- \frac{\hbar}{2\ell_B^2}x$, in terms of the magnetic length $\ell_B=\sqrt{\hbar/eB}$. The Hamiltonian can be transformed into diagonal form $K=\frac{\hbar\omega_c}{2} \left(a^\dagger a + \frac{1}{2} \right)$ with the help of ladder operators $a \propto \pi_x + i\pi_y$ and $a^\dagger \propto \pi_x - i \pi_y$. However, for each value of $a^\dagger a$, there is residual degeneracy equal to the number of the magnetic flux quanta $N_\phi$. This degeneracy is resolved by a second pair of operators $b, b^\dagger$ that commute with $a, a^\dagger$ and depend on the *guiding center* coordinates of the electron, $R^a = r^a - \frac{\epsilon^{ab}}{\hbar} \pi_b \ell_B^2$. Operators $b^\dagger$ create the (unnormalized) single particle eigenstates of the lowest LL, $$\phi_l^{\alpha=1, \phi=0} (z) = z^l e^{-z^* z/4\ell_B^2},$$\[singleparticleorbital\] with $z=x+iy$ being the complex coordinate of an electron in the plane (and $z^*$ denoting its complex-conjugate). The quantum number $l$ is an eigenvalue of the angular momentum $L_z$ and the single particle states $\phi_l$ are localized on concentric rings around the origin.

To illustrate the effect of mass anisotropy, we take the principal axes of the mass tensor to be along the $x$ and $y$ directions ($\phi=0$), with different masses along the two directions ($\alpha \neq 1$). Via simple rescaling $x\rightarrow x/\sqrt{\alpha}, y\rightarrow y\sqrt{\alpha}$, 