I am a little confused! Some people are arguing that the gapless edge state of Topological insulator is robust as long as the time reversal symmetry is not broken,while other people say that it is not stable for lack of topological order。Please help me out!
[Physics] Is edge state of topological insulator really robust?
condensed-matterquantum mechanicstopological-insulators
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Topological insulator, by definition, cannot exist in magnetic field. This is because the topological insulator is NOT topological. A topological insulator is a material with time reversal symmetry and particle number conservation. Without time-reversal symmetry, topological insulators cannot exist, since they become the same as trivial band insulators. So a magnetic field destroys the topological insulator. True topological phases (ie phases with non-trivial topologically orders) are robust against any perturbations, including magnetic field.
Your following statement is almost accurate:
“I am aware of the argument saying time reversal invariant perturbation can only couple edge states pairwise, …”
The above statement would be more accurate if you replace “couple” by “annihilate.” The use of the former versus the latter has a physically distinct interpretation. If I start with a diagonal Hamiltonian (say) $H_{0}$, expressed as a matrix in the basis of an arbitrary number of (even) gapless edge modes, then the electrons in these modes do not scatter into one another if the matrix is diagonal. In other words, the modes are decoupled. Using your example of three pairs of edge states, and assuming a Dirac-like dispersion for them, we can write their energy-momentum dispersion as $$E_{n,\pm}(k)=\pm\hbar v_{F,n}|k| \; ,$$ where $n=1,2,3$ labels the pairs and $\pm$ identifies the Kramers partner. Assuming $v_{F,n}= v_{F}$ for all $n$, the Hamiltonian can be expressed in the matrix form (in the basis you defined above) as $$H_{0}(k) = \hbar v_{F}|k|\left(\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & -1 \end{array}\right)$$
Now, if we introduce a perturbation $V$ (say from your example), and use a traffic metaphor, then electrons in the same “lane” (color-coded as red or blue) can travel in either direction. As you correctly pointed out, electrons in the $| 1, 1 \rangle$ and $| 2, 1 \rangle$ (red) lanes can take a “U-turn” by switching to the $| 3, 2 \rangle$ lane. In other words, non-magnetic impurities can backscatter right-moving electrons and vice-versa. An equivalent argument can be made for the blue lanes. These are, by construction, time-reversal symmetric processes. Here’s a quick sanity check: with no magnetic impurities (and assuming $s_{z}$ conservation) there will be no spin flips. Hence decoupling the red and blue lanes implicitly assumes time reversal symmetry. The full Hamiltonian now becomes $$ H_{1}(k) = H_{0}(k) + V \; .$$
The reconstructed edge dispersion can be obtained by diagonalizing the above Hamiltonian independently for each $k$. For analytic simplicity let’s focus on $k=0$. Besides, we are interested in the gaplessness of the edge states anyway. Note: $$\begin{align} H_{1}(0) &= H_{0}(0) + V \\ &= V \end{align} \; .$$ Diagonalizing $V$ in a straightforward manner gives doubly degenerate (i.e. total six) eigenvalues $\varepsilon = 0, \; \sqrt{2}a, \; - \sqrt{2}a$. Note that two pairs of edge states have acquired a $2\sqrt{2}a$ band gap. This can be viewed as a momentum space annihilation of a pair of edge states.
From the above example, it would seem as if only two pairs of states absorbed the scattering while leaving the third one intact. From the explicit expression for $V$, however, it is clear that scattering is occurring in all three pairs. That’s because we’re looking at the $V$ matrix in the original basis; i.e. before the perturbation was introduced. As you may already know, in band theory it is customary to label bands in a basis diagonal in $k$-space. Hence, in the new basis, scattering occurs between only two Kramers pair (gapped) bands while leaving the third (gapless) one intact. Another way to see it is this: we can view (contrary to band theory custom) the edge state bandstructure in the new basis before the perturbation is introduced. In other words, we can redefine the edge states as linear combinations of the old basis using the eigenvector components from the $V$ diagonalization. Furthermore, linear combination of a gapless basis will again be gapless (now with different $ v_{F,n}$’s). In this basis scattering will only occur between two pairs of edge states (for $|E| < |\sqrt{2}a|$) upon slowly turning on $V$.
In the Hasan-Kane paper, the authors are discussing a general theory of band topological insulators. Hence Fig. 3 could potentially represent a slice (in momentum space) of the bands $E(k_{x},k_{y},k_{z})$ along a line connecting two Time-Reversal Invariant Momentum (TRIM) points $\Gamma_{a}$ and $\Gamma_{b}$. For the case of the quantum spin Hall effect, $\Gamma_{a} = 0$ and $\Gamma_{b} = \pi$. Taking a mirror image with respect to $\Gamma_{a}=0$ and plotting in the domain $k\in[-\pi,\pi)$ you can see the Dirac cones at $k=0$ and $k=\pi$ clearly. For the Bernevig-Hughes-Zhang (BHZ) model, the 1D Dirac cone appears at $\Gamma_{a}=0$ for $0 < M/B < 4$ and at $\Gamma_{b}=\pi$ for $4 < M/B < 8$ but not both simultaneously. Please see details in the reference:
Shijun Mao, Yoshio Kuramoto, Ken-Ichiro Imura, and Ai Yamakage. “Analytic theory of edge modes in topological insulators.” Journal of the Physical Society of Japan 79, no. 12 (2010). (arXiv)
Note: they use $\Delta/B$ instead of $M/B$. To get multiple Dirac points, as in Fig. 3, we need a mathematically more complex model than the BHZ.
Best Answer
I see how that can be confusing. Unfortunately understanding how to reconcile these statements will require a lot of background. I will try to answer this as concisely as I can (hopefully) without relying on concepts that are too advanced.
Well, topological insulators do not possess a so-called intrinsic topological order. It means that the bulk states of a topological insulator are not entangled quantum mechanically over a long range. Topological insulators are, in fact, short-range entangled just like trivial insulators. However, topological insulators and trivial insulators are clearly not the same phases. Therefore short-range entangled phases are further broken down into subcategories. Two such subcategories are: symmetry protected topological phases (topological insulators) and symmetry-breaking phases (trivial insulators).
The reason the word “topological” appears in the distinction between of topological insulators and trivial insulators is that they can be assigned a distinct “topological invariant.” The notion of a topological invariant comes from topology. For example a sphere and a torus have different topological invariants. Just as you cannot deform a torus into a sphere without cutting it, in the same way you cannot deform the band structure of a topological insulator into that of a trivial insulator without closing the bulk gap. As a consequence of this subtle difference in the two types of band structures the number of edge states will either be even (trivial insulators) or odd (topological insulators). Now this is where time reversal symmetry comes in. If any kind of perturbation, which itself obeys time reversal symmetry, acts on these edge states then it can destroy these edge states only in pairs. Therefore if you had odd number of edge states to begin with then you will end up with at least one edge state even if the perturbation destroys all the remaining edge states (in pairs). Hence time reversal symmetry is responsible for the protection of these edge states in topological insulators. You can find a more detailed explanation here:
What conductance is measured for the quantum spin Hall state when the Hall conductance vanishes?
Just scroll all the way down until you see the question in the block quote “Also: Why is there only a single helical edge state per edge? Why must we have at least one and why can't we have, let's say, two states per edge?” To give the above analogy with topology a firm footing I suggest you take a look at Berry curvature and the Chern number (if you haven't already). The topological invariants are closely connected to these.
So to summarize, gapped phases of matter can be divided into two categories: long-range entangled (with intrinsic topological order) and short-range entangled (without intrinsic topological order). Two subcategories of short-range entangled phases are: symmetry protected topological phases (topological insulators) and symmetry-breaking phases (trivial insulators).
In case you are wondering about long-range entangled phases and what it means to have (intrinsic) topological protection then I recommend a little more background reading on the principle of emergence, the fractional quantum Hall effect, string-net condensation (in that order). There are some excellent posts on physics stackexchange on the topic of string-net condensation. Some of them are even answered by Prof. Xiao-Gang Wen who, as a matter of fact, developed the theory of string-net condensation along with Michael Levin (I don’t know if he’s here).