Preparation of Entangled and Antiferromagnetic States by Dissipative Rydberg Pumping

We propose and analyze an approach for preparation of high fidelity entanglement and antiferromagnetic states using Rydberg mediated interactions with dissipation. Using asymmetric Rydberg interactions the two-atom Bell singlet is a dark state of the Rydberg pumping process. Master equation simulations demonstrate Bell singlet preparation fidelity $\mathcal{F}=0.998$. Antiferromagnetic states are generated on a four-spin plaquette in agreement with results found from diagonalization of the transverse field Ising Hamiltonian.

Figure 1

Level diagram for pumping into an asymmetric ground state using unequal Rydberg interactions. The boxed inset shows how $\pi$ polarized light can be used to recycle all other Zeeman ground states in the example of Cs atoms. See text for details.

Figure 2

Comparison of $P_{\mathrm{AF}}$ from Eq. (2) (solid line) with singlet state fidelity $F_{\mathrm{Bell}}$ from numerical solution of Eq. (3) (red dots). Parameters: $\Omega /2\pi =0.01\mathrm{MHz}$, $\gamma =1./(0.73\mathrm{ms})$, ${\Delta }_{34}/{\Delta }_{33}=0.2$, ${\Omega }_R=2{\Omega }^2/{\Delta }_{33}$, and ${\Omega }_g=0.5{\Omega }_R$. The atomic parameters correspond to the Cs $125p_{1/2}$ state with $|3⟩$, $|4⟩$ $m_j=\pm 1/2$ and the quantization axis perpendicular to the line containing the atoms. The inset shows the time dependence for ${\Delta }_{33}/2\pi =1\mathrm{MHz}$ (solid lines are populations and the dashed line is coherence).

Figure 3

Population of ${\mathrm{AF}}_{\pm }$ states from Monte Carlo simulations of Eqs. (3, 4) on a square plaquette and averaged over 100 trajectories. The blue (lower) and red (upper) curves give the populations of $|1212⟩$ and $|2121⟩$. The inset shows the total population in states $|1212⟩$ and $|2121⟩$ as a function of $B/J$ from numerical diagonalization of (5) (solid blue line) and from Monte-Carlo simulations of Eq. (3) (red dots). Numerical parameters were $\Omega /2\pi =0.01\mathrm{MHz}$, $\gamma =1./(0.3\mathrm{ms})$, ${\Delta }_{33}/2\pi =0.4\mathrm{MHz}$, ${\Delta }_{34}/{\Delta }_{33}=0.85$, $\Delta ={\Delta }_{33}/2$, ${\Omega }_R=2{\Omega }^2/{\Delta }_{33}$, and ${\Omega }_g=0$. The coupling strengths between opposite corners were ${\Delta }_{33}^{\prime }={\Delta }_{33}/8$ and ${\Delta }_{34}^{\prime }={\Delta }_{34}/8$.