I am trying to work through this paper which goes through the Atom Optics Kicked Rotor. Starting with the Hamiltonian:

$$ H = \frac{p^2}{2m}+ K \cos(2k_Lx)\sum \delta(t-nT)$$

This corresponds to a floquet time evolution operator $U = U_{free}U_{kick}$ Where $U_{free} = \exp[-i T\ p^2/2m]$ and $U_{kick} = \exp[-i \tau K \cos(2k_Lx)]$ ($\tau$ is pulse length).

The paper states that it uses a split step method with the initial wavefunction: $\psi\propto \exp(-x^2/\sigma^2)\exp(-ik_i x)$, which is a plane wave with momentum $k_i$ with a gaussian position dependence. They then state that the action of $U_{free}$ on a momentum eigenstate $|k\rangle$ is $\rho|k\rangle = \hbar\bar{k}|k\rangle$.

I am confused as to how figure 1 would be generated from these operators. They say an initial zero momentum state ($k_i = 0$) and resonance case ($\bar k = 2\pi$). After an odd number of steps, there are only peaks at $k = 0,\pm k_L$. If I do 0 kicks and FFT the initial wave function, I will get a gaussian in momentum space as well.

My primary questions are (a): When they state $U_{free}$ acts as $\rho|k\rangle = \hbar \bar k|k\rangle$, those $\bar k$ are actually ln(U), correct? So the proper expression for $\bar k$ is $T k^2/2m$

(b): How does an initial state of momentum k evolve to $k\pm n k_L$ based on $U_{kick}$?


(a) I don't quite understand what you mean by the $\rho$ operator acting on $\left| k \right\rangle$, but the free evolution operator $U_{free} = \exp [-i T \hat{p}^2 / 2m]$ acting on $\left| k \right\rangle$ just tags the basis-ket with a phase factor.

Therefore, $U_{free}\left| k \right\rangle = \exp [-i T k^2 / 2m]\left| k \right\rangle$

(b)If you apply the kick operator $U_{kick}=\exp [-i \tau K \cos(2k_L \hat{x})]$, to a plane-wave state in position-space ($\exp(-i2k_Lx)$) and observe its Fourier-transform, you'll find the momentum space populated similar that observed in the referred article.

Nice working code for the split-operator method can be found at http://albi3ro.github.io/M4/Time-Evolution.html.


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