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Band filling counts occupied one-particle states in an energy band. With spin degeneracy two, a crystal of $N$ primitive unit cells has $2N$ states in each orbital-derived energy band, as in state counting in a Bloch band. Full occupation therefore means two electrons per primitive cell in that band.

In the group velocity approximation, a fully occupied band contributes current densityThe integral vanishes because $E$ is periodic across opposite faces of the Brillouin zone; inversion symmetry is unnecessary.^[ref] A partially occupied band leaves nearby empty states available for ordinary longitudinal conduction. A filled set of bands with a positive band gap instead gives a band insulator at zero temperature.

The vanishing group velocity term does not exclude every electrical response. A filled Chern insulator band can have a transverse quantum Hall effect governed by Berry curvature.^[ref]

In a rigid spectrum, source and drain threshold lines in a bias-versus-gate map satisfy $E_n(V_g)={\mu }_S$ or ${\mu }_D$. Where adjacent thresholds align with opposite edges of the bias window, their spacing equals the window width $e|V|$. For symmetric bias and spin degeneracy, neighboring channel counts at those edges give a differential electrical conductance halfway between integer $2e^2/h$ plateaus. The dc ratio $I/V$ is a different observable; barrier motion and heating can modify the interpretation.

Compare uniform three-dimensional ideal gases at the same number density n, with fermion mass $m_F$, spin degeneracy $g_F$, boson mass $m_B$ and a single boson spin component. The Fermi temperature has $k_B{T}_F={\hslash }^2(6{\pi }^{2}n/g_F{)}^{2/3}/(2m_F)$, while Bose saturation gives $k_B{T}_B=2\pi {\hslash }^2[n/\zeta (3/2){]}^{2/3}/m_B$. Eliminating density yieldsThe degeneracy is essential; using a spinless Fermi sea changes the numerical estimate. This is an ideal-gas comparison and does not remove interaction effects in a dense liquid.

Consider two overlapping energy bands on a square lattice with spin degeneracy two and primitive reciprocal period $Q$. Let $n$ count the areal electron density in this pair, $n_e$ the occupied density in upper-band electron pockets, and $n_h$ the missing density in lower-band hole pockets. A completely filled lower band provides the reference densityCounting filled states gives $n=n_0-n_h+n_e$, hence $n_e-n_h=n-n_0$. Compensation, meaning equal electron pocket and hole pocket carrier densities, therefore occurs at $n=n_0$. For total pocket areas $A_e,A_h$, the same state counting in a Bloch band gives $n_{e,h}=2A_{e,h}/(2\pi {)}^2$; all equivalent pockets must be counted.

If the geometry is parameterized by $Q=\alpha \sqrt{2\pi n}$ with $\alpha >0$, compensation requires $\alpha =\sqrt{\pi }$. The real-space period is $a=2\pi /Q$, so the filling is $n{a}^2=2$ electrons per primitive unit cell. Compensation can coexist with both kinds of charge carriers; a global band gap is additionally required for a band insulator.

For $s=Q/(2R)$, each face cap has area $C=R^2[\arccos s-s\sqrt{1-s^2}]$. The total second-band electron area is $4C$; the first-band hole area is $Q^2-\pi R^2+4C$. Multiplying either area by the spin degeneracy two and dividing by $(2\pi {)}^2$ gives the carrier densities. This is an infinitesimal-potential count, not an exact finite-gap formula.

A scalar effective mass chosen to reproduce carrier state counting near a band edge. For $g_v$ equivalent ellipsoidal conduction valleys, $m_{\mathrm{DOS}}^{3/2}=g_v(m_l{m}_t^2{)}^{1/2}$ when the displayed factor two accounts for spin degeneracy. The valley degeneracy is then already included and must not be multiplied in again. A conductivity mass weights inverse curvatures and is generally different; for an isotropically averaged set of ellipsoids, $3/m_{\mathrm{cond}}=1/m_l+2/m_t$.

The circular zero-temperature occupation boundary of an isotropic two-dimensional electron gas. The occupied disk has area $\pi k_F^2$; converting its area to particle number requires the reciprocal-space state density and a separately specified spin degeneracy.

For independent spin-one-half adsorbates whose two spin states are split only at the surface, the one-site partition function contains $2\cosh ({\mu }_{\mathrm{B}}B/k_{\mathrm{B}}T)$. Relative to the unsplit bulk spin degeneracy, this increases the Langmuir adsorption model affinity by a hyperbolic-cosine factor. At fixed temperature and gas pressure, increasing the magnitude of the magnetic field increases surface coverage in this model.

For a repulsive single orbital at temperatures small compared with U, chemical potentials strictly between the first and second addition energies select a singly occupied defect. Its mean occupation is nearly one while its charge-number variance is exponentially small away from the thermally broadened step regions. The remaining spin degeneracy does not itself produce charge-number fluctuations.

A quantum dot confines electronic motion in all three directions. A semiconductor dot can be thin enough that only one transverse subband is occupied, with discrete lateral states controlled by gates. In a rectangular hard-wall model,A smooth circular lateral confinement is often better approximated by $m^{\ast }{\omega }_0^2{r}^2/2$, giving ${ϵ}_{n_x{n}_y}=\hslash {\omega }_0(n_x+n_y+1)$ above the transverse ground energy. Shell $q=n_x+n_y$ contains $q+1$ orbitals, twice that many electrons before interactions split the spin degeneracy. Hence closed-shell electron numbers are $2,6,12,20,\dots$. These are single-particle quantum numbers; an interacting dot's states are many-electron eigenstates, not simply one orbital occupied by a whole charged island.

A perpendicular magnetic field changes the harmonic spectrum to the Fock-Darwin spectrum. For an electron of charge $-e$, take ${\omega }_c=eB/m^{\ast }>0$ and $L_z=\ell \hslash$ in the usual symmetric gauge. ThenReversing the definition of the angular label changes the displayed sign of its orbital term, not the set of energies. The Zeeman term lifts spin degeneracy, while exchange can favor parallel-spin filling in partially filled shells. These effects are distinguished by how peaks evolve with field rather than by identifying every irregular peak spacing with one-particle confinement.

For electrostatics, let $C_{\Sigma }=C_s+C_d+C_g$ and $Q_g=C_s{V}_s+C_d{V}_d+C_g{V}_g+Q_0$, where $Q_0$ represents a fixed offset charge in the chosen convention. The constant interaction model givesThe charging energy is conventionally $E_C=e^2/(2C_{\Sigma })$, whereas adjacent charge-state potentials differ byThis addition energy separates a capacitive contribution from orbital spacing. Exchange, nonuniform screening and correlations are corrections to the constant interaction model; a change in dot shape or effective capacitance can also alter spacings.

Connect the island weakly to source and drain reservoirs through tunnel barriers. sequential tunnelling is suppressed in a charge-stable region when neither adding nor removing an electron is energetically accessible. Sweeping the gate brings a charge electrochemical potential through the reservoir Fermi level, giving a Coulomb-blockade conductance peak. The gate lever arm is ${\alpha }_g=C_g/C_{\Sigma }$, so $\Delta V_g=E_{\mathrm{add}}/(e{\alpha }_g)$ locally if the capacitances and level spectrum do not vary appreciably. At finite source-drain bias, alignment with either reservoir defines the boundaries of a Coulomb diamond. Its slopes constrain capacitance ratios, and lines parallel to its edges can reveal excited-state transport. The conversion from a diamond's bias height to an energy depends on bias convention; a full-height/half-height factor must not be assumed without declaring the circuit.

A clear charging spectrum requires $k_{B}T$ and tunnel broadening $\hslash \Gamma$ smaller than the relevant addition energy; resolving individual orbital levels additionally requires them smaller than the orbital spacing. Tunnel resistances large compared with $h/e^2$ suppress strong charge fluctuations in the usual weak-junction regime. Blockade is not an assertion of mathematically zero current: higher-order cotunnelling can remain, and strong coupling or low-temperature spin correlations can invalidate the sequential model. An experimental check therefore combines gate sweeps with bias spectroscopy, temperature dependence and barrier tuning to separate level structure from broadening or uncontrolled charge traps.

In a primary vertical-dot experiment, gate-controlled current oscillations were converted into addition energies using finite-bias diamond spectroscopy; varying magnetic field then followed shell and spin filling. Reproducing enhanced addition energies at electron numbers 2 and 6 in several devices strengthened the shell assignment. The electron temperature was independently estimated above the refrigerator base temperature, an essential distinction when judging spectral resolution. Field-dependent reordering also revealed exchange effects beyond a purely spin-degenerate orbital model.^[ref]

For a fixed number of non-interacting molecules, $Z_N=q^N/N!$ in the dilute ideal gas limit. The temperature-independent factorial drops out of the internal energy derivative:Differentiate once more at fixed volume to obtainPrimes denote fixed-volume derivatives. If $q$ factorizes into translational, rotational and other independent contributions, their heat capacities add. In particular, the three-dimensional translational factor contributes $3R/2$, while a constant spin degeneracy factor contributes zero. The formula also follows from the positive canonical energy variance, $C_V=\mathrm{Var}(E)/(k_{\mathrm{B}}{T}^2)$.

Let $\mathcal{A}$ be the total two-dimensional area. There are $\mathcal{A}/(2\pi {)}^2$ wavevector states per reciprocal-space area per spin. Including spin degeneracy two and filling a disk givesFor areal number density $n_e=N_e/\mathcal{A}$ this yieldsThe disk is centred at zero wavevector, and its boundary is a fermi circle. Omitting the two spin states would incorrectly double the energy inferred for a given number density.

There are three useful temperature regimes.

At low temperature, dopant freeze-out keeps most donor electrons bound. Warming ionizes donors and increases conductivity through the rising mobile-electron concentration. For an uncompensated dilute donor system, negligible holes and spin degeneracy two, charge neutrality with partial donor ionization givesDeep in freeze-out, $n\ll N_D$, soThe activation scale is set by the shallow donor binding, rather than the whole band gap.

In the extrinsic regime, donors are largely ionized and $n\simeq N_D$ is nearly constant. The temperature dependence of conductivity then mainly follows the electron mobility. Increasing phonon scattering can make conductivity fall as temperature rises, even though donors remain ionized; impurity scattering can give a different mobility trend at lower temperature.

At sufficiently high temperature, thermally generated electron–hole pairs overwhelm the dopant population. The intrinsic, gap-activated increase in conductivity returns. The boundaries depend on donor concentration, compensation and scattering, so no one exponential describes the entire doped-sample curve.

For the adsorbed lattice gas, the Helmholtz free energy isUse Stirling's approximation with $M,N,M-N$ large, then differentiate at fixed $M,T,B$. Writing surface coverage as $\theta =N/M$ gives the adsorption chemical potentialThe occupation ratio comes from site-mixing entropy; it is essential near saturation.

Let $N_g$ be the number of free atoms and $n_g=N_g/V$. In the field-free gas, both spin states have zero energy, so the one-atom partition function is $q_g=2V/{\Lambda }^3$, where $\Lambda$ is the thermal de Broglie wavelength. For a dilute ideal gas,ThereforeThe last equality uses $p=n_g{k}_{\mathrm{B}}T$. The bulk spin degeneracy is two even though the bulk magnetic field vanishes: degeneracy is not the absence of spin states.

The density of states $D(E)$ counts available one-electron states: $D(E)dE$ is the number with energies in $[E,E+dE]$. It describes the spectrum, irrespective of whether those states contain electrons.

The density of occupied states additionally weights each state by the Fermi-Dirac distribution:Thus $Z(E)dE$ is the mean number of electrons in that energy interval, and $N_e=\int Z(E)dE$. State and electron densities must use the same normalization, including whether they are extensive or per volume and whether $D$ already includes spin degeneracy. The symbol $Z(E)$ here denotes occupied density, not a canonical partition function.

There are $L$ primitive cells and therefore $L$ lattice points. The specified reciprocal-space density $L{A}_c/(2\pi {)}^2$ counts distinct allowed wavevectors; each supports two electron spin states. Filling the disk at zero temperature givesUsing $A_c=2a^2$ and $k_F^2={\pi }^2/(2a^2)$,The circle fills a fraction $\pi /4$ of the first-zone area, while one spin-degenerate band has two states per primitive cell; these give the same result. A single-spin count would instead be $\pi /4$, omitting the physical twofold spin degeneracy of this model. The circle has only just reached the sides, so the zone corners remain unfilled: first contact does not mean filling the entire first band.

With spin degeneracy, each energy band contains two electrons per primitive unit cell. Divalency therefore permits a filled first band, but only if its maximum lies below the second band's minimum. With negligible ionic periodic potential, the circular Fermi surface and the rectangular zone have equal enclosed area, not equal shape. Electrons occupy some states outside the first zone while some higher-energy first-zone states remain empty. Folded bands overlap, and the material is a metal.

A periodic potential mixes plane waves differing by a reciprocal lattice vector. At a two-wave Bragg point, the nearly-free electron model gives a direct gap $2|V_{\mathbf{G}}|$. Increasing appropriate Fourier coefficients can remove the band overlap. With the first band filled, the system is then a band insulator, provided the temperature is small compared with the insulating band gap in energy units.

The rectangular free-electron band-overlap scale follows from two different points in reciprocal space. Put $C={\hslash }^2/(2m_e)$. Before gap opening, the highest first-zone energy is at the corner, while the lowest second-zone energy is at the midpoint of the nearer face:Thus the downward shift ${\delta }_v$ of the first-band maximum plus the upward shift ${\delta }_c$ of the second-band minimum must overcomeIn a rigid-splitting estimate, an opening $\Delta$ lowers the first band by $\Delta /2$ and raises the second by $\Delta /2$, giving the familiar threshold $\Delta >{\hslash }^2{\pi }^2/(2m_e{b}^2)$. Equality is only band touching, not a positive insulating gap. This expression is the overlap that must be removed, under that explicit band-shift approximation.

Positive direct gaps can coexist with band overlap: a local direct splitting and the global indirect band gap are not interchangeable. The shape and Fourier coefficients of the ionic potential are unspecified, so there is no universal exact critical direct-gap width determined by $a,b$ alone. For a concrete demonstration, a separable potential $2V_x\cos (2Kx)+2V_y\cos (2Ly)$ has axial direct gaps ${\Delta }_x=2|V_x|$ and ${\Delta }_y=2|V_y|$. To first order, the corner's lowest level is $C(K^2+L^2)-|V_x|-|V_y|$, and the two candidate second-band minima are $C{K}^2+|V_x|$ and $C{L}^2+|V_y|$. Therefore, within this model and neglecting higher-order band shifts, nonoverlap requiresEqual axial gaps would give the first-order estimate $\Delta >2C{L}^2/3$, rather than the rigid-splitting estimate. Such a large opening can exceed the quantitative range of weak-potential perturbation theory, but the example identifies the missing model assumption: corner levels feel both couplings. The exact general answer is the positive indirect-gap criterion above; a geometry-only direct-gap threshold would claim more than the supplied potential information supports.

Take one conduction electron per atom and a three-dimensional free electron gas, with spin degeneracy two and electron mass $m_e$. State counting inside the occupied Fermi sphere givesThe highest occupied energy at zero temperature is $E_F={\hslash }^2{k}_F^2/(2m_e)$, soThis relates the atomic number density to the Fermi temperature only because monovalency and the free-electron dispersion have been assumed. A real energy band can require a different density of states and band effective mass.

For the isotropic Debye model, count three acoustic polarizations up to one spherical wavevector cutoff $q_D$. The number of vibrational modes per volume isWith $\omega =cq$ and $\hslash {\omega }_D=k_B{\theta }_D$,The effective $c$ is the Debye sound speed: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical Brillouin zone boundary.

Use periodic boundaries in a box of side $L$ and include two spin states for each wavevector. One orbital state occupies $(2\pi /L{)}^d$ in wavevector space. Let $v_d$ be the volume of the unit ball in dimension $d$, namely $2,\pi ,4\pi /3$. Filling the ball of radius $k_F$ givesSince $E_F={\hslash }^2{k}_F^2/(2m)$,The density $n$ means number per length, area or volume as appropriate; it is not assigned three-dimensional units in every case.

For the same spin-summed states, the cumulative number per $d$-dimensional volume below energy $E$ is $n(E)=2v_d(2mE/{\hslash }^2{)}^{d/2}/(2\pi {)}^d$. Differentiating yields $g(E)=(d/2)n(E)/E$. Evaluating at the filling energy,In one, two and three dimensions the density of states varies respectively as $E^{-1/2}$, a constant and $E^{1/2}$. The normalization counts both signs of momentum and both spin states exactly once. A single-spin density of states or extra valley degeneracy would require a different counting convention, not an extra unexplained factor in these constants.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/7/Physics%20Papers%202018.pdf?cookie_passthrough=1

The cover establishes Part B, Trinity Term 2019, 13 June 2019, A13485W1; the annual bundle's 2018 filename does not determine the examination year. This is distinct from the Part C major option. All four questions are solved. Use SI magnetic susceptibility, $\chi =M/H$, and write $e>0$ for the elementary electric charge. In question 3, $n,p$ and $n_d$ are densities per area, and the band model has one parabolic extremum per carrier species with twofold spin degeneracy.