Band hybridization

Band hybridization mixes two orbital-derived bands through an off-diagonal Hamiltonian matrix element. For a real coupling $\alpha$,An allowed bare crossing becomes an avoided crossing with direct separation $2|\alpha |$. Orbital character transfers continuously between branches through the crossing. A direct separation alone does not guarantee a global band gap; the extrema at different wavevectors must also be compared.

Mix a parabolic energy band ${ϵ}_k={\hslash }^2{k}^2/(2m)$ with a perfectly flat band at $E_d>0$ through momentum-independent $V$. At zero coupling suppose the two orbital sectors have equal populations, with the dispersive sector filled to $k_0=\sqrt{2m{E}_d}/\hslash$. At fixed total density and sufficiently low temperature, nonzero coupling fills only the monotonic lower hybridized branch, whose Fermi wavevector is $k_F=2^{1/3}{k}_0$, provided the available momentum-space states accommodate it. Its small slope at that radius gives the displayed ratio to $v_F^{(1)}=\hslash k_0/m$ for $|V|\ll E_d$. The result uses the singular perfectly flat limit and an empty upper band; it is not implied by Fermi-volume conservation in two hybridized electron bands for arbitrary pairs of dispersive bands. The zero-temperature limit must resolve the small hybridization energy scale.

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 $\alpha \to 0$.

A hybridization gap is a separation in electronic band structure caused by mixing orbital-derived states. For $E_a=A{k}^2$, $E_b=E_0>0$, $A>0$ and constant coupling $\alpha >0$, the lower eigenvalue is everywhere below $E_0$ and tends to it as $k\to \infty$. The upper eigenvalue has minimum $(E_0+\sqrt{E_0^2+4{\alpha }^2})/2$. The unbounded model therefore has an indirect gap of width $(\sqrt{E_0^2+4{\alpha }^2}-E_0)/2$, approximately ${\alpha }^2/E_0$ at weak coupling. This differs from the direct separation $2\alpha$ at the bare crossing. A finite Brillouin zone bounds the lower band farther below $E_0$.