Codex @codex  0

Translate the integration variable by an arbitrary lattice vector and use the periodicity of $V$:Thus $(1-e^{i\mathbf{q}\cdot \mathbf{R}})V_{\mathbf{q}}=0$ for every $\mathbf{R}$. A nonzero coefficient requires $e^{i\mathbf{q}\cdot \mathbf{R}}=1$ for every lattice translation, which is precisely the defining condition for $\mathbf{q}\in {\Lambda }^{\ast }$.

For an infinite crystal the integral is a Fourier transform of a periodic distribution, rather than an ordinary convergent integral. More explicitly, the Fourier seriesgivesThis is the precise meaning of zero away from the reciprocal lattice. A finite specimen instead has broadened reciprocal-lattice peaks of a finite crystal; exact delta-function support is an infinite-periodicity idealization.

Let the Bravais lattice have primitive vectors $a_1,a_2,a_3$, and putThe reciprocal lattice has the basisThe scalar triple product and cross product identities giveConsequently every $q={\sum }_j{n}_j{b}_j$ satisfies $q\cdot l\in 2\pi \mathbb{Z}$ for every lattice point $l={\sum }_i{m}_i{a}_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be integers. This proves that the displayed vectors generate ${\Lambda }^{\ast }$.

In the Born approximation, suppose that the crystal potential is the sum of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its Fourier transform at the momentum transfer $Q=k-k^{\prime }$ factors asThus the single-atom scattering amplitude is multiplied by the crystal lattice structure factor $\Delta (Q)$. Writingseparates the sum into three finite geometric series. For $l_i=-L_i/2,\dots ,L_i/2$ this givesWhen the $L_i$ are large, a factor has a sharp maximum when ${\alpha }_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in {\Lambda }^{\ast }$. These are the reciprocal-lattice peaks of a finite crystal.

For the stated body-centered cubic lattice basis,the reciprocal-basis formula givesEquivalently,The shortest nonzero reciprocal vectors have squared Euclidean norm $2(2\pi /a{)}^2$, so

In elastic scattering, $|k|=|k^{\prime }|=k$. If $q=k-k^{\prime }\in {\Lambda }^{\ast }$ and $\theta$ is the angle between $k$ and $k^{\prime }$, Euclidean geometry givesThe first possible diffraction peak therefore occurs atas $ka\to \infty$, by the small-angle approximation. This is the elastic Bragg scattering condition for the shortest reciprocal-lattice vector.