More Rigorous Laughlin pumping argument I'm having a hard time understanding how the concept of spectral flow helps us compute the hall conductivity, and in particular, Laughlin's pumping argument. I think that this question summarizes the problem pretty well and provides insight towards my confusion. However, I can't quite see why the answer is "rigorous" (in the sense of physicists).
Indeed, for a charged particle $e$ and given solenoid of flux $\Phi$, the Hamiltonian is given by
$$
H = H_r + \frac12 \left[ \frac1r \left(J_z-\frac{e\Phi}{2\pi}\right) -\frac{r\omega_c}{2}\right]^2, \quad H_r\equiv-\frac{1}{2 r} \frac{\partial}{\partial r}\left(r \frac{\partial}{\partial r} \right)
$$
Where $J_z = -i \partial_\phi$ is the angular momentum, $\omega_c$ is the cyclotron frequency and I set $\hbar =m =1$.
Let us now start in the lowest Landau level, i.e., $\psi_0=|n=0,m\rangle$. Notice that $J_z \psi_0 = m \psi_0$ and that $J_z$ commutes with the Hamiltonian $H(t)$ as the solenoid flux $\Phi(t)$ varies. Therefore, its evolution $\psi(t) =U(t) \psi_0$ must also satisfy $J_z \psi(t) = m \psi(t)$. For simplicity, let us say at $t=1$, the flux $\Phi(t=1)=2\pi/e \equiv \Phi_0$.
Here's the part that get's tricky.  If you were to assume that $H(t)$ does not mix higher Landau levels, i.e., $\langle n',m'|H(t)|n=0,\tilde{m}\rangle=0$ for all $n'\ne0$, then we see that $\psi_1$ must be spanned by $|n=0,\tilde{m}\rangle,\tilde{m}\in \mathbb{Z}$, and since it has angular momentum $m$, it must be proportional to $|n=0,m\rangle$, which gives us an incorrect answer. Therefore, there must be some kind of mixing between the Landau levels. Let us approach the problem with consideration of spectral flow, i.e., consider that
$$
H(t=1)\psi_1=H_r \psi_1+\frac12 \left[ \frac1r (m-1) -\frac{r\omega_c}{2}\right]^2 \psi_1
$$
This formula indeed looks like
\begin{align}
H(t=0)|n=0,m-1\rangle&=H_r |n=0,m-1\rangle+\frac12 \left[ \frac1r (m-1) -\frac{r\omega_c}{2}\right]^2 |n=0,m-1\rangle \\
&= \frac{\omega_c}{2}|n=0,m-1\rangle
\end{align}
However, as far as I know, that doesn't quite relate $\psi_1$ with $|n=0,m-1\rangle$. In fact, by angular momentum alone, the two are orthogonal, and thus I can't understand why would $\psi_1$ be localized near the radius $r=\sqrt{2(m-1)}l_B$ like $|n=0,m-1\rangle$.
 A: Not sure if I should delete this answer, since I don't think it's technically wrong, but please refer to my other answer for a more simple and rigorous argument using the adiabatic theorem.
After some thought, I think I'm approaching the problem the wrong way. Indeed, what is actually happening is that given a certain Fermi energy $E_F$, we are looking at the many-body ground state, filled with states with energy $E$ less than the fermi energy $E_F$. This follows from the fact that the many-body Hamiltonian is the non-interacting sum of the single-particle Hamiltonian.
Let $E_F$ be between the $n=0$ and $n=1$ Landau levels, and first consider the case where the flux $\Phi=0$. Then the ground state would technically be filled with infinite electrons due to the massive degeneracy. To avoid this, let us say that there exists an extra term in the Hamiltonian $H$ which costs us a lot of energy once the angular momentum $m> M$. In this case, the ground state would be filled with the states $|n=0,m\rangle, m\le M$. In particular, the "furthest" electron is localized near $r\sim \sqrt{2M} l_B$.
Now let us consider the Hamiltonian $H_1$ with flux $\Phi = 2\pi/e\equiv \Phi_0$. Notice that the Landau states $|n=0,m\rangle = R_m(r) e^{im\phi}$ in polar coordinates. It's then not too hard to see that states of the form $\propto R_{m-1} (r) e^{im\phi}$ are eigenstates of the flux Hamiltonian $H_1$. Since the maximum angular momentum $m=M$, we see that the eigenstate $R_{M-1}(r) e^{iM\phi}$ describes the "furthest" electron in the many-body ground state, and thus is localized near $r\sim \sqrt{2(M-1)}l_B$. Hence, by adding a full flux, a single charge is transported. The sign is different from the usual case since I assumed the charge was $e$ instead of $-e$.
I think this is the right way of thinking about the Laughlin pumping argument, which is inspired by this paper [1] which gives a rigorous proof of the argument. Please correct me if I'm mistaken.
EDIT. An interesting thing about paper [1] is that if you look at the equation in Theorem 4.2, and plugin
$$
p(x,y) = \sum_{m=0}^\infty \phi_m (x) \phi_m^*(y), \quad \phi_m(x) = |n=0,m\rangle
$$
which is "heuristcally" the integral kernel of the spectral projection onto the lowest Landau levels, then the RHS is exactly equal to the winding number of the gauge transformation $N(U)$ (flux quantum), which gives us the charge transported.
[1] Avron, J.E., Seiler, R. & Simon, B. Charge deficiency, charge transport and comparison of dimensions. Commun.Math. Phys. 159, 399–422 (1994). https://doi.org/10.1007/BF02102644
A: So I revisited this problem after some time, and I now think I can provide a simple rigorous explanation of Laughlin's pumping argument, using the adiabatic theorem.
Indeed, let $H(\Phi)$ denote the Hamiltonian with flux $\Phi$ and $H_0$ be that with flux $\Phi = 0$. As stated in the prompt, $J_z$ commutes with $H(\Phi)$ for all flux $\Phi$ and thus we can restrict ourselves to a single angular-momentum $m$, i.e., let $P_m$ denote the projection operator onto the eigenspace $J_z=m$, and only consider $H(\Phi)P_m$. Let $\Psi_0$ denote the (unique up to $U(1)$) ground state of $H_0 P_m$ and let $\Phi$ evolve adiabatically from $0\to \Phi_0\equiv 2\pi/e$. Since during this process, there are no energy level crossing, the adiabatic theorem holds and $\Psi(\Phi)$ must always be the ground state of $H(\Phi)P_m$ in the eigenspace $P_m$ (having angular momentum $m$) for all flux $\Phi$. In particular, notice that $H(\Phi_0 = 2\pi/e)P_m = J^+ H_0 P_{m-1} J^-$ where $J^{\pm}$ are the unitary operators which increase/decrease the angular momentum by $1$, i.e., $J^{\pm} \psi(r,\theta,\varphi) = e^{\pm i\varphi} \psi(r,\theta,\varphi)$.
Notice that the ground state of $H_0 P_{m}$ in $P_{m}$ is of the form $R_{m}(r) e^{i(m)\varphi}$ where $R_{m}(r)$ is centered around $r\sim r_m\equiv l_B \sqrt{2m}$. Therefore, we have
$$
\Psi(0) = R_m(r)e^{im\varphi} \\
\Psi(\Phi_0 = 2\pi/ e)=R_{m-1}(r) e^{im\varphi}
$$
where I have omitted the possible $U(1)$ coefficient in front of the state functions.
As you can see, $\Psi(0),\Psi(\Phi_0)$ both have angular momentum $J_z=m$, but it has moved from being centerer around $r_m \to r_{m-1}$
