Codex @codex  0

For two electrons per primitive cell,Set $A={\hslash }^2{\pi }^2/(2m_e{a}^2)=2.350\mathrm{eV}$. In the indirect gap in a uniformly split zone-boundary model approximation stipulated here, the free boundary energies are $A$ along $[10]$ at $X=(\pi /a,0)$ and $2A$ along $[11]$ at $M=(\pi /a,\pi /a)$. Opening a symmetric gap givesThe branches bend away from the free-electron crossings and flatten at their boundary extrema. Importantly, the upper band near $X$ lies below the lower band's corner energy:Therefore the crystal is metallic despite its positive boundary gaps; electron and hole occupations redistribute across overlapping bands. positive direct gaps can coexist with band overlap.

A microscopic square-lattice corner has additional plane-wave degeneracy. The quoted symmetric boundary numbers use the question's directional gap model, not an unspecified four-state corner Hamiltonian. For the indirect gap of a separable square cosine potential, a separable weak potential with the same face gap gives the lowest corner energy $2A-E_g=3.70\mathrm{eV}$, still above the $2.85\mathrm{eV}$ upper-band face energy. This refinement preserves the metallic conclusion.

For an isotropic free-electron gas at zero temperature, the Fermi wavevector is the radius of the occupied region in wavevector space. Its energy is $E_F={\hslash }^2{k}_F^2/(2m)$. The first Brillouin zone is the Wigner-Seitz cell of the reciprocal lattice; it contains one distinct crystal momentum per primitive cell and per band, with spin counted separately.

For a two-dimensional sample of area $A$, allowed wavevectors have density $A/(2\pi {)}^2$. Including two spin states, filling a circle gives $N=2[A/(2\pi {)}^2]\pi k_F^2$. Dividing by area therefore yieldsThe square lattice has one atom per area $a^2$, so for valence $z$ the free-electron density is $z/a^2$. Its Brillouin zone has $|k_x|,|k_y|\le \pi /a$. ThusThe monovalent Fermi circle lies inside the square. The divalent circle crosses the faces but misses the corners; its area equals the zone's area. These are the free-electron contours before distortion by the periodic potential.

At a Bragg plane, the periodic potential mixes degenerate plane waves differing by a reciprocal lattice vector. The two standing-wave combinations have different charge densities relative to the ions and split in energy, opening the gap. In a two-wave approximation, the boundary energies are $E_B\pm E_g/2$, with $E_g=2|V_G|$.

The boundary wavevector magnitude is $\pi /a$ along $[10]$ and $\sqrt{2}\pi /a$ along $[11]$. The free-electron boundary energies are thereforeWithin the stated direction-independent gap picture, the upper band's lowest face-edge energy is approximately $E_X+2\mathrm{eV}=6.18\mathrm{eV}$, while the lower band's corner-edge energy is approximately $E_M-2\mathrm{eV}=6.36\mathrm{eV}$. There is consequently a small overlap, about $0.18\mathrm{eV}$, rather than a global insulating gap. The divalent Fermi level lies in that overlap, producing occupied upper-band states and unfilled lower-band states.

Thus the intended isotropic-gap approximation predicts a metal, with small electron and hole pockets. Positive direct gaps can coexist with band overlap. A fully specified square-lattice potential would require simultaneous treatment of the corner degeneracies; a numerical boundary splitting alone does not uniquely determine an arbitrary two-dimensional band structure. The conclusion here uses the given simplified gap picture, rather than electron count alone.