The Hamiltonian of the honeycomb lattice is $$ H=\sum_{k\sigma}t(k) a_{k\sigma}^\dagger b_{k\sigma}+h.c $$ Where $t(-k)=t^*(k)$.

If we do a time reversal transformation(according the answer to this post): $a_{k\uparrow}\to a_{-k\downarrow}$, $a_{k\downarrow}\to -a_{-k\uparrow}, t(k)\to t(-k)$. 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 $\mathcal{T}=i\sigma_y K$, time reversal symmetry means the Hamiltonian commutes with the time reversal operator. This leads to $\sigma_y H \sigma_y =H^*$.

My question is 1) How is $\sigma_y$ 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 $\vec{A}=A(\sin \Omega t,0)$, similarly to this article.

The Hamiltonian can be written as: $$ H=\sum_{k\sigma}t(k-A) a_{k\sigma}^\dagger b_{k\sigma}+h.c $$

Again if we do a time reversal transformation: $a_{k\uparrow}\to a_{-k\downarrow}$, $a_{k\downarrow}\to -a_{-k\uparrow}, t(k-A)\to t(-k+A)$. We can also show that the Hamiltonian is invariant.

However, if we represent the light by a vector potential $\vec{A}=A(\cos \Omega t,0)$, then a time reversal transformation leads to $t(k-A)\to t(-k-A)$. 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.

  • $\begingroup$ How is your $w(k)$ related to $t(k)$? $\endgroup$ – Everett You Apr 28 '15 at 19:41
  • $\begingroup$ @EverettYou They are same, I've corrected the typos $\endgroup$ – an offer can't refuse Apr 29 '15 at 2:40

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