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For a multiplication operator $A_j(x)$, $[p_i,A_j]=-i\hslash {\partial }_i{A}_j$. Therefore the mechanical momentum satisfiesThe displayed ladder operators implicitly require $qB>0$, so that their denominator is real and the second is the adjoint of the first. Then $[a,a^{†}]=1$ andThe free longitudinal motion gives a continuous spectrum in $k_z$, and the transverse Landau levels have guiding-centre degeneracy of a Landau level. For arbitrary sign of $qB$, use $|qB|$ in the energy and $a=({\pi }_x+i\mathrm{sgn}(qB){\pi }_y)/\sqrt{2\hslash |qB|}$. The question's normalizable wavefunctions likewise assume $qB>0$.

For the stated symmetric-gauge wavefunction, $|\psi {|}^2\propto r^{2M}{e}^{-qB{r}^2/(2\hslash )}$, whose peak for $M>0$ is at $r_M^2=2\hslash M/(qB)$. Thus the largest useful angular index is about $qB{R}^2/(2\hslash )$, givingThe $M=0$ state should also be included in the usual lowest Landau level; its omission by “positive integer” changes only an order-one term. Including the radial measure shifts the most probable radius to $r^2=(2M+1)\hslash /(qB)$, again only an order-one counting correction. Boundary conditions and the tails of the wavefunctions make this a large-flux estimate, not an exact finite-disc count.

For a bounded linear operator $T$ on a complex Hilbert space, the spectrum of a bounded operator consists of $\lambda \in \mathbb{C}$ for which $T-\lambda I$ does not have a bounded everywhere-defined inverse. A bijective bounded operator has a bounded inverse by the bounded inverse theorem. Using the disjoint spectral convention,These are respectively the point spectrum, continuous spectrum and residual spectrum, and partition the spectrum. Some texts let residual spectrum include noninjective operators with nondense range; the displayed convention excludes overlap with point spectrum.

In the disjoint convention for a bounded linear operator $T$, the residual spectrum consists of $\lambda$ for which $T-\lambda I$ is injective but has nondense range. Together with the point spectrum and continuous spectrum it partitions the spectrum of a bounded operator. Some conventions also include noninjective operators with nondense range, so the convention must be stated.