In the section “Hamiltonians, topology, and symmetry”, it has mentioned the topological invariant Q{BdG} of BdG Hamiltonian under particle-hole symmetry, which is defined as Q{BdG} = sign[Pf(i H{BdG})]. However, the given H{BdG} is not skew-symmetric so that the Pfaffian of H{BdG} can not be calculated. According to the topological analysis demonstrated in this section, I think it should be \tilde{H}{BdG}, namely, Q{BdG} = sign[Pf(i \tilde{H}{BdG})]. Is that right?
akhmerov
commented
2 years ago
In the section “Hamiltonians, topology, and symmetry”, it has mentioned the topological invariant Q{BdG} of BdG Hamiltonian under particle-hole symmetry, which is defined as Q{BdG} = sign[Pf(i H{BdG})]. However, the given H{BdG} is not skew-symmetric so that the Pfaffian of H{BdG} can not be calculated. According to the topological analysis demonstrated in this section, I think it should be \tilde{H}{BdG}, namely, Q{BdG} = sign[Pf(i \tilde{H}{BdG})]. Is that right?