# Simulation of the Delta Kicked Rotor in Momentum Space

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.