Question

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

Answer 1

By the qft theorem you can show time reversal symmetry which does not break in the tight binding of Hamiltonnin for the honeycomb lattice because the qft theorem is totally program for the binding of hamiltonian for the Honeycomb

Answer 2

The Hamiltonian of the honeycomb lattice is

H=∑kσt(k)a†kσbkσ+h.c

Where t(−k)=t∗(k).

If we do a time reversal transformation(according the answer to this post): ak↑→a−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=iσyK, time reversal symmetry means the Hamiltonian commutes with the time reversal operator. This leads to σyHσy=H∗.

My question is 1) How is σy applied to H; can anyone show me how to do this calculation? 2) How can I combine these two view points, i.e. the transformation shown above and the commutation relation below?

Another confusion is: Does linearly polarized light break time reversal symmetry?

If we represent homogenous linearly polarized light by a vector field A⃗ =A(sinΩt,0), similarly to this article.

The Hamiltonian can be written as:

H=∑kσt(k−A)a†kσbkσ+h.c

Again if we do a time reversal transformation: ak↑→a−k↓, ak↓→−a−k↑,t(k−A)→t(−k+A). We can also show that the Hamiltonian is invariant.

However, if we represent the light by a vector potential A⃗ =A(cosΩt,0), then a time reversal transformation leads to t(k−A)→t(−k−A). The Hamiltonian seems not time-reversal invariant. How to solve this contradiction, since from a physical point of view, linearized light has no angular momentum, thus should not break time-reversal symmetry.