 Solution

One orbital per primitive cell supplies two states per wave vector when spin is included. A monovalent metal therefore half-fills this band. Nearest-neighbor hopping on the bipartite rectangular lattice has $E(\mathbf{k}+\mathbf{Q})=2E_0-E(\mathbf{k})$ for $\mathbf{Q}=(\pi /a,\pi /b)$, fixing the half-filled Fermi energy atFor $t_x=2t_y>0$, the two Fermi surface obeyThey are corrugated open sheets in an extended-zone description, with $|k_{x}a|$ ranging from $\pi /3$ to $2\pi /3$. They are not a small circular pocket centered at the band bottom. The same $\mathbf{Q}$ maps one Fermi line onto its nested partner. A density-wave or Peierls-like modulation at that wave vector can open additional gaps at the Fermi energy. The unperturbed single-orbital band does not acquire those new gaps without a symmetry-breaking modulation or another coupling; ordinary lattice Bragg effects are already built into its dispersion.

For the three-dimensional tetragonal version set $t_x=t_y=t$, $t_z=t/2$, $a=b<c$. The corresponding half-filled surface isIt is a corrugated periodic surface, not a sphere. At $k_z=\pi /c$ its regular closed contour is electron-like around $\Gamma$, with $\cos (k_{x}a)+\cos (k_{y}a)=1/2$; at $k_z=0$ the closed contour is hole-like around $M$, with the sum $-1/2$. The topology switches through the square-lattice separatrix at $k_{z}c=\pm \pi /2$. Drawing one simple weakly warped cylindrical pocket across every section misses that feature of the half-filled cosine band.

Angle-resolved photoemission can directly map the dispersion and occupied Fermi surface. Angular quantum-oscillation measurements test its closed extremal cross sections and cyclotron masses; momentum-density probes such as positron annihilation or Compton scattering give a complementary bulk test. The different techniques constrain different features rather than merely providing interchangeable pictures.