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research32 min studyLesson 542 in Physics

Chern numbers and quantized Hall conductance

Normalize the occupied-band curvature integral and connect its integer to a signed two-dimensional electrical response.

Before you begin
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The question for today

What you’ll understand

  • Compute an occupied Chern number with the correct 2π normalization.
  • Relate its sign to explicitly oriented Hall current and electric field.
  • Count resolved Landau-level states and identify the gap assumptions behind quantization.

A two-dimensional insulating band structure can have a nonzero integer obtained by integrating occupied-state curvature across its Brillouin torus. That integer controls a transverse electrical response under specified low-temperature and gap conditions. Quantization does not mean that curvature is uniform or integer at every momentum. It means that all local contributions, including the gauge patching between them, combine into a global obstruction that cannot change during an admissible gapped deformation.

The full Brillouin torus supplies the integer

Use the positive-i connection defined previously and orient momentum area by dkx followed by dky. Dividing total curvature flux by two pi gives the first Chern number. The integer follows from consistent phase transition functions between eigenvector patches on the closed torus. A curvature integral over only part of the zone is generally noninteger. Nor should a finite numerical mesh be expected to give an exact integer without a gauge-consistent discretization and convergence checks.

For several isolated occupied bands, add their Chern numbers. If occupied bands touch internally, use the trace of the occupied-subspace curvature instead of insisting on separately smooth eigenvectors. The occupied subspace must remain separated from empty states to define this band-insulator invariant continuously. A gap closing is necessary to change it along a continuous noninteracting bulk path, unless the defining assumptions or the occupied-state selection are changed.

Cocc=12π∫BZtr⁡Ωxy(k) dkx dky∈ZC_{\mathrm{occ}}=\frac{1}{2\pi}\int_{\mathrm{BZ}}\operatorname{tr}\Omega_{xy}(\mathbf k)\,dk_x\,dk_y\in\mathbb Z
The occupied projector is smooth and separated from empty bands across the full two-dimensional periodic Brillouin zone. Ω uses A=i⟨u|∇ku⟩ and the (kx,ky) orientation. Dimensional curvature times momentum area is dimensionless; the trace handles an occupied band group.↔ Scroll wide equations horizontally.

Conductance needs an orientation as well as an integer

Let electrons carry charge −e with e positive. Define the Hall coefficient here by current density along y responding to electric field along x. With our curvature sign this coefficient equals the occupied Chern number times e squared over h. The opposite off-diagonal tensor entry has the opposite sign. Specifying this convention avoids treating a sign difference between published formulas as a disagreement about a measured current.

A simple derivation uses the anomalous wavepacket velocity and fills the entire occupied band. The ordinary group-velocity contribution integrates to zero over a periodic filled band, while the curvature contribution remains. In two dimensions the electrical current density has units of amperes per metre, so the resulting coefficient has conductance units. Quantization assumes appropriate zero-temperature insulating linear response, negligible longitudinal conduction, and a measurement geometry whose contacts do not obscure the response.

jy=σHEx,σH≡σyx=Cocce2h,σxy=−σHj_y=\sigma_H E_x,\qquad \sigma_H\equiv\sigma_{yx}=C_{\mathrm{occ}}\frac{e^2}{h},\qquad \sigma_{xy}=-\sigma_H
e>0 is the elementary-charge magnitude, h=2πℏ, jy is sheet current density in A/m, Ex in V/m, and σH in S. The sign is tied to the previous positive-i Berry convention and ordered axes. This is the charge-conserving noninteracting filled-band result, with gapped insulating linear-response assumptions.↔ Scroll wide equations horizontally.

Landau counting and disorder qualify the band picture

One resolved spin species in a Landau level has one orbital per magnetic flux quantum, giving a degeneracy proportional to sample area and magnetic-field magnitude. A separate unresolved spin or valley multiplicity must be included explicitly, rather than inserted as an automatic factor of two. Integer filling can therefore be counted without referring to a lattice band, although Hall sign still depends on charge, field, axes, and the coefficient convention.

Disorder can remove crystal momentum as a useful label while a Hall plateau survives through localized states and a mobility gap. Real-space or twisted-boundary formulations then replace the clean Brillouin integral. Interacting systems can also exhibit fractional quantization, which cannot be explained by pretending a partially filled single-electron band has a fractional first Chern number. The filled-band theorem is a precise starting point, not a universal account of every Hall phase.

Work it through

Sum occupied bands before predicting current

A clean two-dimensional insulator has occupied-band Chern numbers +3 and −1 in the stated convention. A uniform Ex is produced by Vx=1.0 mV across sample width Lx. Find σH and the current Iy=jyLx, using e²/h≈38.74 μS.

  1. The occupied integer is Cocc=3−1=2C_{\mathrm{occ}}=3-1=2. Individual occupied-band contributions combine before computing the total electrical response; their opposite signs do not imply that both cancel completely.

  2. The signed Hall coefficient is σH=2(38.74)=77.48\sigma_H=2(38.74)=77.48 microSiemens. Under the declared convention, a positive Ex therefore produces a positive y-directed sheet current density.

  3. Since Ex=Vx/LxE_x=V_x/L_x, integrating the sheet current gives Iy=σHVx=77.48×10−6×10−3=77.48I_y=\sigma_HV_x=77.48\times10^{-6}\times10^{-3}=77.48 nA. The width cancels in this uniform ideal geometry.

The result

The occupied Chern number is +2, σyx is +77.48 μS, and Iy is +77.48 nA. The other tensor component σxy is negative in the same axes.

The living notebook

Explore the model

Interactive lab

Interactive lab · fields

Local curl is not automatically an integer

Reset and set both controls to Source strength q=0 and Rotation Ω=0.5; predict Fy=0.5 and magnitude 0.5 at (1,0). For a fresh comparison, Reset and set both controls to q=0 and Ω=1; those readouts become 1 and 1. The analytic planar curl changes continuously from 1 to 2. This illustrates local differential geometry only, not a periodic Bloch connection.

Local curl is not automatically an integerThe softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Current readings: Fx at (1, 0): 0.833; Fy at (1, 0): 0.5; Field magnitude there: 0.972. All parameters are adjustable with the labeled controls below.−3.2−1.601.63.2−1.5−0.7500.751.5xyq = 1
Local field directionSource strength
Fx at (1, 0)
0.833
Fy at (1, 0)
0.5
Field magnitude there
0.972
t = 0 s

The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn integer Chern invariant requires a complete band bundle and global integration. The softened planar field lab has neither a Brillouin torus nor a Hall-conductance or Chern-number readout; arrow lengths are compressed.

Pause & reason

Check your understanding

Two occupied isolated bands have Chern numbers +1 and −1. What follows for their total charge Hall response in the stated filled-band setting?

Your turn

Try it yourself

A spin-resolved Landau level covers area A=2.0 μm² in a field with |B|=4.0 T. Taking h/e=4.136×10⁻¹⁵ T m², estimate its orbital degeneracy and the electron count for three completely filled resolved levels.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use Nϕ=∣B∣A/(h/e)N_\phi=|B|A/(h/e) and convert square micrometres to square metres. No extra spin factor is present in a resolved species.

Compare with the solution

The area is 2.0×10−12 m22.0\times10^{-12}\,\mathrm{m^2}, so Nϕ=(4.0)(2.0×10−12)/(4.136×10−15)≈1934N_\phi=(4.0)(2.0\times10^{-12})/(4.136\times10^{-15})\approx1934. Three resolved filled levels contain approximately 5803 electrons in this macroscopic counting estimate. A finite sample has integer state counts with boundary-dependent corrections, so the rounded values are estimates rather than exact degeneracies for the stated decimal flux. An additional spin multiplicity would require an additional physical assumption.

Keep the idea

Take it with you

  • Normalize complete occupied curvature flux by 2π to obtain a Chern integer.
  • Hall signs require explicit axes, charge, and Berry-connection conventions.
  • Resolved Landau degeneracy has no automatic spin factor, and disorder may require a mobility-gap formulation.

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