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?

  • $\begingroup$ You can use \langle \rangle $\langle\rangle$ instead of < > $\endgroup$ – Superfast Jellyfish Mar 2 at 16:43
  • 1
    $\begingroup$ thanks for the reminder, edited $\endgroup$ – 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: 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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.