# Why Integer Quantum Hall Effect (IQHE) can only happen in even dimensions?

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?

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.

• stacking opposite copies of "IQH" states is weak TI (WTI) – Wenjie Zhang Dec 30 '20 at 16:21