Dissipation of Gaussian-Enhenced Chaotic State

Abstract

We investigate how an optical Gaussian-enhenced chaotic state (GECS) evolves in an amplitude dissipation channel. We have used the integration within ordered product of operators technique to derive its evolution law and show that the dacay of photon number. Also both the normally ordered and anti-normally ordered of the evolution law are given.

Appendix

From (20) we can further derive the normally ordered form of ρ(t), using

$$ \int \frac{d^{2}z}{\pi }|z\rangle \langle z|=\int \frac{d^{2}z}{\pi }:\exp \left[ -|z|^{2}+za^{\dag }+z^{\ast }a-a^{\dag }a\right] := 1 $$

(28)

and (22) we see

$$ \rho (t) = \mathfrak{A}^{\prime }\int \frac{d^{2}z}{\pi }:\exp \left[\! e^{2\kappa t}\frac{(f^{2}\!-4|E|^{2}-\!f)|z|^{2} + Ez^{2} + E^{\ast }z^{\ast 2}}{ f^{2} - 4|E|^{2}}-\!|z|^{2} + za^{\dag } + z^{\ast }a - a^{\dag }a\!\right] : $$

(29)

Performing this integration within : : and usin (24), we obtain

$$\begin{array}{@{}rcl@{}} \rho (t) &=&\mathfrak{A}^{\prime }\frac{f^{2}-4|E|^{2}}{e^{2\kappa t}} \\ &&\times \int \frac{d^{2}z}{\pi } :\exp \left[ \left[ \left( f^{2}-4|E|^{2}\right) T-f\right] |z|^{2}\right.\\ &&\left.+Ez^{2}+E^{\ast }z^{\ast 2}+e^{-\kappa t}\sqrt{f^{2}-4|E|^{2}}\left( za^{\dag }+z^{\ast }a\right) -a^{\dag }a\right] : \\ &=&\sqrt{\frac{(f^{2}-4|E|^{2}-f)^{2}-4|E|^{2}}{\left[ \left( f^{2}-4|E|^{2}\right) T-f\right]^{2}-4|E|^{2}}} \\ &&\times :\exp\! \left[ e^{-2\kappa t}\!\left( f^{2}\!-4|E|^{2}\right) \frac{\left[ f - \left( f^{2} - 4|E|^{2}\right) T\right] a^{\dag }a + \left( Ea^{2} + E^{\ast }a^{\dag 2}\right) }{\left[ \left( f^{2}-4|E|^{2}\right) T-f\right]^{2}-4|E|^{2}}-a^{\dag }a\right] : \end{array} $$

(30)

here T = 1 − e^− 2κt, this is the normally ordered outstate in the channel. When t = 0, e^− 2κt = 1,T = 0, from (24) we see that it reduces to

$$\begin{array}{@{}rcl@{}} \rho (t) &\rightarrow &\sqrt{\frac{(f^{2}-4|E|^{2}-f)^{2}-4|E|^{2}}{ f^{2}-4|E|^{2}}}:\exp \left[ \left( f-1\right) a^{\dag }a+\left( Ea^{2}+E^{\ast }a^{\dag 2}\right) \right] : \\ &=&\sqrt{(f-1)^{2}-4|E|^{2}}\exp (E^{\ast }a^{\dagger 2})\exp (a^{\dagger }a\ln f)\exp (Ea^{2})=\rho_{0}=\rho_{g} \end{array} $$

(31)

which is just the initial state, as expected.

By comparing (30) with (31) we can see clearly the evolution law of the Gaussian-enhenced chaotic state(GECS) in an amplitude damping channel.

1.1 A.1 Discussion

We now discuss the special case when f = 0, and meanwhile we take

$$ E=\frac{\tanh \lambda }{2}=E^{\ast } $$

(32)

in this case the initial state ρ_g in (8) becomes to

$$ \rho_{g} = \mathfrak{A}\exp (E^{\ast }a^{\dagger 2})\exp (a^{\dagger }a\ln f)\exp (Ea^{2})\!\rightarrow\! \sec h\lambda \exp \left( \frac{\tanh \lambda }{2} a^{\dagger 2}\right)\left\vert 0\right\rangle \left\langle 0\right\vert \exp \left( \frac{\tanh \lambda }{2}a^{2}\right) $$

(33)

which is a pure squeezed vacuum state, and |0〉 〈0| is the vacuum state. Then according to (30) we see that the final state becomes to

$$ \rho (t)|_{f = 0}=\frac{\sec h\lambda }{\sqrt{1-T^{2}\tanh^{2}\lambda }}\exp \left( \frac{\beta }{2}a^{\dag 2}\right) :\exp \left[ \left( \beta T\tanh \lambda -1\right) a^{\dag }a\right] :\exp \left( \frac{\beta }{2} a^{2}\right) $$

(34)

which is a mixed state in a Gaussian enhanced form, here

$$ \beta \equiv \frac{e^{-2\kappa t}\tanh \lambda }{1-T^{2}\tanh^{2}\lambda } $$

(35)

Therefore we conclude that when a pure squeezed vacuum state passing through an amplitude damping channel, it will evolve into a Gaussian enhanced form density operator.