Band filling

Band filling counts occupied one-particle states in an energy band. With spin degeneracy two, a crystal of $N$ primitive unit cells has $2N$ states in each orbital-derived energy band, as in state counting in a Bloch band. Full occupation therefore means two electrons per primitive cell in that band.

In the group velocity approximation, a fully occupied band contributes current densityThe integral vanishes because $E$ is periodic across opposite faces of the Brillouin zone; inversion symmetry is unnecessary.^[ref] A partially occupied band leaves nearby empty states available for ordinary longitudinal conduction. A filled set of bands with a positive band gap instead gives a band insulator at zero temperature.

The vanishing group velocity term does not exclude every electrical response. A filled Chern insulator band can have a transverse quantum Hall effect governed by Berry curvature.^[ref]

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.

A band insulator has an integer number of completely filled energy bands, with the Fermi energy inside a positive band gap. At zero temperature, no arbitrarily low-energy charged excitation is then available.

In an independent-electron crystal with two spin states and no spin splitting, each completely filled energy band contains two electrons per primitive cell. An odd electron count per primitive cell therefore cannot fill an integer number of such bands. Enlarging the actual primitive cell, breaking the relevant degeneracy or including correlated-electron physics changes the premises. Even filling is necessary in this model but not sufficient: bands can overlap.

A Kondo insulator has a bulk insulating state associated with mixing between relatively localized correlated electronic states and itinerant conduction states. A renormalized two-band picture captures a narrow hybridization gap, although an uncorrelated orbital-mixing model does not explain the formation and temperature dependence of the coherent low-energy state. Samarium hexaboride is an example.

Consider a noninteracting band model in which all states of one energy band would be occupied and the next energy band empty. Define the global indirect band gapIf ${\Delta }_{\mathrm{ind}}>0$, this filling can have a positive bulk excitation gap. If ${\Delta }_{\mathrm{ind}}<0$, transferring an electron from the highest occupied states of the lower energy band to sufficiently low states of the upper energy band lowers the total energy. The nominal complete-band filling is not the equilibrium state: electron and electron hole populations can occupy the overlapping bands, precluding a gapped band insulator at that filling.

The equality case requires care. At ${\Delta }_{\mathrm{ind}}=0$, the excitation gap closes, but contact only at isolated wavevectors need not create finite-volume carrier pockets or a finite carrier density at zero temperature. It is therefore incorrect to infer strictly energy-lowering redistribution or partially occupied pockets merely from a band touching. Electron interactions and symmetry breaking can alter this single-particle description.

Suppose a model replaces a set of free-electron boundary energies $E^{(0)}$ by lower and upper edges $E^{(0)}-V$ and $E^{(0)}+V$, with $V>0$. If these are the relevant global band edges, the indirect band gap is $(E_{\min }^{(0)}+V)-(E_{\max }^{(0)}-V)$, giving the displayed expression. A positive direct gap $2V$ at each crossing need not be enough: a band insulator needs $2V>E_{\max }^{(0)}-E_{\min }^{(0)}$. This is a band-edge estimate, not a general three-dimensional nearly-free-electron theorem; multiple Bragg planes at a corner can require a larger coupling matrix.