean field supplied separately.

For the electronic and lattice heat-capacity crossover in a monovalent metal, to obtain a numerical estimate from $E_F$ alone, assume monovalent potassium is a three-dimensional free electron gas with a parabolic band. Including both spins, the molar density of states is $g(E_F)=3N_A/(2E_F)$. Using the supplied approximate coefficient $3$, rather than silently replacing it by ${\pi }^2/3$, givesEquating it to the stated low-temperature Debye heat capacity yields the nonzero crossoverWe used $k_B=8.61733\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$ and $E_F=2\mathrm{eV}$. This is self-consistently far below ${\theta }_D=100\mathrm{K}$ and the Fermi temperature. Using the exact Sommerfeld coefficient instead gives $0.954\mathrm{K}$; that is a change of approximation, not a discrepancy in the algebra. Without a band/DOS assumption, the supplied Fermi energy does not by itself determine $g(E_F)$.

identical particles have the same intrinsic properties. Exchanging their bookkeeping labels cannot distinguish a physical preparation or change a measurable prediction. This is stronger than merely failing to track classical trajectories. For ordinary bosons and fermions, the many-particle wavefunction has respectively bosonic exchange symmetry or fermionic exchange symmetry under every particle exchange operator, acting on spatial and spin coordinates together. The Spin-statistics theorem connects integer spin to bosons and half-integer spin to fermions; this connection uses relativistic quantum-field assumptions, rather than following from the nonrelativistic Schrödinger equation alone.

For one-particle states $u,v$ satisfying orthonormality, the two alternatives form a quantum superposition:Its probability density contains an exchange interference term, absent from an incoherent mixture of labelled alternatives. The antisymmetric expression vanishes when both one-particle states are identical: one complete one-particle state can contain at most one fermion. This Pauli exclusion principle explains electron shell filling, the electronic Fermi sea and degeneracy pressure. Opposite spins are different one-particle states, so two electrons may share an atomic orbital. Spatially separated wave packets with negligible overlap can often be treated as distinguishable for local measurements; the underlying exchange symmetry remains.

In thermal equilibrium, occupation of a level of energy $ϵ$ obeysbosons can accumulate macroscopically in one state, giving Bose-Einstein condensation; fermions instead fill successive states up to the Fermi energy at zero temperature. When occupation is dilute, both expressions reduce to the Maxwell-Boltzmann distribution. The identical-particle classical partition function contains $1/N!$, avoiding spurious entropy from permutations and the Gibbs paradox. Exchange effects become conspicuous when the thermal de Broglie wavelength is comparable to the interparticle spacing.

An operational test is the Hong-Ou-Mandel effect. Down-converted photon pairs enter separate ports of a balanced beam splitter; indistinguishable alternatives for one photon per output cancel. Scanning the relative arrival delay restores coincidences away from overlap. Matching polarization, spectral filtering and spatial alignment controls distinguishability; measuring delayed accidental coincidences controls background. The observed dip tests two-photon interference, and its finite depth must not automatically be interpreted as a failure of bosonic symmetry. The original apparatus varied beam-splitter position and subtracted a separately measured accidental background.^[ref]

spin normalization in two-dimensional free-electron state counting. Treat the mobile electrons as independent particles in a constant interior potential, with parabolic dispersion $ϵ={\hslash }^2{k}^2/(2m)$, and impose confinement or periodic boundary conditions. electron–electron interactions and the detailed periodic ionic potential are neglected, while Fermi statistics are retained. Each spatial state occupies area $(2\pi {)}^2/A$ in two-dimensional wavevector space. For spin degeneracy $g_s$,The printed coefficient $Am/(2\pi {\hslash }^2)$ is the per-spin density of states. Unpolarized spin-one-half electrons have $g_s=2$, doubling it. Actual graphene near its Dirac points has approximately linear dispersion, so its energy-dependent density of states is not this parabolic-model result; the plane geometry alone does not establish the free-electron dispersion.^[ref]

Occupied disk and Fermi energy. At zero temperature fill the lowest states up to a disk of radius $k_F$. State counting givesThus the physical spin-degenerate result is ${ϵ}_F=\pi {\hslash }^{2}N/(Am)$. If $N$ instead counts one spin species, the per-spin convention gives $2\pi {\hslash }^{2}N/(Am)$. Each occupied wavevector can contain one electron of each allowed spin, never two in the same complete one-particle state.

Occupation probability and chemical potential of a two-dimensional ideal Fermi gas. A single fermion state permits occupation zero or one. In a grand canonical ensemble their relative weights are $1$ and $e^{-\beta (ϵ-\mu )}$. Normalizing givesThe chemical potential is fixed by the total particle number. Here the total density of states $D=g_{s}Am/(2\pi {\hslash }^2)$ is constant above the band bottom. Integrating the occupation gives $N=D{k}_{B}T\ln (1+e^{\mu /(k_{B}T)})$, and thereforeThe approach is exponentially close in this ideal two-dimensional constant-density-of-states model; a three-dimensional $T^2$ correction should not be inserted here. A narrow interval of width of order $k_{B}T$ around $\mu$ is thermally smeared.

The free electron model replaces the microscopic ionic potential and interacting electronic motion by independent mobile electrons in a uniform neutralizing background, with approximately parabolic dispersion $ϵ={\hslash }^2{k}^2/(2m^{\ast })$. Its quantum version fills states according to Fermi-Dirac statistics and the Pauli exclusion principle. Scattering is represented by a momentum relaxation time $\tau$, rather than derived from the model's ideal free trajectories. This model explains a Fermi energy much larger than room-temperature thermal energy, an electronic heat capacity proportional to temperature at low temperature, and a Drude electrical conductivity. With appropriate scattering assumptions it also yields the Wiedemann-Franz law. It cannot by itself explain band gaps, detailed band anisotropy, strong correlations, or the microscopic temperature dependence of $\tau$.

The inertial term is mass times the rate of change of mean drift velocity. The electric force is $eE$, taking the printed $e$ as the signed electron charge, and $-\gamma ⟨v⟩=-m^{\ast }⟨v⟩/\tau$ describes loss of directed momentum to scattering. electrons retain large microscopic velocities; this damping is of their mean directed motion. In a steady DC field,Both charge factors are negative for electrons, so the conductivity is positive. If $e$ instead denotes a positive charge magnitude, the force and current equations each require a minus sign and give the same result.

For AC response choose physical fields as real parts of phasors proportional to $e^{-i\omega t}$. Substitution givesThusThe real part decreases from ${\sigma }_0$ to zero. The imaginary part starts at zero, reaches ${\sigma }_0/2$ at $\omega \tau =1$, and falls as $1/\omega$. The real part falls as $1/{\omega }^2$ at high frequency. With an $e^{+i\omega t}$ convention, only the sign of the imaginary part changes. Positive $\mathrm{Re}\sigma$ gives positive cycle-averaged dissipation $⟨jE⟩=\mathrm{Re}\sigma |\hat{E}{|}^2/2$.

The supplied DC conductivity determines the transport time:Inertia matters when $\omega \tau \sim 1$, giving ${\omega }_c\simeq 4.02\times 10^{13}\mathrm{rad}{\mathrm{s}}^{-1}$, or ${\nu }_c={\omega }_c/(2\pi )\simeq 6.40\times 10^{12}\mathrm{Hz}$. Quoting the former number as hertz would introduce an erroneous factor of $2\pi$.

The free electron model replaces the detailed periodic potential by a constant background, treating conduction electrons as mobile particles with approximately parabolic dispersion $E={\hslash }^2{k}^2/(2m)$. It is useful to distinguish the classical Drude model from the quantum free electron gas. They share a simple transport picture, but differ fundamentally in occupation statistics.

With a momentum-relaxation time $\tau$, both lead in a simple isotropic one-band approximation toThis explains approximately Ohmic longitudinal conduction and a carrier-density scale from the Hall effect. The relaxation time is phenomenological: introducing it does not itself predict scattering from defects or phonons. A four-probe resistance measurement separates the sample's longitudinal response from current-contact voltage drops; field and current reversals isolate the transverse signal.

The classical model incorrectly assigns roughly $3N{k}_B/2$ of electronic heat capacity, far too large in a degenerate metal. Fermi-Dirac statistics fixes this: most occupied states cannot change occupation after a small thermal perturbation because nearby low-energy states are already filled. Only a fraction of order $k_{B}T/E_F$ near the Fermi energy participates. For $T\ll T_F$,The quantum model also gives a finite Fermi velocity even at zero temperature and Pauli paramagnetism. Under predominantly elastic scattering, its electronic heat and charge conductivities obey the Wiedemann-Franz law,This is a quantum success; classical equipartition does not give the correct small heat capacity and Lorenz number together.

Its limitations are equally instructive. Without the periodic potential it has no band gaps, so it cannot distinguish a filled-band insulator from a metal or explain semiconductor doping. The simplest single-electron Hall formula cannot explain positive Hall coefficients; band curvature and electron holes, or a multiband transport treatment, are needed. It neglects much Fermi-surface anisotropy and detailed optical interband absorption. Noninteracting electrons alone do not explain ferromagnetism or superconductivity, and lattice thermal behavior needs phonons. The nearly-free electron model, which allows reciprocal-lattice Fourier components of the potential to mix plane waves, provides a controlled first correction near Bragg planes.

Its main success is a simple quantitative baseline for degenerate conduction electrons; its main failure is treating the crystal and interactions as too featureless. A fit of conductivity alone is not a test of the whole model: compare heat capacity, Hall sign, optical response and temperature dependence using independently constrained parameters.

Impose periodic boundary conditions on a large cubic box. The allowed wavevectors occupy a lattice of cell volume $(2\pi {)}^3/V$ in wavevector space. Including the two spin states, the number of states with magnitude at most $k$ isUsing $k=\sqrt{2m_eϵ}/\hslash$ and differentiating gives the total, spin-summed three-dimensional free-electron density of states,It counts states per unit energy; divide by $V$ for states per unit volume per unit energy. At zero temperature, the Fermi-Dirac distribution fills all states up to the Fermi energy, givingWriting $n=N/V$ givesThe energy zero is the bottom of the free-electron band. The electron Fermi temperature is $T_F=E_F/k_B$.

Only a shell of energy width of order $k_{B}T$ near the Fermi energy can change occupation: much lower states are blocked by the Pauli exclusion principle. The thermal-shell estimate of electronic heat capacity countsTo recover the rough coefficient requested in the paper, assign each of these active particles an excitation energy of order $3k_{B}T/2$. ThenThe differentiation matters: the active shell itself widens as temperature rises. This argument explains the small linear-in-$T$ electronic heat capacity; its shell width and average excitation energy are heuristic, so it does not establish a precise numerical prefactor.

A controlled calculation from the Fermi-Dirac distribution gives a nearby but different result. At fixed particle number, the Sommerfeld expansion yieldsThus $9/2$ is a rough estimate of the exact free-electron coefficient ${\pi }^2/2\simeq 4.935$, not an exact consequence of the stated density of states. Both require $T\ll T_F$.

Assume sodium supplies one conduction electron per atom. The supplied Fermi energy givesThe gas is strongly degenerate at room temperature. For one mole, $N=N_A$ and $N_A{k}_B=R=8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. Using the requested estimate,The supplied mass density determines the answer per unit volume: $N/V=(970/0.023)N_A=2.54\times 10^{28}{\mathrm{m}}^{-3}$. ThereforeThe density is not needed for the molar heat capacity once $E_F$ is supplied. Using the precise Sommerfeld expansion coefficient instead gives $0.327\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

Either electronic estimate is only about $1\%$ of the stated measured total. Most of the total heat capacity comes from lattice phonons, whose high-temperature Dulong-Petit law limit is $3R=24.94\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The remaining difference from $28\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ need not be electronic: anharmonic lattice effects and the difference between heat capacity at constant pressure and heat capacity at constant volume also contribute. The paper does not specify the measurement constraint, so the two totals should not be equated exactly.

The quantum free electron model treats conduction electrons as independent particles moving in an approximately constant potential, with parabolic dispersion relation $E={\hslash }^2{k}^2/(2m)$. Fermi-Dirac statistics successfully explains important features such as a Fermi energy and a small electronic heat capacity, but a constant potential gives no band gaps and cannot by itself explain why different band fillings produce metals, semiconductors and band insulators. It also misses lattice-dependent effective masses, detailed directional responses and the topology of real Fermi surfaces. Scattering and finite electrical resistivity require additional physics.

The nearly-free electron model introduces a weak periodic lattice potential. Bloch's theorem replaces individual free plane waves by Bloch states, and Fourier components of the potential mix plane waves whose wavevectors differ by a reciprocal lattice vector. Near degeneracy at a Brillouin zone boundary, this coupling opens a band gap; for a two-wave crossing its size is $2|V_G|$. Away from the crossing, energies remain close to free-particle values. Band curvature produces band effective masses, and band filling determines whether the Fermi energy lies inside an allowed band or a gap.

The improvement is the inclusion of lattice periodicity, rather than a switch from classical to quantum statistics: the quantum free electron model already uses Fermi-Dirac statistics. The strong delta barriers in the later numerical example are treated exactly by the Kronig-Penney model, rather than by assuming that their strength is perturbatively small.

With length measured in units of $a=1$, plot the real functionOnly the portions with $|f(k)|\le 1$ can equal $\cos \varphi$, so the strip between the horizontal lines $+1$ and $-1$ identifies allowed energy bands. The band edges of a repulsive delta-comb potential have the following approximate first intervals:Each interval corresponds to energy $E={\hslash }^2{k}^2/(2m)$; the intervening intervals are forbidden. The exact upper endpoints follow from $f(j\pi )=(-1{)}^j$. The lower endpoints solve $f(k)=(-1{)}^{j-1}$ inside $((j-1)\pi ,j\pi )$. The divergence suggested by $\sin k/k$ at the origin is removable, and $f(0)=11>1$, so there is no zero-energy state in the repulsive model.

The periodic barriers produce separated energy bands and band gaps, unlike the continuous free-particle spectrum. A finite $N$-atom loop has discrete states within these bands. Whether the material conducts still depends on band filling: one electron per primitive cell half-fills a spin-degenerate band, while two electrons per cell can fill the first band and place the Fermi energy in a gap. Band gaps alone do not establish that every filling is insulating. At high $k$, the perturbing term $10\sin k/k$ becomes relatively small and the spectrum approaches free-electron behaviour in relative terms.

Semiconductor doping introduces controlled impurities to alter equilibrium carrier populations and electrical conductivity. In an intrinsic semiconductor, thermally excited electrons in the conduction band and electron holes in the valence band have equal densities $n=p=n_i$. A shallow donor impurity, such as phosphorus in silicon, supplies an electron from an impurity level just below the conduction-band edge. Once ionized it is positively charged, and the material is n type. An acceptor impurity, such as boron in silicon, accepts a valence electron into a level just above the valence-band edge, leaving a mobile hole and a negatively charged ionized impurity; the material is p type.

With dopant densities $N_D,N_A$, charge neutrality requiresFor fully ionized, nondegenerate dopants at equilibrium, the semiconductor mass-action law gives $np=n_i^2$. In a substantially n-type sample, $N_D-N_A\gg n_i$, soThe corresponding p-type result is $p\simeq N_A-N_D$ and $n\simeq n_i^2/(N_A-N_D)$. The majority population grows while the minority population decreases; adding donors does not simply add equal numbers of free electrons and holes.

The Fermi energy shifts toward the conduction band for n-type material and toward the valence band for p-type material. In the nondegenerate approximation,where $N_C,N_V$ are effective band-edge state densities. At very low temperature, dopants may not be ionized and carrier freeze-out occurs. At intermediate temperatures, ionized dopants dominate the carrier density; at sufficiently high temperature, intrinsic excitation can dominate. Heavy doping can produce degenerate carrier statistics, invalidating these Boltzmann expressions.

The electrical conductivity is $\boxed{\sigma =e(n{\mu }_e+p{\mu }_h)}$. Doping usually raises the majority carrier density, but ionized impurity scattering can lower mobility, so conductivity need not scale indefinitely with dopant concentration. Joining p-type and n-type regions creates a p-n junction, with a depletion region and built-in electric field established by carrier diffusion and compensating drift. The fixed ionized dopants, not an absence of all charge, account for the depleted region's net charge density.

In a metal, a partially filled energy band or overlapping energy bands supplies electrons with nearby unoccupied states. At thermal equilibrium, oppositely directed group velocities are equally populated and the net electric current vanishes. An applied electric field distorts the occupation distribution, producing a small mean drift velocity. A filled energy band alone has no such low-field intraband response because its velocity contributions cancel over the entire Brillouin zone.

In the Drude model, let $n$ be the mobile electron number density, let $e>0$ be the elementary charge, and let $\tau$ be the momentum relaxation time. The mean motion obeysThe steady state is ${\mathbf{v}}_d=-e\tau \mathbf{E}/m$. The current density is opposite to the electron drift velocity:This gives microscopic Ohm's law, with electrical conductivity $\sigma$ in $\mathrm{S}{\mathrm{m}}^{-1}$ and electrical resistivity ${\rho }_{\mathrm{el}}$ in $\Omega \mathrm{m}$. For a uniform wire of length $\ell$ and cross-sectional area $A$, its electrical resistance is $R={\rho }_{\mathrm{el}}\ell /A$. The dissipated power per volume is $\mathbf{J}\cdot \mathbf{E}=\sigma E^2$, the Joule heating supplied by the external source.

The free electron model with Fermi-Dirac statistics improves the microscopic picture. The Pauli exclusion principle fills states up to the Fermi energy at low temperature. For a spherical, parabolic energy band with two spin states,Only states near the Fermi surface change their occupation in a weak-field response; the large equilibrium Fermi velocity is not the small drift velocity. The same $n{e}^2\tau /m$ formula results for this isotropic parabolic model when $\tau$ is evaluated for transport near the Fermi surface. In a crystal, the energy band group velocity, effective mass and possibly direction-dependent scattering replace the simplest bare-particle picture. The mean free path is approximately ${\ell }_{\mathrm{mfp}}=v_F\tau$.

A perfectly periodic static lattice is already included in Bloch states and does not by itself give a finite electrical resistivity. Momentum relaxation comes from defects, impurities, boundaries, and electron-phonon scattering. Independent scattering mechanisms approximately add their rates, giving Matthiessen's rule,The approximately temperature-independent ${\rho }_0$ is the residual resistivity. In ordinary simple metals, the phonon contribution is approximately proportional to temperature at high temperature; a three-dimensional acoustic-phonon transport model gives a $T^5$ contribution well below the relevant Debye temperature. Other mechanisms can dominate in particular materials or regimes, so neither power law is universal.

The classical Drude model incorrectly treats the equilibrium carrier statistics. Fermi-Dirac statistics explains the small electronic heat capacity, while electrons carrying both charge and heat give the approximate Wiedemann-Franz law ${\kappa }_e/(\sigma T)={\pi }^2{k}_B^2/(3e^2)$ for degenerate carriers with appropriate nearly elastic scattering. Both the simple electrical conductivity formula and this thermal relation depend on the assumptions about energy bands and scattering, rather than being exact for all metals.

Let $d=1,2,3$ and take a specimen of length, area or volume $L^d$. One monovalent atom per unit cell supplieselectrons. At zero temperature, the Pauli exclusion principle fills the lowest energy states, with two spin states per spatial wavevector. The occupied region is an interval, disk or ball of radius $k_F$. Using the $k$-space density of states gives, in one dimension,In two dimensions,In three dimensions,Each Fermi energy follows from the free electron model dispersion relation $E_F={\hslash }^2{k}_F^2/(2m)$. The Fermi wavevectors have units ${\mathrm{m}}^{-1}$ and the Fermi energies have units joules for $a$ in metres and $m$ in kilograms. No numerical lattice spacing is supplied, so numerical energies in electronvolts cannot be uniquely calculated. The spin factor of two is necessary here for actual monovalent crystals; the spinless auxiliary count of part (a) was not an instruction to remove electron spin from these crystals.

Keep the electron number fixed, soand $\mu (0)=E_F$. Heating smooths the Fermi-Dirac distribution across the Fermi energy, creating holes below $\mu$ and particles above it. The sign of the low-temperature adjustment of chemical potential follows from the local slope of the density of states.

In one dimension, $g\propto {ϵ}^{-1/2}$ decreases with energy. At fixed $\mu$, smoothing removes more electrons below it than it adds above it, so $\mu$ initially rises. In three dimensions, $g\propto {ϵ}^{1/2}$ increases, so $\mu$ falls. In two dimensions the constant density of states makes the leading changes cancel, so $\mu$ stays extremely close to $E_F$ in the degenerate regime.

For quantitative checks, a Sommerfeld expansion gives, for $k_{B}T\ll E_F$,For a constant two-dimensional density of states above an unbounded parabolic band bottom, the number integral can be done exactly:soIt actually decreases, but its low-temperature correction is exponentially small rather than quadratic. “Constant in two dimensions” is therefore a low-temperature approximation, not an exact statement for every $T$.

The one-dimensional rise is also only an initial trend. In the high-temperature Maxwell-Boltzmann distribution limit, the ideal $d$-dimensional electron gas has $\mu \simeq k_{B}T\ln [n_d{\lambda }_T^d/2]$, where $n_d=N_e/L^d$ and ${\lambda }_T=\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$ is the thermal de Broglie wavelength. It eventually becomes negative and decreases in every dimension. The one-dimensional chemical potential thus passes through a maximum. This high-temperature statement refers to the stipulated unbounded free-electron dispersion relation, not a claim that a real finite-width crystal band remains parabolic indefinitely.

For the usual nonrelativistic dispersion, the three-dimensional density of states is $g(ϵ)=C{ϵ}^{1/2}$. Filling all states through the Fermi energy givesDegeneracy factors cancel. More generally $ϵ\propto p^s$ gives $U/N=3{ϵ}_F/(3+s)$; an ultrarelativistic gas would give $3{ϵ}_F/4$. The nonrelativistic result is the standard massive ideal-gas interpretation.

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]

Use $e>0$, $B>0$ along $+z$, and $U(y)$ for the electron's confinement potential energy. The spin-resolved Landau level branches in the smooth-edge approximation areThe spin labels denote upper and lower energies; the sign of a material's effective $g$ factor is not inferred from the stated positive splitting. In units of $\hslash {\omega }_c$, the four occupied bulk energies are $1/3,2/3,4/3,5/3$. A chemical potential in $(5/3,7/3)$ therefore gives filling factor four. The branches rise at both confining edges and cross the common Fermi energy once at each edge. Each crossing is one spin-resolved quantum Hall edge state; there are four on either side. The two spin branches within an orbital level are separated by $\hslash {\omega }_c/3$, while the intervening gap to the next orbital pair is $2\hslash {\omega }_c/3$.

In Landau gauge for a uniform magnetic field $\mathbf{A}=(-By,0,0)$, the guiding centre is $Y=\hslash k_x/(eB)$. The group velocity isThus the upper edge in the original trajectory diagram carries electrons to the right and the lower edge to the left. Reversing the magnetic field reverses both arrows. Semiclassically these are skipping trajectories; quantum mechanically the guiding centres label propagating quantum Hall edge states. The larger orbital-plus-spin energy crosses the Fermi energy farther toward the interior, because less confinement energy is needed. Opposite spins in one orbital branch are therefore spatially separated.

In a clean bulk with the Fermi energy between Landau levels, there are no extended bulk states at that energy. Occupied bulk cyclotron orbits lie below it; one must not draw them as extra conducting Fermi-level channels. Smooth disorder can instead produce localized bulk contours without adding a direct source-to-drain channel.