$$\Psi_{+,0}(\bm{r})=\frac{N_+}{\Omega}\left(\begin{array}{c}e^{F(\bm{r})}\\0\end{array}\right),
\hspace{3mm}
\Psi_{-,0}(\bm{r})=\frac{N_-}{\Omega}\left(\begin{array}{c}0\\e^{-F(\bm{r})}\end{array}\right),
\label{eq_sols2}$$ with $N_\pm=\left[\int_{UC}dr^2\,\exp\pm2F(\bm{r})\right]^{-\tfrac{1}{2}}$ ensuring normalization of $\Psi_{\pm,0}$ to unity over the sample. The two eigenstates have opposite sublattice polarization: In valley $K$, $\Psi_{+,0}$ is $A$ sublattice polarized while $\Psi_{-,0}$ is $B$-sublattice polarized. Since the valley isotropic representation is employed, the roles of the sublattices are reversed in the $K'$ valley. Hence, the state $\Psi_{+,0}$ in the $K'$ valley is $B$ polarized while the $K'$ state $\Psi_{-,0}$ is $A$ polarized. The result is consistent with the Atiyah Singer index theorem. It states that in a given valley the difference between the number of zero energy eigenstates that are $A$ polarized and $B$ polarized equals the total flux through the sample.[@Pac07] Note also that because $F$ is $UC$-periodic, the same is true for $\Psi_{\pm,0}$, even though only sample-periodicity is imposed on $\Psi_{\pm,0}$ by the boundary condition \[Eq. (\[eq\_bc\])\].

The solutions in Eq. (\[eq\_sols2\]) were found without invoking Bloch’s theorem. I now briefly re-examine them from this point of view. Since the Hamiltonian $H$ is periodic, an eigenbasis exists in which all eigenfunctions of $H$ are of the form $$\Psi_{n,\bm{k}}(\bm{r})=\frac{1}{\Omega}e^{i\bm{k}\cdot\bm{r}}\psi_{n,\bm{k}}(\bm{r}),$$ with $\bm{k}\in BZ$ and $n$ a discrete index. The functions $\psi_{n,\bm{k}}(\bm{r})$ are $UC$-periodic and will be referred to as Bloch 