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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.

Choose the Landau gaugeIt preserves translation invariance in the $y$-direction, so $p_y=\hslash k$ is conserved. With $p_z=0$, magnetic minimal coupling giveswhereThis is a quantum harmonic oscillator, so the Landau levels areindependent of $k$.

The periodic boundary condition in $y$ quantizes $k=2\pi r/L_y$. Requiring the guiding centre to lie in a strip of width $L_x$ gives a $k$-interval of width $|qB|L_x/\hslash$. The number of allowed integers is therefore approximatelythe degeneracy of a Landau level, up to boundary corrections.

The radial factoris sharply peaked, for large $n$, nearRequiring the orbit to lie inside a disk of radius $R$ permitsThus the total degeneracy at fixed $k_z$ is, up to the negligible boundary correction,Since the disk has area $\pi R^2$, the degeneracy of a Landau level per unit area isFor either sign of the charge-field product, replace $eB$ by $|eB|$.

Take $B>0$ and use the Landau gauge $\mathbf{A}=(0,Bx,0)$. With zero $z$-momentum, the Hamiltonian becomesFor a state $e^{i{k}_{y}y}\chi (x)$, this is a quantum harmonic oscillator in $x$, centred atwith cyclotron frequencyThe Landau levels are therefore

Periodicity in $y$ quantizes $k_y=2\pi \ell /R_y$. Requiring the corresponding oscillator centre to lie in an $x$ interval of length $R_x$ leavesallowed centres for each $n$. This is the degeneracy of a Landau level, equal to the magnetic flux through the rectangle divided by the magnetic flux quantum $h/|e|$. Consistent magnetic periodic boundary conditions make $D$ an integer.