Are band structures non-analytic only at degenerate points?

Question

The electronic properties of (crystalline) solids is typically described in terms of the electronic band structure, which reveals many properties of the electronic structure such as the band gap, electron- and hole effective masses etc. For some of these concepts to be well-defined, however, it is necessary that the band structure be analytic near the point of interest.

In the review paper Maximally localized Wannier functions: Theory and applications (2012) the authors seem to claim that band structures are always analytic except at points of degeneracy, i.e. when two (or more) bands cross. However, they do not provide a proof nor a reference supporting this statement. Searching the web, the only resource I could find was this question from this site, which unfortunately did not receive any answers.

Recently, I came across a paper titled Two-Layer High-Throughput: Effective Mass Calculations Including Warping (2022). The authors of that paper claim to have found computational evidence supporting the statement that band structures are only non-analytic at points of degeneracy. However, this still does not prove or even explain why we might expect such a result.

In posting this question I am hoping that someone can help provide either (ideally) a proof, or an intuitive / heuristic explanation to help me understand why electronic band structures are only non-analytic at points of degeneracy.

Edit: I found another paper titled Analytic properties of band functions (1978) which may be relevant to the question. Unfortunately, I find that paper quite hard to follow, and it is not clear to me whether they actually address this question.

Answer 1

Let us consider some general one-dimensional tight-binding model: \begin{equation} \hat{H} = \sum_{i,l} t_l (c_i^\dagger c_{i+l} + c_{i+l}^\dagger c_i). \end{equation} Its energy spectrum is given by \begin{equation} \epsilon(p) = \sum_l 2t_l \cos{pl}, \end{equation} which is nondegenerate in $k$ space but doesn't need to be analytic. With appropriate choice of $t_k$, the energy $\epsilon(p)$ can be any continuous periodic function which can have any (at least finite) set of nonanalyticity points in $k$ space.

However, I doubt if such tight-binding models are realistic. One reason for that is the slow decay of $t_l$ needed for nonanalyticity. In particular, $t_l$ should decay as $\sim 1/l^{s+1}$ if $\epsilon(p)$ doesn't have an $s$-th derivative. On the contrary, for the exponential decay of $t_l$, the energy is analytic (see this question). I guess that long-range couplings are rare in realistic tight-binding models, whereas the exponential decay is a reasonable assumption.

Answer 2

I have been thinking some more about this, and have come up with something that I do not consider a full answer to the question, but may be useful for further discussion.

I will assume that the system we are interested in is described by a finite-dimensional Hamiltonian, i.e. an $N\times N$ matrix, and that this Hamiltonian is a smooth function of $k$ in the sense that all the matrix elements $H_{ij}(k)$ are smooth functions of $k$, which I here assume to be 1-dimensional.

The eigenvalues of this Hamiltonian are the roots of the characteristic polynomial $P_k( \lambda) = \mathrm{det}(H(k) - \lambda I)$. Assume that for a given $k = k_0$, the eigenvalue $\lambda_0$ is doubly degenerate, i.e. $P_{k_0}(\lambda)$ has a double root at $\lambda = \lambda_0$.

$P_k(\lambda)$ is a polynomial in $\lambda$, so it is certainly analytic in $\lambda$. We can therefore Taylor-expand $P_{k}$ in $\lambda - \lambda_0$ to the second order, which gives $$P_{k}(\lambda) \approx a(k)(\lambda - \lambda_0)^2 + b(k) (\lambda - \lambda_0) + c(k),$$

valid when $\lambda$ is sufficiently close to $\lambda_0$. From the Taylor expansion we can estimate the eigenvalues (roots of P) as

\begin{equation}\tag{1}\label{eq:lambdapm} \lambda^\pm(k) = \frac{1}{2 a(k)} \left(-b(k) \pm \sqrt{b^2(k) - 4 a(k)c(k)}\right), \end{equation}

though this estimate will only be good when the predicted roots are sufficiently close to $\lambda_0$ as to make the Taylor expansion valid.

Now, near $k = k_0$ we know that $\lambda_0$ is a root, so we can expect eq. (\ref{eq:lambdapm}) to hold. We may then try to find the eigenvalues near $k_0$ by expanding (\ref{eq:lambdapm}) in $k-k_0$. However since $\lambda_0$ is a double root at $k=k_0$, both $P_{k_0}$ and its first derivative must vanish at this point, which means that $b(k_0) = c(k_0) = 0$. Therefore $\lambda^\pm$ is non-analytic at $k_0$, as the square-root cannot be expanded around $0$.

This seems to be to be a reasonably promising start, however the above leaves several things wanted. For instance it applies only to finite-dimensional Hamiltonians, a 1-dimensional $k$-space, and doubly degenerate roots (not higher order roots). In addition, it does not explain why the band structure should be analytic at non-degenerate points (although perhaps the argument can be extended to say something about this, using that near such a point we must have $b \neq 0$). In addition, the assumption that the Hamiltonian is a smooth function of $k$ is not justified, although to me it seems physically reasonable. Further discussion and answers would be most welcome.