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In BdG Hamiltonians, the particle-hole symmetry is not a true symmetry but rather a redundancy of description. In my oppinion, saying with the presence of particle-hole symmetry is just saying: hey, we are talking about superconductors!

However, almost in all the literatures, people are using BdG Hamiltonians. Since it is not the original Hamiltonian, why the topological property of it exactly reflect the original Hamiltonian. Is there any rigorious proof?

I think it would be perfectly OK to discuss the topological properties (and other physical properties) using $H_{BCS}$ rather than $H_{BdG}$, is it much more complicated to do so? I am always concerned about the side effect of using this "fake Hamiltonian". i.e. what seems right derived by using BdG Hamiltonian are actually wrong by carefully examing the original Hamiltonian?

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    $\begingroup$ There is nothing wrong. We perfectly understand how to translate between the BCS and the BdG Hamiltonian, and we understand that after we diagonalize the BdG Hamiltonian, we only need to keep track of the positive (or negative) energy eigenstates due to the particle-hole symmetry, and construct the ground state by "filling" all the negative energy states just like what we did in band insulator. The classification is about this ground state wavefunction as a vector bundle over the Brillouin zone and in this regard it is not so much different from band insulators. $\endgroup$
    – Meng Cheng
    Commented Jan 9, 2016 at 2:18
  • $\begingroup$ @MengCheng Thanks for the comment. In the ten-fold periodic table, are entries with non-zero particle-hole symmetry all represent superconductors with this "fake symmetry"? Or some of them are real symmetry for some kinds of Hamiltonians/systems? $\endgroup$
    – an offer can't refuse
    Commented Jan 9, 2016 at 3:01
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    $\begingroup$ The ten-fold way deals with the matrix Hamiltonians (some people call it "first-quantized" Hamiltonians) and does not care about the physical origin of the symmetry (only that the ground state is a vector bundle which is the mathematical model for the "filled bands below Fermi energy".). So yes, you can have non-superconducting Hamiltonians with a "real" particle-hole symmetry, falling into the same class, and the classification still applies. But then the interpretation at the level of second-quantized Hamiltonian is very different. $\endgroup$
    – Meng Cheng
    Commented Jan 9, 2016 at 3:09
  • $\begingroup$ As far as I can see, you're asking an interesting question using wrong arguments. There is no reason to question the particle-hole redundancy, it's inherent to the Cooper pairing, not to the superconducting state, though both are intimately related. The BdG Hamiltonian is just the mean-field version of the BCS model in the Cooper channel. Usually, you can prove that the mean-field approximation is consistent in the case of superconductors : there are no critical fluctuations detrimental to the BCS ground states. $\endgroup$
    – FraSchelle
    Commented Jan 11, 2016 at 9:24
  • $\begingroup$ So the interesting question I can see from your approach would be : is the non-trivial topological ground state of a superconductor correctly defined by the mean-field limit ? The question of the symmetry is of no interest in itself. For instance for the classification you're questioning, the presence of the particle-hole symmetry is not mandatory to get non-trivial topological states. $\endgroup$
    – FraSchelle
    Commented Jan 11, 2016 at 9:27

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I think when we talk about the symmetry of the first-quantized Hamiltonian, we could talk about it based on its matrix form alone.

But when we talk about the symmetry of the second-quantized Hamiltonian, it is insufficient to deduce the symmetry only from the Hamiltonian matrix form. We also need to define the ground state or the annihilation operators precisely.

In the BdG Hamiltonian, we are talking about the fermion system, so that the ground state is the filled fermi sea. In this case, we will have to redefine the annihilation and creation operators when the energy of the corresponding states is lower than the fermi level since we could go below the ground state energy.

By redefining the creation and annihilation operators we effectively "enlarge" the Hamiltonian matrix and equip it with particle-hole symmetry.

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