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  A photonic topological insulator and homotopy classification of a class of network operators

+ 5 like - 0 dislike

The full account of this question is going to be long, so I'll first extract the mathematical problem in a self contained way and then give the backgrounds afterwards. 

The math problem

I have the following maps from the 2-torus \(T^2\) to a subset of 4 by 4 unitary matrices \(U(4)\), explicitly given as


where \(k_x, k_y\in[0,2\pi)\), and \(\begin{bmatrix}t_{11}&t_{12}\\t_{21}&t_{22}\end{bmatrix}\)is any 2 by 2 unitary matrix, that is, the maps are parametrized by these four numbers. I have some good reasons(see background below) to speculate that these maps can be associated with some topological integer(s) which distinguish these maps into homtopy classes, here homotopy refers to the ones that don't close any band gap.

What methods/directions should I pursue to find out the topological invariant?

I'm aware of the classification algorithm when the range of the maps is the whole \(U(4)\), but it does not help since it gives 0 for all the maps in form of (1), which isn't really surprising since this submanifold may have a more complicated structure that the whole \(U(4)\), especially when requiring the band gap must not be closed. In fact, as an example, it can be shown that for maps of the form of (1), any homotopy that connects the two \(U(k_x,k_y)\) parametrized by \(\begin{bmatrix}t_{11}&t_{12}\\t_{21}&t_{22}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)and \(\begin{bmatrix}t_{11}&t_{12}\\t_{21}&t_{22}\end{bmatrix}=\begin{bmatrix}0&1\\1&0\end{bmatrix}\)will have to close some band gap.


I'm studying a lattice network system made of optical ring resonators, the network aspect is explained in this paper by my colleagues, and it is found numerically that there exist phases with roubust edge states which wind nontrivially across the spectrum(notice in such systems the spectrum lives in \(S^1\), since they are eigenvalues of unitary matrices. This is an fundamental difference with ordinary static condensed matter system), and these winding states disappear when we consider the bulk operator of the same system, which is just the unitary operator presented in (1). This is consistent with the fact that the aforementioned "whole U(4) classification algorithm" gives null result, since a non-zero result would indicate a winding exists even in bulk spectrum.

In addition, Chern number cannot distinguish the trivial insulator and topological insulator phase.(As a digression, interestingly, on a honeycomb lattice, trivial phase, non-zero Chern number phase, and nonzero winding number phase can all be realized,see this.)

By the conventional wisdom, whenever there are robust edge states, there should be some characterizing bulk topological invariant. I'm trying to work out a bulk-edge correspondence, lacking a bulk invariant as the guidance, I've tried to use some bottom-to-top approach. that is, start with edge states counting and try hard to get a bulk invariant. I've tried and failed to mimic the mathematics used in the classic paper by Hatsugai, and a recent paper by Rudner,Levin et al., both of which have the advantage of already knowing a bulk invariant as the guidance(Of course there's always the possiblity that I'm not enlightened enough to get the gist). So I really would like to attack the problem in a top-to-bottom approach, that is, first know a bulk invariant and then show it is related to number of edge states.

Some probably not-so-important observations

I feel the difficulty here is very much like classifying systems without any symmetry(like quantum hall insulator) is easier than the ones with symmetries(like time reversal invariant \(Z_2\) topological insulator). The block structure of matrix (1) can be written as a anticommuator equation


where \(\gamma_5\) is the fifth gamma matrix in Weyl basis, 


This reminds me a lot of the chiral symmetry of massless Dirac operator. This could be completely superficial, but who knows...

asked Apr 8, 2014 in Theoretical Physics by Jia Yiyang (2,640 points) [ revision history ]
edited Apr 10, 2014 by Jia Yiyang
Most voted comments show all comments

The space of these matrices looks to me like the orbit of a certain permutation under the action of U(2) on U(4) that puts the U(2) matrix as the 2 blocks of a U(4) matrix and then right multiplies. If we can compute the stabilizer of the permutation, then the orbit is the same space as U(4) mod the stabilizer. Perhaps then we can apply Neil Strickland's arguments. I have to think about it more. Nice question though!

EDIT: Are we only allowed to vary the defining U(2) matrix?

Yes, but since square lattice is probably the simplest one, it should be helpful to work this out first. 

Could you please specify in mathematical terms (rather than physical ones) what it means in this comment to have non-zero winding edge states when $\theta>\pi/4$?

@ArnoldNeumaier, so when $\theta>\frac{\pi}{4}$, if we truncate the infinite lattice, say from $\mathbb{Z}\times \mathbb{Z}$ to $\mathbb{Z}\times \mathbb{H}$, where $\mathbb{H}$ is the half space, then on this lattice with edge(and with largely arbitrary boundary conditions), there will appear modes that are exponentially localized on the edge, this is called the edge states. If we look at the bands $E(k_x)$(note we only have periodicity in one direction now, hence the Bloch momentum only has one component $k_x$), in addition to the bulk bands(i.e. the bands that would be already present if the lattice were $\mathbb{Z}\times \mathbb{Z}$ ), there will appear lines traversing the band gaps, precisely representing the edge states. The above are generally true for many lattice systems, but in our case what is special is that, the domain of the spectrum is a circle $S^1$ since the matrices are unitary, hence the traversing lines may(or may not) wind around the whole $S^1$, this is why I call these edge states "winding edge states". This is impossible for ordinary time-independent electronic condensed matter system, where the domain of the spectrum is $\mathbb{R}$.  

Thanks for the explanation. I have no intuition for how to solve your problem, but it looks interesting.

Most recent comments show all comments

So there are only two phases? It seems your problem is done!

Well, the square lattice only has two phases, and yes I can just label them 0 and 1 if I want, but this is completely ad hoc and won't teach me anything. Once I start dealing with other lattices, that means I still have to do it numerically on a edged system to check which phase is topologically nontrivial(honeycomb lattice has three phases, c.f. my main post, and many many phase boundaries), this just isn't bulk-edge correspondence. I need to learn what is really going on.

1 Answer

+ 2 like - 0 dislike

I'd like to give a partial answer (or a partial failure, depending on perspective:)) to my own question. The strategy used for the homotopy classification is derived from the classic paper by Avron, Seiler and Simon, discussing the homotopy perspective of IQHE, and the method is very succinctly described in the lecture notes by Moore(see discussions around equation (39)). Let me summarize it here, following the lines by Moore:

Mathematically, to classify n-band IQHE systems, we are homotopically classifying (gapped) maps from $T^2$ to $H_n$, where $H_n$ is the space of $n\times n$ Hermitian matrices(i.e. Hamiltonians). We can diagonalize the Hamiltonian $H$ using unitary matrices, i.e. $H=UDU^\dagger$. Since the spectrum of the Hamiltonian is gapped, we can homotopically flatten and shift the bands within $D$, keeping $U$ fixed, without band crossing, to a standard matrix, say diag$[1,2,\ldots, n]$. Hence, all the information about homotopy class should be encoded in the unitary matrices $U$. However, we can always right-multiply $U$ by any diagonal unitary matrix without changing $H$, hence such equivalence must be quotiented(not proper English, I know) away. In the end, we need to homotopically classify maps $T^2\to U(n)/\sim$. Such maps are precisely classified by n-1 Chern numbers.   

Now come to my own problem, I slightly enlarge the space of maps to make standard techniques applicable, namely, instead of the specific maps in the form of (1) in the main post, I consider the maps of the form 

$$U(k)=\begin{bmatrix} 0&U_1(k)\\ \\U_2(k)&0\end{bmatrix}\cdots\cdots (2),$$

with the only requirement that $U(k)$ is unitary. Now as a spoiler to the readers not patient enough for the would-be longwinded discussion, there exist nontrivial homotopy classification for such maps, and the classification is strictly finer than a Chern-number classification, however the classification is not good enough to distinguish the topological phase established numerically in my collegues' work. 

Using the observation made in the main post that $\gamma_5 U=-U\gamma_5$, it is easily derived that $U$ can be diagonalized as  

$$U=\begin{bmatrix}a_1&b_1&a_1&b_1\\a_2&b_2&a_2&b_2\\a_3&b_3&-a_3&-b_3\\a_4&b_4&-a_4&-b_4 \end{bmatrix}\begin{bmatrix}d_1& & & \\ &d_2& & \\ & & -d_1 & \\ & & & -d_2 \end{bmatrix}\begin{bmatrix}a_1&b_1&a_1&b_1\\a_2&b_2&a_2&b_2\\a_3&b_3&-a_3&-b_3\\a_4&b_4&-a_4&-b_4 \end{bmatrix}^{\dagger},$$

By the same reasoning as outlined in Moore's notes, the diagonal matrix diag$[d_1,d_2,-d_1,-d_2]$ can be deformed to a standard one, say diag$[1,i,-1,-i]$. Now we only need to worry about the similarity transformations. Note our similarity transformations have the form 



$$A=\begin{bmatrix}a_1&b_1\\a_2&b_2\end{bmatrix}, \ B=\begin{bmatrix}a_3&b_3\\a_4&b_4\end{bmatrix}.$$

Hence A and B are independent matrices, satisfying $AA^\dagger=BB^\dagger=\frac{1}{2}I$ because of the unitarity of the similarity transformation. Also, the right-multiplications on (3) allowed are of the form

$$\begin{bmatrix}\Lambda& \\ &\Lambda\end{bmatrix},$$

to maintain the form (2), where $\Lambda$ is any 2 by 2 unitary diagonal matrix.

Therefore, we need to deal with maps $T^2\to M$, where $M$ is the quotient space of the pair of $2\times 2$ matrices $(A,B)$, with the equivalence relation $(A,B)\sim(A\Lambda,B\Lambda)$, that is, $M=(U(2)\times U(2))/\sim$. In fact, we can perform a diffeomorphism before doing quotient: $(A,B)\mapsto (A,BA^{-1})$, then the equivalence relation is simplified to $(A,BA^{-1})\sim (A\Lambda,BA^{-1})$, and so $M\approx (U(2)/\sim)\times U(2)$. (I learnt this trick from this mathoverflow post

In the end, we have the homotopy classes $[T^2, (U(2)/\sim)\times U(2)]=[T^2, U(2)/\sim]\times [T^2, U(2)]$, the former is characterized by a Chern number as usual and the latter is characterized by two winding numbers. Therefore, in this setup, we have a $\mathbb{Z}^3$ classification scheme, and it is easy to prove such classification is finer than the Chern number classification implemented on (2). The honeycomb lattice has more or less the same mathematical structure, and it has a $\mathbb{Z}^5$(1 Chern number and two pairs of winding numbers) classification since the size of the matrix is $6\times 6$. 

As I said, this scheme is not able to distinguish some topological phase in my college's work(the AFI phase mentioned in their paper). It is not very hard to understand the failure: In my collegues' work, the ring resonator system posses edge states which are robust against the missing of rings, whereas the constraint I imposed by $\gamma_5$ (which in position space represent a $\pi$ phase delay on 2 of the 4 amplitudes in a ring) is not, hence I'm probably barking at the wrong "symmetry" constraint. It seems the symmetry constraint of ring resonator model is less straightfoward to find than that of electronic condensed matter system, since it is from the beginning constructed on a very phenomenlogical level, while in electronic condensed matter system one can think of the fundamental symmetries that we are used to, like time reversal etc.

answered Aug 28, 2014 by Jia Yiyang (2,640 points) [ revision history ]
edited Aug 29, 2014 by Jia Yiyang

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