Temperature Variation of Chaotic Optical Field in Quantum Dissipation Channel and Diffusion Channel

Abstract

In this paper we discuss the temperature variation of a new chaotic optical field after it passes through a single-mode quantum dissipation and diffusion channel respectively. Using the entangled state representation and the technique of integration within ordered product (IWOP) of operators, we solve the master equations and obtain the new output fields. We show the final explicit forms of the new output fields by detailed derivation. By virtue of thermo dynamics theory at finite temperature, we finally conclude the results of temperature variation of chaotic optical field both in quantum dissipation channel and diffusion channel.

Appendix

Derivation of Eq. ( 16 )

$$ {\displaystyle \begin{array}{c}\rho (t)=\sum \limits_{n=0}^{\infty}\frac{T^n}{n!}{e}^{-\kappa t{a}_2^{\dagger }{a}_2}{a}_2^n{\rho}_0{a}_2^{\dagger n}{e}^{-\kappa t{a}_2^{\dagger }{a}_2}\\ {}=\sum \limits_{n=0}^{\infty}\frac{T^n\mathit{\sec}{h}^2\theta }{n!}{e}^{-\kappa t{a}_2^{\dagger }{a}_2}{a}_2^n{e}^{a_1^{\dagger }{a}_2^{\dagger } tanh\theta}\left|00\right\rangle \left.\left\langle 0\right.0\right|{e}^{a_1{a}_2 tanh\theta}{a}_2^{\dagger n}{e}^{-\kappa t{a}_2^{\dagger }{a}_2}\end{array}} $$

(A1)

By using unitary transformation, we have

$$ \rho (t)=\sum \limits_{n=0}^{\infty}\frac{T^n\mathit{\sec}{h}^2\theta }{n!}{e}^{-\kappa t{a}_2^{\dagger }{a}_2}{a}_2^n{e}^{a_1^{\dagger }{a}_2^{\dagger } tanh\theta}{e}^{\kappa t{a}_2^{\dagger }{a}_2}{e}^{-\kappa t{a}_2^{\dagger }{a}_2}\left|00\right\rangle \left.\left\langle 0\right.0\right|{e}^{-\kappa t{a}_2^{\dagger }{a}_2}{e}^{\kappa t{a}_2^{\dagger }{a}_2}{e}^{a_1{a}_2 tanh\theta}{a}_2^{\dagger n}{e}^{-\kappa t{a}_2^{\dagger }{a}_2}. $$

(A2)

According to identity transformation of operator

$$ \exp \left(\gamma {a}^{\dagger }a\right)f\left({a}^{\dagger },a\right)\exp \left(-\gamma {a}^{\dagger }a\right)=f\left({ae}^{-\gamma },{a}^{\dagger }{e}^{\gamma}\right), $$

(A3)

we can get

$$ {\displaystyle \begin{array}{c}\rho (t)=\sum \limits_{n=0}^{\infty}\frac{T^n\mathit{\sec}{h}^2\theta }{n!}{\left({a}_2{e}^{kt}\right)}^n{e}^{a_1^{\dagger }{a}_2^{\dagger }{e}^{- kt} tanh\theta}\left|00\right\rangle \left.\left\langle 0\right.0\right|{e}^{a_1{a}_2{e}^{- kt} tanh\theta}{\left({a}_2^{\dagger }{e}^{kt}\right)}^n\\ {}=\sum \limits_{n=0}^{\infty}\frac{T^n\mathit{\sec}{h}^2\theta }{n!}{\left({e}^{kt}\right)}^{2n}{e}^{a_1^{\dagger }{a}_2^{\dagger }{e}^{- kt} tanh\theta}{e}^{-{a}_1^{\dagger }{a}_2^{\dagger }{e}^{- kt} tanh\theta}{\left({a}_2\right)}^n{e}^{a_1^{\dagger }{a}_2^{\dagger }{e}^{- kt} tanh\theta}\left|00\right\rangle \\ {}\times \Big\langle 00{e}^{a_1{a}_2{e}^{- kt} tanh\theta}{\left({a}_2^{\dagger}\right)}^n{e}^{-{a}_1{a}_2{e}^{- kt} tanh\theta}{e}^{a_1{a}_2{e}^{- kt} tanh\theta}.\end{array}} $$

(A4)

Further, by using the following operator identity

$$ {e}^{-{a}_1^{\dagger }{a}_2^{\dagger }{e}^{- kt}\tanh \theta }{\left({a}_2\right)}^n{\mathrm{e}}^{a_2^{\dagger }{a}_2^{\dagger }{e}^{- kt}\tanh \theta }={\left({a}_2+{a}_1^{\dagger }{e}^{- kt}\tanh \theta \right)}^n, $$

(A5)

and binomial theorem, we can deduce

$$ \rho (t)=\sum \limits_{n=0}^{\infty}\;\frac{T^n{\mathit{\operatorname{sech}}}^2\theta }{n!}{\left({e}^{kt}\right)}^{2n}{e}^{a_1^{\dagger }{a}_2^{\dagger }{e}^{- kt}\tanh \theta }{\left({a}_2+{a}_1^{\dagger }{e}^{- kt}\tanh \theta \right)}^n\left|00\right\rangle \left\langle 00\right|{\left({a}_2^{\dagger }+{a}_1{e}^{- kt}\tanh \theta \right)}^n{e}^{a_1{a}_2{e}^{- kt}\tanh \theta } $$

$$ =\sum \limits_{n=0}^{\infty}\;\frac{T^n{\mathit{\operatorname{sech}}}^2\theta }{n!}{\left({e}^{kt}\right)}^{2n}{e}^{a_1^{\dagger }{a}_2^{\dagger }{e}^{- kt}\tanh \theta }{\left({a}_1^{\dagger }{e}^{- kt}\tanh \theta \right)}^n\left|00\right\rangle \left\langle 00\right|{\left({a}_1{e}^{- kt}\tanh \theta \right)}^n{e}^{a_1{a}_2{e}^{- kt}\tanh \theta }. $$

(A6)

Finally, we can derive the explicit form of ρ(t)

$$ \rho (t)=\sum \limits_{n=0}^{\infty}\;\frac{T^n\tan {\mathrm{h}}^{2n}\theta {\mathit{\operatorname{sech}}}^2\theta }{n!}{e}^{a_1^{\dagger }{a}_2^{\dagger }{e}^{-\kappa t}\tanh \theta }{a}_1^{\dagger n}\left|00\right\rangle \left\langle 00\right|{a}_1^n{e}^{a_1{a}_2{e}^{-\kappa t}\tanh \theta }. $$

(A7)