Collapses and revivals of entanglement in phase space in an optomechanical cavity

Abstract

We analyse the dynamics of an optomechanical cavity considering two modes of the electromagnetic field and a mechanical resonator. As a result of this interaction, the initial non-entangled state evolves into a final entangled state. In this work, we use the $Q(\alpha )$ function to observe the behavior of the system in the phase space, finding that such entanglement is a function of time showing collapses and revivals. To carry out this analysis, we use the generalization of the single-party $Q(\alpha )$ function towards three-party systems $Q(\alpha ,\mu ,\nu )$ and we ascertain that the entanglement perfectly maps into the $Q(\alpha ,\mu ,\nu )$ function because it becomes non-separable too. To prove that the Q function accurately certifies entanglement, we compare its behaviour with the plots of the van Loock inequalities. Additionally, as the optomechanical systems were proposed to carry out quantum information tasks, we investigate the kind of phase gate that is produced when we keep the Kerr term that arises in the evolution operator.

Appendix A: Factorization of time evolution operator

In this appendix the time evolution operator will be factorized, but before that it is convenient to write the Hamiltonian in equation (4) as follows:

$$\begin{aligned} {\hat{H}}=\xi ({\hat{A}}+{\hat{B}}+{\hat{C}}+{\hat{D}}), \end{aligned}$$

(A1)

where we apply the following definitions:

$$\begin{aligned} \xi= & {} -\frac{it}{\hbar }, \end{aligned}$$

(A2)

$$\begin{aligned} {\hat{A}}= & {} \hbar \omega _1{\hat{a}}^{\dagger }_1{\hat{a}}_1, \end{aligned}$$

(A3)

$$\begin{aligned} {\hat{B}}= & {} \hbar \omega _2{\hat{a}}^{\dagger }_2{\hat{a}}_2, \end{aligned}$$

(A4)

$$\begin{aligned} {\hat{C}}= & {} \hbar \omega _m{\hat{b}}^{\dagger }{\hat{b}}, \end{aligned}$$

(A5)

$$\begin{aligned} {\hat{D}}= & {} \hbar (g_1 {\hat{a}}_1^{\dagger }{\hat{a}}_1+g_2 {\hat{a}}_2^{\dagger }{\hat{a}}_2)({\hat{b}}+{\hat{b}}^{\dagger }). \end{aligned}$$

(A6)

From the above definitions, the following commutation relations are deduced:

$$\begin{aligned} {[}{\hat{A}},{\hat{B}}]=[{\hat{A}},{\hat{C}}]=[{\hat{A}},{\hat{D}}]=[{\hat{B}},{\hat{C}}]=[{\hat{B}},{\hat{D}}]=0, \end{aligned}$$

(A7)

and the time evolution operator can be written in a more simple way:

$$\begin{aligned} {\hat{U}}(t)=e^{\xi ({\hat{A}}+{\hat{B}}+{\hat{C}}+{\hat{D}})}. \end{aligned}$$

(A8)

Now, by using the commutation relations and the well-known Baker–Cambell–Hausdorff formula we get the following expression:

$$\begin{aligned} e^{{\hat{A}}+{\hat{B}}}=e^{-\frac{1}{2}[{\hat{A}},{\hat{B}}]}e^{{\hat{A}}}e^{{\hat{B}}}, \end{aligned}$$

(A9)

it is also possible to write the temporal evolution operator in the following way:

$$\begin{aligned} {\hat{U}}(t)=e^{\xi {\hat{A}}}e^{\xi {\hat{B}}}e^{\xi ({\hat{C}}+{\hat{D}})}. \end{aligned}$$

(A10)

The last factor in (A10) involves the operators ${\hat{C}}$ and ${\hat{D}}$, then it is necessary to find their commutation relation, which are given by the following expression:

$$\begin{aligned}{}[{\hat{C}},{\hat{D}}]=\hbar \omega _{m}{\hat{k}}({\hat{b}}^{\dagger }-{\hat{b}})={\hat{E}}, \end{aligned}$$

(A11)

where for convenience of this calculation the following operator is defined ${\hat{k}}=k_1{\hat{A}}+k_2{\hat{B}}$ and $d_i=g_i/\omega _i$ is a constant. Taking into account the result (A11) the factorization method reviewed in Ref. [30] could be applied, because $[{\hat{C}},{\hat{D}}]\ne 0$. For this purpose, the following functions are defined

$$\begin{aligned} {\hat{F}}(\xi )= & {} e^{\xi ({\hat{C}}+{\hat{D}})}, \end{aligned}$$

(A12)

$$\begin{aligned} {\hat{F}}(\xi )= & {} e^{f_0(\xi ){\hat{k}}^2}e^{f_1(\xi )[{\hat{C}},{\hat{D}}]}e^{f_2(\xi ){\hat{D}}}e^{f_3(\xi ){\hat{C}}}. \end{aligned}$$

(A13)

The first expression is an auxiliary function, while the second one corresponds to the proposed factorization. These definitions belong to the first step of the method.

The second step consists in differentiating (A12) and (A13) with respect to the parameter $\xi$ to obtain

$$\begin{aligned} \frac{\mathrm{d} }{\mathrm{d} \xi } F(\xi )=({\hat{C}}+{\hat{D}})e^{\xi ({\hat{C}}+{\hat{D}})}=({\hat{C}}+{\hat{D}})F(\xi ), \end{aligned}$$

(A14)

and on the other hand

$$\begin{aligned} \frac{\mathrm{d} }{\mathrm{d} \xi } F(\xi )= & {} \frac{\mathrm{d} f_0(\xi )}{\mathrm{d} \xi } {\hat{k}}^2 e^{f_0(\xi ){\hat{k}}^2}e^{f_1(\xi )[{\hat{C}},{\hat{D}}]}e^{f_2(\xi ){\hat{D}}}e^{f_3(\xi ){\hat{C}}}\nonumber \\&+\, \frac{\mathrm{d} f_1(\xi )}{\mathrm{d} \xi } e^{f_0(\xi ){\hat{k}}^2}[{\hat{C}},{\hat{D}}]e^{f_1(\xi )[{\hat{C}},{\hat{D}}]}e^{f_2(\xi ){\hat{D}}}e^{f_3(\xi ){\hat{C}}}\nonumber \\&+\,\frac{\mathrm{d} f_2(\xi )}{\mathrm{d} \xi } e^{f_0(\xi ){\hat{k}}^2}e^{f_1(\xi )[{\hat{C}},{\hat{D}}]}{\hat{D}}e^{f_2(\xi ){\hat{D}}}e^{f_3(\xi ){\hat{C}}}\nonumber \\&\, + \frac{\mathrm{d} f_3(\xi )}{\mathrm{d} \xi } e^{f_0(\xi ){\hat{k}}^2}e^{f_1(\xi )[{\hat{C}},{\hat{D}}]}e^{f_2(\xi ){\hat{D}}}{\hat{C}} e^{f_3(\xi ){\hat{C}}}. \end{aligned}$$

(A15)

It is necessary to move to the right the function $F(\xi )$, like in Eq. (A14). In order to make this arrangement, the next relations must be used (see [46])

$$\begin{aligned}&e^{\xi {\hat{A}}}{\hat{B}}=(e^{\xi {\hat{A}}}{\hat{B}}e^{-\xi {\hat{A}}})e^{\xi {\hat{A}}}, \end{aligned}$$

(A16)

$$\begin{aligned}&e^{\gamma {\hat{A}}}{\hat{B}}e^{-\gamma {\hat{A}}}={\hat{B}}+\gamma [{\hat{A}},{\hat{B}}]+\frac{\gamma ^2}{2!}[{\hat{A}},[{\hat{A}},{\hat{B}}]]+\cdots . \end{aligned}$$

(A17)

After some algebra and with the help of the commutations relations

$$\begin{aligned} {[}{\hat{C}},{\hat{E}}]= & {} \hbar ^2\omega _m^2{\hat{D}}, \end{aligned}$$

(A18)

$$\begin{aligned} {[}{\hat{D}},{\hat{E}}]= & {} 2\hbar \omega _m{\hat{k}}^2, \end{aligned}$$

(A19)

Eq. (A15) becomes

$$\begin{aligned} \frac{\mathrm{d} }{\mathrm{d} \xi } F(\xi )= & {} \Bigg \{ \frac{\mathrm{d} f_3(\xi ) }{\mathrm{d} \xi }{\hat{C}}+\bigg (\frac{\mathrm{d} f_2(\xi ) }{\mathrm{d} \xi }-\hbar ^2\omega ^2_m f_1(\xi )\frac{\mathrm{d} f_3(\xi ) }{\mathrm{d} \xi }\bigg ){\hat{D}}\nonumber \\&\Bigg (\frac{\mathrm{d} f_1(\xi ) }{\mathrm{d} \xi }-f_2(\xi )\frac{\mathrm{d} f_3(\xi ) }{\mathrm{d} \xi }\Bigg ){\hat{E}} \nonumber \\&\Bigg (\frac{\mathrm{d} f_0(\xi )}{\mathrm{d} \xi }-2\hbar \omega _mf_1(\xi )\frac{\mathrm{d} f_2(\xi )}{\mathrm{d} \xi } +\hbar ^2\omega ^3_m f_1^2(\xi )\nonumber \\&-\hbar \omega _m f_2^2(\xi )\frac{\mathrm{d} f_3(\xi ) }{\mathrm{d} \xi }\Bigg ){\hat{k}}^2\Bigg \}F(\xi ). \end{aligned}$$

(A20)

The third step consists in comparing the coefficients of (A14) and (A20); from this comparison, the following system of differential equations is obtained

$$\begin{aligned}&\frac{\mathrm{d} f_3(\xi ) }{\mathrm{d} \xi }=1,\nonumber \\&\frac{\mathrm{d} f_2(\xi ) }{\mathrm{d} \xi }-\hbar ^2\omega ^2_m f_1(\xi )\frac{\mathrm{d} f_3(\xi ) }{\mathrm{d} \xi } = 1, \nonumber \\&\frac{\mathrm{d} f_1(\xi ) }{\mathrm{d} \xi }-f_2(\xi )\frac{\mathrm{d} f_3(\xi ) }{\mathrm{d} \xi } = 0,\nonumber \\&\frac{\mathrm{d} f_0(\xi )}{\mathrm{d} \xi }-2\hbar \omega _mf_1(\xi )\frac{\mathrm{d} f_2(\xi )}{\mathrm{d} \xi } +\hbar ^2\omega ^3_m f_1^2(\xi )-\hbar \omega _m f_2^2(\xi )\frac{\mathrm{d} f_3(\xi ) }{\mathrm{d} \xi }=0. \end{aligned}$$

(A21)

The initial condition is $F(0)=1$, which implies $f_0(0)=f_1(0)=f_2(0)=f_3(0)=0$ and gives the following solutions:

$$\begin{aligned} f_0(\xi )= & {} \frac{\sinh (\hbar \omega _m\xi )\cosh (\hbar \omega _m\xi )-\hbar \omega _m \xi }{\hbar ^2\omega _m^2}, \end{aligned}$$

(A22)

$$\begin{aligned} f_1(\xi )= & {} \frac{\cosh (\hbar \omega _m\xi )-1}{\hbar ^2\omega _m^2}, \end{aligned}$$

(A23)

$$\begin{aligned} f_2(\xi )= & {} \frac{\sinh (\hbar \omega _m\xi )}{\hbar \omega _m}, \end{aligned}$$

(A24)

$$\begin{aligned} f_3(\xi )= & {} \xi . \end{aligned}$$

(A25)

Accordingly, the operator ${\hat{U}}(t)$ is expressed as:

$$\begin{aligned} {\hat{U}}(t)=e^{\xi {\hat{A}}}e^{\xi {\hat{B}}}e^{f_0(\xi ){\hat{k}}^2}e^{f_1(\xi )[{\hat{C}},{\hat{D}}]}e^{f_2(\xi ){\hat{D}}}e^{f_3(\xi ){\hat{C}}}. \end{aligned}$$

(A26)

Since the time evolution operator expressed in (A26) will be applied to the initial state, it is necessary to have a more practical expression; to do this, the factors $e^{f_1(\xi )[{\hat{C}},{\hat{D}}]}$ and $e^{f_2(\xi ){\hat{D}}}$ are conveniently factorized by using again this method. After performing the corresponding algebra, the final expression for ${\hat{U}}(t)$ is given by

$$\begin{aligned} {\hat{U}}(t)=e^{\xi {\hat{A}}}e^{\xi {\hat{B}}}e^{F_1{\hat{k}}^{2}} e^{F_2{\hat{k}}{\hat{b}}^{\dagger }}e^{F_3{\hat{k}}{\hat{b}}}e^{\xi {\hat{C}}}, \end{aligned}$$

(A27)

where ${\hat{k}}$ is previously defined as ${\hat{k}}=k_1{\hat{A}}+k_2{\hat{B}}$ and $F_1(\xi )$, $F_2(\xi )$ and $F_3(\xi )$ are the following functions:

$$\begin{aligned} F_1(\xi )= & {} f_0(\xi )-\frac{1}{2}\big [f_1^2(\xi )\hbar ^2 \omega _m^2\nonumber \\&+\,2\hbar \omega _m f_1f_2-f_2^2(\xi )\big ] , \end{aligned}$$

(A28)

$$\begin{aligned} F_2(\xi )= & {} f_1(\xi )\hbar \omega _m + f_2(\xi ), \end{aligned}$$

(A29)

$$\begin{aligned} F_2(\xi )= & {} -f_1(\xi )\hbar \omega _m + f_2(\xi ). \end{aligned}$$

(A30)

Such functions, after substituting $\xi =-it/\hbar$ and some algebra, take the following form:

$$\begin{aligned} F_1(t)= & {} \frac{ e^{-i\omega _m t }-i\omega _m t-1}{ \hbar ^2 \omega _m ^2}, \end{aligned}$$

(A31)

$$\begin{aligned} F_2(t)= & {} \frac{e^{-i\omega _m t}-1}{\hbar \omega _m }, \end{aligned}$$

(A32)

$$\begin{aligned} F_3(t)= & {} \frac{1-e^{i\omega _m t}}{\hbar \omega _m }. \end{aligned}$$

(A33)