Energy Levels in a Crystalline Solid — Question 5

Question 5

A filled symmetric band has zero current even though individual states move

At zero temperature every k state in an isolated band is occupied by both spins. The band obeys $E(-k)=E(k)$ and $v(k)=\hbar^{-1}dE/dk$. For a ring of length L, define current $I=-(e/L)\sum_{k,s}v(k)$.

1. Show that the band current vanishes.

2. Remove one electron from a state of velocity $v_0$. Find the change in current and its hole interpretation.

Question 5 - Solution

A filled symmetric band has zero current even though individual states move

(a) Method and calculation

$$E(-k)=E(k)\Rightarrow\boxed{v(-k)=-v(k)}.$$$$\sum_{k,s}v(k)=0\Rightarrow\boxed{I_{\mathrm{full}}=0}.$$

(b) Interpretation and further calculation

$$\Delta I=-\left(-\frac{e}{L}v_0\right)=\boxed{+ev_0/L}.$$$$I_{\mathrm{hole}}=q_hv_h/L\Rightarrow\boxed{q_h=+e,\quad v_h=v_0}.$$

Physical check

The hole represents a missing contribution from an otherwise full band. This cancellation is a property of occupied band states, not a claim that every occupied electron is stationary.