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


1 Answer 1


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:

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

  • $\begingroup$ How do you prove that if the increment is an integral multiple of $\frac{\hbar c}{e}$ the physics should remain the same? Is this applicable for this particular case only? I have searched google for the Arahanov-Bohm effect, but I couldn't prove this claim. $\endgroup$
    – user7757
    Apr 22, 2013 at 11:58
  • $\begingroup$ What happens when the flux is in between two integral multiples? Won't this cause the physics to change? But no change is observed until the next quanta is reached. $\endgroup$
    – user7757
    Apr 22, 2013 at 14:06
  • $\begingroup$ @ramanujan_dirac: The Schroedinger equation a particle moving on a circular ring is invariant under the gauge transformation $\psi\rightarrow e^{i\phi} \psi$, $ A_{\theta }= A_{\theta } + \frac{c \hbar}{eR} \partial_{\theta}\phi$. $\endgroup$ Apr 22, 2013 at 14:53
  • $\begingroup$ @ramanujan_dirac: (cont.) The gauge transformation $\phi$ must be a true function on the circle, i.e. $\phi(0) = \phi(2\pi)$. Thus we can choose a (large i.e., nonhomotopic to the identity) gauge transformation $\phi = n \theta$ to remove the integer part of the flux, such that only its fractional part will enter the Schroedinger equation. $\endgroup$ Apr 22, 2013 at 14:53
  • $\begingroup$ Thanks! After thinking about the question at hand and going through Laughlins original paper, I have developed a slightly different understanding of the question. The IQHE system has two quantum numbers, one can be taken to be that of the landau level, and the other is the y coordinate of the guiding center which is responsible for the degeneracy. The change in flux would act on the extended states through the center of the cyclotron motion(the guiding center). However, as the flux change approaches to that of one quanta, the y coordinate edge states would no longer be able to increase $\endgroup$
    – user7757
    Apr 22, 2013 at 15:44

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