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Nov 21 at 6:08 comment added Vokaylop No, the assumption is incorrect. I should have explicitly said that my Hamiltonian is normal. For example, physically one would expect a spinless fermion with only nearest neighbor hopping on a bipartite lattice without magnetic flux to always has T and S, right? Regarding your "one can't guarantee that the Hamiltonian, after performing an anitunitary transfo., remains a bilinear form between the same two vectors": of course not, and you have a symmetry only when it does. Isn't that the whole point of this symmetry thing?
S Nov 19 at 16:19 review First answers
Nov 19 at 16:28
S Nov 19 at 16:19 history answered Bailey Bussiere CC BY-SA 4.0