# Helicity and Pseudospin in Graphene

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?

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Note that as opposed to the case of neutrinos where the Dirac-Weyl equation is unambiguous, the effective equation for electrons in graphene has some ambiguity. Specifically it depends on the orientation of the axis with respect to the graphene lattice, and on the implied basis (which is often not explicitly written). So people just redefine the helicity operator in the other valley or they change the basis. For example if you happen to choose the $x$ axis along the zigzag direction, and your basis as $\left(\Psi_{AK},\Psi_{BK},\Psi_{BK^{\prime}},\Psi_{AK^{\prime}}\right)^{T}$ you get a neat Hamiltonian (arXiv:1004.3396)
$$H=\hbar v_F\tau_z\otimes\boldsymbol{\sigma}\cdot\mathbf{k}$$
where $\tau_z$ is the Pauli matrix in the valley space. Then you can use the same definition for the helicity operator in both valleys
$$h=\frac{\boldsymbol{\sigma}\cdot\mathbf{k}}{|\mathbf{k}|}$$