Periodic boundary condition and Bloch function

The first thing to be distinguished is that, the wavefunction $\psi(r)$ has a (much larger) period of $Na$ while the electron density $\rho(r)$ has a period of $a$, in which the $a$ is the lattice constant. Then we have two relations: \(\begin{align} \psi(r) &= \psi(r+Na), \\ \rho(r) &= \rho(r+a). \end{align}\) Considering the relation between $\psi(r)$ and $\rho(r)$: \(\begin{align} \rho(r) = \psi^*(r)\psi(r), \end{align}\) the wavefunction at $r$ and $r+a$ should only differ by a phase factor: \(\begin{align} \psi(r+a) = e^{i\theta}\psi(r). \end{align}\) Combine the Eq. (1) and Eq. (4), the phase factor $\theta$ should be: \(\begin{align} \theta = \frac{2\pi n}{N}, \end{align}\) in which the $n$ is an integer. Tyically, we define the wave vector $k=(n/N)[(2\pi)/a]$ and the phase factor $\theta$ becomes $\exp(ika)$. Without loss of generality, the wavefunction $\psi(r)$ can be written as: \(\begin{align} \psi(r) = e^{ikr}u(r), \end{align}\) in which the $u(r)$ is a periodic function with the period of $a$.

Note that the $k$ is can be defined with any integer number $n$, but there are only $N$ valid $k$. Normally, we choose $k$ to be in the range of $[-\pi/a, \pi/a]$, which is the so-call the first Brillouin zone.

There are two limited cases:

  1. $k=0$, giving that $\psi(r+a)=\psi(r)$. This means that the wavefurnctions at the neighboring atoms are in phase;
  2. $k=\pi/a$, giving that $\psi(r+a)=-\psi(r)$. This means that the wavefurnctions at the neighboring atoms are out of phase.

In general, when $k$ deviates from 0, nodes are introduced into the wavefunction, and the energy will rise.

Electronic conductivity

Without the perturbation of the external field, the energy levels of $k$ and $-k$ are degenerated. This means that two electrons with the same energy but opposite momentum can be found. As a result, there can be no net motion and no net current flowing.

If we want to have a net current, we need to break the degeneracy of the energy levels of $k$ and $-k$ by, for example, shifting all ground-state energy levels with $\delta k$. The current can be written as: \(\begin{align} i &= -ne\delta v=-ne\frac{\delta p}{m^*}=-ne\hbar\frac{\delta k}{m^*}\\ &= \frac{ne^2\tau}{m^*}\mathcal{E}=\sigma\mathcal{E}, \end{align}\) in which the $n$ is the electron density, the $e$ is the electron charge, the $\tau$ is the relaxation time, the $m^*$ is the effective mass, and the $\sigma$ is the electronic conductivity. The smaller the effective mass, the higher the conductivity. This picture can be intuitively understood by imagining the mass of a real particle. The smaller the mass, the easier the particle can be accelerated by an external force. It is also informative that he effective mass depends on the curvature of the E(k) plot. If the curvature is zero, the shift of $\delta k$ would cost nothing and the effective mass would be zero.