Why does lattice potential scattering cause band gaps?

Question

In my condensed matter course my professor writes: "The lattice potential causes scattering that connects k states, and when these states are degenerate a gap opens. For example, the $\cos(2\pi/a)$ term connects the degenerate states at $k = −\pi/a$ and $+\pi/a$ (i.e. $⟨\psi_{k=-\pi/a}|\hat{V}|\psi_{k=+\pi/a}\rangle \neq 0$), and so a gap opens at these k values." Where by lattice potential he's referring to the 1-D nearly-free electron lattice potential described by a Fourier series:

$V(x) = V_0 + 2V_1cos(\frac{2\pi}{a}x) + 2V_2cos(\frac{4\pi}{a}x) + ...$

which perturbs the eigenenergies in accordance with non-degenerate perturbation theory.

However I don't understand why the scattering "connects" k states somehow and why this leads to a band gap. It was mentioned that the sign of the second order perturbation to the eigenenergies is the culprit for this effect but again, I don't understand why. Any help in understanding this would be greatly appreciated.

Answer 1

You need to look at the potential in momentum space. The Fourier transform gives: \begin{align} \hat V(k) &:= \int dx V(x) e^{-ikx} \\ &= 2\pi\left(V_0\delta(k)+V_1\left(\delta\left(k-\frac{2\pi}{a}\right)+\delta\left(k+\frac{2\pi}{a}\right)\right)+...\right) \end{align} By analogy with the discrete case, the $V_1$ Dirac deltas give "off diagonal" coefficients in Fourier space. In particular, starting from the usual Hamiltonian: $$ H_0 = \frac{k^2}{2m} $$ the states $|k=\pm\pi/a\rangle$ are degenerate eigenvectors with $H_0$, but get mixed due to $V$, lifting the degeneracy. This will generate an energy splitting.

The splitting leads to the creation of bands. However, there are two possible cases, and without further calculation, it's hard to determine which one is relevant. In one case, this energy splitting at the edge of the Brillouin zone corresponds to a separation of energy bands (lower level of the split connects with the energies of $|k|<\pi/a$). In another case case, it corresponds to an overlap of energy bands (higher level of the split connects with the energies of $|k|<\pi/a$). It turns out that the former is correct, and can be checked by doing a second order perturbation calculation in $V$. Check out for example D. Tong's Lecture notes for the full computation.

Hope this helps.