Question

How to show time reversal symmetry does not break in the tight binding Hamiltonian for the honeycomb lattice?

Answer 1

The Hamiltonian of the honeycomb lattice is H=∑kσt(k)a†kσbkσ+h.c H=∑kσt(k)akσ†bkσ+h.c Where t(−k)=t∗(k)t(−k)=t∗(k).  If we do a time reversal transformation(according the answer to this post): ak↑→a−k↓ak↑→a−k↓, ak↓→−a−k↑,t(k)→t(−k)ak↓→−a−k↑,t(k)→t(−k). We can show that the Hamiltonian is invariant.  On the other hand, if we recall that the time reversal operator for the spin-1/2 particles can be written as T=iσyKT=iσyK, time reversal symmetry means the Hamiltonian commutes with the time reversal operator. This leads to σyHσy=H∗σyHσy=H∗.