Fermi-volume conservation in two hybridized electron bands

For two isotropic electron-like bands with equal spin degeneracy, conserving total carrier density conserves the sum of their occupied momentum-space volumes. If both uncoupled radii equal $k_0$,Only when the upper pocket is empty does this imply a single Fermi wavevector $k_F=2^{1/3}{k}_0$; a nonzero mixing matrix element alone does not imply that emptying.

For a counterexample at arbitrarily weak coupling, take $E_a=A{k}^2$, $E_b=C+B{k}^2$, $A>B>0$, and $C=(A-B)k_0^2$. At the uncoupled chemical potential ${\mu }_0=A{k}_0^2$, both pockets have radius $k_0$. Let $\mu ={\mu }_0+\delta \mu$ and $y=k_F^2-k_0^2$. The characteristic equation at the Fermi level becomesFixed density cancels the first-order chemical-potential shift. Thus $\delta \mu =O({\alpha }^2)$ and $k_{F,\mp }=k_0\pm \alpha /(2k_0\sqrt{AB})+O({\alpha }^2)$: both pockets remain occupied at small coupling. This expansion assumes fixed $A,B>0$ before taking $\alpha \to 0$.