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

The primitive translations connect the two centred-rectangular sublattices. In the orientations marked in the source,

a_{p}

points right and down, whereas

b_{p}

points right and up. Thus

a_{p} = \frac{1}{2} (a_{r} - b_{r}), b_{p} = \frac{1}{2} (a_{r} + b_{r}) .

Their parallelogram has half the area of the rectangular cell, confirming that the latter contains two primitive unit cells.

ii

G = h_{r} a_{r}^{*} + k_{r} b_{r}^{*} = h_{p} a_{p}^{*} + k_{p} b_{p}^{*}

, with each dual pair obeying

a \cdot a^{*} = b \cdot b^{*} = 2 π

and zero cross products. Its coefficients are obtained by dotting with the corresponding direct-space primitive lattice translation vectors:

h_{p} k_{p} = \frac{G \cdot a _{p}}{2 π} = \frac{h _{r} - k _{r}}{2}, = \frac{G \cdot b _{p}}{2 π} = \frac{h _{r} + k _{r}}{2} .

Hence

h_{p} = \frac{1}{2} h_{r} - \frac{1}{2} k_{r}, k_{p} = \frac{1}{2} h_{r} + \frac{1}{2} k_{r} .

For actual integer-indexed reciprocal lattice vectors, this also explains the centred-rectangular condition that

h_{r}

and

k_{r}

Comparing the two coefficients in the reciprocal-coordinate transformation gives

α = β = \frac{1}{2} .

The inverse transformation is

h_{r} = h_{p} + k_{p}

k_{r} = k_{p} - h_{p}

. This provides an independent check on the signs associated with the two primitive arrows in the source.

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