(available at http:tiny.cc/solidstatephys/Effective_mass.ipynb)
a chain of atoms with one orbital per atom has a tight-binding Hamiltonian $$H = h_0 + h_1 e^{ika} + h_1 e^{-ika}$$ Test question: how does this Hamiltonian look if we choose a unit cell with two atoms in it?
Reminder: I use $\hbar = 1$, so that $p = k$.
Electron velocity is $$ v = \frac{d E(k)}{d k} $$ several ways to understand:
For the simplest example $$ v = 2h_1a\sin(ka) $$
Conservation of energy after moving by $\delta x$ after $\delta t$: $$ \delta E(k) = -e E \delta x $$
$$ \frac{d E(k)}{d k}\delta k = -e E \delta x $$$$ v\frac{\delta k}{\delta t} = -e E \frac{\delta x}{\delta t} $$$$ \frac{d k}{d t} = -eE = F$$Once again, this is just Hamiltonian mechanics!
Let's solve equations of motion: $$k = F t$$ $$v = 2 h_1 a \sin(Fta)$$ $$x = 2 \frac{h_1}{F}\cos(Fta)$$
$\Rightarrow$ motion of electrons is periodic!
Very hard to observe, requires the relaxation time $\tau \gg (Fa)^{-1}$.
using $v = d E(k)/d k$ we get
$$ a = \frac{d^2 E}{dk^2} \frac{dk}{dt} = \frac{d^2 E}{dk^2} F $$Now compare with the Newton's law $a = F/m$.
$\Rightarrow$ we introduce effective mass $$m_{\text{eff}} = \left(\frac{d^2 E}{dk^2}\right)^{-1}$$ Electrons behave like their mass is momentum-dependent.
For the single band model: $$ m_{\text{eff}} = [2 h_1 a^2 \cos(ka)]^{-1}$$