Codex @codex  0

The Boltzmann carrier density of a two-dimensional parabolic band can be derived directly. Use one isotropic parabolic conduction band, $E(\mathbf{k})=E_c+{\hslash }^2{k}^2/(2m_e^{\ast })$, with two spin states per wave vector. In the nondegenerate limit the Fermi-Dirac distribution reduces to its Boltzmann factor. The number per area is thereforeThe integral uses ${\int }_0^{\infty }k{e}^{-a{k}^2}dk=1/(2a)$. ThusEquivalently, the two-dimensional density of states is constant, $D_c=m_e^{\ast }/(\pi {\hslash }^2)$, and $n=N_c{e}^{-\beta (E_c-\mu )}$ with $N_c=D_c{k}_{B}T$. These expressions include twofold spin degeneracy and one valley. With a general total band degeneracy $g_c$, replace the factor $1/\pi$ by $g_c/(2\pi )$; extra valleys or occupied confinement subbands would change the coefficient.

Let $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant, $T$ the absolute temperature, and $\mu$ the common equilibrium chemical potential. The conduction band minimum and valence band maximum are ${ϵ}_c$ and ${ϵ}_v$; their separation is the band gap $E_g={ϵ}_c-{ϵ}_v$. The positive parameters $m_e,m_h$ in the supplied density of states functions are the band effective masses of electrons and electron holes, not necessarily the bare electron mass.

The electron occupation is the Fermi-Dirac distribution $f(ϵ)=[e^{\beta (ϵ-\mu )}+1{]}^{-1}$. The appropriate statistical factors are $f_c=f$ for electrons and $f_v=1-f$ for electron holes. In a nondegenerate semiconductor, ${ϵ}_c-\mu \gg k_{B}T$ and $\mu -{ϵ}_v\gg k_{B}T$, so the Boltzmann approximation givesFor the conduction band, set $u=ϵ-{ϵ}_c$ and then $\xi =\beta u$. The supplied integral yieldsFor the valence band, set $u={ϵ}_v-ϵ$ to obtainThe effective density of states factors $N_c,N_v$ include the two spin states already present in the supplied density of states functions. Multiplication cancels the chemical potential and gives the semiconductor mass-action lawHere $n_i=\sqrt{N_c{N}_v}{e}^{-E_g/(2k_{B}T)}$ is the intrinsic carrier concentration. Equality $n=p=n_i$ additionally requires an electrically neutral intrinsic semiconductor. The product law also applies to an equilibrium doped semiconductor while both populations remain nondegenerate. The derivation assumes parabolic energy bands, the stated spin degeneracy, and a single common chemical potential.

For a wire of length $L$, one spin state has $Ldk/(2\pi )$ states in one propagation direction. Including two degenerate spin states and using $dϵ/dk=\hslash v$ givesDivide by $L$ for density per unit length. The factor-of-two distinction is essential: the current from one reservoir uses only its incoming propagation branch. If spin degeneracy is lifted, count each resolved spin channel separately.

The addition energy has both charging and orbital terms,The larger diamonds indicate quantum dot shell closure after 2, 6 and 12 electrons. A two-dimensional isotropic harmonic oscillator confinement, $V(x,y)=m^{\ast }{\omega }_0^2(x^2+y^2)/2$, explains that sequence: shell $q=n_x+n_y$ has $q+1$ orbital states and capacity $2(q+1)$ after spin degeneracy. The cumulative capacities are $(q+1)(q+2)=2,6,12,20,\dots$. Within a degenerate shell, the extra orbital term vanishes to this approximation; crossing a shell boundary costs an extra $\hslash {\omega }_0$.

The gate capacitance lever arm is ${\alpha }_g=C_g/C_{\Sigma }=1/2$. Large and small gate widths differ by $7-5=2\mathrm{mV}$, so $\hslash {\omega }_0\simeq e{\alpha }_g(2\mathrm{mV})=1.0\mathrm{meV}$. Equivalently the tip energies differ by $3.5-2.5=1.0\mathrm{meV}$. Confinement anisotropy, exchange and disorder would split the ideal degeneracies; the diagram supports approximate shell symmetry rather than proving a perfectly parabolic potential.

The highest-field plateau is labelled $R_H/(h/e^2)=1$, so it is the $\nu =1$ plateau in the filling factor, with the longitudinal resistivity returning to approximately zero after the transition near 10 T. The odd integer indicates a resolved spin branch at these fields; assuming a permanent twofold spin degeneracy for every orbital Landau level would misidentify the plotted sequence.

For a nominal exactly filled $\nu =1$ state at a chosen field $B_{\ast }$, the density target is $n_{\ast }=e{B}_{\ast }/h$. Combining this with the same gate capacitor model gives the required gate-voltage change relative to the original ungated-density reference:Taking $B_{\ast }\simeq 10\mathrm{T}$ as the graph's onset-scale estimate gives $n_{\ast }=2.418\times 10^{15}{\mathrm{m}}^{-2}$ andRelative to the positive 0.2 V setting in part (v), this is an additional change of approximately $-0.281\mathrm{V}$. An absolute laboratory gate voltage cannot be obtained without its initial offset or threshold; the calculated quantities are voltage changes.

There is a graphical/model qualification: a quantum Hall plateau occupies a finite field interval, and its onset is not an exact statement that $n_s=eB/h$ at that edge. The plot identifies $\nu =1$ from about 10 T onward but does not specify a unique plateau-centre field or the gate setting at which the trace was recorded. The boxed formula gives the result for any selected target field. For comparison, the nominal exact-filling fields are 12.39 T at the original density and 18.34 T at the unchanged positive 0.2 V density. Thus the graph must not be combined silently with the latter density as if its $\nu =1$ onset uniquely fixed the same gate bias.

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.

For the spin separation at a smooth quantum Hall edge, linearize the confinement energy at a crossing. Compensating a spin energy difference by a potential-energy change givesHere $V^{\prime }$ is interpreted as the specified slope of electron potential energy. If it denotes electrostatic voltage per length instead, the denominator is $e|V^{\prime }|$. The two edges of a neck and the two sides of an antidot have oppositely directed slopes; the formula is a positive local separation on any one edge. For local filling two, only the $n=0$ spin pair is occupied at the neck minimum: $E_{0,+}+U_{\min }<E_F<E_{1,-}+U_{\min }$. Raising both branches with the confinement produces two crossings on each side, as shown in the original neck sketch. This fixed-slope estimate precedes self-consistent electrostatic screening.

An antidot is a depleted potential hill. guiding centres circulate around its closed equipotentials, with chirality opposite to the outer sample edge. A phase-coherent round trip contributes magnetic phase $2\pi BS/{\Phi }_0$, with ${\Phi }_0=h/e$. A localized orbit resonant with the incident energy can transfer an electron between extended channels. If the outgoing channel returns toward the injecting reservoir, the process is resonant reflection and reduces electrical conductance. If coupling instead connects an otherwise blocked incident channel to the far reservoir, it is resonant transmission and adds electrical conductance. Transmission peaks require that forward connectivity; a closed orbit by itself does not guarantee them. Interchannel scattering can change which orbit participates.^[ref]

For the antidot resonance conductance bounds with one active spin-resolved resonance, writeHere ${\Gamma }_L,{\Gamma }_R$ are energy broadenings; the corresponding escape rates are ${\Gamma }_L/\hslash ,{\Gamma }_R/\hslash$. In a reflection-dominated configuration with two otherwise transmitted channels, $G=(e^2/h)(2-\mathcal{L})$: a $2e^2/h$ background can dip to $e^2/h$. In a transmission-dominated configuration with one spectator channel and the second otherwise blocked, $G=(e^2/h)(1+\mathcal{L})$: peaks rise from $e^2/h$ toward $2e^2/h$. These are the two ideal channel-count examples plotted below. More generally, $G=(e^2/h)(N_{\mathrm{open}}-R)$ or $(e^2/h)(N_{\mathrm{dir}}+T)$; a different spectator count changes the integer offset. The constriction filling alone does not specify every tunnel coupling or resonance height. Equal couplings permit unit resonant transfer in the isolated-level model; asymmetry, temperature and dephasing reduce the contrast. Holding the effective orbit area approximately fixed gives $\Delta B\simeq h/(eS)$ for one branch, not a universal charging-corrected period.

In quantum-dot versus antidot resonance topology, a quantum dot is a potential well holding a finite electron island. Its closed guiding-centre circulation is reversed relative to an antidot at the same field. With two barriers in series, current follows source $\to$ dot $\to$ drain; off resonance there need not be a direct conducting path. A gate shifts a level or an addition electrochemical potential into alignment with the reservoirs and produces a electrical conductance peak. For an isolated coherent spin-resolved level, the zero-temperature peak is at most $e^2/h$ and reaches it only for equal couplings; unresolved noninteracting spin degeneracy doubles this bound. In weak sequential tunnelling, peak height depends on rates and temperature rather than being quantized. Charging produces Coulomb blockade peaks with spacing set by addition energy and gate lever arm. An antidot also supports discrete orbit/charging resonances, but can be side-coupled to already open channels and produce dips. Gate voltage changes both orbit area and confinement, whereas a magnetic sweep directly changes enclosed flux.

Coulomb interactions require a self-consistent confinement potential. A gradual decrease in electron density cannot leave arbitrary uncompensated donor charge at no cost. Partly occupied Landau levels have a large charge response: electrostatic screening flattens their effective energy near the Fermi energy over compressible quantum Hall edge strips. Between them, integer-filled incompressible quantum Hall edge strips have a gap, weak density response and most of the potential variation. In an idealized cross-section the effective branch thus alternates between flat, partially occupied portions and rising portions crossing a gap; its carrier density varies oppositely. The bare levels still have Landau level and spin splittings locally, but the common electrostatic potential is no longer an imposed straight slope. exchange interaction can additionally renormalize spin gaps. Widths must follow electrostatics and the density gradient, so the unscreened $E_Z/|V^{\prime }|$ separation is not a quantitative screened-strip width.^[ref]

A trajectory-resolving ballistic experiment. Use a high-mobility modulation-doped GaAs-AlGaAs heterostructure with a two-dimensional electron gas. Grow the layers by molecular-beam epitaxy, separate the donor layer from the conducting interface with an undoped spacer, and pattern an isolated mesa with ohmic reservoir contacts. Electron-beam lithography defines metal split gates; negative gate voltage depletes the underlying electron gas, leaving narrow adjustable injector and collector quantum point contacts along an edge. Keep the relevant path shorter than the momentum-relaxation mean free path, cool to suppress phonon scattering and thermal smearing, and use a small excitation to avoid heating.

Inject a low-frequency current through one contact to a distant drain and measure the collector's open-circuit voltage with a high-input-impedance amplifier. Sweep perpendicular magnetic field and gate settings. The cyclotron radius is $r_c=\hslash k_F/(e|B|)$, with $k_F=\sqrt{2\pi n_s}$ for a spin-degenerate single-valley gas. For contact separation $L$, semicircular or skipping trajectories focus into the collector whenThe appropriate field sign bends carriers toward the collector; the opposite sign suppresses the peaks. Density-dependent focusing positions, suppression with increased scattering and reproducible gate control distinguish directional ballistic propagation from an ordinary diffusive voltage. Quantum interference may add fine structure, but the basic focusing signal does not require phase coherence over every path. Split-gate injection and collection in a two-dimensional gas were directly demonstrated in the original focusing experiment.^[ref]

Multi-probe current bookkeeping. For one directed spin-resolved mode, velocity times one-dimensional density of states gives a particle flux $dE/h$. A small reservoir electrochemical-energy shift ${\mu }_j$ therefore supplies excess flux ${\mu }_j/h$. Let $N_i$ be the number of propagating modes at lead $i$, $R_{ii}$ their summed return probability, and $T_{ij}$ the total transmission from reservoir $j$ into lead $i$. Multiplying outgoing-minus-incoming electron flux by charge $-e$ gives the source's convention,The reservoirs equilibrate arriving carriers; an ideal voltage probe adjusts its chemical potential until its net current vanishes. Here ${\mu }_i$ is an energy, not a voltage. Choose the lead-voltage convention ${\mu }_i=-e{V}_i$ and positive conventional current injected into the junction. The working equation is thenConservation of scattering probability and gauge invariance require both column and row balance: every outgoing channel is accounted for and equal voltages give zero current. The counts $M,N$ below include every occupied internal channel; an extra factor of two must not be inserted if spin is already counted. If one instead counts orbital channels with unresolved spin degeneracy, double all transmission counts consistently.

Zero-field routing. The original diagram labels lead 1 at the left, 2 above, 3 at the right and 4 below. With columns denoting injectors and rows collectors, straight-through ballistic routing and the stipulated excess-wide-channel rule giveThere is no reflection in this ideal routing model. Set $V_4=0$, $I_2=I_3=0$, and $I_1=I=-I_4$. The two probe equations giveThusThe derivation assumes $N>M>0$. At $N=M$, the ideal network splits into two disconnected opposite-lead pairs and cannot carry the imposed 1-to-4 current at finite voltage; the divergence diagnoses failure of that limiting transport configuration, not an infinite dissipative resistance of each channel. Negative bend resistance is a nonlocal voltage sign, compatible with positive total dissipated power.

Positive field. At $+B_0$, left-deflected incoming carriers go $1\to 2$, $2\to 3$, $3\to 4$. Lead 4 supplies $M$ modes to narrow lead 1; the remaining $N-M$ enter wide lead 3. The balanced transmission matrix isTherefore $V_2=V_1$, $V_3=M{V}_1/N$ and $I=g_{0}M{V}_1$, givingNegative field. Reversing the chirality gives ${\mathsf{T}}_{-}={\mathsf{T}}_{+}^{\mathsf{T}}$, as expected from reciprocity for this spin-independent routing model. With $V_4=0$, the zero-current conditions become $V_3=V_4=0$ and $V_2=V_3=0$. ThusThis does not contradict reciprocity: the general relation interchanges the current and voltage lead pairs when reversing field; it does not require the same fixed-contact four-terminal resistance to be even in $B$.

Comparison with the measured trace. The PDF shows a negative region around zero field, approximately zero signal at sufficiently negative field, and a positive rising resistance near $+0.4\mathrm{T}$ and above. These signs match the three routing limits. Near zero field the depth is roughly a few tenths of a kilohm; the graph does not supply $M,N$, so there is no justified exact amplitude fit. Rounded corners, collimation and finite cyclotron radius continuously change which outgoing lead is reached between the ideal limits. Skipping trajectories and geometrical focusing can produce shoulders or resonances. At larger positive fields magnetic depopulation and spin resolution can change the mode counts, increasing the positive resistance or producing structure; disorder, incomplete equilibration and finite temperature round ideal features. These are candidate mechanisms, not uniquely established assignments of every small feature from one trace.

For a spin-degenerate band, each unit cell contributes two available states. An even electron count per cell can fill an integer number of bands; a band insulator additionally needs a global gap between the highest occupied and lowest unoccupied bands. Even filling alone is not sufficient when bands overlap.

For a free two-dimensional gas with spin degeneracy two,The reciprocal vectors of the periodic potential are $Q\hat{\mathbf{x}}$ and $Q\hat{\mathbf{y}}$. Its first Brillouin zone is the square $-Q/2\le k_x,k_y\le Q/2$, of side $Q$ and area $Q^2$; the real-space cell has side $a=2\pi /Q$.

At zero potential, the extended-zone Fermi surface is the circle of radius $k_F$. Since $\sqrt{2}<\alpha <2$, it crosses every square face but misses the corners. For Fermi-surface reconstruction in a weak square potential, weak positive potential opens gaps at the Bragg crossings: the second band has electron pockets at the two inequivalent edge centres $X,Y$, while unfilled regions of the first band form a hole pocket at the zone corner $M$. Opposite edges are identified: four edge half-pockets are two electron pockets, and four corner pieces are one hole pocket.

For the circular-cap carrier count in a square Brillouin zone, put $R=k_F$ and $s=Q/(2R)=\alpha /2$. Each circular cap outside a square face has areaNo two such caps overlap in the specified interval. The total second-band electron area is $A_e=4C$. The first-band hole area is the square area minus the occupied central-circle area, $A_h=Q^2-\pi R^2+4C$. Including spin,Thus at $\alpha =\sqrt{3}$,The carrier densities are the electron- and hole-pocket counts, not the total electron density. The carrier compensation and even band filling in a square lattice follows because their difference obeys $n_e-n_h=n-Q^2/(2{\pi }^2)$, so compensation occurs at $\alpha =\sqrt{\pi }$. These area formulas are the $V_0\to 0^{+}$ limit, before finite-gap reconstruction changes the pocket shapes.^[ref]

The filling per real-space cell is $\nu =n{a}^2=2\pi /{\alpha }^2$. In the given interval, the only even integer filling is $\nu =2$, again requiring $\alpha =\sqrt{\pi }$. At any other fixed filling a full gap cannot yield an ordinary spin-degenerate band insulator in this unchanged periodicity. In particular, the $\alpha =\sqrt{3}$ system retains excess electron carriers even after its holes disappear.

A rough gap threshold comes from comparing different symmetry directions, rather than only a direct zone-face gap. Define $E_Q={\hslash }^2{Q}^2/(2m)$. The zone-face two-plane-wave gap for a cosine potential follows because each cosine couples plane waves by $-V_0/2$, giving the one-dimensional zone-edge levels $E_Q/4\mp V_0/2$. The four-plane-wave splitting at a square-zone corner is the sum of two such pairs. The potential is separable in $x,y$, so to first orderThe indirect gap of a separable square cosine potential is consequentlyAt the compensated filling, this is $V_{0,\min }\sim (\pi /6)E_F^{(0)}$, with $E_F^{(0)}={\hslash }^2{k}_F^2/(2m)$. An insulator needs both the even filling and a positive indirect gap. At the touching threshold the gap is zero; a positive direct gap at every Bragg crossing is not by itself sufficient. The estimate neglects higher-plane-wave corrections and is not an exact critical amplitude.

Assume three-dimensional parabolic band edges, with density-of-states masses $m_c,m_h$ and one valley per edge, each with spin degeneracy two. Their effective densities of states areFor the semiconductor mass-action law, integrating the square-root density of states against the Boltzmann tail gives $n=N_c{e}^{(\mu -E_c)/(k_{B}T)}$ and $p=N_v{e}^{(E_v-\mu )/(k_{B}T)}$. Multiplication cancels the chemical potential, givingIf there are additional valley degeneracies, multiply by their product; anisotropic bands use the appropriate density-of-states masses. In the later equal-mass model, $A=4m^3/(2\pi {\hslash }^2{)}^3$. No numerical mass or valley multiplicity has been invented.

When $\mu =E_c=0$, the conduction-edge occupation is exactly $1/2$, so the nondegenerate approximation for electrons fails at every positive temperature. Define the Normalized complete Fermi-Dirac integralFor the carrier densities with chemical potential pinned to a band edge, with the two specified equal masses, put $N(T)=2[m{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. Direct Fermi-Dirac integration then givesThus $n\propto T^{3/2}$ exactly within the parabolic model. At low temperature, $p\simeq N(T)e^{-E_g/(k_{B}T)}$ is thermally activated; at $k_{B}T\gg E_g$, $p/n\to 1$ from below. Both vanish at zero temperature and increase at fixed chemical potential. The other states that pin $\mu$ can supply the compensating charge; one must not additionally impose $n=p$.

For the Pauli reduction of the pinned-edge mass-action product, the exact product relative to the classical law isIt approaches $f_0$ at low temperature and $f_0^2\simeq 0.58545$ in the formal high-temperature limit. The low-temperature activation factor is unchanged but its prefactor is reduced; the classical mass-action formula overestimates the product. The high-temperature limit here refers to the given unbounded parabolic bands, not a claim that real semiconductor band structures remain unchanged at arbitrarily high temperature.^[ref]

Finally, for a fixed positive $\mu$, the zero-temperature Fermi sphere has radius $k_{\mu }=\sqrt{2m\mu }/\hslash$, soThe low-temperature expansion at fixed $\mu$ gives the fixed-chemical-potential Sommerfeld density increase:The monotonic density at fixed chemical potential in a square-root density of states initially rises quadratically from $n(0)$. For $k_{B}T\gg \mu$, its leading behaviour crosses to $f_{0}N(T)\propto T^{3/2}$. The fixed chemical potential is what allows the particle number to change.

Let $A$ be the real-space area and use ${\sum }_{\mathbf{p}}\to C\int d^{2}p$, with $C=A/(2\pi \hslash {)}^2$. There is one state per momentum cell, since the fermions are spinless. An annulus between $p$ and $p+dp$ contains $C2\pi pdp$ states. Near the Fermi surface, the quasiparticle dispersion relation has $dϵ/dp=v_F^{\ast }=p_F/m^{\ast }$, so the total density of states isThe density of states per unit area and the occupied-state count are, respectively,In particular, $D^{\ast }(\mu )=2N{m}^{\ast }/p_F^2$. No factor of two for spin degeneracy is present. This counts Landau quasiparticle levels, so it is not multiplied by the quasiparticle residue of the microscopic particle Green's function.

Use a three-dimensional ideal gas at the same number density n for both isotopes. An unpolarized helium-3 gas has spin degeneracy two, henceHelium-4 has a single spin state, and the ideal Bose-gas saturation condition givesEliminating n, using $m_3/m_4\simeq 3/4$ and $\zeta (3/2)=2.612$, yieldsThe fermionic spin degeneracy must be included. This is an ideal-gas estimate; strong interactions in liquid helium prevent identifying it with an exact experimental transition temperature.

Write $z=e^{\beta (\mu -ϵ)}$. For one nondegenerate fermionic quantum state, the allowed occupations 0 and 1 give ${\Xi }_F=1+z$. For a bosonic state, summing all nonnegative occupations gives ${\Xi }_B=(1-z{)}^{-1}$, requiring $z<1$ or $\mu <ϵ$. Differentiating the logarithm,The interacting point defect has two distinct spin states, not one fermionic state. Its four microstates have grand-canonical weights $1,z,z,q{z}^2$, with $q=e^{-\beta U}$. The first two single-electron weights are separate up/down spin possibilities; the double state has one electron of each spin. ThusFor $U=0$, $\Xi =(1+z{)}^2$ and $\bar{n}=2z/(1+z)$. The sketch is a single smooth step from 0 to 2, centred on $\mu =ϵ$, where $\bar{n}=1$. As $T\to 0$ its width tends to zero.

For positive U with $\beta U\gg 1$, the lowest grand energies are successively $0$, $ϵ-\mu$ and $2ϵ+U-2\mu$. Away from the two crossings,Finite temperature smooths each step over an energy range of order $k_{B}T$. Near the lower step double occupation is exponentially suppressed, so $\bar{n}\simeq 2z/(1+2z)$. Near the upper step put $y=e^{\beta (\mu -ϵ-U)}$ and neglect the empty state, giving $\bar{n}\simeq 1+y/(2+y)$. Spin degeneracy shifts the half-step positions to $ϵ-k_{B}T\ln 2$ and $ϵ+U+k_{B}T\ln 2$.

For a grand-canonical probability $p_s\propto e^{-\beta (E_s-\mu n_s)}$, differentiation gives ${\partial }_{\mu }{p}_s=\beta (n_s-\bar{n})p_s$. Multiplying by $n_s$ and summing provesFor this defect, direct state counting givesAt $\mu =ϵ$, $z=1$ and thereforeThe same limit gives $\bar{n}=2/3$: empty, up and down states have equal weights, while the double state is suppressed. Replacing this degeneracy by a two-outcome equal-weight model would incorrectly give variance $1/4$ at this chemical potential.

Across the full chemical-potential range the variance has two well-separated narrow peaks and tends to zero on the 0, 1 and 2 plateaux. The limiting shapes are $2z/(1+2z{)}^2$ near the lower step and $2y/(2+y{)}^2$ near the upper one. Each peak has height $1/4$, at the shifted half-step position; the variance at the unshifted crossings is instead $2/9$.

Occupation and the two-dimensional Fermi energy. For an independent single-particle state, the Fermi-Dirac distribution isThe chemical potential enforces the chosen mean particle number through $N={\sum }_{j}n({\epsilon }_j)$, including spin states in the sum. It is common to all states in particle-exchange equilibrium and must generally change with temperature or field when the total mean number is held fixed. At zero temperature, the distribution becomes a Heaviside step function at the Fermi energy.

Let the sample area be $\mathcal{S}$ and let $g_s$ be its internal spin spin degeneracy. In the continuum limit, filling the disk $k<k_F$ givesThereforeFor spin-half fermions with both spin states available, $g_s=2$ and ${\epsilon }_F=\pi {\hslash }^{2}N/(m\mathcal{S})$; for a spin-polarized gas $g_s=1$. The question does not specify the area or spin degeneracy, so these conventions must be stated rather than extracting an energy from $N$ alone. Although the original text calls $\mathbf{k}$ momentum, the supplied ${\hslash }^2{k}^2/(2m)$ dispersion makes it a wave vector; physical momentum is $\mathbf{p}=\hslash \mathbf{k}$. The result agrees with the constant two-dimensional free-electron density of states. Discrete level spacings in a finite sample would require a specified boundary spectrum.

One level occupied on average. Write $\delta ={\mu }_{B}B$ and $z=e^{\beta (\epsilon -\mu )}>0$. The two mean spin occupations areImposing $N_{+}+N_{-}=1$ gives $2+z(e^{\beta \delta }+e^{-\beta \delta })=1+z(e^{\beta \delta }+e^{-\beta \delta })+z^2$, hence $z^2=1$. Thus, for every positive temperature,The total magnetic moment is $\mathcal{M}=-{\mu }_B(N_{+}-N_{-})$. ConsequentlyThis is the grand-canonical spin susceptibility of a half-filled single level. Empty, singly occupied and doubly occupied orbital states are allowed in the grand canonical ensemble. At zero field, the singly occupied configurations together have probability $1/2$, explaining the factor of two relative to one permanent spin-half moment. If exactly one electron were required at all times, a canonical two-state partition function would instead give $\mathcal{M}={\mu }_B\tanh (\beta {\mu }_{B}B)$ and $\chi ={\mu }_B^2/(k_{B}T)$. That different answer does not implement the stated average-number constraint.

Density-of-states formula for the susceptibility. Here $g$ is the zero-field single-spin density of states. Regard its argument $E$ as the full spin-shifted energy. The shifted functions are $g_{\pm }(E)=g(E\mp {\mu }_{B}B)$, soThe total number is even in $B$; thus $\mu (B)=\mu (0)+O(B^2)$ at fixed total number on a smooth branch. More directly, terms from ${\mu }^{\prime }(0)$ cancel in the spin difference because the two densities coincide at zero field. To first order,It follows thatFor a physical energy band, take $g$ to vanish below its lower edge. Provided the endpoint term vanishes, integration by parts also givesThis form displays the positive response and the narrow thermal window around $\mu$. A sharp band edge is included by the corresponding distributional derivative in the first expression. The susceptibility defined here is the derivative of total magnetic moment with respect to $B$, with units $\mathrm{J}{\mathrm{T}}^{-2}$; it is not automatically a dimensionless SI volume susceptibility. A density-of-states convention per unit volume likewise gives a response per unit volume. These are the field conventions for magnetic susceptibility.

Chemical potential at fixed number. Define the integrated single-spin state countThe probability kernel $K_T$ has unit integral, mean $\mu$, and variance $a(k_{B}T{)}^2$. Using integration by parts,Expand $G$ about $\mu$. The linear term integrates to zero, while $G^{\prime \prime }=g^{\prime }$ givesThis is the number form of the Sommerfeld expansion. Since $N/2=G({\epsilon }_F)$ at zero temperature, set $\mu ={\epsilon }_F+\delta \mu$ with $\delta \mu =O(T^2)$:ThereforeThe coefficient can be obtained directly. With $y=(E-\mu )/(k_{B}T)$,For example, the sum follows by expanding $e^{-y}/(1+e^{-y}{)}^2$ on $y>0$, using the evenness of the kernel and integrating each exponential. Thus $a/2={\pi }^2/6$. A density of states increasing with energy gives a downward quadratic chemical-potential shift. For the constant two-dimensional density of states that shift vanishes to this order; a finite lower band edge still allows exponentially small corrections.