Carrier compensation and even band filling in a square lattice

Consider two overlapping energy bands on a square lattice with spin degeneracy two and primitive reciprocal period $Q$. Let $n$ count the areal electron density in this pair, $n_e$ the occupied density in upper-band electron pockets, and $n_h$ the missing density in lower-band hole pockets. A completely filled lower band provides the reference densityCounting filled states gives $n=n_0-n_h+n_e$, hence $n_e-n_h=n-n_0$. Compensation, meaning equal electron pocket and hole pocket carrier densities, therefore occurs at $n=n_0$. For total pocket areas $A_e,A_h$, the same state counting in a Bloch band gives $n_{e,h}=2A_{e,h}/(2\pi {)}^2$; all equivalent pockets must be counted.

If the geometry is parameterized by $Q=\alpha \sqrt{2\pi n}$ with $\alpha >0$, compensation requires $\alpha =\sqrt{\pi }$. The real-space period is $a=2\pi /Q$, so the filling is $n{a}^2=2$ electrons per primitive unit cell. Compensation can coexist with both kinds of charge carriers; a global band gap is additionally required for a band insulator.

A band filling constraint counts electrons per primitive unit cell of the actual translation symmetry. If an ordered state doubles the cell, a filling of one electron per original cell becomes two per new cell. The reduced Brillouin zone then permits reconstruction into different energy bands.

For spin-degenerate independent electrons, a filled band holds two electrons per primitive cell. Doubling can therefore remove the odd-filling obstruction to a band insulator, provided the reconstruction opens a global band gap. It does not guarantee a gap: folded bands can still overlap or retain symmetry-protected crossings. The original counting argument remains correct for its original symmetry assumptions; applying it after translation symmetry changes requires recounting using the new cell.