1
$\begingroup$

As I understand, if the Chern number which is obtained by integrating Berry curvature over a surface with a boundary is an integer, then the Chern number is a topological invariant. So when does Chern number become integer? If the berry phase is well defined along the closed curve will that be enough for the system to be topological?

$\endgroup$
  • 4
    $\begingroup$ When you integrate the Berry curvature over the Brillouin zone (which is a torus topologically, so it is closed), you always get an integer. $\endgroup$ – Meng Cheng Sep 28 '15 at 15:13
1
$\begingroup$

Any system is topological. The question is, rather, whether it has non-trivial topology.

When considering translation invariant systems, we make a Bloch decomposition and so a system is specified by maps from the Brillouin zone into some space of Hamiltonians. Via the Bloch decomposition, this space of Hamiltonians will simply be the space of Hermitian $n\times n$ matrices which have a gap.

For simplicity let's assume we have no extra symmetries, and we are in two dimensions, and we are only considering two level systems. Then the Brillion zone is the $2$-torus, the space of Hamiltonians is $\mathbb{R}^3-{0}\cong S^2$ (think of the set of Hermitian invertible $2\times 2$ matrices, this space is spanned by the three Pauli matrices, with real coefficients, and such that not all three coefficients are zero), and a system is specified as a continuous map between these two spaces.

Then it turns out (after some study of topology) that these maps $$ \mathbb{T}^2\to S^2 $$ fall into classes indexed by $\mathbb{Z}$, where classes are defined by continuous deformations of the maps (i.e., homotopy).

For general $d$-dimensions and general $N$-level systems, the $2$-torus is replaced by a $d$-torus and the $2$-sphere is replaced by the Grassmannian manifold. Taking symmetries into account one have to then classify $G$-equivariant maps, rather than all continuous maps, which leads to a finer classification.

In this sense, a system (i.e. a map $\mathbb{T}^2\to S^2$) is topological when it is not a map homotopic to a constant map (the constant is irrelevant if your target space is path-connected).

Next, for non-translation invariant systems, one does not have a Brillouin zone anymore. So a Hamiltonian is just an operator in Hilbert space (with some conditions to make physical sense). One then classifies not a set of maps as before, but the set of possible Hamiltonians. The equivalence classes is now not homotopy as before, but path-connectedness.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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