Concept of effective mass in a crystal

The effective mass $m^*$ of an electron in a crystal is the mass it appears to have when it responds to an external force, after the effect of all the internal forces from the lattice ions and other electrons has been included in it. It lets us treat a crystal electron as if it were a free particle with mass $m^*$.

Why it is needed

Inside a crystal an electron feels the external force $F_{ext}$ (for example $-eE$ from an applied field) and also the internal periodic force $F_{int}$ from the lattice. Newton's law gives

$$F_{ext} + F_{int} = m_e a$$

$F_{int}$ is very complicated and changes from point to point. We therefore write

$$F_{ext} = m^* a$$

and put all the effect of the lattice into $m^*$.

Expression for $m^*$

The electron is a wave packet that moves with the group velocity

$$v_g = \frac{d\omega}{dk} = \frac{1}{\hbar}\frac{dE}{dk}$$

Its acceleration is

$$a = \frac{dv_g}{dt} = \frac{1}{\hbar}\frac{d^2E}{dk^2}\frac{dk}{dt}$$

In time $dt$ the external force does work $dE = F_{ext}v_g\,dt$, so $\frac{dE}{dk}dk = F_{ext}\frac{1}{\hbar}\frac{dE}{dk}dt$, giving

$$F_{ext} = \hbar\frac{dk}{dt}$$

Substituting $dk/dt = F_{ext}/\hbar$:

$$a = \frac{1}{\hbar^2}\frac{d^2E}{dk^2}F_{ext} \quad\Rightarrow\quad m^* = \frac{\hbar^2}{d^2E/dk^2}$$

Meaning

• $m^*$ depends on the curvature of the E–k curve. A sharply curved band gives a small $m^*$ (light, fast electron); a flat band gives a large $m^*$.
• For a free electron $E = \hbar^2k^2/2m_e$, so $d^2E/dk^2 = \hbar^2/m_e$ and $m^* = m_e$.
• Near the bottom of a band $m^*$ is positive; near the top it is negative (this is why holes are used); at the inflection point it is infinite.

Typical values

The small $m^*$ of electrons in GaAs is why GaAs devices are faster than Si devices.