Reciprocal Lattice from the Bravais Lattice

Statement

Given a Bravais lattice \(\{\vec{R} = n_1\vec{a}_1 + n_2\vec{a}_2 + n_3\vec{a}_3 : n_i \in \mathbb{Z}\}\) with primitive vectors \(\vec{a}_1,\vec{a}_2,\vec{a}_3\), the reciprocal lattice is the set of wavevectors \(\vec{G}\) for which the plane wave \(e^{i\vec{G}\cdot\vec{r}}\) has the periodicity of the lattice, i.e. \(e^{i\vec{G}\cdot\vec{R}} = 1\) for every \(\vec{R}\). This set is itself a Bravais lattice, spanned by primitive vectors \(\vec{b}_i = 2\pi\,\dfrac{\vec{a}_j\times\vec{a}_k}{\vec{a}_1\cdot(\vec{a}_2\times\vec{a}_3)}\) (with \(i,j,k\) cyclic), satisfying \(\vec{a}_i\cdot\vec{b}_j = 2\pi\,\delta_{ij}\).

Why it matters

The reciprocal lattice is the natural arena for everything wave-like in a crystal. Because any lattice-periodic function admits a Fourier expansion whose only surviving wavevectors are the reciprocal lattice vectors, the reciprocal lattice controls X-ray and neutron diffraction (the Laue and Bragg conditions are statements about \(\vec{G}\)), the classification of electronic states by crystal momentum, and the geometry of Brillouin zones.

Conceptually it is the Fourier dual of the direct lattice: distances in real space map to inverse distances in \(\vec{k}\)-space, and the point-group symmetry of the crystal is faithfully reproduced. Master the reciprocal lattice once and the same object reappears in band structure, phonons, and pseudopotentials.

Assumptions

Derivation

Result

Limiting cases

• Simple cubic, edge \(a\). \(\vec{a}_i = a\hat{e}_i\Rightarrow\vec{b}_i = (2\pi/a)\hat{e}_i\): reciprocal lattice is again simple cubic with edge \(2\pi/a\). Large real-space spacing \(\to\) small reciprocal spacing.
• Orthorhombic. \(\vec{a}_i=a_i\hat{e}_i\Rightarrow\vec{b}_i=(2\pi/a_i)\hat{e}_i\): each axis inverts independently; the reciprocal cell is stretched where the direct cell is squeezed.
• FCC \(\leftrightarrow\) BCC. The reciprocal of a face-centred cubic lattice is body-centred cubic (and vice versa); the conventional-cube edges relate as \(a_{\text{rec}} = 4\pi/a\).
• Self-dual scaling. Applying the construction twice returns the original lattice up to the factor \((2\pi)^2\): the reciprocal of the reciprocal lattice is the direct lattice.
• 2D limit. For a planar lattice one embeds an auxiliary unit normal \(\hat{n}\) as \(\vec{a}_3=\hat{n}\); then \(\vec{b}_1 = 2\pi(\vec{a}_2\times\hat{n})/[\hat{n}\cdot(\vec{a}_1\times\vec{a}_2)]\), the standard 2D result.

Breaks when

Failure modes

• Dropping the \(2\pi\). Mixing the physicist convention \(\vec{a}_i\cdot\vec{b}_j=2\pi\delta_{ij}\) with the crystallographer convention \(\vec{a}_i\cdot\vec{b}_j=\delta_{ij}\) within one calculation. Diffraction conditions then come out wrong by factors of \(2\pi\).
• Inverting component-wise. Believing \(\vec{b}_i = 2\pi/\vec{a}_i\) component by component. This is only true for orthogonal axes; for oblique cells \(\vec{b}_i\) is generally not parallel to \(\vec{a}_i\).
• Wrong cyclic partner. Writing \(\vec{b}_1\propto\vec{a}_1\times\vec{a}_2\) instead of \(\vec{a}_2\times\vec{a}_3\). Always pair \(\vec{b}_i\) with the cross product of the two direct vectors it is NOT dual to.
• Sign of the triple product. Using \(|v|\) blindly for a left-handed set \(\vec{a}_i\); the signed \(v\) keeps \(\vec{a}_i\cdot\vec{b}_i=+2\pi\), whereas \(|v|\) can flip a sign and corrupt orientation.
• Confusing conventional and primitive cells. Building \(\vec{b}_i\) from the conventional cubic vectors of FCC/BCC rather than the primitive rhombohedral ones, producing a reciprocal lattice with spurious "extra" points that are actually forbidden reflections.

Discussion

The reciprocal lattice is the shadow the direct lattice casts in Fourier space. Its deepest justification is the one used above: a lattice-periodic function has a Fourier series, and periodicity restricts the sum to run over precisely those wavevectors \(\vec{G}\) with \(e^{i\vec{G}\cdot\vec{R}}=1\). Every physical consequence flows from this. In diffraction, the scattered amplitude from a crystal is the Fourier transform of its density, which is a sum of delta functions located at the \(\vec{G}\); constructive interference (the Laue condition \(\Delta\vec{k}=\vec{G}\)) happens only at reciprocal lattice points, and the equivalent Bragg law \(2d\sin\theta=n\lambda\) is recovered because \(|\vec{G}_{hk\ell}| = 2\pi/d_{hk\ell}\) relates reciprocal vector length to interplanar spacing.

The construction also carries symmetry faithfully. Because \(\vec{b}_i\) are built from \(\vec{a}_i\) by operations (cross product, scalar triple product) that commute with rotations, any point-group operation that leaves the direct lattice invariant leaves the reciprocal lattice invariant too. This is why the Brillouin zone — the Wigner–Seitz primitive cell of the reciprocal lattice — has the full point-group symmetry of the crystal, and why band structures and phonon dispersions are plotted along its high-symmetry directions.

A more abstract reading: the \(\vec{b}_i\) are the dual basis of the \(\vec{a}_i\) with respect to the scalar product, up to the factor \(2\pi\). In matrix form, if \(A=[\vec{a}_1\ \vec{a}_2\ \vec{a}_3]\) has the direct vectors as columns and \(B=[\vec{b}_1\ \vec{b}_2\ \vec{b}_3]\), then \(A^{\mathsf T}B = 2\pi\,\mathbb{1}\), so \(B = 2\pi (A^{\mathsf T})^{-1} = 2\pi (A^{-1})^{\mathsf T}\). The cross-product formulas are simply Cramer's rule applied to this inverse, which is why they generalise cleanly to any dimension via the metric tensor \(g_{ij}=\vec{a}_i\cdot\vec{a}_j\): \(\vec{b}_i = 2\pi\sum_j (g^{-1})_{ij}\,\vec{a}_j\). This viewpoint makes it manifest that the reciprocal-of-reciprocal returns the direct lattice, since \(((A^{-1})^{\mathsf T})^{-1\,\mathsf T}=A\).

Common misconceptions. The reciprocal lattice is not "the crystal in momentum space" in the sense of where electrons live — that is the Brillouin zone, a single primitive cell of it. Nor is a reciprocal lattice vector the same as a general crystal momentum \(\vec{k}\); the \(\vec{G}\) are the special \(\vec{k}\) that are lattice-periodic and connect physically equivalent states \(\vec{k}\) and \(\vec{k}+\vec{G}\). Finally, the reciprocal lattice depends only on the Bravais lattice, not on the basis of atoms decorating it — the atomic basis enters diffraction through the structure factor (which can extinguish spots) but never changes the positions of the reciprocal lattice points.

Worked examples

Problems

1. (A) Cubic reciprocal. A simple cubic lattice has \(a=4.00\ \text{Å}\). Find \(|\vec{b}_i|\) and the volume of the reciprocal primitive cell.
   Solution

   \(\vec{b}_i=(2\pi/a)\hat e_i\), so \(|\vec{b}_i| = 2\pi/(4.00\times10^{-10}) = 1.571\times10^{10}\ \text{m}^{-1} = 1.57\ \text{Å}^{-1}\). Reciprocal cell volume \(= (2\pi/a)^3 = (2\pi)^3/v\). With \(v=a^3=64\ \text{Å}^3\): \(V_{\text{rec}} = (1.571)^3 = 3.88\ \text{Å}^{-3}\). Check: \(V_{\text{rec}} = (2\pi)^3/v = 248.05/64 = 3.88\ \text{Å}^{-3}\). ✓

2. (A) Reciprocal cell volume identity. Prove that \(V_{\text{rec}} = \vec{b}_1\cdot(\vec{b}_2\times\vec{b}_3) = (2\pi)^3/v\) for any Bravais lattice.
   Solution

   Using \(\vec b_i = 2\pi(A^{\mathsf T})^{-1}\) columns, \(\det B = (2\pi)^3\det(A^{\mathsf T})^{-1} = (2\pi)^3/\det A = (2\pi)^3/v\). Since \(\det B = \vec b_1\cdot(\vec b_2\times\vec b_3)\), the result follows. Alternatively substitute the cross-product forms and use the vector identity \((\vec a_2\times\vec a_3)\cdot[(\vec a_3\times\vec a_1)\times(\vec a_1\times\vec a_2)] = v^2\), giving \(V_{\text{rec}} = (2\pi)^3 v^2/v^3 = (2\pi)^3/v\). ✓

3. (B) Interplanar spacing. Show that the family of lattice planes \((h k \ell)\) has spacing \(d_{hk\ell} = 2\pi/|\vec{G}_{hk\ell}|\), and evaluate \(d_{111}\) for copper's FCC lattice using the reciprocal vectors found above (\(a=3.61\ \text{Å}\)).
   Solution

   \(\vec G_{hk\ell}=h\vec b_1+k\vec b_2+\ell\vec b_3\) is perpendicular to the \((hk\ell)\) planes, and successive planes differ in phase by \(2\pi\), so their spacing is \(d=2\pi/|\vec G|\). For the \((111)\) plane referred to the FCC primitive basis, \(\vec G_{111} = \vec b_1+\vec b_2+\vec b_3 = \tfrac{2\pi}{a}(1,1,1)\), giving \(|\vec G_{111}| = 2\pi\sqrt3/a\) and \(d_{111} = 2\pi/|\vec G_{111}| = a/\sqrt3 = 3.61/\sqrt3 = 2.08\ \text{Å}\). (This is the spacing between the primitive-basis \((111)\) planes.) ✓

4. (B) Oblique 2D lattice. A 2D lattice has \(\vec a_1 = (3,0)\ \text{Å}\) and \(\vec a_2 = (1,2)\ \text{Å}\). Find \(\vec b_1,\vec b_2\) and confirm \(\vec a_i\cdot\vec b_j = 2\pi\delta_{ij}\).
   Solution

   Cell area \(v = a_{1x}a_{2y}-a_{1y}a_{2x} = 3\cdot2-0\cdot1 = 6\ \text{Å}^2\). In 2D, \(\vec b_1 = \tfrac{2\pi}{v}(a_{2y},-a_{2x}) = \tfrac{2\pi}{6}(2,-1) = \tfrac{\pi}{3}(2,-1)\ \text{Å}^{-1}\); \(\vec b_2 = \tfrac{2\pi}{v}(-a_{1y},a_{1x}) = \tfrac{2\pi}{6}(0,3) = (0,\pi)\ \text{Å}^{-1}\). Check: \(\vec a_1\cdot\vec b_1 = (3)(\tfrac{2\pi}{3})+0 = 2\pi\) ✓; \(\vec a_1\cdot\vec b_2 = 3\cdot0+0\cdot\pi = 0\) ✓; \(\vec a_2\cdot\vec b_2 = 1\cdot0+2\pi = 2\pi\) ✓; \(\vec a_2\cdot\vec b_1 = \tfrac{2\pi}{6}(1\cdot2+2\cdot(-1)) = 0\) ✓.

5. (C) BCC reciprocal is FCC. Starting from BCC primitive vectors \(\vec a_1=\tfrac{a}{2}(-1,1,1),\ \vec a_2=\tfrac{a}{2}(1,-1,1),\ \vec a_3=\tfrac{a}{2}(1,1,-1)\), derive the reciprocal vectors and identify the lattice type.
   Solution

   Compute \(v = \vec a_1\cdot(\vec a_2\times\vec a_3)\). \(\vec a_2\times\vec a_3 = \tfrac{a^2}{4}\big((-1)(-1)-(1)(1),\,(1)(1)-(1)(-1),\,(1)(1)-(-1)(1)\big) = \tfrac{a^2}{4}(0,2,2)\). Then \(v = \tfrac{a}{2}(-1,1,1)\cdot\tfrac{a^2}{4}(0,2,2) = \tfrac{a^3}{8}(0+2+2) = \tfrac{a^3}{2}\). Now \(\vec b_1 = 2\pi\tfrac{\vec a_2\times\vec a_3}{v} = 2\pi\tfrac{(a^2/4)(0,2,2)}{a^3/2} = \tfrac{2\pi}{a}(0,1,1)\). Cyclically, \(\vec b_2 = \tfrac{2\pi}{a}(1,0,1)\), \(\vec b_3 = \tfrac{2\pi}{a}(1,1,0)\). These are FCC primitive vectors with conventional edge \(4\pi/a\): the reciprocal of BCC is FCC, confirming the duality of Worked Example 2. ✓