If we write the BdG equation for any model, and double the degrees of freedom (e.g. 4N*4N matrix for a N site chain), then we are guaranteed the particle-hole symmetry. Is there any constraints to do this procedure for every model? Thanks!

PS: in topology class of BdG classes, this corresponds to class D, which is of the least symmetry in the BdG class.

  • $\begingroup$ This question is very tacit as it stands. Could you elaborate by adding some details? $\endgroup$ – AlQuemist Dec 15 '15 at 15:04
  • $\begingroup$ Yes, BCS Hamiltonians always have this symmetry. $\endgroup$ – Meng Cheng Dec 15 '15 at 16:26
  • $\begingroup$ The particle-hole symmetry is built in automatically by the doubling of degrees of freedom in order to diagonalize the mean-field Hamiltonian. It is an exact symmetry of any BdG Hamiltonian. $\endgroup$ – Praan Dec 15 '15 at 17:13
  • $\begingroup$ thanks, do we need some condition to write the BdG Hamiltonian, e.g. superconductivity? $\endgroup$ – Tian Binbin Dec 15 '15 at 17:44

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