The Hamiltonian for graphene at $\vec{k}$ away from the $K$ point is proportional to
$$ \vec{\sigma} \cdot \vec{k} =\begin{pmatrix} 0 & k_x - i k_y \\ k_x + i k_y & 0 \\ \end{pmatrix} = k \begin{pmatrix} 0 & e^{-i \theta} \\ e^{i \theta} & 0 \\ \end{pmatrix} $$ where $\theta=\tan^{-1}(k_y/k_x)$ and $k^2=k_x^2+k_y^2$
The helicity operator is $\vec{\sigma} \cdot \vec{k}/k$, which is proportional to the Hamiltonian for a fixed k, so the eigenvectors of the Hamiltonian are also helicity eigenstates. These eigenstates are $(e^{-i \theta/2},\pm e^{i \theta/2})$. So, in graphene lingo, we say that the pseudospin is locked parallel or antiparallel to the momentum.
The Hamiltonian away from the $K'$ point is proportional to $$ \vec{\sigma}^* \cdot \vec{k} =\begin{pmatrix} 0 & k_x + i k_y \\ k_x - i k_y & 0 \\ \end{pmatrix} = k \begin{pmatrix} 0 & e^{i \theta} \\ e^{-i \theta} & 0 \\ \end{pmatrix} $$
Then, it's eigenvectors are $(e^{i \theta/2},\pm e^{-i \theta/2})$. The literature claims that these are helicity eigenstates. However, these aren't eigenvectors of the matrix $\vec{\sigma} \cdot \vec{k}$ above, so they can't be helicity eigenstates. What stupid mistake am I making here?
Also, does anyone have a geometric picture of what the direction of the pseudospin even means? I know the angle $\theta$ is the phase of sublattice B relative to A, but does anyone have any intuition for what it means for the pseudospin to point in the same direction as the momentum?
