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I am encoutering some problems in understading the argument in David Tong's QHE lecture notes that the spectral flow of extended states give rise to currents in IQHE. First, Tong argues in page 36 that a particle moving aroung a solenoid has its $n=0$ state "transformed into the state that we previously lablled $n=1$" under spectral flow. Secondly, in page 52 by the same spirit, the symmetric gauge Landau level wavefunctions "shift from $m$ to $m+1$".

My questions are:

  1. Does the transformation of state in first part refer to a change of wavefunction and energy, or to a change of just energy? I think the later is more plausible since the set of wave fuctions are the eigenstates during the whole process of spectral flow. But the argument of the second part seems to be established on the former.

  2. By the "flux undoing" argument in page 53, it is shown that the set of extended eigenstates is the same when $\Psi$ is an integer multiply of $\Psi_0$, thus confirming the existence of spectral flow. However, I see no evidence that the spectral flow occurs in such a way as to tranform a $m$ state to a $m+1$ state. The situation here differs from section 1.5.3 in that the set of extended eigenstates are not the same during the whole process of spectral flow. So I think such an argument based on the situation of 1.5.3 seems ungrounded.

  3. A bit off topic: by the usual band analysis, electrons in a filled band have no states to scatter into, thus should give rise to no current. Why the electrons on filled Landau level contradict this analysis?

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To answer 1, let us consider the Hamiltonian $H_\Phi$ on page 53, and ignore $V(r,\theta)$ for now. Then you can easily check that the wavefunction in the lowest Landau level takes the form $$ e^{im\phi}r^{m+\frac{e\Phi}{2\pi}}e^{-\frac{r^2}{4l_B}} $$ Thus as $\Phi$ adiabatically increases by $\Phi_0$, the energy does not change (always in the LLL), but the wavefunction continuously evolves from $m$-th state to the $(m+1)$-th.

The analogy with the solenoid example in 1.5.3 is actually a bit tricky: in the solenoid, during $\Phi\rightarrow \Phi+2\pi$, the energy levels become degenerate at $\Phi=\pi$, where the adiabaticity breaks down. So there is really a level crossing and the ground state wavefunction jumps from the $n=0$ one to $n=1$ one, neither of which depends on $\Phi$. The situation is quite different in LLL: the energy stays the same, and the wavefunction evolves continuously.

Now for 2, we can no longer write down the wavefunction exactly, so have to rely on intuition. The heuristic argument is that the extended state is the one that more or less "looks like" the eigenstate without disorder, even though the $\phi$ dependence becomes more complicated. So if we imagine turning off the disorder potential slowly, these extended states should go back to the LLL wavefunctions discussed earlier. Then we can expect that the same spectral flow occurs for the extended states. The "flux undoing" argument shows that localized states do not care about $\Phi$.

For 3, note that LLL is an insulator, in the sense that there is no longitudinal current (i.e. in the direction of the applied electric field). However, the Hall current is very different. It is not dissipative, so no need to have empty states to scatter into. Later in Tong's note he derived the TKNN formula for the Hall conductance.

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  • $\begingroup$ Thanks for your answer! If I denote the wave function by $$\psi_{\Phi, m}=e^{im\phi}r^{m+\frac{e\Phi}{2\pi\hbar}}e^{-\frac{r^2}{4l^2_B}}$$, then I can see the functional form of $\psi_{\Phi+\Phi_0, m}$ and $\psi_{\Phi, m+1}$ only differ by an overall factor $e^{i\phi}$, thus confirming the transformation of state. However, I don't know why you say that "as Φ adiabatically increases by $\Phi_0$, the energy does not change''? I calculated the energy and get $E_{\Phi ,m}=\frac{2(m+{\Phi\over\Phi_0})+1}{2m_e}\frac{\hbar^2}{2l^2_B}$, thus the energy monotonically increase with $\Phi$? $\endgroup$
    – Cheng Tao
    May 23, 2022 at 10:46
  • $\begingroup$ Secondly, I am thinking about a situation where there are only 2 extended states on a filled Landau level, one on the inner ring and one on the outer ring. If I start with an inner ring state, then your intuition will say that it lands on the outer ring state after one spectral flow. However, if now I turn off the disorder, I should expect it to be close to where it would have been without the disorder, that is, a little bit outward than the inner ring but still far from the outer ring. So I am wandering if this intuition still stands from this perspective? $\endgroup$
    – Cheng Tao
    May 23, 2022 at 10:56
  • $\begingroup$ @ChengTao For your first comment, I think there is a wrong sign in Tong's note. I'll edit the answer to clarify this. But the conclusion is that the energy should not depend on $m$, otherwise there would be no degenerate Landau levels to begin with. $\endgroup$
    – Meng Cheng
    May 23, 2022 at 14:44
  • $\begingroup$ @ChengTao For your other comment, without disorder the spectral flow occurs for all $m$, so it is like shifting all the orbitals by $1$. $\endgroup$
    – Meng Cheng
    May 24, 2022 at 5:03
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    $\begingroup$ Indeed, Tong made a sign error on $\phi$ in page 52, and the spectrum would have no $m$ and $\Phi$ dependence if the wave function is corrected to $$ e^{-im\phi} \space r^{m-\frac{e\Phi}{2\pi\hbar}} \space\space e^{-\frac{r^2}{4l^2_B}} $$ Here I made another correction on the sign of $\Phi$, which shows the electrons actually move inwards the ring. I think it makes sense because charged particles generally move in the direction of $E\times B$ irrespective of charge. $\endgroup$
    – Cheng Tao
    May 25, 2022 at 11:19

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