What is the time reversal operator in graphene (do not consider spin)?

Question

For graphene in one valley, the low energy Hamiltonian writes as:

$$ H_K(q)=\sigma_xq_x+\sigma_y q_y $$

The Hamiltonian on the other valley I remember has two ways of writing(not 100 percent sure), one is: $$ H_{K'}(q)=-\sigma_xq_x+\sigma_y q_y $$

the other is: $$ H_{K'}(q)=-\sigma_xq_x-\sigma_y q_y $$

The two $H_{K'}(q)$ are related to $H_K(q)$ by time reversal operation, I found that the operator for first one is $T_1=\sigma_zK$, for the second is $T_2=-i\sigma_yK$, where $K$ is the complex conjugate operation.

However $T_1^2=I$ while $T_2^2=-I$, which seems to have contradiction, since that spinless fermion has time reversal operator $T^2=I$ and 1/2 spin fermion has $T^2=-I$.

So what is going wrong? What is the time reversal operator for graphene?

Answer 1

For the pristine graphene, $$ d_x(\vec{k}) = - t_1 \cos k_y - 2 t_2 \cos \frac{\sqrt{3} k_x }{2} \cos \frac{k_y}{2}\,, d_y(\vec{k}) = + t_1 \sin k_y - 2 t_2 \cos \frac{\sqrt{3} k_x }{2} \sin \frac{k_y}{2},$$ at the two valleys, $\vec{K}^\pm = ( \pm \frac{4 \pi}{3 \sqrt{3}},0)$, we have $$H_{K^\pm}(q)=v_F(\pm\sigma_xq_x+\sigma_y q_y),$$ if we do the derivation from the tight-binding honeycomb lattice model. Thus, your first form, the correct one, seems consistent with $\mathcal{T}^2=1$ you expected.

I feel doubtful about the second one you gave since the chirality remains the same as $H_{K}(q)$.