[Physics] Aharonov-Bohm Effect and Integer Quantum Hall Effect

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What is the relationship between Aharonov-Bohm effect and Integer Quantum Hall effect?

Best Answer

There is a famous argument by R. Laughlin Phys. Rev. B 23, 5632–5633 (1981) explaining the integer quantum Hall effect based on the Aharonov-Bohm effect. This argument is explained in the following lecture notes by Manfred Sigrist (page 70, please see figure 3.17). The argument goes as follows:

Consider a system of electrons moving on a two dimensional annulus subject to a very large magnetic field so, they are constrained to the (degenerate) lowest Landau levels, because the energy they need to shift to an excited level is very large. In addition suppose that an electric field $E$ is applied in the radial direction Suppose that an increment of a uniform magnetic flux is applied in the annulus hole. According to Aharonov-Bohm, if this increment is an integer multiple of $\frac{\hbar c}{e}$, the physics should remain the same because in this case the increment can be removed by a gauge transformation.

Suppose, now that this increment is applied adiabatically, in this case the (mean) radius of the Landau level will increase adiabatically and when the flux becimes an integer, the electron will necessarily occupy the next lowest Landau level of the original Lagrangian because the Lagrangian is the same up to a gauge transformation. In particular, the net change in the magnetic and electric energies should be zero. The net change in the magnetic energy is:

$ \delta E_{M} = \frac{e}{mc} \mathbf{p}.\mathbf{\delta A} = \frac{1}{c} I_{\phi} \delta \Phi = \frac{1}{c} j_{\phi} b \frac{\hbar c}{e} $

Where $I_{\phi}$ is the angular current and $\delta \Phi$ is the flux increment $j_{\phi}$ is the angular current density and $b$ is the average radial distance between two Landau Levels. The electric energy is: $ \delta E_{E} =-e E b$ Equating the two contributions one gets the contribution of a single electron to the hall conductivity:

$\sigma_H = \frac{j_{\phi}}{E} = \frac{e^2}{\hbar}$

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