is to consider the two zero-energy solutions $\Psi_{\pm,0}\propto e^{\pm F}$ that are obtained when $U=0$. A key feature of these wave functions is that the sign of $F$ is anti-correlated with that of $B$ i.e. $F$ is negative on average where the magnetic field points in the positive $z$ direction and positive on average where the magnetic field points in the negative $z$ direction. To prove this assertion I show that the number $P=(e/\hbar)\int_{UC}dr^2\,B(\bm{r})F(\bm{r})$ is negative. The proof is elementary. From Eq. (\[eq\_d2F\]) follows that $P=\int_{UC}dr^2\,B(\bm{r})\bm{\partial}_{\bm{r}}^2B(\bm{r})$. Using integration by parts, and invoking Gauss’s theorem I find $$P=\underbrace{\oint_{\partial UC}d\bm{r}_\perp\cdot B\bm{\partial}_{\bm{r}}B}_{=0}
-\int_{UC}dr^2\,|\bm{\partial}_{\bm{r}}B|^2<0.\label{eq_p_negative}$$ The integration in the first term is around the border of $UC$ and $d\bm{r}_\perp$ is normal to the border (in the $x-y$ plane) pointing outwards and yields zero because $B$ is $UC$-periodic. The result \[Eq. (\[eq\_p\_negative\])\] implies that a particle in the state $\Psi_{-,0}$ ($\Psi_{+,0}$) is largely confined to regions where $\bm{B}$ points in the positive (negative) $z$ direction. In Sec. \[two\] it was noted that the eigenstates $\Psi_{\pm,0}$, have opposite sublattice polarization. The correlation between $B$ and $F$ then implies that sublattice polarization is correlated with the sign of the magnetic field. For instance, in valley $K$, regions where 