---
title: The Reciprocal Lattice and Brillouin Zones
module: Crystal Structure
moduleNumber: 3
lessonNumber: 3
order: 303
summary: >
  Every periodic crystal has a dual lattice in wavevector space. This lesson
  defines the reciprocal lattice through the condition b_i dot a_j equals two pi
  delta, derives its properties, shows the reciprocal of fcc is bcc, links
  reciprocal vectors to families of lattice planes, and builds the first Brillouin
  zone as the Wigner-Seitz cell of the reciprocal lattice, including the higher
  zones.
topics: [Crystal Structure]
draft: false
sources:
  - book: Ashcroft & Mermin
    ref: "Ch. 5 — The Reciprocal Lattice"
  - book: Kittel
    ref: "Ch. 2 — Wave Diffraction and the Reciprocal Lattice"
  - book: Simon
    ref: "Ch. 13 — Reciprocal Lattice, Brillouin Zone"
---

Any function with the periodicity of a crystal — the electron density, the
potential, the charge distribution — is a sum of plane waves, and only certain
wavevectors appear in that sum. Those allowed wavevectors form a lattice of their
own, the reciprocal lattice. It is the natural arena for diffraction, for Bloch's
theorem, and for the whole of band theory: a wave scatters off a crystal only when
its wavevector change is a reciprocal-lattice vector, and an electron's crystal
momentum is defined only up to one. This lesson constructs the reciprocal lattice
from the [direct (Bravais) lattice](/condensed-matter/crystal-structure/bravais-lattices-and-crystal-systems)
and partitions reciprocal space into Brillouin zones.

## Definition and construction

Let a crystal property have the periodicity of a Bravais lattice with primitive
vectors $\vec a_1, \vec a_2, \vec a_3$. Expand it in plane waves $e^{i\vec k \cdot
\vec r}$. Periodicity $f(\vec r + \vec R) = f(\vec r)$ for every lattice vector
$\vec R$ requires $e^{i\vec k \cdot \vec R} = 1$, i.e. $\vec k \cdot \vec R = 2\pi
\times \text{integer}$. The set of $\vec k$ satisfying this for all $\vec R$ is the
reciprocal lattice.

> **Definition (Reciprocal lattice).** The **reciprocal lattice** of a Bravais
> lattice $\{\vec R\}$ is the set of vectors $\vec G$ with
> $$
> e^{i\vec G \cdot \vec R} = 1 \quad\text{for all } \vec R.
> $$
> It is itself a Bravais lattice, generated by primitive vectors $\vec b_1, \vec
> b_2, \vec b_3$ satisfying $\vec b_i \cdot \vec a_j = 2\pi\,\delta_{ij}$.

The generating vectors follow explicitly. With $v_{\text{c}} = \vec a_1 \cdot (\vec
a_2 \times \vec a_3)$ the primitive cell volume,

$$
\vec b_1 = 2\pi\,\frac{\vec a_2 \times \vec a_3}{v_{\text{c}}}, \qquad
\vec b_2 = 2\pi\,\frac{\vec a_3 \times \vec a_1}{v_{\text{c}}}, \qquad
\vec b_3 = 2\pi\,\frac{\vec a_1 \times \vec a_2}{v_{\text{c}}}.
$$

Each $\vec b_i$ is perpendicular to two of the direct primitive vectors, and the
$2\pi$ normalization gives $\vec b_i \cdot \vec a_j = 2\pi\,\delta_{ij}$ directly.
A general reciprocal vector is $\vec G = h\vec b_1 + k\vec b_2 + l\vec b_3$ with
integer $h,k,l$; then $\vec G \cdot \vec R = 2\pi(h n_1 + k n_2 + l n_3)$ is
$2\pi$ times an integer, as required.

Three properties follow immediately:

- The reciprocal of the reciprocal lattice is the original direct lattice.
- The reciprocal-lattice primitive cell has volume $(2\pi)^3/v_{\text{c}}$.
- The units of $\vec G$ are inverse length; the reciprocal lattice lives in
  wavevector (momentum) space.

$$
% caption: A direct lattice (left) and its reciprocal (right) in two dimensions.
% The oblique direct cell with primitive vectors a1, a2 maps to a reciprocal cell
% with b1 perpendicular to a2 and b2 perpendicular to a1; a denser direct lattice
% gives a sparser reciprocal one.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % ---- direct lattice (oblique) ----
  \begin{scope}
    \foreach \i in {0,1,2,3}{
      \foreach \j in {0,1,2,3}{
        \pgfmathsetmacro{\px}{1.1*\i + 0.4*\j}
        \pgfmathsetmacro{\py}{0.95*\j}
        \fill[black] (\px,\py) circle (1.8pt);
      }
    }
    \draw[acc, very thick, ->] (0,0) -- (1.1,0) node[midway, below] {$a_1$};
    \draw[acc, very thick, ->] (0,0) -- (0.4,0.95) node[midway, left] {$a_2$};
    \node at (1.5,-0.8) {direct lattice};
  \end{scope}
  % ---- reciprocal lattice ----
  \begin{scope}[xshift=6.5cm]
    \foreach \i in {0,1,2,3}{
      \foreach \j in {0,1,2,3}{
        \pgfmathsetmacro{\px}{1.2*\i}
        \pgfmathsetmacro{\py}{-0.5*\i + 1.15*\j}
        \fill[black] (\px,\py) circle (1.8pt);
      }
    }
    \draw[acc, very thick, ->] (0,0) -- (1.2,-0.5) node[midway, below] {$b_1$};
    \draw[acc, very thick, ->] (0,0) -- (0,1.15) node[midway, left] {$b_2$};
    \node at (1.6,-1.0) {reciprocal lattice};
  \end{scope}
\end{tikzpicture}
$$

## Reciprocal lattices of the cubic families

For a **simple-cubic** lattice with $a_i = a\,\hat e_i$, the reciprocal vectors are
$b_i = (2\pi/a)\hat e_i$: another simple-cubic lattice with constant $2\pi/a$.

The centered cubics swap. Applying the cross-product formula to the fcc primitive
vectors $\vec a_1 = \tfrac{a}{2}(\hat y + \hat z)$, $\vec a_2 = \tfrac{a}{2}(\hat z
+ \hat x)$, $\vec a_3 = \tfrac{a}{2}(\hat x + \hat y)$ gives reciprocal vectors of
the form $(2\pi/a)(\hat y + \hat z - \hat x)$, which generate a **body-centered
cubic** lattice with cubic constant $4\pi/a$. Conversely the reciprocal of bcc is
fcc. This duality controls diffraction: X-rays scattered from an fcc crystal
produce spots on a bcc reciprocal lattice.

| Direct lattice | Reciprocal lattice |
| --- | --- |
| Simple cubic, constant $a$ | Simple cubic, constant $2\pi/a$ |
| Body-centered cubic | Face-centered cubic |
| Face-centered cubic | Body-centered cubic |

## Reciprocal vectors and lattice planes

The reason Miller indices use reciprocals is that each reciprocal-lattice vector is
tied to a family of direct-lattice planes.

> **Theorem (Reciprocal vectors and lattice planes).** For any family of parallel
> lattice planes separated by spacing $d$, there is a reciprocal-lattice vector
> $\vec G$ perpendicular to the planes, and the shortest such $\vec G$ has length
> $\lvert \vec G\rvert = 2\pi/d$. Conversely every reciprocal vector $\vec G$ is
> normal to a family of lattice planes with spacing $2\pi/\lvert\vec G\rvert$.

> **Proof.** The planes are surfaces of constant phase of the wave $e^{i\vec G
> \cdot \vec r}$: adjacent planes differ in phase by $2\pi$, so their normal
> spacing is $d = 2\pi/\lvert\vec G\rvert$ when $\vec G$ is the shortest reciprocal
> vector in that direction. That $\vec G = h\vec b_1 + k\vec b_2 + l\vec b_3$ is
> perpendicular to the plane with intercepts $a_1/h, a_2/k, a_3/l$ follows because
> $\vec G$ dotted into any two in-plane difference vectors vanishes: $\vec G \cdot
> (\vec a_1/h - \vec a_2/k) = 2\pi - 2\pi = 0$. Hence the $(hkl)$ plane is normal
> to $\vec G_{hkl}$, and in a cubic crystal $\lvert\vec G_{hkl}\rvert = (2\pi/a)
> \sqrt{h^2+k^2+l^2}$ reproduces $d_{hkl} = a/\sqrt{h^2+k^2+l^2}$.

$$
% caption: A reciprocal-lattice vector G is normal to its family of direct
% lattice planes, and its length fixes the plane spacing through d = 2 pi over
% the magnitude of G. Denser planes (larger index) mean a longer G.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % three parallel planes (lines) with spacing d, tilted
  \foreach \s in {0,1,2}{
    \draw[black, thick] ({0.9*\s},{-0.6*\s}) -- ({0.9*\s+2.4},{-0.6*\s+3.2});
  }
  % G vector perpendicular to planes
  \draw[acc, very thick, ->] (1.2,0.4) -- (2.55,1.35) node[midway, above left] {$G$};
  % spacing indicator d
  \draw[black, <->] (0.9,-0.6) -- (1.8,-1.2);
  \node[black, anchor=north] at (1.35,-1.0) {$d$};
  \node[black, anchor=west] at (2.7,2.6) {lattice planes};
\end{tikzpicture}
$$

## The first Brillouin zone

The Wigner-Seitz cell of the reciprocal lattice is the **first Brillouin zone**. It
is built exactly as in the direct lattice: draw vectors from a reciprocal point to
its neighbors, bisect each with a perpendicular plane, and take the enclosed
volume.

> **Definition (Brillouin zones).** The **first Brillouin zone** is the set of
> points in reciprocal space closer to a chosen reciprocal-lattice point than to
> any other — the Wigner-Seitz cell of the reciprocal lattice. The **$n$-th
> Brillouin zone** is the region reachable from the origin by crossing exactly
> $n-1$ Bragg planes (the perpendicular bisectors of reciprocal vectors).

Because the reciprocal of fcc is bcc, the first Brillouin zone of an fcc crystal is
the Wigner-Seitz cell of the bcc lattice: a **truncated octahedron** with six
square faces (along the cube axes) and eight hexagonal faces (along the body
diagonals). Its high-symmetry points carry standard labels used throughout band
theory: $\Gamma$ at the center, $X$ at a square face center, $L$ at a hexagonal
face center, and $K$ on an edge.

$$
% caption: The first Brillouin zone of the fcc lattice is a truncated
% octahedron: six square faces along the cube axes, eight hexagonal faces along
% the body diagonals. Gamma is the zone center, X a square-face center, L a
% hexagon-face center, K a hexagon edge midpoint.
\begin{tikzpicture}[scale=1.5, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % schematic truncated octahedron outline (hexagonal silhouette)
  \coordinate (t) at (0,1.5);
  \coordinate (tr) at (1.3,0.75);
  \coordinate (br) at (1.3,-0.75);
  \coordinate (b) at (0,-1.5);
  \coordinate (bl) at (-1.3,-0.75);
  \coordinate (tl) at (-1.3,0.75);
  \draw[acc, thick] (t) -- (tr) -- (br) -- (b) -- (bl) -- (tl) -- cycle;
  % inner edges suggesting square + hexagon faces
  \draw[black] (t) -- (0.55,0.5) -- (tr);
  \draw[black] (bl) -- (-0.55,-0.5) -- (b);
  \draw[black] (0.55,0.5) -- (0.55,-0.5) -- (br);
  \draw[black] (0.55,0.5) -- (-0.55,0.5) -- (tl);
  \draw[black] (0.55,-0.5) -- (-0.55,-0.5);
  \draw[black] (-0.55,0.5) -- (-0.55,-0.5);
  % labeled points
  \fill[acc] (0,0) circle (1.6pt);
  \node[anchor=north east] at (0.05,0.02) {Gamma};
  \fill[black!70] (0.55,0.0) circle (1.4pt);
  \node[anchor=west] at (0.62,0.05) {X};
  \fill[black!70] (0.9,0.75) circle (1.4pt);
  \node[anchor=south west] at (0.9,0.78) {L};
  \fill[black!70] (1.3,0.0) circle (1.4pt);
  \node[anchor=west] at (1.36,0.0) {K};
\end{tikzpicture}
$$

## Higher zones

The perpendicular bisectors of _all_ reciprocal vectors — the Bragg planes — carve
reciprocal space into concentric regions. The first zone is the innermost cell;
the second zone is the set of fragments lying between the first and second Bragg
planes; the third lies beyond those. Every zone has the same total volume,
$(2\pi)^3/v_{\text{c}}$, since each is a full primitive cell reassembled from
pieces. Translating the fragments of the $n$-th zone by suitable reciprocal
vectors folds them back into the first zone, which is why band structures can
always be drawn in the reduced-zone scheme.

$$
% caption: First, second, and third Brillouin zones of a two-dimensional square
% lattice. Zone 1 is the central square (nearest Bragg planes); zone 2 is the
% four triangles beyond it; zone 3 is the outer fragments. Each has equal area
% and folds back onto zone 1 by a reciprocal-lattice translation.
\begin{tikzpicture}[scale=0.85, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % reciprocal lattice points
  \foreach \i in {-2,-1,0,1,2}{
    \foreach \j in {-2,-1,0,1,2}{
      \fill[black] (2*\i,2*\j) circle (1.6pt);
    }
  }
  % zone 1: central square (bisectors to (±1,0),(0,±1)) -> square side 2 centered
  \fill[acc!18] (-1,-1) rectangle (1,1);
  \draw[acc, very thick] (-1,-1) rectangle (1,1);
  \node[acc] at (0,0) {1};
  % zone 2: diagonals bisectors to (±1,±1): lines x±y=±2 forming a rotated square
  % draw the rotated square (corners at (±2,0),(0,±2)) minus zone1
  \draw[black, thick] (2,0) -- (0,2) -- (-2,0) -- (0,-2) -- cycle;
  \node[black!70] at (1.3,1.3) {2};
  % zone 3: outer square edges bisectors to (±2,0),(0,±2): lines x=±2, y=±2
  \draw[black, dashed] (-2,-2) rectangle (2,2);
  \node[black] at (1.9,0.4) {3};
\end{tikzpicture}
$$

The reciprocal lattice and its first zone are the coordinate system for
everything that follows. Diffraction spots sit at reciprocal-lattice points, and
the [Bragg and Laue conditions](/condensed-matter/crystal-structure/diffraction-and-structure-factors)
are two statements of the same requirement, that the scattering vector be a
reciprocal-lattice vector.