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A normal Fermi liquid describes an interacting fermion system whose low-energy excitations remain continuously connected to adding particles to or removing particles from a Fermi gas. The elementary long-lived excitation is a Landau quasiparticle: adding a particle also changes the surrounding many-body state. Its energy, velocity and interactions are renormalized, even though its electric charge and spin remain those appropriate to the excitation.

Near an isotropic Fermi surface,Residual interactions make the quasiparticle energy depend on the occupations of other quasiparticles, expressed by the Landau quasiparticle energy functional. Its angular components give the Landau parameters controlling density and spin responses. Thus the successful particle picture does not mean that interactions have become zero.

Pauli blocking restricts the final states available for low-energy collisions. In an ordinary three-dimensional Fermi liquid with regular interactions, a representative decay law isAt $T=0$, the width vanishes faster than $|\xi |$, so an excitation becomes increasingly sharp as it approaches the Fermi surface. Its coherent spectral weight is the quasiparticle residue $Z$, and the physical electron occupation has a discontinuity of height $Z$ at zero temperature. Incoherent spectral weight accounts for the remainder. The quasiparticle occupations themselves obey a Fermi-Dirac distribution.

Several leading thermodynamic signatures retain their gas-like form but have interaction-dependent coefficients. In particular,for an isotropic spin-one-half model, with $N^{\ast }(0)$ the spin-summed quasiparticle density of states per volume. effective mass changes $N^{\ast }(0)$, while the dimensionless interaction parameters $F_0^a,F_0^s$ separately change the spin susceptibility and compressibility. Collisionless collective density motion can propagate as zero sound, distinct from the single-particle excitations.

The $T^2$ quasiparticle scattering rate often underlies a low-temperature $T^2$ contribution to metallic electrical resistivity, but a mechanism relaxing electrical electric current is needed. In a clean Galilean-invariant continuum, electron-electron collisions conserve total momentum and do not alone give a finite dc electrical resistivity. Crystal Umklapp scattering or other current-relaxing processes change that conclusion.

Normal liquid helium-3 and many metals motivate the Fermi-liquid framework. Pairing, magnetic ordering or other transitions can reconstruct the low-energy states. Generic interacting one-dimensional fermion systems instead form a different low-energy state, so the three-dimensional quasiparticle conclusions should not simply be transferred to a chain.

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

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

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

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

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

Applied bias. Define positive $V$ as forward bias, raising the p-side terminal potential relative to the n-side terminal. It lowers the remaining barrier from $\Delta {\varphi }_0$ to $\Delta {\varphi }_0-V$. Reverse bias has $V<0$ and raises that barrier. The same Poisson integration applies, with this replacement:Forward bias narrows the depleted region; reverse bias widens it. Under bias there need not be one equilibrium chemical potential across the device, so the equilibrium flat-level argument must not be reapplied unchanged. The barrier/width change is illustrated in the next figure.

Differential capacitance and electric charge sign. Put $B=\Delta {\varphi }_0-V>0$ and $N_{\mathrm{eff}}=N_A{N}_D/(N_A+N_D)$. The signed p-side fixed electric charge per unit area isDifferentiating with respect to forward electric potential difference gives the positive capacitance per unit areaThe n-side electric charge is $Q_n=-Q_p$, so $d{Q}_n/dV=-C_A$ instead. If $Q$ denotes a positive space-charge magnitude, the positive capacitance is $-dQ/dV$ for this forward-voltage convention. This resolves the sign implicit in the paper's positive $dQ/dV$. The total device capacitance is $C=\mathcal{A}{C}_A$, where $\mathcal{A}$ is its cross-sectional area.

Finite device and numerical sketch. For the symmetric device, $N_A=N_D=N$ and each side has thickness $d$. Before full depletion,Using ${\epsilon }_s=12{ϵ}_0=1.0625\times 10^{-10}\mathrm{F}{\mathrm{m}}^{-1}$, the zero-bias width on each side is $4.07\mu \mathrm{m}$, less than $d=10\mu \mathrm{m}$. The total zero-bias capacitance is $\boxed{C(0)\simeq 1.30\mathrm{pF}}$.

Full depletion occurs when $w=d$, soAt larger reverse bias the depleted thickness cannot exceed $2d$. In the ideal fully depleted slab, incremental terminal electric charge changes the field uniformly through that fixed thickness, giving the full-depletion capacitance of a symmetric junction slabThus the idealized piecewise sketch over the specified range isIt is continuous at full depletion, rises through $1.30\mathrm{pF}$ at zero bias, and has a formal inverse-square-root divergence as $V\to 0.5\mathrm{V}$ from below.

There are two physical limits to this extension. Once all dopants are exposed, their fixed space charge no longer increases, but the terminal electrodes can still acquire incremental electric charge; differentiating only the fixed ionized electric charge would incorrectly predict zero capacitance. Near $V=\Delta {\varphi }_0$, the depletion approximation itself fails as mobile carriers and forward injection become important, so the formal infinite endpoint is not a prediction of an actual infinite measured capacitance. The fully depleted plateau assumes ideal contacts and no breakdown, injection storage or extra series capacitance.

Why degenerate perturbation theory is needed. If several unperturbed states have equal or nearly equal energy, a perturbation can mix them at leading order. In that subspace, diagonalize the perturbation matrix rather than using nondegenerate formulas with small energy denominators. The resulting eigenvectors identify the combinations selected by the coupling, and the eigenvalues give the leading splitting. Other remote states can then be included perturbatively.

For the linear combination of atomic orbitals description of molecular hydrogen, equivalent atomic $1s$ states on the two nuclei supply the nearly degenerate starting functions. In an orthonormal two-state model, equal diagonal energy $E_a$ and real hopping $t$ give $E_a\pm t$ and symmetric/antisymmetric combinations. For actual overlapping atomic functions with overlap $S$ and Hamiltonian coupling $h$, one instead solves a generalized eigenvalue problem:The bonding state has enhanced density between the nuclei and the antibonding state a separating node. Two opposite-spin electrons occupying the bonding molecular orbital describe the simplest closed-shell hydrogen-molecule bond. Electron-electron interaction and nuclear repulsion are needed to calculate a realistic total binding energy; the two-state splitting alone is not the full molecular energy.

In the nearly-free electron model, the analogous degenerate states are plane waves $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$ at a Bragg scattering. A Fourier coefficient $U_{\mathbf{G}}$ of the periodic potential couples them. At exact degeneracy $E_B$, diagonalizing their two-by-two matrix givesThe associated standing waves place electronic density differently relative to the ionic potential. These are the same gap-opening combinations illustrated in Q1(b). The labels of the two starting basis states survive as mixed orbital or plane-wave character, rather than independent crossing eigenstates.

Origin of broad and narrow bands. Spatially extended s/p-derived orbitals have substantial overlap and therefore large hopping amplitudes and broad bands. More localized d orbitals overlap less strongly and produce narrower bands. A multiorbital tight-binding calculation assigns the on-site orbital energies and hopping matrices, forms Bloch sums, and diagonalizes the resulting matrix at each wavevector. Different hopping ranges and orbital symmetry determine the bandwidths and allowed interband coupling. The two-band description retains one broad combination and one nearly flat d-derived combination; a real transition metal contains several d orbitals and more complicated dispersions.

Hybridized eigenvalues and eigenstates. Let $|{\psi }_{\mathbf{k}}⟩=a|{\varphi }_{\mathbf{k}}⟩+b|{\chi }_{\mathbf{k}}⟩$, in an orthonormal basis. Hermitian operator fixes the reverse coupling to be $V^{\ast }$. Projection onto the two components givesSolving this quadratic yields the band hybridization dispersionFor nonzero $V$, one eigenvector representation is $(a,b)\propto (V,E-E_1)$, followed by normalization. Thus the wavefunctions exchange broad-band and d-band character through the avoided crossing; the levels are equal mixtures at the uncoupled crossing.

At $V=0$, the parabola crosses the horizontal level at $k=\pm k_0$, with $k_0=\sqrt{2m{E}_d}/\hslash$. For small nonzero $|V|$, the curves avoid crossing with minimum direct separation $2|V|$. The lower branch follows the parabola below the crossing and flattens toward $E_d$ above it; the upper does the opposite. The lower branch satisfies $E_{-}<E_d$ everywhere and tends to $E_d$ from below as $k\to \infty$, while the upper branch always lies above $E_d$. This is the perfectly flat-band continuum approximation, with zone-boundary intersections excluded as requested.

Fixed particle number and the larger Fermi sphere. Denote the uncoupled broad-band Fermi radius by $k_0$ to distinguish it from the new one $K$. With two spin states, the broad-band number density at $V=0$ is $k_0^3/(3{\pi }^2)$. The stipulated equal population in the flat sector makes the total density twice that value. At zero temperature and any nonzero hybridization, all occupied states fit into the monotonic lower branch below $E_d$, while the upper branch is empty. ThereforeThis is the stated factor increase. It uses fixed total particle number, equal spin counting, an exactly flat unhybridized band and enough available states below any Brillouin-zone restriction. In generic pairs of dispersive bands both hybridized pockets can remain occupied; then Fermi-volume conservation in two hybridized electron bands fixes a sum of volumes rather than this one radius. The exactly flat band is degenerate at $V=0$, so its initial occupation must be specified, as it is here.

Fermi-velocity suppression. Set $\Delta (k)=E_1(k)-E_d$. The exact derivative of the lower branch isAt $k=K$, $E_1(K)=2^{2/3}{E}_d$ and $\Delta (K)=(2^{2/3}-1)E_d>0$. For $|V|\ll \Delta (K)$, the square bracket divided by two is $|V{|}^2/\Delta (K{)}^2+O(|V{|}^4/\Delta (K{)}^4)$. Since group velocity is ${\hslash }^{-1}dE/dk$ and the broad-band reference velocity is $v_F^{(1)}=\hslash k_0/m$,so the flat-band hybridization Fermi-velocity suppression isThe reference velocity is that of the original broad-band Fermi sphere, not the broad-band slope at the larger $K$; omitting the radius change loses the factor $2^{1/3}$.

The low-temperature qualification becomes more stringent as $V\to 0$: the new chemical potential approaches $E_d$ with $E_d-\mu \simeq |V{|}^2/\Delta (K)$, and the relevant thermal energy must be small compared with this hybridization scale. At fixed nonzero temperature the singular flat-band redistribution is not automatically resolved for arbitrarily small coupling. A small but finite narrow-band dispersion can also change the occupation topology.

Density of states and magnetism. For an isotropic three-dimensional branch with spin degeneracy two,A small Fermi velocity therefore produces a large density of states. The localized, partially filled d-derived bands in iron, cobalt and nickel place many states close to the Fermi level; broad s/p-derived conduction bands in copper, silver and gold coexist with filled d bands lying below it. This explains the substantially larger Fermi-level density of states of the former group within the band picture. An exact factor of ten requires the actual band parameters and is not fixed by the two-band toy model.

A high density of states lowers the kinetic-energy cost of spin polarization. A repulsive exchange interaction can then satisfy the Stoner criterion and stabilize ferromagnetism. With a single-spin density of states $g_s$ and an on-site interaction normalization, the criterion is $U{g}_s(E_F)>1$; with spin-summed $g=2g_s$, it is $Ug(E_F)>2$. The large d-state contribution favors this instability in Fe, Co and Ni. The noble metals do not satisfy the corresponding instability condition in the same ordinary state; this does not imply that they have no weak magnetic response whatsoever.

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.

Adopt the usual orthogonal tight-binding approximation, so the three localized spatial states are orthonormal. Each site has two neighbors, with the same real hopping matrix element. For coefficients $c_n$, the Schrödinger equation at site $n$ isThe constant-coefficient state has $c_1=c_2=c_3$, giving energy $E_0+2t$. Its normalization therefore yieldsTo verify that it is the ground state, the three discrete ring momenta are $\theta =0,\pm 2\pi /3$, with $c_n\propto e^{i\theta (n-1)}$ and energies $E_0+2t\cos \theta$. The other two states both have energy $E_0-t$. Since $t<0$, $E_0+2t<E_0-t$ and the gap is $3|t|$. This is the three-site tight-binding ring spectrum.

The spatial ground state is unique and has equal probability $1/3$ on every site; its common overall phase is arbitrary. Spin would supply a separate twofold degeneracy when no spin-dependent terms are included, but does not change the requested spatial state. If the localized functions were instead nonorthogonal raw atomic orbitals, their overlap matrix would be additional data needed for normalization and a generalized eigenvalue problem.

The energy favoring spin polarization must compete with the increase in Fermi kinetic energy. Assume the usual three-dimensional zero-temperature parabolic conduction band at fixed $N$ and $V$. Write the spin polarization as $\zeta =(N_{\uparrow }-N_{\downarrow })/N$, so $N_{\uparrow ,\downarrow }=N(1\pm \zeta )/2$. The unpolarized Fermi energy isThe two spin populations have Fermi energies $E_F(1\pm \zeta {)}^{2/3}$. Summing their individual $3E_{F,\sigma }/5$ mean kinetic energies givesThe specified interaction is an energy per electron, $U(1-{\zeta }^2)/4$. Thus the energy change per electron near the unpolarized state isThe quadratic coefficient changes sign at the Stoner threshold for an interaction normalized per electron,The positive quartic term gives a continuous onset in this model. More globally, $f^{\prime \prime }(\zeta )=(E_F/3)[(1+\zeta {)}^{-1/3}+(1-\zeta {)}^{-1/3}]-U/2$. Convexity of $x^{-1/3}$ makes this nonnegative for $U\le 4E_F/3$, so no lower polarized minimum appears before that threshold.

At fixed number density $N/V$, the critical $U$ is independent of the total sample size. A comparison restricted to a fully polarized state gives a different numerical coefficient, $U>\frac{12}{5}{E}_F(2^{2/3}-1)$, but the same requested density scaling; it misses the partially polarized onset. One must also distinguish this $U$, with units of energy, from a contact coupling multiplying an interaction-energy density $g{n}_{\uparrow }{n}_{\downarrow }$: here $g=U/(N/V)$. Importing a differently normalized Stoner criterion would introduce an erroneous extra density factor.

A covalent bond is stabilized by shared electronic states extending across neighboring atomic nuclei. In a simple two-centre description, combining comparable atomic orbitals produces a bonding state with enhanced electron density between the nuclei and an antibonding state with a node there. For equivalent normalized orbitals ${\varphi }_A,{\varphi }_B$ with real overlap $S$,Their energy ordering follows from the molecular Hamiltonian; constructive interference and reduced gradients help explain why the bonding combination can be lower. Opposite-spin electrons can occupy the bonding orbital together. The familiar molecular-hydrogen bond is a basic example, while extended covalent networks form solids such as diamond. Orbital shape and hybridization make covalent bonding strongly directional and constrain bond angles. Occupying antibonding states reduces the net bond strength.

Ionic bonding instead emphasizes electric charge transfer and the electrostatic attraction of the resulting oppositely charged ions. Removing an electron from one atom costs ionization energy and adding it to another gains electron affinity; these isolated-atom terms alone need not favor transfer. In a crystal, the electrostatic lattice energy can stabilize the ionic arrangement. For a schematic lattice of singly charged ions at nearest-neighbor distance $r$,where the geometrical Madelung constant $A_M$ sums attractions and repulsions throughout the lattice. Short-range overlap repulsion prevents collapse and gives an equilibrium separation. Sodium chloride is an example of an extended ionic crystal, rather than isolated NaCl molecules in the solid.

Ideal ionic bonding is comparatively nondirectional, so packing, ion sizes and electric charge balance strongly influence coordination. ions cannot freely migrate in the ordinary solid; melting or dissolution can make ionic electric charge transport possible. Covalent materials do not obey one universal electrical conductivity rule: their electronic bands and filling determine whether mobile electronic carriers are available.

The two descriptions are limiting models, not a sharp classification of every bond. Unequal electronegativities can produce polar covalent bonds with partial electric charge transfer, and a material may contain both covalent groups and ionic interactions between those groups. Chemical bonding reflects a balance of electronic kinetic energy, electron-nucleus attraction, electron-electron repulsion and nuclear repulsion; it cannot be reduced to a force between two assigned point charges in every material.

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.

Diagonalization and conventions. The spin-dependent part is $\lambda (-k_y{\sigma }_x+k_x{\sigma }_y)$. For a Hermitian Hamiltonian $\lambda$ is real; the supplied lower-band spinor uses the convention $\lambda >0$, which will be adopted below. The squared spin matrix is ${\lambda }^2(k_x^2+k_y^2)I$. Hence the two ordered energies areIn the $k_z=0$ plane, let $k=\sqrt{k_x^2+k_y^2}$ and $q_R=2m\lambda /{\hslash }^2>0$. ThenIf $\lambda <0$, use $q_R=2m|\lambda |/{\hslash }^2$ to retain the energy ordering, and interchange the spin directions assigned below. This is an isotropic Rashba effect; $q_R$ has units of inverse length, whereas $\lambda$ has units of energy times length.

Band minima. Define $k_R=q_R/2$ and $E_R={\hslash }^2{q}_R^2/(8m)$. Completing the square givesAlong $k_y=k_z=0$, the radial coordinate is $|k_x|$, so the lower branch has two minima:In the full plane these are points on a circle of minima $k=k_R$. Both branches meet at $k=0$, $E=0$. Their ordered energy curves contain $|k_x|$: the lower curve descends from the crossing on either side, while the upper rises. Treating them as globally labeled smooth spin branches without switching labels across $k_x=0$ would obscure the lower-band minima.

Occupied regions and Fermi contours. At zero temperature each spin-split eigenstate with energy below $E_F$ is occupied. PutFor $-E_R<E_F<0$, $0<D<k_R$. The lower-band roots are $k_R-D$ and $k_R+D$, while the upper band is entirely empty. Thus the central disk $k<k_R-D$ is unoccupied, the annulus $k_R-D<k<k_R+D$ is singly occupied, and the exterior is unoccupied. Both Fermi circles belong to the lower band; there is no doubly occupied region.

At $E_F=0$, the lower band is occupied for $0<k<2k_R=q_R$, giving a singly occupied disk with outer Fermi circle at $q_R$. The two bands also meet the Fermi energy at the origin. That isolated degenerate point is not a finite doubly occupied disk, and occupation exactly at the Fermi energy is a boundary convention of zero area. Outside the outer circle the states are empty.

For $E_F>0$, $D>k_R$. The lower branch has outer radius $k_{-}=D+k_R$, while the upper branch has inner radius $k_{+}=D-k_R$. The disk $k<k_{+}$ is doubly occupied, with one state in each band at every momentum. The annulus $k_{+}<k<k_{-}$ is singly occupied, in the lower band only, and the exterior is unoccupied. These are the Fermi contours of an isotropic Rashba band; double occupancy here refers to two orthogonal spinor states at the same momentum, not two electrons occupying one identical quantum state.

Inferring the splitting. Above the crossing, the two positive radii differ by $2k_R=q_R$:The minimum radius is therefore $k_R=0.05{\mathring{\mathrm{A}}}^{-1}$. It is the difference of the measured radii that is required for $E_F>0$; below the crossing the two lower-band radii instead add to $q_R$.

Spinors and spin direction. Write $a=(k_y+i{k}_x)/k$, which has $|a|=1$ for $k>0$. An orthogonal normalized partner to the supplied lower-band state isIndeed $u_{-}^{†}{u}_{+}=(-|a{|}^2+1)/2=0$. Direct multiplication by the spin matrix gives eigenvalues $-\lambda k$ and $+\lambda k$ for $u_{-}$ and $u_{+}$, respectively, so orthogonality has selected the other band rather than merely another arbitrary spin basis.

For a normalized spinor $(\alpha ,\beta {)}^T$, the Pauli expectation values are $(2\mathrm{Re}({\alpha }^{\ast }\beta ),2\mathrm{Im}({\alpha }^{\ast }\beta ),|\alpha {|}^2-|\beta {|}^2)$. Consequently the Rashba spin texture isThe spin lies in the plane and is perpendicular to momentum. On the outer circle $k_1$, which belongs to $E_{-}$, it points clockwise as viewed from $+z$; at $\mathbf{k}=(k_1,0)$ it points along $-y$. On the inner circle $k_2$, belonging to $E_{+}$, it points counterclockwise; at $\mathbf{k}=(k_2,0)$ it points along $+y$. Opposite momenta have opposite spins, consistent with time-reversal symmetry. At the origin the eigenstates are degenerate and no unique spin orientation is defined.

Nondegenerate carrier product. Assume three-dimensional parabolic energy bands near their extrema, spin degeneracy two, and no additional uncounted valley multiplicity. Near the conduction edge, a Boltzmann approximation to the Fermi-Dirac distribution givesAngular integration and the Gaussian radial integral yieldFor valence electron holes, the empty-state probability is $1-f(E)\simeq e^{(E-\mu )/(k_{B}T)}$. Measuring electron hole kinetic energy downwards from $E_v$ similarly givesThe factors two count spin once in each band. Multiplying cancels the chemical potential and proves the semiconductor mass-action law:In an intrinsic neutral semiconductor, $n=p=n_i$. In a doped equilibrium semiconductor the same product holds under the stated nondegenerate assumptions, even when $n\ne p$. Extra valleys or several valence branches must be included in the effective density-of-states factors rather than counted inconsistently.

Equilibrium junction. Put the p region on the left and the n region on the right. Before contact their majority carriers are electron holes and electrons, respectively. Diffusion initially sends electrons from n to p and electron holes from p to n. This leaves negative ionized acceptors on the p side and positive ionized donors on the n side, creating a depletion region and an electric field directed from n to p. At equilibrium this field balances diffusion, so the net electric current vanishes.

For a homojunction the gap remains constant while the conduction band and valence band edges bend downwards from p to n by $e{V}_{\mathrm{bi}}$. The equilibrium chemical potential is flat. Since $E_{c,v}(x)=E_{c,v}^{\mathrm{ref}}-e\varphi (x)$, the electrostatic potential rises from p to n. The carrier profiles satisfyWith fully ionized nondegenerate dopants, the neutral p side has $p_p\simeq N_A$, $n_p\simeq n_i^2/N_A$, while the neutral n side has $n_n\simeq N_D$, $p_n\simeq n_i^2/N_D$. Thus the electron concentration rises across the junction and the electron hole concentration falls, with both small relative to the far-side majority densities in much of the depleted region. The ideal depletion approximation neglects these mobile populations when calculating the fixed space electric charge, but the actual equilibrium profiles are smooth.

The built-in potential follows from these Boltzmann factors:There is no applied terminal bias; the internal electrostatic potential variation does not contradict a constant electrochemical potential.

Opposing carrier fluxes under reverse bias. Use $V>0$ for the magnitude of reverse bias and $I>0$ for reverse electric current, as in the PDF. The junction barrier increases by $eV$. The Shockley diode equation in that orientation can be writtenThe reverse-directed contribution is supplied by minority electrons on the p side and minority electron holes on the n side. They reach the depletion edge by diffusion and are swept across by the built-in field. The opposing contribution is associated with majority carriers thermally crossing the barrier in the forward direction; reverse bias suppresses it by the Boltzmann factor. At zero bias detailed balance makes the two equal, while at large enough reverse bias the opposing flux becomes negligible.

In the ideal neutral-region diffusion model,Here $D_n,D_p$ are minority-carrier diffusion coefficients, $L_n,L_p$ their diffusion lengths, and $\mathcal{A}$ is junction area. The two terms inside this positive prefactor are the electron and electron hole contributions; both flow in the same reverse-current direction. This separates the two carrier-species contributions from the two opposing fluxes in the Shockley diode equation. Calling the saturation term a drift current refers to its passage across the depleted region, and does not eliminate the minority-carrier diffusion needed to supply it.

Reading the measured curves. At $eV\ll k_{B}T$, the ideal equation gives $I\simeq I_{\mathrm{sat}}eV/(k_{B}T)$, initially proportional to reverse electric potential difference. At $eV\gg k_{B}T$, it approaches a plateau $I\simeq I_{\mathrm{sat}}$. The thermal electric potential differences at the three temperatures are about $21$–$26\mathrm{mV}$, consistent with a rise at small reverse biases and near saturation at larger ones. Higher temperature produces a much larger plateau because thermal electron-hole excitation raises $n_i^2$ exponentially.

The lower-temperature curve has a noticeable residual electric potential difference dependence rather than a perfectly flat plateau. Depletion-region generation, surface leakage or other departures from the ideal diffusion model can contribute. The sharp upturn at the largest reverse electric potential differences is reverse breakdown, outside the unextended Shockley diode equation; the plot alone does not identify a unique microscopic breakdown mechanism. The original horizontal axis duplicates the final decade label, so a precise breakdown electric potential difference cannot be read reliably. Neither issue prevents estimating the gap from the intermediate-bias saturation region.

Band-gap estimate. Use plateau electric currents at a common intermediate reverse bias, away from both the small-bias rise and the breakdown upturn. Approximate readings are $I_{\mathrm{sat}}(298\mathrm{K})\sim 2.5\times 10^{-6}\mathrm{A}$, $I_{\mathrm{sat}}(273\mathrm{K})\sim 1.5$–$2\times 10^{-7}\mathrm{A}$, and $I_{\mathrm{sat}}(248\mathrm{K})\sim 10^{-8}\mathrm{A}$. Kelvin temperatures are required in the activation law.

For fixed effective masses, $n_i^2\propto T^3{e}^{-E_g/(k_{B}T)}$. If the remaining transport prefactor is approximately constant, extracting a semiconductor band gap from diode saturation current therefore usesUsing the widest temperature separation and the rounded readings givesNeglecting all algebraic prefactors and fitting $\ln I_{\mathrm{sat}}$ instead gives about $0.70\mathrm{eV}$. These differ by roughly $0.07\mathrm{eV}$, a material correction on the graph's precision scale. Thus an honest graph-based conclusion isThe nonflat low-temperature plateau, rough electric current readings, unprovided temperature dependence of $D/L$, and possible gap variation prevent a more precise extraction. If depletion-region generation dominated rather than the specified $I_{\mathrm{sat}}\propto n_i^2$ mechanism, its activation slope could involve a different energy scale; the displayed estimate explicitly uses the diffusion-dominated saturation model requested here.

For nondegenerate electrons in a parabolic conduction band,when the density-of-states effective mass and valley multiplicity are fixed. The intrinsic carrier extrapolation across a crossover must therefore specify how the activation energy changes with temperature.

The usual intended estimate treats the quoted $0.9\mathrm{eV}$ as a constant intrinsic activation energy between the two temperatures, and identifies the crossover by $n_i(T_c)\simeq n_{\mathrm{ext}}$. ThenFor $T_1=350\mathrm{K}$, $T_c=280\mathrm{K}$ and $k_B=8.6173\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$, the exponent is $-7.46$ and the prefactor is $0.8^{3/2}=0.7155$. HenceEquivalently, a fixed-gap intrinsic model with $E_g/2=0.9\mathrm{eV}$ gives this result. A crossover is gradual: if $n_{\mathrm{ext}}$ denotes the net ionized donor density and $n_i(T_c)=n_{\mathrm{ext}}$, charge neutrality gives the actual electron population $n(T_c)=(1+\sqrt{5})n_{\mathrm{ext}}/2$, rather than exactly $n_{\mathrm{ext}}$.

There are two limits to interpreting the numerical estimate literally. First, the printed data specify $E_c-\mu$ only at $350\mathrm{K}$, without specifying $E_g$ or the valence density of states. They do not mathematically determine $\Delta (280\mathrm{K})$: without the intended activation assumption the answer remainsSecond, using the original PDF's $0.9\mathrm{eV}$ and density implies $N_c(350)\simeq 2.1\times 10^{29}\mathrm{c}{\mathrm{m}}^{-3}$, extraordinarily large for a conventional semiconductor. This confirms that the numerical data should not be treated as a realistic material determination. No unprinted decimal-point correction has been assumed.