When reading this paper I came across the following claim:

From a band-theory point of view, the flat bands should have localized wavefunction profiles in real space

Is there a rigorous proof of this statement? So far I've only seen a hand-waving argument that goes like this: a flat band means the effective mass of the particle is infinite, i.e. it's very heavy and can't move - hence the localization in real space.

Presumably, this can be shown with a Fourier transform. I know that if we start with a localized wavefunction, i.e. something like $\Psi(x,t)=\delta(x-a)$ (where $a$ is the site) we arrive at the following momentum representation:

$$ \Phi\left(p,t\right)=\frac{1}{\sqrt{2\pi\hbar}}e^{-\frac{ipa}{\hbar}} $$

However I'm not sure how this representation leads to the dispersion relation of a flat band ($E(k)=\mathrm{const}$). Doesn't it depend on the Hamiltonian of the system?

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    $\begingroup$ If all the modes making up a band are degenerate then arbitrary rotations amongst the modes produce eigenmodes. The statement that the band is localised is surely not a statement about the eigenmodes (which you are free to choose), but simply that the group velocity is zero. $\endgroup$ Commented May 31, 2021 at 14:18
  • $\begingroup$ i did not read the paper but i guess they talk about the gapless edge states am I rigth? and if it is the case, the explanation is not that simple as flat band=localized state it is more sophisticated $\endgroup$
    – physshyp
    Commented Jun 2, 2021 at 0:07
  • $\begingroup$ As already commented below by @ComptonScattering, bands with non-trivial Chern number cannot have (exponentially) localized Wannier functions, so at least if the statement is read in this way, it seems incorrect. (See, e.g., arxiv.org/abs/1608.04696) Of course, the quote is out of context, so it is hard to tell what is meant. $\endgroup$ Commented Jun 6, 2021 at 12:00

2 Answers 2


The quoted statement is not precise. A better formulation might be

From a band-theory point of view, flat bands lead to dispersionless evolution of wavefunctions. In most cases, they admit localized wavefunction profiles in real space.

Flat bands are said to be dispersionless in the following sense: the group velocity for wave packets is $dE(k)/dk = 0$ and therefore each wave packet preserves its form under time evolution. If it is initially localized it stays localized forever. This can be easily shown using the Fourier transform:

$$ \psi(t,\vec x) = \int dk\, e^{-i\vec k\vec x} \tilde \psi(t, \vec k) = \int dk\, e^{-i\vec k\vec x} e^{-i E(\vec k) t} \tilde \psi(0, \vec k) $$ Now if $E(\vec k) = E_0$ then $$ \psi(t,\vec x) = e^{-i E_0 t} \int dk\, e^{-i\vec k\vec x} \tilde \psi(0, \vec k) = e^{-i E_0 t} \psi(0, \vec x) $$ which means that the form of $|\psi(t,\vec x)|$ stays constant in time. The flat band is a property of the Hamiltonian. But once the Hamiltonian admits a flat band then this flat band is always dispersionless in the same way, independently of further details of the Hamiltonian.

The question about the localization in real space goes further and is related to the fact whether well localized Wannier functions can be constructed for the given Hamiltonian. In most cases, Wannier functions are exponentially localized but the general criterion is not known. The papers arXiv:2104.01236 and arXiv:2010.06782, however, offer several examples of systems with flat bands and localization.

  • $\begingroup$ Could you comment on how non trivial topological indices effect your more refined claim? Specifically, consider a band with a non trivial Chern number. A band with a non trivial Chern number does not admit a localised basis. Moreover, the band may be flattened by adding localised terms to the Hamiltonian and without closing a band gap, and thus preserving the non trivial Chern number. Is this a counter example to your claim? $\endgroup$ Commented Jun 2, 2021 at 14:41
  • $\begingroup$ Can you show that the wavepacket stays localized forever if $dE(k)/dk=0$? $\endgroup$
    – grjj3
    Commented Jun 3, 2021 at 13:27
  • $\begingroup$ @grjj3 I have edited my answer and added the calculation you are asking for. $\endgroup$
    – Nikodem
    Commented Jun 3, 2021 at 15:20
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    $\begingroup$ The incompatibility of non trivial Chern numbers, and an exponentially localised Wannier basis is well known (see eg Bernevig and Hughes). The reason being that (by Paley Weiner theorem) an exponentially localised Wannier basis implies the existence of a smooth gauge in k-space, and hence therefore that the Chern number is zero. Thus a band with non zero Chern number is a counter example to your reformulated claim, that a flat band admits a basis of localised states, but is not a counter example to your calculation, which shows a different claim, that wavepackets do not evolve. $\endgroup$ Commented Jun 3, 2021 at 19:52
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    $\begingroup$ @Nikodem How does re-editing the statement without changing the proof make the latter more correct? Where are you using that the band has a trivial Chern number. -- Also, it seems that your answer has not much to do with the question: You argue that states stay dispersionless, which might be necessary, but logically is not sufficient, for localized eigenstates. $\endgroup$ Commented Jun 6, 2021 at 12:00

If we look at this from the point of view of the tight-binding approximation, the band width is proportional to the hopping integral, i.e., in one dimension $$ H=E_0\sum_n|n\rangle\langle n | - \Delta\sum_n\left(|n\rangle\langle n+1| + |n+1\rangle\langle n|\right),\\ E(k)=E_0-2\Delta\cos(ka) $$ If the band is flat, it means that $\Delta=0$, i.e., there is no hopping, that is the wave functions are localized.

Note that this argument can be trivially generalized to a lattice in any number of dimensions: without hopping all the sites have the same energy - thus, we can construct the Bloch states which have the same energy for any $\mathbf{k}$. In case we have atoms with different site energies in a unit cell, we well have several flat bands.

  • $\begingroup$ This is a single basis tight-binding picture. If you include multi-bases, the argument will be different. $\endgroup$
    – ytlu
    Commented Jun 3, 2021 at 18:44
  • $\begingroup$ @ytlu please expand your comment a bit - it is not clear what you mean. $\endgroup$
    – Roger V.
    Commented Jun 3, 2021 at 19:10
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    $\begingroup$ @RogerVadim Your answer is alright, of course. $\endgroup$
    – ytlu
    Commented Jun 6, 2021 at 16:52
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    $\begingroup$ This answer is pretty wrong. First, also Hamiltonians with non-zero hopping integral can have flat bands, see e.g. Lieb lattices. Second, the phenomenon of wavefunction localization is not due to vanishing of the hopping parameter $\Delta$ because this would be totally trivial. See my answer for an alternative attempt to address this question. $\endgroup$
    – Nikodem
    Commented Jun 7, 2021 at 12:56
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    $\begingroup$ @RogerVadim I understand. The two papers I have added to my answer address exactly this -- localization in sublattices. They give many examples. But I don't know any general theory. $\endgroup$
    – Nikodem
    Commented Jun 8, 2021 at 19:20

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