# Helicity in Graphene

In Graphene, there are two independent points, the Dirac points, where the conduction and the valence band touch. Let's call these points $K_+$ and $K_-$. In a low-energy description, the Hamiltonians at $K_+$ and $K_-$ are given by

$$H_+ = v_F \vec{\sigma}^*\cdot\vec{p}\qquad\text{and}\qquad H_- = v_F \vec{\sigma}\cdot\vec{p}$$

with $v_F$ the Fermi velocity of the electrons and $\vec{\sigma} = (\sigma_x,\sigma_y)$ the vector of Pauli matrices. $\vec{\sigma}^*$ is the complex conjugate of $\vec{\sigma}$, so $\vec{\sigma}^*=(\sigma_x^*,\sigma_y^*)=(\sigma_x,-\sigma_y)$. Let the eigenfunctions of the Hamiltonians be $\Psi_+$ and $\Psi_-$, defined by

$$\Psi_\pm=\left(\begin{array}{c}e^{\pm\frac{i \phi_p}{2}}\\ \xi e^{\mp \frac{i\phi_p}{2}} \end{array} \right)e^{\frac{i}{\hbar}p\cdot x},$$

where $\xi=\pm$ stands for the valence band ($\xi=-1$) and for the conduction band ($\xi=+1$).

In the literature it is often said that for the helicity operator $$h = \frac{\vec{\sigma}\cdot\vec{p}}{|\vec{p}|}$$ with eigenvalues $\pm1$ the following equations hold:

$$h\Psi_+ = \pm\Psi_+\qquad\text{and}\quad h\Psi_-=\mp\Psi_-.\qquad\qquad(1)$$

This is interesting because scattering of electrons from one Dirac point to the other is forbidden because helicity is a conserved quantity. The thing is that I don't understand why the signs of the eigenvalues of $h$ are reversed. I would like to know how to mathematically derive the relations in $(1)$ from the Hamiltonians given above. Hopefully someone can show me this.

• Since you wrote $h\Psi_+ = \pm \Psi_+$ and $h\Psi_- = \mp \Psi_-$, $\Psi$ always refers to 2 eigenfunctions, right? Otherwise, they could not produce both eigenvalues... Can you please clarify the difference between $\vec{\sigma}$ and $\vec{\sigma}*$ Is it the complex conjugate? This will then give you the eigenfunctions and explain equations (1). – lmr Jun 5 '18 at 11:58
• @lmr I edited my post and clarified the points you mentioned. – MeMeansMe Jun 5 '18 at 12:29
• I updated the answer. Looks like we need your original source though... – lmr Jun 5 '18 at 19:57

We'll start with the Hamiltonians $H_+$ and $H_-$. For a given momentum $|\vec{p}| = p$, they are proportional to the helicity operator $h_- = \vec{\sigma} \cdot \vec{p} /p$ in case of $H_-$ and $h_+ = \vec{\sigma}^* \cdot \vec{p} /p$ in case of $H_+$. That means, we will find the eigenstates of $H_\pm$ by finding the eigenstates of $h_\pm$. Note that the helicity operator $h$ you defined actually corresponds to $h_-$ and is not proportional to both $H_+$ and $H_-$.

Using the Pauli matrices, $h_\pm$ can be written in matrix notation as: $$\begin{equation} h_\pm = \frac{1}{p} \begin{pmatrix} 0 & p_x \pm ip_y \\ p_x \mp ip_y & 0 \end{pmatrix} \end{equation}$$ with the momentum components $p_x$ and $p_y$. For easier handling, let's express momentum in the polar form:

$$\begin{equation} h_\pm = \frac{1}{p} p \begin{pmatrix} 0 & e^{\pm i\phi_p} \\ e^{\mp i\phi_p} & 0 \end{pmatrix} = \begin{pmatrix} 0 & e^{\pm i\phi_p} \\ e^{\mp i\phi_p} & 0 \end{pmatrix} \end{equation}$$ with the polar angle $\phi$. With this, we can easily determine the eigenfunctions and eigenvalues. To avoid confusion concerning the $\pm$ sign, I'll write them down seperately.

The eigenfunctions $\psi_+$ and eigenvalues $\lambda_+$ to $h_+$ are: $$\begin{equation} \psi_+ = \begin{pmatrix} e^{i\phi_p/2} \\ \pm e^{-i\phi_p/2} \end{pmatrix} , \quad \lambda_+ = \pm 1 \end{equation}$$

The eigenfunctions $\psi_-$ and eigenvalues $\lambda_-$ to $h_-$ are: $$\begin{equation} \psi_- = \begin{pmatrix} e^{-i\phi_p/2} \\ \pm e^{i\phi_p/2} \end{pmatrix} , \quad \lambda_- = \pm 1 \end{equation}$$

You can easily verify this plugging all this into the eigenequation $h_\pm \psi_\pm = \lambda_\pm \psi_\pm$. Note that the eigenvectors $\psi_\pm$ are proportional to the $\Psi_\pm$ you defined in your post. So this should answer the first part of your question on how to derive all the relevant relations mathematically.

Now on to your equation (1): While the eigenvalues $\lambda_-$ and $\lambda_+$ are identical, the eigenvectors are not. $\psi_+$ is not an eigenvector of $h_-$. But your equation (1) seems to imply exactly that: $$\begin{equation} h_- \Psi_+ = \pm \Psi_+ \end{equation}$$ Which does not make sense with the above definitions. There are several possibilities that can be investigated. Where exactly did you get the expression? As was stated in this answer, people like to redefine the helicity operator in the different valleys. In order to answer why the eigenvalues might be reversed, it is necessary to know what exactly $h$ and the $\Psi_\pm$ refer to. Please be careful with the $\pm$ definition. In this case here, it is difficult to distinguish between eigenvalues and the sign change that corresponds to the different valleys. So as I've shown above, equations (1) do not work with your initial definitions. There needs to be some extra information... Maybe your original source can help?

"I don't understand why the signs of the eigenvalues of h are reversed." The eigenstates of the helicity operator are the ones where $\vec{\sigma}$ and $\vec{p}$ are either parallel or antiparallel. (Physically speaking: The pseudospin is parallel or antiparallel to electron/hole momentum. ) As a consequence, the dot product of $\vec{\sigma}$ and $\vec{p} / |\vec{p} |$ will yield $\pm 1$. So in your notation, $\Psi_-$ is the antiparallel case while $\Psi_+$ is the parallel case.
• This interpretation is clear to me, but how can I derive the equations in $(1)$ from the two Hamiltonians? I edited my post to make clear that that's what my question is about. – MeMeansMe Jun 5 '18 at 10:21
• I would like to make an edit to answer this question but I am not sure whether I understand your problem. Do you want to know how to go from $H$ to $h$ and its eigenvalues from a mathematical point of view? Do you want to know how to obtain the eigenvalues for $h$? The operator $h$ has got two eigenvalues: + and - 1 . Are you confused why one of the eigenfunctions yields $\pm 1$ and the other one yields $\mp 1$? You look for a mathematical answer, right? – lmr Jun 5 '18 at 11:37