Past exam of the condensed matter physics course of the University of Oxford / 2019 / 1

A Bravais lattice is the set of pointswhere the primitive lattice translation vectors ${\mathbf{a}}_i$ are linearly independent. Every lattice point has the same translational environment.

An atomic basis is the motif of atoms attached to each point of the Bravais lattice, specified by positions ${\mathbf{r}}_p$ and atomic species. The full crystal consists of atoms at $\mathbf{R}+{\mathbf{r}}_p$, with $\mathbf{R}\in \Lambda$. The Bravais lattice describes translations; the atomic basis describes what is repeated. A primitive cell contains one lattice point but can contain several atoms.

Put $v={\mathbf{a}}_1\cdot ({\mathbf{a}}_2\times {\mathbf{a}}_3)\ne 0$. DefineThe scalar triple product gives ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$. Consequently every integer combination $\mathbf{G}={\sum }_j{m}_j{\mathbf{b}}_j$ satisfiesThis proves that the proposed vectors generate points in the reciprocal lattice, but primitiveness also requires the converse. Let $\mathbf{G}$ satisfy the exponential condition for every lattice vector, and in particular for $\mathbf{R}={\mathbf{a}}_i$. ThenSince the ${\mathbf{b}}_i$ form a basis of three-dimensional space, the vector ${\sum }_i{m}_i{\mathbf{b}}_i$ has the same dot products with each ${\mathbf{a}}_j$ as $\mathbf{G}$. Their difference is orthogonal to three independent vectors and is zero. Henceand no additional reciprocal lattice vectors lie outside that integer span. Thus the ${\mathbf{b}}_i$ are primitive, not merely three possible reciprocal translations.

Translate the integration variable by an arbitrary lattice vector and use the periodicity of $V$:Thus $(1-e^{i\mathbf{q}\cdot \mathbf{R}})V_{\mathbf{q}}=0$ for every $\mathbf{R}$. A nonzero coefficient requires $e^{i\mathbf{q}\cdot \mathbf{R}}=1$ for every lattice translation, which is precisely the defining condition for $\mathbf{q}\in {\Lambda }^{\ast }$.

For an infinite crystal the integral is a Fourier transform of a periodic distribution, rather than an ordinary convergent integral. More explicitly, the Fourier seriesgivesThis is the precise meaning of zero away from the reciprocal lattice. A finite specimen instead has broadened reciprocal-lattice peaks of a finite crystal; exact delta-function support is an infinite-periodicity idealization.

Separate the entire-system atom index into a cell translation $\mathbf{R}$ and an index $p$ within a single primitive unit cell. The potential becomesThe same $u_p$ is used in every translated cell because the atomic basis repeats. Normalize plane waves in volume $\Omega$ by $⟨\mathbf{x}|\mathbf{k}⟩={\Omega }^{-1/2}{e}^{i\mathbf{k}\cdot \mathbf{x}}$. For $\mathbf{G}={\mathbf{k}}^{\prime }-\mathbf{k}$, substitution and $\mathbf{s}=\mathbf{x}-\mathbf{R}-{\mathbf{r}}_p$ giveFor a real static scalar potential, define the atomic scattering factor in the positive-exponent convention of the question byThe second bracket in the matrix element is $S(\mathbf{G}{)}^{\ast }$. At a reciprocal lattice vector every term of the first bracket is one, giving the number of cells $N_c$. Therefore the stated transition-rate rule, equivalently the first Born approximation and Fermi's golden rule up to the final-state kinematic factors, yieldsThe harmless complex conjugation records the distinction between the source's positive-exponent Fourier transform and the usual positive-exponent incoming plane wave. Reversing every Fourier sign gives the equivalent amplitude convention. Only the cell-translation sum ranges over the full crystal; the crystal basis structure factor contains the atoms of one cell.

In cubic coordinates write $\mathbf{G}=(2\pi /a)(h,k,l)$. The face-centred cubic selection rule allows integer triples only when $h,k,l$ are all even or all odd. One way to derive it is to sum the four conventional-cell lattice positions:This vanishes for mixed parity and equals four otherwise. The two-atom diamond atomic basis adds the factorFor all-odd indices this is nonzero. For all-even indices it vanishes when $h+k+l=2(\mathrm{mod}4)$ and survives when the sum is divisible by four: this is diamond-basis diffraction extinction.

Exclude the forward beam $000$ and order the remaining families by $h^2+k^2+l^2$. The first is $111$, with squared length three. The candidate $200$, with squared length four, is extinguished by the diamond atomic basis. The next allowed family is $220$, with squared length eight. ThusFor fixed incident wavelength, $|\mathbf{G}|=2k\sin \theta$ increases with scattering angle, so these are the two lowest-angle powder rings. The two atoms are counted per primitive cell; a conventional silicon cube contains four lattice points and eight atoms.