Past exam of the chemistry A course of the University of Cambridge / 2023 / Paper 2 / A9 / a / iv

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- Solution iv

Solution

A visibly different primitive parallelogram is generated by

a_{1}^{'} = a_{1} = a (1, 1), a_{2}^{'} = a_{1} + a_{2} = a (0, 2) .

Its vertices, relative to one lattice point, are

(0, 0)

,

(a, a)

,

(a, 3 a)

and

(0, 2 a)

. Its area is

A_{c}^{'} = det (a a 0 2 a) = 2 a^{2} .

The basis change has integer matrix

(1011)

with determinant one. It is invertible over the integers, so the new vectors generate every original lattice point, not merely a sublattice. The cell is therefore primitive despite being sheared relative to cell B.

Figure 1.
Two primitive bases generate the same centered-square lattice
. Original coordinate construction from the derived translation vectors, with axes measured in units of $a$ . The left cell uses the diagonal orthogonal basis; the right uses a determinant-one shear. Both highlighted parallelograms have area $2 a^{2}$ and one lattice point per cell. Details.

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