Past exam of the condensed matter physics course of the University of Oxford / 2019 / 1 / c

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- Solution c

Solution

Translate the integration variable by an arbitrary lattice vector and use the periodicity of

V

:

V_{q} = \int d^{3} x e^{i q \cdot x} V (x) = e^{i q \cdot R} \int d^{3} y e^{i q \cdot y} V (y) = e^{i q \cdot R} V_{q} .

Thus

(1 - e^{i q \cdot R}) V_{q} = 0

for every

R

. A nonzero coefficient requires

e^{i q \cdot R} = 1

for every lattice translation, which is precisely the defining condition for

q \in Λ^{*}

.

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 (x) = \sum_{G \in Λ^{*}} v_{G} e^{- i G \cdot x}, v_{G} = \frac{1}{Ω _{c}} \int_{cell} d^{3} x e^{i G \cdot x} V (x)

gives

V_{q} = (2 π)^{3} G \in Λ^{*} \sum v_{G} δ^{(3)} (q - 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.

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