Non-classical properties of a mechanical resonator coupled to a qubit

Abstract

We have studied the squeezing properties of a Nanomechanical resonator coupled to a qubit and driven by two resonant lasers of unequal intensities. The squeezing can be transferred from X quadrature to P quadrature by selecting the qubit states. Squeezing can be increased by increasing the parameter $\eta =g/\omega _0$, where g is the resonator-qubit coupling strength and $\omega _0$ is the resonator frequency. The strong laser does not play any role on squeezing. We have also shown the anti bunching effect of the phonon field by plotting the second order correlation function.

Appendix

1.1 Appendix A

In general the squeezing operator is given by,

$$\begin{aligned} s(\xi )=e^{\frac{1}{2}(\xi ^* {\hat{a}}^2-\xi {{\hat{a}}^\dag 2})} , \end{aligned}$$

(11)

This is an unitary evolution operator with $\xi =re^{i\theta }$, r is the squeezing parameter varies from $-\infty$ to $+\infty$ and $\theta$ varies from 0 to $2\pi$. We see that the squeezing operator corresponds to the Hamiltonian $H \propto ({\hat{a}}^2+{{\hat{a}}^\dag 2})$ which contains non linear terms in $a,a^{\dag }$. This non-linearity will create some dynamical link between ${\hat{a}}$ and ${\hat{a}}^\dag$ as shown in our paper [14] where we see that the time evolution of ${\hat{a}}(t)$ depends on ${\hat{a}}^\dag (t)$ and vice versa. This type of dynamics is necessary to create squeezed states.

1.2 Appendix B

The Hamiltonian of the system in rotating frame is given by,

$$ H = \hbar \omega a^{\dag } a + \hbar \Omega _{1} (\sigma _{ + } + \sigma _{ - } ) + i\hbar \Omega _{2} (\sigma _{ + } - \sigma _{ - } ) + \hbar g|e\rangle \langle e|(a + a^{\dag } ){\text{ }} $$

(B.1)

where $\omega ,a,(a^\dag ),\Omega _1,(\Omega _2),g,\sigma _-(\sigma _+)$ are frequency of NMR mode, annihilation (creation) operator for phonon field, coupling constants for qubit with first (second) external field, coupling constant for qubit with NMR mode, lowering (raising) operator for qubit, respectively. In order to get a clear picture of the interaction of the Hamiltonian (1), we perform a time-independent unitary transformation. ${\tilde{H}}= e^sHe^{-s}$, where $s=\eta |e\rangle \langle e|(a^\dag -a)$ and $\eta =g/\omega$. The term $e^s=e^{\eta (a^\dag -a)|e\rangle \langle e|}$, is the rotation operator for the qubit whereas it will act as the displacement operator in phonon mode a.

$$\begin{aligned}&\sigma _+\rightarrow \sigma _+e^{\eta (a^\dag -a)} , \sigma _-\rightarrow \sigma _-e^{-\eta (a^\dag -a)} \, \hbox {and} \\&\quad a\rightarrow a+\eta |e\rangle \langle e|. \end{aligned}$$

The above Hamiltonian takes the form,

$$\begin{aligned} \begin{array}{c} {\tilde{H}}=\hbar \omega a^\dag a+\hbar \Omega _1 (\sigma _+e^{\eta (a^\dag -a)} +\sigma _-e^{-\eta (a^\dag -a)})\\ +i\hbar \Omega _2(\sigma _+e^{\eta (a^\dag -a)}-\sigma _-e^{-\eta (a^\dag -a)}) \end{array} \end{aligned}$$

(B.2)

For most practical cases $\eta \sim 10^{-3}$, which is very small. That’s why we expand the above Hamiltonian up to 2nd order in $\eta$ and then rearranging the terms,

$$ \begin{aligned} \tilde{H} & = \hbar \omega a^{\dag } a + \hbar \Omega _{1} (\sigma _{ + } + \sigma _{ - } ) + \hbar \Omega _{1} \eta (\sigma _{ + } - \sigma _{ - } )(a^{\dag } - a) \\ & \quad + \frac{{\hbar \Omega _{1} \eta ^{2} }}{2}(\sigma _{ + } + \sigma _{ - } )(a^{\dag } - a)^{2} \\ & \quad + i\hbar \Omega _{2} (\sigma _{ + } - \sigma _{ - } ) + i\hbar \Omega _{2} \eta (\sigma _{ + } + \sigma _{ - } )(a^{\dag } - a) \\ & \quad - \frac{{i\hbar \Omega _{2} \eta ^{2} }}{2}(\sigma _{ + } - \sigma _{ - } )(a^{\dag } - a)^{2} \\ \end{aligned} $$

(B.3)

We further rewrite the above Hamiltonian in the interaction picture in which the interaction with the stronger field has been diagonalized. In this picture, the state of the system ${\tilde{\psi }}$ and the Hamiltonian ${\tilde{H}}$ are transformed to,

$$\begin{aligned} \begin{array}{c} |{\bar{\psi }} \rangle =e^{iht} |{\tilde{\psi }} \rangle {\bar{H}}=e^{iht}{\tilde{H}} e^{-iht} \end{array} \end{aligned}$$

(B.4)

Where $h=\Omega _1(\sigma _++\sigma _-)$. The qubit spin operators $\sigma _\pm$ transform as,

$$\begin{aligned} \begin{array}{c} e^{iht}\sigma _\pm e^{-iht}=\sigma _\pm cos^2\Omega _1t+\sigma _\mp sin^2\Omega _1t \\ \mp i\sigma _z sin(2\Omega _1t) \end{array} \end{aligned}$$

(B.5)

As we consider that the qubit is driven strongly such that $\Omega _1$ is very large($\Omega _1\gg g,\Omega _2 ,\omega ,\Gamma$), and therefore we can neglect the highly oscillating terms i.e. $sin(2\Omega _1t)\approx 0$

$$ \begin{aligned} & \sigma _{ + } \to \sigma _{ + } cos^{2} \Omega _{1} t + \sigma _{ - } sin^{2} \Omega _{1} t \\ & \quad = \sigma _{ + } (\frac{{1 + cos2\Omega _{1} t}}{2}) + \sigma _{ - } (\frac{{1 - cos2\Omega _{1} t}}{2}) \approx \frac{{\sigma _{ + } + \sigma _{ - } }}{2} \\ \end{aligned} $$

as $cos(\Omega _1 t)\approx 0$. Similarly, $\sigma _-\rightarrow \sigma _-cos^2\Omega _1t+\sigma _+sin^2\Omega _1t\approx \frac{\sigma _++\sigma _-}{2}$. Thus $\sigma _++\sigma _- \rightarrow \sigma _++\sigma _-$ and $\sigma _+-\sigma _- \rightarrow 0$. Under this approximation, the effective Hamiltonian becomes,

$$\begin{aligned} \begin{array}{c} H_{eff}=\hbar \omega a^\dag a +i\hbar \Omega _2 \eta \sigma _x(a^\dag - a)\\ +\frac{\hbar \Omega _1 \eta ^2}{2}\sigma _x(a^\dag -a)^2 \end{array} \end{aligned}$$

(B.6)

After simplification the above Hamiltonian becomes,

$$ \begin{aligned} \overline{{\hat{H}}} _{{eff}} & = \hbar \Omega _{{1^{\prime}}} \sigma _{x} + \hbar \omega _{0} \hat{a}^{\dag } \hat{a} + i\hbar \Omega _{2} \eta \hat{\sigma }_{x} \left( {\hat{a}^{\dag } - \hat{a}} \right) \\ & \quad + \frac{{\hbar \Omega _{1} \eta ^{2} }}{2}\hat{\sigma }_{x} \left( {\hat{a}^{2} + \hat{a}^{{\dag 2}} } \right) - \hbar \Omega _{1} \eta ^{2} \sigma _{x} \widehat{{a^{\dag } }}\hat{a} \\ \end{aligned} $$

(B.7)