Tuesday, March 20, 2018

graphene - Topological invariant in 1D


In 2D, with state $\psi(k_x, k_y)$, it is common to calculate measure of topology of material:



1 - Calculate Berry connection $a = -i <\psi | \partial_{\boldsymbol{k}} | \psi>$.


2 - Calculate Berry curvature $F = \partial_{k_x} a_y -\partial_{k_y} a_x $.


3 - Define topological invariant (Chern number) $C = \frac{1}{2 \pi} \int F(k_x, k_y) \mathrm{d}k_x \mathrm{d}k_y$.


How to calculate an analogous topological invariant for a 1D state $\psi(k_x)$?


Naively, the above recipe can never give a non-zero topological invariant since the Berry curvature vansihes. However, 1D topological insulators are well known to exist.



Answer



There are similar topological invariants for band structures in one dimension, but an important difference is that these invariants always require some symmetry in the band Hamiltonians, for example particle-hole symmetry. In such cases, typically the invariant is given by


$$ C=\int\frac{dk}{2\pi} a_k \text{ mod }1 $$


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