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A Bravais lattice is a set of equivalent points generated by integer combinations of three independent translation vectors. A crystal motif is the arrangement of atoms attached identically to every lattice point. Together they specify the periodic crystal structure: lattice translations repeat the motif without changing its internal arrangement.

A unit cell is a region whose translations tile the structure. A primitive unit cell contains one lattice point after sharing boundary points with neighboring cells. A conventional cell can contain more than one lattice point to display the crystal's symmetry with simple axes and angles; for example, a conventional face-centered cube has four lattice points but a less visually symmetric primitive cell has one.

Rotational symmetry compatible with a periodic three-dimensional lattice has order $1,2,3,4$ or $6$, as expressed by the crystallographic restriction theorem. A crystallographic fivefold axis is excluded from an ordinary periodic crystal. This statement concerns exact lattice symmetry; it does not exclude local fivefold atomic arrangements or nonperiodic ordered structures.

The zinc blende structure has an FCC sulfur sublattice and zinc on half its tetrahedral sites. One conventional unit cell uses S at $(0,0,0),(1/2,1/2,0),(1/2,0,1/2),(0,1/2,1/2)$ and Zn at those positions displaced by $(1/4,1/4,1/4)$, reduced modulo one. The wurtzite structure has ABAB sulfur layers, instead of ABCABC layers. One hexagonal-cell convention uses S at $(0,0,0),(2/3,1/3,1/2)$ and Zn at $(0,0,u),(2/3,1/3,1/2+u)$, with ideal $u=3/8$. Looking along $[001]$ superposes S and Zn in that latter coordinate convention; the height labels distinguish them.

The closest-packed sulfur planes are $\{111\}$, with four distinct plane orientations, for zinc blende structure, and $\{001\}$, with one orientation, for wurtzite structure. Opposite normals describe the same unoriented plane orientation. Hexagonal four-index notation writes the latter basal family as $\{0001\}$.

Reduce the supplied fractional coordinates modulo one. The rutile structure has Ti at $(0,0)$ with height $z=0$ and at $(1/2,1/2)$ with height $z=1/2$. The oxygen projections are $(0.30,0.30),(0.70,0.70)$ at $z=0$, and $(0.80,0.20),(0.20,0.80)$ at $z=1/2$. Include equivalent corner sites to show the boundary of a complete unit cell; the cell edges are $a$ along $[100]$ and $[010]$.

The fluorite structure has Ce on the corner and face-centre sites of a face-centered cubic lattice. Oxygen occupies all eight positions $(x,y,z)$ with each coordinate independently $1/4$ or $3/4$. In projection, there are four oxygen positions $(1/4,1/4),(1/4,3/4),(3/4,1/4),(3/4,3/4)$, each representing two heights. Accounting for shared boundary sites gives four Ce and eight O per conventional unit cell.

There are four cation sites per conventional unit cell, hence $0.4$ Gd substitutions per cell on average. Two substitutions of Gd${}^{3+}$ for Ce${}^{4+}$ require one oxygen vacancy for charge neutrality, so the cell contains $0.2$ vacancies. Equivalently, the composition is $\mathrm{C}{\mathrm{e}}_{0.9}\mathrm{G}{\mathrm{d}}_{0.1}{\mathrm{O}}_{1.95}$. ThereforeThis uses the supplied lattice parameter literally and treats it as the conventional fluorite-cell edge.

Each missing oxide ion leaves two units of excess positive effective charge, compensated by reducing two Ce${}^{4+}$ ions to Ce${}^{3+}$. For $0.2$ oxygen vacancies per conventional unit cell, $0.4$ of its four cerium ions must be trivalent: 10% Ce${}^{3+}$. This assumes the same number of cation sites per unit volume as in the doped calculation and that oxygen loss is compensated entirely by this cerium-valence change. If reduction changes the cell volume appreciably, matching the absolute vacancy density requires correcting the percentage for that change.

An exciton is a Coulomb-correlated electron-hole pair, electrically neutral overall. Three useful properties are:

In many molecular organic semiconductors, limited screening and localization make exciton binding important, so absorption need not immediately produce collectable free charges.

In a conventional face-centered cubic lattice cell of side $a_F$, touching nearest-neighbour hard spheres lie along a face diagonal. Four radii span that diagonal, so $\sqrt{2}{a}_F=4r$ and $a_F=2\sqrt{2}r$. In a conventional body-centered cubic lattice cell, the contact lies on the body diagonal: $\sqrt{3}{a}_B=4r$, giving $a_B=4r/\sqrt{3}$. ThereforeThese are the nonprimitive cubic unit cells requested. They contain four and two atoms respectively, so their volume ratio should not be confused with the volume-per-atom ratio, which is half as large, $3\sqrt{6}/8\simeq 0.9186$. The common sphere radius cancels.

A Bravais lattice is the set of positions generated by integer combinations of independent primitive translation vectors, $\mathbf{R}=n_1{\mathbf{a}}_1+n_2{\mathbf{a}}_2+n_3{\mathbf{a}}_3$. Every lattice point has the same translational environment. A crystal may additionally have a basis of several atoms attached to each lattice point; those atoms are not automatically separate points of its Bravais lattice.

A unit cell is a region whose translated copies fill the crystal without gaps or overlaps. It can be chosen for convenient symmetry, and a conventional cell may contain several lattice points. A primitive unit cell is a smallest such repeating cell, containing exactly one lattice point when boundary sharing is counted. Its shape is not unique, but its volume $\Omega =|{\mathbf{a}}_1\cdot ({\mathbf{a}}_2\times {\mathbf{a}}_3)|$ is fixed for a given lattice. For a monatomic Bravais crystal, each primitive cell contains one atom.

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.

The length, area and volume of a first Brillouin zone are $(2\pi /a{)}^d$. The corresponding occupied $k$-space measures areHenceThis is state counting, independent of the interval/disk/ball geometry. One Brillouin zone contains one spatial state per unit cell in a single energy band, and that band can hold two electrons per cell because of spin. One electron per cell therefore occupies half of it. The “volume enclosed” is the occupied Fermi sea, not the lower-dimensional area of the Fermi surface itself. Its half-volume does not mean half the distance to the zone boundary in two or three dimensions.

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.

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 $m/m^{\ast }$.

At nonzero band hybridization, the crossing becomes an avoided crossing. The separation at $k_c$ is $2\alpha$. If $\alpha \ll E_{\mathrm{fl}}$, appreciable mixing occurs only where $|A{k}^2-E_{\mathrm{fl}}|\lesssim \alpha$; away from that region each branch retains nearly its original orbital character. For $\alpha \sim E_{\mathrm{fl}}$, the avoided crossing is broad and both branches are substantially reshaped, as shown in the figure.

Use $E_{-}$ for the lower branch and $E_{+}$ for the upper one. For $\alpha >0$,The lower branch increases towards $E_{\mathrm{fl}}$ from below. Thus an indirect gap exists in the exactly flat-band model. In the unbounded-parabola limit its lower bound isA finite Brillouin zone gives a finite maximum of $E_{-}$ below $E_{\mathrm{fl}}$ and hence a slightly larger indirect gap. This hybridization gap differs from the local direct gap $2\alpha$.

The material can therefore be a band insulator when the lower band is full and the upper band empty. With two spin-degenerate orbitals per unit cell, the required filling isWith spin degeneracy removed, replace two by one for each independent spin sector. Other fillings are metallic in this noninteracting model.

The sum needs a stated interpretation. The original equation leaves the second bond outside explicit parentheses, with its index then formally free. The natural translation-invariant, equal-bond interpretation used here is the energy $-J{\sum }_j{\tau }_j({\sigma }_j+{\sigma }_{j+1})$.

Let $K=\beta J$. Sum over each decorating spin before summing over the remaining spins. The decimation of an alternating mixed-spin chain gives the effective transfer matrix:No on-site entropy has been omitted: the term $1$ is the weight of the zero decorating spin. The normalized eigenvectors are $v_{\pm }=(1,\pm 1{)}^{\mathsf{T}}/\sqrt{2}$, with eigenvaluesThere are $L$ transfer steps but $2L$ physical sites, soThe free energy per two-site unit cell is twice this value. At $J=0$, the formula gives $Z_L=6^L$, as required by two states on one sublattice and three on the other. As $T\to 0$, the free energy per physical site tends to $-J$, the energy of an aligned chain with two satisfied bonds per unit cell. These checks also fix the factor of two in the site count.

Fermion occupation and ordinary metallic heat capacity. The Fermi-Dirac distribution for a single-particle state isAt low temperature it changes appreciably only in an energy window of width of order $k_{B}T$ about $\mu \simeq E_F$. If the full density of states $D(E)$ is smooth and nonzero there, only about $D(E_F)k_{B}T$ states change population. Each excited particle or hole carries energy of order $k_{B}T$ relative to the Fermi surface, so $U(T)-U(0)$ is of order $D(E_F)(k_{B}T{)}^2$. Differentiation therefore gives $C_{\mathrm{el}}\propto T$.

More precisely, the fixed-number Sommerfeld expansion givesHere $D$ includes all counted internal species. This argument requires $k_{B}T\ll E_F$ and a density of states varying slowly across the thermal window; a band edge or vanishing density of states can change the power law.

Two-dimensional linear dispersion at zero chemical potential. Let $g$ count identical positive-energy branches, keeping the state-counting convention explicit. The number of wavevector states in a ring of radius $k$ is $gA(2\pi kdk)/(2\pi {)}^2$. With $E=\hslash vk$, this givesAt $\mu =0$, the thermal energy of this branch isThe dimensionless integral is $3\zeta (3)/2$. Differentiating givesThis is the heat capacity for a two-dimensional linear fermion dispersion. A physical neutral particle–hole-symmetric Dirac cone also has electron holes in its filled negative-energy band. Their excitation energy gives an equal contribution, doubling $C_{+}$ for the same cone count. The $T^2$ scaling is unchanged. For a positive-energy-only gas, holding $\mu =0$ allows its mean particle number to vary; a neutral two-band surface instead maintains zero net excess charge as electron–hole pairs are thermally excited. These conventions should not be conflated when specifying an exact coefficient.

Nonzero chemical potential. The ordinary metallic regime isThe thermal window then lies far from the cone tip. Since $D(E)\propto E$, its fractional change across that window is only of order $k_{B}T/\mu$. Treating it as locally constant restores the Sommerfeld result $C_{\mathrm{el}}\propto D(\mu )T\propto \mu T$. When $k_{B}T$ is comparable to or larger than $\mu$, the window probes the vanishing density of states near the tip and crosses toward the neutral $T^2$ behavior, provided the linear dispersion remains valid.

Bulk phonons and the surface-to-bulk comparison. For the topological insulator model, assume bulk electronic excitations are frozen out while the conducting surface remains active. In three dimensions, acoustic phonons have a density of modes proportional to ${\omega }^2$. Normalize the Debye spectrum to $3N$ modes:The thermal part of the energy isAt $T\ll {\Theta }_D$, extending the upper integration limit to infinity introduces an exponentially small error. With $x=\hslash \omega /(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$,Temperature-independent zero-point energy makes no contribution to this heat capacity. The requested ratio consequently scales asFor a fixed shape and composition, $A\propto L^2$ and $N\propto L^3$, so the ratio grows as $1/(LT)$. Smaller samples and lower temperatures favor the surface contribution. It dominates if the dimensionless right-hand side, including its state-counting prefactor, exceeds unity. This is the surface-electron and bulk-phonon heat-capacity crossover.

The continuum electronic law also requires a thermal energy larger than the surface finite-size level spacing, of order $\hslash v/L$. The phonon continuum has its own finite-size cutoff. Therefore dominance obtained by formally taking $T\to 0$ cannot always be inferred from the two continuum power laws alone.

There is a numerical-normalization qualification in the paper. Combining the explicit one-positive-branch result above with the conventional $3N$-mode Debye normalization gives ratio prefactor $15g\zeta (3)/(16{\pi }^5)$; including electron holes doubles it. For $g=1$, this is approximately $0.00368$ before that doubling, rather than the supplied $5/(12{\pi }^2)\simeq 0.0422$. The problem specifies neither a cone/spin multiplicity that equates these nor another mode normalization. Its supplied prefactor is therefore used as the stipulated approximation for the requested numerical estimate, rather than claimed as an exact consequence of the single-branch count.

Numerical estimate. A cubic sample of volume $1\mathrm{m}{\mathrm{m}}^3$ has side $L=10^{-3}\mathrm{m}$. Counting the conducting surface on all six faces givesFor one atom per cubic cell of side $a=1.5\times 10^{-9}\mathrm{m}$,Use $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ and $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$. Inserting the stated $v=5\times 10^5\mathrm{m}{\mathrm{s}}^{-1}$, ${\Theta }_D=500\mathrm{K}$ and $T=1\mathrm{K}$ into the paper's supplied prefactor givesThus bulk phonons dominate at this temperature in this estimate. If only one face contributes, the ratio is $1.22\times 10^{-6}$; if there are $\nu$ chemical atoms per unit cell, divide the corresponding estimate by $\nu$. These geometrical and crystallographic factors are not supplied explicitly.

With the same prefactor and geometry, extrapolating $1/T$ would suggest equality near $7.3\mu \mathrm{K}$. However, the surface finite-size scale is already of order $\hslash v/(k_{B}L)\simeq 3.8\mathrm{mK}$ for this specimen. That extrapolated crossover lies outside the continuum electronic regime, so it is not a reliable prediction of the sample's ultimate lowest-temperature heat capacity.

Consider a chosen primitive cubic lattice $a{\mathbb{Z}}^3$ with $a=0.400\mathrm{nm}$ and one identical point scatterer per primitive unit cell. Retain the incident wavelength $\lambda =0.154\mathrm{nm}$. This is an ideal point model. Its reciprocal lattice vectors are $q=(2\pi /a)(h,k,l)$ for integer triples other than $(0,0,0)$. Write $S=h^2+k^2+l^2$. The associated interplanar spacing and first-order Bragg angle satisfyThe reflection labels $(hkl)$ use the Miller index convention, keeping unreduced triples to encode diffraction order. With that convention, set $n=1$ in Bragg's law for each label. The $(200)$ reflection is equivalently the second diffraction order of $(100)$: $d_{200}=a/2$ does not imply extra scatterers halfway between the occupied $(100)$ planes. Counting both descriptions as distinct reflections would count the same direction twice.

For $(100)$, $(110)$, $(111)$ and $(200)$, respectively, $S=1,2,3,4$ and the calculated interplanar spacings are $0.400000$, $0.282843$, $0.230940$ and $0.200000\mathrm{nm}$. Their calculated deflections $2\theta$ are $22.197435^{\circ }$, $31.594744^{\circ }$, $38.952948^{\circ }$ and $45.287481^{\circ }$. Equal steps in $S$ give equal steps in ${\sin }^2\theta$, not equal steps in $2\theta$. This regularity makes an angular pattern a test of a proposed crystal structure.

A real crystal structure has a Bravais lattice together with an atomic basis of a crystal. The crystal lattice structure factor selects the periodic directions; the crystal basis structure factor combines the scattering amplitudes of atoms within a unit cell and can cancel a reflection. Consequently Bragg's law alone predicts neither allowed reflections nor their wave intensities. ^[ref] Systematic absences further restrict observed families; for example, face-centered cubic lattice centering permits only indices all even or all odd in conventional cubic indexing. The four strips above apply to the stated primitive cubic lattice, not to sodium chloride or a face-centered cubic lattice. ^[ref]

Consider a spin-degenerate two-dimensional free electron model with rectangular periods $a>b$ and two electrons per unit cell. Write $K=\pi /a$, $L=\pi /b$, and $k_F^2=4\pi /(ab)$. If $K<k_F<L$, the circular Fermi contour crosses only the vertical boundaries of the first Brillouin zone. Folding translates its protruding caps back by a reciprocal lattice vector $(2K,0)$. A weak periodic potential mixes the intersecting plane waves, reconnecting the contours into two open first-band sheets and a closed second-band pocket. The pocket is centred on a zone boundary: its halves on opposite sides of a reduced-zone plot are identified periodically and belong to one closed sheet.

The rock salt structure of ordinary sodium chloride has a coordination number of six for each kind of ion: six nearest neighbours of the opposite sign surround each ion along three perpendicular directions. Its conventional cubic unit cell contains four formula units; sharing corner, face, and edge sites between adjacent cells must be included in that count. Sixfold local coordination is compatible with $1:1$ overall stoichiometry.^[ref]

The checkerboard plane in the diagram is a slice through this three-dimensional arrangement. It leaves out the two nearest neighbours above and below the plane. Repeating a complete even-by-even patch gives equal numbers of signs; the displayed odd-by-odd crop has a boundary imbalance and is not a complete neutral ionic crystal or a unit cell. Other salts need not have the rock salt structure or the same coordination number.

The coherent scattering amplitude associated with a specified spatial arrangement, obtained by summing the site amplitudes with their relative wave phases. Distinguish the crystal lattice structure factor, which sums over lattice sites, from the crystal basis structure factor, which sums the chemical atoms inside a unit cell. In a periodic crystal the two factors multiply. A basis factor can vanish even at an allowed reciprocal lattice vector, causing an extinct reflection.