We can now see explicitly how the Laughlin pumping argument works, starting from the microscopic description of electrons in terms of Landau levels. Starting from the formulas we derived, it is a little difficult to do so in the Corbino geometry, which has an angular symmetry rather than a translational symmetry. It is very easy if we consider the Laughlin pump for electrons in a cylinder:

In fact the cylinder drawn above and the Corbino disk are completely equivalent - you can imagine deforming one into the other. The advantage of the cylinder is that we get to keep our \((x, y)\) coordinates. The Hall cylinder that we considered for Laughlin’s argument is in fact equivalent to a ribbon in the \((x, y)\) plane, with periodic boundary conditions \(x\equiv x+L\) in the \(x\) direction (\(L\) is the circumference of the cylinder). The periodic boundary conditions along the \(x\) direction discretize the allowed values of \(k\) as \(k=2\pi n/L\).

For the Laughlin pumping argument, we need to introduce a flux through the cylinder. Using Stokes’ theorem, we know that the line integral of the vector potential around the cylinder must be equal to the flux passing through it, \(\oint \textbf{dr}\cdot\textbf{A(r)}=\Phi\). So we can introduce a flux through the cylinder by choosing our vector potential \(\bf{A}\) as

\[\textbf{A}(x,y)=(B y +\Phi/L)\,\hat{\textbf{x}}\,,\]

very similar to the previous calculation. The resulting Hamiltonian for the states labeled by \(n\) is

\[H=p_y^2+\left(\frac{\hbar 2\pi n}{L}-e B y-\frac{e\Phi}{L}\right)^2\,.\]

Comparing the above equation to the quantum harmonic oscillator, we see that the harmonic oscillator levels must be centered at

\[y_0(n) = \left(n-\frac{\Phi}{\Phi_0}\right)\frac{h}{e B L}\,.\]

We see from this that the Landau level wave-functions are centered around a discrete set of rings at \(y_0(n)\) on the cylinder axis that are labelled by the integer \(n\). As \(\Phi\) is increased we see that the centers \(y_0\) move so that after one flux quantum \(\Delta\Phi=\Phi_0=h/e\) all the electrons have moved down by one step along \(y\), i.e. \(n \rightarrow n-1\). If \(n\) Landau levels are filled then a total charge of \(\Delta Q=n e\) will be transferred between the edges, in exact accordance with the Laughlin argument.

We can now look again at the Laughlin pump, monitoring at the same time the Landau levels. You can see that the total pumped charge jumps in integer steps each time a Landau level passes through the Fermi level.