Why does the scattering wave vector and reciprocal lattice vector have to be the same?

Question

In Introduction to Solid state physics 8th edition by Charles Kittel. The scattering amplitude is defined as, $$F=int dV n_G exp[i(bf{G-Delta k}).bf{r}]$$ where G is the reciprocal lattice vector and $Delta k$ is the scattering vector. It goes on to state that when $Delta k neq G $ the integral goes to zero. Can someone help with the proof for the same? Thanks in advance.

Michael Seifert · Answer

This is a consequence of the distributional identity
$$
int_{-infty}^{infty} e^{i k x} , dx = 2 pi delta(k).
$$
In particular, this implies that
$$
iiint e^{i vec{k} cdot vec{x}} , dV = left[ int_{-infty}^{infty} e^{i k_x x} , dx right] left[ int_{-infty}^{infty} e^{i k_y y} , dy right] left[ int_{-infty}^{infty} e^{i k_z z} , dz right] = (2 pi)^3 delta(k_x) delta(k_y) delta(k_z) = (2pi)^3 delta^{(3)}(vec{k}).
$$
This distribution vanishes for all $vec{k} neq 0$.
As to why this identity is valid, there's a fairly good discussion over at Mathematics StackExchange.  At a "physics" level of rigor, it is probably best seen as a logical consequence of the properties of the Fourier transform and inverse Fourier transform.

Why does the scattering wave vector and reciprocal lattice vector have to be the same?

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