# Dirac equation of the surface states, 3D Bernevig-Hughes-Zhang model¶

## Introduction¶

Joel Moore from the University of California, Berkeley will introduce this week’s topic, by telling us how the idea of a two-dimensional topological insulator was generalized to three dimensions.

## Making 3D topological invariants out of 2D ones¶

Let us follow the direction explained by Joel Moore and construct a three-dimensional topological state from the two-dimensional topological state. This time, we’ll do this by studying the system in momentum space rather than in real space as we did before. As with two dimensional systems, time-reversal invariant momenta (TRIMs) play an important role in three dimensions.

For illustrative purposes, consider the three dimensional irreducible Brillouin Zone (i.e. $$k_j\in [0,\pi]$$) of a cubic system shown below. Fixing one of the three momenta $$k_{x,y,z}$$ to a TRIM, say $$k_x=0$$ without loss of generality, we can think of the Hamiltonian in the $$(k_y,k_z)$$ plane as a two dimensional Hamiltonian, which may either be topologically trivial ($$\mathbb{Z}_2$$-index $$=0$$) or non-trivial ($$\mathbb{Z}_2$$-index $$=1$$). So for every side of the cube shown above we can compute a QSHE topological invariant, which gives us 6 numbers. However not all of them are independent. Specifically, there is a constraint $$Q(k_x=0)\,Q(k_x=\pi) \equiv Q(k_y=0)\,Q(k_y=\pi) \equiv Q(k_z=0)\,Q(k_z=\pi)$$.

This product is called the strong topological invariant. Accordingly, the topological insulators where this invariant is non-trivial are called strong topological insulators. For the remaining three invariants, we can choose $$Q(k_x=\pi),\,Q(k_y=\pi),\,Q(k_z=\pi)$$.

Very frequently the topological invariants of a compound are written as $$(1;010)$$, where the first number corresponds to the strong invariant, and the remaining three to the weak invariants along each axis. For example, the first predicted topological insulator, the alloy Bi$$_x$$Sb$$_{1-x}$$ is $$(1;111)$$, and the second generation topological insulators Bi$$_2$$Te$$_3$$ and Bi$$_2$$Se$$_3$$ are $$(1;000)$$.

Just by using the bulk-edge correspondence for $$Q$$ we know that the strong topological invariant means that there is an odd number of helical states going in each direction on each facet of the topological insulator. We will see later why this is special, but before that let’s construct a model for a 3D TI.

## BHZ model of a 3D topological insulator¶

Our goal in this unit is to derive an effective three-dimensional Hamiltonian $$H(\mathbf{k})$$ for a strong topological insulator.

We follow the same logic that led us to defining the three-dimensional topological invariant in the previous unit, building up on our knowledge of 2D topological insulators. Our first step is therefore to set $$k_z=0$$ and start from a two-dimensional Bloch Hamiltonian which describes a non-trivial 2D topological insulator. Of course, we choose a model we already know for this 2D Hamiltonian, the Bernevig-Hughes-Zhang (BHZ) model.

Let’s recapitulate what we said about the BHZ model last week. It is a four band model, which has two electron bands (spin up and spin down) and two hole bands (spin up and down). It has inversion symmetry, with electron and hole bands having opposite parity. We will not need more bands for our 3D topological insulator model.

Copying the BHZ Hamiltonian of last week, at $$k_z=0$$ we have

$\begin{split} H(k_z=0) = \epsilon(\mathbf{k})\cdot\mathbb{1} + \begin{pmatrix} M_0(\mathbf{k}) & A k_+ & 0 & 0 \\ A k_- & -M_0(\mathbf{k}) & 0 & 0 \\ 0 & 0 & M_0(\mathbf{k}) & -Ak_- \\ 0 & 0 & -Ak_+ & -M_0(\mathbf{k}) \end{pmatrix}\,, \end{split}$

where $$k_\pm = k_x \pm i k_y$$. This Hamiltonian is written in a basis given by the states $$\left|E\uparrow\right\rangle$$, $$\left|H\uparrow\right\rangle$$, $$\left|E\downarrow\right\rangle$$, $$\left|H\downarrow\right\rangle$$, in that order. The block structure of the Hamiltonian reminds you that it is a doubled version of a Chern insulator, with two diagonal blocks for up and down spins. The particular form of $$\epsilon(\mathbf{k})$$ can be important to describe the band structure of a given material, but will not play a role in what follows. The effective mass is given by $$M_0(\mathbf{k}) = M - B(k_x^2+k_y^2)$$, and the transition between the topological and trivial insulating phases in this 2D model happens when $$M$$ changes sign.

To get a strong topological insulator, we would like the two-dimensional $$\mathbb{Z}_2$$ invariant applied to the $$(k_x, k_y)$$ plane to take different values at $$k_z=0$$ and $$k_z=\pi$$. It is easy to achieve this by adding a $$k_z$$-dependent term to the effective mass, for instance in the following way

$M_0(\mathbf{k})\,\to\,M(\mathbf{k}) = M - B(k_x^2+k_y^2+k_z^2)\,.$

Compared to the 2D model above, we can make the sign of the ‘new’ mass $$M-Bk_z^2$$ effectively change at fixed $$k_x$$ and $$k_y$$ by choosing $$k_z$$ large enough.

There is now a problem, however. Looking at the Hamiltonian at different values of $$k_z$$, we see that if we fix $$k_z$$ it still has time-reversal symmetry. Since it is topologically nontrivial at $$k_z=0$$ and becomes trivial at $$k_z = \pi$$, it should have a topological phase transition somewhere in between.

In other words, if we just add the $$B k_z^2$$ term, the Hamiltonian becomes gapless! Of course, we would like to have a gapped Hamiltonian in the whole 3D Brillouin zone instead.

How can we avoid the gap closing? We definitely need to couple the two spin blocks in $$H(\mathbf{k})$$, since otherwise each block is undergoing a Chern insulator transition. Recall that spin is odd under time-reversal, so to couple the spins we need a coupling which is odd in momentum, in order to maintain time-reversal invariance in the system. The simplest thing is to pick a coupling that is linear in $$k_z$$.

We then arrive at the following 3D Hamiltonian,

$\begin{split} H(\mathbf{k}) = \epsilon(\mathbf{k})\cdot\mathbb{1} + \begin{pmatrix} M(\mathbf{k}) & A k_+ & 0 & \tilde{A}k_z \\ A k_- & -M(\mathbf{k}) & \tilde{A}k_z & 0 \\ 0 & \tilde{A}k_z & M(\mathbf{k}) & -Ak_- \\ \tilde{A}k_z & 0 & -Ak_+ & -M(\mathbf{k}) \end{pmatrix}\,. \end{split}$

This Hamiltonian is known as the 3D BHZ model. It is gapped at finite $$M$$, and a transition between the trivial and strong topological insulator phases is achieved by changing the sign of $$M$$. Just like its two-dimensional counterpart, the 3D BHZ model can be used as a prototype for a strong topological insulator, as well as a starting point to model real materials.

The above derivation makes one important point evident: a necessary ingredient to have a strong topological insulator is to break spin conservation. Above, we achieved this by adding coupling between the spins, to avoid the undesirable gap closing at finite $$k_z$$

## Dirac surface states¶

What is the dispersion of the surface state of the $$3D$$ topological insulator?

We know that if we fix one momentum (say $$k_x$$) to zero, the Hamiltonian of the remaining system is that of a quantum spin Hall insulator. For this system we know that the Hamiltonian of the edge states is just that of a pair of counter-propagating modes, so

$H = v \sigma_y k_y.$

Here, the matrix $$\sigma_y$$ acts on the degrees of freedom of these two surface modes, and doesn’t correspond to particle spin.

Since time-reversal symmetry changes the sign of $$k_y$$, it must also change the sign of $$\sigma_y$$, so the time-reversal operator must be $$\mathcal{T} = i \sigma_y K$$.

What if we consider a nonzero $$k_x$$? Generically, the two modes are then coupled by an extra term in the Hamiltonian. This term should be proportional to $$k_x$$, and since it couples the modes it must also include a Pauli matrix, which we can just choose to be $$\sigma_x$$.

So if the surface of the topological insulator is isotropic, its Hamiltonian is merely

$H=v \mathbf{\sigma} \cdot \mathbf{k}.$

Let’s have a quick look at it to get a more concrete understanding: