What about the degeneracy in the brillouin zone as the energy at $k$ and $-k$ are the same?

Question

When we add weak periodic potential $V=\sum_{G} V_G e^{iGx}$ to the free electron theory and use perturbation theory we get second order energy as \begin{equation} E^2=\frac{|<k|V|k'>|^2}{E_k-E_{-k}} = \frac{|V_G|^2}{E_k-E_k'} \end{equation}

When K'=K+G, $E_k=E_{k+G}$ and second order energy blows up.So, we use degenerate perturbation theory and we get gaps at these points which is equal to $2|V_G|$, But Why do we only consider $K$ and $K+G$, What about when $K'=-K$? We know that $E_K=E_{-K}$ due to parabolic nature of energy. So, How do we explain this degeneracy? Is there something fundamental I am missing?

Answer 1

The matrix element of this weak periodic potential has the property that: $$ <k'|V|k> = \frac{1}{L^3} \int d^3 \mathbf{r} e^{i (\mathbf{k}-\mathbf{k'})\cdot \mathbf{r}} V(\mathbf{r}) = V_{\mathbf{k'}-\mathbf{k}} $$

This is nonzero only when $\mathbf{k}-\mathbf{k'}$ is reiprocal lattice vector. That means that the first order of perturbation theory gives just a constant shift in energy $<\mathbf{k}|V|\mathbf{k}> = V_0$ which is irrelevant, while the second order perturbation theory gives nonzero corrections only when both of the following two conditions are fulfilled: \begin{eqnarray} E_0(\mathbf{k}) &=& E_0 (\mathbf{k'}) \\ \mathbf{k'} &=& \mathbf{k} + \mathbf{G} \end{eqnarray}

where $\mathbf{G}$ is reciprocal lattice vector. These two conditions are satisfied only near the Brillouin zone boundary.

In the case mentioned $\mathbf{k} = \mathbf{-k'}$ we have that $\mathbf{k} - \mathbf{k'} = 2\mathbf{k}$. That is not reciprocal lattice vector in general case, so $ V_{\mathbf{k'}-\mathbf{k}}$ can be zero.