Effective mass can be positive or negative

The effective mass $m^*$ is the mass an electron in a crystal appears to have when it is accelerated by an external force; it includes the effect of the periodic lattice potential. It is given by $m^* = \hbar^2/(d^2E/dk^2)$.

Derivation

An electron in a crystal is a wave packet. Its velocity is the group velocity:

$$v_g = \frac{d\omega}{dk} = \frac{1}{\hbar}\frac{dE}{dk} \qquad (E = \hbar\omega)$$

If an external force $F$ (e.g. $-eE$) acts for time $dt$, the work done raises the electron's energy:

$$dE = F\,v_g\,dt = F\frac{1}{\hbar}\frac{dE}{dk}dt$$

Since $dE = \frac{dE}{dk}dk$, this gives

$$F = \hbar\frac{dk}{dt}$$

The acceleration is

$$\begin{aligned} a &= \frac{dv_g}{dt} = \frac{1}{\hbar}\frac{d}{dt}\left(\frac{dE}{dk}\right) = \frac{1}{\hbar}\frac{d^2E}{dk^2}\frac{dk}{dt} \\ &= \frac{1}{\hbar^2}\frac{d^2E}{dk^2}\,F \end{aligned}$$

Comparing with $F = m^*a$:

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

Positive and negative effective mass

In a crystal the E–k curve of a band is not a simple parabola. In the tight-binding picture a band can be written as

$$E(k) = E_0 - 2\gamma\cos(ka), \qquad \gamma > 0$$

Then

$$\frac{d^2E}{dk^2} = 2\gamma a^2\cos(ka), \qquad m^* = \frac{\hbar^2}{2\gamma a^2\cos(ka)}$$

   E
   ^      top of band: curve bends down
   |  ___                         ___
   | /   \  m* < 0          m*<0 /   \
   |      \                     /
   |       \ <- inflection:    /
   |        \   m* = infinite /
   |         \               /
   |          \__         __/ <- bottom: m* > 0
   |             '--___--'
   +---|---------|---------|-------> k
     -pi/a       0       +pi/a

• Near the bottom of the band ($k \approx 0$): $\cos ka \approx 1$, the curve is concave up, $d^2E/dk^2 > 0$, so $m^*$ is positive. Here $E \approx E_{min} + \hbar^2k^2/2m^*$, like a free electron.
• Near the top of the band ($k \approx \pm\pi/a$): $\cos ka \approx -1$, the curve is concave down, $d^2E/dk^2 < 0$, so $m^*$ is negative. An applied force makes the electron accelerate opposite to the force, because it is being Bragg-reflected by the lattice.
• At the inflection points ($k = \pm\pi/2a$): $d^2E/dk^2 = 0$, so $m^* \to \infty$; the electron does not accelerate at all.

Significance

An electron with negative mass and negative charge near the top of the valence band behaves like a particle with positive mass and positive charge. This is the idea of the hole, used to describe conduction in the valence band of semiconductors.