# Introduction¶

This topic is special, since in order to meaningfully discuss experimental progress we need to do something we didn't do before in the course: we will show you the measurements and compare them with the *simple* theoretical expectations. Like this we will see what agrees and what doesn't.

All the figures showing the experiments are copyright Physical Society of Japan (2008), published in J. Phys. Soc. Jpn. 77, 031007 (2008) by Markus König, Hartmut Buhmann, Laurens W. Molenkamp, Taylor Hughes, Chao-Xing Liu, Xiao-Liang Qi, and Shou-Cheng Zhang. They are available under CC-BY-NC-SA 4.0 International license.

# Two limits: Mexican hat and weak pairing¶

We just learned that topological insulators with inversion symmetry were simpler to think about. We will now use the topological invariant to find a simple recipe for finding topological insulators. All we need to do is somehow vary the parity of the occupied states. One fact of nature that comes to our aid in this is that electrons in semiconductors typically occupy even parity $s$-orbitals and odd parity $p$-orbitals.

If we look up the bandstructure of a typical "non-topological" semiconductor, the highest valence-band is of odd parity and the lowest conduction band is even parity. As one moves down the periodic table to heavier elements with larger spin-orbit coupling the odd parity orbital switches spots with the even parity orbital. This *band inversion* is the domain where we can hope to find topological insulators.

Now you might think that all we have to do is go down the periodic table to heavier elements and just pick some material like HgTe (actually used in the creation of QSHE), but that's not all yet. We still need to make a quantum well out of this semiconductor to make the system two-dimensional. This leads to two dimensional bands derived from the three dimensional band structure.

By carefully choosing the widths, it is possible to invert the odd and even parity bands. We saw from the last unit, that such a band-inversion leads to a topologically non-trivial value of the parity invariant. Right around the topological transition, the even and odd parity bands are degenerate. Thus, we can follow the discussion in the last unit to derive domain wall states at the edges.

We can write down the simplest Hamiltonian for an even and an odd parity band in a basis $|e,\sigma\rangle$ and $|o,\sigma\rangle$ in a block form

$$H({\bf k})=\left(\begin{array}{cc}\epsilon_e({\bf k})&\Delta({\bf k})\\\Delta^\dagger({\bf k})&\epsilon_o({\bf k})\end{array}\right),$$where $\Delta({\bf k})$ is the $2\times 2$ hybridization matrix. Inversion and time-reversal symmetries imply that $\Delta({\bf k})=-\Delta(-{\bf k})$ is odd under inversion and even under time-reversal. Here we will focus on one such model, $\Delta({\bf k})=\alpha\sigma_z(k_x+i k_y)$, which we call the Bernevig-Hughes-Zhang model.

Since the even band is electron-like, we approximate the even-band dispersion $\epsilon_e({\bf k})$ as $\epsilon_e({\bf k}) = \delta_e + m_e k^2$, while we take the odd parity dispersion to be $\epsilon_o({\bf k})= \delta_o - m_o k^2$ for simplicity. The band inversion happens when $\delta_e < \delta_o$.

The spectrum of this Hamiltonian is very similar to that of a Chern insulator (after all we essentially just doubled the degrees of freedom). Just like in most topological systems, the shape of the band structure depends on the relative strength of band inversion and inter-band coupling.

So below we see a qualitative band structure of one of the QSHE insulators, HgTe/CdTe quantum well, compared with the band structure of InAs/GaSb quantum well.

In the last unit, we understood the nature of the edge modes near the topological phase transition, where a doubled Dirac model was appropriate. Deep in the strongly band-inverted topological regime, the bulk band structure has a mexican hat structure with the gap proportional to $\alpha$.

The edge modes in this regime are quite different in structure from those near the topological transition. To see this, let us first set $k_y=0$ in the Hamiltonian. If we set $\alpha=0$ then there are two fermi points where the dispersion is roughly linear - let us label these points by $\tau_z=\pm 1$. We can describe the edge of the system, by assigning boundary conditions to the $k_x=\pm k_F$ modes in terms of time-reversal invariant phase-shifts.

The bulk solutions near $k_x\sim\pm k_F$ can be written as $\psi_\pm(x)=e^{-x/\xi}\psi_\pm(0)$. Matching boundary conditions, we find that a zero energy pair of edge solutions exists in the case of inverted bands. These solutions differ from the ones in the Dirac limit by the presence of the oscillating part of the wave function.

# Quantized conductance and length dependence¶

Unlike in the case of Majoranas, not much thinking is required to figure out the relevant signature of the quantum spin Hall effect. There is a pair of modes on each edge of the sample that is protected from backscattering. All the other modes are gapped or backscattered, so the edge states are the only ones to carry current. This current will not suffer from backscattering.

If we consider the simplest case, a sample with only two terminals, then Landauer's formula together with the absence of backscattering gives the conductance $G_0=2 e^2/h$.

When we move the Fermi level outside of the bulk gap, the bulk becomes conducting, and so the conductance increases.

We end up with this situation: