Band filling, metals and insulators

Impose Born-von Karman boundary conditions on a crystal of N primitive cells and exactly N distinct values of k fit into the first Brillouin zone — one per cell, whatever shape the dispersion takes. Spin doubles the capacity, so one band holds 2N electrons and z valence electrons per primitive cell fill z/2 bands. Sodium is body-centred cubic with one atom per primitive cell and one 3s electron, so z = 1: the band is half full and sodium is a metal before any energy is evaluated. Silicon has two atoms per primitive cell and four valence electrons each, so z = 8: four bands can fill exactly, and in silicon a gap does open above them. The commonest slip is counting per atom or per conventional cell: silicon's conventional cubic cell contains eight atoms and its primitive cell two, and only the primitive count decides anything.

The current from a band is j = −(e/V) times the sum of v(k) over its occupied states, with v = (1/ℏ) dE/dk. Time-reversal symmetry gives E(k) = E(−k), so v(−k) = −v(k): in a full band every state is paired with its opposite and the sum vanishes exactly. A field cannot repair that — short of one strong enough to drive electrons across the gap into the next band, which is Zener breakdown. The semiclassical equation ℏ dk/dt = −eℰ displaces every k by the same amount, and in a full band the displaced set of occupied states is the same set — a state pushed past the zone boundary re-enters at the opposite edge by Bragg reflection — so the occupation, and the current, are unchanged. A partly filled band behaves entirely differently: the whole Fermi sea slides by |Δk| = eℰτ/ℏ, which for copper at 10 mV m⁻¹ is 0.38 m⁻¹ against a Fermi wavevector of 1.36 × 10¹⁰ m⁻¹. Conduction is that one-part-in-3.6 × 10¹⁰ asymmetry.

An even z fills bands only if a gap survives in every direction of k. The energy at the zone boundary depends on direction, so in three dimensions the maximum of band n along one direction can lie above the minimum of band n+1 along another, and the two bands overlap in energy without ever touching at the same k. Calcium is face-centred cubic with one atom per primitive cell and z = 2, so the count alone predicts an insulator; its resistivity is 34 nΩ m, a better conductor than iron. The 4s band overlaps the 3d, electrons spill from one to the other, and the metal ends up with a hole Fermi surface in the lower band and an electron Fermi surface in the upper, carrying equal numbers of each. Squeeze the overlap towards zero and you reach a semimetal: bismuth carries about 3 × 10²³ carriers per cubic metre, a few millionths of copper's.

Summing 10²⁸ occupied velocities to get a current controlled by a handful of empty states is absurd, so subtract the identically zero full-band sum from it. What remains is j = +(e/V) times the sum of v(k) over the empty states: the same current, now written as a sum over the few vacancies, each behaving as a carrier of charge +e. Near a band maximum the curvature d²E/dk² is negative, so the electron effective mass there is negative and the hole mass m*ₕ = −m*ₑ comes out positive; a hole therefore accelerates along ℰ like an ordinary positive particle. Nothing physical has been added — this is one algebraic identity applied to one sum — but it is what explains the positive Hall coefficient measured in zinc, and in aluminium in strong magnetic fields, which no free-electron model can produce.

Empty states nearby make conduction possible, and σ = ne²τ/m* = neμ says how much of it you get. The point worth pausing on is that a perfect static periodic lattice contributes no resistance at all: Bloch states are stationary states of that lattice, an electron placed in one propagates forever, and σ would be infinite. Resistivity measures departures from periodicity — phonons, impurities, vacancies, surfaces — which combine by Matthiessen's rule as 1/τ = 1/τₚₕ + 1/τᵢₘₚ, giving ρ = ρ₀ + ρₚₕ(T). Copper's measured conductivity implies τ = 25 fs and a mean free path of 39 nm, about 110 lattice spacings between events. The mass is not the free-electron mass but the curvature of the band: 1/m* = (1/ℏ²) d²E/dk², so a tight-binding band E = E₀ − 2t cos ka with hopping energy t = 1.0 eV and a = 0.30 nm gives m* = 0.42 mₑ.

Insulator and semiconductor are not different structures; they are the same structure differing in one number. Diamond, silicon and germanium share the diamond lattice and z = 8, and their gaps are 5.47, 1.12 and 0.66 eV. At 300 K the factor exp(−E_g/2kT) that populates the conduction band is 1.1 × 10⁻⁴⁶, 3.9 × 10⁻¹⁰ and 2.9 × 10⁻⁶ respectively: forty orders of magnitude of behaviour from a factor of eight in the gap. Draw the line wherever you like, but know it is a line drawn on a continuum. The deeper limit is the independent-electron assumption. NiO leaves a partly filled 3d band, so band filling predicts a metal; NiO is an insulator with a gap near 4 eV, because the on-site Coulomb repulsion U exceeds the bandwidth W and the electrons localise. Mott insulators are where band filling stops being the right question to ask.