Einstein-Podolsky-Rosen Entanglement Strategies in Two-Well Bose-Einstein Condensates

Criteria suitable for measuring entanglement between two different potential wells in a Bose-Einstein condensation are evaluated. We show how to generate the required entanglement, utilizing either an adiabatic two-mode or a dynamic four-mode interaction strategy, with techniques that take advantage of $s$-wave scattering interactions to provide the nonlinear coupling. The dynamic entanglement method results in an entanglement signature with spatially separated detectors, as in the Einstein-Podolsky-Rosen paradox.

Figure 1

(a) Two spatial modes $a$ and $b$; (b) two pairs of modes, each with two internal spin configurations, giving operators $a_1$, $a_2$ and $b_1$, $b_2$.

Figure 2

Adiabatic entanglement with interactions in a two-well potential. Solid line: Entropic entanglement [$E_{\mathrm{entropic}}=1-\epsilon (\rho )/{ϵ}_{\max }<1$] at $T=0\mathrm{K}$, with a number state. Dashed and dash-dotted lines: Entanglement signature ($E_{\mathrm{PE}}<1$) at $T=0\mathrm{K}$ and 50 nK, respectively, with a Poissonian mixture of numbers. In all cases, $⟨N⟩=100$.

Figure 3

(a) Squeezing of Schwinger spin operators $S_{\mathrm{dB}}$: $S_{+}=10{log}_{10}[{\Delta }^2({\widehat{J}}_A^{\theta }-{\widehat{J}}_B^{\theta })/n_0]$ (solid line), $S_{-}=10{log}_{10}[{\Delta }^2({\widehat{J}}_A^{\theta +\pi /2}+{\widehat{J}}_B^{\theta +\pi /2})/n_0]$ (dashed line), and $n_0=\frac{1}{2}(|⟨{\widehat{J}}_A^Y⟩|+|⟨{\widehat{J}}_B^Y⟩|)$ is shot noise (dotted line). (b) Entanglement ($E_{\mathrm{product}}$) based on the criterion (3) by the solid curve and $E_{\mathrm{sum}}$ in sum criterion (4) by the dashed curve. Here the parameters correspond to Rb atoms at magnetic field $B=9.131\mathrm{G}$, with scattering lengths $a_{11}=100.4a_0$, $a_{22}=95.5a_0$, and $a_{12}=80.8a_0$. $a_0=53\mathrm{pm}$. The coupling constant $g_{ij}\propto 2w_{\perp }{a}_{ij}$. Here $N_A=200$ and $\tau =g_{11}{N}_{A}t$.

Figure 4

The same as Fig. 3 but assuming $g_{12}=0$.