As just mentioned, topological invariants in higher dimensions are often difficult to write down and evaluate in the general case. Luckily, in the presence of inversion symmetry - which reverses the lattice coordinates with respect to a symmetry center - the topological condition can be stated in rather simple terms. This turns out to be quite useful to describe most topological materials, which happen to have crystal structure with inversion symmetry.

From our earlier discussion, we know that a system is a time-reversal invariant topological insulator if it has an odd number of helical edge states. We will now see how we can find an expression for the bulk topological invariant, using inversion symmetry and bulk-boundary correspondence.

So let’s consider a two-dimensional Bloch Hamiltonian \(H(\mathbf{k})\) with both inversion and time-reversal symmetry. Inversion symmetry has a unitary operator \(\mathcal{P}\) which maps \(\mathbf{k}\rightarrow -\mathbf{k}\) and satisfies \(\mathcal{P}^2=1\). If we have both inversion symmetry \(\mathcal{P}\) and time-reversal \(\mathcal{T}\), we get an anti-unitary symmetry \(\mathcal{T}\otimes\mathcal{P}\), which preserves \(\mathbf{k}\) and squares to \(-1\).

These are precisely the conditions needed for Kramers theorem to apply - only this time, every point \(\mathbf{k}\) is mapped to itself because inversion symmetry is included as well. We conclude that every eigenstate at any \(\mathbf{k}\) is two-fold degenerate. We may label these two eigenstates with an index \(\sigma=\pm\). If spin is a good quantum number, \(\sigma\) labels two states with opposite spin. However, this may not be the case so we will just refer to it as a pseudospin associated with Kramers degeneracy.

Note that the simplification obtained by adding inversion symmetry is that the spectrum is two-fold degenerate at all \(\mathbf{k}\) in the Brillouin zone. Time-reversal symmetry alone cannot guarantee that, because it maps \(\mathbf{k}\) to \(-\mathbf{k}\).

Our next step is to calculate the effective description of helical edge states at a domain wall between a topological phase and a non-topological phase. This is something we already know how to do thanks to our experience with domain walls in the Kitaev chain and in Chern insulators. It will give us insight into the topological transition and the bulk topological invariant.

Study of a domain wall

Let’s imagine that the helical edge runs along the \(y\) direction, and that the domain wall is described by a mass profile \(M(x)\) along the \(x\) direction, which is zero at the domain wall:

In this configuration, \(k_y\) is still a good quantum number, and we can study the energy dispersion of states bound to the domain wall as a function of \(k_y\). If the edge is gapless there must be a momentum, say \(\bar{k}_y\), where counterpropagating modes cross at the Fermi level. Let’s fix \(k_y=\bar{k}_y\), and write down an effective Hamiltonian for the motion transverse to the domain wall.

We have in total four states, distinguished by two quantum numbers: their direction of propagation, which we denote with \(b=\pm\), and their pseudospin \(\sigma\). Inversion symmetry \(\mathcal{P}\) flips the direction of propagation \(b\), while the pseudo-spin degeneracy \(\sigma\) is related to the combination of inversion and time-reversal \(\mathcal{T}\otimes\mathcal{P}.\) To lowest order in the momentum \(k_x\) perpendicular to the domain wall, the states at the transition point disperse linearly with \(k_x\), and are two-fold degenerate. In fact, as we noted from Kramers degeneracy, the Hamiltonian must be chosen such that none of the terms break the two-fold degeneracy associated with the pseudospin \(\sigma\). This means that the domain wall cannot couple states with different values of \(\sigma\), which leads us to an effective Hamiltonian

\[ H(\bar{k}_y)=\sum_{\sigma,b}\,k_x b\,|b,\sigma\rangle\langle b,\sigma|+M(x)(|+,\sigma\rangle\langle -,\sigma|+h.c.)]\,. \]

where the factor \(b\) is odd under time-reversal symmetry so that \(k_x b\) is even under time-reversal symmetry.

We are back to an old friend, the one dimensional Dirac Hamiltonian with a position-dependent mass \(M(x)\). Adapting our arguments from the first week, we can immediately say that the domain-wall hosts a pair of zero modes only if \(M(x)\) changes sign.

As interesting as this sounds, we must remember that this pair of zero modes is present for \(k_y=\bar{k}_y\). Because of inversion symmetry, there is necessarily an identical pair at \(-\bar{k}_y\). So we get a total of 4 degenerate domain wall states from this type of gap closing - an even number of pairs. As we know form before, such pairs of gap closings do not affect the value of the topological invariant on the two sides of the domain walls. To change the value of the topological invariant, we would need an odd number of pairs crossing zero energy.

However, there are points in momentum space which are mapped onto each other by time-reversal symmetry, up to a reciprocal lattice vector. For these values, the above counting does not hold. People refer to these momenta as “time-reversal invariant momenta” or TRIMs. In the simple case of a square Brillouin zone, they are the points \((k_x, k_y) = (0,0), (0,\pi), (\pi,0),(\pi,\pi)\).

Since TRIMs are their own time-reversed partners, it is still possible for a gap closing at \(\bar{k}_y=0\) or \(\bar{k}_y=\pi\) to change the topology of the system. In this case, our doubling problem in momentum space is solved, and we can produce just one pair of edge modes at the domain wall. If we move the momentum \(k_y\) slightly away from \(\bar{k}_y\), the degenerate pair of modes splits linearly to form a single helical mode that produces a non-trivial fermion parity pump.

To make the distinction clear between a gap closing at a finite \(\bar{k}_y\) and at a time-reversal invariant point, let’s draw a sketch of the edge dispersion in the two cases.

Kramers pairs are colored in red and blue and have the same linestyle. On the left, you have two pairs of Kramers partners, which however never meet at zero energy. On the right, there is a single Kramers pair meeting at zero energy. This argument summarizes the simplification that inversion symmetry brings to time-reversal invariant topological insulators.

We can determine the topological invariant for the inversion symmetric topological insulators entirely from the bulk Hamiltonian at time-reversal invariant momenta, since gap closings at any other point can only add domain wall states in multiples of four.

Does this mean that any gap closing at a TRIM is a topological transition? The states \(|b,\sigma\rangle\) are Bloch states with definite values of \(k_y\). We are considering a time-reversal invariant value of \(k_y\), and since \(b\) is flipped by inversion symmetry, we can apply inversion symmetry to conclude that the states \(|\pm,\sigma\rangle\) transform into each other under inversion i.e. \(\mathcal{P}|\pm,\sigma\rangle= |\mp,\sigma\rangle\). By combining these states into symmetric and anti-symmetric superpositions

\[|e,\sigma\rangle=\frac{1}{\sqrt{2}}\left[|+,\sigma\rangle\, + \,|-,\sigma\rangle\right],\,\quad |o,\sigma\rangle=\frac{i}{\sqrt{2}}\left[|+,\sigma\rangle\, - \,|-,\sigma\rangle\right],\]

we obtain states that are even (\(e\)) and odd (\(o\)) under inversion—they are eigenstates of \(\mathcal{P}\) with eigenvalue \(+1\) or \(-1\). They are also eigenstates of \(M\) at \(k_x=k_y=0\). The factor of \(i\) in \(|o,\sigma\rangle\) ensures a consistency under the time-reversal transformation, such that \(\mathcal{T}|(e,o),\sigma\rangle=\sigma|(e,o),\sigma\rangle\).

Every gap closing at a TRIM is an even parity state crossing with an odd parity state. The effective Hamiltonian of such a gap closing must also add an extra Kramers pair of states at the domain wall, and therefore indeed every gap closing at a TRIM is a topological phase transition, while gap closings at all the other momenta are unimportant due to inversion symmetry.

This leads to a simplified way of computing a topological invariant of quantum spin Hall insulators with inversion symmetry:

To compute a bulk topological invariant for a two-dimensional topological state with time reversal and inversion symmetry we need to keep track of the parity \(P\) of all the occupied eigenstates of \(H(\mathbf{k})\) at the different time-reversal invariant momenta in the Brillouin zone. We may write such a bulk topological invariant as a product

\[Q=\prod_{n,j}P_{n,j}\,,\]

where \(P_{n,j}\) is the parity, \(n\) runs over the occupied bands of \(H(\mathbf{k})\) and \(j\) over the time-reversal invariant momenta.

You might now worry whether this definition of the invariant relied on having a smooth domain wall. From the fermion parity pump argument, bulk-edge correspondence implies that the bulk must be topologically non-trivial once you have edge states for any termination. Reversing this argument, we know that once we have a topologically non-trivial bulk, we must have helical edge states for any termination.

Thus, by looking at smooth domain walls we are able to establish a connection between the topological invariant in the presence of time-reversal and inversion symmetry, and the existence of helical edge states and fermion-parity pumping that characterizes the two dimensional topological insulator.

As a bonus, thanks to the previous arguments we can begin to understand how to look for two-dimensional topological insulators among real materials, or how to create them. The main idea is to generate a “band-inversion” between an even and an odd parity band at a TRIM.

Such a band inversion is not impossible to achieve in real materials, and can be captured using the BHZ model. But let’s leave this to the next lecture.

What is the value of the parity invariant $Q$ if you stack together two quantum spin Hall systems in the topological phase (i.e., both with $Q=-1$)?

The system has edge states and is therefore topologically non-trivial.
The total number of odd parity occupied orbitals must be even, so you get $Q=1$.
It depends on whether the helical states in the two layers have same or opposite spin for a given direction.
The invariant depends on the number of edge Dirac points at $k$ away from 0.