 Solution

For photon absorption or emission, electron energy changes by $\pm \hslash \omega$ and crystal wavevector by $\pm \mathbf{q}$ modulo a reciprocal lattice vector. Optical photon momentum is normally tiny on the Brillouin zone scale, so interband optical transitions are approximately vertical on an energy-wavevector diagram. A direct-gap semiconductor can therefore absorb a photon between valence and conduction states near the same wavevector; an indirect transition usually needs an additional phonon or another source of momentum. Occupation and optical matrix elements impose further restrictions.

Within one ideal parabolic band of a clean metal, such a small photon momentum cannot normally supply the required finite energy difference. Intraband optical absorption then relies on momentum relaxation by impurities, phonons or boundaries, as in the finite-scattering-time Drude response. Interband transitions can still occur. The first panel shows the near-vertical direct transition.