I read that Integer Quantum Hall Effect (IQHE) can only exist in even dimensions, while
Quantum Spin Hall Effect (QSHE) can be generalized to 3D (or rather any dimensions?). Does anyone have a hand-waving physical argument to understand this?
[Physics] Why Integer Quantum Hall Effect (IQHE) can only happen in even dimensions
condensed-matterquantum-hall-effect
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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.
If we solve the Schrödinger equation for an electron in a magnetic field in Landau gauge, the solutions are discrete bands, infinitely degenerate in transverse momentum. (In addition, there is free movement along the direction of the field, which in QHE is frozen by confining the electrin gas to two dimensions.) Since there are plenty of available momentum states, the system is conducting: it is not gapped, but rather the band is flat. This is conventional Hall effect.
Now, if there is an additional confining potential constraining the momentum states, or, if the magnetic field is so strong that we have to consider the sample edges as such confinement potential, we are left with one-dimensional edge channels. These exhibit conductance quantization, similar to that of the quantum point contacts, hence the QHE.
Best Answer
The simplest answer is that the generalizations of IQH states in odd dimensions are equivalent to trivial band insulators, so IQH states only exist in even dimensions. The QSH state can not be generalized to 3D, because you can not even define a spin Hall conductance in 3D in the first place. The proper generalization of QSH state in 3D is called a topological insulator (TI) which has the same symmetry as a QSH insulator in 2D. In 2D, you may construct a QSH state by stacking opposite layers of IQH states. But in 3D, a TI is not constructed by stacking opposite copies of "IQH" states, so there is no corresponding IQH state in 3D.
Let me rephrase your question more precisely. First let us restrict our scope to free fermion systems without interaction. Both IQH and QSH states are free Fermion Symmetry Protected Topological (fFSPT) states, a general concept for all topological insulators and superconductors. The IQH states are the fFSPT states with $U(1)$ symmetry, and QSH states should be generalized to the fFSTP states with $U(1)\rtimes Z_2^T$ symmetry where the time reversal squares to the fermion parity $\mathcal{T}^2=(-)^F$. These symmetries are ascribed to different symmetry classes (A and AII respectively). So your question is what is the classification of A class and AII class fFSPT states in all dimensions, and why.
The classification problem was solved in arXiv:0901.2686, arXiv:0905.2029, arXiv:0912.2157, and I copy the conclusions here for A and AII classes. $$\begin{array}{ccrrrrrrrrc} \text{class} & \text{symmetry} & d=0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \cdots\\ \text{A} & U(1) & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \cdots \\ \text{AII} & U(1)\rtimes Z_2^T (\mathcal{T}^2=(-)^F) & \mathbb{Z} & 0 & \mathbb{Z}_2 & \mathbb{Z}_2 & \mathbb{Z} & 0 & 0 & 0 & \cdots \end{array} $$ The A class (IQH) states are non-trivial (and therefore exist) only in even dimensions, and the classification has a periodicity of 2. The AII class (QSH,TI) states are non-trivial (and therefore exist) only in 0,2,3,4 (mod 8) dimensions, with the periodicity of 8, however the classification in 2,3 dimensions is only $\mathbb{Z}_2$ instead of $\mathbb{Z}$.
All these has to do with the properties of Clifford algebra. In the momentum space, the low-energy effective Hamiltonian for the IQH phase always takes the form of the Dirac Hamiltonian (under $k\cdot p$ approximation) $$H=\sum_k \psi_k^\dagger\left(\sum_{i=1}^d k_i \gamma_i+m \gamma_{d+1}\right)\psi_k,$$ where $\psi_k$ is the electron at momentum $k$ (which is a $d$-dimensional vector). $\gamma_i$ are matrices that anti-commute with each other, i.e. $\{\gamma_i,\gamma_j\}=\delta_{ij}$. Only if the matrices are all anti-commuting, the single-particle dispersion relation can take the form $\epsilon_k^2=k^2+m^2$ which is fully gapped (with the band gap $2|m|$). Therefore we must be able to find $(d+1)$ anti-commuting matrices in order to write down the Dirac Hamiltonian in $d$ dimensional space.
For example, in 2D ($d=2$), we need three anti-commuting matrices, which can be taken as $\gamma_i=\sigma^1,\sigma^2,\sigma^3$. In this case, $$H=\sum_k \psi_k^\dagger\left(k_1\sigma^1+k_2\sigma^2+m \sigma^3\right)\psi_k.$$ $m>0$ and $m<0$ respectively corresponds to the trivial and non-trivial fFSPT states, and the non-trivial one is an IQH state. By saying that the IQH state exists, we mean it is a distinct phase that can not be smoothly tuned to the trivial phase without closing the band gap and without breaking the symmetry. The symmetry here is $U(1):\psi_k\to e^{i\theta}\psi_k$. If we want to go from the IQH phase to the trivial phase without breaking the $U(1)$ symmetry, the only way is to tune the mass term $m$ from positive to negative, which must go through $m=0$ where the band gap closes (this is the point that the IQH phase goes through a bulk transition to the trivial phase). So we see the phase transition is inevitable, this makes the IQH phase "exists".
However if we go to 3D ($d=3$), we need to find four anti-commuting matrices. They can be chosen as $\gamma_i=\sigma^{11},\sigma^{12},\sigma^{13},\sigma^{20}$, where $\sigma^{ij}\equiv\sigma^i\otimes\sigma^j$. Then we can write down the Dirac Hamiltonian $$H=\sum_k \psi_k^\dagger\left(k_1\sigma^{11}+k_2\sigma^{12}+k_3\sigma^{13}+m \sigma^{20}\right)\psi_k.$$ But now $m>0$ and $m<0$ belongs to the same phase. Because I can always introduce the additional mass term $m'\sigma^{30}$ (due to the fact that there exist the 5th anti-commuting matrix $\sigma^{30}$), such that the dispersion relation reads $\epsilon_k^2=k^2+m^2+m'^2$. Then the band gap is bounded from below by $2|m'|$ as $m$ is tuned from positive to negative. So the candidate topological state can actually be smoothly connected the trivial state, and therefore there is no nontrivial A class fFSPT (or your so-called IQH) state in 3D.
If you go further to 4D, the anti-commuting matrices are used up in the Dirac Hamiltonian again, without any additional mass to be added. So you can have IQH states in 4D again. The fact that the $2^n\times 2^n$ matrix space supports at most $(2n+1)$ anti-commuting matrices tells us the IQH states only exist when $d+1=2n+1$, so $d$ must be even.