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For degenerate carriers with approximately elastic scattering and a common transport relaxation time, $\sigma =n{e}^2\tau /m^{\ast }$ and ${\kappa }_e=(C_e/V)v_F^2\tau /3$. Using $C_e/V={\pi }^{2}n{k}_B^{2}T/(2E_F)$ and $v_F^2=2E_F/m^{\ast }$ givesThis is the Wiedemann–Franz law. The same mobile electrons carry charge and heat, but Pauli statistics set the electronic heat capacity and hence the Lorenz number. The relation applies to electronic thermal conductivity, not automatically to total electrical conductivity including phonons. Inelastic electron-phonon scattering can relax energy and charge currents differently; nondegeneracy and strong energy dependence of scattering also limit the simple form.

A meaningful test combines a four-terminal electrical measurement with a steady thermal-gradient measurement on the same sample and temperature range. Contact heat resistance, parasitic heat leaks, phonon conduction and sample geometry must be controlled before interpreting a measured Lorenz ratio as an electronic departure from the law.

A carrier in a locally parabolic isotropic band obeys $m^{\ast }\dot{\mathbf{v}}=q\mathbf{v}\times \mathbf{B}$, soIlluminate a high-mobility crystal with a transverse alternating electric field, sweep a calibrated static magnetic field at known microwave or far-infrared frequency, and detect resonant absorbed power. Cooling lengthens the relaxation time; resolving the resonance generally needs ${\omega }_c\tau \gg 1$. Varying field direction measures band anisotropy, while circular polarization distinguishes the two senses of rotation for opposite carrier charges. This is orbital motion rather than electron spin resonance.

The classic silicon/germanium measurements used single crystals near $4\mathrm{K}$, light modulation and circularly polarized microwaves, with directional scans to infer anisotropic carrier-energy surfaces.^[ref] Carrier generation, calibration, scattering linewidth and inhomogeneity must be separated from a genuine mass shift.

In a general band the cyclotron mass is an orbit property, $m_c={\hslash }^2(\partial A_k/\partial E)/(2\pi )$ for the appropriate field-normal orbit. It need not equal every component of the band-curvature mass tensor. For an ellipsoidal parabolic band with field along a principal $z$ direction it is $\sqrt{m_x{m}_y}$. Thus a single resonance does not determine a whole effective-mass tensor without angular information.

The mass dependence of fixed-barrier tunnelling follows by holding energy and the potential-energy barrier fixed: quantum tunnelling has the leading WKB approximationThe alpha-particle mass is about $7.3\times 10^3$ electron masses, so its decay exponent is about 85 times larger for the same $V-E$ profile. Exponential suppression therefore makes transmission much smaller. This comparison holds the potential energy and incident energy fixed: the same electrostatic potential is not the same potential-energy barrier for charges of different magnitude and sign.

The remaining cyclic commutators are $[S_y,S_z]=i\hslash S_x$ and $[S_z,S_x]=i\hslash S_y$. In a spin-$s$ irreducible representation,For spin one-half, $S_i=\hslash {\sigma }_i/2$, appropriate to an electron's spin. Each $S_i^2={\hslash }^{2}I/4$, so this Hamiltonian is $H=A{\hslash }^{2}I/4$. The orthonormal vectors $(1,0{)}^T$ and $(0,1{)}^T$ are eigenstates with the same energy $A{\hslash }^2/4$; every normalized superposition is also an eigenstate. The $B$ term vanishes identically because spin-one-half quadratic anisotropy is constant.

For the supplied three-dimensional basis one consistent choice isThese Hermitian matrices obey the cyclic commutators with the given $S_x$, and $S^2=2{\hslash }^{2}I$, representing spin one. Their squares giveThus the three normalized Cartesian basis vectors $e_1,e_2,e_3$ have energies ${\hslash }^2(A-B)$, ${\hslash }^2(A+B)$ and $0$, respectively; degeneracies permit arbitrary orthonormal choices in the corresponding eigenspaces.

The parallel and antiparallel $S_x$ states are $|+x⟩=(0,1,-i{)}^T/\sqrt{2}$ and $|-x⟩=(0,1,i{)}^T/\sqrt{2}$. Evolution from the first givesThis spin-one reversal under quadratic anisotropy is complete whenprovided $A+B\ne 0$. If $A+B=0$, its initial state is stationary up to phase and complete reversal never occurs. Assume real $A,B$ for a Hermitian Hamiltonian.

The anisotropic quantum harmonic oscillator is a sum of three commuting one-coordinate oscillator Hamiltonians. Substituting $\psi =X(x)Y(y)Z(z)$ and dividing by $XYZ$ gives a sum of three functions of independent coordinates, so each equals a separation constant. ConsequentlyFor a time-independent operator, conservation for arbitrary states follows from $[H,Q]=0$, since $d⟨Q⟩/dt=(i/\hslash )⟨[H,Q]⟩$. The first labelled operator is $i{L}_z/\hslash$; the second is $-L^2/{\hslash }^2$. The kinetic energy is rotationally invariant. The potential is invariant under rotations about $z$ precisely when (i) ${\omega }_x^2={\omega }_y^2$; full rotational invariance requires (ii) ${\omega }_x^2={\omega }_y^2={\omega }_z^2$. For positive oscillator frequencies these are equality of the frequencies themselves. Acting with the commutators on arbitrary test functions excludes anisotropic potentials in case (ii).

In the isotropic case let $b=m\omega /\hslash$. ThenThe polynomial $x$ is the first excited oscillator factor and the other two factors are ground-state Gaussians. The first excited spatial level has three independent choices, proportional to $x,y,z$. Odd parity gives $⟨x⟩=0$, and Gaussian integration yieldsFor oscillator uncertainty and allocation of classical energy, the $x$ mode has energy $3\hslash \omega /2$, whose classical amplitude is $A_x=\sqrt{3\hslash /(m\omega )}$, giving $\Delta x=A_x/\sqrt{2}$. If “same energy” instead means assigning the entire three-dimensional energy $5\hslash \omega /2$ to one classical coordinate, its amplitude is $\sqrt{5\hslash /(m\omega )}$ and the ratio is $\sqrt{3/10}$. A general three-dimensional classical orbit's $x$ amplitude is not fixed by total energy without specifying its allocation among modes.

The capacity of an isotropic oscillator shell, including the two electron spin states, is $(N+1)(N+2)$ electrons. Nine non-interacting electrons occupy shells as $2,6,1$ for $N=0,1,2$. ThusThe least positive excitation promotes an electron by one oscillator quantum. Rearrangements within the partly filled degenerate shell remain ground states and cost no energy.

A pure quantum state is a unit ray in a complex Hilbert space: normalization fixes total probability, a common phase is unobservable, and superpositions retain relative phases. Observables are self-adjoint operators. For an ideal projective measurement with spectral projector $P_a$, the Born rule gives $p(a)=⟨\psi |P_a|\psi ⟩$ and the Lüders rule gives conditional state $P_a\psi /\sqrt{p(a)}$. This preserves an eigenspace, not necessarily a unique vector when the outcome is degenerate. Between measurements, a closed system evolves unitarily according to $i\hslash {\partial }_t\psi =H\psi$. Composite systems use tensor products; quantum statistics additionally restrict identical-particle states. These are a framework supported by many experiments, not separate logical deductions from any one experiment.

In the photoelectric effect, monochromatic light illuminates a clean metal and a retarding potential measures the largest emitted electron kinetic energy. frequency controls the stopping energy, $e{V}_s=h\nu -\Phi$, while in the one-photon regime intensity mainly controls the emission rate. A threshold frequency and a stopping-energy slope $h/e$ support quantized energy exchange rather than gradual absorption of arbitrary classical-wave energy by each electron. Contact potentials, surface contamination, dark current and spectral impurities require controls. Robert Andrews Millikan tested the frequency–stopping-potential relation with careful surface and electrical control.^[ref]

The Stern-Gerlach experiment sends a collimated neutral silver beam through an inhomogeneous magnetic field onto a detector. The force is approximately $F_z={\mu }_z{\partial }_{z}B$ in an appropriate strong-field electronic description. Two resolved components rather than a continuous classical orientation spread support discrete angular-momentum projections. Field-off beam profiles and magnetic-gradient/beam-width controls distinguish splitting from ordinary broadening. In the modern description the silver ground electronic state has $J=1/2$; the 1922 experiment preceded the electron-spin interpretation. Ideal sequential analyzers along different axes illustrate incompatible observables and state preparation, but were not the original apparatus.^[ref]

In the Davisson–Germer experiment, accelerated electrons scatter from a nickel single crystal and an angularly scanned collector measures the scattered intensity. Changing voltage and crystal orientation changes narrow diffraction maxima. The inferred wavelength agrees with $\lambda =h/p$, with nonrelativistic $p=\sqrt{2meV}$. vacuum, crystal cleaning, energy selection and background measurements matter: adsorbed surface gas can produce additional diffraction features. This supports matter-wave amplitudes and interference, with measured intensities related to their squared magnitude; it does not mean an electron must follow a classical orbit through the apparatus.^[ref] Together these experiments motivate quantized outcomes, wave amplitudes and probabilistic measurement while leaving a distinction between observed evidence and the general postulates used to organize it.

In the Hall effect, the Lorentz force deflects drifting charge carriers until a transverse electric field balances it: ${\mathbf{E}}_H=-{\mathbf{v}}_d\times \mathbf{B}$. Choose current along $x$, field along $z$ and define $R_H=E_y/(j_x{B}_z)$. Since $j_x=nq{v}_x$,The sign is negative for electrons with this convention. If the sample's mass density is ${\rho }_m$ and its molar mass is $M$, the atomic number density is $n_{\mathrm{at}}={\rho }_m{N}_A/M$, soMeasure the transverse voltage, specimen thickness, current and magnetic field; reversing field and current helps remove longitudinal-voltage pickup and offsets. This free-electron count from a Hall coefficient assumes one approximately free carrier band. Multiple bands, unequal mobilities or a complicated Fermi surface can make a Hall coefficient differ from $-1/(ne)$.

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.

Write the one-electron Coulomb energies as $E_n=-C/n^2$, $C>0$. The Pauli exclusion principle allows $2n^2$ electrons in shell $n$, including spin. The ground configuration therefore has two electrons in $n=1$ and seven in $n=2$, giving $E_g=-C(2+7/4)=-15C/4$. The least energy-increasing promotion is from $n=2$ to $n=3$:Rearranging the partly occupied degenerate $n=2$ shell does not produce a higher energy. Both total energies are negative, so this ratio is below one even though $E_e>E_g$.

Set $k=\sqrt{2mE}/\hslash$ and $\kappa =\sqrt{2m(U-E)}/\hslash$. For unit incident amplitude, use $\psi =e^{ikx}+r{e}^{-ikx}$ on the left and $\psi =t_s{e}^{-\kappa x}$ on the right; the growing exponential is excluded at infinity. Continuity at zero gives $1+r=t_s$ and $ik(1-r)=-\kappa t_s$. ThusThe phase is chosen in $(-\pi ,0)$ for $0<E<U$. The evanescent wave is $\psi (x)=2k{e}^{-\kappa x}/(k+i\kappa )$ for $x>0$. Its probability density is nonzero, but its probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ is zero. Penetration into a forbidden region therefore does not require transmitted flux through a semi-infinite step.

For a finite rectangular barrier, matching both interfaces givesThe approximation requires $\kappa a\gg 1$, or a width much greater than the evanescent penetration length. The tunnelling exponent is $-2a\sqrt{2m(U-E)}/\hslash$; its factor of two converts an amplitude decay to a probability decay. For the electron parameters, $\kappa =1.984\times 10^9{\mathrm{m}}^{-1}$ and $\kappa a=19.84$. ThereforeKeeping only the exponential gives $5.83\times 10^{-18}$; the matched-interface prefactor accounts for the factor of three.

The Pauli exclusion principle confines thermally excited conduction electrons to an energy interval of order $k_{B}T$ around the Fermi surface. Their excess energy is therefore proportional to $T^2$, giving an electronic heat capacity $\gamma T$. More precisely, $\gamma =({\pi }^2/3)k_B^2{g}_{\mathrm{mol}}({ϵ}_F)$. In a three-dimensional Debye model, acoustic phonons have density of states proportional to ${\omega }^2$; their thermal energy scales as $T^4$, givingper mole of atoms. This assumes a normal metal without a lower-temperature superconducting or magnetic anomaly.

For ${\theta }_D\ll T\ll T_F$, the lattice approaches the Dulong-Petit value, so $C_V\simeq 3R+\gamma T$, usually dominated by $3R$. Each classical vibrational mode contributes $k_B$ through kinetic and potential equipartition. This is a high-temperature solid-state regime, not a claim that the solid persists to arbitrarily high temperature; anharmonicity and $C_P-C_V$ can matter.

The body-centred cubic cell has two atoms, so $n_{\mathrm{at}}=2/a^3$. Counting three acoustic branches gives $k_D^3=6{\pi }^2{n}_{\mathrm{at}}$. By the Debye-temperature definition,This is an effective Debye sound speed; separate longitudinal and transverse speeds are not determined. For isotropic branches, $3/v_D^3=1/v_L^3+2/v_T^3$.

For divalent atoms, $k_{F}a=\sqrt{4\pi }$. The circle has the same area as the first Brillouin zone but crosses its faces and leaves its corners unfilled. Thus the free-electron case remains conducting: filling capacity alone does not remove band overlap. A nearly free-electron conductor has reconstructed electron and hole pockets; a divalent band insulator instead has one completely filled band separated from all higher bands, with no Fermi surface in the gap.

For two electrons per primitive cell,Set $A={\hslash }^2{\pi }^2/(2m_e{a}^2)=2.350\mathrm{eV}$. In the indirect gap in a uniformly split zone-boundary model approximation stipulated here, the free boundary energies are $A$ along $[10]$ at $X=(\pi /a,0)$ and $2A$ along $[11]$ at $M=(\pi /a,\pi /a)$. Opening a symmetric gap givesThe branches bend away from the free-electron crossings and flatten at their boundary extrema. Importantly, the upper band near $X$ lies below the lower band's corner energy:Therefore the crystal is metallic despite its positive boundary gaps; electron and hole occupations redistribute across overlapping bands. positive direct gaps can coexist with band overlap.

A microscopic square-lattice corner has additional plane-wave degeneracy. The quoted symmetric boundary numbers use the question's directional gap model, not an unspecified four-state corner Hamiltonian. For the indirect gap of a separable square cosine potential, a separable weak potential with the same face gap gives the lowest corner energy $2A-E_g=3.70\mathrm{eV}$, still above the $2.85\mathrm{eV}$ upper-band face energy. This refinement preserves the metallic conclusion.

Donor doping supplies conduction electrons to n-type semiconductor; acceptor doping supplies holes to p-type semiconductor. In the nondegenerate, fully ionized limit, majority densities are approximately $n_n=N_D$ and $p_p=N_A$, while minority densities follow $np=n_i^2$. On contact, majority carriers diffuse across the junction, exposing positive donor ions on the n side and negative acceptor ions on the p side. The depletion-region electric field points from n to p and opposes further diffusion. At thermal equilibrium the Fermi level is uniform and drift balances diffusion, with no net current.

For an abrupt junction with uniform doping and permittivity ${\epsilon }_s$, charge neutrality gives $N_D{x}_n=N_A{x}_p$. Poisson's equation then givesPositive applied $V$ means forward bias: it lowers the barrier and narrows the depletion region. Most depletion width lies in the more lightly doped side. The formula assumes the depletion approximation and fails once its operating assumptions break down.

forward bias increases minority densities at the depletion edges by the factor $e^{qV/k_{B}T}$. In the neutral material, the steady diffusion–recombination equation gives exponentially decaying excess minorities, with lengths $L_n=\sqrt{D_n{\tau }_n}$ and $L_p=\sqrt{D_p{\tau }_p}$. Taking the carrier gradients at the junction yieldsThis ideal diode law assumes low injection, nondegenerate statistics, negligible depletion-region recombination and series resistance, and sufficiently long neutral regions. Its exponential comes from the bias-dependent injection barrier; there is no universal abrupt “turn-on” voltage. reverse bias suppresses injection, giving an approximate $-I_S$ current until additional leakage or breakdown matters.^[ref]

Rectification uses the asymmetric current-voltage characteristic; controlled reverse breakdown enables voltage references, with current limiting. Bias-dependent depletion capacitance enables varactors. photodiodes and solar cells collect optically generated carriers through the junction field and diffusion toward it; efficient light-emitting diodes use radiative electron–hole recombination, preferably in a direct-gap material. Series resistance, high injection, trap-assisted recombination, heating and finite carrier storage explain departures from the ideal steady-state law.

Exchange constraint. Exchange of two identical particles cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the exchange operator obeys $P_{12}^2=1$, so its allowed eigenvalues are $\pm 1$. bosons use the symmetric overall wavefunction, $P_{12}\Psi =+\Psi$; fermions use the antisymmetric overall wavefunction, $P_{12}\Psi =-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the Spin-statistics theorem.

Factored fermion states. If $\Psi ={\psi }_{\mathrm{space}}{\chi }_{\mathrm{spin}}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$; the latter factors are the three triplet states. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.

hydrogen spin isomers. The two protons are identical spin-one-half fermions. parahydrogen has total nuclear spin $I_{\mathrm{n}}=0$, the antisymmetric singlet; orthohydrogen has $I_{\mathrm{n}}=1$, the symmetric triplet. Here $I_{\mathrm{n}}$ is a spin quantum number, distinct from the moment of inertia used below.

Rotational restriction. In the electronic and vibrational ground state, these two factors are symmetric under nuclear exchange. Exchanging the protons reverses the internuclear axis, $(\theta ,\varphi )\mapsto (\pi -\theta ,\varphi +\pi )$. The spherical harmonic rotational state contributes the exchange sign $(-1{)}^{\ell }$. Overall antisymmetry therefore requiresHence parahydrogen has even $\ell$; orthohydrogen has odd $\ell$. The $\ell$ here labels molecular rotation, rather than the angular momentum obtained by also coupling the nuclear spins.

Conversion heat. The ortho-para conversion heat follows from the linear rigid rotor approximation, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ with $I=\mu r^2\simeq m_p{r}^2/2$. Thus conversion of one molecule from $\ell =1$ to $\ell =0$ releasesNeglecting the small electron contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ molecules. Only the initial orthohydrogen fraction releases this energy, givingThe estimate uses $r=74$ pm, $m_p\simeq 1.67262\times 10^{-27}$ kg and $\hslash \simeq 1.05457\times 10^{-34}$ J s. Handbook-level rounding and the rigid rotor approximation justify about three significant figures, not a high-precision thermochemical prediction.^[ref]

Storage consequence. The internal conversion is a delayed heat source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external heat transfer but does not remove this source. Converting to predominantly parahydrogen during liquefaction, while removing the released heat, reduces the subsequent storage heat load. No evaporation fraction can be calculated without a latent heat and a model for how the heat is removed.

With two coherent paths, the probability amplitude at a screen is ${\psi }_1(x)+e^{i\varphi }{\psi }_2(x)$. The Born rule givesThe final term shifts when the relative phase changes, producing interference fringes. For slit separation $d$ in the far field, the phase difference is approximately $2\pi d\sin \theta /\lambda$; constructive interference occurs at $d\sin \theta =n\lambda$. Finite slit widths add a diffraction envelope, so an ideal equally bright, infinitely narrow slit model does not predict the full observed envelope.

single-electron interference pattern buildup starts with isolated localized detector events at sufficiently low source intensity. Repeated events accumulate the interference probability distribution. Thus detection is discrete while propagation is described by coherent probability amplitudes. The effect cannot be explained by interference between different particles if the apparatus is occupied by at most one particle at a time. A motivating realization used a field-emission electron source, an electron biprism to recombine two paths, electron lenses and a position-sensitive electron counter; it recorded individual arrivals building up the fringes. This was a biprism realization of two-path interference, not a plate with two literal slits.^[ref]

The crucial comparison is between both coherent paths open and a preparation that supplies which-path information. If the paths become correlated with detector states $|D_1⟩,|D_2⟩$, the joint state is ${\psi }_1|D_1⟩+e^{i\varphi }{\psi }_2|D_2⟩$. Ignoring that detector multiplies the interference term by $⟨D_1|D_2⟩$. Orthogonal detector states remove the term completely, while identical detector states preserve it. This is a consequence of entanglement and the partial trace; a human need not read the detector.

In a real test, check the single-path distributions, detector background, source coherence and stability of the relative phase. Classical phase drift can wash out fringes without supplying which-path information. Conversely, localized detections alone do not establish coherent two-path propagation; the controlled phase dependence of their accumulated distribution provides the discriminating evidence.

shot noise arises from fluctuations in discrete transfer events. For equal charges $q$ arriving independently with a constant Poisson rate $\lambda$, the charge collected during $\tau$ is $Q=qN$, withThus the relative counting uncertainty is $1/\sqrt{\lambda \tau }$. Longer integration improves the precision of the mean current even when the microscopic arrivals remain irregular.

In the frequency range where independent arrivals can be treated as instantaneous pulses, the one-sided current power spectral density isFor an ideal unit-gain rectangular bandwidth $B$, this gives $i_{\mathrm{rms}}=\sqrt{2qIB}$. boxcar averaging of Poisson shot noise over duration $\tau$ has effective noise bandwidth $B_{\mathrm{n}}=1/(2\tau )$, reproducing the counting result $i_{\mathrm{rms}}=\sqrt{qI/\tau }$. The two-sided spectral convention instead assigns $qI$ to each sign of frequency. Stating that convention prevents a factor-of-two error.

A photodiode illuminated at a stable rate provides a useful measurement: subtract the mean current, amplify its fluctuations through a calibrated linear filter, and compare the variance with the mean photocurrent. Independent detected photoelectrons predict a variance proportional to mean current. Measure the dark and amplifier backgrounds separately, check the filter's effective noise bandwidth and vary illumination to distinguish this term from classical intensity noise, whose variance can scale quadratically with mean current. The shot noise model concerns counting statistics, whereas thermal electrical noise can exist in an unbiased resistor. Walter Schottky's motivating analysis treated amplifier limits and independent electron transfers in thermionic discharge tubes.^[ref]

The Poisson result is a model, not a universal consequence of charge quantization. Correlated arrivals, detector dead time, avalanche multiplication or a non-Poisson light source can change the fluctuations. Filtering reduces integrated noise at the cost of temporal resolution. Increasing the detected event rate reduces relative shot noise but raises its absolute variance; balanced photodetection cancels shared classical fluctuations, not the independent shot noise of two uncorrelated detectors.

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 $2\pi$.

A completely filled valence band carries no net current: its occupied crystal-momentum states have velocities that cancel. Near a valence-band maximum, an almost full band is conveniently described by the few missing electrons, or electron holes. Removing an electron of charge $-e_0$, with $e_0>0$, changes the charge relative to the filled band by $+e_0$. This is a quasiparticle description of electronic motion, not a mobile positive ion or a Positron.

For an isotropic parabolic maximum, ${ϵ}_v(\mathbf{k})=E_v-{\hslash }^2{k}^2/(2m_h^{\ast })$. Assign the missing-electron state momentum ${\mathbf{k}}_h=-{\mathbf{k}}_e$ and hole energyIt has positive effective mass $m_h^{\ast }$. Its velocity equals the velocity of the missing electron state, and its positive-charge current replaces the negative of that state's electronic current. Semiclassical motion is thus $\hslash {\dot{\mathbf{k}}}_h=+e_0(\mathbf{E}+{\mathbf{v}}_h\times \mathbf{B})$. With momentum relaxation time ${\tau }_h$, the low-field mobility is ${\mu }_h=e_0{\tau }_h/m_h^{\ast }$ andIf electrons also contribute, $\sigma =e_0(p{\mu }_h+n{\mu }_e)$: the two species' longitudinal conductivities add.

A Hall bar separates carrier concentration from mobility. Apply current density $j_x$, magnetic field $B_z$, and measure transverse electric field $E_y$. For one isotropic hole population, transverse force balance gives $E_y=v_x{B}_z$, soFor thickness $t$, the Hall-voltage magnitude is $|V_H|=IB/(e_{0}pt)$. A longitudinal four-terminal resistance measurement supplies $\sigma$ independently. Use ohmic contacts, known conducting thickness, controlled temperature and low enough current to avoid self-heating. Reversing magnetic field removes even-in-field longitudinal offsets from contact misalignment; reversing current also helps reject thermoelectric offsets. These are central controls, rather than merely taking one transverse voltage reading.^[ref]

The simple inference assumes one carrier type and a known Hall scattering factor. More generally $R_H=r_H/(e_{0}p)$, so $\sigma R_H$ measures Hall mobility rather than automatically the drift mobility. With two carrier populations at weak field,Its sign alone need not identify the more numerous carrier. A Hall experiment also cannot determine $m_h^{\ast }$ and ${\tau }_h$ separately: their ratio sets mobility. cyclotron resonance supplies an independent mass measurement by sweeping frequency or magnetic field through $\omega =e_{0}B/m_c$ and detecting resonant absorption. The measured cyclotron effective mass can depend on band anisotropy and need not equal every curvature mass; the isotropic parabolic model identifies them. Its linewidth provides a scattering-time estimate only with a specified damping model.

semiconductor doping generates holes by adding acceptor impurities. In silicon, a substitutional group-III atom such as boron provides one fewer valence electron than silicon. A shallow acceptor level lies slightly above the valence-band edge: accepting a valence electron makes the dopant negatively ionized and leaves a mobile hole in the valence band. The impurity stays fixed in the lattice. charge neutrality and the nondegenerate equilibrium mass-action law giveFor fully ionized, uncompensated acceptors in the extrinsic regime, $p\simeq N_A$ and $n\simeq n_i^2/N_A$; with fully ionized donor compensation, $p\simeq N_A-N_D$ while holes remain dominant. The Fermi level moves towards the valence band. At low temperature, semiconductor carrier freeze-out suppresses mobile electron holes; at high temperature intrinsic electron-hole generation can dominate. Increasing dopant density also changes scattering, so a conductivity increase alone cannot distinguish increased carrier density from changed mobility. Temperature-dependent Hall and conductivity measurements separate these effects, subject to the carrier and scattering qualifications above.

The effective mass follows from the curvature of the electronic band structure. Differentiating twice givesAt either first Brillouin zone boundary $k=\pm \pi /a$,The boundary is a band maximum, with negative curvature and zero group velocity. In the semiclassical single-band description, an applied force changes $\hslash k$; the resulting acceleration is opposite the force near that maximum. This is a band-response coefficient, not a negative microscopic electron rest mass. A nearly filled band is often described more conveniently through positively charged electron holes near its maximum.