I'm just doing a small ugrad assignment where I have to present a talk on twisted bilayer graphene. I'm having trouble understanding the meaning of a flat band.

As far as I understand, a flat band means that the charge carriers have infinite mass. What I don't understand is how this supports superconductivity in twisted bilayer graphene. Also, what does the degeneracy of the bands imply?


1 Answer 1


A flat band does not mean infinite mass, but rather that energy is independent of momentum. It means that $\partial E / \partial k = 0$. If we insist on $E=\hbar^2 k^2 / 2m$ then this indeed is infinite mass. However a more convenient form is to look at a Dirac equation (for zero-mass particles) that implies $E=v_F p$ and then we get that $v_F$ is zero at the magic angle.

For superconductivity we need to pair electrons and couple them into a Cooper-pair. This is more easily done when they have the same energy, as can be seen when considering perturbation theory: pairing within a degenerate subspace is linear in the perturbation, and outside the degenerate subspace is quadratic. In case of flat bands all the band is degenerate, as all excitations have the same energy.

  • 1
    $\begingroup$ could you please provide a reference? $\endgroup$ Dec 24, 2019 at 8:08
  • 2
    $\begingroup$ @ArtemAlexandrov to which part? The relation between flat band and sc can be read here: arxiv.org/abs/1803.08799 or here: arxiv.org/abs/1404.5482 The general nature of the vanishing of Fermi velocity in twisted bilayer graphene near the magic angle has been extensively researched. Here is one of the original papers arxiv.org/abs/1009.4203 and here is a recent one arxiv.org/abs/1808.05250 $\endgroup$
    – user245141
    Dec 25, 2019 at 13:36
  • $\begingroup$ $v_F$ is the Fermi velocity, correct? And k is.. A wave vector? Or the wavenumber? M is mass (rest mass?) and p is... momentum? $\endgroup$
    – Kurt Hikes
    Mar 17, 2021 at 23:40

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.