= Fermi-volume conservation in two hybridized electron bands
{c}
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 $k_0$,
$$
k_{F,-}^3+k_{F,+}^3=2k_0^3.
$$
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 $E_a=Ak^2$, $E_b=C+Bk^2$, $A>B>0$, and $C=(A-B)k_0^2$. At the uncoupled <chemical potential> $\mu_0=Ak_0^2$, 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> becomes
$$
(\delta\mu-Ay)(\delta\mu-By)=\alpha^2.
$$
Fixed 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 $\alpha\to0$.
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