Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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

$$ h=\frac{\boldsymbol{\sigma}\cdot\mathbf{k}}{|\mathbf{k}|} $$

As far as the interpretation of the pseudospin goes, it is a degree of freedom related to the presence of the two sublattices as you say. It is helpful to consider it since in the absence of sublattice symmetry breaking term it is conserved, hence Klein tunneling. However despite the often claimed fact that it just resembles spin, it seems that pseudospin actually does carry some real angular momentum (arXiv:1003.3715)! Apparently this happens because 2D Dirac-Weyl equation leaves out certain angular momenta, i.e. graphene is not exactly 2D as we treat it to be.

share|improve this answer
    
Thanks for the reply! If we can tweak the basis to make both $K$ and $K'$ fit the definition of the helicity operator, does that mean the concept of helicity/chirality is not so useful when discussing scattering between the two valleys? –  ChickenGod Dec 6 at 23:25
1  
I wouldn't say helicity is useful when talking about intervalley scattering. The character of the scattering potential is crucial there; short-range varying potentials can cause intervalley scattering, while potentials varying slowly on the lattice scale cannot. Helicity as a concept is more important in the latter case, where it tells us that normal, intravalley, backscattering is forbidden since it would require flipping the pseudospin. –  mgphys Dec 18 at 12:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.