The periodic table of topological insulators and superconductors suggests that there can be topological non-trivial phases in zero dimension in non-interacting system with certain symmetries. A 0D system can be thought as a single atom, a quantum dot, or any system with discrete energy levels (no bands, no Brillouin zone).

Are there physical 0D systems which are topologically non trivial, at least theoretically? How one defines in this case the topological invariant, and what is its physical meaning?

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Due to Bott periodicity, dimension $d=0$ has the same symmetry classification as $d=8$.

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    $\begingroup$ A 0D system can be seen as a single atom. It can have time-reversal / particle-hole symmetry, and so it can be topologically non trivial. There is neither position nor momentum dependency in such a system by definition, and so it's difficult to have non-trivial properties associated to transport. $\endgroup$
    – FraSchelle
    Commented Jun 23, 2015 at 14:02
  • $\begingroup$ For example, a quantum dot can be considered as 0D so you can apply the classification. But the physical interpretation of the invariants are mostly really obvious: for example, for class A (i.e. IQHE in 2D), the 0D $\mathbb{Z}$ is just the $\mathrm{U}(1)$ charge of the system. I think the same to class AII (i.e. TI in two and three dimensions). For class D, the $\mathbb{Z}_2$ is the fermion parity. I do not immediately know the interpretation of $\mathbb{Z}$ invariant for BDI class. $\endgroup$
    – Meng Cheng
    Commented Jun 23, 2015 at 14:58
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    $\begingroup$ Be careful applying Bott periodicity here. Topological systems like this generally have time-evolution operators, NLSM fermions, and so forth that lie on a manifold dependent on the number of sites, and Bott periodicity works because the dimension of this manifold goes to infinity in the thermodynamic limit of an infinite number of sites. A 0D system probably only has one site, so unless your model squeezes a lot onto that dot, Bott periodicity isn't applicable. $\endgroup$
    – calavicci
    Commented Jul 22, 2016 at 23:21
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    $\begingroup$ Additionally, Bott periodicity is in the dimension of the sphere mapped into the target space ("the dimension of the homotopy group," in a certain but not uncommon abuse of language), not in the dimension of the physical system. $\endgroup$
    – calavicci
    Commented Jul 22, 2016 at 23:35
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    $\begingroup$ The size of the Hamiltonian has nothing to do with the dimensions of course. But the dimension in the periodic table does not refer strictly to the physical space dimension. Imagine a Hamiltonian of a one dimensional system and parametrized as a function of a continuous and periodic variable $\theta$. This Hamiltonian $H(k,\theta)$ is clearly two-dimensional. Take for example the 4D quantum Hall effect, which is defined in two spatial dimensions plus two "synthetic" dimensions. $\endgroup$
    – sintetico
    Commented Aug 1, 2016 at 18:56

1 Answer 1


There seems to be a physical realization of a quantum dot that can be in two insulating phases. Somewhat arbitrarily we can call one phase ordinary and the other topological. The real point is that one cannot deform one phase to the other without closing the gap. My reading of the following papers (I am not a physicist) tells me that what happens in practice is one sees a superconducting phase between the two phases of the quantum dot.

Szombati, D. B., et al.``Josephson ϕ0-junction in nanowire quantum dots'' Nature Physics 12.6 (2016): 568.

Marra, Pasquale, Roberta Citro, and Alessandro Braggio. "Signatures of topological phase transitions in Josephson current-phase discontinuities." Physical Review B 93.22 (2016): 220507.

The reason I say it is somewhat arbitrary how one assigns one of the phases the label topological is that there are oddities in defining the $K_2$ group of $C^*$-algebras. These go back to the arbitrary choice one makes in defining the Pfaffian of a skew-symmetric matrix.

There is no boundary here. What one is seeing is the same basic phenomenon as when one perturbs a Chern insulator into an ordinary insulator. One gets something like metallic behavior in the bulk.

So my answer is: yes.

  • $\begingroup$ Thank you for your very useful answer! Can you explain a little more why the choice is arbitrary on a mathematical level? For what I can understand from the papers above, the difference between topological and nontopological phase is justified by the fact that the Pfaffian definition can be continuously extended from the 0D to the 1D case (the Kitaev model). $\endgroup$
    – sintetico
    Commented Oct 29, 2018 at 3:58
  • $\begingroup$ I defer to those papers I listed in my answer. I had not read them in a while. If the limit of the 1D case tells us which phase is topological, great. I was thinking of Kitaev's paper ``Periodic table for topological insulators and superconductors.'' His definition of the trivial 0D insulator in class D seems, to me, a bit arbitrary. Contrast that the case in higher dimensions. There we generally identify the trivial gapped insulators on a system with periodic boundary conditions as the ones that remain gapped when the Hamiltonian is modified to introduce open boundary conditions. $\endgroup$ Commented Oct 29, 2018 at 5:17
  • $\begingroup$ As to the mathematics, one issue is that it is hard to distinguish the Pfaffian with the negative of the Pfaffian. For the determinant, we are happy to see the identity matrix has positive determinant. For the Pfaffian, we are only looking at skew symmetric matrices. Why does [0 1; -1 0] get a positive Pfaffian while [0 -1; 1 0] is negative? Neither is really a simpler matrix. $\endgroup$ Commented Oct 29, 2018 at 5:21

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