Quantum synchronization of chaotic oscillator behaviors among coupled BEC–optomechanical systems

Abstract

We consider and theoretically analyze a Bose-Einstein condensate (BEC) trapped inside an optomechanical system consisting of single-mode optical cavity with a moving end mirror. The BEC is formally analogous to a mirror driven by radiation pressure with strong nonlinear coupling. Such a nonlinear enhancement can make the oscillator display chaotic behavior. By establishing proper oscillator couplings, we find that this chaotic motion can be synchronized with other oscillators, even an oscillator network. We also discuss the scheme feasibility by analyzing recent experiment parameters. Our results provide a promising platform for the quantum signal transmission and quantum logic control, and they are of potential applications in quantum information processing and quantum networks.

Appendices

Appendix

Simplification of the Bose-Einstein condensate Hamiltonian

The total Hamiltonian corresponding to such a system can be expressed as $\hat{H}=\hat{H}_\mathrm{{c}}+\hat{H}_\mathrm{{a}}+\hat{H}_\mathrm{{d}}$, where

$$\begin{aligned} \begin{aligned} \hat{H}_\mathrm{{c}}=\hbar \omega _\mathrm{{c}}\hat{a}^\dagger \hat{a}-\hbar {g}_0\hat{a}^\dagger \hat{a}\hat{q}+\dfrac{\hbar \omega _\mathrm{{m}}}{2}\left( \hat{p}^2+\hat{q}^2\right) +i\hbar E(\hat{a}^\dagger e^{-i\omega _\mathrm{{p}}t}-\hat{a}e^{i\omega _\mathrm{{p}}t}) \end{aligned} \end{aligned}$$

(10)

is the standard Hamiltonian of optomechanical system with the radiation pressure term $\hbar {g}_0\hat{a}^\dagger \hat{a}\hat{q}$ [28, 29]. $\hat{H}_\mathrm{{a}}$ describes the BEC and its interaction with the light field and correspondingly, $H_\mathrm{{d}}$ denotes the Hamiltonian of the thermal reservoir which hence to dissipation. In Eq. (1), $\hat{a}$ ($\hat{a^{\dagger }}$) is the bosonic annihilation (creation) operator for the optical cavity field, $\hat{q}$ ($\hat{p}$) is the position (momentum) operator for the oscillator. $\omega _\mathrm{{c}}$, $\omega _\mathrm{{p}}$ and $\omega _\mathrm{{m}}$ are the intrinsic frequencies of the cavity field, pump field and the oscillator, respectively. $g_0$ is the single-photon coupling coefficient satisfying $g_0=(\omega _\mathrm{{c}}/L)\sqrt{\hbar /m\omega _\mathrm{{m}}}$, where L and m are the cavity length and the oscillator mass, respectively [28]. E is the driving intensity of the pump field. In the rotating frame, $\hat{H}_\mathrm{{c}}$ is given explicitly by

$$\begin{aligned} \begin{aligned} \hat{H}_\mathrm{{c}}=-\hbar \Delta _\mathrm{{c}}\hat{a}^\dagger \hat{a}-\hbar {g}_0\hat{a}^\dagger \hat{a}\hat{q}+\dfrac{\hbar \omega _\mathrm{{m}}}{2}\left( \hat{p}^2+\hat{q}^2\right) +i\hbar E(\hat{a}^\dagger -\hat{a}), \end{aligned} \end{aligned}$$

(11)

via introducing cavity-pump detuning $\Delta _\mathrm{{c}}=\omega _\mathrm{{p}}-\omega _\mathrm{{c}}$.

The BEC trapped in a cavity optomechanical system can be simplified as a one-dimensional motion model. Previous works have proved that the internal excited state dynamic of the atom can be adiabatically eliminated if the light-atom detuning $\Delta _\mathrm{{a}}$ is large enough [44]. Here, we assume that the atom spontaneous emission can be ignored, in addition, two-body interaction can also be neglected if atomic density is low enough. Therefore, $\hat{H}_\mathrm{{a}}$ becomes [42]

$$\begin{aligned} \begin{aligned} \hat{H}_\mathrm{{a}}=\int \hat{\Psi }^{\dagger }(x)\left( -\dfrac{\hbar ^2}{2m_\mathrm{{a}}}\dfrac{{\text{ d }}^2}{{\text{ d }}x^2}+\hbar U_0\hat{a}^\dagger \hat{a}\cos ^2 kx\right) \hat{\Psi }(x){\text{ d }}x, \end{aligned} \end{aligned}$$

(12)

where $\hat{\Psi }(x)$ ($\hat{\Psi }^{\dagger }(x)$) is a bosonic field annihilation (creation) operator and $m_\mathrm{{a}}$ is the atomic mass. In this expression, $U_0$ is the far off-resonant vacuum Rabi frequency and $k=\omega _\mathrm{{p}}/c$ is the wave number of the light field. Here, we consider the optical field is so weak that the momentum side modes of the BEC are generated at $\pm 2\hbar k$. Then, the field operator $\hat{\Psi }(x)$ can be expanded as

$$\begin{aligned} \begin{aligned} \hat{\Psi }(x)\simeq [\hat{c}_0+\sqrt{2}\cos (2kx)\hat{c}_2]/\sqrt{L}. \end{aligned} \end{aligned}$$

(13)

Here, the bosonic annihilation operators $\hat{c_0}$ and $\hat{c_2}$ are, respectively, the zero-momentum state and side-mode component. By substituting Eq. (13) into Eq. (12) and finishing the integral, the Hamiltonian $\hat{H}_\mathrm{{a}}$ can be simplified as

$$\begin{aligned} \begin{aligned} \hat{H}'_\mathrm{{a}}=\dfrac{\hbar U_0}{2}\hat{a}^\dagger \hat{a}\left( \hat{c}^\dagger _0\hat{c}_0+\hat{c}^\dagger _2\hat{c}_2\right) +\dfrac{\sqrt{2}\hbar U_0}{4}\hat{a}^\dagger \hat{a}\left( \hat{c}^\dagger _0\hat{c}_2+\hat{c}^\dagger _2\hat{c}_0\right) +4\hbar \omega _\mathrm{{r}}\hat{c}^\dagger _2\hat{c}_2, \end{aligned} \end{aligned}$$

(14)

where $\omega _\mathrm{{r}}=\hbar k^2/2 m_\mathrm{{a}}$ is the effective frequency and $\hat{c}^\dagger _0\hat{c}_0$ ($\hat{c}^\dagger _2\hat{c}_2$) is the corresponding particle number operator of the zero-momentum mode (the side mode). Here, we assume that $\hat{c}^\dagger _0\hat{c}_0=N-\hat{c}^\dagger _2\hat{c}_2$, i.e., the particle loss is ignored and the population of zero-momentum mode is far greater than that of the side mode. If N is large enough, the zero-momentum mode can be regarded as a classical field, i.e., $\hat{c}_0$ and $\hat{c}^\dagger _0$ are thought as the c-numbers $\sqrt{N}$. With this condition, Eq. (14) becomes

$$\begin{aligned} \begin{aligned} \hat{H}'_\mathrm{{a}}=\dfrac{\hbar U_0N}{2}\hat{a}^\dagger \hat{a}+\dfrac{\sqrt{2N}\hbar U_0}{4}\hat{a}^\dagger \hat{a}\left( \hat{c}_2+\hat{c}^\dagger _2\right) +4\hbar \omega _\mathrm{{r}}\hat{c}^\dagger _2\hat{c}_2, \end{aligned} \end{aligned}$$

(15)

and the total Hamiltonian of the system can be gained

$$\begin{aligned} \hat{H}_\mathrm{{c}}= & {} -\hbar \Delta \hat{a}^\dagger \hat{a}-\hbar {g}_0\hat{a}^\dagger \hat{a}\hat{q}+\dfrac{\hbar \omega _\mathrm{{m}}}{2}\left( \hat{p}^2+\hat{q}^2\right) +i\hbar E(\hat{a}^\dagger -\hat{a})\nonumber \\&\quad +4\hbar \omega _\mathrm{{r}}\hat{c}^\dagger _2\hat{c}_2-\hbar g_\mathrm{{sm}}\hat{a}^\dagger \hat{a}\hat{Q}. \end{aligned}$$

(16)