Why the Quantum Hall Plateau Needs Dirt: A Researcher's Notebook — Epoche C1
Working notes, kept while preparing three lectures on the integer quantum Hall effect. Question for the week: why is the most exact electrical measurement in physics performed on samples full of impurities? 3 February — a resistance too exact for its sample Re-read von Klitzing, Dorda and Pepper (1980). The device is an ordinary silicon field-effect transistor, cooled to $1.5\,\mathrm{K}$ in a magnetic field of about $18\,\mathrm{T}$. Its electrons form a two-dimensional electron gas — a sheet in which electrons move freely within a plane but are confined so tightly perpendicular to it that no excited state of the perpendicular motion is thermally accessible. As the gate voltage raises the areal density $n_e$, the Hall resistance $R_{xy} = V_H/I$ — the transverse voltage $V_H$ divided by the current $I$ driven along the sample — does not fall smoothly. It locks onto flat steps at $R_{xy} = h/(Ne^{2})$ with $N$ a whole number, $h$ Planck's constant and $e$ the elementary charge. From the SI values $h = 6.626\,070\,15\times10^{-34}\,\mathrm{J\,s}$ and $e = 1.602\,176\,634\times10^{-19}\,\mathrm{C}$, the first step is $h/e^{2} = 25\,812.807\,\Omega$. Devices made of silicon, gallium arsenide and graphene agree on this value to parts in $10^{9}$ — a universality no property of any individual sample could account for, since these three materials share almost nothing but dimensionality. What bothers me is the sample. A transistor interface is crowded with charged impurities in the oxide and at the interface, and everything in my training says disorder degrades precision: the cleanest sample ought to quantise best. Plan for the week: calculate what a perfectly clean sample would show, then see what the dirt changes. 5 February — the perfect crystal makes no steps The clean calculation is ten lines long, and it reversed the plan. In crossed electric and magnetic fields a free charge drifts sideways at the velocity for which the electric and magnetic forces balance: $e\mathcal{E} = ev_dB$, so $v_d = \mathcal{E}/B$, independent of charge and mass. Every electron therefore drifts identically. The sheet current density is $J_x = n_e e v_d$ and the transverse field is $\mathcal{E}_y = v_dB$, so the measured ratio is $$R_{xy} = \frac{V_H}{I} = \frac{\mathcal{E}_yW}{J_xW} = \frac{v_dB}{n_e e v_d} = \frac{B}{n_e e},$$ where the sample width $W$ cancels — in two dimensions the Hall resistance is a resistance, not a resistivity multiplied by a shape factor, which is one reason the effect is useful for metrology at all. Notice what this formula is not: a staircase. It falls smoothly as $1/n_e$, with no feature at any density. And the only assumption behind it is that nothing in the sample singles out a rest frame; with no fixed obstacles one may transform to the frame drifting with the electrons, in which the in-plane electric field vanishes and nothing at all happens, and the linear response follows from that change of viewpoint alone. So a perfect crystal shows no plateau. The causal arrow points the opposite way from my expectation: the steps must be caused by the very thing the ideal sample lacks. One thing the clean formula does supply, which I had not noticed before. Define the filling factor $\nu = n_e/(eB/h)$, the electron density measured in units of the density $eB/h$ derived tomorrow. Then $B/(n_e e) = h/(\nu e^{2})$ identically. The quantised value is not a different formula from the classical one; it is the classical one evaluated at integer $\nu$. What needs explaining is therefore not the value but why $\nu$ should get stuck. 9 February — what disorder does to a Landau level The mechanism, step by step. In a magnetic field an electron runs in circles, and quantisation of the closed orbit permits only discrete energies — the Landau levels $E_\ell = (\ell + \tfrac12)\hbar\omega_c$ for $\ell = 0,1,2,\dots$, with $\omega_c = eB/m^{*}$ the cyclotron frequency and $m^{*}$ the effective mass in the crystal. In gallium arsenide, where $m^{*} = 0.067$ of the free-electron mass, $\hbar\omega_c$ at $10\,\mathrm{T}$ is $1.05\times10^{-34}\times 2.6\times10^{13} \approx 2.8\times10^{-21}\,\mathrm{J}$, or about $17\,\mathrm{meV}$. The degeneracy of each level is the number that makes everything else work, so it is worth deriving rather than quoting. The natural length in a magnetic field is the magnetic length $\ell_B = \sqrt{\hbar/eB}$, which at $10\,\mathrm{T}$ is $8.1\,\mathrm{nm}$ — the radius scale of the quantum orbit. The independent states within one Landau level are distinguished by the position of the orbit's guiding centre, and those centres are quantised so that each state occupies an area $2\pi\ell_B^{2}$. The number of states per unit area is therefore $$\frac{1}{2\pi\ell_B^{2}} = \frac{eB}{2\pi\hbar} = \frac{eB}{h} \approx 2.4\times10^{15}\,\mathrm{m^{-2}} \quad\text{at } 10\,\mathrm{T},$$ which is exactly the number of flux quanta $h/e$ passing through unit area: one state per quantum of flux, a count fixed by the field alone and by nothing about the material. That is why $\nu$ is the ratio of electrons to flux quanta, and why an integer $\nu$ means an exactly filled set of levels. Now the disorder. A random impurity potential broadens each razor-sharp level into a band of finite width, and Anderson's 1958 study of electrons in random lattices says what that band contains: beyond a threshold of randomness the eigenstates are not extended waves but exponentially localised, $|\psi| \sim e^{-|r-r_0|/\xi}$, so that an electron in such a state circles one hill or hollow of the impurity landscape and its wavefunction connects to nothing far away. It cannot carry current between contacts however long one waits, which is the content of the paper's title: absence of diffusion. The two-dimensional case sharpens the puzzle before resolving it. Scaling theory says that in two dimensions, without a magnetic field, all states are localised for arbitrarily weak disorder (Abrahams, Anderson, Licciardello