Why and how Dirac cones are "tilted"?

[devmessias]

Given a Weyl Hamiltonian, at rest,

\begin{align}
H = \vec \sigma \cdot \vec{p}
\end{align}

A Lorentz boost in the x-direction returns

\begin{align}
H = \vec\sigma\cdot\vec{p} - \gamma\sigma_0 p_x
\end{align}

The second term gives rise to a tilt in the "light" cone of graphene. My doubts are:

How I can derive such term given a Lorentz transformations in x direction? If the cone of "light" is "tilted" then the Fermi velocity has changed?


[Tilted anisotropic Dirac cones in quinoid-type graphene and α−(BEDT-TTF)2I3][1]

 

[eljose79]

To derive the second term in the Hamiltonian, we can use the transformation properties of the Weyl Hamiltonian under Lorentz boosts. The Weyl Hamiltonian is a spinor operator, so it transforms under boosts as a spinor. In particular, under a boost in the x-direction, the Weyl Hamiltonian transforms as:

\begin{align}
H' = \Lambda H \Lambda^{-1} = \begin{pmatrix} \gamma & -\gamma v_x & 0 & 0 \\ -\gamma v_x & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & \sigma_x p_x + \sigma_y p_y + \sigma_z p_z \\ \sigma_x p_x + \sigma_y p_y + \sigma_z p_z & 0 \end{pmatrix} \begin{pmatrix} \gamma & \gamma v_x & 0 & 0 \\ \gamma v_x & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}
\end{align}

Expanding this out and simplifying, we get:

\begin{align}
H' = \begin{pmatrix} 0 & \gamma(\sigma_x p_x + \sigma_y p_y + \sigma_z p_z) - \gamma v_x(\sigma_x p_x + \sigma_y p_y + \sigma_z p_z) & 0 & 0 \\ \gamma(\sigma_x p_x + \sigma_y p_y + \sigma_z p_z) - \gamma v_x(\sigma_x p_x + \sigma_y p_y + \sigma_z p_z) & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}
\end{align}

We can see that the second term in the Hamiltonian, $\gamma\sigma_0 p_x$, arises from the transformation of the $p_x$ term in the Weyl Hamiltonian. This term is responsible for the tilt in the "light" cone in graphene.

The tilt in the