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Angle-resolved photoemission spectroscopy maps occupied energy bands: ultraviolet or x-ray photons eject electrons, and an energy- and angle-resolving analyzer measures $E_{\mathrm{kin}}$ and emission direction. The photoelectric effect gives the binding energy $E_B=h\nu -\Phi -E_{\mathrm{kin}}$; the parallel momentum follows from $k_{\parallel }=\sqrt{2m{E}_{\mathrm{kin}}}\sin \theta /\hslash$, up to surface reciprocal vectors. Clean surfaces and controlled photon energy are needed, and an occupied-state measurement alone does not map an empty conduction band.

Scanning tunneling spectroscopy instead brings a conducting tip close to a surface and modulates the bias while recording the differential tunnel current. At low temperature with a smooth tip density of states and tunnelling matrix element, $dI/dV$ probes the local density of states near energy $E_F+eV$. The signal is thermally broadened and depends on tunnelling selection, so it is not an unqualified map of the whole bulk band structure.

Bulk probes complement these surface measurements. In the de Haas-van Alphen effect, a low-temperature magnetization or torque measurement versus magnetic field gives quantum-oscillation frequencies $F=\hslash A_{\mathrm{ext}}/(2\pi e)$, measuring extremal Fermi surface sections; temperature damping gives a cyclotron effective mass. The linear electronic heat capacity gives the density of low-energy excitations at the Fermi level. Optical absorption measures transition thresholds and matrix-element-weighted joint densities, rather than a unique full dispersion by itself.

A partly filled one-dimensional energy band is especially sensitive to a lattice modulation at $2k_F$. At half filling, $2k_F=\pi /a$, so the Peierls distortion alternates the bond lengths and doubles the primitive unit cell. Mixing states at the Fermi points opens a band gap, lowering the energy of occupied states. In the ideal weak-coupling, zero-temperature chain, the electronic gain behaves as $-u^2\ln (1/|u|)$ while the elastic cost is proportional to $u^2$, favouring a small distortion. The dimerized tight-binding chain gives an explicit band calculation.

The structural signature is extra diffraction intensity at the modulation wavevector; optical absorption can detect the gap and transport can detect loss of metallic conduction. A dynamic precursor is a Kohn anomaly, with phonon softening near $2k_F$. Inelastic neutron scattering of TTF-TCNQ, using a deuterated sample over $60$–$300\mathrm{K}$, found an anomaly near $2k_F$ developing below about $150\mathrm{K}$ as the $54\mathrm{K}$ transition was approached. This probes a lattice mode coupled to the electronic instability rather than inferring distortion from resistance alone.^[ref]

Away from half filling, the modulation need not simply double the lattice. Finite-temperature ordering in real quasi-one-dimensional solids also involves interchain coupling and fluctuations; the ideal isolated-chain argument alone does not establish a finite-temperature transition.

A wave packet in an energy band has group velocity $v={\hslash }^{-1}dE/dk$. The semiclassical force law $\hslash \dot{k}=F$ then gives $\dot{v}={\hslash }^{-2}(d^{2}E/d{k}^2)F$. The curvature therefore defines the one-dimensional effective mass by $1/m^{\ast }=E^{\prime \prime }/{\hslash }^2$. Here $E^{\prime \prime }(k)=A{a}^2\cos (ka)$, so at the band minimum $k=0$,It measures acceleration within the band, rather than a change in the electron's rest mass. At a band maximum the electron curvature mass can be negative; a missing electron is then described by a positively charged hole with a positive curvature mass.

Use two plane waves, $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$, as a basis near their Bragg degeneracy. The potential has Fourier coefficients $V_{\pm \mathbf{G}}$; its zero Fourier coefficient vanishes. All other plane waves are off resonance and are neglected in the leading nearly-free electron model. With ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$, the Schrödinger equation reduces to the Hermitian matrix problemSetting the characteristic polynomial equal to zero givesSolving this quadratic equation givesFor the real cosine potential its coefficient $V_G$ is real; the absolute-value form also allows an arbitrary choice of origin and corresponding Fourier phase.

At the Brillouin-zone boundary, $\mathbf{k}\cdot \mathbf{G}=G^2/2$, the free energies are equal. The two standing waves have different probability densities relative to the lattice potential and are split byRepeating this avoided crossing at the reciprocal lattice boundaries joins smooth branches into energy bands. In one dimension a nonzero Fourier coefficient opens a gap at the corresponding crossing. In higher dimensions a direct gap at one wavevector need not give a common forbidden energy interval at every wavevector: different bands may still overlap in energy. The present two-dimensional example remains metallic even though the crossings are locally gapped.

For the rectangular lattice, choose real-space primitive vectors $a\hat{\mathbf{x}}$ and $b\hat{\mathbf{y}}$. The reciprocal lattice vectors areDefine the half-widths of the first Brillouin zone by $K=\pi /a$ and $L=\pi /b$. Their values and the zone areA primitive reciprocal cell can be drawn with a corner at the origin; its centred translate is the first Brillouin zone here. Its area is $4{\pi }^2/(ab)=37.9600\mathrm{n}{\mathrm{m}}^{-2}$.

Including both spin states, counting the occupied free-electron states per unit area givesConsequentlyThe occupied disk has precisely the area of the first Brillouin zone, but their shapes differ: $K<k_F<L$. It protrudes through both vertical zone boundaries and does not reach the horizontal ones. Thus two electrons per cell do not imply a filled first band in the absence of a separating band gap.

Fold the protruding caps by $\pm {\mathbf{G}}_x$. The original circle and the translated circle intersect at $k_x=K$, $k_y=\pm \sqrt{k_F^2-K^2}$. The potential mixes these states and replaces each crossing by an avoided crossing. The lower branch reconnects into two open sheets running across successive zones in the $k_x$ direction; the upper branch forms a closed lens-shaped pocket about a vertical zone boundary. In a reduced-zone sketch its two halves sit at opposite edges, but periodic identification makes them one closed sheet. This is rectangular-lattice Fermi-surface reconstruction.

The largest pocket in this weak-potential construction is the lens-shaped intersection of two disks enclosed by the intersecting free circles. Each has radius $k_F$ and their centres are separated by $2K$. If $\theta =\arccos (K/k_F)$, two sectors minus their two triangles give the lens bound on a reconstructed Fermi pocket:Opening a gap raises the upper branch at fixed chemical potential and shrinks its pocket. The reported quantum oscillation area satisfiesIt is therefore consistent with this closed sheet. Onsager quantization gives $F=\hslash A_k/(2\pi e)\simeq 601\mathrm{T}$. An orbit on this closed contour can return to itself under the Lorentz force and occupy quantized Landau levels. The open sheets do not enclose a finite orbit area for a perpendicular field in this two-dimensional model. The observation is consistent with the reconstructed pocket, without uniquely determining a completely arbitrary band structure.

For a quantitative rough coupling estimate, use the supplied fixed-chemical-potential approximation and retain the two waves coupled by ${\mathbf{G}}_x$. WriteTheir upper eigenvalue isSetting it equal to $C{k}_F^2$ gives the pocket boundaryIts intersection with $k_y=0$ occurs at $q=\pm q_m$, whereThe two-wave estimate of a Bragg-gap coupling from a pocket area therefore usesAt $v=0$ this reproduces the lens bound on a reconstructed Fermi pocket; the pocket disappears at $v=C(k_F^2-K^2)=0.23786\mathrm{eV}$. Numerical integration gives $A(0.075\mathrm{eV})=6.22\mathrm{n}{\mathrm{m}}^{-2}$ and $A(0.100\mathrm{eV})=5.54\mathrm{n}{\mathrm{m}}^{-2}$. Matching the measured area gives $v\simeq 0.0931\mathrm{eV}$, so the justified rough result isThis estimate neglects remote plane-wave admixtures and other Fourier coefficients. It also holds the chemical potential fixed as requested; an exact fixed-density calculation would determine it self-consistently after reconstruction. The numerical root is a way to implement the approximate model, not a measurement of $V_G$ to four significant figures.

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.

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 $m/m^{\ast }$.

For a lattice-periodic one-electron Hamiltonian, Bloch's theorem givesHere $\mathbf{R}$ is any lattice translation and $n$ labels an energy band. Under that translation the wavefunction gains phase $e^{i\mathbf{k}\cdot \mathbf{R}}$. The crystal momentum $\hslash \mathbf{k}$ labels this translation eigenvalue and is defined modulo $\hslash \mathbf{G}$ for a reciprocal lattice vector $\mathbf{G}$.

The mechanical momentum is different. In zero magnetic field its operator is $-i\hslash \nabla$, which acts asA general Bloch state is therefore not a mechanical-momentum eigenstate. Expanding its periodic factor gives plane-wave components with momenta $\hslash (\mathbf{k}+\mathbf{G})$, with probabilities determined by their Fourier coefficients. For a local Hamiltonian $p^2/(2m_e)+V(\mathbf{r})$, its mean momentum satisfiesnot generally $\hslash \mathbf{k}$. The lattice exchanges actual momentum with the electron; crystal-momentum conservation in scattering is instead modulo reciprocal-lattice vectors, as in Umklapp scattering.

In the nearly-free electron model, a weak periodic potential has Fourier components $V_{\mathbf{G}}$ that couple plane waves at $\mathbf{k}$ and $\mathbf{k}-\mathbf{G}$. Far from a degeneracy their admixture gives small perturbative energy shifts. At a Bragg plane, however,so degenerate perturbation theory is required. In one dimension the first such crossing is at $k=\pm G/2$, the boundary of the first Brillouin zone.

Within the degenerate pair the eigenvalues are ${ϵ}_{G/2}\pm |V_G|$. ThusThe eigenstates are standing-wave combinations of counterpropagating waves. Their charge densities sit differently relative to the ionic potential, giving different potential energies and removing the free-electron degeneracy. The resulting avoided crossing produces continuous allowed energy bands separated by a forbidden interval. A Fourier component or symmetry-allowed coupling is needed: geometry alone does not require every possible gap to be nonzero.

For state counting in a Bloch band, take a primitive cubic lattice of spacing $a$ and impose periodic boundary conditions on lengths $L_i=N_{i}a$. The allowed wavevectors are $k_i=2\pi l_i/L_i$. The first Brillouin zone has side $2\pi /a$, so it contains $N_i$ distinct values in each direction and hence $N_x{N}_y{N}_z=N$ wavevectors. Each orbital-derived energy band has one state per wavevector per spin orientation. ThereforeThe phrase “per atom” attached to $2N$ in the PDF is a counting error: $2N$ is the total, not the per-atom count.

By the Pauli exclusion principle, a monovalent crystal supplies $N$ valence electrons and half-fills a spin-degenerate band; its Fermi level lies within allowed states, giving a metal in the independent-electron picture. Two electrons per atom can fill one band. If a band gap separates it from all empty bands, the crystal is a band insulator. An even valence count only permits insulation: overlapping bands can still produce a metal. Conversely, electron interactions can produce a Mott insulator, and symmetry breaking can permit insulation outside this simple spin-degenerate electronic band structure criterion. The relevant count in a crystal with a basis is per primitive unit cell.

For a Hamiltonian with a lattice-periodic potential $V(\mathbf{r}+\mathbf{R})=V(\mathbf{r})$, its simultaneous energy eigenstates and lattice-translation eigenstates can be chosen in the Bloch wave formEquivalently, ${\psi }_{n\mathbf{k}}(\mathbf{r}+\mathbf{R})=e^{i\mathbf{k}\cdot \mathbf{R}}{\psi }_{n\mathbf{k}}(\mathbf{r})$. The index $n$ labels an energy band; crystal momentum $\hslash \mathbf{k}$ is defined modulo a reciprocal lattice vector. These are the content of Bloch's theorem.

The electronic bandwidth is $W(a)=24E_1{e}^{-a/a_1}$. With $a$ reduced from $a_1$ to $0.99a_1$,Thus the energy band widens by approximately 1.005 percent, or 1 percent to first order. Compression increases the orbital coupling.

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.

Low-temperature susceptibility at fixed number. Apply the same kernel expansion to $g$ itself:The shift of $\mu$ must be retained in the first term. Writing $g_F=g({\epsilon }_F)$ and similarly for its derivatives,ConsequentlyThis is the fixed-number low-temperature Pauli susceptibility. The $g_F^{\prime \prime }$ contribution is thermal broadening at fixed chemical potential; the $-g_F^{\prime 2}/g_F$ contribution enforces fixed total number. Both are needed. If the density of states were instead defined with both spin directions included, the prefactor $2$ would be absorbed into that definition. Only the spin response has been calculated; no orbital magnetic response is included.

The expansion assumes $g_F>0$, sufficient smoothness near the Fermi energy, and a temperature small compared with the energy scale over which $g$ changes or encounters a band edge. It does not apply unmodified at a singular density of states. For the ideal two-dimensional parabolic band, $g_F^{\prime }=g_F^{\prime \prime }=0$, so the leading Pauli response is temperature independent, consistently with the earlier state count.

A Peierls distortion modulates the atomic positions and chemical bonds of a one-dimensional chain with a partly filled energy band. In the half-filled example, alternating short and long bonds double the primitive unit cell. The distortion is the structural change; the associated Peierls transition is the change in electronic conduction.^[ref]

A direct band gap compares two energy bands at the same crystal momentum. A global gap instead requires the lowest energy of the upper band to exceed the highest energy of the lower band, allowing the two extrema to occur at different momenta. For example, the periodic toy bands $E_1(k)=A\cos k$ and $E_2(k)=A\cos k+\delta$, with $A>0$ and $0<\delta <2A$, have positive direct separation $\delta$ everywhere but global overlap $2A-\delta$. A positive direct splitting at each Brillouin zone boundary therefore does not alone establish an insulating filled band. At a suitable electron count the overlap can produce both electron and hole pockets.

The curvature tensor of a nondegenerate energy band, determining the linear relation between force and acceleration in semiclassical band dynamics. From group velocity $v_i={\hslash }^{-1}{\partial }_{k_i}E$ and $\hslash {\dot{k}}_j=F_j$, one obtains ${\dot{v}}_i=(M^{-1}{)}_{ij}{F}_j$. Near a parabolic minimum, $E-E_0={\hslash }^2\delta {\mathbf{k}}^T{M}^{-1}\delta \mathbf{k}/2$. The principal values are reciprocal curvature effective masses; a cyclotron experiment measures a direction-dependent combination rather than every tensor entry separately.

For one orthonormal localized orbital per simple-cubic site and real equal nearest-neighbour hopping integral $t$, the six neighbours give the displayed energy band. Each cosine ranges from minus one to one, so the positive bandwidth is $12|t|$, independently of the sign convention for $t$. Near the zone centre, $E=E_0+6t-t{a}^2{k}^2+O(k^4)$ and the band effective mass is $-{\hslash }^2/(2t{a}^2)$. It is positive there for $t<0$; for $t>0$ the zone centre is a maximum instead.

For a divalent square Bravais lattice, let $E_X={\hslash }^2{\pi }^2/(2m_e{a}^2)$ and real periodic potential coefficients $V_{10}<V_{11}<0$. The leading two-wave edge levels are $E_X\pm V_{10}$. The four-wave corner levels are $2E_X+V_{11}\pm 2V_{10}$ and a doublet at $2E_X-V_{11}$. Comparing the lowest corner level with the upper edge level gives the displayed indirect band gap. The first spin-degenerate energy band can hold the two electrons per primitive unit cell; it becomes isolated when $-E_X-V_{11}-3V_{10}>0$ in this truncation. This follows from resonant degenerate perturbation theory, and higher plane waves can shift the threshold.

A crystal with $N$ primitive unit cells has $N$ distinct wavevectors in one Brillouin zone. For a rectangular periodic sample, $k_i=2\pi l_i/L_i$, and the product of the numbers of allowed values along the three directions equals the number of cells. Each orbital-derived energy band therefore holds $N$ states per spin orientation, or $2N$ with twofold spin degeneracy. This means two states per primitive cell per spin-degenerate band, independently of how many atoms the cell contains.

For symmetric touching parabolic dispersions with constant density of states $D$ per energy band, including spin degeneracy, charge neutrality fixes $\mu =0$. The Fermi-Dirac distribution gives $N_c=D{k}_{B}T\ln 2$ and the thermal internal energy, including electrons and holes, is $\Delta U=({\pi }^2/6)D(k_{B}T{)}^2$. Thus $C=({\pi }^2/3)D{k}_B^{2}T$. Its large ratio to $N_c$ is compatible with the equipartition theorem because this excitation count grows with temperature and the heat capacity includes both species.

A tunnel diode is a p-n junction with sufficiently heavy semiconductor doping on both sides to make a thin depletion region and permit substantial interband tunnelling. The neutral regions are degenerate semiconductors. Its distinctive forward current-voltage characteristic contains a peak followed by a valley: over part of the curve, increasing the applied electric potential difference decreases the electric current. The Esaki tunnel-diode experiment made this behavior experimentally concrete.

The useful quantity is the local slope, not the ratio of total voltage to total electric current. With conventional electric current $I$ entering the positive-voltage terminal, defineNegative differential resistance means $g_d<0$, hence $r_d<0$ where defined. A forward operating point can still have $V>0$, $I>0$ and absorbed power $VI>0$. A small voltage increase then gives a negative current increment; this does not make the total electric current negative or turn the device into an unlimited source of energy. The external supply sets the operating point and supplies the energy available for an electronic amplifier.

An original idealized construction explains the shape. Put the n-type semiconductor on the left and the p-type semiconductor on the right. Take $q>0$ as the elementary charge and $V=V_p-V_n$. Increasing forward bias lowers the right-hand electron energy levels relative to the left. Denote the two reservoir electrochemical potentials by ${\mu }_n,{\mu }_p$, so that ${\mu }_n-{\mu }_p=qV$. At zero temperature, occupied left-hand conduction band states extend up to ${\mu }_n$; empty right-hand valence band states start at ${\mu }_p$. For transfer conserving electron energy, the available interval isIt is nonempty only when its upper endpoint exceeds its lower endpoint. Both an occupied initial quantum state and an empty final quantum state must exist at the same energy. The Pauli exclusion principle prevents transfer into an already occupied final quantum state.