# Evolution of the single-mode squeezed vacuum state in amplitude dissipative
channel

Hong-Yi Fan1, Shuai Wang1 and Li-Yun Hu2∗ E-mail<EMAIL_ADDRESS>1Department of Physics, Shanghai Jiao Tong University, Shanghai 200240,China
2College of Physics & Communication Electronics, Jiangxi Normal University,
Nanchang 330022, China
$\ast$Corresponding author.E-mail<EMAIL_ADDRESS>
###### Abstract

Using the way of deriving infinitive sum representation of density operator as
a solution to the master equation describing the amplitude dissipative channel
by virtue of the entangled state representation, we show manifestly how the
initial density operator of a single-mode squeezed vacuum state evolves into a
definite mixed state which turns out to be a squeezed chaotic state with
decreasing-squeezing. We investigate average photon number, photon statistics
distributions for this state.

##  I Introduction

Squeezed states are such for which the noise in one of the chosen pair of
observables is reduced below the vacuum or ground-state noise level, at the
expense of increased noise in the other observable. The squeezing effect
indeed improves interferometric and spectroscopic measurements, so in the
context of interferometric detection and of gravitational waves the squeezed
state is very useful 01 ; 02 . In a very recently published paper, Agarwal r1
revealed that a vortex state of a two-mode system can be generated from a
squeezed vacuum by subtracting a photon, such a subtracting mechanism may
happen in a quantum channel with amplitude damping. Usually, in nature every
system is not isolated, dissipation or dephasing usually happens when a system
is immersed in a thermal environment, or a signal (a quantum state) passes
through a quantum channel which is described by a master equation 03 . For
example, when a pure state propagates in a medium, it inevitably interacts
with it and evolves into a mixed state 04 . Dissipation or dephasing will
deteriorate the degree of nonclassicality of photon fields, so physicists pay
much attention to it 05 ; 06 ; 07 .

In this present work we investigate how an initial single-mode squeezed vacuum
state evolves in an amplitude dissipative channel (ADC). When a system is
described by its interaction with a channel with a large number of degrees of
freedom, master equations are set up for a better understanding how quantum
decoherence is processed to affect unitary character in the dissipation or
gain of the system. In most cases people are interested in the evolution of
the variables associated with the system only. This requires us to obtain the
equations of motion for the system of interest only after tracing over the
reservoir variables. A quantitative measure of nonclassicality of quantum
fields is necessary for further investigating the system’s dynamical behavior.
For this channel, the associated loss mechanism in physical processes is
governed by the following master equation 03

$\frac{d\rho\left(t\right)}{dt}=\kappa\left(2a\rho
a^{\dagger}-a^{\dagger}a\rho-\rho a^{\dagger}a\right),$ (1)

where $\rho$ is the density operator of the system, and $\kappa$ is the rate
of decay. We have solved this problem with use of the thermo entangled state
representation 08 . Our questions are: What kind of mixed state does the
initial squeezed state turns into? How does the photon statistics
distributions varies in the ADC?

Thus solving master equations is one of the fundamental tasks in quantum
optics. Usually people use various quasi-probability representations, such as
P-representation, Q-representation, complex P-representation, and Wigner
functions, etc. for converting the master equations of density operators into
their corresponding c-number equations. Recently, a new approach 08 ; 09 ,
using the thermal entangled state representation 10 ; 11  to convert operator
master equations to their c-number equations is presented which can directly
lead to the corresponding Kraus operators (the infinitive representation of
evolved density operators) in many cases.

The work is arranged as follows. In Sec. 2 by virtue of the entangled state
representation we briefly review our way of deriving the infinitive sum
representation of density operator as a solution of the master equation. In
Sec. 3 we show that a pure squeezed vacuum state (with squeezing parameter
$\lambda)$ will evolves into a mixed state (output state), whose exact form is
derived, which turns out to be a squeezed chaotic state. We investigate
average photon number, photon statistics distributions for this state. The
probability of finding $n$ photons in this mixed state is obtained which turns
out to be a Legendre polynomial function relating to the squeezing parameter
$\lambda$ and the decaying rate $\kappa$. In Sec. 4 we discuss the photon
statistics distributions of the output state. In Sec. 5 and 6 we respectively
discuss the Wigner function and tomogram of the output state.

##  II Brief review of deducing the infinitive sum representation of
$\rho\left(t\right)$

For solving the above master equation, in a recent review paper 12  we have
introduced a convenient approach in which the two-mode entangled state 10 ; 11

$|\eta\rangle=\exp(-\frac{1}{2}|\eta|^{2}+\eta
a^{{\dagger}}-\eta^{\ast}\tilde{a}^{{\dagger}}+a^{{\dagger}}\tilde{a}^{{\dagger}})|0\tilde{0}\rangle,$
(2)

is employed, where $\tilde{a}^{{\dagger}}$ is a fictitious mode independent of
the real mode $a^{\dagger},$ $[\tilde{a},a^{\dagger}]=0$. $|\eta=0\rangle$
possesses the properties

$\displaystyle a|\eta$
$\displaystyle=0\rangle=\tilde{a}^{{\dagger}}|\eta=0\rangle,$ $\displaystyle
a^{{\dagger}}|\eta$ $\displaystyle=0\rangle=\tilde{a}|\eta=0\rangle,$ (3)
$\displaystyle(a^{{\dagger}}a)^{n}|\eta$
$\displaystyle=0\rangle=(\tilde{a}^{{\dagger}}\tilde{a})^{n}|\eta=0\rangle.$

Acting the both sides of Eq.(1) on the state
$|\eta=0\rangle\equiv\left|I\right\rangle$, and denoting
$\left|\rho\right\rangle=\rho\left|I\right\rangle$, we have

$\displaystyle\frac{d}{dt}\left|\rho\right\rangle$
$\displaystyle=\kappa\left(2a\rho a^{\dagger}-a^{\dagger}a\rho-\rho
a^{\dagger}a\right)\left|I\right\rangle$
$\displaystyle=\kappa\left(2a\tilde{a}-a^{\dagger}a-\tilde{a}^{\dagger}\tilde{a}\right)\left|\rho\right\rangle,$
(4)

so its formal solution is

$\left|\rho\right\rangle=\exp\left[\kappa
t\left(2a\tilde{a}-a^{\dagger}a-\tilde{a}^{\dagger}\tilde{a}\right)\right]\left|\rho_{0}\right\rangle,$
(5)

where $\left|\rho_{0}\right\rangle\equiv\rho_{0}\left|I\right\rangle,$
$\rho_{0}$ is the initial density operator.

Noticing that the operators in Eq.(5) obey the following commutative relation,

$\left[a\tilde{a},a^{\dagger}a\right]=\left[a\tilde{a},\tilde{a}^{\dagger}\tilde{a}\right]=\tilde{a}a$
(6)

and

$\left[\frac{a^{\dagger}a+\tilde{a}^{\dagger}\tilde{a}}{2},a\tilde{a}\right]=-\tilde{a}a,$
(7)

as well as using the operator identity 13

$e^{\lambda\left(A+\sigma B\right)}=e^{\lambda
A}e^{\sigma\left(1-e^{-\lambda\tau}\right)B/\tau},$ (8)

(which is valid for $\left[A,B\right]=\tau B$), we have

$e^{-2\kappa
t\left(\frac{a^{\dagger}a+\tilde{a}^{\dagger}\tilde{a}}{2}-a\tilde{a}\right)}=e^{-\kappa
t\left(a^{\dagger}a+\tilde{a}^{\dagger}\tilde{a}\right)}e^{T^{\prime}a\tilde{a}},$
(9)

where $T^{\prime}=1-e^{-2\kappa t}.$

Then substituting Eq.(9) into Eq.(5) yields 12

$\displaystyle\left|\rho\right\rangle$ $\displaystyle=e^{-\kappa
t\left(a^{\dagger}a+\tilde{a}^{\dagger}\tilde{a}\right)}\sum_{n=0}^{\infty}\frac{T^{\prime
n}}{n!}a^{n}\tilde{a}^{n}\left|\rho_{0}\right\rangle$
$\displaystyle=e^{-\kappa ta^{\dagger}a}\sum_{n=0}^{\infty}\frac{T^{\prime
n}}{n!}a^{n}\rho_{0}a^{{\dagger}n}e^{-\kappa
t\tilde{a}^{\dagger}\tilde{a}}\left|I\right\rangle$
$\displaystyle=\sum_{n=0}^{\infty}\frac{T^{\prime n}}{n!}e^{-\kappa
ta^{\dagger}a}a^{n}\rho_{0}a^{{\dagger}n}e^{-\kappa
ta^{\dagger}a}\left|I\right\rangle,$ (10)

which leads to the infinitive operator-sum representation of$\ \rho$,

$\rho=\sum_{n=0}^{\infty}M_{n}\rho_{0}M_{n}^{\dagger},$ (11)

where

$M_{n}\equiv\sqrt{\frac{T^{\prime n}}{n!}}e^{-\kappa ta^{\dagger}a}a^{n}.$
(12)

We can prove

$\displaystyle\sum_{n}M_{n}^{\dagger}M_{n}$
$\displaystyle=\sum_{n}\frac{T^{\prime n}}{n!}a^{{\dagger}n}e^{-2\kappa
ta^{\dagger}a}a^{n}$ $\displaystyle=\sum_{n}\frac{T^{\prime n}}{n!}e^{2n\kappa
t}\colon a^{{\dagger}n}a^{n}\colon e^{-2\kappa ta^{\dagger}a}$
$\displaystyle=\left.:e^{T^{\prime}e^{2\kappa
t}a^{\dagger}a}:\right.e^{-2\kappa ta^{\dagger}a}$
$\displaystyle=\left.:e^{\left(e^{2\kappa
t}-1\right)a^{\dagger}a}:\right.e^{-2\kappa ta^{\dagger}a}=1,$ (13)

where $\colon\colon$ stands for the normal ordering. Thus $M_{n}$ is a kind of
Kraus operator, and $\rho$ in Eq.(11) is qualified to be a density operator,
i.e.,

$Tr\left[\rho\left(t\right)\right]=Tr\left[\sum_{n=0}^{\infty}M_{n}\rho_{0}M_{n}^{\dagger}\right]=Tr\rho_{0}.$
(14)

Therefore, for any given initial state $\rho_{0}$, the density operator
$\rho\left(t\right)$ can be directly calculated from Eq.(11). The entangled
state representation provides us with an elegant way of deriving the
infinitive sum representation of density operator as a solution of the master
equation.

##  III Evolving of an initial single-mode squeezed vacuum state in ADC

It is seen from Eq.(11) that for any given initial state $\rho_{0}$, the
density operator $\rho\left(t\right)$ can be directly calculated. When
$\rho_{0}$ is a single-mode squeezed vacuum state,

$\rho_{0}=\text{sech}\lambda\exp\left(\frac{\tanh\lambda}{2}a^{{\dagger}2}\right)\left|0\right\rangle\left\langle
0\right|\exp\left(\frac{\tanh\lambda}{2}a^{2}\right),$ (15)

we see

$\displaystyle\rho\left(t\right)$ $\displaystyle=$
$\displaystyle\text{sech}\lambda\sum_{n=0}^{\infty}\frac{T^{\prime
n}}{n!}e^{-\kappa
ta^{\dagger}a}a^{n}\exp\left(\frac{\tanh\lambda}{2}a^{{\dagger}2}\right)\left|0\right\rangle$
(16) $\displaystyle\times\left\langle
0\right|\exp\left(\frac{\tanh\lambda}{2}a^{2}\right)a^{{\dagger}n}e^{-\kappa
ta^{\dagger}a}.$

Using the Baker-Hausdorff lemma 14 ,

$e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}}=\hat{B}+\lambda\left[\hat{A},\hat{B}\right]+\frac{\lambda^{2}}{2!}\left[\hat{A},\left[\hat{A},\hat{B}\right]\right]+\cdots.$
(17)

we have

$\displaystyle
a^{n}\exp\left(\frac{\tanh\lambda}{2}a^{{\dagger}2}\right)\left|0\right\rangle$
$\displaystyle=$ $\displaystyle
e^{\frac{\tanh\lambda}{2}a^{{\dagger}2}}e^{-\frac{\tanh\lambda}{2}a^{{\dagger}2}}a^{n}e^{\frac{\tanh\lambda}{2}a^{{\dagger}2}}\left|0\right\rangle$
(18) $\displaystyle=$ $\displaystyle
e^{\frac{\tanh\lambda}{2}a^{{\dagger}2}}\left(a+a^{\dagger}\tanh\lambda\right)^{n}\left|0\right\rangle.$

Further employing the operator identity 15

$\left(\mu a+\nu
a^{\dagger}\right)^{m}=\left(-i\sqrt{\frac{\mu\nu}{2}}\right)^{m}\colon
H_{m}\left(i\sqrt{\frac{\mu}{2\nu}}a+i\sqrt{\frac{\nu}{2\mu}}a^{\dagger}\right)\colon,$
(19)

where $H_{m}(x)$ is the Hermite polynomial, we know

$\displaystyle\left(a+a^{\dagger}\tanh\lambda\right)^{n}$ (20)
$\displaystyle=$
$\displaystyle\left(-i\sqrt{\frac{\tanh\lambda}{2}}\right)^{n}\colon
H_{n}\left(i\sqrt{\frac{1}{2\tanh\lambda}}a+i\sqrt{\frac{\tanh\lambda}{2}}a^{\dagger}\right)\colon.$

From Eq.(18), it follows that

$\displaystyle
a^{n}e^{\frac{\tanh\lambda}{2}a^{{\dagger}2}}\left|0\right\rangle$
$\displaystyle=$
$\displaystyle\left(-i\sqrt{\frac{\tanh\lambda}{2}}\right)^{n}e^{\frac{\tanh\lambda}{2}a^{{\dagger}2}}$
(21) $\displaystyle\times
H_{n}\left(i\sqrt{\frac{\tanh\lambda}{2}}a^{\dagger}\right)\left|0\right\rangle.$

On the other hand, noting $e^{-\kappa ta^{\dagger}a}a^{\dagger}e^{\kappa
ta^{\dagger}a}=a^{\dagger}e^{-\kappa t},e^{\kappa ta^{\dagger}a}ae^{-\kappa
ta^{\dagger}a}=ae^{-\kappa t}$ and the normally ordered form of the vacuum
projector $\left|0\right\rangle\left\langle 0\right|=\colon
e^{-a^{\dagger}a}\colon,$ we have

$\displaystyle\rho\left(t\right)$
$\displaystyle=\text{sech}\lambda\sum_{n=0}^{\infty}\frac{T^{\prime
n}}{n!}e^{-\kappa
ta^{\dagger}a}a^{n}e^{\frac{\tanh\lambda}{2}a^{{\dagger}2}}\left|0\right\rangle$
$\displaystyle\times\left\langle
0\right|e^{\frac{\tanh\lambda}{2}a^{2}}a^{{\dagger}n}e^{-\kappa
ta^{\dagger}a}$
$\displaystyle=\text{sech}\lambda\sum_{n=0}^{\infty}\frac{\left(T^{\prime}\tanh\lambda\right)^{n}}{2^{n}n!}e^{\frac{e^{-2\kappa
t}a^{{\dagger}2}\tanh\lambda}{2}}$ $\displaystyle\times
H_{n}\left(i\sqrt{\frac{\tanh\lambda}{2}}a^{\dagger}e^{-\kappa
t}\right)\left|0\right\rangle\left\langle 0\right|$ $\displaystyle\times
H_{n}\left(-i\sqrt{\frac{\tanh\lambda}{2}}ae^{-\kappa
t}\right)e^{\frac{e^{-2\kappa t}a^{2}\tanh\lambda}{2}}$
$\displaystyle=\text{sech}\lambda\sum_{n=0}^{\infty}\frac{\left(T^{\prime}\tanh\lambda\right)^{n}}{2^{n}n!}\colon
e^{\frac{e^{-2\kappa
t}\left(a^{2}+a^{{\dagger}2}\right)\tanh\lambda}{2}-a^{\dagger}a}$
$\displaystyle\times
H_{n}\left(i\sqrt{\frac{\tanh\lambda}{2}}a^{\dagger}e^{-\kappa
t}\right)H_{n}\left(-i\sqrt{\frac{\tanh\lambda}{2}}ae^{-\kappa
t}\right)\colon$ (22)

then using the following identity 16

$\displaystyle\sum_{n=0}^{\infty}\frac{t^{n}}{2^{n}n!}H_{n}\left(x\right)H_{n}\left(y\right)$
(23) $\displaystyle=$
$\displaystyle\left(1-t^{2}\right)^{-1/2}\exp\left[\frac{t^{2}\left(x^{2}+y^{2}\right)-2txy}{t^{2}-1}\right],$

and $e^{\lambda a^{{\dagger}}a}=\colon
e^{\left(e^{\lambda}-1\right)a^{{\dagger}}a}\colon,$ we finally obtain the
expression of the output state

$\rho\left(t\right)=We^{\frac{\text{\ss}}{2}a^{{\dagger}2}}e^{a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)}e^{\frac{\text{\ss}}{2}a^{2}},$
(24)

with $T^{\prime}=1-e^{-2\kappa t}$ and

$W\equiv\frac{\text{sech}\lambda}{\sqrt{1-T^{\prime
2}\tanh^{2}\lambda}},\text{\ss}\equiv\frac{e^{-2\kappa
t}\tanh\lambda}{1-T^{\prime 2}\tanh^{2}\lambda}.$ (25)

By comparing Eq.(15) with (23) one can see that after going through the
channel the initial squeezing parameter $\tanh\lambda$ in Eq.( 15) becomes to
ß$\equiv\frac{e^{-2\kappa t}\tanh\lambda}{1-T^{\prime 2}\tanh^{2}\lambda},$
and $\left|0\right\rangle\left\langle
0\right|\rightarrow\frac{1}{\sqrt{1-T^{\prime
2}\tanh^{2}\lambda}}e^{a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)},$
a chaotic state (mixed state), due to $T^{\prime}>0,$ we can prove
$\frac{e^{-2\kappa t}}{1-T^{\prime 2}\tanh^{2}\lambda}<1,$ which means a
squeezing-decreasing process. When $\kappa t=0$, then $T^{\prime}=0$ and ß
$=\tanh\lambda$, Eq.(22) becomes the initial squeezed vacuum state as
expected.

It is important to check: if Tr$\rho(t)=1$. Using Eq.(22) and the completeness
of coherent state $\int\frac{d^{2}z}{\pi}\left|z\right\rangle\left\langle
z\right|=1$ as well as the following formula 17

$\int\frac{d^{2}z}{\pi}e^{\zeta\left|z\right|^{2}+\xi z+\eta
z^{\ast}+fz^{2}+gz^{\ast
2}}=\frac{1}{\sqrt{\zeta^{2}-4fg}}e^{\frac{-\zeta\xi\eta+f\eta^{2}+g\xi^{2}}{\zeta^{2}-4fg}},$
(26)

whose convergent condition is Re$\left(\zeta\pm f\pm g\right)<0$ and$\
\mathtt{Re}\left(\frac{\zeta^{2}-4fg}{\zeta\pm f\pm g}\right)<0$, we really
see

$\displaystyle\text{Tr}\rho\left(t\right)$ $\displaystyle=$ $\displaystyle
W\int\frac{d^{2}z}{\pi}\left\langle
z\right|e^{\frac{\text{\ss}}{2}a^{{\dagger}2}}e^{a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)}e^{\frac{\text{\ss}}{2}a^{2}}\left|z\right\rangle$
(27) $\displaystyle=$
$\displaystyle\frac{W}{\sqrt{\left(\text{\ss}T^{\prime}\tanh\lambda-1\right)^{2}-\text{\ss}^{2}}}=1.$

so $\rho\left(t\right)$ is qualified to be a mixed state, thus we see an
initial pure squeezed vacuum state evolves into a squeezed chaotic state with
decreasing-squeezing after passing through an amplitude dissipative channel.

##  IV Average photon number

Using the completeness relation of coherent state and the normally ordering
form of $\rho\left(t\right)$ in Eq. (22), and using
$e^{\frac{\text{\ss}}{2}a^{2}}a^{\dagger}e^{-\frac{\text{\ss}}{2}a^{2}}=a^{\dagger}+$ß$a$,
as well as
$e^{a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)}a^{\dagger}e^{-a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)}$=$a^{\dagger}$ß$T^{\prime}\tanh\lambda,$
we have

$\displaystyle\mathtt{Tr}\left(\rho\left(t\right)a^{\dagger}a\right)$
$\displaystyle=W\int\frac{d^{2}z}{\pi}\left\langle
z\right|e^{\frac{\text{\ss}}{2}a^{{\dagger}2}}e^{a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)}e^{\frac{\text{\ss}}{2}a^{2}}a^{\dagger}a\left|z\right\rangle$
$\displaystyle=W\int\frac{d^{2}z}{\pi}\left\langle
z\right|e^{\frac{\text{\ss}}{2}a^{{\dagger}2}}e^{a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)}e^{\frac{\text{\ss}}{2}a^{2}}za^{\dagger}\left|z\right\rangle$
$\displaystyle=W\int\frac{d^{2}z}{\pi}z\left\langle
z\right|e^{\frac{\text{\ss}}{2}a^{{\dagger}2}}e^{a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)}\left(a^{\dagger}+\text{\ss}a\right)e^{\frac{\text{\ss}}{2}a^{2}}\left|z\right\rangle$
$\displaystyle=W\text{\ss}\int\frac{d^{2}z}{\pi}ze^{\frac{\text{\ss}}{2}\left(z^{\ast
2}+z^{2}\right)}\left\langle
z\right|\left(a^{\dagger}T^{\prime}\tanh\lambda+z\right)e^{a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)}\left|z\right\rangle$
$\displaystyle=W\text{\ss}\int\frac{d^{2}z}{\pi}\left(|z|^{2}T^{\prime}\tanh\lambda+z^{2}\right)$
$\displaystyle\times\exp\left[\left(\text{\ss}T^{\prime}\tanh\lambda-1\right)|z|^{2}+\frac{\text{\ss}}{2}\left(z^{\ast
2}+z^{2}\right)\right].$ (28)

In order to perform the integration, we reform Eq.(28) as

$\displaystyle\mathtt{Tr}\left(\rho\left(t\right)a^{\dagger}a\right)$
$\displaystyle=$ $\displaystyle
W\text{\ss}\left\\{T^{\prime}\tanh\lambda\frac{\partial}{\partial
f}+\frac{2}{\text{ \ss}}\frac{\partial}{\partial s}\right\\}$ (29)
$\displaystyle\times\int\frac{d^{2}z}{\pi}\exp\left[\left(\text{\ss}T^{\prime}\tanh\lambda-1+f\right)|z|^{2}\right.$
$\displaystyle+\left.\frac{\text{\ss}}{2}\left(z^{\ast
2}+\left(1+s\right)z^{2}\right)\right]_{f=s=0}$ $\displaystyle=$
$\displaystyle\frac{1-\text{\ss}T^{\prime}\tanh\lambda}{\left(\text{\ss}T^{\prime}\tanh\lambda-1\right)^{2}-\text{\ss}^{2}}-1$

in the last step, we have used Eq.(27). Using Eq.(29), we present the time
evolution of the average photon number in Fig. 1, from which we find that the
average photon number of the single-mode squeezed vacuum state in the
amplitude damping channel reduces gradually to zero when decay time goes.

Figure 1: (Color online) The average $\bar{n}\left(\kappa t\right)$ as the
function of $\kappa t$ for different values of squeezing parameter $\lambda$
(from bottom to top $\lambda=0,0.1,0.3,0.5,1$.)

##  V Photon statistics distribution

Next, we shall derive the photon statistics distributions of
$\rho\left(t\right)$. The photon number is given by
$p\left(n,t\right)=\left\langle
n\right|\rho\left(t\right)\left|n\right\rangle$. Noticing
$a^{{\dagger}m}\left|n\right\rangle=\sqrt{(m+n)!/n!}\left|m+n\right\rangle$
and using the un-normalized coherent state
$\left|\alpha\right\rangle=\exp[\alpha a^{{\dagger}}]\left|0\right\rangle$, 18
; 19  leading to
$\left|n\right\rangle=\frac{1}{\sqrt{n!}}\frac{\mathtt{d}^{n}}{\mathtt{d}\alpha^{n}}\left|\alpha\right\rangle\left|{}_{\alpha=0}\right.,$
$\left(\left\langle\beta\right.\left|\alpha\right\rangle=e^{\alpha\beta^{\ast}}\right)$,
as well as the normal ordering form of $\rho\left(t\right)$ in Eq. (22), the
probability of finding $n$ photons in the field is given by

$\displaystyle p\left(n,t\right)$ (30) $\displaystyle=$
$\displaystyle\left\langle n\right|\rho\left(t\right)\left|n\right\rangle$
$\displaystyle=$
$\displaystyle\frac{W}{n!}\frac{\mathtt{d}^{n}}{\mathtt{d}\beta^{\ast
n}}\frac{\mathtt{\
d}^{n}}{\mathtt{d}\alpha^{n}}\left.\left\langle\beta\right|e^{\frac{\text{\ss}}{2}\beta^{\ast
2}}e^{a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)}e^{\frac{\text{\ss}}{2}\alpha^{2}}\left|\alpha\right\rangle\right|_{\alpha,\beta^{\ast}=0}$
$\displaystyle=$
$\displaystyle\frac{W}{n!}\frac{\mathtt{d}^{n}}{\mathtt{d}\beta^{\ast
n}}\frac{\mathtt{\
d}^{n}}{\mathtt{d}\alpha^{n}}\left.\exp\left[\beta^{\ast}\alpha\text{
\ss}T^{\prime}\tanh\lambda+\frac{\text{\ss}}{2}\beta^{\ast
2}+\frac{\text{\ss}}{2}\alpha^{2}\right]\right|_{\alpha,\beta^{\ast}=0}.$

Note that

$\left[e^{\frac{\text{\ss}}{2}a^{\dagger
2}}e^{a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)}e^{\frac{\text{\ss}}{2}\alpha^{2}}\right]^{\dagger}=e^{\frac{\text{\ss}}{2}a^{\dagger
2}}e^{a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)}e^{\frac{\text{\ss}}{2}\alpha^{2}}$

so

$\left\langle
n\right|\rho\left(t\right)\left|n\right\rangle^{\ast}=\left\langle
n\right|\rho\left(t\right)^{\dagger}\left|n\right\rangle=\left\langle
n\right|\rho\left(t\right)\left|n\right\rangle$
$\displaystyle\frac{\partial^{n+n}}{\partial t^{n}\partial t^{\prime
n}}\exp\left[2xtt^{\prime}-t^{2}-t^{\prime 2}\right]_{t=t^{\prime}=0}$ (31)
$\displaystyle=$ $\displaystyle
2^{n}n!\sum_{m=0}^{[n/2]}\frac{n!}{2^{2m}\left(m!\right)^{2}(n-2m)!}x^{n-2m},$

we derive the compact form for $\mathfrak{p}\left(n,t\right)$, i.e.,

$\displaystyle p\left(n,t\right)$ (32) $\displaystyle=$
$\displaystyle\frac{W}{n!}\left(-\frac{\text{\ss}}{2}\right)^{n}\frac{\mathtt{d}^{n}}{\mathtt{d}\beta^{\ast
n}}\frac{\mathtt{d}^{n}}{\mathtt{d}\alpha^{n}}\left.e^{-2T^{\prime}\tanh\lambda\beta^{\ast}\alpha-\beta^{\ast
2}-\alpha^{2}}\right|_{\alpha,\beta^{\ast}=0}$ $\displaystyle=$ $\displaystyle
W\left(\text{\ss}T^{\prime}\tanh\lambda\right)^{n}\sum_{m=0}^{[n/2]}\frac{n!\left(T^{\prime}\tanh\lambda\right)^{-2m}}{2^{2m}\left(m!\right)^{2}(n-2m)!}.$

Using the newly expression of Legendre polynomials found in Ref. 20

$x^{n}\sum_{m=0}^{[n/2]}\frac{n!}{2^{2m}\left(m!\right)^{2}(n-2m)!}\left(1-\frac{1}{x^{2}}\right)^{m}=P_{n}\left(x\right),$
(33)

we can formally recast Eq.(32) into the following compact form, i.e.,

$p\left(n,t\right)=W\left(e^{-\kappa
t}\sqrt{-\text{\ss}\tanh\lambda}\right)^{n}P_{n}\left(e^{\kappa
t}T^{\prime}\sqrt{-\text{\ss}\tanh\lambda}\right)$

note that since $\sqrt{-\text{\ss}\tanh\lambda}$ is pure imaginary, while
$p\left(n,t\right)$ is real, so we must still use the power-series expansion
on the right-hand side of Eq.(32) to depict figures of the variation of
$p\left(n,t\right)$. In particular, when $t=0$, Eq.(32 ) reduces to

$\displaystyle p\left(n,0\right)$ $\displaystyle=$
$\displaystyle\text{sech}\lambda\left(\tanh\lambda\right)^{n}\lim_{T^{\prime}\rightarrow
0}\sum_{m=0}^{[n/2]}\frac{n!\left(T^{\prime}\tanh\lambda\right)^{n-2m}}{2^{2m}\left(m!\right)^{2}(n-2m)!}$
(36) $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{cc}\frac{\left(2k\right)!}{2^{2k}k!k!}\text{sech}\lambda\tanh^{2k}\lambda,&n=2k\\\
0&n=2k+1\end{array}\right.,$

which just correspond to the number distributions of the squeezed vacuum state
21 ; 22 . From Eq.(36) it is not difficult to see that the photocount
distribution decreases as the squeezing parameter $\lambda$ increases. While
for $\kappa t\rightarrow\infty,$ we see that $p\left(n,\infty\right)=0.$ This
indicates that there is no photon when a system interacting with a amplitude
dissipative channel for enough long time, as expected. In Fig. 2, the photon
number distribution is shown for different $\kappa t$.

Figure 2: (Color online) Photon number distribution of the squeezed vacuum
state in amplitude damping channel for $\lambda=1$, and different $\kappa t$:
($a)$ $\kappa t=0$, ($b$) $\kappa t=0.5,$ ($c$) $\kappa t=1$ and ($d$) $\kappa
t=2$.

##  VI Wigner functions

In this section, we shall use the normally ordering for of density operators
to calculate the analytical expression of Wigner function. For a single-mode
system, the WF is given by 23

$W\left(\alpha,\alpha^{\ast},t\right)=e^{2\left|\alpha\right|^{2}}\int\frac{d^{2}\beta}{\pi^{2}}\left\langle-\beta\right|\rho\left(t\right)\left|\beta\right\rangle
e^{-2\left(\beta\alpha^{\ast}-\beta^{\ast}\alpha\right)},$ (37)

where $\left|\beta\right\rangle$ is the coherent state 18 ; 19  . From Eq.(22)
it is easy to see that once the normal ordered form of $\rho\left(t\right)$ is
known, we can conveniently obtain the Wigner function of $\rho\left(t\right)$.

On substituting Eq.(24) into Eq.(37) we obtain the WF of the single-mode
squeezed state in the ADC,

$\displaystyle W\left(\alpha,\alpha^{\ast},t\right)$ (38) $\displaystyle=$
$\displaystyle
We^{2\left|\alpha\right|^{2}}\int\frac{d^{2}\beta}{\pi^{2}}\exp\left[-\left(1+\text{\ss}T^{\prime}\tanh\lambda\right)\left|\beta\right|^{2}\right.$
$\displaystyle\left.-2\left(\beta\alpha^{\ast}-\beta^{\ast}\alpha\right)+\frac{\text{\ss}}{2}\beta^{\ast
2}+\frac{\text{\ss}}{2}\beta^{2}\right]$ $\displaystyle=$
$\displaystyle\frac{W}{\pi\sqrt{\left(1+\text{\ss}T^{\prime}\tanh\lambda\right)^{2}-\text{\ss}^{2}}}\exp\left[2\left|\alpha\right|^{2}\right]$
$\displaystyle\times\exp\left[2\frac{-2\left(1+\text{\ss}T^{\prime}\tanh\lambda\right)\left|\alpha\right|^{2}+\text{\ss}\left(\alpha^{\ast
2}+\alpha^{2}\right)}{\left(1+\text{\ss}T^{\prime}\tanh\lambda\right)^{2}-\text{\ss}^{2}}\right]$

In particular, when $t=0$ and $t\rightarrow\infty$, Eq.(38) reduces to
$W\left(\alpha,\alpha^{\ast},0\right)=\frac{1}{\pi}\exp[-2\left|\alpha\right|^{2}\cosh
2\lambda+\left(\alpha^{\ast 2}+\alpha^{2}\right)\sinh 2\lambda]$, and
$W\left(\alpha,\alpha^{\ast},\infty\right)=\frac{1}{\pi}\exp\left[-2\left|\alpha\right|^{2}\right]$,
which are just the WF of the single-mode squeezed vacuum state and the vacuum
state, respectively. In Fig. 3, the WF of the single-mode squeezed vacuum
state in the amplitude damping channel is shown for different decay time
$\kappa t$.

Figure 3: (Color online) Wigner function of the squeezed vacuum state in
amplitude damping channel for $\lambda=1.0$, different $\kappa t$: ($a)$
$\kappa t=0.0$, ($b$) $\kappa t=0.5$, ($c$) $\kappa t=1$, and ($d$) $\kappa
t=2$.

##  VII Tomogram

As we know, once the probability distributions
$P_{\theta}\left(\hat{x}_{\theta}\right)$ of the quadrature amplitude are
obtained, one can use the inverse Radon transformation familiar in tomographic
imaging to obtain the WF and density matrix 24 . Thus the Radon transform of
the WF is corresponding to the probability distributions
$P_{\theta}\left(\hat{x}_{\theta}\right)$. In this section we derive the
tomogram of $\rho\left(t\right)$.

For a single-mode system, the Radon transform of WF, denoted as $\mathcal{R}$
is defined by 25

$\displaystyle\mathcal{R}\left(q\right)_{f,g}$ $\displaystyle=$
$\displaystyle\int\delta\left(q-fq^{\prime}-gp^{\prime}\right)Tr\left[\Delta\left(\beta\right)\rho\left(t\right)\right]dq^{\prime}dp^{\prime}$
(39) $\displaystyle=$ $\displaystyle Tr\left[\left|q\right\rangle_{f,g\text{
}f,g}\left\langle q\right|\rho\left(t\right)\right]=_{f,g}\left\langle
q\right|\rho\left(t\right)\left|q\right\rangle_{f,g}$

where the operator $\left|q\right\rangle_{f,g\text{ }f,g}\left\langle
q\right|$ is just the Radon transform of single-mode Wigner operator
$\Delta\left(\beta\right)$, and

$\left|q\right\rangle_{f,g}=A\exp\left[\frac{\sqrt{2}qa^{{\dagger}}}{B}-\frac{B^{\ast}}{2B}a^{{\dagger}2}\right]\left|0\right\rangle,$
(40)

as well as $B=f-ig,$
$A=\left[\pi\left(f^{2}+g^{2}\right)\right]^{-1/4}\exp[-q^{2}/2\left(f^{2}+g^{2}\right)]$.
Thus the tomogram of a quantum state $\rho\left(t\right)$ is just the quantum
average of $\rho\left(t\right)$ in $\left|q\right\rangle_{f,g}$ representation
(a kind of intermediate coordinate-momentum representation) 26 .

Substituting Eqs.(24) and (40) into Eq.(39), and using the completeness
relation of coherent state, we see that the Radom transform of WF of
$\rho\left(t\right)$ is given by

$\displaystyle\mathcal{R}\left(q\right)_{f,g}$ (41) $\displaystyle=$
$\displaystyle W_{f,g}\left\langle
q\right|e^{\frac{\text{\ss}}{2}a^{{\dagger}2}}e^{a^{\dagger}a\ln\left(\text{\ss}T^{\prime}\tanh\lambda\right)}e^{\frac{\text{\ss}}{2}a^{2}}\left|q\right\rangle_{f,g}$
$\displaystyle=$
$\displaystyle\frac{WA^{2}}{\sqrt{E}}\exp\left\\{\frac{q^{2}\text{\ss}}{E\left|B\right|^{4}}\left(B^{2}+B^{\ast}{}^{2}\right)\right.$
$\displaystyle+\left.\frac{2q^{2}\text{\ss}}{E\left|B\right|^{2}}\left(T\tanh\lambda+\text{\ss}-\text{\ss}T^{2}\tanh^{2}\lambda\right)\right\\},$

where we have used the formula (26) and
$\left\langle\alpha\right|\left.\gamma\right\rangle=\exp[-\left|\alpha\right|^{2}/2-\left|\gamma\right|^{2}/2+\alpha^{\ast}\gamma]$,
as well as

$\displaystyle E$ $\displaystyle=$
$\displaystyle\left(1+\text{\ss}\frac{B}{B^{\ast}}\right)\left(1+\frac{B^{\ast}}{B}\text{\ss}-B^{\ast}\frac{\left(\text{\ss}T^{\prime}\tanh\lambda\right)^{2}}{B^{\ast}+\text{\ss}B}\right)$
(42) $\displaystyle=$
$\displaystyle\left|1+\frac{\text{\ss}B}{B^{\ast}}\right|^{2}-\left(\text{
\ss}T^{\prime}\tanh\lambda\right)^{2}.$

In particular, when $t=0,$ ($T=0$), then Eq.(41) reduces to
($\frac{B}{B^{\ast}}=e^{2i\phi}$)

$\displaystyle\mathcal{R}\left(q\right)_{f,g}$ $\displaystyle=$
$\displaystyle\frac{A^{2}\text{sech}\lambda}{\left|1+e^{2i\phi}\tanh\lambda\right|}$
(43)
$\displaystyle\times\exp\left\\{\frac{q^{2}\left(B^{2}+B^{\ast}{}^{2}+2\left|B\right|^{2}\tanh\lambda\right)\tanh\lambda}{\left|1+e^{2i\phi}\left|B\right|^{4}\tanh\lambda\right|^{2}}\right\\},$

which is a tomogram of single-mode squeezed vacuum state; while for $\kappa
t\rightarrow\infty,$($T=1$), then $\mathcal{R}\left(q\right)_{f,g}=A^{2},$
which is a Gaussian distribution corresponding to the vacuum state.

In summary, using the way of deriving infinitive sum representation of density
operator by virtue of the entangled state representation describing, we
conclude that in the amplitude dissipative channel the initial density
operator of a single-mode squeezed vacuum state evolves into a squeezed
chaotic state with decreasing-squeezing. We investigate average photon number,
photon statistics distributions, Wigner functions and tomogram for the output
state.

## Acknowledgments

This work is supported by the National Natural Science Foundation of China
(Grant No.11175113 and 11047133), Shandong Provincial Natural Science
Foundation in China (Gant No.ZR2010AQ024), and a grant from the Key Programs
Foundation of Ministry of Education of China (Grant No. 210115),as well as
Jiangxi Provincial Natural Science Foundation in China (No. 2010GQW0027).

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