Evolution of entanglement under an Ising-like Hamiltonian with particle losses

We present an analytical compact solution for the density matrix and all correlation functions of two collective-macroscopic spins evolving via an Ising-like Hamiltonian in the presence of particle losses. The losses introduce nonlocal phase noise which destroys highly entangled states arising in the evolution. On the other hand, the states appearing at relatively short time scales, possessing Einstein-Podolsky-Rosen (EPR)-like entanglement, will survive. Applying our solutions to the recently proposed scheme to entangle two Bose-Einstein condensates, we estimate the optimal number of atoms for EPR correlations.

Figure 1

Negativity as a function of time in case. Without losses ${\mathrm{\Gamma }}_1={\mathrm{\Gamma }}_0=0$ (blue dashed line); with losses ${\mathrm{\Gamma }}_1={\mathrm{\Gamma }}_0=0.01{\chi }_{ab}$ (red solid line). Other parameters are as follows: ${\chi }_{ab}=\chi$ and $N=M=10$.

Figure 2

The entanglement criterion $E_{\mathrm{EPR}}^2$ for the various numbers of particles as a function of rescaled time. For comparison, $E_{\mathrm{EPR}}^2$ calculated with losses is shown as dashed lines. The thickness of the line refers to $N$: from least to most thick, $N={10}^2,{10}^3,{10}^4,5\times {10}^4$. Other parameters are as follows: ${\mathrm{\Gamma }}_1={\mathrm{\Gamma }}_0={\chi }_{ab}$ and $\chi =0$. The optimal angles are $\alpha =\beta =0$.

Figure 3

Cuts of the Husimi $Q$ functions and partial contributions for $N=10,{\chi }_{ab}=\chi , \mathrm{\Gamma }=0.01\chi$, in the case where particles are lost only in the mode $|0\rangle$ of condensate “a”. The Husimi functions are computed in ${\chi }_{ab}t=\pi$. (a) $Q({\varphi }_a,{\varphi }_b)$ in the subspace where no particle is lost. (b) $Q^{\prime }({\varphi }_a,{\varphi }_b)$ in the subspace where one particle in the mode 0 of condensate “a” is lost. The Husimi functions of selected partial contributions to (b), from (c) the trajectory in which the transition occurred in $t_1=\pi /4$ and (d) the trajectory in which the transition occurred in $t_1=\pi /4$.

Figure 4

The scheme showing the central stage of the protocol to create the EPR-entangled states based on four Bose-Einstein condensates trapped in state-dependent potentials. As sketched in the plot, we assume that all BECs are in the Thomas-Fermi regime, hence they have parabolic shapes. The condensates “a0” and “b1” have lower densities than condensates “a1” and “b0” due to repulsive interactions between different species, i.e., collisions $|0\rangle -|1\rangle$. After the interaction time, the traps holding BECs should be moved to reach two bimodal condensates “a” and “b”. Only then the measurements should be taken to prove the EPR entanglement.

Figure 5

The optimized value of EPR condition $E_{\min }^2=\left({min}_{\alpha ,\beta ,t}{E}_{\mathrm{EPR}}^2\right)$ (minimized with respect to time and angles $\alpha , \beta$; $E_{\text{min}}^2<1$ indicates the presence of EPR entanglement) as a function of the total number of particles. The atoms are assumed to be evenly distributed to four identical harmonic traps with the trap frequency: (top) $\omega =2\pi \times 200\mathrm{Hz}$ and (bottom) $\omega =2\pi \times 1000\mathrm{Hz}$. The lines correspond to the effective model, with one-body losses artificially enhanced to mimic higher-order losses according to (41). The symbols are results of a full numerical solution of the master equation with all kinds of jumps, solved with the help of the quantum trajectory method (200 Monte Carlo realizations). One-body losses (pluses), one- and three-body losses together (crosses), and all jumps together (stars). In both cases the parameters $\chi , {\chi }_{\mathrm{ab}}$, and loss rates were calculated from the formulas (37), (38), and (43), respectively. The atomic decay constants are $K_2=8\times {10}^{-20}{\mathrm{m}}^3/\mathrm{s}$ and $K^{\left(3\right)}=6\times {10}^{-42}{\mathrm{m}}^6/\mathrm{s}$. The rate of one-body losses is ${\mathrm{\Gamma }}_{\epsilon \sigma }=K^{\left(1\right)}=0.5\mathrm{Hz}$, irrespectively, on $\epsilon$ and $\sigma$. The scattering length is $a=100$ Bohr radii and mass is $m=87$ in atomic units.