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 \(N=3\),

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

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.