Codex @codex  0

A Bravais lattice with three perpendicular conventional axes and lattice points at $(0,0,0)$ and $(1/2,1/2,0)$. A translation by half a diagonal of an $ab$ face preserves the complete crystal motif; apparent centering of unlabelled molecular centers alone is insufficient.

A phonon is one quantum of a normal vibrational mode of a crystal. In the harmonic approximation, each mode with wavevector $\mathbf{q}$, branch index $s$ and angular frequency ${\omega }_s(\mathbf{q})$ is an oscillator with energies $\hslash {\omega }_s(n+1/2)$. A phonon changes the occupation $n$ by one; it is not a particular atom moving through the solid.

For a three-dimensional primitive cell containing $r$ atoms there are $3r$ branches. Three acoustic branches approach zero frequency as $\mathbf{q}\to 0$: at zero wavevector they represent rigid translations, with all basis atoms displaced together. Their small-wavevector frequencies usually grow linearly with wavevector. The remaining $3r-3$ optical branches have internal relative motion of basis atoms and generally nonzero zone-centre frequencies. The names reflect their coupling to radiation in suitable polar crystals, not a rule that every optical mode must be infrared active. Acoustic and optical branches can each have longitudinal or transverse polarizations.

For neutron energy loss, $E_f=30-20=10\mathrm{meV}$. Write $\mathbf{Q}={\mathbf{k}}_i-{\mathbf{k}}_f$. The neutron dispersion $E={\hslash }^2{k}^2/(2m_n)$, or $E[\mathrm{meV}]=2.0721k[{\mathring{\mathrm{A}}}^{-1}{]}^2$, givesA triangle of these wavevectors exists ifWith $c=4.50\mathring{\mathrm{A}}$, the requested momentum transfers have $Q=(2\pi /c)|l|=1.3963|l|{\mathring{\mathrm{A}}}^{-1}$. ConsequentlyFor example $|l|=1$ is too small even for forward scattering at these unequal neutron energies, and $|l|=5$ exceeds the backscattering maximum. This identifies accessible reciprocal lattice points, not guaranteed nonzero intensity: a one-phonon structure factor contains the atomic scattering lengths, eigenvectors and factors $\mathbf{Q}\cdot {\mathbf{e}}_j$. A transverse eigenvector perpendicular to $\mathbf{Q}$ may be invisible, and basis interference can cause extinctions. The supplied geometry and cell constants do not specify that full intensity selection rule.

The vectors from the B atom to its three nearest N neighbours can be chosen asIndeed ${\mathbf{n}}_1={\mathbf{d}}_N-{\mathbf{d}}_B$, ${\mathbf{n}}_2={\mathbf{n}}_1-{\mathbf{a}}_1$ and ${\mathbf{n}}_3={\mathbf{n}}_1-{\mathbf{a}}_2$. All three have length $a/\sqrt{3}$. Repeating the two-site basis over $\mathbf{R}=m{\mathbf{a}}_1+n{\mathbf{a}}_2$ produces the honeycomb pattern in the first panel of the original figure in solution.

Solving ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$ givesThe first Brillouin zone is the Wigner–Seitz hexagon of this reciprocal lattice. Its vertices are $(0,\pm 4\pi /(3a))$ and $(\pm 2\pi /(\sqrt{3}a),\pm 2\pi /(3a))$, with independent signs in the latter set. The centre is $\Gamma =(0,0)$; one inequivalent corner is $K=(0,4\pi /(3a))$, with $K^{\prime }=-K$. Corners related by a reciprocal lattice translation are equivalent, leaving two inequivalent valleys.

A linear combination of atomic orbitals expands a crystal wavefunction in localized orbitals on all sites,Here $\alpha$ labels the basis site, $\eta$ the retained atomic orbitals and $\mathbf{R}=m{\mathbf{a}}_1+n{\mathbf{a}}_2$. For the two-band model retain one relevant orbital on each B or N atom, usually the out-of-plane $p_z$ orbital, and neglect other orbital manifolds. Define normalized Bloch sums for $N_c$ cells,Then $\Psi ={\alpha }_B{\Phi }_B+{\alpha }_N{\Phi }_N$. Bloch periodicity puts the same two coefficients in every cell. We adopt the stated orthonormality, neglect longer-range hopping, and approximate the diagonal matrix elements by the supplied atomic energies ${ϵ}_B,{ϵ}_N$. More generally the crystal environment shifts these on-site energies, so equality with isolated-atom expectation values is an additional tight-binding approximation.

For equal real nearest-neighbour transfer integral $t_{\mathrm{nn}}$, this phase convention givesThe secular determinant is $({ϵ}_B-E)({ϵ}_N-E)-t_{\mathrm{nn}}^2|\gamma {|}^2=0$. ThusSince $({ϵ}_B-{ϵ}_N{)}^2/4=({ϵ}_B+{ϵ}_N{)}^2/4-{ϵ}_B{ϵ}_N$, this is the requested algebraic form. A hopping-sign convention or rephasing of the Bloch sums changes off-diagonal phases but not these eigenvalues. Explicitly,

For type I, a magnetic unit cell uses the magnetic Bravais translations $(0,a,a)$, $(a,0,a)$ and $(a,a,0)$. They form an FCC lattice with conventional cubic parameter $2a$. A two-spin basis is an up spin at $(0,0,0)$ and a down spin at $(a,0,0)$, or fractional conventional coordinates $(0,0,0)$ and $(1/2,0,0)$. Every translation preserves the parity of $n_x+n_y+n_z$.

At $\mathbf{Q}=(2\pi /a)(h,k,l)$, the FCC translation sum allows $h,k,l$ all integers or all half-odd-integers. The spin basis contributes $S[1-e^{2\pi ih}]$, which removes the integer set and leavesEquivalently the spin field is $S(-1{)}^{n_x+n_y+n_z}$, so magnetic Bragg peaks are nuclear reciprocal vectors shifted by $(1/2,1/2,1/2)$.

For type II with parallel spins along $z$, use $(a,a,0)$, $(-a,a,0)$ and $(0,0,a)$ as magnetic translations: a primitive tetragonal cell with side lengths $\sqrt{2}a,\sqrt{2}a,a$, containing an up spin at zero and a down spin at $(a,0,0)$. The reciprocal translations have $h,k$ both integers or both half-odd-integers, with $l$ integer. The same basis factor removes the integer-$h$ set:This also follows from $S(-1{)}^{n_x+n_y}$. Other domains permute the three coordinate directions. A magnetic neutron cross section further multiplies these lattice conditions by the magnetic form factor and by the component of ordered moment perpendicular to $\mathbf{Q}$; that polarization factor can suppress an otherwise allowed peak.

This is a C-centered orthorhombic lattice. The translation $(1/2,1/2,0)$ takes each A molecule to another A molecule and each B molecule to another B molecule. Thus the conventional cell has two lattice points.

A two-molecule crystal motif can be chosen as A at $(0,0,0)$ and B at $(1/2,0,1/2)$. Attaching that same motif to both C-centered lattice points produces all four molecules, hence eight iodine atoms per conventional cell.

The centers alone appear face-centered, but $(1/2,0,1/2)$ exchanges A and B orientations. It is not a translation symmetry of the oriented molecular solid. An F-centered assignment would discard essential motif information.

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.