= Solution
{c}
Translate the integration variable by an arbitrary <lattice vector> and use the periodicity of $V$:
$$
V_{\mathbf q}=\int d^3x\,e^{i\mathbf q\cdot\mathbf x}V(\mathbf x)
=e^{i\mathbf q\cdot\mathbf R}\int d^3y\,e^{i\mathbf q\cdot\mathbf y}V(\mathbf y)
=e^{i\mathbf q\cdot\mathbf R}V_{\mathbf q}.
$$
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 $\mathbf q\in\Lambda^*$.
For an infinite crystal the integral is a <Fourier transform of a periodic distribution>, rather than an ordinary convergent integral. More explicitly, the <Fourier series>
$$
V(\mathbf x)=\sum_{\mathbf G\in\Lambda^*}v_{\mathbf G}e^{-i\mathbf G\cdot\mathbf x},\qquad
v_{\mathbf G}=\frac1{\Omega_c}\int_{\mathrm{cell}}d^3x\,e^{i\mathbf G\cdot\mathbf x}V(\mathbf x)
$$
gives
$$
\boxed{V_{\mathbf q}=(2\pi)^3\sum_{\mathbf G\in\Lambda^*}v_{\mathbf G}\delta^{(3)}(\mathbf q-\mathbf G).}
$$
This 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.
Back to article page
