Negative effective mass at top of valence band

Effective mass $m^*$ is the apparent mass of an electron in a crystal when it is acted on by an external force, including the effect of the lattice:

$$m^* = \frac{\hbar^2}{d^2E/dk^2}$$

It depends on the curvature of the E–k curve.

Negative effective mass at the top of the valence band

  E
  ^            Conduction band
  |   \                          /
  |    \          m* > 0        /
  |     `-.__              __.-'
  |           ''--....--''        bottom of CB
  |                               (concave up)
  |  - - - - - - Eg  - - - - - - - - -
  |           .--~~~~--.         top of VB
  |       _.-'  m* < 0  '-._     (concave down)
  |     /                    \
  |    /   Valence band       \
  +---------------|------------------> k
                  0

• At the top of the valence band the E–k curve is an inverted parabola: $E \approx E_v - \dfrac{\hbar^2k^2}{2|m^*|}$.
• So $\dfrac{d^2E}{dk^2} = -\dfrac{\hbar^2}{|m^*|} < 0$, and therefore $m^* = \dfrac{\hbar^2}{d^2E/dk^2}$ is negative.

Physical meaning

An electron near the top of the band has a wave vector close to the Bragg reflection condition. When an electric field pushes it, the lattice reflects it so strongly that the electron accelerates opposite to the direction expected for a normal negative particle.

An electron with negative charge ($-e$) and negative mass ($-|m^*|$) accelerates in the same direction as a particle with positive charge and positive mass:

$$a = \frac{(-e)E}{-|m^*|} = \frac{(+e)E}{+|m^*|}$$

So instead of tracking the many electrons with negative mass at the top of the valence band, we track the few empty states there and treat each one as a hole with charge $+e$ and positive effective mass $m_h^*$. This is why holes in p-type semiconductors move in the direction of the field and carry current like positive particles.