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Let $N=L/a$ be the integer number of primitive unit cells. A Bloch state can be written as ${\psi }_k(x)=e^{ikx}{u}_k(x)$, with $u_k(x+a)=u_k(x)$. The periodic boundary condition ${\psi }_k(x+L)={\psi }_k(x)$ therefore requiresThe spacing of allowed wavevectors is $\Delta k=2\pi /L$. The first Brillouin zone has width $2\pi /a$, soFor an exact endpoint convention take a half-open Brillouin zone, such as $[-\pi /a,\pi /a)$. Its two endpoints differ by a reciprocal lattice vector and must not both be counted as distinct values of crystal momentum. There are exactly $N$ inequivalent integers $j$ modulo $N$, irrespective of whether $N$ is even or odd. This is state counting in a Bloch band: one allowed wavevector per primitive unit cell per energy band. A separate twofold spin multiplicity gives $2N$ electronic states per energy band, without doubling the number of wavevectors.

The crystal has periodic potential $V(x+a)=V(x)$. By Bloch's theorem, an energy eigenstate may be chosen asExpand the periodic factor in a Fourier series:Multiplication by $e^{ikx}$ gives the stated plane-wave expansion. Here $x$ is position, $a$ is the primitive unit cell length, $G_1$ is the primitive reciprocal lattice wavenumber, $n$ runs over every integer, and $C_{k,n}$ are complex numbers specifying the component wave amplitudes. The wavevector $k$ labels the Bloch state modulo $G_1$; $\hslash k$ is its crystal momentum, rather than generally its measured free-space momentum.

The component $e^{i(k+n{G}_1)x}$ is a plane wave of wavevector $k+n{G}_1$. All these components acquire the same phase factor $e^{ika}$ under $x\mapsto x+a$, since $e^{in{G}_{1}a}=1$. Their mixing therefore respects the crystal's translation symmetry. The coefficients and energy eigenvalue follow from the Schrödinger equation; the expression is not a claim that the coefficients are equal or that only finitely many are nonzero. The energy band index is suppressed in the notation, so different energy bands at a common $k$ generally have different coefficient sets.

The Debye model approximates a three-dimensional crystal as an elastic continuum supporting quantized normal modes, or phonons. At long wavelengths, the acoustic phonon branches have approximately linear dispersion relations ${\omega }_s(k)=v_{s}k$: one longitudinal branch and two transverse branches. Replace their complicated actual spectra by these linear forms, then introduce a cutoff chosen to preserve the total $3N$ vibrational degrees of freedom for $N$ atoms. This is an approximation to the spectrum, not a claim that a real crystal has no non-acoustic branches.

For periodic boundary conditions in volume $V$, each allowed wavevector occupies $(2\pi {)}^3/V$ in the space of wavevectors. Counting a spherical shell for each branch gives the density of states per unit angular frequencyAn effective speed of sound $v_D$ can be defined by $3/v_D^3=1/v_l^3+2/v_t^3$. In the single-speed Debye model, choose ${\omega }_D=v_D{k}_D$ so thatThus the Debye wavevector, Debye frequency and Debye temperature satisfyHere $g$ has units of seconds when angular frequency is measured in ${\mathrm{s}}^{-1}$; its integral is the dimensionless number of normal modes. The spherical cutoff replaces the real Brillouin zone by an equal-count continuum construction.

Each phonon mode is a quantum harmonic oscillator. Since phonon number is not conserved at thermal equilibrium, the chemical potential is zero and its Bose-Einstein distribution gives mean thermal energy $\hslash \omega /[e^{\hslash \omega /(k_{B}T)}-1]$. The zero-point energy is independent of temperature in the harmonic, fixed-volume approximation and does not contribute to the heat capacity at constant volume. Excluding that constant term,Differentiating the first integral at fixed volume avoids losing the temperature-dependent upper limit in the second one. The resulting Debye heat capacity isFor $T\ll {\Theta }_D$, extend the upper limit to infinity. Expanding $1/(e^x-1)={\sum }_{r\ge 1}{e}^{-rx}$ and integrating term by term gives ${\int }_0^{\infty }{x}^3/(e^x-1)dx=6\zeta (4)={\pi }^4/15$. ThereforeThis $T^3$ law follows from the three-dimensional ${\omega }^2$ density of states and the availability only of low-frequency modes at low temperature. It is a central success of the Debye model.

For $T\gg {\Theta }_D$, the Bose-Einstein distribution reduces to the classical equipartition theorem: every mode contributes thermal energy $k_{B}T$. Consequentlythe Dulong-Petit law, or $3R$ per mole of atoms. The simple model omits detailed phonon dispersion relation, distinct optical phonon branches, and anharmonic effects such as thermal expansion. In a metal, the low-temperature total heat capacity additionally contains a term approximately proportional to $T$ from electrons, so the lattice $T^3$ term should not be mistaken for the whole measurement.

Let $d=1,2,3$ and take a specimen of length, area or volume $L^d$. One monovalent atom per unit cell supplieselectrons. At zero temperature, the Pauli exclusion principle fills the lowest energy states, with two spin states per spatial wavevector. The occupied region is an interval, disk or ball of radius $k_F$. Using the $k$-space density of states gives, in one dimension,In two dimensions,In three dimensions,Each Fermi energy follows from the free electron model dispersion relation $E_F={\hslash }^2{k}_F^2/(2m)$. The Fermi wavevectors have units ${\mathrm{m}}^{-1}$ and the Fermi energies have units joules for $a$ in metres and $m$ in kilograms. No numerical lattice spacing is supplied, so numerical energies in electronvolts cannot be uniquely calculated. The spin factor of two is necessary here for actual monovalent crystals; the spinless auxiliary count of part (a) was not an instruction to remove electron spin from these crystals.

To determine a crystal structure, scatter probes whose wavelength is comparable to atomic spacings, typically $0.1\mathrm{nm}$. X-ray diffraction, neutron diffraction and electron diffraction probe periodic order through coherent elastic scattering. With incident and outgoing wavevectors ${\mathbf{k}}_i,{\mathbf{k}}_f$, define $\mathbf{Q}={\mathbf{k}}_i-{\mathbf{k}}_f$. Elastic scattering has $|{\mathbf{k}}_i|=|{\mathbf{k}}_f|$ and constructive interference occurs when $\mathbf{Q}=\mathbf{G}$, a reciprocal lattice vector. The equivalent Bragg's law iswhere the detector scattering angle is $2\theta$, not $\theta$. Peak positions reveal plane spacings and the unit cell; systematic absent reflections constrain centring and the atomic basis.

For X-ray diffraction, a laboratory tube or a synchrotron supplies a collimated beam. In a monochromatic single-crystal experiment a goniometer rotates the specimen and an area detector records reflection positions and integrated wave intensities. A powder experiment uses many randomly oriented crystallites; the reflections form cones recorded as rings or peaks versus $2\theta$, permitting indexing and determination of lattice constants without selecting one crystallite. For a cubic cell, for example, $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$. The white-beam Laue method is especially useful for crystal orientation.

The crystal basis structure factor isafter geometric and thermal corrections, including the Debye–Waller factor. For X-rays, $f_j$ describes scattering from the atomic electron density. Thus positions provide lattice geometry while relative wave intensities constrain atomic positions and species. Wave intensities do not directly give the phases of $F_{\mathbf{G}}$: structure refinement or other constraints must address the phase problem.

Neutron diffraction uses a reactor or spallation source, wavelength selection and a detector bank or time-of-flight detection. Neutrons interact with nuclei through nuclear scattering lengths that vary between elements and isotopes rather than simply with atomic number. This makes them useful for light atoms, isotope contrast and distinguishing some neighbours in the periodic table; their magnetic interaction also probes magnetic order. Electron diffraction in a transmission electron microscope gives strong signals from very small thin crystallites, although strong multiple scattering often complicates quantitative wave intensity analysis. These techniques complement one another.

A phonon dispersion relation requires both momentum and energy transfer, so elastic diffraction alone is insufficient. In coherent inelastic neutron scattering, the measured transfers for phonon creation areHere $\mathbf{q}$ is the phonon wavevector reduced into a Brillouin zone, and $s$ labels an acoustic phonon branch or optical phonon branch. Negative energy transfer corresponds to annihilating an already occupied phonon. For neutrons $E={\hslash }^2{k}^2/(2m_n)$, so wavelength and energy selection are linked. Typical thermal-neutron energies of a few to tens of millielectronvolts ($1\mathrm{meV}=10^{-3}$ electronvolts) are well matched to many lattice vibrations.

A triple-axis spectrometer uses a monochromator crystal to select ${\mathbf{k}}_i$, the sample as the second scattering axis, and an analyser crystal to select ${\mathbf{k}}_f$ before detection. Rotate these elements and the specimen to hold $\mathbf{Q}$ fixed while scanning $E_i-E_f$; inelastic peaks locate ${\omega }_s(\mathbf{q})$. Repeat at different $\mathbf{q}$ along symmetry directions to map the phonon dispersion relation. The dynamic structure factor governs peak weights, including the selection of phonon polarization by the scattering geometry. A time-of-flight spectrometer instead infers outgoing neutron energy from travel time over a known flight path and records many directions simultaneously.

Inelastic X-ray scattering can also resolve phonons when very high energy resolution is available. Optical Raman scattering and Brillouin scattering measure vibrations near the zone centre because photon momentum transfers are small compared with a typical reciprocal lattice spacing; they complement, rather than generally map the entire zone like, momentum-resolved neutron measurements. Ultrasound measurements give the long-wavelength sound speed, the initial slope $d\omega /dq$ of an acoustic phonon branch.

For a homogeneous isotropic solid with small displacement $\mathbf{u}$, linear elasticity gives ${\sigma }_{ij}={\lambda }_L(\nabla \cdot \mathbf{u}){\delta }_{ij}+2\mu e_{ij}$. Momentum balance becomesA plane wave with displacement parallel to its wavevector is longitudinal, whereas perpendicular displacement is transverse. Writing $K={\lambda }_L+2\mu /3$ for the bulk modulus and $\mu$ for the shear modulus, their bulk speeds areThere are two independent transverse polarizations. The bulk modulus supplies resistance to compression; shear rigidity supplies transverse restoring stress. An ordinary fluid has no static shear rigidity and cannot sustain these propagating bulk shear waves. A thin rod with freely contracting sides instead has longitudinal speed $\sqrt{Y/\rho }$, because its lateral boundary conditions differ from a bulk plane wave. Reflection, refraction and mode conversion occur at interfaces; a free surface also supports Rayleigh waves.

An incident electromagnetic wave generally produces reflected and transmitted waves because the two media have different electromagnetic response. Consider a flat interface between homogeneous, linear, lossless media with permittivities ${ϵ}_j$ and magnetic permeabilities ${\mu }_j$. Their wave speeds and wave impedances areFor each plane electromagnetic wave, $\mathbf{H}=\hat{\mathbf{k}}\times \mathbf{E}/Z_j$. With no free surface charge or current, the interface conditions for electromagnetic fields require continuous tangential $\mathbf{E},\mathbf{H}$ and normal $\mathbf{D},\mathbf{B}$. Equality of the phase at every point on the boundary gives equal angular frequencies and tangential wavevectors. Hence the reflection angle equals the incidence angle and Snell's law iswhere $n_j=c/v_j$ and each angle is measured from the normal.

At normal incidence use the same tangential axis for all electric amplitudes. The two independent boundary equations areSolving gives the amplitude Fresnel equationsFor nonmagnetic media, these become $r=(n_1-n_2)/(n_1+n_2)$ and $t=2n_1/(n_1+n_2)$. Reflection into a medium of larger refractive index therefore reverses the electric amplitude's sign at normal incidence.

The normal Poynting vector determines the reflected and transmitted fractions of incident power. At normal incidence,It is the power fractions, rather than the squared electric amplitudes alone, that add to unity.

At oblique incidence separate transverse electric polarization ($s$, electric field perpendicular to the incidence plane) from transverse magnetic polarization ($p$, electric field in that plane). For nonmagnetic media, writing $c_i=\cos {\theta }_i$ and $c_t=\cos {\theta }_t$, the Fresnel equations areHere choose a common $s$ direction and, for each ray, the $p$ direction as $\hat{\mathbf{s}}\times \hat{\mathbf{k}}$. This makes the reflected $p$ basis reverse its tangential direction at normal incidence, explaining why the displayed $r_p$ then has the opposite sign to $r_s$. Physical reflectance is independent of this sign convention. For either polarization with propagating transmission,For $p$ polarization, zero reflection occurs at the Brewster angle $\tan {\theta }_B=n_2/n_1$. If $n_1>n_2$, incidence above the critical angle for total internal reflection $\sin {\theta }_c=n_2/n_1$ produces an evanescent field in total internal reflection and $\mathcal{R}=1$ for an isolated lossless interface. A lossy second medium requires complex response coefficients and an absorption term in the energy balance.

A cubic reflecting cavity supports standing waves with positive integer mode indices and wavevector components $k_i=\pi n_i/L$. A cell in mode space has volume $(\pi /L{)}^3$. In the continuum approximation, a spherical shell of radius $k$ and thickness $dk$ has one-eighth of its volume in the positive octant. There are two independent transverse photon polarizations for a generic wavevector. HenceThe photon energy is $ϵ=\hslash \omega =\hslash ck$, so $dk=dϵ/(\hslash c)$. ThereforeThis is the leading bulk density of states, valid when the cavity is large compared with the relevant wavelengths and many modes lie in the counting interval. A finite cavity has a discrete mode spectrum, with boundary corrections beyond this approximation. Modes with vanishing indices require separate boundary counting but do not change the leading $L^3$ term. Here $\hslash$ is the reduced Planck constant, as printed in the original PDF; replacing it by $h$ would introduce an erroneous factor of $(2\pi {)}^3$.

The displacement is along $x$, transverse to propagation in the membrane's $yz$ plane. Write $A=|A|e^{i\varphi }$; then $d=|A|\cos (\omega t-k_{y}y-k_{z}z+\varphi )$. Here $|A|$ is the displacement amplitude, $\varphi$ the initial phase, $\omega$ the angular frequency and $\mathbf{k}$ the wavevector normal to constant-phase lines. Its magnitude is $2\pi /\lambda$, and its direction specifies phase propagation for the chosen sign convention.

For equal-phase two-wave membrane modes, two equal complex amplitudes with opposite $k_y$ sum toThe edge condition gives $\cos (k_{y}a/2)=0$, hence $k_y=(2j+1)\pi /a$, $j=0,1,\dots$, for this even-in-$y$ family. Opposite relative phase gives the sine family; together they yield all Dirichlet transverse modes $k_y=n\pi /a$, $n\ge 1$. For the equal-phase family,The lowest mode has $k_y=\pi /a$. ThereforeIt approaches $c$ for $\lambda \to 0$ and diverges as $\lambda \to 2a^{-}$; there is no real travelling-mode $k_z$ beyond the cutoff. This is an axial phase velocity, not the group velocity: $v_g=c\sqrt{1-(\lambda /2a{)}^2}<c$. Here $c$ denotes the membrane wave speed, not the speed of light.

For the evanescent membrane response below cutoff, below $f_c=c/(2a)$, put $k_z=-i\kappa$ for decay along positive $z$, with $\kappa =\sqrt{(\pi /a{)}^2-(\omega /c{)}^2}$. A localized central drive creates an evanescent, sinusoidally oscillating deformation decaying away in both axial directions. Far enough from the drive the leading profile is proportional to $\cos (\pi y/a)e^{-\kappa |z-z_0|}\cos (\omega t+{\varphi }_0)$. Higher evanescent modes contribute near the point source. The pattern has no steadily advancing axial phase front or far-field propagating wave; decreasing frequency increases its spatial localization.

Take a Northern-Hemisphere cyclonic vortex, with positive relative vorticity. For a fixed-depth incompressible layer, conservation of absolute vorticity givesThe anticlockwise flow on the east side moves parcels northward, increasing their planetary vorticity and creating negative relative vorticity anomalies. On the west side it moves parcels southward and creates positive relative vorticity anomalies. Initially,so the positive anomaly grows on the western flank while the eastern flank loses positive vorticity. The cyclonic pattern consequently shifts westward. The induced anomalies themselves drive meridional flow and further relative vorticity anomalies, transmitting a Rossby wave pattern through the surrounding fluid. The restoring process is this exchange between planetary vorticity and relative vorticity, rather than a parcel executing a permanently westward translation.

The sketch shows the initial tendency, with the generated signs separated from the initial positive core. An isolated disturbance contains many wavevectors and disperses; westward phase velocity does not imply that all its wave energy travels westward. A Southern-Hemisphere cyclonic vortex reverses the signs of the initial and generated relative vorticities, while the same beta plane mechanism remains.

For an isotropic free-electron gas at zero temperature, the Fermi wavevector is the radius of the occupied region in wavevector space. Its energy is $E_F={\hslash }^2{k}_F^2/(2m)$. The first Brillouin zone is the Wigner-Seitz cell of the reciprocal lattice; it contains one distinct crystal momentum per primitive cell and per band, with spin counted separately.

For a two-dimensional sample of area $A$, allowed wavevectors have density $A/(2\pi {)}^2$. Including two spin states, filling a circle gives $N=2[A/(2\pi {)}^2]\pi k_F^2$. Dividing by area therefore yieldsThe square lattice has one atom per area $a^2$, so for valence $z$ the free-electron density is $z/a^2$. Its Brillouin zone has $|k_x|,|k_y|\le \pi /a$. ThusThe monovalent Fermi circle lies inside the square. The divalent circle crosses the faces but misses the corners; its area equals the zone's area. These are the free-electron contours before distortion by the periodic potential.

At a Bragg plane, the periodic potential mixes degenerate plane waves differing by a reciprocal lattice vector. The two standing-wave combinations have different charge densities relative to the ions and split in energy, opening the gap. In a two-wave approximation, the boundary energies are $E_B\pm E_g/2$, with $E_g=2|V_G|$.

The boundary wavevector magnitude is $\pi /a$ along $[10]$ and $\sqrt{2}\pi /a$ along $[11]$. The free-electron boundary energies are thereforeWithin the stated direction-independent gap picture, the upper band's lowest face-edge energy is approximately $E_X+2\mathrm{eV}=6.18\mathrm{eV}$, while the lower band's corner-edge energy is approximately $E_M-2\mathrm{eV}=6.36\mathrm{eV}$. There is consequently a small overlap, about $0.18\mathrm{eV}$, rather than a global insulating gap. The divalent Fermi level lies in that overlap, producing occupied upper-band states and unfilled lower-band states.

Thus the intended isotropic-gap approximation predicts a metal, with small electron and hole pockets. Positive direct gaps can coexist with band overlap. A fully specified square-lattice potential would require simultaneous treatment of the corner degeneracies; a numerical boundary splitting alone does not uniquely determine an arbitrary two-dimensional band structure. The conclusion here uses the given simplified gap picture, rather than electron count alone.

The mean interface height fluctuation is zero; its root mean square roughness is obtained by summing the mode variances. Replacing the two-dimensional wavevector sum by its continuum density gives the thermal roughness of a tension-dominated interface:Take $q_{\min }=2\pi /L$ with $L=\sqrt{A}=10^{-2}\mathrm{m}$, and $q_{\max }=2\pi /d$ with $d=2\times 10^{-10}\mathrm{m}$. At $T=300\mathrm{K}$,The logarithm makes the answer only weakly sensitive to order-one choices of cutoff. Gravity has been omitted as stipulated; the fixed mean height removes the zero wavevector mode.

Put $N=N_x{N}_y{N}_z$ primitive cells in a periodic box. In each reciprocal-space direction, the allowed Bloch wavevectors are spaced by $2\pi /L_i$, giving $N_i$ independent values in the first Brillouin zone. There are consequently $N$ wavevectors per band and, including two electron spin states at each wavevector,The total and per-atom counts are different; a factor $N$ cannot remain in the per-atom count for this monatomic lattice.

In independent-electron band theory, a partially occupied band permits low-energy changes of occupancy and gives a metal. Filled bands separated from empty bands by a finite band gap give a band insulator at zero temperature. An odd number of electrons per primitive cell cannot fill spin-degenerate bands, whereas an even number can; even filling alone does not ensure an insulator if bands overlap. Magnetic symmetry breaking and strong correlations, such as a Mott insulator, go beyond this simple counting criterion.

For photon absorption or emission, electron energy changes by $\pm \hslash \omega$ and crystal wavevector by $\pm \mathbf{q}$ modulo a reciprocal lattice vector. Optical photon momentum is normally tiny on the Brillouin zone scale, so interband optical transitions are approximately vertical on an energy-wavevector diagram. A direct-gap semiconductor can therefore absorb a photon between valence and conduction states near the same wavevector; an indirect transition usually needs an additional phonon or another source of momentum. Occupation and optical matrix elements impose further restrictions.

Within one ideal parabolic band of a clean metal, such a small photon momentum cannot normally supply the required finite energy difference. Intraband optical absorption then relies on momentum relaxation by impurities, phonons or boundaries, as in the finite-scattering-time Drude response. Interband transitions can still occur. The first panel shows the near-vertical direct transition.

An optical phonon often has approximately constant energy $\hslash {\omega }_0$ over the relevant wavevector range. Its absorption or emission therefore moves the electron between energy surfaces separated by this finite amount, with the phonon supplying the associated crystal momentum modulo a reciprocal vector. The last panel shows emission between two points on a parabolic conduction band.

Emission from a band with a lower edge requires at least $\hslash {\omega }_0$ of available kinetic energy and an unoccupied final state. For nondegenerate hot carriers this produces a threshold and can lead to rapid energy loss in successive phonon emissions. Absorption is weighted by the Bose-Einstein distribution $n_B({\omega }_0)$, while emission has the factor $n_B+1$; absorption is strongly suppressed at $k_{B}T\ll \hslash {\omega }_0$. In a metal, Pauli blocking also prevents an electron near the Fermi energy from simply dropping into the occupied Fermi sea.

Let $\alpha ={\hslash }^2/(2m^{\ast })$, so the initial conduction-electron kinetic energy is $E_1=\alpha k_1^2$. In the normal momentum-conserving process with the final hole fixed at zero wavevector, the two final conduction electrons obey ${\mathbf{k}}_1={\mathbf{k}}_2+{\mathbf{k}}_3$. Their total kinetic energy satisfiesEnergy conservation for impact ionization is $E_1=E_g+\alpha (k_2^2+k_3^2)$. Consequently $E_1\ge E_g+E_1/2$, givingThe bound is attained when ${\mathbf{k}}_2={\mathbf{k}}_3={\mathbf{k}}_1/2$; opposite spins permit the two electrons to share that wavevector. Fixing the hole momentum is essential: a mobile final hole gives a different threshold. This is the threshold for impact ionization with a hole at zero momentum.

The two hot electrons subsequently redistribute energy by electron-electron collisions and transfer it to the lattice by electron-phonon scattering. Optical-phonon emission can provide large discrete losses while the electrons are hot; acoustic phonons dominate later small energy exchanges. Electron-electron collisions alone conserve the electron system's total energy and do not cool it into the lattice. The electron-hole pair eventually recombines radiatively or through nonradiative channels, such as defect-assisted recombination or Auger recombination. Recombination is distinct from the preceding intraband cooling.

Only states within an energy window of order $k_{B}T$ around the Fermi energy can change occupation in a degenerate metal. A Sommerfeld expansion gives electronic energy above its zero-temperature value proportional to $g(E_F)(k_{B}T{)}^2$, hence $C_e=\gamma T$. Three-dimensional linear-dispersion acoustic phonons have density of states proportional to ${\omega }^2$; integrating their Bose occupations gives thermal energy proportional to $T^4$ and the Debye heat capacity $C_{mph}=\beta T^3$. Thus $C/T=\gamma +\beta T^2$ at sufficiently low temperature in the normal state.

Reading the straight low-temperature fit gives approximately $\gamma \simeq 33\mathrm{mJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-2}$ and $\beta \simeq 3.8\mathrm{mJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-4}$. These estimates agree with the reported fit to this material in the original experiment.^[ref] Treating the graph values as approximate, the molar density of states isThe stipulated two electrons per formula unit give $n=2N_A/V_m\simeq 2.62\times 10^{27}{\mathrm{m}}^{-3}$. For a single spin-degenerate parabolic band, the molar density of states is $g(E_F)=3(2N_A)/(2E_F)=3N_A/E_F$. HenceThese are model-dependent effective parameters inferred from the measured heat capacity and assumed carrier count, rather than an independent determination of a unique band structure.

Using the Debye prefactor stated in the question givesThis prefactor counts three effective modes per formula unit. Consistently choosing the same count for the Debye wavevector gives $k_D=(6{\pi }^2{N}_A/V_m{)}^{1/3}\simeq 4.3\times 10^9{\mathrm{m}}^{-1}$. The corresponding average sound speed isThere is a normalization choice here: a formula unit has 31 atoms. With the conventional full atomic-mode count, the molar Debye coefficient contains an additional factor 31, and both ${\Theta }_D$ and $k_D$ increase by $31^{1/3}$. That convention gives ${\Theta }_D\simeq 251\mathrm{K}$ and $k_D\simeq 1.34\times 10^{10}{\mathrm{m}}^{-1}$, while leaving the inferred sound speed unchanged. The mode count in the Debye model must be used consistently; mixing the printed temperature with an atomic-count wavevector gives an erroneous sound speed.

The upward curvature above about $3\mathrm{K}$ indicates low-energy vibrations missing from a pure acoustic $T^3$ model. In this cage compound, the original analysis fits an additional localized mode with $\hslash \Omega /k_B\simeq 32\mathrm{K}$, assigned to a rattling mode of a barium guest atom.^[ref] At temperatures high compared with all relevant phonon energies, the full lattice approaches the Dulong-Petit law, $C_{mlattice}\to 3(31)R=93R\simeq 773\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The reduced three-mode model alone approaches $3R$ and omits the other branches. An approximately linear electronic term can still be added while the carriers remain degenerate, but cannot be extrapolated beyond their Fermi temperature.

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.

In the nearly-free electron model, the free parabola ${\epsilon }_k^0={\hslash }^2{k}^2/(2m)$ is coupled to copies shifted by reciprocal wavevectors $G=2\pi /a$. The first degeneracy is at $k=\pm \pi /a$, where the counterpropagating waves have equal kinetic energy. Retaining their coupling gives two branchesConsequently a nonzero Fourier component $V_G$ opens a gap $\boxed{\Delta E=2|V_G|}$ at the zone edge. The dispersion remains approximately free-electron-like away from the crossing, but the branches avoid crossing and their slopes flatten at the band edges. The horizontal axis can be labeled either wavevector $k$, with edges at $\pm \pi /a$, or physical momentum $p=\hslash k$, with edges at $\pm \pi \hslash /a$.

At the edge, an appropriate choice of spatial origin makes $V_G$ real. The eigenstates are equal-weight standing waves. In a normalization length $L$ containing an integer number of periods, representative real forms areThey come from the sum and difference of $e^{i\pi x/a}$ and $e^{-i\pi x/a}$, up to a global phase. Their densities have maxima and nodes displaced by $a/2$ relative to one another. Both wavefunctions change sign under translation by $a$, as required by their zone-edge Bloch phase, while their densities are lattice-periodic.

For the figure's choice $V(x)=2V_G\cos (2\pi x/a)$ with $V_G<0$, the potential minima occur at $x=na$. The cosine state puts density there and is lower, while the sine state puts a node there and is higher. Reversing the potential's phase relative to the coordinate origin exchanges that assignment. Since the question supplies only the spatial period, it fixes the zone-edge locations and qualitative standing-wave structure, but not a numerical gap or an origin-independent choice of which drawn sine/cosine is lower.

A Peierls distortion is a spontaneous lattice modulation that can lower the energy of a one-dimensional metal by opening a gap at its Fermi points. The associated change from metallic to gapped electronic behavior is a Peierls transition. The key wavevector is $2k_F$: it couples states near $+k_F$ and $-k_F$, which otherwise have the same energy.

Consider a spin-degenerate chain with one orbital per atom at spacing $a$ and one electron per atom. The band is half filled and $k_F=\pi /(2a)$. A distortion of period $2a$ alternates short and long bonds, doubles the primitive cell, and folds the original Fermi points onto the new zone boundary. The distortion can therefore mix the degenerate states precisely where the occupied and empty states meet.

An explicit dimerized tight-binding chain has alternating hopping amplitudes $-(t+\delta )$ and $-(t-\delta )$, where $t>0$. In the reduced zone $|k|\le \pi /(2a)$,There are two electrons per doubled cell, including spin, so the lower band is filled and the upper one empty. Splitting the degenerate states pushes occupied levels downward and empty ones upward, giving an electronic energy gain.