 Solution

Keep $\mathbf{A}=(0,Bx,0)$ and choose the scalar potential $\varphi =-Ex$. ThenFor $p_z=0$, completing the square givesHence the spectrum on ${\mathbb{R}}^3$ is

For a translationally invariant dispersion, the group velocity is $v_y={\hslash }^{-1}\partial E_n/\partial k$. Thereforein agreement with the electric-cross-magnetic-field drift $\mathbf{E}\times \mathbf{B}/B^2$.

In the rectangle, $k=2\pi r/L_y$ as before. The energy now changes with $r$, so the guiding-centre degeneracy is lifted, while the number of states in each tilted band remains approximately $|qB|A/(2\pi \hslash )$. Adjacent states within a formerly degenerate level have the electric-field splitting of a Landau level in a rectangleThus this is the ground-to-first-excited gap when the lowest Landau band contains at least two allowed guiding centres and $\Delta E_{\mathrm{electric}}<\hslash {\omega }_c$. Without that implicit weak-field or large-sample condition, the exact full spectral gap is