I'm reading a paper (Rev. Mod. Phys. 84, 1419 (2012) - Maximally localized Wannier functions: Theory and applications p 1423) and I'm trying to understand a certain part. They show this figure:
Then they say:
[..] This band 4 becomes degenerate at $\Gamma$ with band 2 & 3 as show in Fig 3. As a result band 4 in non-analytic in k in the vicinity of $\Gamma$.
I couldn't find anything on why degenerate bands become non-analytic. Is it because of perturbation theory?
I will try to give a "physicist's proof" for the 1-dimensional case (the approach can be easily extended to higher dimensional cases).
Let's say there is a single degenerate point of order 2. Then we can define the 2 bands intersecting at this point in two different ways
If we start from a point on the RHS of the blue band in either cases, we cannot find the "value" of the band on the other side of the degenerate point using Taylor expansion, simply because it is not clear how we should extend the blue band beyond the point of intersection. Now you may say its clear which picture is the correct one, the one that looks "more smooth". Then you should read below.
When we talk about a band structure, we refer to the solutions of the Bloch equation \begin{equation}\label{bloch_eq} \frac{k^2}{2m}\psi_k +\sum_{G}U_G \,\psi_{k-G}=E_k \,\psi_k, \end{equation} where $k=\frac{2\pi}{Na}n$ is a wave vector in the first Brillouin zone (FBZ) while $G=\frac{2\pi}{a}m$ are "large" wave vectors. For every number $k$, the Bloch equation is a system linear system of (an infinite number of) equations to be solved to get the energy levels for that $k$ together with their eigenvectors.
Now, lets take two neighbouring wave vectors $k_n$ and $k_{n+1}$. For each one, we have a set of energy levels and eigenvectors and we want to assign them to different bands. The most straightforward way to do it is to take an arbitrary energy level for $k_n$ and then, find the closest energy level to it for $k_{n+1}$ in the thermodynamics limit where the difference between the two closest energy levels of these adjacent wave vectors goes to zero. We can apply the same rule to every pair of adjacent wave vectors to take the continuum limit where the notion of a band becomes meaningful.
But, the above prescription becomes useless when two or more of the energy levels at a wave vector are degenerate, simply because we do not have a unique way to "glue" these energy levels to the rest of the points in the BZ. Any ordering of any linearly independent superposition of degenerate levels can be assigned to the bands which are well defined in the rest (= non degenerate points) of the BZ. Therefore, the energy bands are not well-defined at degenerate points.
Another explanation which has a more mathematical spirit is the following: Let's assume that the band structure (by which we mean not just the energies, but also the eigenvectors) is analytic everywhere. This means that, I can Taylor expand the eigenvectors around every point in the BZ. To do this, I need to know all of the derivatives of the components of eigenvectors at every point. These derivatives can be obtained using their finite difference definitions and subsequently, taking the limit $dk \to 0$ which is equivalent to the thermodynamics regime $N \to \infty$. Again, we face a problem when we try to do this procedure at the points of degeneracy. Simply because we do not have a unique way to assign values to the eigenvectors at degenerate point.
