# Compute $Z_2$ Invariant of 2D Topological Insulators without Computing the Eigenstates

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For 2D Time-Reversal Invariant systems ($T H(\vec{k}) T^{-1} = H(-\vec{k})$), there is a formula by Fu-Kane-Mele in order to determine whether the system belongs to either one of distinct topological phases: simple insulator or topological insulator. The formula can be found in this first paper from 2005 (see also arXiv:cond-mat/0607699) for instance.

All the formulas I have found so far to compute this $\mathbb{Z}_2$ index require that one computes the eigenstates of $H(\vec{k})$ first. Then, for example, one has to count the zeros of a certain Pfaffian of some overlap matrix of the states with the time-reversed states, or compute certain integrals of the Berry-curvature and connection over part of the BZ. But still, one has to know the eigenstates in order to proceed.

However, in order to compute the Chern number, for example, one only has to know the projectors onto the occupied states.

So my question is: suppose that all I am given is the Hamiltonian ($H(\vec{k})$, in the form of some $N\times N$ Hermitian matrix with entries which are continuous functions over the BZ) so that there is no clear way to compute eigenfunctions, is there a direct formula that gives the $\mathbb{Z}_2$ invariant index from the entries of $H(\vec{k})$? If not, is there a formula that uses only the projector onto the occupied states, $P(\vec{k})$?

This post imported from StackExchange Physics at 2015-03-02 12:49 (UTC), posted by SE-user PPR
asked Feb 28, 2015
Perhaps you should change the title of this question, since it is pretty vague in the actual form. I think you always need to know the wave-function to compute the topological invariant, the reason is partially explained there: physics.stackexchange.com/a/70074/16689

This post imported from StackExchange Physics at 2015-03-02 12:49 (UTC), posted by SE-user FraSchelle
What is difficult about computing the eigenstates of $H(k)$? The Fu-Kane formula expressing the $Z_2$ invariant in terms of parity eigenvalues is quite easy to use, since you just need to diagonalize the Hamiltonian at the high symmetry points, instead of doing an integral over the whole Brillouin zone. Even the Hamiltonian can not be diagonalized analytically, you can always do it numerically.

This post imported from StackExchange Physics at 2015-03-02 12:49 (UTC), posted by SE-user Meng Cheng
@MengCheng, I believe that is misleading. Indeed the formula uses the overlap matrix only in the 4 TRIM symmetric points, but it relies crucially on the relative difference in sign between these points (the continuous selection of the branch of the square root), for which you need to know the Hamiltonian at least also on a continuous path between TRIMs. I am not so interested in numerical analysis, and in general I cannot diagonalize my Hamiltonian because I don't want to assume anything about its particular form besides its size and time-reversal symmetry.

This post imported from StackExchange Physics at 2015-03-02 12:49 (UTC), posted by SE-user PPR

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