Sublattice symmetry vs Particle hole symmetry

Question

Sublattice symmetry and particle hole symmetry generally constrain a system's energy spectrum to be symmetric with respect to fermi level. My understanding is that they are both represented by an operator with the property $OHO^{-1}=-H$. For sublattice symmetry, $O$ is unitary and linear while for particle hole symmetry, $O$ is antinunitary and antililear. Here I have two questions:

1. Sometimes in literature, if a system's energy spectrum is symmetric with respect to fermi level, it is said to have particle hole symmetry. However, according to my understanding, this can also be attributed to sublattice symmetry. Does the term "particle hole symmetry" have some generalizations here?

2. I have never seen particle hole symmetry(in my definition) in a condensed matter system other than superconductors. Can anyone offer me an example?

Thanks!

Answer 1

@1, Sublattice symmetry also goes by a different name: Chiral symmetry, but I've not heard of it referred to as a particle-hole symmetry before, I think wherever you came across this, might have been mistaken, I'd need more context to answer fully.

The best(simple) introduction to the three fundamental symmetries (Chiral, Time and True Particle-hole symmetry) in terms of the symmetry of some energy spectrum can be found here: https://topocondmat.org/w1_topointro/0d.html

@2, There are different kinds of particle hole-symmetry. The one you refer to in superconductors arises from doubling the degrees of freedom of your Hamiltonian in the Bogoliubov approach, where the mean field description of the quasi-holes and quasi-electrons are bound together by the superconducting coupling. In that case the particle-hole symmetry exists between the "cooper-pair" or if you prefer the composite boson. Where as in a single particle description of a normal conductor you may have particle-hole symmetry between the quasi-electrons and quasi-holes, themselves, a particle-hole symmetry acting on the fermionic operators.

Answer 2

Particle hole symmetry is charge conjugation symmetry. Charge conjugation implies $C H^* C = -H$ .Sublattice symmetry is Chiral symmetry. Chiral symmetry implies What you are thinking, namely $O H O^\dagger = -H$ with the additional fact that $O^2 = 1$.

Answer 3

If you think of it in momentum space it is easier to see the differences between them. There are three symmetries that are used to classify topological insulators: time reversal ($\hat{\mathcal{T}}$), particle-hole, also known as charge conjugation ($\hat{\mathcal{C}}$), and the combination of both, known as sublattice or chiral symmetry ($\hat{\mathcal{S}}$). For the second-quantized Hamiltonian, $\hat{H}=\sum_k \hat\psi_k^\dagger \mathcal{H}_k\hat\psi_k$, these operators fulfill a commutation relation, $\left[\hat{H},\;\hat{U}\right]=0$, with $\hat{U}$ any of the operators above. These operators act on the fermionic fields in different manners, but typically one can express their action as a transformation on the fields (e.g. for particle hole $\hat{C}$, it flips between creation and annihilation operators) plus a matrix mixing different internal states, e.g. spin states. This matrix, which I shall denote $\mathcal{U}_T$, $\mathcal{U}_C$ and $\mathcal{U}_S$, can be used to express the symmetry conditions using the "first-quantized" or single-particle Hamiltonian $\mathcal{H}$. The resulting conditions are

• $\mathcal{U}_T \mathcal{H}_k^{*}\mathcal{U}_T^\dagger=\mathcal{H}_{-k}$
• $\mathcal{U}_C \mathcal{H}_k^{*}\mathcal{U}_C^\dagger=-\mathcal{H}_{-k}$
• $\mathcal{U}_S \mathcal{H}_k\mathcal{U}_S^\dagger=-\mathcal{H}_{k}$

Using this, one finds that the restriction that these symmetries impose on the energy bands are

• Time reversal symmetry: $E_k=E_{-k}$.
• Particle-hole symmetry: $E_k=-E_{-k}$.
• Sublattice symmetry: $E_k=-E_{k}$.

This explains the differences between both symmetries.