Note, that there is an extra shift of the Landau level index from $n$ to $n-1$ for the $B$ sublattice part of the wave function.In the braket notation we write the states as spinors in the $A/B$ basis and perform a  rotation such that the Hamiltonian becomes diagonal\begin{align}\begin{pmatrix}\Ket{n,k_z,p,a} \\\Ket{n-1,k_z,p,b}\end{pmatrix}    &= U(n,k_z,p)    \begin{pmatrix}         \Ket{n,k_z,p,A} \\        \Ket{n-1,k_z,p,B} \end{pmatrix}\label{basistrafo}\end{align}where $U$ is a unitary transformation that is easily obtained from \eqref{weyl2-ll-hamiltonian}.The energies, labeled by $s\in a,b$ for  $\Ket{n,p,k_z,s}$ are given by \begin{align}    E_{n,a/b}(k_z) &= \hbar v_F \left(  -\eta k_z  + \gamma k_z^3 \pm \sqrt{(k_z + \beta k_z^3)^2 + \frac{2}{l_B^2}|n|} \right)    \label{ll-dispersion}\end{align}for $n\neq 0$.For $n=0$, the energy of the chiral level is\begin{align}    E_{0,k_z} &= \hbar v_F \left((1 - \eta) k_z + (\beta + \gamma) k_z^3 \right).\end{align}The Landau level degeneracy becomes apparent as $E$ does not depend on $p$.Note that the chiral mode corresponds to the $(1 \; 0)^T$ spinor already in the original basis.Therefore, there is no basis transformation $U(n=0)$ for the chiral level.Following our recent work Ref.\cite{trescher_charge_2017}, we