 Solution

ID: past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2021/2/b/solution

a_{1}, a_{2}, a_{3}

V_{p} = a_{1} \cdot (a_{2} \times a_{3}), b_{1} = 2 π \frac{a _{2} \times a _{3}}{V _{p}}, b_{2} = 2 π \frac{a _{3} \times a _{1}}{V _{p}}, b_{3} = 2 π \frac{a _{1} \times a _{2}}{V _{p}} .

Then

a_{i} \cdot b_{j} = 2 π δ_{ij}

G = h b_{1} + k b_{2} + l b_{3}, h, k, l \in Z,

equivalently all wavevectors for which

e^{i G \cdot R} = 1

for every direct lattice vector

R

a_{1} = \frac{a}{2} (0, 1, 1), a_{2} = \frac{a}{2} (1, 0, 1), a_{3} = \frac{a}{2} (1, 1, 0) .

a^{3} /4

, and direct evaluation of the cross products gives

b_{1} = \frac{2 π}{a} (- 1, 1, 1), b_{2} = \frac{2 π}{a} (1, - 1, 1), b_{3} = \frac{2 π}{a} (1, 1, - 1) .

Every integer combination has coordinates

(2 π / a) (H, K, L)

with

H, K, L

all even or all odd. Conversely, for such coordinates the coefficients

(K + L) /2, (H + L) /2, (H + K) /2

are integers, so every such point occurs. The even-coordinate points are the corners of cubes of edge

4 π / a

, and the odd-coordinate points are their body centres. Therefore

reciprocal Bravais lattice: body-centred cubic, a_{reciprocal} = \frac{4 π}{a} .

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