Is there a way to justify formally the Laughlin pumping argument? It is often argued that the spectral flow of Landau levels while varying the flux inserted through the Corbino ring should account for the quantized condutivity. In the symmetric gauge we classify Landau levels with two quantum number $n$ and $m$. $n$ identifies the level, while $m$ is basically the radius of the wavefunction. The wavefunction are effectively ring-shaped. Laughlin argues that upon the insertion of one quantum flux the states flow as:

$$ \psi_{n, m} \rightarrow \psi_{n, m+1} $$

Thus one electron per filled Landau level moves outward.

Now the spectral flow is often introduced in the context of the one particle on a ring problem, where we have a flow of non degenerate states. Here the adiabatic theorem will save us from state mixing and after the insertion of one quantum of flux the each state will gain one excitation, like we have raised the ladder.

In the context of the Quantum Hall effect all the $m$ states are degenerate. Thus the analogy with the previous case doesn't carry over: we don't have any adiabatic arguument that protects the state while raising the flux.

Is there a rigorous way to justify Laughlin assumption?

Edit: in the comments it was pointed out that the particle on a ring problem suffer of similar ambiguity. The energy levels when $\Phi$ is the inserted flux are $E_n = \frac{\hbar^2}{2 m r^2} \left( n + \frac{\Phi}{\Phi_0} \right)^2$. The spectrum is two-fold degenerate when $\frac{\Phi}{\Phi_0} = \frac{1}{2}$. Thus invaliding the adiabaticity argument. How can we overcome this issue?

  • $\begingroup$ I just want to point out there is a similar ambiguity for the one-particle-on-a-ring problem. For MOST of the adiabatic evolution the energy levels are non-degenerate. However, there is always one point in the flux-pumping where the energy levels become twofold degenerate. See, e.g, the middle of figure 2.1 here. $\endgroup$ May 30, 2018 at 18:39

1 Answer 1


I solved the issue. I will explain in detail how the existence of an operator conserved by the adiabatic evolution allows us to make sense of the spectral flow argument. Let's start with the problem of the particle on a ring. Here the Hamiltonian (for a static inserted flux $\Phi$) is:

$$ \mathcal{H} = \frac{\hbar^2}{2m r^2} \left( -i \frac{\partial}{\partial \phi} + \frac{\Phi}{\Phi_0} \right)^2 $$

We can always diagonalize both $p_{\phi}$ and $\mathcal{H}$ and obtain a complete basis of eigenstate for the Hilbert space of the single particle on a ring. The eigenstate at $\Phi = 0$ are $\lbrace e^{i m \phi} \rbrace$. with energy $E_n = \frac{\hbar^2}{2m r^2} m^2$. If we insert a time dependent flux $\Phi (t)$ we get:

$$ \mathcal{H} = \frac{\hbar^2}{2m r^2} \left( -i \frac{\partial}{\partial \phi} + \frac{\Phi (t)}{\Phi_0} \right)^{2} $$

The momentum operator (up to an $\hbar$) is $p_{\phi} = - i \frac{\partial}{\partial \phi}$ which is conserved by the adiabatic switching on of the flux: $\left[ \mathcal{H}(t), \frac{\partial}{\partial \phi} \right] = 0$.

If we start the evolution with a eigenstate ${\psi (t = 0)}$ of the conserved operator:

$$p_{\phi} {\psi(t = 0)} = m {\psi(t = 0)}$$ the adiabatic evolution $U(t)$ conserves this status:

$$ U(t) p_{\phi} U^{-1}(t) U(t) {\psi(t = 0)} = m U(t) {\psi(t = 0)} $$

$$ p_{\phi} {\psi(t)} = m {\psi(t)} $$

Thus even if for $\frac{\Phi}{\Phi_0} = \frac{1}{2}$ there is a double degeneracy, the eigenstates with different $m$ values don't get mixed up. In other words after the adiabatic insertion of the flux the number $m$ is conserved but the energy has raised: $E_{n} = \frac{\hbar^2}{2m r^2} \left(m + 1 \right)^2$. Basically the state $\psi_{m}$ has moved to $e^{-i \phi} {\psi_{m+1}}$, where the gauge transformation has been made explicit.

We now see the Laughlin pumping argument. Here the momentum $p_{\phi}$ is conserved by the evolution and indeed in the symmetric gauge its eigenvalue is the quantum number internal to every Landau level. We start in a eigenstate ${\psi_m}$ of the momentum operator in the lowest Landau level. There is no mixing with higher Landau levels and the final state must have eigenvalue $m$. The spectrum of the Hamiltonian after the insertion of the flux can be linked to the spectrum of it before the insertion through a gauge transformation: $e^{-i\phi} \phi_{m}$. The eigenvector with eigenvalue $m$ is $e^{-i\phi} \phi_{m+1}$ since the exponential $e^{-i \phi}$ lower the eigenvalue by one. The $r$ dependence of $\phi_{m+1}$ is $r^{m+1}$ thus the state moves outward after an adiabatic insertion, while keeping its $\phi$ dependence.

  • $\begingroup$ I'm with you up until your last paragraph, which seems to be a little heuristic, and I can't quite grasp the logic. Indeed, I summarized my confusion here (physics.stackexchange.com/questions/650234/…). Any help would be appreciated. $\endgroup$ Jul 8, 2021 at 8:02
  • $\begingroup$ And I'm also a little confused about your gauge transform, since the gauge/angle $\phi$ is not well-defined when you go around the circle. $\endgroup$ Jul 8, 2021 at 8:04

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