Higher-order squeezing oscillations in Jaynes–Cummings model of a pair of cold atoms

Abstract

We study interaction between a pair of indistinguishable two-level atoms and the single-mode cavity field. It is supposed that the pair of two-level atoms is laser cooled and trapped into the ground vibrational state, in which vibrational quantum number $\langle n_v\rangle =0$. Two Jaynes–Cummings models are investigated. One is the Jaynes–Cummings model with intensity-dependent coupling and the another is the two-photon Jaynes–Cummings model of a pair of indistinguishable two-level atoms. It should be noted that in the present model, at initial moment $t=0$, in intensity-dependent Jaynes–Cummings model the cavity field is prepared in Holstein–Primakoff SU(1,1) CS, while in two-photon Jaynes–Cummings model it is prepared in the squeezed vacuum state. Moreover, at initial moment $t=0$, pair of atoms is supposed to be in the first excited state $\vert e_1\rangle$ in both models. By using exact analytical solutions for state-vectors of the coupled atom-field systems amplitude-squared squeezing of the quantized cavity field is examined as a function of the $\vert \xi \vert$ parameter. In this situation, in both models higher-order squeezing has the tendency towards oscillations, but exact periodicity of these oscillations is violated by the analogy with the second-order squeezing.

Appendix

In order to study squeezing properties of quantized cavity field by using exact analytical solutions given by Eqs. (8) and (9), we have found mean values $\langle a^2\rangle$, $\langle a^4\rangle$, $\langle a^{\dagger 2}\rangle$, $\langle a^{\dagger 4}\rangle$, $\langle a^{\dagger 2}a^2\rangle$ and $\langle a^2a^{\dagger 2}\rangle$, which are written through functions $A_{2}(t)$, $A_{4}(t)$, $B_{1}(t)$ and $B_{2}(t)$. In intensity-dependent JCM these functions are given by

$$\begin{aligned} A_2(t)&= \vert \xi \vert ^2\sum _{n=0}^{\infty }P_n \Bigg [\cos \left(\sqrt{2n^2+2n+1}\tau \right)\nonumber \\&\quad \times \cos\left (\sqrt{2(n+2)^2+2(n+2)+1}\tau \right) \nonumber \\&\quad \times \sqrt{(n+1)(n+2)} + \{(n+3)^{3/2}\sqrt{n+2}\nonumber \\&\quad \times (n+1) + n^{3/2}\sqrt{n+1}(n+2)\}\nonumber \\&\quad \times \frac{\sin \left(\sqrt{2n^2+2n+1}\tau \right)}{\sqrt{2n^2 + 2n + 1}}\nonumber \\&\quad \times \frac{\sin \left(\sqrt{2(n+2)^2+2(n+2)+1}\tau \right)}{\sqrt{2(n+2)^2 + 2(n+2)+1}}\Bigg ]\end{aligned}$$

(17)

$$\begin{aligned} A_4(t)&= \vert \xi \vert ^4\sum _{n=0}^{\infty }P_n\Bigg [\cos \left(\sqrt{2n^2+2n+1} \tau \right)\nonumber \\&\quad \times \cos \left(\sqrt{2(n+4)^2+2(n+4)+1}\tau \right) \nonumber \\&+\{(n+5)^{3/2}\sqrt{n+1} + n^{3/2}\sqrt{n+4}\}\nonumber \\&\quad \times \frac{\sin \left(\sqrt{2n^2+2n+1}\tau \right)}{\sqrt{2n^2+2n+1}}\nonumber \\&\quad \times \frac{\sin \left(\sqrt{2(n+4)^2+2(n+4)+1}\tau \right)}{\sqrt{2(n+4)^2 + 2(n+4)+1}}\Bigg ]\nonumber \\&\quad \times \sqrt{(n+1)(n+2)(n+3)(n+4)},\end{aligned}$$

(18)

$$\begin{aligned} B_{1}(t)&= \sum _{n=0}^{\infty }P_n\Bigg [\cos ^2\left(\sqrt{2n^2+2n+1} \tau \right)n(n-1) \nonumber \\&\quad + n\{(n+1)^3 + n(n-1)(n-2)\}\nonumber \\&\quad \times \frac{\sin ^2\left(\sqrt{2n^2+2n+1}\tau \right)}{2n^2+2n+1}\Bigg ],\end{aligned}$$

(19)

$$\begin{aligned} B_{2}(t)&= \sum _{n=0}^{\infty }P_n\Bigg [\cos ^2\left(\sqrt{2n^2+2n+1}\tau \right)\nonumber \\&\quad \times (n+1)(n+2) \nonumber \\&\quad + \{(n+1)^2(n+2)(n+3) + n^3(n+1)\}\nonumber \\&\quad \times \frac{\sin ^2\left(\sqrt{2n^2+2n+1}\tau \right)}{2n^2+2n+1}\Bigg ], \end{aligned}$$

(20)

where $P_n=\vert Q_n\vert ^2$. In two-photon JCM these functions have the following form

$$\begin{aligned} A_2(t)&= \vert \xi \vert ^2(1 - \vert \xi \vert ^2)^{1/2}\sum _{n=0}^{\infty }\vert \xi \vert ^{2n}\nonumber \\&\quad \times \Bigg [\cos \left(\sqrt{n^2+n+1}\tau \right) \nonumber \\&\quad \times \cos \left(\sqrt{(n+2)^2+(n+2)+1}\tau \right)\nonumber \\&\quad +\{(n+3)(n+4) +n(n-1)\}\nonumber \\&\quad \times \frac{\sin \left(\sqrt{n^2+n+1}\tau \right)}{2\sqrt{ n^2+n+1}} \nonumber \\&\quad \times \frac{\sin \left(\sqrt{(n+2)^2+(n+2)+1}\tau \right)}{\sqrt{(n+2)^2+(n+2)+1}}\Bigg ] \nonumber \\&\quad \times \sqrt{(n+1)(n+2)}\nonumber \\&\quad \times \left[ \frac{\Gamma (n+\frac{1}{2})\Gamma (n+\frac{5}{2})}{n!(n+2)! \Gamma ^2(\frac{1}{2})}\right] ^{1/2},\end{aligned}$$

(21)

$$\begin{aligned} A_4(t)&= \vert \xi \vert ^4(1 - \vert \xi \vert ^2)^{1/2}\sum _{n=0}^{\infty }\vert \xi \vert ^{2n}\nonumber \\&\quad \times \Bigg [\cos \left(\sqrt{n^2+n+1}\tau \right) \nonumber \\&\quad \times \cos \left(\sqrt{(n+4)^2+(n+4)+1}\tau \right)\nonumber \\&\quad + \{(n+5)(n+6) + (n-1)n\}\nonumber \\&\quad \times \frac{\sin\left (\sqrt{n^2+n+1}\tau \right)}{2\sqrt{ n^2+n+1}} \nonumber \\&\quad \times \frac{\sin \left(\sqrt{(n+4)^2+(n+4)+1}\tau \right)}{\sqrt{(n+4)^2+(n+4)+1}}\Bigg ] \nonumber \\&\quad \times \sqrt{(n+1)(n+2)(n+3)(n+4)}\nonumber \\&\quad \times \left[ \frac{\Gamma (n+\frac{1}{2})\Gamma (n+\frac{9}{2})}{n!(n+4)! \Gamma ^2(\frac{1}{2})} \right] ^{1/2}, \end{aligned}$$

(22)

$$\begin{aligned} B_{1}(t)&= \sum _{n=0}^{\infty }P_n\Bigg [\cos ^2\left(\sqrt{n^2+n+1}\tau \right)n(n-1) \nonumber \\&\quad + n\{(n+1)^2(n+2) + (n-1)^2(n-2)\}\nonumber \\&\quad \times \frac{\sin ^2\left(\sqrt{n^2+n+1}\tau \right)}{2(n^2+n+1)}\Bigg ],\end{aligned}$$

(23)

$$\begin{aligned} B_{2}(t)&= \sum _{n=0}^{\infty }P_n\Bigg [\cos ^2\left(\sqrt{n^2+n+1}\tau \right)\nonumber \\&\quad \times (n+1)(n+2)\nonumber \\&\quad + \{(n+1)(n+2)(n+3)(n+4)\nonumber \\&\quad + n^2(n-1)^2\} \frac{\sin ^2\left(\sqrt{n^2+n+1}\tau \right)}{2(n^2+n+1)}\Bigg ] \end{aligned}$$

(24)