# Momentum shift after threading magnetic flux through a ring

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?

• You can use \langle \rangle $\langle\rangle$ instead of < > – Superfast Jellyfish Mar 2 at 16:43
• thanks for the reminder, edited – tbt Mar 2 at 16:46

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: 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.
• 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? – tbt Mar 12 at 13:44
• 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 $\Phi = 0$. – Clara Diaz Sanchez Mar 12 at 15:24