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ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2022/paper-2/section-c/c8/e/solution

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Past exam of the physics A course of the University of Cambridge / 2022 / Paper 2 / Section C / C8 / e / Solution by

Codex  0 2026-10-18

The length, area and volume of a first Brillouin zone are

(2 π / a)^{d}

. The corresponding occupied

k

-space measures are

2 k_{F}^{(1)} = \frac{π}{a} = \frac{1}{2} \frac{2 π}{a},

π (k_{F}^{(2)})^{2} = \frac{2 π ^{2}}{a ^{2}} = \frac{1}{2} (\frac{2 π}{a})^{2},

\frac{4 π}{3} (k_{F}^{(3)})^{3} = \frac{4 π ^{3}}{a ^{3}} = \frac{1}{2} (\frac{2 π}{a})^{3} .

Hence

\frac{occupied k -space measure}{one Brillouin-zone measure} = \frac{1}{2} in all three dimensions .

This is state counting, independent of the interval/disk/ball geometry. One Brillouin zone contains one spatial state per unit cell in a single energy band, and that band can hold two Electrons per cell because of spin. One Electron per cell therefore occupies half of it. The “volume enclosed” is the occupied Fermi sea, not the lower-dimensional area of the Fermi surface itself. Its half-volume does not mean half the distance to the zone boundary in two or three dimensions.

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