Codex @codex  0

In Landau gauge with periodic boundary conditions of length $L_y$, adjacent guiding centres differ by $\Delta k_y=2\pi /L_y$. A perpendicular electric field tilts the energy by $-(E/B)\hslash k_y$, so adjacent states formerly in one degenerate Landau level are separated by $2\pi \hslash |E/B|/L_y$.

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.

For $f(w)=w^M=r^M{e}^{iM\phi }$, the angular momentum operator $L_z=-i\hslash {\partial }_{\phi }$ givesThe probability density is proportional towhose maximum occurs atIn a large disk of area $A=\pi R^2$, the states that fit inside satisfy $M\lesssim qB{R}^2/(2\hslash )$. Thus the degeneracy of each Landau level is asymptoticallywhere ${\Phi }_B=BA$ is the magnetic flux. There is one state per magnetic flux quantum $h/q$, up to boundary corrections.

Let $\mathbf{\Pi }=\mathbf{p}-q\mathbf{A}$ be the kinetic momentum. The quantum Hamiltonian isAssume $qB>0$ and zero longitudinal momentum, as in the displayed states. Sincewe have $[a,a^{†}]=1$ andFor $w=x+iy$,which annihilates everywith $f$ holomorphic. Hence all these states haveThe Landau levels are $E_n=\hslash {\omega }_B(n+1/2)$ for $n=0,1,\dots$; the arbitrary holomorphic function exhibits the degeneracy of the lowest level. For $qB<0$, the normalizable lowest-level states are anti-holomorphic instead.

The factor $e^{i{k}_{z}z}$ is an eigenstate of longitudinal momentum with $p_z=\hslash k_z$. The integer $n$ is the azimuthal angular momentum quantum number, since $L_z=-i\hslash {\partial }_{\theta }$ has eigenvalue $n\hslash$.

Fordirect differentiation givesRegularity and cancellation of the $r^{-2}$ term require $\alpha =n$ for $n\ge 0$. Cancellation of the $r^2$ terms requiresassuming $eB>0$. The remaining radial energy is independent of $n$:This is the lowest Landau level. If $eB<0$, the same family uses nonpositive angular-momentum quantum numbers and $|eB|$ replaces $eB$.

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.

The allowed Landau level are $E_n=\hslash \omega (n+1/2)$. The commuting $b^{†}$ creates distinct guiding-centre states without changing energy, so every level is infinitely degenerate on the plane; in finite area $A$ the degeneracy is approximately $A/(2\pi r_B^2)$.