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Robert Laughlin's 1983 variational method uses pairwise zeros to suppress close electron encounters within the lowest Landau level. For aligned spin, odd $m$ gives fermionic exchange symmetry.^[ref]

Here $\mathcal{N}$ enforces wavefunction normalization, ${\ell }_B$ is the magnetic length, and complex coordinates $z_j=x_j-i{y}_j$ describe electrons in a magnetic field $B\hat{\mathbf{z}}$, $B>0$, with the symmetric gauge. Reversing the orientation conjugates the polynomial without changing probability densities. The Gaussian function confines the wavefunction.

The exchange check is especially simple for two electrons. Writing $w=z_1-z_2$, exchange sends $w\mapsto -w$, so $w^m\mapsto (-1{)}^m{w}^m$. An aligned two-electron spin state is symmetric, requiring the spatial wavefunction to change sign. Hence $m=1,3,5,\dots$ works, whereas $m=2$ would give the wrong sign. At a close encounter, the probability density vanishes as $|w{|}^{2m}$: $m=3$ imposes a sixth-order zero, stronger than the quadratic zero at $m=1$. This is the local geometric mechanism for avoiding a large repulsive Coulomb potential energy.

A useful large-$N$ counting outline follows directly from the polynomial. Each complex coordinate has maximum degree $m(N-1)$. A lowest Landau level orbital proportional to $z^k{e}^{-|z{|}^2/(4{\ell }_B^2)}$ has its areal probability density maximum at $r^2=2k{\ell }_B^2$, by differentiating its logarithm. This suggests droplet area $A\simeq 2\pi mN{\ell }_B^2$, neglecting the boundary correction. Dividing $N/A$ by the degeneracy of a Landau level per area, $1/(2\pi {\ell }_B^2)$, gives filling factor $\nu \simeq 1/m$. This counting is a motivation for the fraction, not an independent proof that the wavefunction minimizes the interacting Hamiltonian.

A uniform magnetic field quantizes the transverse cyclotron motion of an electron. With electron charge $-e$ and the Landau gauge for a uniform magnetic field $\mathbf{A}=(0,Bx,0)$, minimal coupling givesFor an eigenstate of $p_y$ with eigenvalue $\hslash k_y$, this is a quantum harmonic oscillator centered at $x_c=-\hslash k_y/(eB)$, with cyclotron frequency ${\omega }_c=eB/m$. Its Landau levels areThe oscillator width is set by the magnetic length ${\ell }_B=\sqrt{\hslash /(eB)}$. In three dimensions, the free motion parallel to the magnetic field adds ${\hslash }^2{k}_z^2/(2m)$. The electron spin adds a Zeeman splitting $\pm g_s{\mu }_{B}B/2$. In a crystal the appropriate band effective mass and effective spin factor replace the free-electron values.

Changing $k_y$ moves the orbit center without changing the energy. In a rectangle $L_x{L}_y$, adjacent allowed centers differ by $2\pi {\ell }_B^2/L_y$. Requiring the center to lie in the sample gives the degeneracy of a Landau level, per spin,This large state degeneracy, together with the increasing level spacing, rearranges the occupied states as the magnetic field changes. To resolve the Landau levels, thermal broadening $k_{B}T$ and disorder broadening should both be small compared with $\hslash {\omega }_c$.

The de Haas-van Alphen effect is an oscillation of magnetization as successive Landau levels pass the Fermi energy. The Shubnikov-de Haas effect is the corresponding oscillation in longitudinal magnetoresistance. For a simple orbit, successive crossings occur periodically in $1/B$; the general relation to the extremal orbit area is Onsager quantization. These measurements reveal the geometry and effective masses of a Fermi surface.

In a two-dimensional electron system, filling an integer number $\nu$ of resolved levels leads to the integer quantum Hall effect, withand very small longitudinal resistivity on a plateau. Disorder localization helps produce finite plateau widths. Cyclotron resonance provides another direct signature: radiation is absorbed at the adjacent-level spacing $\hslash {\omega }_c$.

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.

For interface area $\mathcal{A}$, each resolved spin branch has one state per flux increment $h/e$, so its degeneracy of a Landau level isIf $\nu$ such branches are occupied, including fractional occupation of the highest, the filling factor givesThe filling counts spin and any other internal branches separately; it is an integer when a whole number of resolved levels are full.

From the visible PDF axes, approximately ten unsplit periods lie between $1/B=1.0$ and $2.5{\mathrm{T}}^{-1}$, giving $\Delta (1/B)\simeq 0.15{\mathrm{T}}^{-1}$. Consecutive unsplit orbital peaks differ by two occupied spin branches, so carrier density from spin-degenerate quantum oscillations givesThe precision is limited by reading the plotted trace, roughly a few percent. The factor two assumes only spin remains unresolved in that part of the graph; counting individual spin-split peaks would instead use a one-branch filling increment.

The degeneracy of a Landau level is $eBA/h$ per resolved internal branch in area $A$. The filling factor $f=n_{s}h/(eB)$ varies continuously. Additional localized eigenstates need not carry bulk transport, allowing an integer plateau over an interval of $f$; a mobility gap need not be an empty spectral gap.^[ref]

The plot is an original phenomenological construction, not measured data or a calculation of Anderson localization. It stipulates dimensionless transverse sheet conductivity $k(f)=1+\frac{1}{2}{\sum }_{m=1}^5[1+\tanh ((f-m-1/2)/w)]$ and longitudinal sheet conductivity $s(f)=a{\sum }_{m=1}^5{\mathrm{sech}}^2((f-m-1/2)/w)$, with chosen $w=0.06$, $a=0.08$. Both are measured in units $e^2/h$. The quantum Hall conductivity tensor inversion produces $|{\rho }_{yx}|/R_K=k/(s^2+k^2)$ and ${\rho }_{xx}/R_K=s/(s^2+k^2)$. Regions near $k=N$, $s=0$ yield plateaus near $R_K/N$; the model alone does not explain their microscopic robustness. Each step represents one resolved branch, without assuming unresolved spin or valley multiplicities.