Every Bravais lattice has a reciprocal lattice. The Bravias lattice can be specified by giving three primitive lattice vectors $\vec{a}_1$, $\vec{a}_2$, and $\vec{a}_3$. A translation vector is a vector that points from one Bravais lattice point to some other Bravais lattice point. The translation vectors are,

$$\vec{T}_{hkl} = h\vec{a}_1 + k\vec{a}_2 + l\vec{a}_3,$$

where $h$, $k$, and $l$ are integers. The reciprocal lattice is a set of points connected by three primitive reciprocal lattice vectors $\vec{b}_1$, $\vec{b}_2$, and $\vec{b}_3$. The reciprocal lattice vectors point from one reciprocal lattice point to some other reciprocal lattice point,

$$\vec{G}=\nu_1\vec{b}_1+\nu_2\vec{b}_2+\nu_3\vec{b}_3\hspace{1.5 cm}\nu_1,\nu_2,\nu_3 = \cdots ,-2,-1,0,1,2,\cdots.$$

These reciprocal lattice vectors are needed to construct a Fourier series. The primitive reciprocal lattice vectors can be determined from the real space primitive lattice vectors with the formula,

$$\vec{a}_i\cdot \vec{b}_j=2\pi\delta_{ij},$$

where $\delta_{ij}$ is the Kronecker delta,

$$\delta_{ij}= \begin{cases} 1 & \mbox{for } i=j \\ 0, & \mbox{for } i\ne j \end{cases}.$$

Another way to determine the reciprocal primitive lattice vectors from the real space primitive lattice vectors (or vice versa) is,

\begin{eqnarray} \vec{b}_1=2\pi \frac{\vec{a}_2\times\vec{a}_3}{\vec{a}_1\cdot\left(\vec{a}_2\times\vec{a}_3\right)}&\hspace{2cm}&\vec{a}_1=2\pi \frac{\vec{b}_2\times\vec{b}_3}{\vec{b}_1\cdot\left(\vec{b}_2\times\vec{b}_3\right)} \\ \vec{b}_2=2\pi \frac{\vec{a}_3\times\vec{a}_1}{\vec{a}_1\cdot\left(\vec{a}_2\times\vec{a}_3\right)}&\hspace{2cm}&\vec{a}_1=2\pi \frac{\vec{b}_3\times\vec{b}_1}{\vec{b}_1\cdot\left(\vec{b}_2\times\vec{b}_3\right)} \\ \vec{b}_3=2\pi \frac{\vec{a}_1\times\vec{a}_2}{\vec{a}_1\cdot\left(\vec{a}_2\times\vec{a}_3\right)}&\hspace{2cm}&\vec{a}_3=2\pi \frac{\vec{b}_1\times\vec{b}_2}{\vec{b}_1\cdot\left(\vec{b}_2\times\vec{b}_3\right)} \end{eqnarray}

Notice that because of the cross product, $\vec{b}_1$ is perpendicular to $\vec{a}_2$ and $\vec{a}_3$ so $\vec{a}_2\cdot\vec{b}_1 = \vec{a}_3\cdot\vec{b}_1=0$ and $\vec{a}_1\cdot\vec{b}_1= 2\pi \vec{a}_1\cdot\left(\vec{a}_2\times\vec{a}_3\right)/\left( \vec{a}_1\cdot\left(\vec{a}_2\times\vec{a}_3\right)\right) = 2\pi$ so that the expressions with the cross products are the same as $a_ib_j=2\pi\delta_{ij}$.

By applying these rules one can show that a simple cubic lattice has a simple cubic reciprocal lattice.

$$\text{sc:}\qquad\vec{a}_1=a\hat{x},\quad \vec{a}_2=a\hat{y},\quad\vec{a}_3=a\hat{z},\\ \vec{b}_1=\frac{2\pi}{a}\hat{k}_x,\quad \vec{b}_2=\frac{2\pi}{a}\hat{k}_y,\quad\vec{b}_3=\frac{2\pi}{a}\hat{k}_z.$$

An fcc lattice has a bcc reciprocal lattice.

$$\text{fcc:}\qquad\vec{a}_1=\frac{a}{2}(\hat{x}+\hat{z}),\quad \vec{a}_2=\frac{a}{2}(\hat{x}+\hat{y}),\quad\vec{a}_3=\frac{a}{2}(\hat{y}+\hat{z}),\\ \vec{b}_1=\frac{2\pi}{a}(\hat{k}_x-\hat{k}_y+\hat{k}_z),\quad \vec{b}_2=\frac{2\pi}{a}(\hat{k}_x+\hat{k}_y-\hat{k}_z),\quad\vec{b}_3=\frac{2\pi}{a}(-\hat{k}_x+\hat{k}_y+\hat{k}_z).$$

A bcc lattice has an fcc reciprocal lattice.

$$\text{bcc:}\qquad\vec{a}_1=\frac{a}{2}(\hat{x}+\hat{y}-\hat{z}),\quad \vec{a}_2=\frac{a}{2}(-\hat{x}+\hat{y}+\hat{z}),\quad\vec{a}_3=\frac{a}{2}(\hat{x}-\hat{y}+\hat{z}),\\ \vec{b}_1=\frac{2\pi}{a}(\hat{k}_x+\hat{k}_y),\quad \vec{b}_2=\frac{2\pi}{a}(\hat{k}_y+\hat{k}_z),\quad\vec{b}_3=\frac{2\pi}{a}(\hat{k}_x+\hat{k}_z).$$

A hexagonal lattice has a hexagonal reciprocal lattice.

$$\text{hex:}\qquad\vec{a}_1=a\hat{x},\qquad\vec{a}_2=\frac{a}{2}\hat{x}+\frac{\sqrt{3}a}{2}\hat{y},\qquad\vec{a}_3=c\hat{z},\\\vec{b}_1=\frac{2\pi}{\sqrt{3}a}\left(\sqrt{3}\hat{k}_x-\hat{k}_y\right),\qquad\vec{b}_2=\frac{4\pi}{\sqrt{3}a}\hat{k}_y,\qquad\vec{b}_3=\frac{2\pi}{c}\hat{k}_z.$$

The form below will calculate the primitive reciprocal lattice vectors given the real space primitive lattice vectors.

Primitive lattice vectors:  $\vec{a}_1=$ $\hat{x}+$ $\hat{y}+$ $\hat{z}$ [m]   $\vec{a}_2=$ $\hat{x}+$ $\hat{y}+$ $\hat{z}$ [m]  $\vec{a}_3=$ $\hat{x}+$ $\hat{y}+$ $\hat{z}$ [m]

Primitive reciprocal lattice vectors

$\vec{b}_1=2\pi\frac{\vec{a}_2\times\vec{a}_3}{\vec{a}_1\cdot\left(\vec{a}_2\times\vec{a}_3\right)}=$ $\hat{k}_x+$ $\hat{k}_y+$ $\hat{k}_z$ [m-1]

$\vec{b}_2=2\pi\frac{\vec{a}_3\times\vec{a}_1}{\vec{a}_1\cdot\left(\vec{a}_2\times\vec{a}_3\right)}=$ $\hat{k}_x+$ $\hat{k}_y+$ $\hat{k}_z$ [m-1]

$\vec{b}_3=2\pi\frac{\vec{a}_1\times\vec{a}_2}{\vec{a}_1\cdot\left(\vec{a}_2\times\vec{a}_3\right)}=$ $\hat{k}_x+$ $\hat{k}_y+$ $\hat{k}_z$ [m-1]

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• Wikipedia: Reciprocal lattice