I want to get more intuition on topologically nontrivial band structures.

There's this popular 2D two-band model for a topological insulator where $H=\sum_{k}h(\boldsymbol{k})$ (see Qi, Hughes, and Zhang, PRB 78, 195424 (2008)),

\begin{eqnarray*} h(\boldsymbol{k})=\left(\begin{array}{cc} m+\mbox{cos}k_{x}+\mbox{cos}k_{y} & \mbox{sin}k_{x}-i\mbox{sin}k_{y}\\ \mbox{sin}k_{x}+i\mbox{sin}k_{y} & -m-\mbox{cos}k_{x}-\mbox{cos}k_{y} \end{array}\right) \end{eqnarray*}

Let me write its real space representation by Fourier transforming $h(\boldsymbol{k})$. This gives two sets of non-identical orbitals $a$, $b$ per site $n$ in a square lattice:

\begin{eqnarray*} H & = & \sum_{mn}t_{mn}^{(a)}a_{m}^{\dagger}a_{n}+t_{mn}^{(b)}b_{m}^{\dagger}b_{n}+t_{mn}^{a\leftarrow b}a_{m}^{\dagger}b_{n}+t_{mn}^{b\leftarrow a}b_{m}^{\dagger}a_{n}\\ & = & \sum_{n}\frac{1}{2}\Big[ma_{n}^{\dagger}a_{n}+a_{n}^{\dagger}a_{n+\hat{x}}+a_{n}^{\dagger}a_{n+\hat{y}}\\ & & -mb_{n}^{\dagger}b_{n}-b_{n}^{\dagger}b_{n+\hat{x}}-b_{n}^{\dagger}b_{n+\hat{y}}\\ & & -ia_{n+\hat{x}}^{\dagger}b_{n}+ia_{n-\hat{x}}^{\dagger}b_{n}\\ & & -a_{n+\hat{y}}^{\dagger}b_{n}+a_{n-\hat{y}}^{\dagger}b_{n}^{\dagger}\Big]+\mbox{h.c.} \end{eqnarray*}

A rudimentary code in Mathematica allows me to numerically calculate the Chern number and get what the authors say (that for $-2<m<2$, I get a topollogically nontrivial band structure, and a trivial otherwise).

However, I would like some further intuition on how this works. People normally talk about calculating a Berry-like phase from the hopping amplitudes $t_{mn}$ of the tight-binding model and seeing that for a closed circuit they get a phase that is not 0, $\pi$, or $-\pi$, indicating breaking of time-reversal symmetry.

I don't understand how that works in the context of this simple model:

Suppose I go through all the hopping in a square plaquette, completing the closed circuit $b_{(1,1)}\rightarrow a_{(1,2)}\rightarrow b_{(2,2)}\rightarrow a_{(2,1)}\rightarrow b_{(1,1)}$ where $n=(1,1)$ indicates the $x,y$ coordinates of the site in the lattice, then $t_{(1,1)(2,1)}^{b\leftarrow a}t_{(2,1)(2,2)}^{a\leftarrow b}t_{(2,2)(1,2)}^{b\leftarrow a}t_{(1,2)(1,1)}^{a\leftarrow b}=\left(\frac{1}{2}\right)^{4}(i)(+1)(-i)(-1)=-\frac{1}{16}$ which has an argument of $\pi$. What's going on? Do I also need to consider the loops of the hopping amplitudes between identical sites? If so, why?

Any intuitions will be appreciated. (I'm not a string theorist, so the least abstract, the simplest qualitative explanation, the better).

Thanks a lot.


1 Answer 1


Just to give a rough intuitive explanation as to why we want to look at loops. Time-reversal symmetry is roughly complex conjugation, so we want to know if this matrix elements are non-real. However, just by redefining the electron creation operators $a_R \rightarrow a_Rexp(i\theta)$ I can change the phases of the hopping matrix elements, which is just gauge symmetry. If instead I at product loops of hopping elements these are gauge invariants. These are physical - they are the Ahranov-Bohm phases and basically measure the amount of "magnetic field" which pierces the loop (which may not be a real magnetic field, but to the electrons the effect is indistinguishable)/.

In your case I think you just picked the wrong loop. If you pick the circuit $b_{(1,1)}\rightarrow b_{(1,2)} \rightarrow a_{(2,2)} \rightarrow a_{(2,1)}$ I believe you get a factor of $i$ which is what you expect from a material "apparently" breaking time-reversal symmetry. Again since it is not really a magnetic field the material in toto does not have to actually break time reversal - the phases from different bands can cancel. But if we use symmetry to prevent different bands from interacting than each bands thinks its in a magnetic field and we get fun effects.


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