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Both models respect Bloch's theorem and describe electrons in a periodic lattice, but they start from different useful limits.

In the nearly-free electron model, start with extended plane waves and free-electron energies ${\epsilon }_{\mathbf{k}}^0={\hslash }^2{k}^2/(2m)$. A weak periodic potential has Fourier coefficients $V_{\mathbf{G}}$ at reciprocal lattice vectors and couples $\mathbf{k}$ to $\mathbf{k}-\mathbf{G}$. Away from degeneracies its corrections are small. Near a Bragg scattering, retain the two resonant plane waves and diagonalizeThe resulting eigenvalues areAt the crossing the gap is $\boxed{2|V_{\mathbf{G}}|}$. The two standing-wave combinations place electronic density differently relative to the periodic potential and therefore acquire different energies. Folding the many free parabolas into the first Brillouin zone produces the repeated band labels; the nonzero potential opens avoided crossings between coupled states. A Fourier component forbidden by symmetry need not open its corresponding first-order gap.

In the tight-binding model, instead start with localized orbitals on well-separated atoms or sites, with discrete on-site energies. Their overlap permits tunneling represented by hopping integrals. Bloch sums of these orbitals diagonalize the lattice translation symmetry. For one orthonormal orbital per site in a one-dimensional nearest-neighbor chain,The many initially degenerate copies of one atomic orbital broaden into a band. Other atomic orbitals or multiple sites per primitive cell produce additional bands; their hybridization creates bonding and antibonding combinations and may open gaps. For two orbitals per cell, one illustrative Hamiltonian iswith eigenvalues $(E_A+E_B)/2\pm \sqrt{(E_A-E_B{)}^2/4+|t_1+t_2{e}^{-ika}{|}^2}$. This shows explicitly how two bands arise from the two cell degrees of freedom, even when the isolated sites have equal on-site energies.

For each orbital degree of freedom per primitive cell there is one band per spin; band filling and the presence of a gap at the Fermi level determine metallic or insulating behavior. The nearly-free limit emphasizes weak Bragg scattering of extended states, while tight binding emphasizes hopping between localized states. They are complementary approximations to electronic band structure, rather than two incompatible explanations of why periodic solids have bands.

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.

A junction field-effect transistor has a conducting channel between source and drain, with gate regions forming reverse-biased p-n junctions to that channel. In an n-channel device, a more negative gate-source electric potential difference increases the reverse bias and widens the depletion region. The conducting cross-section shrinks, reducing the drain electric current. The gate ideally draws only a small leakage electric current, giving a high input resistance; the source-drain supply provides the controlled output power. Carrier motion in a p-channel device reverses the relevant polarities.

At small positive $V_{DS}$, the channel conducts approximately as a resistor. As the potential rises from source to drain, the local gate-channel reverse bias increases toward the drain. The depletion layers are therefore thicker there. For a symmetric n channel of half-thickness $a$, uniformly doped at density $N_D$, the one-sided depletion approximation giveswhere $w$ is the depletion width from each side and $V_{\mathrm{bi}}$ the built-in junction potential. The neutral channel closes locally when $w=a$. Define the positive geometrical depletion electric potential difference $U_P=e{N}_D{a}^2/(2{\epsilon }_s)$. The approximate drain-end pinch-off condition isMore negative gate bias makes this saturation occur at a smaller drain electric potential difference.

Pinch-off in a junction field-effect transistor does not mean that the drain electric current vanishes. Electrons arriving through the remaining neutral channel are swept across the high-field depleted region into the drain. Further increases in $V_{DS}$ mostly fall across that region, so the electric current changes only weakly: this is the saturation regime used for amplification. Channel shortening gives a finite output slope in real devices. Excessive reverse electric potential difference eventually causes breakdown, beyond the normal saturation model.

A sufficiently negative gate electric potential difference can instead deplete the entire channel from the source end and produce cutoff. In the simplified geometry its threshold is $V_{GS,\mathrm{off}}\simeq V_{\mathrm{bi}}-U_P<0$. A commonly used approximate transfer law in saturation isHere $I_{DSS}$ is the saturated drain electric current at zero gate-source bias. The square law is a useful device approximation, not an exact result for arbitrary channel profiles. It connects field control of depletion to a measurable transfer characteristic while keeping drain-end pinch-off distinct from complete gate cutoff.

Direct exchange is a quantum-mechanical spin-dependent energy arising from the Coulomb interaction and the required antisymmetry of a two-electron state. It requires spatial overlap of the relevant orbitals. It is not the classical magnetic dipole force between two small magnets, and it can be much stronger than that force.

For an explicit fixed-orbital illustration, take orthonormal spatial orbitals $a,b$. Orthonormality means $⟨a|b⟩=0$; it does not require their spatial densities to occupy disjoint regions. The normalized spatial states areThe symmetric spatial state combines with the antisymmetric spin-one-half singlet state, and the antisymmetric spatial state combines with a symmetric spin-one-half triplet state, so that the complete fermionic state is antisymmetric.

For Coulomb potential $V_{12}=e^2/(4\pi \epsilon |{\mathbf{r}}_1-{\mathbf{r}}_2|)$, distinguish the direct electrostatic integral and exchange integral:The Coulomb exchange splitting of two orbitals givesThe repulsive Coulomb potential makes $K\ge 0$ in this model. Thus the triplet is favored when the orbitals have nonzero exchange overlap. Its antisymmetric spatial wavefunction suppresses close electron encounters and reduces their repulsion. Since the Coulomb interaction contains no explicit spin operator, the spin dependence enters through the allowed spatial symmetry.

For dimensionless spin operators ${\mathbf{s}}_i={\mathbf{\sigma }}_i/2$, whose pair eigenvalues are $1/4$ in the triplet and $-3/4$ in the singlet, the same splitting is represented byThis explicitly states the sign convention: positive $K$ here favors parallel spins. A convention $H=J{\mathbf{s}}_1\cdot {\mathbf{s}}_2$ would instead have $J=-2K$.

In a crystal, direct exchange therefore depends sensitively on orbital extent, overlap and interatomic spacing. It often decreases rapidly as localized orbitals separate. The fixed-orbital example does not prove that every material must be ferromagnetic: orbital rearrangement and competing kinetic or virtual-hopping processes alter the singlet-triplet balance. In particular, antiferromagnetic superexchange through an intermediate ligand is a different mechanism and should not be confused with direct overlap exchange.

mass as band curvature. In a periodic potential, a weakly forced electron is usefully described by a wave packet confined to one energy band. Its group velocity isDifferentiating the velocity givesThis effective-mass tensor describes the band response, not a change of the electron's bare rest mass. Along a principal direction, a steep upward curvature gives a small positive mass and a flatter curvature a larger one. Near a band maximum the electron curvature and effective mass are negative; the corresponding missing-electron description uses positive electron hole masses. Anisotropy produces distinct principal masses, and strong nonparabolicity makes them momentum dependent. The single-band semiclassical description assumes fields weak and slowly varying enough to avoid substantial interband transitions.

Lattice symmetry and Bloch's theorem. Use the active spatial translation operator $T_{\mathbf{a}}\psi (\mathbf{r})=\psi (\mathbf{r}-\mathbf{a})$. A lattice-periodic scalar potential satisfies $V(\mathbf{r}+\mathbf{a})=V(\mathbf{r})$, while the kinetic term is translation invariant. Therefore $[H,T_{\mathbf{a}}]=0$. The lattice translations also commute with one another, so they can be diagonalized simultaneously with $H$, including within a degenerate energy subspace.

For a common eigenstate, write $T_{\mathbf{a}}\psi =t(\mathbf{a})\psi$. Unitarity gives $|t(\mathbf{a})|=1$, and the group law gives $t(\mathbf{a}+\mathbf{b})=t(\mathbf{a})t(\mathbf{b})$. On primitive Bravais vectors ${\mathbf{a}}_j$, choose $t({\mathbf{a}}_j)=e^{-i\mathbf{k}\cdot {\mathbf{a}}_j}$. Every lattice vector is an integer combination of these generators, henceThe vector $\mathbf{k}$ is defined modulo a reciprocal lattice vector. The band index labels the different energy eigenstates with the same translation character. Define $u_{n\mathbf{k}}(\mathbf{r})=e^{-i\mathbf{k}\cdot \mathbf{r}}{\psi }_{n\mathbf{k}}(\mathbf{r})$. The phase factors cancel under a lattice translation, proving Bloch's theorem in its usual form:This is the Translation-character proof of Bloch theorem; the commutator guarantees a basis of such states, even when an arbitrary superposition of different translation characters is not itself a translation eigenstate.

Periodic Hamiltonian. Acting once with momentum gives $\mathbf{p}(e^{i\mathbf{k}\cdot \mathbf{r}}u)=e^{i\mathbf{k}\cdot \mathbf{r}}(\mathbf{p}+\hslash \mathbf{k})u$. Applying it again and cancelling the common plane-wave factor yieldsThe Bloch Hamiltonian acts on lattice-periodic functions in a unit cell. Expanding its square separates a zero-momentum Hamiltonian, a term linear in $\mathbf{k}$ and an identity term quadratic in $\mathbf{k}$.

Expansion about the zone centre. Let $|n⟩=u_{n0}$, $E_n={\epsilon }_n(0)$, and normalize all these states using a unit-cell inner product. For the real scalar Hamiltonian in the question, a nondegenerate zone-centre eigenfunction can be chosen real. Thenbecause opposite unit-cell faces cancel. Equivalently, time-reversal symmetry forbids a nonzero electric current in this nondegenerate spinless state. This explains why no linear energy term appears; it should not simply be omitted without a symmetry or extremum argument.

The first-order energy shift is therefore ${\hslash }^2{k}^2/(2m)$. In the second-order sum the quadratic term is proportional to the identity and has no off-diagonal matrix elements. Only the linear momentum vertex contributes. Second-order nondegenerate perturbation theory gives the momentum matrix-element expansion of band energyHermitian operator makes the two reciprocal matrix elements complex conjugates. All other periodic eigenstates at $\mathbf{k}=0$ enter the sum. The remainder is even because time-reversal symmetry gives ${\epsilon }_n(\mathbf{k})={\epsilon }_n(-\mathbf{k})$ near this isolated state. At a degeneracy the denominators signal the need for degenerate perturbation theory instead.

mass ratio and the printed formula. Write $p_{nl}^i=⟨n|p_i|l⟩$. Differentiating the quadratic energy yieldsFor a specified unit direction $\mathbf{e}$, the resulting scalar mass ratio isIn an isotropic three-dimensional band, averaging equivalent directions givesCoupling to lower bands contributes positively to inverse mass; coupling to higher bands contributes negatively. Their signs and matrix elements determine the actual curvature.

The original PDF prints $m^{\ast }={\hslash }^2[{\nabla }_{\mathbf{k}}^2{\epsilon }_n{]}^{-1}$. Taken literally as a three-dimensional Laplacian, this misses a factor of three for an isotropic scalar mass. The correct identities areThe printed expression would return $m/3$ even for the free dispersion ${\hslash }^2{k}^2/(2m)$. It is correct in one dimension, or if its second derivative is interpreted directionally. For completeness, its literal three-dimensional trace ratio would instead be $3+(2/m){\sum }_{l\ne n}|⟨n|\mathbf{p}|l⟩{|}^2/(E_n-E_l)$; that is the trace of the dimensionless inverse-mass tensor, not the usual scalar $m/m^{\ast }$.

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.

Choose the $e^{-i\omega t}$ convention. The steady displacement in the Lorentz oscillator model has complex amplitudeThe induced electric dipole moment is $p_{\omega }=q{u}_{\omega }$, so the microscopic electric polarizability is $\alpha (\omega )=q^2/[m({\omega }_T^2-{\omega }^2-i\gamma \omega )]$. For $N$ identical oscillators per unit volume, $P_{\omega }=N{p}_{\omega }={ϵ}_0{\chi }_{\omega }{E}_{\omega }$. ConsequentlyHere the dimensionless quantity called polarizability in the paper is the electric susceptibility; the single-electron polarizability $\alpha$ has different units. The sign of the electron charge disappears because both the driving force and dipole moment contain $q$.

Multiplying by the complex conjugate of the denominator givesFor positive damping and positive frequency, ${ϵ}_r^{\prime \prime }\ge 0$, as required for passive wave absorption in this Fourier convention. At zero frequency, ${ϵ}_r^{\prime }=1+{\omega }_p^2/{\omega }_T^2$ and ${ϵ}_r^{\prime \prime }=0$. Near a weakly damped resonance the imaginary part has a narrow positive peak, while the real part rises on the low-frequency side and then falls steeply through ${ϵ}_r^{\prime }=1$ at ${\omega }_T$. It can become negative above resonance for sufficiently strong oscillator strength and weak damping. At large frequency, ${ϵ}_r^{\prime }=1-{\omega }_p^2/{\omega }^2+O({\omega }^{-4})$ and ${ϵ}_r^{\prime \prime }\sim {\omega }_p^2\gamma /{\omega }^3$: both approach the vacuum values. The absorption maximum is close to, but need not be exactly at, ${\omega }_T$ for finite damping.

A self-assembled InAs quantum-dot single-photon source uses small indium arsenide islands inside a larger-gap gallium arsenide host. Strain during epitaxial growth favors islands after a thin wetting layer has formed. Their characteristic diameter is of order $10$–$30\mathrm{nm}$ and their height a few nanometres, for example $2$–$5\mathrm{nm}$. Growth conditions determine the areal density: low-density samples suitable for isolating dots can be around $10^8$–$10^9\mathrm{c}{\mathrm{m}}^{-2}$, while ordinary dense layers can be much denser. The particular electrically driven device underlying the supplied image had about $5\times 10^8\mathrm{c}{\mathrm{m}}^{-2}$. These are characteristic scales, not dimensions deducible from the emission image.^[ref]

Quantum confinement in the dot traps both electrons and electron holes in three directions, giving discrete energy levels. In a junction device, injected carriers relax into these levels. Electron-hole recombination then emits a photon, and the dot must be repopulated before emitting again on that same transition. Spatial selection of one dot, spectral selection of one line, and well-separated injection pulses make this a single-photon source. Photon antibunching, rather than low average brightness alone, establishes suppression of simultaneous photons. In particular, collecting both photons of a cascade without selecting a line would not produce exactly one collected photon per excitation.

The lower-current feature $X$ is the transition from one confined exciton to the empty dot. The higher-current feature $X_2$ is the first step of the biexciton cascade $XX\to X\to 0$: it becomes strong when the dot can contain two electron-hole pairs. In the weak-pumping limit the probabilities of one and two pairs scale approximately as $I$ and $I^2$, respectively. Coulomb and confinement energies shift the two transitions relative to one another; $X_2$ is not a photon of twice the $X$ energy. In this particular image the $X_2$ photon is about $4$–$5\mathrm{meV}$ higher in energy. With $B_{XX}=2E_X-E_{XX}$, this is a negative biexciton binding energy, since $E_{XX}-E_X=E_X-B_{XX}$.

For the mass estimate, the original plot places the $X$ line at approximately $E_X=1.3945\mathrm{eV}$, with graph-reading precision of order $0.5\mathrm{meV}$. Under the prescribed identification of photon energy with the full lowest exciton excitation energy,The three-dimensional screened Wannier-Mott exciton has the hydrogenic exciton binding energyThereforeReading the line as $1.395\mathrm{eV}$ instead gives $0.331m_0$; about $0.33$–$0.34m_0$ is an appropriate precision. This inference uses the supplied pair-creation energy and a three-dimensional screened hydrogenic model. In a real dot, strain and confinement also alter transition energies, so it is an effective model estimate and does not independently determine either individual carrier mass.