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

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.