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Since energy and angular frequency are related by $E=\hslash \omega$, the given eigenvalue relation yieldsWith $E_P=4.3\mathrm{eV}$,The ${\lambda }_i$ are dimensionless modal eigenvalues, not optical wavelengths despite the symbol. The same $\hslash$ multiplies both mode and bulk angular frequencies and cancels in the energy ratio. These are bare-core predictions of the stipulated lossless Drude model; adding a dielectric/metallic shell, interband response or damping changes the actual resonances.

With a positive charge magnitude e, the relaxation-time equation for electrons is $m_e^{\ast }\dot{\mathbf{v}}=-e\mathbf{E}-m_e^{\ast }\mathbf{v}/\tau$. Steady drift is opposite the field, ${\mathbf{v}}_d=-e\tau \mathbf{E}/m_e^{\ast }$, so the positive electron mobility and electrical conductivity areThis is the weak-field Drude model; the mobility is a magnitude, not the signed velocity-to-field ratio.

For an intrinsic nondegenerate semiconductor, $n_i=p_i=\sqrt{N_c{N}_v}{e}^{-E_g/(2k_{B}T)}$. Equal electron and hole masses give $N_c=N_v$ and intrinsic chemical potential midway between the band edges. At300K, the exponential is $e^{-19.34}=3.98\times 10^{-9}$. Using the later supplied mass $0.015m_e$ givesThis is many orders below a typical metal's carrier density. Since ${\sigma }_i=e{n}_i({\mu }_e+{\mu }_h)$, ordinary finite carrier mobilities leave its intrinsic conduction negligible on a metallic scale; the gap alone does not specify an exact conductivity without the mobilities.

Under the explicitly stipulated full donor ionization, $n\simeq N_d=10^{24}{\mathrm{m}}^{-3}$ and the supplied mobility is $0.0100{\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. ThereforeThe band diagram places the intrinsic chemical potential at midgap and a shallow isolated donor level just below the conduction edge. The doped chemical potential moves upward. Quantitatively, the degeneracy test for a fully ionized light-mass donor gas gives $N_d/N_c=21.7$, so a nondegenerate logarithmic chemical-potential formula is inconsistent. The parabolic-band zero-temperature estimate is $E_{F,d}-E_c={\hslash }^2(3{\pi }^2{N}_d{)}^{2/3}/(2m^{\ast })=0.243\mathrm{eV}$; its300K value is about0.241eV. The diagram accordingly shows the doped chemical potential inside the conduction band, not between an isolated donor level and the edge.

The hydrogenic donor state scaling using the supplied hydrogen values isThe binding is much smaller than $k_{B}T=25.9\mathrm{meV}$, indicating a shallow donor on the isolated-state estimate. However, the mean donor spacing is only $N_d^{-1/3}=10\mathrm{nm}$, well below the orbit radius. The donor-state overlap estimate invalidates an independent isolated-level ionization model at this concentration. Shallow binding supports abundant mobile carriers, but the full ionization versus isolated donor occupation distinction means full ionization cannot be justified solely by comparing binding with temperature while simultaneously retaining isolated levels and a degenerate chemical potential. The conductivity above follows the question's carrier-density assumption; an overlapping impurity/conduction-band description is needed for a self-consistent microscopic account. ^[ref]

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.

Under the one-carrier Drude model, the magnitude of the Hall coefficient gives an effective carrier densityConvert the material mass density to $1850\mathrm{kg}{\mathrm{m}}^{-3}$ and its molar mass to $0.009\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. The atomic number density isHere $N_A=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$ is the Avogadro constant. The positive Hall coefficient gives a hole-like response in a one-carrier interpretation. The Hall carrier count versus valence distinction matters: beryllium has two valence electrons, but the result is neither two free electrons nor a literal fractional atom donating charge. Its energy bands and Fermi surface invalidate the simple isotropic one-carrier inference. Electron-like and hole-like contributions with different charge carrier mobilities can yield this sign and effective count.

Let $e>0$ denote the elementary charge, so an electron's charge is $-e$. In the relaxation-time Drude model with positive transport effective mass,The steady drift velocity is ${\mathbf{v}}_d=-e\tau \mathbf{E}/m^{\ast }=-\mu \mathbf{E}$, defining positive electron mobility $\mu =e\tau /m^{\ast }$. The current density is opposite to the electron drift:By the definition of electrical conductivity, $\mathbf{J}=\sigma \mathbf{E}$, soIts units are $({\mathrm{m}}^{-3})(\mathrm{C})({\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1})=\mathrm{S}{\mathrm{m}}^{-1}$. The mobility definition alone establishes the result in the scalar linear-response regime; the relaxation-time equation supplies a microscopic approximation for $\mu$.

Treat sodium as monovalent, with one mobile electron per atom. Its mass density is ${\rho }_m=0.97\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}=970\mathrm{kg}{\mathrm{m}}^{-3}$, and its molar mass is $M=0.023\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Using the Avogadro constant,The electron mobility $\mu =e\tau /m_e$ relates the magnitude of drift velocity to electric field. The Drude model then givesUsing $m_e=9.109\times 10^{-31}\mathrm{kg}$,The inferred time is the momentum relaxation time. Identifying it with the mean time between collisions is the intended simple isotropic Drude model assumption; small-angle scattering need not relax directed momentum in a single event.

In a metal, mobile electrons transport heat as well as electric charge. A temperature gradient creates an imbalance between carriers arriving from warmer and colder regions. The resulting heat current obeys Fourier's law ${\mathbf{j}}_Q=-\kappa \nabla T$ for an isotropic material, where thermal conductivity $\kappa$ has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. For a bar of area $A$ carrying steady heat flow $P$, $P=-\kappa AdT/dx$ when parasitic losses are negligible.

A kinetic estimate for the electronic contribution iswhere $C_e$ is the electronic heat capacity per volume, $v_F$ the Fermi velocity, and ${\ell }_{\kappa }=v_F{\tau }_{\kappa }$ the heat-transport mean free path. Fermi-Dirac statistics and the Pauli exclusion principle restrict thermal rearrangements to states near the Fermi surface, rather than allowing every electron to contribute a classical heat capacity. A degenerate parabolic band givesUsing the Drude model form $\sigma =n{e}^2{\tau }_{\sigma }/m$ for electrical transport, the relation between thermal and electrical relaxation times in metals becomesFor scattering that relaxes heat and charge currents similarly, ${\tau }_{\kappa }\simeq {\tau }_{\sigma }$ and the Wiedemann-Franz law follows:Thus good electrical conductors are usually good thermal conductors. The correct quantum Lorenz number differs from a naive classical electron-gas estimate.

Scattering from impurities, boundaries and phonons limits the heat-transport path. In an impurity-dominated low-temperature regime with approximately constant relaxation time, ${\kappa }_e$ is approximately proportional to $T$. Where phonon-limited electrical resistivity is proportional to $T$ and the Wiedemann-Franz approximation is applicable, ${\kappa }_e$ can be roughly temperature independent. Inelastic processes can relax heat and charge currents differently, so the Lorenz ratio is not universal at all temperatures. Lattice phonons add a separate positive thermal contribution; the measured total is $\kappa ={\kappa }_e+{\kappa }_{\mathrm{ph}}$, and the electronic Wiedemann-Franz expression should not be applied blindly to that total.

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.

State a time convention before using complex response: take fields proportional to $e^{i(kz-\omega t)}$, with $k=k_r+i\kappa$ and $\kappa \ge 0$ for a passive wave entering the material. In a homogeneous Ohmic medium, Maxwell equations with $\mathbf{J}=\sigma \mathbf{E}$ giveThe ohmic-conductor complex wavenumber describes both phase advance and amplitude decay. Real constant conductivity is only a low-frequency approximation. In the Drude model, $m\dot{\mathbf{v}}+m\gamma \mathbf{v}=-e\mathbf{E}$ giveswith a fixed positive ionic background and no additional bound-electron contribution in this simple model. The plasma frequency marks the inertial electron-response scale. More elaborate band structure, interband transitions and nonlocal transport limit this scalar treatment.^[ref]

In the collisionless nonmagnetic limit, the dispersion relation is ${\omega }^2={\omega }_p^2+c^2{k}^2$. For $\omega >{\omega }_p$,A crest can move faster than $c$ without carrying a new signal faster than light. Below cutoff the ideal wave is evanescent, with penetration length $c/\sqrt{{\omega }_p^2-{\omega }^2}$; it does not propagate as an ordinary real-$k$ bulk wave. Evanescence in the lossless limit is not Joule dissipation. Nonzero collision rate gives a dissipative conductivity and smooths this ideal cutoff picture.

For a good conductor with approximately real $\sigma \gg \omega ϵ$, the spatial branch becomesA field amplitude falls as $e^{-z/\delta }$, while intensity falls as $e^{-2z/\delta }$. The phase velocity is $\omega \delta$ in this limit. Magnetic and electric amplitudes are not related by the vacuum ratio: $Z_s=E/H=(1-i)\sqrt{\mu \omega /(2\sigma )}$ with the chosen time convention. Its positive real part is the surface resistance. The local mean heating is $\frac{1}{2}\mathrm{Re}\sigma |E{|}^2$, consistent with electromagnetic energy flowing into the metal. skin depth is an amplitude length; interpreting it as an intensity length gives a factor-of-two error.

A transmission-line mode propagates chiefly through the dielectric between conductors, even though finite metal conductivity causes loss at its boundaries. With per-length parameters $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$, the telegrapher's equations give, for the same $e^{-i\omega t}$ convention,Writing a forward wave as $e^{-\Gamma z}$, $\Gamma =\alpha -i\beta$, gives ${\Gamma }^2=(R^{\prime }-i\omega L^{\prime })(G^{\prime }-i\omega C^{\prime })$. In the low-loss regime $R^{\prime }\ll \omega L^{\prime }$ and $G^{\prime }\ll \omega C^{\prime }$, set $Z_0\simeq \sqrt{L^{\prime }/C^{\prime }}$ to obtainThis low-loss transmission-line attenuation separates series conductor loss from dielectric leakage loss. Average forward power falls as $e^{-2\alpha z}$. In the normal skin-effect regime, reduced conducting area makes the AC conductor resistance increase roughly as $\sqrt{\omega }$; dielectric loss can supply a separate frequency dependence. Reflection from a mismatched termination changes delivered power without being, by itself, dissipation in the line.^[ref]

To test loss, compare several lengths of otherwise identical line with matched terminations, calibrate connectors and incident/reflected power, and monitor temperature. Separate amplitude attenuation from phase delay and from standing-wave minima. Bulk penetration, line attenuation and dielectric propagation speed are related electromagnetic quantities but describe different experimental geometries.

Electromagnetic waves drive mobile charges in a conducting medium. Their current changes both the phase and attenuation of the wave. A description in terms of a real, frequency-independent refractive index is therefore generally insufficient: the material response and the propagation constant are complex and frequency dependent.

Use phasors proportional to $e^{i(kz-\omega t)}$, with $k=k^{\prime }+i{k}^{\prime \prime }$ and $k^{\prime \prime }\ge 0$ for a passive medium. In a homogeneous, isotropic, linear material, let ${ϵ}_b$ represent the permittivity associated with bound-charge electric polarization, $\mu$ the magnetic permeability and $\sigma (\omega )$ the mobile-charge conductivity. The macroscopic Maxwell equations giveFor a transverse plane wave, taking another curl yieldsThis division of the response avoids counting mobile electrons both in ${ϵ}_b$ and in $\sigma$. Define the complex refractive index $\tilde{n}=ck/\omega =n+i\kappa$. For a nonmagnetic medium, ${\tilde{n}}^2={ϵ}_{\mathrm{eff}}/{ϵ}_0$. Its real part determines the phase variation, while the positive optical extinction coefficient $\kappa$ gives amplitude attenuation $e^{-\omega \kappa z/c}$ and intensity attenuation $e^{-2\omega \kappa z/c}$. The amplitude penetration length is

A simple microscopic explanation comes from the Drude model. Let electrons of charge $-e$, mass $m$ and number density $n_e$ move against an effectively stationary positive background. Collisions relax their mean velocity at rate $\nu$:ConsequentlyWith a vacuum-like bound-charge background, this givesIts imaginary part is positive for positive $\omega ,\nu$, consistent with dissipation under the chosen phasor convention. Real materials can require an effective electron mass, a nontrivial bound-charge response, or additional absorption associated with the electronic band structure.

A plasma is a conducting ionized medium with mobile electrons and ions, approximately neutral on large scales because of plasma screening. For the electron-frequency transverse waves considered here, the heavy ions can often be treated as stationary. In a dilute weakly collisional plasma, $\nu \ll \omega$, soFor $\omega >{\omega }_p$, $k$ is real in this collisionless model and the wave propagates. Its phase velocity is $v_{\mathrm{ph}}=c/n>c$, while its group velocity is $v_g=cn<c$; their product is $c^2$. A phase crest exceeding $c$ does not imply superluminal information transport. For $\omega <{\omega }_p$, the refractive-index square is negative andThe transmitted field is evanescent rather than a traveling bulk wave. A sufficiently thick collisionless plasma reflects such radiation without Ohmic absorption. Collisions make the response complex and permit absorption. This accounts for a plasma cutoff and, qualitatively, the reflection of some radio frequencies by ionized atmospheric layers. The simple model describes transverse waves; plasmas also support longitudinal collective modes.

In a metal, positive ions usually form a crystal lattice. Mobile conduction electrons occupy states described by the electronic band structure and normally form a degenerate Fermi gas. The conduction-electron density is usually much higher than in a dilute gas plasma. Thus its plasma frequency is commonly in the optical or ultraviolet range, rather than the lower radio-frequency values possible in dilute plasmas. Electron scattering from the lattice, defects and other excitations gives a finite relaxation rate. The distinction is not that every plasma is collisionless and every metal collisional: the controlling ratio for either material is $\omega /\nu$, together with its carrier density and bound-charge response.

At sufficiently low frequency, $\omega \ll \nu$, the metal's conductivity is approximately real, ${\sigma }_0=n_e{e}^2/(m\nu )$. In a good conductor the conduction current dominates the displacement current, ${\sigma }_0\gg \omega {ϵ}_b$. The wave equation then givesFor example, a tangential electric field varies asThe current $\mathbf{J}={\sigma }_0\mathbf{E}$ is therefore concentrated within a few skin depths of the surface: this is the skin effect. As frequency increases within this regime, $\delta \propto {\omega }^{-1/2}$. In a wire whose radius greatly exceeds $\delta$, only a thin outer region carries appreciable alternating current, increasing the effective resistance. In a conducting shield, the same penetration suppresses the field transmitted through a sufficiently thick wall.

The Poynting theorem identifies the loss as Joule heating. The cycle-averaged heating density for a complex conductivity isThe wave impedance is $Z=\mu \omega /k$. For incidence from vacuum at normal incidence, the electric reflection amplitude is $r=(Z-Z_0)/(Z+Z_0)$, where $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$. A good conductor has small $|Z|$ and therefore strong reflection, while the part of the incident power entering it is dissipated over the skin depth.

At high enough frequency that $\omega \gg \nu$, a metal's free-electron response also approaches the collisionless plasma form. The low-frequency formula $\delta \propto {\omega }^{-1/2}$ must not simply be extrapolated into that regime. Real optical behavior additionally depends on the electronic band structure and transitions between bands. Both plasmas and metals are therefore described by the same electromagnetic response framework, but their relevant densities, scattering rates and frequency regimes differ.

Use two collinear quantum point contacts in a clean two-dimensional electron gas. Surface split gates deplete the electron gas beneath them, leaving two short constrictions facing one another across an open region. The first injects a directional beam from a biased reservoir; the second collects carriers into a downstream reservoir, with electric current flowing in the same spatial direction through both constrictions. Additional surrounding contacts can absorb carriers that miss the collector. A direct collector signal at zero magnetic field, sensitive to alignment and to obstruction of the intervening path, demonstrates ballistic electron transport across the separation.

An illustrative GaAs device uses gate gaps of a few hundred nanometres and electrically tunable constriction widths of tens to a few hundred nanometres, with injector–collector separation of a few micrometres. These are typical design scales, not dimensions supplied for a particular chip in the question. The electrical width can be much smaller than the lithographic gap because depletion extends laterally. A one-mode opening has width comparable to roughly half a Fermi wavelength; it is not simply any opening below one micrometre. Low-temperature operation suppresses inelastic scattering and resolves transverse modes. The direct trajectory length should be comfortably below the relevant transport mean free path.

Collimation is provided by the smooth exit from a constriction. In adiabatic electron collimation, transverse mode action is approximately conserved, giving $p_{\perp }W\simeq \mathrm{constant}$. As the channel opens from width $W_0$ to $W_{\mathrm{out}}$, $p_{\perp ,\mathrm{out}}\simeq p_{\perp ,0}{W}_0/W_{\mathrm{out}}$. At the same time the bottom of the electrostatic potential falls, increasing longitudinal kinetic energy. Thusand the angular distribution narrows. The reciprocal collector selects this forward-directed set of trajectories. A narrow abrupt aperture by itself may diffract into a broad fan; the smooth funnel and its potential profile are important. Here “focusing” refers primarily to collimation and efficient injector–collector coupling, not a claim that every emitted trajectory converges to one exact geometric point.

In scanning gate microscopy, a conducting tip biased sufficiently negatively creates a local repulsive potential and, when strong enough, a small depleted region in the electron gas beneath it. Raster the tip and record the collector current or differential conductance. Placing the perturbation on a strongly contributing path deflects or backscatters electrons and changes collection, producing spatial contrast. A primary scanning-probe experiment records conductance changes as a negatively charged tip depletes and backscatters the two-dimensional gas.^[ref]

The resulting image is a map of transport sensitivity to the tip, not a literal nonperturbing photograph of an electron trajectory. Broad electrostatic gating of either constriction should be distinguished from local beam disruption. Weak smooth disorder can create branches even in a high-mobility sample. Where electron phase coherence is preserved, interference between alternative paths can add fringes, often with a scale comparable to half a Fermi wavelength for a round-trip path-length change. Ballistic motion and phase-coherent interference are related but distinct requirements.

A small perpendicular magnetic field curves the imaged paths through the Lorentz force, $\mathbf{F}=-e\mathbf{v}\times \mathbf{B}$. Reversing $B$ reverses the bending, and a fixed aligned collector generally loses the initially straight beam unless a curved path connects it. For a circular orbit at the Fermi velocity,For one valley with two nearly equally occupied spin species at small field, state counting in the Fermi disc gives $n_s=2\pi k_F^2/(2\pi {)}^2=k_F^2/(2\pi )$. HenceSpin polarization or extra valley degeneracy changes the density-to-wavevector relation; the stated low-field GaAs estimate uses the ordinary spin-degenerate case.

The electron mobility gives the momentum relaxation time in a parabolic-band Drude model, $\mu =e{\tau }_{\mathrm{tr}}/m^{\ast }$. ThereforeThe effective mass cancels from both ${\ell }_{\mathrm{tr}}$ and $r_c$ when expressed in terms of density and mobility. For the supplied density,With $m^{\ast }=0.067m_0$, $m_0\simeq 9.1093837139\times 10^{-31}\mathrm{kg}$, one also obtains $v_F\simeq 1.937\times 10^5\mathrm{m}{\mathrm{s}}^{-1}$ and ${\tau }_{\mathrm{tr}}\simeq 38.1\mathrm{ps}$. The requested lengths areA useful independent ratio is ${\ell }_{\mathrm{tr}}/r_c=\mu |B|=3$. The standard constants used here are explicitly tabulated, rather than inferred from the sample.^[ref]

Mobility measures momentum relaxation. Predominantly small-angle elastic scattering can occur more frequently than that transport time, and the phase-coherence time is different again. Thus the calculation supports a several-micrometre transport scale, without proving that every electron experiences literally no perturbation over that distance.

As temperature rises, the directed ballistic collector signal decreases. Increased electron-phonon scattering and other inelastic processes reduce survival of the directed beam; thermal averaging broadens the mode and angular selection, and interference contrast also diminishes. A diffuse background current can remain, so disappearance of beam contrast need not mean all current through the device vanishes.

In a direct optical transition, a photon promotes an electron between the valence band and conduction band at essentially the same crystal wavevector: optical photon momentum is small on the scale of the Brillouin zone. An indirect optical transition also needs a phonon or another momentum source. Near an indirect gap, phonon emission requires photon energy approximately $E_g+\hslash \Omega$, while phonon absorption permits $E_g-\hslash \Omega$ with a thermally occupied phonon.

Strong direct interband absorption needs both many allowed transitions and a nonzero optical matrix element. The joint density of states counts pairs with $E_c(\mathbf{k})-E_v(\mathbf{k})=\hslash \omega$; nearly parallel bands and critical points of this energy difference can enhance it. Dipole selection rules and polarization weight that count, so a large joint density of states alone does not guarantee a strong absorption peak.

The supplied infrared spectrum lies below the interband threshold and instead probes free-carrier response. In the Drude model, $ϵ(\omega )={ϵ}_{\infty }-{\omega }_p^2/[\omega (\omega +i/\tau )]$, with ${\omega }_p^2=n{e}^2/({ϵ}_0{m}^{\ast })$. At long wavelengths the mobile electrons screen the field and the reflectivity is high. As wavelength decreases through the screened plasma edge, the real dielectric function becomes positive and propagating light can enter. The reflectivity reaches its minimum near index matching, $ϵ\simeq 1$, then rises toward the bound-electron background reflectance.

For background ${ϵ}_{\infty }=16$, its refractive index is four and the normal-incidence Fresnel equations give $R_{\infty }=[(4-1)/(4+1){]}^2=9/25=0.36$. The short-wavelength side of the figure is approaching a reflectance of this order, consistent with that background; its shortest plotted wavelength still has a carrier correction. Here “infinity” means high frequency relative to the free-carrier scale while remaining below the remote interband resonances.

The Drude-Lorentz reflectivity minimum is near $ϵ=1$, not the plasma-edge zero $ϵ=0$. Reading the deep minimum as ${\lambda }_{\min }\simeq 22\mu \mathrm{m}$ and neglecting damping there gives ${\omega }_p^2=({ϵ}_{\infty }-1)(2\pi c/{\lambda }_{\min }{)}^2$. The carrier effective mass from a reflectivity minimum is thereforeThe graph supports only a rough read-off: shifting the minimum by $0.5\mu \mathrm{m}$ changes this mass by about five percent. The deep narrow minimum supports the weak-loss assumption $\omega \tau \gg 1$ in this region, although finite scattering explains why the long-wavelength reflectivity is below unity. The estimate also assumes one dominant carrier species, a nonmagnetic bulk sample and a nearly constant background dielectric function. If the plasma-edge zero is used instead, its denominator is ${ϵ}_{\infty }$ and its wavelength is slightly longer; the two wavelengths must not be interchanged.

The mobile electrons in a metal screen an incident optical field. In the weak-damping Drude model, $\epsilon (\omega )\simeq {\epsilon }_{\infty }-{\omega }_p^2/{\omega }^2$. Below the screened plasma frequency, the real dielectric response is negative, the wave is evanescent inside the metal, and a clean surface is highly reflective. Above the plasma edge, the response can become positive and propagating waves are allowed, explaining the qualitative transmission window at sufficiently high, often ultraviolet, frequencies.

This is a free-carrier argument, not a statement that every metal is transparent throughout the ultraviolet. Interband absorption, damping and sample thickness can still produce strong absorption above the plasma edge.

In a Hall effect measurement, a current flows along $x$ and a magnetic field is applied along $z$. The transverse Lorentz force initially deflects the mobile charges, building a transverse electric field until the net transverse current vanishes. Defining $R_H=E_y/(j_x{B}_z)$, the one-carrier Drude model gives $R_H=1/(nq)$. Thus electron-like carriers give $R_H=-1/(ne)$ and hole-like carriers give $R_H=+1/(pe)$.

A nearly filled band can conduct through holes: its missing electrons behave as positive mobile charges. Metals can have both electron-like and hole-like Fermi surface sheets. In the weak-field two-carrier approximation the Hall coefficient isThe sign depends on the carrier densities and squared mobilities; it need not identify the most numerous carrier. Experimentally the Hall voltage changes sign when the field or current reverses, allowing subtraction of unwanted longitudinal-voltage pickup. Anisotropic bands and different scattering rates complicate a literal conversion of $R_H$ into a carrier count.

For one isotropic carrier species in the Drude model, the longitudinal electrical resistivity at zero transverse electric current is ${\rho }_{xx}=1/(ne\mu )$, and the Hall electrical resistivity has magnitude $|{\rho }_{yx}|=B/(ne)$. Equivalently, the electric field balances the magnetic deflection through $E_y=v_{x}B$, while $|v_x|=\mu |E_x|$. Therefore the magnitude of the Hall angle obeysThe required ratio givesThe electron density cancels. With electric current along $+x$ and field along $+z$, electrons have negative Hall coefficient and hence $E_y<0$; the requested one-percent condition concerns the magnitude. This calculation assumes electrons dominate transport and the transverse contacts draw no electric current, so a Hall field can develop.

Use a representative conduction-electron density $n\sim 10^{29}{\mathrm{m}}^{-3}$, roughly one mobile electron per atom in a dense metal, and the bare electron mass for this estimate. The plasma frequency is thenThis is an order-of-magnitude answer because no particular metal or carrier density is specified. It corresponds to an ordinary frequency $f_p\simeq 2.8\times 10^{15}\mathrm{Hz}$, excitation energy $\hslash {\omega }_p\simeq 12\mathrm{eV}$, and free-space wavelength near $1.1\times 10^{-7}\mathrm{m}$.

The physical oscillation is collective motion of the electron gas against the positive ionic background. Displacing the electrons relative to the ions produces a restoring electric field. In the simplest uniform-displacement picture, $\ddot{x}+{\omega }_p^{2}x=0$: the frequency increases with the density of mobile electric charge and decreases with carrier inertia. Quantizing this longitudinal collective oscillation gives a plasmon.

The same scale controls the ideal collisionless Drude model optical response. Its cold collisionless plasma dielectric response is ${ϵ}_r(\omega )=1-{\omega }_p^2/{\omega }^2$. Below ${\omega }_p$, the negative relative permittivity makes a transverse electromagnetic wave evanescent, producing strong reflection; above it, the ideal model admits propagation. Real-metal collisions, interband photon absorption and background polarization modify this simple optical edge. Thus ${\omega }_p$ is a collective charge-response scale, not a single-electron scattering rate or a sharp transparency threshold for every real metal.

The Drude model treats mobile electric charge carriers as independent particles with density $n$, signed electric charge $q$, isotropic mass $m$, and momentum relaxation time $\tau$. In uniform steady fields,At zero magnetic field the drift velocity is $q\tau \mathbf{E}/m$, giving electrical conductivity and electrical resistivityThe electrical conductivity is positive for either carrier sign. Scattering dissipates directed drift energy into the environment; a finite steady electric current is compatible with a short-lived drift momentum. The independent-carrier relaxation-time approximation omits detailed band anisotropy, quantum interference and microscopic collision correlations.

For electric current along $x$ and magnetic field $B\hat{\mathbf{z}}$, the magnetic force deflects carriers toward one side of the sample. electric charge accumulates until a transverse Hall field cancels that force in an open transverse circuit. With $J_y=0$, $E_y=v_{x}B$, soElectrons have negative $R_H$ in this convention, whereas positive carriers have positive $R_H$. For a strip of thickness $d$ along $z$, the magnitude of the Hall electric potential difference across its width is $|V_H|=|IB/(nqd)|$. Thus Hall measurements determine carrier sign and, within this one-carrier model, density, while combining $R_H$ with ${\sigma }_0$ determines charge carrier mobility $\mu =|q|\tau /m={\sigma }_0|R_H|$.

Keeping both in-plane electric current components, the resistivity tensor isThis fixes the off-diagonal sign convention and shows that inversion is not componentwise reciprocation. Under the Hall boundary condition $J_y=0$, the ideal one-carrier isotropic model gives $E_x/J_x={\rho }_0$, independent of magnetic field: there is no ordinary longitudinal magnetoresistance in this minimal model. Multiple carrier types or anisotropic bands can change both this conclusion and the simple $R_H=1/(nq)$ inference. Their electrical conductivity must be added before inverting to obtain the total electrical resistivity.

Relaxation-time model. The Drude model treats mobile charge carriers as independent particles accelerated by the Lorentz force between scattering events. Collisions randomize the drift momentum toward local equilibrium and are represented by a uniform, field-independent momentum relaxation time. For the present calculation the carrier density, isotropic mass and relaxation time are fixed, the fields are uniform and stationary, and quantum orbital quantization and intercarrier correlations are neglected. The steady drift velocity obeysHere $n$ is the number of carriers of that species per unit volume, $q$ their signed charge, $m>0$ their inertial or isotropic band mass, $\tau$ the momentum relaxation time, and ${\sigma }_0$ their single-species electrical conductivity at zero field.

One-carrier tensor. With $\mathbf{B}=B\hat{\mathbf{z}}$ and $\mathbf{J}=nq\mathbf{v}$, solving for the electric field givesSince $(\mathbf{J}\times \mathbf{B}{)}_x=B{J}_y$ and $(\mathbf{J}\times \mathbf{B}{)}_y=-B{J}_x$,Thus the resistivity tensor, defined by $\mathbf{E}=\mathbf{\rho }\mathbf{J}$, isIts off-diagonal signs follow from the stated axes and signed $q$. For zero transverse electric current, $E_y/J_x=B/(nq)$ is the Hall electrical resistivity; reversing charge reverses it. The single-species longitudinal electrical resistivity is independent of $B$ in this ideal isotropic model.

Adding electrons and electron holes. Let both species have density $n$ and charges $q=\pm e$, with $e>0$. The stipulated equal zero-field electrical conductivity means equal positive charge carrier mobility $\mu ={\sigma }_0/(ne)$, even if their masses and momentum relaxation times separately differ. Put $r=1/{\sigma }_0$ and $h=B/(ne)$. Inverting each electrical resistivity matrix givesBoth species experience the same electric field, so their electric current densities add. Therefore it is their conductivity tensors, rather than their resistivity tensors, that must be summed. The Hall parts cancel andThis is compensated two-carrier magnetoresistance. Although the net Hall field vanishes, each carrier species is bent in its own direction; maintaining the total longitudinal electric current requires an increasing longitudinal field. Equal carrier numbers alone do not ensure zero Hall response if charge carrier mobilities differ; both equalities given here are used.

Large-field limit and quantum oscillations. Expanding the exact diagonal entry,For $\mu B\gg 1$,Here the last equality applies to either species with its own mass and transport time; the stipulated charge carrier mobilities give the same threshold. This is the classical strong-field condition ${\omega }_c{\tau }_{\mathrm{tr}}\gg 1$: carriers turn substantially before their drift momentum relaxes.

Observable quantum oscillations additionally require sufficiently sharp Landau levels and a sufficiently cold Fermi-Dirac distribution, approximatelyWhen ${\tau }_q\simeq {\tau }_{\mathrm{tr}}$, the disorder field scale is comparable to the onset scale for large classical magnetoresistance. The thermal condition can demand a much larger field, and when ${\tau }_q\ll {\tau }_{\mathrm{tr}}$ even the disorder criterion is stronger. The quantitative damping factors in Q2(c) show that these are crossover scales rather than sharp visibility boundaries. A large smooth $B^2$ magnetoresistance therefore does not itself demonstrate Landau level; it can exist while thermal smearing suppresses oscillations. Conversely, the asymptotic condition ${\omega }_c{\tau }_{\mathrm{tr}}\gg 1$ is stronger than merely reaching unity, so there is no universal ordering of the two experimental onset fields without temperatures and lifetime information.

Scaling function. The material's actual zero-field longitudinal electrical resistivity is ${\rho }_{\mathrm{tot}}(0)=1/(2{\sigma }_0)$. Dividing the exact answer by this value givesThe assumption that $n$ stays constant is essential to eliminating further temperature-dependent parameters. $B{\sigma }_0$ is a dimensionful scaling variable; $B{\sigma }_0/(ne)=\mu B$ is its dimensionless version. The corresponding fractional increase is $[\rho (B)-\rho (0)]/\rho (0)=(\mu B{)}^2$.

What the supplied data test. In the original PDF's left panel, the field dependence is strongly temperature dependent. In the right panel, plotting the fractional magnetoresistance against the squared rescaled field makes the displayed 125, 150, 175, 200 and 225 K curves collapse approximately onto one straight line, over the field intervals where they overlap. This supports the proposed one-parameter scaling and, on that range, the quadratic field law. The 100 K curve does not collapse: it bends away and has a different slope in the rescaled plot. The indicated scaling is therefore supported for the displayed 125–225 K data, not for the entire displayed temperature range. The graph alone does not locate a sharp transition between 100 and 125 K or establish behavior above 225 K.

This empirical collapse is Kohler's rule. It is consistent with a common temperature-dependent charge carrier mobility or relaxation scale and a fixed carrier population; it is not a unique proof of exact compensation or of one microscopic scattering mechanism. A change in relative scattering times, carrier population or Fermi-surface structure could spoil the collapse, and this figure alone does not select the cause of the 100 K departure.

There is a factor-of-two normalization convention in the printed text. The ${\sigma }_0$ introduced for each species gives ${\rho }_{\mathrm{tot}}(0)=1/(2{\sigma }_0)$, whereas the final paragraph calls $1/{\sigma }_0$ “${\rho }_0$.” If ${\rho }_0$ denotes the experimentally measured zero-field electrical resistivity, the consistent slope isIf one instead keeps the literal single-species definition ${\rho }_s=1/{\sigma }_0$, the same model has $[\rho (B)-{\rho }_s]/{\rho }_s=-1/2+[B/{\rho }_s{]}^2/(2n^2{e}^2)$. The choice rescales and offsets the axes by temperature-independent constants, so it does not change which temperature curves collapse, but it must be resolved before inferring a carrier density from the slope.