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The question in short:

A translationally-invariant system living on a ring is in a state of momentum $p_0$. How does the momentum change after threading one magnetic flux quantum through the ring?

Let's add some details. Consider a system living on a $1$D ring with lenght $L_x$ or, if you prefer, an integer number $L_x$ of sites with periodic boundary conditions.
I will call $\hat x$ the direction tangential to the ring.
In the initial state there are no external fields; the system has a translationally-invariant hamiltionian $H_0$ and is in the state $|\psi_0\rangle$, eigenstate of the momentum operator with eigenvalue $p_0$.
At time $t=0$ we turn on a time-dependent magnetic flux $\Phi(t)$ piercing the ring, whose value grows with time from $0$ to one flux quantum $\Phi_f=hc/e$ at the final time $t_f$.
This effect is included in the hamiltonian through the introduction of a vector potential $\vec{A}(t)=\Phi(t) \hat x$ directed tangentially along the ring .

My aim is answering the question: What is the value (eigenvalue or mean value) of the momentum at $t=t_f$ ?

Through $\vec{A}(t)$, the hamiltionan acquires a time-dependence, $H(t)$ and the system evolves to the final state $$ |\psi_f\rangle=\mathcal{T}\left( e^{-i\int_0^{t_f} H(t)} \right) |\psi_0\rangle $$ with $\mathcal{T}$ being the time-ordering.
Let $T_x(a)$ the translation operator along $\hat x$; since the initial state is an eigenstate of momentum we have $$ T_x(a) |\psi_0\rangle = e^{ip_0 a} |\psi_0\rangle $$ Translational invariance is preserved by $H(t)$ so we also have $T_x(a) |\psi_f\rangle = e^{ip_0 a} |\psi_f\rangle $ so I would naively conclude that the momentum is unaltered.

My problem is: I have been reading in various references (and refs therein) who (almost quoting) state that, since the introduction of one flux quantum does not alter the spectrum of the hamiltonian (and I agree) they actually describe "the same physics" so we should transform back $\hat H(t_f)$ to $\hat H_0$ trough a certain operator $U_\Phi$ $$ U_\Phi H(t_f) U^{-1}_\Phi = H_0 $$ Analogously, those papers state that the momentum of the final state should be found by examining $U_\Phi |\psi_f\rangle$, yielding a different answer from the "naive" computation. What is the reasoning behind this final transformation?

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  • $\begingroup$ You can use \langle \rangle $\langle\rangle$ instead of < > $\endgroup$ – Superfast Jellyfish Mar 2 at 16:43
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    $\begingroup$ thanks for the reminder, edited $\endgroup$ – tbt Mar 2 at 16:46
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It might be revealing to plot the eigenstates of the system as a function of the magnetic flux $\Phi$ threading the ring. In the absence of a flux, $\Phi = 0$, the eigenstates of the system are just plane waves, $\psi_n (\theta) = e^{- n \theta}/\sqrt{2 \pi}$, and in suitable units the energy of each eigenstate is just given by $E_n = n^2$. So each state is doubly degenerate corresponding to a clockwise or anticlockwise angular momentum $\pm n \hbar$, except for the ground state which has $n = 0$.

Now consider applying the magnetic flux. The $\psi_n$ remain eigenstates of the Hamiltonian, but with a shifted eigenvalue $E_n = (n - \Phi)^2$. I plot this behavior here: enter image description here

You can see that the flux initially breaks the degeneracy of the $n \neq 0$ states, as one of them circulates in the same direction as the vector potential, while the other one opposes it. The energy of the ground state also increases as $\Phi$ increases. When one flux quantum threads the system, we have actually recovered the flux-free case; just as you state in the question we have the same spectrum as before. The difference is that the new ground state has evolved adiabatically from the $n=1$ state. If we continue increasing the flux the spectrum repeats periodically as expected.

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  • $\begingroup$ I am familiar with the spectrum that you so clearly present; what I am missing is the logic of measuring the momentum (i.e. acting with $T_x(a)$ )of the transformed state $U_\Phi |\psi_f\rangle$ instead of the adiabatically-evolved one, $|\psi_f \rangle$. Would you have any input on that? $\endgroup$ – tbt Mar 12 at 13:44
  • $\begingroup$ I think that the $U_\Phi$ operator transforms to a co-rotating frame. The lowest energy state at $\Phi = 1$ has $n=1$, so if you transform to a rotating frame you measure its momentum as $n=0$, and so recover the physics at $\Phi = 0$. $\endgroup$ – Clara Diaz Sanchez Mar 12 at 15:24

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