One of the simplest models for electronic band structures is a tight binding model on a one dimensional chain with spacing $a$, one atom per cell and interactions only for nearest neighbours. This produces the following spectrum:

$ E(k) = -2t\cos(ka) $

which has period $2\pi/a$ and can be graphed in the Brillouin Zone $[-\pi/a,\pi/a)$.

If we consider instead two atoms per site, with two different hopping terms $t_1,t_2$, the spacing becomes $2a$ and we get two bands:

$E(k) = \pm\sqrt{t_1^2 + t_2^2 + 2t_1t_2\cos(2ka)}$

which lives on the Brillouin Zone $[-\pi/2a,\pi/2a)$, as the solution has in fact period $2\pi/2a$.

When $t_1=t_2$ we should recover the previous result, since the hamiltonian is the same. However if we were to look just at the spectrum, we would find that there are still two bands:

$E(k) = \pm t\sqrt{2+2\cos(2ka)} = \pm 2t|\cos(ka)|$

However this dispersion still has a "higher" periodicity in k-space, since it still has period $\pi/a$.

I understand that this can be obtained by "folding" the initial band around $\pm \pi/2a$, but how can we "unfold" it? Just by looking at the spectrum one might use the following two:

$ -2t\cos(ka) \qquad 2t\cos(ka) $

This can also be accomplished by translating appropriately part of the bands.

From the point of view of the spectrum $k$ is just a label and what matters is just what are the possible values and corresponding eigenvectors. Therefore I can imagine that both unfoldings are valid as long as there is a proper correspondence with the eigenvectors. But clearly the first case is "natural" in some sense, I just don't understand what.

This problem also comes up when looking at the nearly free electron model, considering a periodic perturbation on the kinetic energy. This makes sense in the extended zone scheme, but then we notice that the eigenvectors are the same by translation of momentum of $2\pi/a$, which means we can fold everything outside of the Brillouin Zone inside it (or repeat the dispersion and then take only everything inside the Brillouin Zone). Again, once we've folded the parabola, how can we unfold it? This time it's much more difficult since there infinite bands.

I guess the problem lies with the definition of the crystal momentum $k$, its relation with the band structure $E(k)$ and the eigenvectors $|\psi_k\rangle$. Within the tight binding model I think it's clear what $E(k)$ is, given the expression of $|\psi_k\rangle$ using the atomic orbitals, however when we start manipulating the spectrum alone I get confused and feel like I couldn't repeat the procedure if I already didn't know the answer, namely, how to recover the dispersion $E(k)$ when the space translation symmetry is increased from $2a$ to $a$ at $t_1=t_2$ in the tight binding model with two atoms per cell.


1 Answer 1


This is not a complete answer, but it was too long to fit into a comment.

I'm having troubles with the same question. I haven't found a good explanation about this anywhere yet.

So far what I managed to realize is that if we "fold" the band for this system $n$ times we get a set of eigenvalues $2 t \cos(ka + (2 m \pi / n))$, $m = 0, ..., n-1$, which is a generalization of the result you had above. This can be interpreted as n equivalent shifted bands, each centered around $2 m \pi / n$.

This leads me to think of the "folding/unfolding" of the bands as actually just shifts in momentum space, but I'm not sure if this interpretation is correct or if it breaks down at some point.

I got dragged into this topic by reading this article, where they some some visualization techniques for band unfolding. You could find it or its references useful somehow.

Hopefully I'll update this answer when I find a more satisfying discussion on the topic.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.