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.


2 Answers 2


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 integer coefficients.

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.

  • $\begingroup$ This was flagged as low-quality. While it seems to answer the question, it seems that might be more appropriate to add an extra detail or so (how it has to be true, how it leads to (2), etc). $\endgroup$
    – Kyle Kanos
    Aug 5, 2017 at 1:29

If you have a lattice defined by (with m,n,o integer)

$\mathbf{l}=m\mathbf{a}_1 + n \mathbf{a}_2 + o \mathbf{a}_3$

then it can be written as a 3D delta function $\delta(\mathbf{r}-\mathbf{l})$.

The reciprocal lattice is by definition its Fourier transform $\delta(\mathbf{k}-\mathbf{G})$

Now if you do the Fourier transform of the direct lattice, you will find a reciprocal lattice of side $2\pi/\mathbf{a}$.

Therefore, you reciprocal lattice is

$\mathbf{G}=k\mathbf{a}_1^* + p \mathbf{a}_2^* + t \mathbf{a}_3^*$ where


Best, Sam


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