 Solution

With spin degeneracy, each energy band contains two electrons per primitive unit cell. Divalency therefore permits a filled first band, but only if its maximum lies below the second band's minimum. With negligible ionic periodic potential, the circular Fermi surface and the rectangular zone have equal enclosed area, not equal shape. Electrons occupy some states outside the first zone while some higher-energy first-zone states remain empty. Folded bands overlap, and the material is a metal.

A periodic potential mixes plane waves differing by a reciprocal lattice vector. At a two-wave Bragg point, the nearly-free electron model gives a direct gap $2|V_{\mathbf{G}}|$. Increasing appropriate Fourier coefficients can remove the band overlap. With the first band filled, the system is then a band insulator, provided the temperature is small compared with the insulating band gap in energy units.

The rectangular free-electron band-overlap scale follows from two different points in reciprocal space. Put $C={\hslash }^2/(2m_e)$. Before gap opening, the highest first-zone energy is at the corner, while the lowest second-zone energy is at the midpoint of the nearer face:Thus the downward shift ${\delta }_v$ of the first-band maximum plus the upward shift ${\delta }_c$ of the second-band minimum must overcomeIn a rigid-splitting estimate, an opening $\Delta$ lowers the first band by $\Delta /2$ and raises the second by $\Delta /2$, giving the familiar threshold $\Delta >{\hslash }^2{\pi }^2/(2m_e{b}^2)$. Equality is only band touching, not a positive insulating gap. This expression is the overlap that must be removed, under that explicit band-shift approximation.

Positive direct gaps can coexist with band overlap: a local direct splitting and the global indirect band gap are not interchangeable. The shape and Fourier coefficients of the ionic potential are unspecified, so there is no universal exact critical direct-gap width determined by $a,b$ alone. For a concrete demonstration, a separable potential $2V_x\cos (2Kx)+2V_y\cos (2Ly)$ has axial direct gaps ${\Delta }_x=2|V_x|$ and ${\Delta }_y=2|V_y|$. To first order, the corner's lowest level is $C(K^2+L^2)-|V_x|-|V_y|$, and the two candidate second-band minima are $C{K}^2+|V_x|$ and $C{L}^2+|V_y|$. Therefore, within this model and neglecting higher-order band shifts, nonoverlap requiresEqual axial gaps would give the first-order estimate $\Delta >2C{L}^2/3$, rather than the rigid-splitting estimate. Such a large opening can exceed the quantitative range of weak-potential perturbation theory, but the example identifies the missing model assumption: corner levels feel both couplings. The exact general answer is the positive indirect-gap criterion above; a geometry-only direct-gap threshold would claim more than the supplied potential information supports.