Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2023 / Paper 2 / Section A(ii) / 15

Take a two-dimensional conducting sheet in the $xy$ plane with magnetic flux density $\mathbf{B}=B\hat{\mathbf{z}}$ and electric current density along $x$. Moving carriers feel the transverse Lorentz force $q{\mathbf{v}}_d\times \mathbf{B}$. They accumulate at opposite edges, producing a transverse electric field until its force balances the magnetic deflection:For the transverse-voltage convention $E_y/j_x$, this gives $E_y=v_{x}B$ andThe Hall effect therefore measures carrier sign as well as density in the single-carrier approximation: electrons give the opposite transverse-voltage sign from positive carriers. Some tensor conventions call $E_y/j_x$ ${\rho }_{yx}$ rather than ${\rho }_{xy}$; the physical sign is fixed by the stated lead and field directions.

The Hall voltage is the integral of this transverse electric field across the sample width. In a three-dimensional slab of thickness $t$, its magnitude is $|V_H|=IB/(n|q|t)$; for a sheet, use the sheet number density $n_s$ and omit $t$.

In the classical single-carrier model,for a two-dimensional sheet. The curve is a straight line through the origin, with its sign determined by the carrier charge and the voltage convention. The sheet Hall coefficient has units ${\mathrm{m}}^2/C$, consistent with the numerical slope used later.

The classical transverse response is not quantized. With carrier density and scattering parameters fixed, reversing $B$ reverses the transverse signal. Several carrier populations can produce a nonlinear classical curve, so straight-line behaviour here uses the stated single-population approximation rather than being a property of every conductor.

The integer quantum Hall effect requires a conducting two-dimensional electron gas with well-resolved Landau levels. In the simple parabolic-band model their orbital spacing isLow temperature and sufficiently low disorder broadening are needed so that $k_{B}T$ and the relevant level width $\Gamma$ are small compared with this spacing. A high-mobility sample with ${\omega }_c\tau \gg 1$ is therefore useful.

Some disorder has a constructive role: localized states between extended Landau-level states allow the Fermi energy to move through a range of carrier populations while the transport response stays on a plateau. The longitudinal response is then very small and the transverse response is quantized. Whether odd or only even integer plateaux are resolved also depends on whether spin splitting is resolved.

On an integer plateau the sheet transverse resistivity has magnitudewhere the integer $\nu$ counts occupied resolved quantum Hall edge states. Increasing $B$ at fixed sheet density reduces the Landau filling factor ${\nu }_{\mathrm{fill}}=n_{s}h/(eB)$. The observed trace is therefore a staircase toward larger $|{\rho }_{xy}|$, with transitions between plateaux rather than the classical straight line. In the plateau regions ${\rho }_{xx}\simeq 0$; dissipative peaks occur at transitions.

It is not correct to demand an exactly integer value of $n_{s}h/(eB)$ everywhere along a finite-width plateau. Localization accommodates changes in total population without changing the number of extended current-carrying channels. If spin remains unresolved, the usual spin-degenerate sequence contains even integers rather than every integer.

Using the magnitude of the classical slope,The Landau filling factor at $1\mathrm{T}$ isUnder the convention that the “last Landau level” means the last resolved spin branch, its degeneracy per area is $eB/h$. Setting ${\nu }_{\mathrm{fill}}=1$ givesAbove this threshold all carriers fit into that lowest resolved branch, which is partially filled when ${\nu }_{\mathrm{fill}}<1$.

There is a spin convention to state explicitly: if “Landau level” instead means one orbital level with unresolved twofold spin degeneracy, its capacity is $2eB/h$, and all carriers first fit into orbital index $n=0$ at $B=3.24\mathrm{T}$, where ${\nu }_{\mathrm{fill}}=2$. The slope alone determines $n_s$ and ${\nu }_{\mathrm{fill}}$, but does not specify the spin splitting.