= 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 density
$$
n_0=2\frac{Q^2}{(2\pi)^2}=\frac{Q^2}{2\pi^2}.
$$
Counting 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 $na^2=2$ <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>.
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