 Solution

The first Brillouin zone is the squareAt half filling, translation of the wavevector by $\mathbf{Q}=(\pi /a,\pi /a)$ exchanges the two band energies:Thus the same energy-pair argument as in part c givesAt zero temperature, the Fermi surface satisfies $\cos (k_{x}a)+\cos (k_{y}a)=0$. Within the square Brillouin zone this is the diamondwhose vertices are $(\pm \pi /a,0)$ and $(0,\pm \pi /a)$. The occupied region is the diamond's interior, containing the band minimum at the zone centre. It has half the Brillouin zone area, consistent with one electron per atom and two spin states per wavevector.