Well, the answer is yes and no. The band inversion between the $s$-like (conduction) band $\Gamma_6$ and $p$-like (valence) band $\Gamma_8$ in HgTe is primarily responsible for its topologically nontrivial band structure. The bulk band structure of HgTe with (right) and without (left) spin-orbit coupling is shown in the figure below. There are a total of eight bands (including spin) shown in both figures. Since we’re interested in the physics close to the $\Gamma$ point, we can approximately ignore bulk inversion asymmetry. Under this assumption the spin up and down bands are degenerate as clearly seen from the figure. From this point on I will not consider spin explicitly when talking about bulk band structure; i.e. there are a total of four bands (ignoring spin) in the figures below. Note: Please don’t focus on the quantitative details of the left figure. It is a hypothetical scenario introduced purely for pedagogical purposes.
You can notice that, in the figure on the left (without spin-orbit coupling), the heavy hole (HH) and light hole (LH) bands are degenerate. When you turn on spin-orbit, the $\Gamma_6$ and $\Gamma_8$ bands reverse their order, the $\Gamma_8$ band splits its degeneracy, and the LH band gets inverted. The Fermi energy sits at the intersection point of the LH and HH bands. But notice that, despite LH and HH acting as the conduction and valence bands (right figure) respectively, there is no gap between them! You cannot get a topological insulator without a bulk gap. If you could somehow induce a gap between the LH and HH bands (say) by straining HgTe then it could, in fact, be turned into a 3D topological insulator!
Now, there were several (experimental) advantages in creating a CdTe/HgTe/CdTe quantum well. First of all, since it’s a quantum well you would have sub-bands (not bands unlike bulk materials) due to quantum confinement in the out-of-plane (say $z$) direction. As a result, a single band in the bulk will split up into several sub-bands, each corresponding to a different quantized $k_z$, as you shrink the thickness of the material in the $z$-direction. Now, you can notice (in the figure below), unlike the bulk, the electron (conduction) and hole (valence) sub-bands do have an energy gap.
This plot obviously shows the minima (electron) or maxima (hole) of these sub-bands; they still disperse in k-space. And as you may know the inversion of the sub-bands will occur when you cross the critical thickness (as shown in the figure below).
Another very important advantage of using a quantum well structure in doing your experiments is that, unlike a bulk sample, you can electrically tune your Fermi energy using a gate. You could both tune your Fermi energy to intersect the electron (or hole) sub-band or keep it in the gap, and observe the change in conductance. When you are in the quantum spin Hall regime you will never stop conducting as your Fermi energy goes from the electron (or hole) sub-band to the gap; this is due to the topologically protected (due to time-reversal symmetry) edge states inside the bulk gap (here bulk means not on the edge of the well). In a bulk sample (bulk meaning not quantum confined) you would probably have performed some sort of controlled doping (assuming the gap has already been induced somehow) to control your Fermi energy. In that case you would probably have to fabricate different samples for different values of Fermi energy; that’s certainly very inconvenient.
In summary, you need to somehow induce a gap in HgTe, by either quantum confinement or induced strain to turn it into a 2D or 3D topological insulator. CdTe is not responsible for the key physics, i.e. band inversion, which gives rise to a topologically nontrivial band structure in HgTe. It is interesting to note that the HgTe quantum well was not the first proposal by Bernevig, Hughes, and Zhang. The experimental difficulty of working with strained HgTe led them to revise their proposal and predict a topological insulator in the quantum well instead! This was back in 2006; people have now managed to experimentally create 3D topological insulators out of strained HgTe.
All three questions can be answered by first artificially separating the graphene sheet into two sheets:
- (a) first sheet with only spin up electrons, and
- (b) second sheet with only spin down electrons.
This statement alone should partially answer your third question; for the sake of organization, however, I will repeat a summary of this paragraph (in the end) anyway. This step of artificially separating spin species cannot be done unless $s_{z}$ is conserved. Spin-orbit coupling can be interpreted as a form of “spin scattering” which couples states with different spin. If different spin states are not decoupled then decoupling the sheet into (a) and (b) would not faithfully represent the original system. Hence conservation of $s_{z}$ is a necessary condition.
Now, according to the last paragraph of the left column (same page), the authors (indirectly) say that these two sheets independently realize Haldane’s model for spinless electrons; this is nothing but a lattice realization of the quantum Hall effect with zero net magnetic field. We can now apply Laughlin’s argument to the two sheets independently. There is, however, one thing to watch out for: the signs of the gaps for the spin up ($s_{z}=+1$) and down ($s_{z}=-1$) electrons are opposite. Note: in Eq. (3) you will either get $\pm \Delta_{{\rm so}}$ ($s_{z}=\pm 1$). Hence the transverse pumping of spins will occur in opposite directions for spin up and down electrons. Kane and Mele say the same thing (in different words) just a few lines above Eq. (5). Consequently, an up spin of $\hbar/2$ is pumped from (say) edge 1 to edge 2 for sheet (a) and a down spin of $\hbar/2$ is pumped from edge 2 to edge 1 for sheet (b). Hence a net spin of $\hbar$ is pumped from one edge to the other regardless of which you choose to label as “up” or “down” (or 1 or 2). Note: $\lambda_{R}$ is still assumed to be zero. That should answer your first question.
Note that in the paragraph above Eq. (6) the authors say “...adiabatically insert a quantum $\phi=h/e$ of magnetic flux quantum down the cylinder (slower than $\Delta_{{\rm so}}/\hbar$).” This means that the longitudinal electric field does not impart enough energy, to an electron in the highest occupied Landau level, such that it can overcome the mobility gap (in the case of the integer quantum Hall effect). Hence the only way a state is available for the pumped electron (or spin), on the other edge, is if it had sub-gap states. In other words, the edges are gapless.
I apologize for messing up the order of the questions; my explanation required this order (no pun intended). Anyways, here’s a summary:
- The pumping of spins can be explained by using the same gauge invariance in the Laughlin argument. This is much easier to see once you split your system into two spinless systems with each experiencing opposite effective magnetic fields.
- A system with the lack of over-the-gap excitations, while still permitting sub-gap transport, implies the existence of gapless edge states.
- $s_{z}$ conservation is necessary for decoupling spin up and spin down species.
I hope that helped.
Best Answer
Yes, König et al. did indeed measure charge conductance in CdTe/HgTe/CdTe quantum wells. I think your dilemma is a result of mixing up the description of the properties of the quantum spin Hall insulator with and without the existence of external bias. The intuitive picture of counter propagating edge states with opposite spins, that is repeatedly discussed in the literature, is without external bias. Imagine a HgTe 2D layer (in the inverted regime) just sitting there without anyone doing anything to it. Focusing on (say) the top edge, you have (say) $|\mathbf{k},\uparrow\rangle$ propagating to the right (with conductance $e^2/h$) and it's Kramer's partner $|-\mathbf{k},\downarrow\rangle$ propagating the left. In the absence of an external bias the Fermi levels of both states are equal. That is why the charge conductance is $\sigma^{\text{charge}} = e^2/h + (-e^2/h) = 0$ (where the minus sign comes from the fact that the current from the $|-\mathbf{k},\downarrow\rangle$ state is flowing in the opposite direction). However, a spin up current going in one direction (say $+\hat{x}$) and the same magnitude of spin down current going in the other direction ($-\hat{x}$) is equivalent to twice the amount of spin up current in the $+\hat{x}$ direction. That's why you end up with a spin conductance of $2e^2/h$.
Now, in the König et al. transport experiment the charge currents due to $|\mathbf{k}_{\text{top}},\uparrow\rangle$ and $|-\mathbf{k}_{\text{top}},\downarrow\rangle$ do not cancel each other perfectly. In other words (say) on the top edge the quasi-Fermi level of $|\mathbf{k},\uparrow\rangle$ is greater than the quasi-Fermi level of $|-\mathbf{k},\downarrow\rangle$. This difference in Fermi levels would correspond to a net flow of electrons in the $+\hat{x}$ direction along the top edge. This net flow gives rise to a conductance of $e^2/h$. On the bottom edge, however, quasi-Fermi level of $|\mathbf{k}_{\text{bottom}},\downarrow\rangle$ is greater than the quasi-Fermi level of $|-\mathbf{k}_{\text{bottom}},\uparrow\rangle$. Thus you again have a net flow of electrons in the $+\hat{x}$ direction along the bottom edge. This gives rise to another channel with conductance $e^2/h$. Thus the total contribution would be $2e^2/h$. What I just described above holds for a two-terminal resistance (or conductance) measurement. If I pass a current $I$ between the two contacts then the voltage (proportional to the difference in Fermi levels in the two contacts) would be $V=(h/2e^2)I$. A way to quantify this analysis is using the Landauer-Büttiker formula $$I_{i}=\frac{e}{h}\sum_{j}\left(T_{ji}\mu_{i}-T_{ij}\mu_{j}\right)$$ where the quantities with single subscript index are indicating which contact these quantities belong to. For example consider a six terminal device as follows
You can observe the counter propagation of the spin up (say red) and spin down (blue) along the top and bottom edges. The quantity $T_{ij}$ represents the transmission probability for the electron to go from contact $i \rightarrow j$. As you can observe from the figure only $T_{i,i+1}$ and $T_{i+1,i}$ will be nonzero. As the theory for the quantum spin Hall effect predicts that the edge states are robust to (non-magnetic) disorder, we must have $$T_{i,i+1}=T_{i+1,i}=1$$ i.e. perfect (dissipationless) transmission. Plugging this into the Landauer-Büttiker formula (and assuming a current passing from contact $1 \rightarrow 4$) above you'll get six linear equations in six unknowns ${\mu_i}$: $$\frac{e}{h}\underbrace{\left(\begin{array}{cccccc} -2 & 1 & 0 & 0 & 0 & 1\\ 1 & -2 & 1 & 0 & 0 & 0\\ 0 & 1 & -2 & 1 & 0 & 0\\ 0 & 0 & 1 & -2 & 1 & 0\\ 0 & 0 & 1 & 1 & -2 & 1\\ 1 & 0 & 0 & 0 & 1 & -2 \end{array}\right)}_{A} \underbrace{\left(\begin{array}{c} \mu_{1}\\ \mu_{2}\\ \mu_{3}\\ \mu_{4}\\ \mu_{5}\\ \mu_{6} \end{array}\right)}_{x} = I_{14}\underbrace{\left(\begin{array}{c} 1\\ 0\\ 0\\ -1\\ 0\\ 0 \end{array}\right)}_{b}.$$
However, there is a redundancy in this system of equations (or $\det\left(A\right)=0$). Not all the $\mu_i$'s are really unknowns. We can set $\mu_4 = 0$ (i.e. reference potential or ground). In that case you can reduce the system of equations to $$\frac{e}{h}\left(\begin{array}{ccccc} -2 & 1 & 0 & 0 & 1\\ 1 & -2 & 1 & 0 & 0\\ 0 & 1 & -2 & 0 & 0\\ 0 & 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 & -2 \end{array}\right)\left(\begin{array}{c} \mu_{1}\\ \mu_{2}\\ \mu_{3}\\ \mu_{5}\\ \mu_{6} \end{array}\right)=I_{14}\left(\begin{array}{c} 1\\ 0\\ 0\\ -1\\ 0 \end{array}\right).$$ Now, solving this you get $$\left(\begin{array}{c} \mu_{1}\\ \mu_{2}\\ \mu_{3}\\ \mu_{5}\\ \mu_{6} \end{array}\right)=\frac{I_{14}h}{e}\left(\begin{array}{ccccc} -2 & 1 & 0 & 0 & 1\\ 1 & -2 & 1 & 0 & 0\\ 0 & 1 & -2 & 0 & 0\\ 0 & 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 & -2 \end{array}\right)^{-1}\left(\begin{array}{c} 1\\ 0\\ 0\\ -1\\ 0 \end{array}\right),$$ $$\left(\begin{array}{c} \mu_{1}\\ \mu_{2}\\ \mu_{3}\\ \mu_{5}\\ \mu_{6} \end{array}\right)=\frac{I_{14}h}{e}\left(\begin{array}{c} -3/2\\ -1\\ -1/2\\ -1/2\\ -1 \end{array}\right).$$ Voltage difference across contacts $i$ and $j$ is $$V_{ij}=\frac{1}{\left(-e\right)}\left(\mu_{i}-\mu_{j}\right).$$ You can check that $$V_{14}=\left(\frac{3h}{2e^{2}}\right)I_{14}$$ and $$V_{23}=\left(\frac{h}{2e^{2}}\right)I_{14}.$$
This is exactly what Roth et. al experimentally observed
These values of two- and four-terminal resistance were confirmed to be independent of the sample geometry ($L$ and $W$). As a result, you can rule out any type of conduction other than the edge states. Additionally, you expect these values of resistance only if you assume helical and dissipationless edge states. Therefore these measurements should confirm the existence of the quantum spin Hall state in HgTe.
If you had (say) two pairs of counter propagating helical edge states then such a system is not robust to disorder. Such a situation is shown in part (a) (in the absence of disorder) in the figure below. The states belonging to the red and blue bands at the same height (i.e. same energy $E(\mathbf{k},\uparrow)=E(-\mathbf{k},\downarrow)$) form Kramers' partners. The brown shaded regions represent the bulk bands. Since you have two pairs of Kramers' partners, the respective bands will naturally intersect at two points (as shown in part (a)). Any kind of disorder will result in gapping out of the states as shown in part (b). But note that such a gapping out process is permitted by Kramers' theorem. A quick way to see this is: look at the reflection of any band with respect to the vertical ($k=0$) axis. Under such reflections red should transform into blue and vice versa.
Now, imagine that you had two copies of Dirac-like helical edge states. In other words, two copies of part (d) superimposed on one another. When you gap out the system it will look like part (c). You can observe that in the part (c) time reversal symmetry is still preserved after gapping. In part (d), however, you only have one copy of Kramers' partners. There is only one point of intersection (as opposed to parts (a) and (b)). You can observe that there is no way in which you can open a gap (at $k=0$) while still satisfying time reversal symmetry. More specifically, introducing a gap (in part (d)) will only violate Kramers theorem at the $k=0$ point (i.e. $E(\mathbf{k},\uparrow)$ and $E(-\mathbf{k},\downarrow)$ will not be equal at $k=0$). Hence if the disorder respects time reversal symmetry then such a band intersection is said to be "protected by time reversal symmetry." In realistic systems like the HgTe quantum well, say you had $2n+1$ Kramers' partners. In that case disorder will destroy such Kramers' partners in $n$ pairs such that only one pair is left in the end. The existence of odd number of pairs is guaranteed in a topologically nontrivial phase. In fact, that is how people identify a topologically nontrivial phase.
You should read the lines before the above ones. The authors mentioned that spin is not a good quantum number. When you introduce spin orbit coupling the Hamiltonian is diagonal in the total angular momentum basis. The total angular momentum can be defined as $$\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$$ and you can label the eigenstates as $|j,m_j,s\rangle$ where $\hbar j(j+1)$, $\hbar m_j$, and $\hbar s$ are the eigenvalues of $\hat{\mathbf{J}}^2$, $\hat{J}_z$, and $\hat{S}_z$. In the bulk of HgTe $m_j$ is a good quantum number instead of $s$. On the edge, however, even $m_j$ is not conserved in due to lack of rotational symmetry. I think that what the authors are trying to do is emphasize the difference between the quantized spin Hall effect and the quantum spin Hall effect. As I will describe shortly, quantized spin Hall effect is not possible. For example, say you are trying to observe the spin analogue of the integer quantum Hall effect. You pass a longitudinal current in a ferromagnetic material (where spin is conserved) then you would get the same integer steps $\mathbb{Z}$. There will be both spin and charge accumulation in the transverse direction. Heuristically this is sort of like a hybrid between spin and charge Hall effects. This is also known as the quantum anomalous Hall effect. But note that time reversal symmetry is broken in such a system. The moment you introduce time reversal symmetry $\mathbb{Z}$ collapses to $\mathbb{Z}_2$. If you had a pure spin Hall effect, i.e. there is spin but no charge accumulation in the transverse direction, then time-reversal symmetry is preserved. Therefore you will never observe a quantized (or $\mathbb{Z}$) spin Hall effect.