4.3.3 Quantum Hall Effect#

Prompts

Define the filling factor \(\nu = N_e/N_\Phi\). What does it mean physically when \(\nu\) is an integer, and why is the system incompressible there?
The Hall conductance at integer \(\nu\) is \(\sigma_{xy} = \nu\,e^2/h\). Why is this exact, and why is it independent of microscopic details like sample geometry or disorder?
Walk through Laughlin’s charge-pumping argument on a Corbino disk. How does threading one flux quantum \(\Phi_0\) pump exactly \(\nu\) electrons across, and why does this force quantized \(\sigma_{xy}\)?
The Kubo (linear-response) formula gives \(\sigma_{xy} = \nu\,e^2/h\) from a dynamical calculation. What is the physical content of summing matrix elements between filled and empty Landau levels?
Why are the Hall plateaus wide in \(B\), not just isolated points at integer \(\nu\)? What role do disorder-induced localized states play?

Lecture Notes#

Overview#

The integer quantum Hall effect (IQHE) is the most precisely-quantized macroscopic phenomenon in condensed matter: at integer filling of Landau levels, the Hall conductance of a 2D electron gas takes the value

\[ \sigma_{xy} = \nu\,\frac{e^2}{h} \]

for integer \(\nu = 1, 2, 3, \ldots\), accurate to better than one part in \(10^9\) and independent of sample geometry, disorder, and material details. This section explains why. The story has a clear arc:

At integer filling, the Fermi level lies in a Landau-level gap — the system is incompressible.
Two complementary arguments give the quantization \(\sigma_{xy} = \nu e^2/h\): Laughlin’s charge pumping (topological) and the Kubo formula (linear response). Their agreement reveals the topological origin of the result.
Disorder broadens Landau levels but preserves the quantized plateaus, because only states at the level center carry current.

The same logic extends to fractional fillings, where electron-electron interactions produce richer phases — the fractional QHE (§2.3.2).

Filling Factor and Incompressibility#

§4.3.2 established that a 2D electron in a perpendicular field \(B\) has Landau levels \(E_n = (n+\tfrac{1}{2})\hbar\omega_c\), each with degeneracy \(N_\Phi = BA/\Phi_0 = eBA/h\).

Filling factor

The filling factor is the ratio of electrons to flux quanta,

(158)#\[ \nu \;=\; \frac{N_e}{N_\Phi} \;=\; \frac{n_e\,h}{eB}. \]

When \(\nu\) is an integer, exactly \(\nu\) Landau levels are completely filled and the Fermi level sits in the gap \(\hbar\omega_c\) between the last filled and first empty level. Adding one more electron costs \(\hbar\omega_c\), so the system is incompressible: its electron density cannot change in response to a small chemical-potential shift.

Incompressibility at integer \(\nu\) is the precondition for everything that follows. The gap protects the ground state from local perturbations, and the macroscopic degeneracy \(N_\Phi\) provides the rigidity needed to pump charge coherently across the sample.

The Quantized Hall Conductance#

The central experimental fact is that at integer \(\nu\), the Hall response is exactly

\[\begin{split} \begin{split} \sigma_{xy} &= \nu\,\frac{e^2}{h}, \\ \rho_{xy} &= \frac{1}{\sigma_{xy}} = \frac{h}{\nu e^2}. \end{split} \end{split}\]

The Hall resistivity \(\rho_{xy}\) is the inverse of \(\sigma_{xy}\) on the plateau, where the longitudinal conductivity \(\sigma_{xx}=0\) (the bulk is gapped). In 2D, \(\rho_{xy}\) has units of resistance, so its plateau value \(h/(\nu e^2)\) matches the directly measured transverse resistance.

Compared to the classical prediction \(\rho_{xy}^{\mathrm{cl}} = B/(n_e e)\) (a smooth straight line, §4.3.1), the quantum result is flat plateaus at integer \(\nu\) with sharp jumps between them. We give two derivations.

Argument 1: Laughlin’s Charge Pumping#

Consider a Corbino-disk geometry — a 2D sample shaped as an annulus, threaded by the uniform field \(B\) through its plane plus an extra flux \(\Phi(t)\) through its central hole. Ramp \(\Phi(t)\) slowly from \(0\) to \(\Phi_0\) over a long time \(T \gg 1/\omega_c\) (adiabatic), keeping the sample at integer filling \(\nu\) throughout.

Step 1 — Orbit radii from AB-phase quantization.

In the symmetric gauge, each occupied Landau-level state circulates at a definite radius \(r_m\). Single-valuedness of the wavefunction around the orbit requires the enclosed Aharonov-Bohm phase to be a multiple of \(2\pi\) (same logic as §4.2.2 and §4.2.3),

\[\begin{split} \begin{split} \frac{e\,\Phi_{\mathrm{encl}}}{\hbar} &= 2\pi m, \\ \Phi_{\mathrm{encl}} &= \pi r_m^{2}\,B + \Phi(t), \end{split} \end{split}\]

with the enclosed flux being the uniform-field flux through the orbit’s interior plus the threaded flux through the hole. Solving for \(r_m^{2}\),

(159)#\[ r_m^{2}(t) \;=\; \frac{\Phi_0}{\pi B}\bigl(m - t/T\bigr). \]

Step 2 — Radii shrink as flux is threaded.

As \(\Phi(t)\) ramps from \(0\) to \(\Phi_0\) (so \(t/T\) goes from \(0\) to \(1\)), \(r_m^{2}\) decreases by \(\Phi_0/(\pi B)\) — equivalently \(r_m \to r_{m-1}\). Every orbit shifts inward by one slot. The innermost orbit \(m=0\) collapses onto the inner edge; a new vacant slot opens at the outer edge. Net result: per filled Landau level, one electron is pumped from the outer edge to the inner edge during one flux-quantum cycle.

Step 3 — Hall conductivity from the local current.

At a generic radius \(r\) inside the annulus, one electron crosses the circle of circumference \(2\pi r\) during time \(T\), giving the radial current density

\[ j_r \;=\; \frac{e}{2\pi r\,T}. \]

Faraday’s law converts the changing flux into an azimuthal electric field,

\[ E_\theta \;=\; \frac{1}{2\pi r}\,\frac{\mathrm{d}\Phi}{\mathrm{d}t} \;=\; \frac{\Phi_0}{2\pi r\,T}. \]

Their ratio is the Hall conductivity contribution from one filled Landau level,

\[ \sigma_H \;=\; \frac{j_r}{E_\theta} \;=\; \frac{e}{\Phi_0} \;=\; \frac{e^2}{h}. \]

Step 4 — Sum over filled levels.

For \(\nu\) completely filled Landau levels, each contributes \(e^2/h\) independently:

Charge-pumping result

Threading one flux quantum through the Corbino geometry pumps a quantized charge across the sample.

(160)#\[ \boxed{\;\sigma_{xy} \;=\; \nu\,\frac{e^2}{h}\;} \]

Each filled Landau level contributes one unit \(e^2/h\), so quantized flux insertion leads directly to quantized Hall transport.

The argument is topological: it uses only flux quantization, the Landau-level gap, and integer counting of filled orbits — never the microscopic Hamiltonian. The quantization therefore survives disorder and local deformations of the sample, so long as the gap remains open.

Argument 2: Kubo Formula (Linear Response)#

The same answer emerges from standard linear-response theory. A weak applied electric field \(\boldsymbol{E}\) drives a current \(\boldsymbol{j} = \sigma\boldsymbol{E}\); the antisymmetric (Hall) component of the conductivity tensor is given by the Kubo formula, derived in §5.2.3 from first-order time-dependent perturbation theory in the vector-potential perturbation \(\delta\hat{H} = -\hat{\boldsymbol{j}}\cdot\delta\boldsymbol{A}\). We quote it here and evaluate it on Landau-level eigenstates.

Kubo formula for the Hall conductivity

(161)#\[ \sigma_{xy} = \frac{\mathrm{i}\hbar}{A}\sum_{\alpha\in\mathrm{occ}}\sum_{\beta\notin\mathrm{occ}}\frac{\langle\alpha\vert\hat{j}_x\vert\beta\rangle\langle\beta\vert\hat{j}_y\vert\alpha\rangle - \langle\alpha\vert\hat{j}_y\vert\beta\rangle\langle\beta\vert\hat{j}_x\vert\alpha\rangle}{(E_\beta - E_\alpha)^2}, \]

with \(\hat{\boldsymbol{j}} = q\hat{\boldsymbol{\pi}}/m\) the single-particle current operator. The sum runs over occupied \(\vert\alpha\rangle\) and empty \(\vert\beta\rangle\) single-particle eigenstates of \(\hat{H}_0\) (no Fermi-Dirac function — at \(T=0\), occupancy is sharp). For Landau-level eigenstates \(\vert n,m\rangle\), the formula evaluates to \(\sigma_{xy} = \nu\,e^2/h\) — one unit of \(e^2/h\) per filled Landau level. See §5.2.3 Eq. (201) for the derivation.

Derivation: Kubo evaluation on Landau-level eigenstates

Goal. Evaluate the Kubo formula (161) on Landau-level eigenstates \(\vert n,m\rangle\) from §4.3.2 — Landau index \(n\), energy \(E_n = (n+\tfrac{1}{2})\hbar\omega_c\), guiding-center label \(m\) (\(N_\Phi\) values per level) — and show that for \(\nu\) completely filled levels, \(\sigma_{xy} = \nu\,e^2/h\).

Three observations make the calculation tractable:

Selection rules from the ladder. \(\hat{\pi}_x, \hat{\pi}_y\) are linear in \(\hat{a},\hat{a}^\dagger\) and commute with the guiding-center ladder \(\hat{b},\hat{b}^\dagger\), so \(\langle n',m'\vert\hat{\pi}_a\vert n,m\rangle\) vanishes unless \(m'=m\) and \(n' = n\pm 1\).
Antisymmetrization \(=\) “twice the imaginary part.” The numerator \(\langle\alpha\vert\hat{j}_x\vert\beta\rangle\langle\beta\vert\hat{j}_y\vert\alpha\rangle - (x\leftrightarrow y)\) is \(2\mathrm{i}\,\mathrm{Im}\) of a single matrix-element product, which the ladder identities deliver in closed form.
The constrained occ↔empty sum collapses. Pairs of states both inside the filled set cancel; only the count of occupied states survives.

Step 1 — Matrix elements.

In ladder form (§4.3.2),

\[\begin{split} \begin{split} \hat{\pi}_x &= \sqrt{q\hbar B/2}\,(\hat{a}^\dagger + \hat{a}), \\ \hat{\pi}_y &= \mathrm{i}\sqrt{q\hbar B/2}\,(\hat{a}^\dagger - \hat{a}). \end{split} \end{split}\]

Acting on \(\vert n,m\rangle\) with \(\hat{a}\vert n\rangle = \sqrt{n}\vert n-1\rangle\), \(\hat{a}^\dagger\vert n\rangle = \sqrt{n+1}\vert n+1\rangle\),

\[ \langle n,m\vert\hat{\pi}_x\vert n',m\rangle = \sqrt{\tfrac{q\hbar B}{2}}\,\big(\sqrt{n+1}\,\delta_{n',n+1} + \sqrt{n}\,\delta_{n',n-1}\big), \]

\[ \langle n',m\vert\hat{\pi}_y\vert n,m\rangle = \mathrm{i}\sqrt{\tfrac{q\hbar B}{2}}\,\big(\sqrt{n+1}\,\delta_{n',n+1} - \sqrt{n}\,\delta_{n',n-1}\big). \]

Step 2 — Antisymmetric current product.

Multiplying (cross-terms vanish since different deltas don’t overlap),

\[ \langle n,m\vert\hat{\pi}_x\vert n',m\rangle\,\langle n',m\vert\hat{\pi}_y\vert n,m\rangle = \frac{\mathrm{i}q\hbar B}{2}\,\big[(n+1)\,\delta_{n',n+1} - n\,\delta_{n',n-1}\big]. \]

The swap \((x\leftrightarrow y)\) gives the complex conjugate (Hermiticity of \(\hat{\pi}_x, \hat{\pi}_y\)); subtracting doubles the imaginary part,

\[ \langle n,m\vert\hat{\pi}_x\vert n',m\rangle\langle n',m\vert\hat{\pi}_y\vert n,m\rangle - (x\leftrightarrow y) = \mathrm{i}\,q\hbar B\,\big[(n+1)\,\delta_{n',n+1} - n\,\delta_{n',n-1}\big]. \]

Step 3 — Plug into the Kubo formula.

The energy gap \((E_{n'} - E_n)^2 = (\hbar\omega_c)^2\) is the same for \(n' = n\pm 1\). With \(\hat{j}_a = (q/m)\hat{\pi}_a\), eq. (161) becomes

\[\begin{split} \begin{split} \sigma_{xy} &= \frac{\mathrm{i}\hbar}{A}\cdot\frac{(q/m)^2\,\mathrm{i}\,q\hbar B}{(\hbar\omega_c)^2}\;S, \\ S &\;\equiv\!\sum_{(n,m)\,\mathrm{occ}}\;\sum_{(n',m)\,\mathrm{empty}}\big[(n+1)\,\delta_{n',n+1} - n\,\delta_{n',n-1}\big]. \end{split} \end{split}\]

Step 4 — The constrained sum \(S\) collapses to \(N_\mathrm{occ}\).

Drop the empty-state restriction first: each occupied \((n,m)\) contributes \((n+1) - n = 1\) via the two delta terms, so the unconstrained sum equals \(N_\mathrm{occ}\), the total number of occupied single-particle states.

The empty-state restriction adds nothing because the “blocked” pairs cancel. If both \((n,m)\) and \((n+1,m)\) are occupied, the pair contributes twice to the unconstrained sum, with opposite signs:

\(\alpha=(n,m)\to\beta=(n+1,m)\): weight \((n+1)\) from the \(\delta_{n',n+1}\) term.
\(\alpha=(n+1,m)\to\beta=(n,m)\): weight \(-(n+1)\) from the \(-n_\alpha\,\delta_{n',n_\alpha-1}\) term.

These cancel pairwise, so the constrained sum equals the unconstrained sum: \(S = N_\mathrm{occ}\).

Step 5 — Simplify the prefactor.

Using \(\omega_c = qB/m\) and the Landau-level degeneracy \(N_\Phi = qBA/h\) (§4.3.2),

\[\begin{split} \begin{split} \frac{(q/m)^2\,q\hbar B}{(\hbar\omega_c)^2} &= \frac{q}{\hbar B}, \\ \frac{1}{A} &= \frac{qB}{h\,N_\Phi}. \end{split} \end{split}\]

The two \(\mathrm{i}\) factors out front give \(\mathrm{i}^2 = -1\), which absorbs into the carrier-sign convention. The magnitude is

\[ |\sigma_{xy}| = \frac{q^2}{h}\cdot\frac{N_\mathrm{occ}}{N_\Phi}. \]

For \(\nu\) completely filled Landau levels, \(N_\mathrm{occ}/N_\Phi = \nu\) and \(|q| = e\), so

\[ \sigma_{xy} = \nu\,\frac{e^2}{h}. \]

Each filled Landau level contributes exactly \(e^2/h\), independent of its index \(n\).

The two arguments — pumping and Kubo — agree on the same quantization but spotlight different aspects: pumping reveals the topological nature (no microscopic Hamiltonian needed); Kubo (derived perturbatively in §5.2.3) reveals the dynamical mechanism (interlevel coherence between filled and empty states, with universal cancellation of energy denominators against matrix elements).

Robustness from Disorder#

Real samples have disorder, which broadens each Landau level into a band of width \(\Gamma\). Yet the Hall plateaus persist over wide ranges of \(B\) — they are not razor-thin lines at integer \(\nu\). The reason is the structure of disorder-induced states.

Localized vs extended states

Disorder creates localized states in the tails of each broadened Landau level and extended states near the center. As \(B\) varies (changing \(\nu\) continuously), filling localized states changes the electron count but not the current — they don’t carry charge across the sample. Only the few extended states at the level center contribute to transport, and they do so with the same topological quantization argued above.

Result: the Hall conductance stays at the integer value \(\nu e^2/h\) over a finite range of \(B\) — the plateau width — and jumps only when the Fermi level passes through extended states at the next Landau-level center.

Beyond IQHE: Fractional Filling#

At fractional filling \(\nu = p/q\) (with \(q\) odd), electron-electron interactions are essential. The non-interacting picture predicts a metal — partially filled Landau level, no gap. Interactions open an interaction-induced gap and produce an incompressible quantum fluid with fractional Hall conductance \(\sigma_{xy} = (p/q)\,e^2/h\). The elementary excitations carry fractional charge \(e/q\) and obey anyon statistics.

The full story of the fractional QHE — including the Laughlin wavefunction and anyonic quasiparticles — is developed in §2.3.2 Fractional Quantum Hall Anyons (Chapter 2 on identical particles). The IQHE charge-pumping argument generalizes to fractional \(\nu\), but quantization is now achieved through the same topological mechanism applied to a strongly-correlated incompressible fluid.

Poll: Hall resistivity quantization

In the integer quantum Hall effect, the Hall resistivity \(\rho_{xy} = h/(\nu e^2)\) depends only on fundamental constants and the integer \(\nu\) — not on sample geometry, disorder, or material. Which statement best explains this universality?

(A) Every electron in the filled Landau levels contributes equally to the Hall current, so the quantization just counts how many single-particle states sit below the Fermi level.

(B) The quantized nature of Landau levels means only a fixed number of electrons can carry current.

(C) The Hall response is determined by topological properties of the filled bands, robust against local perturbations as long as the gap remains open.

Summary#

Filling factor \(\nu = N_e/N_\Phi\): integer \(\nu\) means filled Landau levels, gap \(\hbar\omega_c\), and an incompressible state.
Quantized Hall conductance \(\sigma_{xy} = \nu\,e^2/h\) at integer \(\nu\): exact and material-independent.
Charge pumping (Laughlin): topological argument — threading one flux quantum pumps \(\nu\) electrons across the sample, forcing \(\sigma_{xy} = \nu e^2/h\).
Kubo formula: dynamical confirmation — each filled Landau level contributes exactly \(e^2/h\) via inter-level matrix elements.
Robustness: disorder localizes most states in each Landau level; only the central extended states carry current, so Hall plateaus persist over wide ranges of \(B\).
Fractional QHE: at \(\nu = p/q\), interactions open an incompressibility gap and produce fractional plateaus, fractional charge, and anyons (§2.3.2).

Homework#

1. Filling factor. A 2D electron system with density \(n_e = 2.4\times 10^{15}\,\mathrm{m}^{-2}\) is placed in a perpendicular field \(B = 5\,\mathrm{T}\).

(a) Compute the flux-quantum density \(n_\Phi = eB/h\) and the filling factor \(\nu = n_e/n_\Phi\).

(b) How many Landau levels are completely filled? Is the system at integer \(\nu\)?

2. Edge states and Hall transport. At integer \(\nu\), the bulk is gapped but the edges support \(\nu\) chiral 1D channels.

(a) Each chiral channel has conductance \(e^2/h\). Argue that the total Hall conductance is \(\sigma_{xy} = \nu e^2/h\).

(b) “Chiral” means edge electrons travel in only one direction along a given edge. Explain why backscattering by impurities is forbidden, and why this is the microscopic reason the Hall plateau is robust.

(c) Compare an edge state at \(\nu = 2\) to a 1D wire with the same impurity profile. Why does the 1D wire show Anderson localization while the edge state retains quantized conductance?

3. Disorder and plateau width. Without disorder, the Hall conductance jumps at integer \(\nu\) but spends zero “time” at each plateau as \(B\) varies. With disorder, plateaus widen.

(a) Explain how disorder broadens each Landau level into a band, with localized states in the tails and extended states near the center.

(b) As \(B\) decreases from a value where \(\nu = 2\) exactly, why does \(\sigma_{xy}\) remain at \(2\,e^2/h\) over a finite interval rather than dropping continuously?

4. Charge pumping in detail. Laughlin’s argument threads flux \(\Phi\) through a Corbino disk at filling \(\nu\).

(a) Compute the EMF around the inner edge as \(\Phi\) ramps from \(0\) to \(\Phi_0\) over time \(T\).

(b) The Hall response converts the azimuthal EMF into a radial current. Show that the total charge transferred from inner to outer edge over the cycle is \(Q = \sigma_{xy}\,\Phi_0\).

(c) Single-valuedness of the many-body state forces \(Q\) to be an integer multiple of \(e\). Use this to derive \(\sigma_{xy} = N e^2/h\), then identify \(N = \nu\) from particle-counting.

(d) Extend the argument heuristically to \(\nu = 1/3\): why must \(Q = e/3\) per flux quantum, and what does this imply about quasiparticle charge?

5. Landau-level gap and temperature. The cyclotron gap is \(\hbar\omega_c = \hbar eB/m^*\).

(a) For GaAs (\(m^* = 0.067\,m_e\)) at \(B = 10\,\mathrm{T}\), compute \(\hbar\omega_c\) in meV and the equivalent temperature \(T^* = \hbar\omega_c/k_B\).

(b) Why does the IQHE require \(k_B T \ll \hbar\omega_c\)? Why is GaAs preferred over Si (\(m^* \approx 0.2\,m_e\))?

(c) Disorder broadens Landau levels by \(\Gamma\). To resolve plateaus, \(\Gamma < \hbar\omega_c\). Estimate the maximum scattering rate \(1/\tau\) and the minimum mobility \(\mu = e\tau/m^*\).

6. Von Klitzing constant. \(R_K \equiv h/e^2 \approx 25\,812.807\,\Omega\).

(a) Why is the Hall resistivity at \(\nu = 1\) exactly \(\rho_{xy} = R_K\), independent of sample, geometry, or disorder? Cite the topological argument from the lecture.

(b) Since 2019 the SI ohm is realized via the QHE rather than via a physical artifact. Explain the metrological logic.

(c) The fine-structure constant satisfies \(\alpha = e^2/(4\pi\epsilon_0\hbar c)\), so \(R_K = 1/(2\alpha\epsilon_0 c)\). How does a high-precision measurement of \(R_K\) constrain \(\alpha\)?