Past exam of the physics A course of the University of Cambridge / 2011 / Paper 1 / Section A / 5

For a free-electron Fermi circle in a square lattice, include two spin states per wavevector. A square cell of side $a$ supplies $z/a^2$ electrons per unit area; counting the disk of occupied states givesThe first Brillouin zone is $|k_x|,|k_y|\le \pi /a$. For $z=1$, $k_F=2.507/a$, so the circular Fermi surface lies wholly inside this square and encloses half its area. The figure uses the stated free-electron approximation; the lattice symmetry alone does not prescribe an exact circular surface.

For $z=2$, $k_F=\sqrt{4\pi }/a=3.545/a$. The circle encloses one whole zone's area, crosses each face because $k_F>\pi /a$, and misses the corners because $k_F<\sqrt{2}\pi /a$. In a reduced Brillouin zone, fold the portions outside the square into neighbouring band branches. A weak periodic potential opens band gaps at crossings and deforms these pieces; divalence alone does not establish a band insulator.