Introduction

Charles Kane from the University of Pennsylvania will introduce today's lecture on two dimensional topological insulators with time-reversal symmetry.

Adding symmetry to a topological insulator

In general, there are different approaches to discover new types of topological systems.

We have already used a very powerful method to make a Kitaev chain and the Chern insulator model. We started from guessing what kind of model to use for the edge, such that it is impossible to obtain without the bulk. Then we combined many such edges (dots for the Kitaev chain and wires for the Chern insulator) and tailored the coupling between them to leave exactly the type of model that we want on one edge.

A very skilled researcher in topology (or more specifically K-theory) may also just calculate the expected topological classification of a system starting only from its dimensionality and symmetries. This is also a powerful method, but often it's too hard and requires a very high skills in math.

Another approach that we can undertake is to start with one topological Hamiltonian and see what happens if we force the Hamiltonian to have some extra symmetry. This is the approach we will use in this chapter.

Let's start from a simple example involving something which we already know, a quantum dot with the Hamiltonian $H_0$. We know that there is a topological invariant, the number of filled energy levels.

Now we can ask what happens if we force the dot to have a particle-hole symmetry. The Hamiltonian becomes

$$ H_\textrm{BdG} = \begin{pmatrix} H_0 & 0\\ 0 & -H_0^* \end{pmatrix}. $$

This model is clearly topologically trivial from the point of view of the old invariant, since the number of filled states is constant. However, there are still level crossings that appear. We may ask if these crossings stay protected if we also include a finite superconducting pairing $\Delta$ in the Hamiltonian, which couples the two blocks $H_0$ and $-H_0^*$.

Of course we know the answer: the crossings stay protected due to the change in the Pfaffian invariant. So what we did was to construct a topologically non-trivial superconducting dot by adding particle-hole symmetry to a topological Hamiltonian with a lower symmetry.

Let's now apply the same logic to a new system. Specifically, let's add time-reversal symmetry to a Chern insulator. The Chern insulator has chiral edge states whose direction of propagation is flipped by time-reversal symmetry $\mathcal{T}$. So let's consider a Hamiltonian of the form

$$ H = \begin{pmatrix} H_0 & 0\\ 0 & \mathcal{T}H_0\mathcal{T}^{-1} \end{pmatrix}\,. $$

If $H_0$ is the Hamiltonian of a Chern insulator with $N$ edge states, then $H$ will have $N$ pairs of counterpropagating edge states that transform into each other by time-reversal symmetry. Moreover, the full $H$ obeys time-reversal symmetry, which merely exchanges the two blocks.

The following sketch describes the situation in the case $N=1$:

The next task which we now face is to understand if such edges stay topologically protected once we add coupling between the two blocks.

A perfectly transmitted channel and Kramers degeneracy

We could try to see if all the edge states can be removed by adding some terms to the Hamiltonian, but instead we will use a closely related fact.

Let's study transport through such edge states as a function of their total number and let's only use the fact that time-reversal symmetry is present. Imagine, there is a total of $N$ states going in each direction along the edge, and that the edge is composed of a disordered region sandwiched between two clean regions. Again, let's represent the situation for the case $N=1$.

Scattering states

We label incoming states on the left and right with $\left|n,\textrm{L}\right\rangle$ and $\left|n,\textrm{R}\right\rangle$. The index $n$ goes from $1$ to $N$. The outgoing states are the time-reversed partners of the incoming states, so they are given by $\mathcal{T}\left|n,\textrm{L}\right\rangle$ and $\mathcal{T}\left|n,\textrm{R}\right\rangle$. Scattering states in the left and right regions are superpositions of incoming and outgoing states,

$$ \left|\Psi,\textrm{L}\right\rangle = \sum_{n=1}^N \alpha_{n,\textrm{L}}\,\left|n,\textrm{L}\right\rangle + \beta_{n,\textrm{L}}\,\mathcal{T}\left|n,\textrm{L}\right\rangle\,, $$$$ \left|\Psi,\textrm{R}\right\rangle = \sum_{n=1}^N \alpha_{n,\textrm{R}}\,\left|n,\textrm{R}\right\rangle + \beta_{n,\textrm{R}}\,\mathcal{T}\left|n,\textrm{R}\right\rangle\,. $$

We can form vectors out of all the coefficients in the superposition, for instance $\alpha_\textrm{L} = (\alpha_{1,\textrm{L}},\dots,\alpha_{N,\textrm{L}})^T$ for the incoming states on the left side. Incoming and outgoing modes are then related by the scattering matrix $S$ of the disordered region,

$$ \begin{pmatrix} \beta_\textrm{L} \\ \beta_\textrm{R} \end{pmatrix} = S \begin{pmatrix} \alpha_\textrm{L} \\ \alpha_\textrm{R} \end{pmatrix}. $$

There are a total of $2N$ incoming and $2N$ outgoing modes, so $S$ is a $2N\times 2N$ matrix. Since we are including all possible initial and final states, $S$ is also unitary, $S=S^\dagger$. It can be split into reflection and transmission blocks of dimension $N\times N$,

$$ S = \begin{pmatrix} r & t\\ t' & r' \end{pmatrix}\,. $$

If we can gap out the edges by adding some extra terms to the Hamiltonian, or backscatter them by adding disorder, then we should be able to achieve the situation where there is no transmission at all, $t = t' = 0$. In this case, all modes must be reflected back, so the reflection blocks of the scattering matrix become unitary, $r^\dagger r = r'^\dagger r' = 1$.

To see whether this is possible at all, we first have to understand the constraints that time-reversal symmetry imposes on $S$.

Scattering matrices with time-reversal symmetry

Let's recall some basic facts about time-reversal symmetry, which we already studied in the first week. Time-reversal symmetry has an antiunitary operator $\mathcal{T}$ which commutes with the Hamiltonian. Being antiunitary, $\mathcal{T}$ may come in two flavors - either $\mathcal{T}^2=1$ or $\mathcal{T}^2=-1$. The first case applies to systems with no or integer spin, such that $\mathcal{T}=\mathcal{K}$ in the simplest case, where $\mathcal{K}$ is the complex conjugation operator. The second case applies to systems with half-integer spin, and in the simplest case we have $\mathcal{T}=i\sigma_y\mathcal{K}$.

Let's apply the time-reversal operator to our scattering states. We get

$$ \mathcal{T}\left|\Psi,\textrm{L}\right\rangle = \sum_{n=1}^N \alpha^*_{n,\textrm{L}}\,\mathcal{T}\left|n,\textrm{L}\right\rangle + \beta^*_{n,\textrm{L}}\,\mathcal{T}^2\left|n,\textrm{L}\right\rangle\,, $$$$ \mathcal{T}\left|\Psi,\textrm{R}\right\rangle = \sum_{n=1}^N \alpha^*_{n,\textrm{R}}\,\mathcal{T}\left|n,\textrm{R}\right\rangle + \beta^*_{n,\textrm{R}}\,\mathcal{T}^2\left|n,\textrm{R}\right\rangle\,. $$

Now, since time-reversal symmetry does not change the energy of a state, $\mathcal{T}\left|\Psi,\textrm{R}\right\rangle$ and $\mathcal{T}\left|\Psi,\textrm{L}\right\rangle$ are scattering states with the same energy as $\left|\Psi,\textrm{R}\right\rangle$ and $\left|\Psi,\textrm{L}\right\rangle$. Hence, the coefficients of incoming and outgoing modes are still related by the same scattering matrix $S$ as before. Note, however, that applying $\mathcal{T}$ exchanged the role of the $\alpha$'s and $\beta$'s, such that the $\alpha$'s now correspond to outgoing states and the $\beta$'s to incoming states. Hence, we have

$$ S\mathcal{T}^2 \begin{pmatrix}\beta^*_\textrm{L} \\ \beta^*_\textrm{R} \end{pmatrix} = \begin{pmatrix} \alpha^*_\textrm{L} \\ \alpha^*_\textrm{R} \end{pmatrix}\,. $$

Multiplying both sides by $\mathcal{T}^2S^\dagger$ and taking the complex conjugate gives

$$ \begin{pmatrix} \beta_\textrm{L} \\ \beta_\textrm{R} \end{pmatrix} = \mathcal{T}^2\,S^T \begin{pmatrix} \alpha_\textrm{L} \\ \alpha_\textrm{R} \end{pmatrix}. $$

By comparing this equation with the one a few lines above, we finally obtain

$$ S = \mathcal{T}^2 S^T. $$

So if $\mathcal{T}^2=1$, the scattering matrix is symmetric ($S=S^T$), while if $\mathcal{T}^2=-1$, it is antisymmetric ($S=-S^T$).

What does this imply if we try to set $t=t'=0$?

If $S=S^T$, it turns out there is really nothing special we can tell. However, if $S=-S^T$ and $t=t'=0$, the $N\times N$ reflection matrix must be both unitary, $r^\dagger r=1$, and antisymmetric, $r=-r^T$.

If $N$ is odd, this isn't possible at all, since any odd-dimensional antisymmetric matrix must have a single zero eigenvalue, while unitary matrices only have eigenvalues with unit norm!

We are forced to conclude that it is impossible to have $r$ unitary, and therefore it is impossible to have $t=0$ in this case. Furthermore, this zero eigenvalue of $r$ means that there is always a single mode that is transmitted with unit probability.

This is the discovery that Charles Kane described in the introductory video. We can quickly check it by randomly selecting an antisymmetric scattering matrix with odd $N$, like the following one with $N=3$,

\begin{pmatrix}0 & 0.38-0.21i & 0.17+0.39i & 0.24-0.09i & 0.66-0.22i & -0.09+0.26i\\-0.38+0.21i & 0 & -0.63+0.03i & -0.05+0.16i & 0.19-0.34i & 0.46-0.14i\\-0.17-0.39i & 0.63-0.03i & 0 & 0.32+0.21i & -0.48-0.06i & 0.16+0.06i\\-0.24+0.09i & 0.05-0.16i & -0.32-0.21i & 0 & -0.12-0.31i & -0.8+0.12i\\-0.66+0.22i & -0.19+0.34i & 0.48+0.06i & 0.12+0.31i & 0 & -0.05+0.1i\\0.09-0.26i & -0.46+0.14i & -0.16-0.06i & 0.8-0.12i & 0.05-0.1i & 0\end{pmatrix}

and looking at the eigenvalues of $r^\dagger r$ and $t^\dagger t$:

Reflection eigenvalues
\begin{pmatrix} & 0. & 0.77 & 0.77\end{pmatrix}
Transmission eigenvalues
\begin{pmatrix} & 1. & 0.23 & 0.23\end{pmatrix}

We conclude that if $\mathcal{T}^2=-1$ and the number of edge states going in one direction is odd, they cannot be gapped out, and the system is topological. On the other hand, if there is an even number of such edge states, they can be gapped out. Since these are the only two options, the integer invariant of a Chern insulator is reduced to a $\pm 1$ invariant in the presence of time reversal symmetry. These topologically protected, counterpropagating edge states are often referred to as helical edge states.

Helical edge states are Kramers pairs

You might ask yourself what makes $\mathcal{T}^2=-1$ special, leading to the topological protection of the helical edge states.

As was mentioned in the first week, if $\mathcal{T}^2=-1$ then Kramers' theorem applies. Kramers' theorem tells us that given an eigenstate $|\Psi\rangle$ of the Hamiltonian with energy $E$, its time-reversed partner $|\Psi_\mathcal{T}\rangle\equiv\mathcal{T}|\Psi\rangle$ has the same energy, and the two states are orthogonal, $\langle \Psi | \Psi_\mathcal{T}\rangle=0$. These two states form a so-called Kramers pair. As we already know, this leads to the fact that Hamiltonians with spinful time-reversal symmetry have two-fold degenerate energy levels - Kramers degeneracy.

Now, the two counterpropagating helical modes are time-reversed partners of each other, so they form precisely such a Kramers pair. The condition $\langle \Psi | \Psi_\mathcal{T}\rangle=0$ implies that it is impossible to introduce any backscattering between the two states, unless we break time-reversal symmetry. This is the origin of the unit transmission and of the topological protection of helical edge states.

To gain a more intuitive understanding of this fact at a more microscopic level, we can assume that the projection of the electrons' spin along a given axis is conserved, say the axis $z$ perpendicular to the plane. Then at the edge you have, say, a right-moving mode with spin up and a left-moving mode with spin down, and no other modes if $N=1$. Let's draw again the picture of a helical edge state entering the disordered region:

Thus, an electron moving to the right must have spin up by assumption. In order to be reflected, its spin must also be flipped. However, this spin-flip scattering process is forbidden, and again we conclude that the electron is transmitted with probability one.

In the case $\mathcal{T}^2=1$, there is no Kramers' theorem. As a consequence, even though you can construct models which have counterpropagating edge states, you will find that they have no topological protection and can be gapped out without breaking the time-reversal symmetry.

The quantum spin Hall effect

There is no really precise name for the 2D topological insulator with time-reversal symmetry. It is often called "$\mathbb{Z}_2$ topological insulator." However, this simply indicates that there are only two values of the topological invariant, and so it isn't a very specific name.

The most commonly used name for this system is "quantum spin Hall insulator." To understand why, let's analyse a Hall bar made of such a non-trivial insulator. We will only need a Hall bar with four terminals, as shown below:

We have a finite voltage applied to terminal 1, so electrons are injected into the system from there. You can see that because of the helical edge states, there are as many modes connecting terminal 1 to terminal 3 as there are to terminal 4. A moment of thought, or otherwise a quick calculation, should convince you that in this case there is no net current flowing orthogonal to the applied voltage. The Hall conductance is zero, which is the expected result if time-reversal symmetry is preserved, as it is in our system.

However, counterpropagating edge states have to have exactly opposite spin due to Kramers degeneracy. This means that there may be a net spin current across the sample, orthogonal to the applied voltage.

In particular, let's again make the simple assumption that the spin projection along some axis is conserved. Then, in the figure above, all modes colored in red have spin up, and all modes colored in blue have spin down. So terminal 1 distributes electrons coming out of it according to their spin: all electrons with spin up end up in terminal 4, and all those with spin down in terminal 3. The system has a quantized spin current between terminals 3 and 4, hence the name "quantum spin Hall effect".

However, the quantized spin Hall current is not a general property of a quantum spin Hall insulator. Here, it arises because we have combined time reversal symmetry with a spin conservation law, and as we learned in the first week, conservation laws are boring from a topological point of view.

Consider the simple case where spin is conserved. In the quantum spin Hall bar system above, what happens if, instead of applying a voltage between terminals 1 and 2, you manage to apply a *spin-polarized* current between terminals 1 and 2?

The system will develop an opposite spin-polarized current to compensate the effect.
A spin-polarized current will develop between terminals 3 and 4.
A voltage difference will develop between terminals 3 and 4.
It is impossible to apply such a current unless the bulk gap closes.

A model for the quantum spin Hall insulator

There is an important model which can be used to describe quantum spin Hall insulators, known as the Bernevig-Hughes-Zhang model or, in short, BHZ model. In essence, this model is equivalent to two copies of the Chern insulator Hamiltonian on the square lattice that we studied in the fourth week.

The BHZ Hamiltonian takes the form

$$ H_\textrm{BHZ}(\mathbf{k}) = \begin{pmatrix} h(\mathbf{k}) & 0 \\ 0 & h^*(-\mathbf{k}) \end{pmatrix}\,, $$

with

$$ h(\mathbf{k}) = \epsilon(\mathbf{k}) + \mathbf{d}(\mathbf{k})\cdot \pmb{\sigma}\,. $$

Here $\pmb\sigma = (\sigma_x, \sigma_y, \sigma_z)$ is a vector of Pauli matrices acting on the electron/hole degree of freedom (the original two bands of the Chern insulator), $\epsilon(\mathbf{k}) = C - D(k_x^2+k_y^2)$, the vector $\mathbf{d} = [A k_x, -A k_y, M(\mathbf{k})]$, and $M(\mathbf{k}) = M - B(k_x^2+k_y^2)$.

You can see that it is basically two copies of the massive Dirac Hamiltonian we used to study Chern insulators. In particular, there is a linear coupling in momentum between the holes and the electrons. The gap in the Hamiltonian is given by the term $M(\mathbf{k})$, a momentum-dependent effective mass.

By changing the sign of $M$ from negative to positive, you get a gap closing at $\mathbf{k}=\pmb{0}$: