Past exam of the condensed matter physics course of the University of Oxford / 2018 / 2 / e / Solution

The reciprocal primitive vectors are $(2\pi /a_x,0)$ and $(0,2\pi /a_y)$, so the first Brillouin zone isIt is wider in the $k_x$ direction because $a_x<a_y$.

The geometry alone does not fix the hopping amplitudes or a unique Fermi surface. One consistent tight-binding model isrepresenting larger hopping between the closer atoms. It remains a nearest-neighbour bipartite band at half filling, so its chemical potential is ${ϵ}_0$. Its Fermi surface consists of the two sheetsThese meet the top and bottom zone edges and continue periodically into the neighbouring zones. The right panel of the figure gives an explicit example with $t_x=2t_y$ and $a_x=0.65a_y$. The central region between the sheets is occupied and has half the zone area. If the hopping amplitudes instead remained equal, a geometrically stretched diamond would also satisfy the stated information. The metric inequality $a_x<a_y$ alone does not determine which of these shapes occurs.