∣ k_{x} ∣, ∣ k_{y} ∣ \leq π / a

. For one atom per primitive cell and two spin states, a monovalent free-electron gas has

n_{2 D} = \frac{1}{a ^{2}} = \frac{k _{F}^{2}}{2 π}, k_{F} a = 2 π .

Its circular Fermi surface lies inside the zone and encloses half its area: a partially filled band is conducting. Here odd filling obstructs a spin-degenerate band insulator: a monovalent band insulator is impossible in this spin-degenerate one-atom independent-electron model. It would require changed premises, such as unit-cell enlargement or physics beyond the free-electron picture; no insulating free-electron contour should be invented.

ii

 0  0

Solution

 0  0

For divalent atoms,

k_{F} a = 4 π

. The circle has the same area as the first Brillouin zone but crosses its faces and leaves its corners unfilled. Thus the free-electron case remains conducting: filling capacity alone does not remove band overlap. A nearly free-electron conductor has reconstructed electron and hole pockets; a divalent band insulator instead has one completely filled band separated from all higher bands, with no Fermi surface in the gap.

Figure 1.
Original square-zone state counting, monovalent and divalent free contours, and insulating band filling
. **Details**.

Solution

 0  0

For two electrons per primitive cell,

ϵ_{F} = \frac{ℏ ^{2} k _{F}^{2}}{2 m _{e}} = \frac{2 π ℏ ^{2}}{m _{e} a ^{2}} ≃ 2.99 eV .

(1)

Set

A = ℏ^{2} π^{2} / (2 m_{e} a^{2}) = 2.350 eV

. In the indirect gap in a uniformly split zone-boundary model approximation stipulated here, the free boundary energies are

A

along

[10]

X = (π / a, 0)

and

2 A

along

[11]

M = (π / a, π / a)

. Opening a symmetric gap gives

E_{X, \pm} = A \pm E_{g} /2 = 1.85, 2.85 eV, E_{M, \pm} = 2 A \pm E_{g} /2 = 4.20, 5.20 eV .

(2)

The 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:

E_{indirect} ≃ E_{X, +} - E_{M, -} = E_{g} - A = - 1.35 eV < 0 .

(3)

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