The derivation of reciprocal lattice vectors in terms of the direct space lattice vectors starts by applying expanding a translationally invariant lattice function $f(\bf{R_k}+r)$ in plane waves $f_k e^{i G_m \cdot R_k} e^{i G_m \cdot r} $. Then by the translational invariance

$$

e^{i G_m \cdot R_k} = 1

$$

from which we have (1)

$$

G_m \cdot R_k = 2\pi N

$$

where N is an integer.

From this the next step in most derivations says that (2)

$$

\vec{a_i}^* \cdot \vec{a_j} = 2\pi \delta_{i,j}

$$

or in matrix form

$$

(\bf{A^*})^T\bf{A} = 2\pi \bf{I}\\

(\bf{A^*})^T = 2\pi \bf{A}^{-1}.

$$

However, I don't see how we can deduce (2) from (1).

Writing $G_m = h \vec{a_1}^* + k \vec{a_2}^* + l \vec{a_3}^*$ and $R_k = m \vec{a_1} + n \vec{a_2} + o \vec{a_3}$

for

$$

(h \vec{a_1}^* + k \vec{a_2}^* + l \vec{a_3}^*) \cdot (m\vec{a_1} + n \vec{a_2} + o \vec{a_3}) = 2\pi N

$$

I still don't see it immediately.

Any help would be appreciated, thank you.

## Best Answer

So first, the convention in crystallography is to write the Fourier series with a $2\pi$ in the phase, i.e. to replace your $G$ with $2\pi G$, which I will do in the following. I will also drop the index $m$ on $G_m$ because it plays no role. So your equation (1) is equivalent to require that

$$G.(m \vec{a}_1 + n \vec{a}_2 + o\vec{a}_3)$$

is an integer for any integer $m$, $n$, and $o$. By taking the three following $mno$: 100, 010, 001, we get the so-called Laue equations:

$$\begin{aligned} G.\vec{a}_1 &= h\\ G.\vec{a}_2 &= k\\ G.\vec{a}_3 &= l \end{aligned}$$

for some integers $h$, $k$, $l$.

The most traditional way to proceed from here I would say is to use the existence of an unique base $(\vec{a}^*_1, \vec{a}^*_2, \vec{a}^*_3)$ dual to the base $(\vec{a}_1, \vec{a}_2, \vec{a}_3)$, which has the essential property that for any vector $\vec{H}$,

$$\vec{H}=(\vec{H}.\vec{a}_1)\vec{a}^*_1 + (\vec{H}.\vec{a}_2)\vec{a}^*_2 + (\vec{H}.\vec{a}_3)\vec{a}^*_3.\tag{3}$$

The Laue equations then immediately give

$$\vec{G}=h\vec{a}^*_1 + k\vec{a}^*_2 + l\vec{a}^*_3, $$

proving that $\vec{G}$ can is a linear combination of the $\vec{a}^*_i$'s with

integercoefficients.Your eq. (2) holds for this dual base (without the factor $2\pi$),

$$\vec{a}_i\cdot\vec{a}^*_j = \delta_{ij}\tag{2}$$

as this is another characterisation of it, but it does not come as a consequence of (1) in this approach: instead it is a general and fundamental result of linear algebra (as dual bases exist in any dimension). In dimension 3, the simplest approach is to construct the dual base as

$$\vec{a}^*_1 = \frac{\vec{a}_2\times\vec{a}_3}{\det(\vec{a}_1, \vec{a}_2, \vec{a}_3)}$$

and circular permutation of indices. Then (2) easily follow, from which (3) is then obvious.