Quantum dynamics of a driven parametric oscillator in a Kerr medium

In this section, we reconsider the Hamiltonian equation (45) and try to find the corresponding time-evolution operator approximately. For this purpose, we make use of the properties of Heisenberg algebra $\{1,\hat{a}, \hat{a}^\dag \}$. Let us rewrite the Hamiltonian equation (45) in the following form

$$\begin{aligned} \hat{H}_{k=0} (t)= \, & {} \Omega _0\,(\hat{a}^\dag \hat{a}+1/2)+\chi \,(\hat{a}^\dag \hat{a})^2+\frac{e(t)}{\sqrt{2\Omega _0}}\,(\hat{a}^\dag +\hat{a}),\nonumber \\= \, & {} \hat{H}_0+\frac{e(t)}{\sqrt{2\Omega _0}}\,(\hat{a}^\dag +\hat{a}), \end{aligned}$$

(45)

where $\hat{H}_0=\Omega _0 (\hat{n}+1/2)+\chi \hat{n}^2$. The evolution operator $\hat{U}(t)$ corresponding to the Hamiltonian $\hat{H}_{k=0}$ can be written as

$$\begin{aligned} \hat{U}(t)=\hat{U}_0(t)\hat{U}_I(t), \end{aligned}$$

(46)

where $\hat{U}_0(t)$ is the evolution operator corresponding to the time-independent Hamiltonian $\hat{H}_0$ given by

$$\begin{aligned} \hat{U}_0(t)= \, & {} e^{-i\hat{H}_0t},\nonumber \\= \, & {} e^{-i\Omega _0t(\hat{n}+\frac{1}{2})-it\chi \hat{n}^2}. \end{aligned}$$

(47)

The time-dependent part $\hat{U}_I(t)$ fulfills

$$\begin{aligned} i\frac{d\hat{U}_I(t)}{dt}=\hat{V}_I(t)\,\hat{U}_I(t), \end{aligned}$$

(48)

where

$$\begin{aligned} \hat{V}_I(t)= \, & {} \hat{U}_0^\dag (t)\hat{V}\hat{U}_0(t),\nonumber \\= \, & {} \frac{e(t)}{\sqrt{2\Omega _0}}(e^{-i\Omega _0t-i\chi t(2\hat{n}+1)}\hat{a}+\hat{a}^\dag e^{i\Omega _0t+i\chi t(2\hat{n}+1)}),\nonumber \\= \, & {} g(t)\hat{a}+\hat{a}^\dag \bar{g}(t), \end{aligned}$$

(49)

and

$$\begin{aligned} g(t)=\frac{e(t)}{\sqrt{2\Omega _0}}\,e^{-it(\Omega _0+\chi )}e^{|\alpha |^2(e^{-2i\chi t}-1)}. \end{aligned}$$

(50)

In deriving Eq. (49), to simplify the calculations, we assumed that the system is initially prepared in a coherent state $|\alpha \rangle $, and replaced the nonlinear term $e^{[\pm i\Omega (\hat{n})t]}$ by its average value $\langle \alpha |e^{[\pm i\Omega (\hat{n})t]}|\alpha \rangle $, see^{31,32,33,34,35}.

To find $\hat{U}_I(t)$, we assume

$$\begin{aligned} \hat{U}_I(t)=\prod _{n=1}^{3}\exp (X_n(t)\hat{\gamma }_n), \end{aligned}$$

(51)

where we have defined $\hat{\gamma }_1=1, \hat{\gamma }_2=\hat{a}, \hat{\gamma }_3=\hat{a}^\dag $ and the initial conditions are $X_1(0)=X_2(0)=X_3(0)=0$. By inserting Eq. (51) into Eq. (48), we find^36,37

$$\begin{aligned} \dot{X}_1(t)= \, & {} \dot{X}_3(t){X}_2(t), \nonumber \\ \dot{X}_2(t)= \, & {} -i\bar{g}(t), \nonumber \\ \dot{X}_3(t)= \, & {} -i g(t). \end{aligned}$$

(52)

The Eq. (52) can be solved either analytically or numerically and from now on we assume that the solutions are known functions. Therefore, if we denote the temporal evolution of the initial state $|\alpha \rangle $ by $|\alpha ,t\rangle $, we have

$$\begin{aligned} |\psi _\alpha ,t\rangle= \, & {} \hat{U}_0(t)\hat{U}_I(t)|\alpha \rangle ,\nonumber \\= \, & {} G_\alpha \sum _{n=0}^{\infty }e^{-i\Omega _0tn-itn^2\chi }\frac{(X_2(t)+\alpha )^n}{\sqrt{n!}}|n\rangle . \end{aligned}$$

(53)

where $G_\alpha =e^{X_3(t)\alpha +X_1(t)-i\frac{\Omega _0t}{2}-\frac{|\alpha |^2}{2}}$, and from the normalization condition of the wave function $\langle \psi _\alpha ,t |\psi _\alpha ,t\rangle =1$, we find $|G_\alpha |^2=\exp (-|(X_2(t)+\alpha )|^2)$. For convenience, let us define the dimensionless parameters $\xi =\chi t$ and $X_2(t)+\alpha =\eta _t$, then Eq. (53) can be rewritten as

$$\begin{aligned} |\psi _\alpha ,t\rangle= \, & {} e^{\frac{-|\eta _t|^2}{2}}e^{-i\xi \hat{n}^2}\sum _{n=0}^{\infty }\frac{(e^{-i\Omega _0t}\eta _t)^n}{\sqrt{n!}}|n\rangle ,\nonumber \\= \, & {} e^{-i\xi \hat{n}^2}|e^{-i\Omega _0t}\eta _t\rangle ,\nonumber \\= \, & {} |e^{-i\Omega _0 t}\eta _t\rangle _\xi , \end{aligned}$$

(54)

where we have defined $|\zeta \rangle _\xi =e^{-i\xi \hat{n}^2}|\zeta \rangle $ for an arbitrary coherent state $|\zeta \rangle $. Therefore, the evolved state $|\psi _\alpha ,t\rangle $ is of the kind $|\beta \rangle _{\xi }$ where $\beta =e^{-i\Omega _0 t}\eta _t$. In the next section we will study the properties of these states.

Properties of the states $|\beta \rangle _{\xi }$

Let us define a new set of ladder operators as

$$\begin{aligned}{} & {} \hat{B}=\hat{a}\,f(\hat{n})=f(\hat{n}+1)\,\hat{a},\nonumber \\{} & {} \hat{B}^\dag =f^\dag (\hat{n})\,\hat{a}^\dag =\hat{a}^\dag \,f^\dag (\hat{n}+1), \end{aligned}$$

(55)

where the function $f(\hat{n})$ is defined by

$$\begin{aligned} f(\hat{n})=e^{i \xi (2 \hat{n}-1)}. \end{aligned}$$

(56)

The operators $\hat{B}$, $\hat{B}^\dag $ and $\hat{n}$, fulfil the usual Heisenberg algebra

$$\begin{aligned}{} & {} [\hat{B},\hat{B}^\dag ]=1,\nonumber \\{} & {} [\hat{n},\hat{B}]=-\hat{B},\nonumber \\{} & {} [\hat{n},\hat{B}^\dag ]=\hat{B}^\dag . \end{aligned}$$

(57)

The state $|\beta \rangle _{\xi }$ can be expanded in number states basis as

$$\begin{aligned} |\beta \rangle _{\xi }= \, & {} e^{-i\xi \hat{n}^2}\left( e^{-\frac{|\beta |^2}{2}}\sum _{n=0}^\infty \frac{\beta ^n}{\sqrt{n!}}\,|n\rangle \right) ,\nonumber \\= \, & {} e^{-\frac{|\beta |^2}{2}}\sum _{n=0}^\infty \frac{\beta ^n}{\sqrt{n!}}e^{-i\xi n^2}\,|n\rangle ,\nonumber \\= \, & {} e^{-\frac{|\beta |^2}{2}}\sum _{n=0}^\infty \frac{\beta ^n}{\sqrt{n!}}\frac{1}{[f(n)]!}\,|n\rangle , \end{aligned}$$

(58)

where $[f(n)]!=\displaystyle \Pi _{k=1}^n f(n)$, and $[f(0)]!=1$. One can easily show that the state $|\beta \rangle _{\xi }$ is a coherent state for the new annihilation operator $\hat{B}$ with eigenvalue $\beta $

$$\begin{aligned} \hat{B} |\beta \rangle _{\xi }=\beta \,|\beta \rangle _{\xi }. \end{aligned}$$

(59)

If we define the modified displacement operator $\hat{D}_B (\beta )=e^{\beta \,\hat{B}^\dag -\bar{\beta }\,\hat{B}}$, then $|\beta \rangle _{\xi }=\hat{D}_B |0\rangle $, note that $|0\rangle _{\xi }=|0\rangle $, and the parameter $\xi =\chi t$ is hidden in the definition of $\hat{B}$ and $\hat{B}^\dag $.

The state $|\beta \rangle _{\xi }$ can also be considered as a Kerr state if we consider the Hamiltonian

$$\begin{aligned} \hat{H}= \, & {} \hbar \chi \,\hat{a}^\dag \hat{a}+\hbar \chi \,\hat{a}^\dag \hat{a}^\dag \hat{a}\hat{a},\nonumber \\= \, & {} \hbar \chi \hat{n}^2, \end{aligned}$$

(60)

with the corresponding time-evolution operator

$$\begin{aligned} \hat{U} (t)=e^{-it\chi \,\hat{n}^2}=e^{-i\xi \hat{n}^2}. \end{aligned}$$

(61)

If the system is initially prepared in the coherent state $|\beta \rangle $, then the evolved state is the state $|\beta \rangle _{\xi }$ given by

$$\begin{aligned} \hat{U}(t) |\beta \rangle= \, & {} e^{-i\xi \hat{n}^2}|\beta \rangle =|\beta \rangle _{\xi },\nonumber \\= \, & {} e^{-\frac{|\beta |^2}{2}}\sum _{n=0}^\infty \frac{\beta ^n}{\sqrt{n!}} e^{-i\xi n^2} |n\rangle . \end{aligned}$$

(62)

The probability of having n excitation in the evolved state $|\beta \rangle _{\xi }$ is a Poissonian distribution

$$\begin{aligned} P_{\xi }(n)=|\langle n|\beta \rangle _{\xi }|^2=e^{-|\beta |^2}\frac{|\beta |^{2n}}{n!}. \end{aligned}$$

(63)

To study the squeezing effects, let us find the normalized variances of the position $\hat{x}=(\hat{a}+\hat{a}^\dag )/2$ and momentum $\hat{p}= (\hat{a}-\hat{a}^\dag )/2i $ defined by $(\triangle q)_\xi / (\triangle q)_{\xi =0}$ and $(\triangle p)_\xi / (\triangle p)_{\xi =0}$, respectively. We have

$$\begin{aligned}{} & {} \frac{(\Delta q)_\xi }{(\Delta q)_{\xi =0}} \nonumber \\{} & {} = \sqrt{2|\beta |^2+1+2|\beta |^2\,e^{-2|\beta |^2\sin ^2(2\xi )}\cos (2\phi -4\xi -|\beta |^2\sin (4\xi ))-4|\beta |^2\,e^{-4|\beta |^2\sin ^2(\xi )}\cos ^2 (\phi -\xi -|\beta |^2\sin (2\xi ))},\nonumber \\{} & {} \frac{(\Delta p)_\xi }{(\Delta p)_{\xi =0}} \nonumber \\{} & {} = \sqrt{2|\beta |^2+1-2|\beta |^2\,e^{-2|\beta |^2\sin ^2(2\xi )}\cos (-2\phi +4\xi +|\beta |^2\sin (4\xi ))-4|\beta |^2\,e^{-4|\beta |^2\sin ^2(\xi )}\sin ^2 (\phi -|\beta |^2\sin (2\xi ))}. \end{aligned}$$

(64)

where $\beta =|\beta |e^{i\phi }$ and $(\Delta q)_{\xi =0}=1/\sqrt{2\Omega _0}$ and $(\Delta p)_{\xi =0}=\sqrt{\Omega _0/2}$. Note that $|\beta \rangle _{\xi =0}=|\beta \rangle $ ia a coherent state for $\hat{a}$ leading to a minimal uncertainty. In Fig. 1, the variances are depicted for $\beta =0.5$.

Figure 1

The normalized variances of position (left) and momentum (right) depicted for $\beta =0.5$ in dimensionless parameter $\xi =\chi t$. Squeezing occurs only for position in time intervals where $\frac{(\Delta q)_\xi }{(\Delta q)_{\xi =0}}<1$.

In the next section, we will focus on statistical properties of the state $|\beta \rangle _{\xi }$ using (quasi)probability distribution functions.

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Quantum dynamics of a driven parametric oscillator in a Kerr medium

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Quantum dynamics of a driven parametric oscillator in a Kerr medium

Properties of the states \(|\beta \rangle _{\xi }\)

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