 Solution

A band effective mass describes the acceleration of an electronic wavepacket in a periodic lattice through the local band curvature. It incorporates the periodic lattice potential into a useful transport parameter without changing the bare Electron mass. For a Bloch state in an isolated band,Differentiation givesIn one dimension this becomes $\boxed{m^{\ast }={\hslash }^2/(d^2ϵ/d{k}^2)}$. A locally parabolic isotropic band minimum has $ϵ\simeq {ϵ}_c+{\hslash }^2|\mathbf{k}-{\mathbf{k}}_c{|}^2/(2m^{\ast })$ with positive $m^{\ast }$. A flatter band has a larger effective mass. At a band maximum the Electron curvature is negative, so the Electron accelerates oppositely to the applied force. Describing the missing Electron as a positive-charge electron hole with positive mass is often simpler.

In the scalar relaxation-time approximation, a conduction Electron in an electric field reaches ${\mathbf{v}}_d=-e\tau \mathbf{E}/m^{\ast }$. ThusFor a crystal with different directional curvatures, conductivity is a tensor and involves the occupied states and their scattering. Contributions from Electrons and holes add: $\sigma =e(n{\mu }_e+p{\mu }_h)$ in an isotropic two-carrier description. A filled band contributes no ordinary longitudinal current because opposite group velocities cancel; a partially filled band or holes permit a current response.

Effective mass therefore connects measured electrical transport to electronic band structure, but carrier density and momentum relaxation time matter as much as curvature. Ionised impurities and phonons alter $\tau$. A smaller mass does not guarantee a higher conductivity if carrier population or scattering changes. Nor are all reported masses identical: density-of-states, cyclotron and conductivity masses weight anisotropic bands differently. The simple semiclassical formula assumes an isolated band, slowly varying weak fields and negligible interband transitions; near degeneracies or strong fields a single constant scalar mass is inadequate.