Global symmetries corresponding to the Altland-Zirnbauer symmetry classes

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There seems to be two ways of defining the global symmetries corresponding to the 10 Altland-Zirnbauer classes.

(1) via the nature of time reversal ($\mathcal{T}$), charge conjugation ($\mathcal{C}$) and chiral  ($\mathcal{S}$) symmetries that may be present in the system- $\mathcal{T}^2 = (\pm 1)^{N_f}$ and so on as listed in Section 2 of this review.
(2) via a different set of global symmetries which have fermion parity as a $Z_2$ subgroup as listed in this paper

I am confused by this. For example, systems belonging to Class A is considered to have no symmetry according to  (1) but has a U(1) symmetry according to (2) and Class D is considered to have  $\mathcal{C}$ with $\mathcal{C}^2 = 1$ symmetry according to (1) but only fermion-parity "symmetry" according to (2).

Q1: Is there a way in which I can relate the two definitions?
Q2: Can I understand why the 10 global symmetries of (2) are special from the point of view of the AZ classes?

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The difference is whether you consider the symmetries of the *single-particle* or *many-particle* (Fock) Hilbert spaces.

Consider, for example, class A. An example of a free-fermion Hamiltonian in this symmetry class, acting in the Fock space, would be

$$H_{mb} = \sum_{i,j \in \mathcal{A}} t_{i,j} \hat{c}_i^{\dagger} \hat{c}_j$$

where $\hat{c}_i^{\dagger}, \hat{c}_i, i \in \mathcal{A}$ are the fermionic creation and annihilation operators, and $\mathcal{A}$ is a set indexing all the fermionic modes. It is readily seen that this Hamiltonian has a $U(1)$ symmetry, since the total particle number $\sum_i c_i^{\dagger} c_i$ is conserved.

However, because the Hamiltonian $H_{mb}$ is non-interacting, it is equivalent to consider the single-particle Hamiltonian

$$H_{sp} = \sum_{i,j \in \mathcal{A}} t_{i,j} |i\rangle \langle j |,$$

which acts in the single-particle Hilbert space of dimension $|\mathcal{A}|$. The eigenstates of $H_{mb}$ are just indexed by the occupation numbers $n_1, n_2, \cdots$ of the single-particle orbitals (eigenstates of $H_{sp}$). This single-particle Hamiltonian indeed does not have a non-trivial $U(1)$ symmetry, since the particle number in this space is just a scalar, namely 1.

Similarly, in class AII, the single-particle Hamiltonian $H_{mb}$ has an anti-unitary symmetry $\mathbb{T}$ such that $\mathbb{T}^2 = -1$. In the many-body Hilbert space, the corresponding statement is that $\mathbb{T}^2 = (-1)^{\hat{N}_f}$, where $\hat{N}_f$ is the fermion number. The reason is that the sector of Fock space with a fixed particle number $n$ is basically built out out of $n$ tensor products of the single-particle Hilbert space (subject to anti-symmetrization), and therefore you get $n$ factors of $-1$.

Class D and the other ones with "particle-hole symmetry" are more subtle. In many-body language, the statement is that we allow ourselves to add superconducting pairing terms which violate particle number conservation, for example

$$\sum_{i,j} \Delta_{i,j} c_i c_j + h.c.$$

Thus, in the many-body language, class D has only fermion parity symmetry, because the $U(1)$ gets broken down. It might not be obvious at this point why this corresponds to particle-hole symmetry in the single-particle language. (Indeed, it is rather surprising that *reducing* the symmetry in the many-body language could *increase* the symmetry in the single-particle picture!) I don't have a great intuitive understanding here, but mathematically it comes about from the fact that, to convert a many-body (but non-interacting, i.e. quadratic) Hamiltonian containing pairing terms to a single-particle one, we first have to get rid of the pairing terms through a Bogoliubov transformation. This involves introducing redundant degrees of freedom, and as a consequence it turns out that the resulting Hamiltonian always has particle-hole symmetry. You perhaps shouldn't think of this as a "real" symmetry, it's more like a "gauge" symmetry in the sense that it results from a redundancy in a certain description of the system.

In answer to question (2), I think that once you go to interacting systems (as in the second reference in the paper), there is nothing that distinguishes the AZ symmetry classes from any other symmetries. Which is why the classification of interacting SPTs is much richer than free-fermion SPTs.

answered Jan 12, 2018 by Dominic Else
edited Jan 12, 2018

Hey Dominic. Is the particle-hole symmetry you're talking about the same as the Majorana condition on spinors? Or perhaps some more general reality condition?

No, the particle-hole symmetry is an anti-unitary operator which commutes with the Hamiltonian and sends $c^{\dagger} \leftrightarrow c$ as described in https://arxiv.org/pdf/1512.08882.pdf. I don't think it has anything to do with the Majorana spinor condition.

Hmm. I thought that the condition on the Nambu spinor in the BdG equation was equivalent to the Majorana condition in the Dirac basis somehow.

Still confusing also because the T-invariance structure corresponds to the old triality real-complex-quaternionic for representations and the behavior of T on relativistic spinors also reflects Bott periodicity.

Hi Dominic. Thanks. I understand what is happening better. I just want to clarify one thing- does each AZ class correspond to a unique global symmetry? Or are there different symmetry groups that in the non-interacting limit correspond to the same class?

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In a series of lectures, "Aspects of Symmetry and Topology for Free-Fermion Ground States with Disorder", Martin Zirnbauer has emphasized that the "Altland-Zirnbauer" classes used in the periodic table for topological insulators (that would be your reference (1)) are not, in fact, the same classes that Alexander Altland and he originally considered in the context of disordered Hamiltonians!

I forgot the details, but there is a non-trivial mapping from the original Altland-Zirnbauer classes to the Tenfold Way (as he likes to call it) of the periodic table for topological insulators, and it involves extra symmetries, like SU(2) and so on. I haven't looked closely, but this leads me to conjecture that reference (2) might use the original definition of the classes, hence the hubub with additional symmetries.

For the precise relation, I would recommend the lecture notes I have linked to above. I am confident that your question will be answered there.

answered Jan 14, 2018 by (775 points)

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