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A Bravais lattice is the translation set $\Lambda =\{n_1{\mathbf{a}}_1+n_2{\mathbf{a}}_2+n_3{\mathbf{a}}_3:n_i\in \mathbb{Z}\}$ generated by primitive lattice translation vectors. Translating by any member of $\Lambda$ leaves its surroundings equivalent. A crystal structure requires both this lattice and an atomic basis attached to every lattice point. The basis can lower the symmetry of the bare point lattice; a familiar-looking geometrical arrangement of atoms is not necessarily itself a Bravais lattice. For example, the diamond cubic structure is a face-centred cubic lattice with an identical-atom basis at fractional positions $(0,0,0)$ and $(1/4,1/4,1/4)$ in conventional cubic coordinates. The displacement between those two basis atoms is not itself a lattice translation.

The seven three-dimensional crystal systems are triclinic, monoclinic, orthorhombic, tetragonal, trigonal, hexagonal and cubic; the distinct translation types give 14 Bravais lattices. A conventional cell can contain several lattice points, while a primitive unit cell contains one. Simple cubic primitive lattice translation vectors can be chosen along the cube edges; for face-centred cubic they can be $(a/2)(0,1,1)$, $(a/2)(1,0,1)$ and $(a/2)(1,1,0)$. Their primitive volume is $a^3/4$, explaining the four lattice points per conventional face-centred cell.

A crystallographic point group comprises the orthogonal parts of the crystal's spatial symmetry operations. It constrains possible rotations, reflections, inversion and their combinations, while forgetting translations. Lattice preservation restricts proper rotational orders in three dimensions to $1,2,3,4,6$; a fivefold axis cannot be a symmetry of a periodic three-dimensional lattice. One quick argument restricts a rotation's integral lattice representation through $1+2\cos \theta$, an integer trace, producing these orders. There are 32 crystallographic point-group types. Examples are the cubic full octahedral group $O_h$ with Hermann–Mauguin symbol $m\bar{3}m$, the tetrahedral class $T_d$ with symbol $\bar{4}3m$, and the polar tetragonal class $C_{4v}$ with symbol $4mm$.

The space group retains the full isometry. In Seitz notation $\{R\mid \mathbf{t}\}$ acts by $\mathbf{x}\mapsto R\mathbf{x}+\mathbf{t}$, and multiplication isIts translation subgroup is normal and the quotient is the crystallographic point group. A screw axis combines rotation with translation along the axis; for example, applying a $2_1$ screw twice gives one whole lattice translation. A glide plane combines reflection with a translation parallel to the plane; applying the glide twice likewise gives a translation. These operations can preserve the complete crystal without their rotation or reflection alone preserving it.

A symmorphic space group admits an origin and representatives for which the point-group operations have no translation component. A nonsymmorphic space group has essential fractional translations that cannot all be removed by moving the origin. There are 230 three-dimensional space-group types, distinguished more finely than their point groups. These counts refer to ordinary, nonmagnetic periodic space groups; magnetic and aperiodic generalizations are different classifications.^[ref]

For concrete comparisons, a monatomic simple-cubic crystal has space group $Pm\bar{3}m$ and point group $m\bar{3}m$. Rock salt has a face-centred cubic lattice, a two-species basis and symmorphic space group $Fm\bar{3}m$. Diamond has the same cubic point group and translation-lattice type, but its basis leads to nonsymmorphic $Fd\bar{3}m$. Replacing the two diamond basis sites by distinct species gives the zinc blende structure, exemplified by GaAs, with $F\bar{4}3m$ and point group $\bar{4}3m$; inversion that would interchange the species is no longer a symmetry. The lattice specifies translations, the point group specifies the linear symmetry classes, and the space group specifies their complete spatial implementation. These examples show why none alone uniquely describes an atomic structure.

The sites form a face-centered cubic lattice decorated by an identical two-atom atomic basis of a crystal, at fractional displacements $0$ and $\mathbf{\delta }=(1/4,1/4,1/4)$. This is the diamond cubic structure; the eight atoms in its conventional cube are not eight independent Bravais-lattice points.

For identical silicon atomic scattering factors $f$, the conventional-cell structure factor separates into the FCC factor and the two-site motif factor:The first factor vanishes unless $h,k,l$ are all even or all odd. For all-even indices, the motif also vanishes if $h+k+l=4r+2$; it survives if the sum is divisible by four. All-odd indices survive, with motif modulus $\sqrt{2}$. Thus $(100)$ and $(110)$ are absent by centering, $(200)$ is absent by the diamond basis, and $(111)$ and $(220)$ are allowed.

For a cubic cell, $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$. At fixed X-ray wavelength, Bragg's law makes the smallest scattering angles correspond to the smallest allowed positive values of $h^2+k^2+l^2$. These start with $3$ and $8$, soThe braces denote whole powder families, including their symmetry-equivalent plane normals. The next family is $\{311\}$, with squared-index sum $11$; using the FCC rule alone would incorrectly insert a $(200)$ line before $(220)$.

For the four conventional face-centered cubic lattice sites $000$, $0\frac{1}{2}\frac{1}{2}$, $\frac{1}{2}0\frac{1}{2}$ and $\frac{1}{2}\frac{1}{2}0$, their centering factor isThis is the face-centred cubic selection rule. Enumerating the nonzero representatives with $h^2+k^2+l^2\le 16$ givesSymmetry-related permutations and signs belong to the same powder ring, not additional rings. For Al, the basis contains one atom, $S=F_{\mathrm{fcc}}{f}_{\mathrm{Al}}$. ThusFor the stated KCl basis,The all-even reflections add the two atomic factors; all-odd reflections subtract them. Ionic $K^{+}$ and $C{l}^{-}$ each have 18 electrons, giving similar X-ray amplitudes. Their difference is much smaller than their sum, so the 111 and 311 rings are weak. Consequently the four strong rings areThis is an accidental diffraction extinction from equal atomic scattering factors in the equal-factor limit, not an extra exact fcc selection rule. Isoelectronic ions can have different radial electron distributions, so real odd reflections need not vanish exactly. Even the cruder neutral-atom estimate $f_K\simeq 19$, $f_{\mathrm{Cl}}\simeq 17$ makes the odd/even basis-intensity ratio only $(2/36{)}^2\simeq 0.0031$, before other ring-intensity factors.

For Si's displaced identical basis atoms,For all-odd indices, the bracket is $1\pm i$, with squared magnitude 2. For all-even indices it vanishes when $h+k+l=2(\mathrm{mod}4)$, and has squared magnitude 4 when the sum is a multiple of four. Thus 200 and 222 vanish, whereas 111, 220, 311 and 400 survive:These are the basis extinctions of the ideal diamond cubic structure. Intensity strengths within each surviving set can still differ because of atomic factors, multiplicity and geometry.

In the diamond cubic structure of silicon, the longitudinal acoustic phonon branch and longitudinal optical phonon branch meet at $X=(2\pi /a,0,0)$. The two atomic basis sites are identical, and the associated crystal symmetry protects their boundary degeneracy of normal modes. In aluminium phosphide, the two sites contain different chemical atoms. The corresponding zinc blende structure lacks the site-exchange symmetry, and these two longitudinal branches generally split at $X$.

A simple way to see the splitting is a diatomic-chain dispersion relation with equal springs, representing successive $[100]$ atomic planes by alternating masses $M_1,M_2$ at spacing $d=a/4$. With effective spring constant $K$, its angular frequencies obeyAt $X$, $kd=\pi /2$, so the inter-sublattice coupling cancels and the two squared angular frequencies are $2K/M_1$ and $2K/M_2$. Equal masses give a contact; unequal masses giveIn aluminium phosphide, the heavier phosphorus plane moves in the lower-frequency edge mode, and the lighter aluminium plane in the higher-frequency mode in this model. Their similar atomic masses explain why the splitting is relatively small. Different electronic environments can also change the force constant matrix. The chain calculation illustrates the mechanism; a quantitative three-dimensional phonon spectrum requires its full dynamical matrix. It does not assert that all transverse acoustic and optical branches meet in silicon.

In the electron-count estimate $f_{\mathrm{Ga}}\simeq 31$, $f_{\mathrm{As}}\simeq 33$, the structure factors are $F_{111}=4(31-33i)$ and $F_{200}=4(31-33)$. Thus a single-directed-reflection estimate isA powder X-ray diffraction peak collects all equivalent reflections. The powder diffraction multiplicities are eight for $\{111\}$ and six for $\{200\}$, so the multiplicity-and-electron-count estimate isAt $(200)$ the two sublattice amplitudes are opposite in phase and nearly cancel; at $(111)$ their relative phase is a quarter cycle, so this cancellation does not occur. In silicon, the two atomic scattering factors are identical and the diamond cubic structure has an exactly forbidden $(200)$ reflection, while $(111)$ remains allowed in the kinematic approximation.

The numerical ratio neglects the angular dependence of the atomic scattering factors, the Lorentz-polarization factor and thermal-motion corrections. No wavelength or measurement geometry is supplied to refine those factors. This is therefore a stated structural estimate, rather than a unique precision ratio of measured peak areas.

Use fractional coordinates in a conventional cubic cell with an atom at the origin. The diamond cubic structure consists of an FCC Bravais lattice with the two-atom basisThe FCC translations within the conventional cell are $(0,0,0)$, $(0,\frac{1}{2},\frac{1}{2})$, $(\frac{1}{2},0,\frac{1}{2})$ and $(\frac{1}{2},\frac{1}{2},0)$. Adding the basis gives eight atoms per conventional cell. Equivalently, two identical FCC sublattices are displaced by $a[111]/4$. It is not a monatomic close-packed arrangement: every atom has four tetrahedrally arranged nearest neighbors rather than twelve.

In the projection, label each atom by its fractional height. The $(0,0)$ and $(1/2,1/2)$ columns have height0; $(0,1/2)$ and $(1/2,0)$ have height1/2. The additional four columns are $(1/4,1/4)$ and $(3/4,3/4)$ at height1/4, and $(1/4,3/4)$ and $(3/4,1/4)$ at height3/4. Boundary columns repeat periodically; they must not be counted as full extra atoms.

The metallic conventional cell has four tin atoms, whereas the conventional diamond cubic structure cell has eight. Therefore the volume per atom isThe fractional change for metallic to nonmetallic tin isThe atom count is unchanged, so this is also the unconstrained bulk volume change for complete transformation under the model. If it were macroscopically isotropic, the corresponding free linear expansion would be $(1.2627{)}^{1/3}-1\simeq 8.1\%$.

Such a large transformation strain creates stresses where transformed and untransformed regions constrain one another. Cracking, loss of mechanical integrity and disintegration can follow: the phenomenon is tin pest. The calculation establishes the volume mismatch, not the transformation rate; temperature, nucleation and impurities affect when a real specimen changes.