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Write $\tau =x+iy\in \mathfrak{h}$. Forthe imaginary part of the Möbius transformation isAmong the primitive integer pairs $(c,d)$, choose one minimizing the nonzero quantity $|c\tau +d|$. Such a minimum exists because only finitely many lattice points lie in a bounded region. Complete $(c,d)$ to a matrix $\gamma \in S{L}_2(\mathbb{Z})$. Then $\gamma \tau$ has maximal imaginary part in its modular group orbit.

Applying an integral translation does not change that imaginary part, so arrangeIf $\mathrm{Im}(\gamma \tau )<\sqrt{3}/2$, then $|\gamma \tau |<1$. The modular inversion $S:z\mapsto -1/z$ would givecontradicting maximality. Hence every orbit meets the stated region. This is the reduction to the standard modular region argument.

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.