We now move on to the quantum Hall effect, the mother of all topological effects in condensed matter physics.

But let’s start from the classical Hall effect, the famous phenomenon by which a current flows perpendicular to an applied voltage, or vice versa a voltage develops perpendicular to a flowing current.

How does one get a Hall effect? The key is to break time-reversal symmetry. A flowing current breaks time-reversal symmetry, while an electric field doesn’t. Hence, any system with a Hall effect must somehow break time-reversal symmetry.

But wait a minute, you might catch me and ask, what about a normal electric current flowing parallel to an electric field? This is what happens in a metal on a regular basis, and a metal does not break time-reversal symmetry.

The key difference there is that such a longitudinal current breaks time-reversal through energy dissipation, which turns into heat that breaks time-reversal by the second law of thermodynamics. A Hall current is special in that it is dissipationless. We can drive a Hall current without wasting any energy because the current flows perpendicular to the voltage gradient.

Thus to get a Hall effect we must somehow break time-reversal symmetry. We will examine the simplest way to achieve this, an external magnetic field.

How to measure the Hall effect

Let’s consider a two dimensional gas of electrons immersed in a strong, perpendicular magnetic field. In particular, we take the following geometry, which is called a Hall bar and is routinely used in experiments:

The electron gas is contacted by six electrodes, numbered in the figure. We can use this Hall bar geometry set-up to measure the transport characteristics of the gas, as follows.

The transport characteristics are tabulated using the 4 components \(\sigma_{xx},\sigma_{yy},\sigma_{xy}\) and \(\sigma_{yx}\) of the so-called conductivity tensor. Once we know the conductivity tensor, we can use it to calculate how the current density \(\mathbf{j} = (j_x,j_y)\) flows in response to the electric field \(\mathbf{E} = (E_x,E_y)\) in the metal, through the equation

\[j_\alpha=\sum_\beta \sigma_{\alpha\beta}E_{\beta}.\]

By inverting this set of relations between current densities and electric field, we obtain the resistivities \(\rho_{xx}, \rho_{xy}, \dots\), which are more often reported in experimental data. Also, in two-dimensional systems there is no real difference between conductance and conductivity (or resistance and resistivity) - they have the same physical units. So the terms are somehow interchangeable.

The way to use the Hall bar device is to drive a current \(I\) along the \(x\) direction, so that there is a current density \(j_x=(I/W)\) where \(W\) is the width of the sample. There is no current density in the perpendicular direction.

We can measure the electric field using the Hall bar geometry from the voltage drops between the probes with voltages \(V_{1,2,3,4}\). We can then measure the \(x\)-component of the electric field from the longitudinal voltage drop \(V_L\sim (V_1-V_2)\) or \((V_3-V_4)\) according to the averaged equation

\[E_x \equiv \frac{V_1+V_3-V_2-V_4}{2L}.\]

Similarly, we can measure the \(y\)-component of the electric field from the Hall voltage \(V_H=(V_1-V_3)\) or \((V_2-V_4)\). Specifically we can calculate the electric field as:

\[E_y \equiv \frac{V_1+V_2-V_4-V_3}{2W}.\]

The Hall bar can only measure the conductance completely for isotropic or rotationally invariant systems. If we rotate the system by 90 degrees we can transform \(x\rightarrow y\) and \(y\rightarrow -x\). So we expect \(\sigma_{xx}=\sigma_{yy}=\sigma_L\), the longitudinal conductance. If we apply this same rotation transformation we conclude that \(\sigma_{xy}=-\sigma_{yx}=\sigma_H\), the Hall conductance.

So with rotational invariance the 4 component conductance tensor has only 2 independent components i.e. the longitudinal and Hall conductance. We can calculate these using the two electric fields \(E_{x,y}\) that we measure using the Hall bar. To do this, we solve the set of equations \(j_y=\sigma_L E_y - \sigma_H E_x=0\) and \(j_x=\sigma_L E_x+\sigma_H E_y\) to obtain \(\sigma_{L,H}\). We obtain the Hall conductance

\[\sigma_H=\frac{j_x E_y}{E_x^2+E_y^2}.\]