# Quantum repeater protocol in mixed single- and two-mode Tavis-Cummings
models

M Ghasemi1 M K Tavassoly1 1Atomic and Molecular Group, Faculty of Physics,
Yazd University, Yazd 89195-741, Iran<EMAIL_ADDRESS>
(today)

###### Abstract

In this paper we study the production of entanglement between two atoms which
are far from each other. We consider a system including eight two-level atoms
$(1,2,\cdots,8)$ such that any atom with its adjacent atom is in atomic Bell
state, so that we have four separate pairs of maximally entangled states
$(i,i+1)$ where $i=1,3,5,7$. Our purpose is to produce entanglement between
the atomic pair (1, 8), while these two distant atoms have no interaction. By
performing the interaction between adjacent non-entangled atomic pairs (2, 3)
as well as (6, 7), each pair with a two-mode quantized field, the entanglement
is produced between atoms (1, 4) and (5, 8), respectively. Finally, by
applying an appropriate Bell state measurement (BSM) on atoms (4, 5) or
performing an interaction between them with a single-mode field (quantum
electrodynamic: QED method), the qubit pair $(1,8)$ becomes entangled and so
the quantum repeater is successfully achieved. This swapped entanglement is
then quantified via concurrence measure and the effects of coupling
coefficients and detuning on the concurrence and success probability are
numerically investigated. The maxima of concurrence and success probability
and the corresponding time periods have been decreased by increasing the
detuning in asymmetric condition in BSM method. Also, the effects of detuning,
initial interaction time and coupling coefficient on the produced entanglement
by QED method are considered. Increasing (decreasing) of the detuning
(interaction time) has destructive effect on the swapped entanglement in
asymmetric condition.

###### pacs:

03.65.Yz; 03.67.Bg; 42.50.-p; 42.79.Fm

##  1 Introduction

Extending the quantum phenomena to macroscopic distances is of considerable
interest [1, 2]. In particular, entangled quantum states can be transferred
over long distances using quantum repeater protocols [2]. In fact, to prevail
the entanglement attenuation in the process of transferring the entanglement
the quantum repeater is proposed. These transferred states are useful in
quantum communication [3], quantum key distribution [4] and quantum
cryptography [5]. In Ref. [6] the authors have investigated the optimal
probability of creating a maximally entangled state by quantum repeater with
respect to classical communication. The hybrid quantum repeater is
investigated in [7, 8] using bright coherent light. A quantum repeater which
uses photon pair sources in combination with memories has been proposed [9].
Also, it is shown that the local generation of entangled pairs of atomic
excitations, in combination with two-photon detections permits the
implementation of a quantum repeater protocol [10]. A quantum repeater model
using quantum-dot spins and photons with spatial entanglement in optical
microcavities has been studied in [11]. An experimental realization of
entanglement concentration and a quantum repeater has been reported in [12].
As two sticking points in the quantum repeater protocols one may refer to the
entanglement swapping and quantum memories. The entanglement can be swapped in
short parts of a long distance by performing interaction described by Tavis-
Cummings [13] or Jaynes-Cummings [14] models. After producing the
entanglement, by Bell state measurement (BSM) [15, 16, 17] or performing
interaction between separable parts [18] the entanglement is swapped to non-
entangled segments (for instance, see [19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29]).
In this paper we consider a quantum repeater protocol to distribute
entanglement between two distant atoms. To achieve the purpose, by inserting
six atoms between the two far atoms we divide the distance to seven shorter
parts (see figure 1). Then, by performing the interaction between atoms (2,3)
and (6,7) in two two-mode cavities, the entanglement is produced in the two
pairs (1,4) and (5,8), separately. At last, the atoms (1,8) can be entangled
either by operating the BSM (performing interaction) between (on) atoms (4,5)
in a single-mode cavity. The degree of entanglement of the produced entangled
states of atoms (1,8) is evaluated by concurrence [30]. In the continuation,
we have analyzed the influence of atom-field coupling coefficient and detuning
on the concurrence and success probability of the proposed model. Also, the
effect of interaction time on the concurrence has been investigated.
This paper organizes as follows: In the next section we have introduced our
quantum repeater protocol model and presented the detailed calculations. The
related numerical results are collected in Sec. 3. Finally, the paper is ended
by conclusions in Sec. 4.

##  2 2\. Quantum repeater protocol

Figure 1:  Quantum repeater protocol: there exist four atomic Bell states
$(i,i+1)$, where $i=1,3,5,7$, the atomic pairs (2,3) and (6,7) interact in
two-mode cavities, separately, and two atoms (4,5) are contacted by operating
Bell state measurement process or performing an interaction in a single-mode
cavity.

As we demonstrated in the Introduction of the paper, in the model shown in
figure 1 , our aim is the generation of entanglement between two separable
distant atoms 1, 8. At first, we consider the interaction between atoms (2,3),
where the initial state for atoms (1-4) reads as
$\ket{\Psi}_{1,2}\otimes\ket{\Psi}_{3,4}$ with,

$\displaystyle\ket{\Psi}_{i,i+1}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(\ket{eg}-\ket{ge})_{i,i+1},\qquad i=1,3.$ (1)

Two two-level atoms (2,3) with excited (ground) state $\ket{e}$ ($\ket{g}$) in
a two-mode cavity field are interacted by the Tavis-Cummings Hamiltonian
introduced by,

$\hat{H}=\hat{H}_{0}+\hat{H}_{1},$ (2)

where the free and interacting parts are respectively defined as $(\hbar=1)$
[31]:

$\displaystyle\hat{H}_{0}$ $\displaystyle=$
$\displaystyle\omega\hat{a}^{\dagger}\hat{a}+\omega^{\prime}\hat{b}^{\dagger}\hat{b}+\sum_{i=2,3}\frac{\omega_{i}}{2}\hat{\sigma}_{iz},$
(3) $\displaystyle\hat{H}_{1}$ $\displaystyle=$
$\displaystyle\sum_{i=2,3}g_{i}\left(\hat{a}\hat{b}\hat{\sigma}_{i}^{+}+\hat{a}^{\dagger}\hat{b}^{\dagger}\hat{\sigma}_{i}^{-}\right),$

where in fact we have considered the two-photon transition process in (3). It
should be mentioned that the two-photon process [32, 33] is not far from
experimental realization (as a few efforts in this relation see Refs. [34, 35,
36]). In above relations, the creation (annihilation) operators of the two
modes have been shown by $\hat{a}^{\dagger}$, $\hat{b}^{\dagger}$ ($\hat{a}$,
$\hat{b}$) and
$\hat{\sigma}_{i}^{+}=\ket{e}_{i}\bra{g}(\hat{\sigma}_{i}^{-}=\hat{\sigma}_{i}^{+\dagger})$,
$\hat{\sigma}_{iz}$ are raising (lowering) and population inversion operators
of the $i$th atom, respectively. The $i$th atom with frequency $\omega_{i}$
(in this case, $i=2,3$) interacts with two-mode field with frequencies
$\omega$, $\omega^{\prime}$. The corresponding Hamiltonian in the interaction
picture reads as,

$\hat{H}_{int}=\sum_{i=2,3}g_{i}\left(\hat{a}\hat{b}\hat{\sigma_{i}}^{+}e^{i\Delta_{i}t}+H.C.\right),$
(4)

which may be obtained from
$\hat{H}_{int}=e^{i\hat{H}_{0}t}\hat{H}_{1}e^{-i\hat{H}_{0}t}$ by Baker-
Hausdorff formula, where the detuning is defined as
$\Delta_{i}=\omega_{i}-(\omega+\omega^{\prime})$. We assume that the cavity is
initially in vacuum state and also $\Delta_{i}\gg g_{i}$. So, following the
path of Refs. [37, 38] the effective Hamiltonian is obtained as follows:

$\displaystyle\hat{H}_{eff}$ $\displaystyle=$ $\displaystyle
g_{23}\left(\hat{\sigma}_{2}^{+}\hat{\sigma}_{3}^{-}e^{i(\Delta_{2}-\Delta_{3})t}+H.C.\right)+\sum_{i=2,3}\lambda_{i}\hat{\sigma}_{i}^{+}\hat{\sigma}_{i}^{-},$
(5) $\displaystyle\lambda_{i}$ $\displaystyle=$
$\displaystyle\frac{g^{2}_{i}}{\Delta_{i}},\qquad\frac{1}{\bar{\omega}_{23}}=\frac{1}{2}(\frac{1}{\Delta_{2}}+\frac{1}{\Delta_{3}}),\qquad
g_{23}=\frac{g_{2}g_{3}}{\bar{\omega}_{23}}.$ (6)

Now, we calculate the time evolution operator with the help of effective
Hamiltonian (5) and using the relation
$i\hat{\dot{U}}(t)=\hat{H}_{eff}\hat{U}(t)$ in the basis spanned by
$\left\\{\ket{ee}_{2,3},\ket{eg}_{2,3},\ket{ge}_{2,3},\ket{gg}_{2,3}\right\\}$,

$\hat{U}_{j,j+1}(t)=\left(\begin{array}[]{cccc}e^{-i(\lambda_{2}+\lambda_{3})t}&0&0&0\\\
0&e^{-i\lambda_{2}t}e^{i\delta_{23}t/2}\left(\cos(\frac{ft}{2})-i\frac{\delta_{23}}{f}\sin(\frac{ft}{2})\right)&e^{-i\lambda_{2}t}e^{i\delta_{23}t/2}\left(-2i\frac{g_{23}}{f}\sin(\frac{ft}{2})\right)&0\\\
0&e^{-i\lambda_{3}t}e^{-i\delta_{23}t/2}\left(-2i\frac{g_{23}}{f}\sin(\frac{ft}{2})\right)&e^{-i\lambda_{3}t}e^{-i\delta_{23}t/2}\left(\cos(\frac{ft}{2})+i\frac{\delta_{23}}{f}\sin(\frac{ft}{2})\right)&0\\\
0&0&0&1\\\ \end{array}\right)_{j,j+1},$ (7) see eq. (7)

where $j=2,6$ and

$\displaystyle\delta_{23}=\Delta_{2}-\Delta_{3}+(\lambda_{2}-\lambda_{3}),\qquad
f=\sqrt{\delta_{23}^{2}+4g^{2}_{23}}.$ (8)

Then, using the initial state $\ket{\Psi}_{1,2}\otimes\ket{\Psi}_{3,4}$ (Eq.
(1)) and (7), the state of atoms (1,2,3,4) after the interaction at time $t$
can be deduced as below:

$\displaystyle\ket{\Psi(t)}_{1-4}$ $\displaystyle=$
$\displaystyle\hat{U}_{2,3}(t)\ket{\Psi}_{1,2}\otimes\ket{\Psi}_{3,4}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(U_{23}(t)\ket{eegg}+U_{33}(t)\ket{egeg}-\ket{egge}\right.$
$\displaystyle-$
$\displaystyle\left.U_{11}(t)\ket{geeg}+U_{22}(t)\ket{gege}+U_{32}(t)\ket{ggee}\right)_{1-4},$

where $U_{ij}(t)$ is the matrix element of (7) locating in the row $i$ and
column $j$. If the atoms (2,3) be in $\ket{ge}_{2,3}$ or $\ket{eg}_{2,3}$, the
state of atomic pair (1,4) is achieved respectively as follows,

$\displaystyle\ket{\Psi(t)}_{1,4}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{\left|U_{33}(t)\right|^{2}+\left|U_{32}(t)\right|^{2}}}(U_{33}(t)\ket{eg}$
$\displaystyle+$ $\displaystyle U_{32}(t)\ket{ge})_{1,4},$
$\displaystyle\ket{\Psi^{\prime}(t)}_{1,4}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{\left|U_{23}(t)\right|^{2}+\left|U_{22}(t)\right|^{2}}}(U_{23}(t)\ket{eg}$
$\displaystyle+$ $\displaystyle U_{22}(t)\ket{ge})_{1,4}.$

Similarly, atoms (5,6,7,8) are evolved with the effective Hamiltonian (5) as
above procedure, however, with the time evolution operator $\hat{U}_{6,7}(t)$
in (7), by which the entangled states of atoms (5,8) can be obtained as
follows,

$\displaystyle\ket{\Psi(t)}_{5,8}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{\left|U_{33}(t)\right|^{2}+\left|U_{32}(t)\right|^{2}}}(U_{33}(t)\ket{eg}$
$\displaystyle+$ $\displaystyle U_{32}(t)\ket{ge})_{5,8},$
$\displaystyle\ket{\Psi^{\prime}(t)}_{5,8}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{\left|U_{23}(t)\right|^{2}+\left|U_{22}(t)\right|^{2}}}(U_{23}(t)\ket{eg}$
$\displaystyle+$ $\displaystyle U_{22}(t)\ket{ge})_{5,8},$

depending on measuring $\ket{ge}_{6,7}$ and $\ket{eg}_{6,7}$, respectively.
Consequently, we have four possible non-entangled states
$\ket{\Psi(t)}_{1,4}\otimes\ket{\Psi(t)}_{5,8}$,
$\ket{\Psi(t)}_{1,4}\otimes\ket{\Psi^{\prime}(t)}_{5,8}$,
$\ket{\Psi^{\prime}(t)}_{1,4}\otimes\ket{\Psi(t)}_{5,8}$ and
$\ket{\Psi^{\prime}(t)}_{1,4}\otimes\ket{\Psi^{\prime}(t)}_{5,8}$. Considering
all these states we are going to introduce our quantum repeater protocol and
compare the obtained results.

###  2.1 The case $\ket{\Psi(t)}_{1,4}\otimes\ket{\Psi(t)}_{5,8}$

In this stage, we operate the Bell state measurement performed with the state
$\ket{B}_{4,5}=\frac{1}{\sqrt{2}}(\ket{ee}+\ket{gg})_{4,5}$ on state
$\ket{\Psi(t)}_{1,4}\otimes\ket{\Psi(t)}_{5,8}$. As a result, the state of
atoms (1,8) is converted to the Bell state,

$\ket{B}_{1,8}=\frac{1}{\sqrt{2}}(\ket{ee}+\ket{gg})_{1,8},$ (14)

with the success probability

$S_{\ket{B}_{1,8}}(t)=\frac{\left|U_{33}(t)U_{32}(t)\right|^{2}}{(\left|U_{33}(t)\right|^{2}+\left|U_{32}(t)\right|^{2})^{2}}.$
(15)

Now, we perform a BSM with Bell state
$\ket{B^{\prime}}_{4,5}=\frac{1}{\sqrt{2}}(\ket{eg}+\ket{ge})_{4,5}$ on state
$\ket{\Psi(t)}_{1,4}\otimes\ket{\Psi(t)}_{5,8}$, so the atoms (1,8) are
converted to the following entangled state,

$\ket{\gamma}^{1}_{1,8}=\frac{1}{\sqrt{\left|U_{33}(t)\right|^{4}+\left|U_{32}(t)\right|^{4}}}(U^{2}_{33}(t)\ket{eg}+U^{2}_{32}(t)\ket{ge})_{1,8},$
(16)

whose concurrence reads as

$C_{1}(t)=\frac{2\left|U^{*2}_{33}(t)U^{2}_{32}(t)\right|}{\left|U_{33}(t)\right|^{4}+\left|U_{32}(t)\right|^{4}}.$
(17)

Next, we consider the initial state
$\ket{\Psi(t)}_{1,4}\otimes\ket{\Psi(t)}_{5,8}$ and investigate the
interaction between atoms (4,5) in a single-mode cavity using the effective
Hamiltonian,

$\hat{H}^{\prime}_{eff}=\lambda^{\prime}\sum_{i=4,5}\hat{\sigma}_{i}^{+}\hat{\sigma}_{i}^{-}+\lambda^{\prime}\left(\hat{\sigma}_{4}^{+}\hat{\sigma}_{5}^{-}+\hat{\sigma}_{4}^{-}\hat{\sigma}_{5}^{+}\right),$
(18)
$\displaystyle\lambda^{\prime}=\frac{g^{2}}{\delta},\qquad\delta=\omega_{4}-\omega=\omega_{5}-\omega.$
(19)

Notice that the above Hamiltonian has been readily obtained by using the path
of Refs. [37, 38] and the following Hamiltonian

$\hat{H^{\prime}}=\omega\hat{a}^{\dagger}\hat{a}+\sum_{i=4,5}\frac{\omega_{i}}{2}\hat{\sigma}_{iz}+\sum_{i=4,5}g\left(\hat{a}\hat{\sigma}_{i}^{+}+\hat{a}^{\dagger}\hat{\sigma}_{i}^{-}\right).$
(20)

By performing the interaction (18) between atoms (4,5) for the initial state
$\ket{\Psi(t)}_{1,4}\otimes\ket{\Psi(t)}_{5,8}$, the following entangled state
can be achieved for atoms $(1,4,5,8)$,

$\displaystyle\ket{\phi}_{1,4,5,8}$
$\displaystyle=\frac{1}{\left|U_{32}(t)\right|^{2}+\left|U_{33}(t)\right|^{2}}$
(21)
$\displaystyle\left\\{U^{2}_{33}(t)\left[\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1)\ket{eegg}\right.\right.$
$\displaystyle+\left.\left.\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1)\ket{egeg}\right]\right.$
$\displaystyle+\left.U^{2}_{32}(t)\left[\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1)\ket{gege}\right.\right.$
$\displaystyle+\left.\left.\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1)\ket{ggee}\right]\right.$
$\displaystyle+\left.U_{32}(t)U_{33}(t)\left(\ket{egge}\right.\right.$
$\displaystyle+\left.\left.e^{-2i\lambda^{\prime}(\tau-t)}\ket{geeg}\right)\right\\}_{1,4,5,8}.$

? By making a measurement on the state (21), one obtains $\ket{eg}_{4,5}$ and
$\ket{ge}_{4,5}$, the result of which arrives us respectively at

$\displaystyle\ket{\phi}^{1}_{1,8}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{\left|a_{1}(t,\tau)\right|^{2}+\left|b_{1}(t,\tau)\right|^{2}}}\left(a_{1}(t,\tau)\ket{eg}\right.$
$\displaystyle+$ $\displaystyle\left.b_{1}(t,\tau)\ket{ge}\right)_{1,8},$
$\displaystyle a_{1}(t,\tau)$ $\displaystyle=$ $\displaystyle
U^{2}_{33}(t)(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1),$
$\displaystyle b_{1}(t,\tau)$ $\displaystyle=$ $\displaystyle
U^{2}_{32}(t)(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1),$

with the concurrence

$C^{\prime}_{1}(t,\tau)=\frac{2\left|a^{*}_{1}(t,\tau)b_{1}(t,\tau)\right|}{\left|a_{1}(t,\tau)\right|^{2}+\left|b_{1}(t,\tau)\right|^{2}},$
(23)

and

$\displaystyle\ket{\phi^{\prime}}^{1}_{1,8}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{\left|a^{\prime}_{1}(t,\tau)\right|^{2}+\left|b^{\prime}_{1}(t,\tau)\right|^{2}}}\left(a^{\prime}_{1}(t,\tau)\ket{eg}\right.$
$\displaystyle+$
$\displaystyle\left.b^{\prime}_{1}(t,\tau)\ket{ge}\right)_{1,8},$
$\displaystyle a^{\prime}_{1}(t,\tau)$ $\displaystyle=$ $\displaystyle
U^{2}_{33}(t)(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1),$
$\displaystyle b^{\prime}_{1}(t,\tau)$ $\displaystyle=$ $\displaystyle
U^{2}_{32}(t)(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1),$

with the concurrence

$C^{\prime\prime}_{1}(t,\tau)=\frac{2\left|a^{\prime*}_{1}(t,\tau)b^{\prime}_{1}(t,\tau)\right|}{\left|a^{\prime}_{1}(t,\tau)\right|^{2}+\left|b^{\prime}_{1}(t,\tau)\right|^{2}}.$
(25)

###  2.2 The case $\ket{\Psi(t)}_{1,4}\otimes\ket{\Psi^{\prime}(t)}_{5,8}$

By the operation of BSM performed with the state
$\ket{B^{\prime}}_{4,5}=\frac{1}{\sqrt{2}}(\ket{eg}+\ket{ge})_{4,5}$ on state
$\ket{\Psi(t)}_{1,4}\otimes\ket{\Psi^{\prime}(t)}_{5,8}$, the atoms (1,8) are
converted to the Bell state

$\ket{B^{\prime}}_{1,8}=\frac{1}{\sqrt{2}}(\ket{eg}+e^{-2i\varphi}\ket{ge})_{1,8},$
(26)

where
$e^{i\varphi}=\pm\frac{\cos(\frac{ft}{2})+i\frac{\delta_{23}}{f}\sin(\frac{ft}{2})}{\left|\cos(\frac{ft}{2})+i\frac{\delta_{23}}{f}\sin(\frac{ft}{2})\right|}$
with $\left|e^{i\varphi}\right|=1$ and the success probability reads as

$S_{\ket{B^{\prime}}_{1,8}}(t)=\frac{\left|U_{33}(t)U_{23}(t)\right|^{2}+\left|U_{32}(t)U_{22}(t)\right|^{2}}{2(\left|U_{33}(t)\right|^{2}+\left|U_{32}(t)\right|^{2})(\left|U_{23}(t)\right|^{2}+\left|U_{22}(t)\right|^{2})}.$
(27)

Notice that $S_{\ket{B^{\prime}}_{1,8}}(t)=S_{\ket{B}_{1,8}}(t)$. But, by
measuring the Bell state
$\ket{B}_{4,5}=\frac{1}{\sqrt{2}}(\ket{ee}+\ket{gg})_{4,5}$ on state
$\ket{\Psi(t)}_{1,4}\otimes\ket{\Psi^{\prime}(t)}_{5,8}$, the atoms (1,8) are
converted to the entangled state

$\displaystyle\ket{\gamma}^{2}_{1,8}$
$\displaystyle=\frac{1}{\sqrt{\left|U_{22}(t)U_{33}(t)\right|^{2}+\left|U_{23}(t)U_{32}(t)\right|^{2}}}$
$\displaystyle(U_{22}(t)U_{33}(t)\ket{ee}+U_{23}(t)U_{32}(t)\ket{gg})_{1,8},$

with the concurrence

$C_{2}(t)=\frac{2\left|U^{*}_{22}(t)U^{*}_{33}(t)U_{23}(t)U_{32}(t)\right|}{\left|U_{22}(t)U_{33}(t)\right|^{2}+\left|U_{23}(t)U_{32}(t)\right|^{2}}.$
(29)

It can be seen that $C_{2}(t)=C_{1}(t)$.
Now, the entangled state for atoms $(1,4,5,8)$ can be achieved after
performing the interaction according to the Hamiltonian (18) between atoms
(4,5) in a single-mode cavity with initial state
$\ket{\Psi(t)}_{1,4}\otimes\ket{\Psi^{\prime}(t)}_{5,8}$ which results in,

$\displaystyle\ket{\phi^{\prime}}_{1,4,5,8}$
$\displaystyle=\frac{1}{\sqrt{\left|U_{32}(t)\right|^{2}+\left|U_{33}(t)\right|^{2}}\sqrt{\left|U_{22}(t)\right|^{2}+\left|U_{23}(t)\right|^{2}}}$
(30)
$\displaystyle\left\\{U_{33}(t)U_{23}(t)\left[\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1)\ket{eegg}\right.\right.$
$\displaystyle+\left.\left.\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1)\ket{egeg}\right]\right.$
$\displaystyle+\left.U_{32}(t)U_{22}(t)\left[\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1)\ket{gege}\right.\right.$
$\displaystyle+\left.\left.\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1)\ket{ggee}\right]\right.$
$\displaystyle+\left.U_{33}(t)U_{22}(t)\ket{egge}\right.$
$\displaystyle+\left.e^{-2i\lambda^{\prime}(\tau-t)}U_{32}(t)U_{23}(t)\ket{geeg}\right\\}_{1,4,5,8}.$

? By applying a measurement on the state (30), one obtains $\ket{eg}_{4,5}$
and $\ket{ge}_{4,5}$, where the results for atoms (1,8) are respectively as

$\displaystyle\ket{\phi}^{2}_{1,8}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{\left|a_{2}(t,\tau)\right|^{2}+\left|b_{2}(t,\tau)\right|^{2}}}\left(a_{2}(t,\tau)\ket{eg}\right.$
$\displaystyle+$ $\displaystyle\left.b_{2}(t,\tau)\ket{ge}\right)_{1,8},$
$\displaystyle a_{2}(t,\tau)$ $\displaystyle=$ $\displaystyle
U_{23}(t)U_{33}(t)(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1),$
$\displaystyle b_{2}(t,\tau)$ $\displaystyle=$ $\displaystyle
U_{22}(t)U_{32}(t)(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1),$

whose concurrence reads as

$\displaystyle C^{\prime}_{2}(t,\tau)$ $\displaystyle=$
$\displaystyle\frac{2\left|a^{*}_{2}(t,\tau)b_{2}(t,\tau)\right|}{\left|a_{2}(t,\tau)\right|^{2}+\left|b_{2}(t,\tau)\right|^{2}},$
(32)

and

$\displaystyle\ket{\phi^{\prime}}^{2}_{1,8}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{\left|a^{\prime}_{2}(t,\tau)\right|^{2}+\left|b^{\prime}_{2}(t,\tau)\right|^{2}}}\left(a^{\prime}_{2}(t,\tau)\ket{eg}\right.$
$\displaystyle+$
$\displaystyle\left.b^{\prime}_{2}(t,\tau)\ket{ge}\right)_{1,8},$
$\displaystyle a^{\prime}_{2}(t,\tau)$ $\displaystyle=$ $\displaystyle
U_{23}(t)U_{33}(t)(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1),$
$\displaystyle b^{\prime}_{2}(t,\tau)$ $\displaystyle=$ $\displaystyle
U_{22}(t)U_{32}(t)(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1),$

with the concurrence

$C^{\prime\prime}_{2}(t,\tau)=\frac{2\left|a^{\prime*}_{2}(t,\tau)b^{\prime}_{2}(t,\tau)\right|}{\left|a^{\prime}_{2}(t,\tau)\right|^{2}+\left|b^{\prime}_{2}(t,\tau)\right|^{2}}.$
(34)

It is easy to check that
$C^{\prime\prime}_{2}(t,\tau)=C^{\prime}_{2}(t,\tau)$.

###  2.3 The case $\ket{\Psi^{\prime}(t)}_{1,4}\otimes\ket{\Psi(t)}_{5,8}$

The above processes can be repeated for the initial state
$\ket{\Psi^{\prime}(t)}_{1,4}\otimes\ket{\Psi(t)}_{5,8}$. With a BSM using the
maximally entangled state
$\ket{B^{\prime}}_{4,5}=\frac{1}{\sqrt{2}}(\ket{eg}+\ket{ge})_{4,5}$ on state
$\ket{\Psi^{\prime}(t)}_{1,4}\otimes\ket{\Psi(t)}_{5,8}$, the state of atoms
(1,8) is converted to Bell state $\ket{B^{\prime}}_{1,8}$ in Eq. (26) with the
success probability (27). Meanwhile, measuring the Bell state
$\ket{B}_{4,5}=\frac{1}{\sqrt{2}}(\ket{ee}+\ket{gg})_{4,5}$ on state
$\ket{\Psi^{\prime}(t)}_{1,4}\otimes\ket{\Psi(t)}_{5,8}$, the entangled state

$\displaystyle\ket{\gamma}^{3}_{1,8}$
$\displaystyle=\frac{1}{\sqrt{\left|U_{32}(t)U_{23}(t)\right|^{2}+\left|U_{22}(t)U_{33}(t)\right|^{2}}}$
$\displaystyle(U_{32}(t)U_{23}(t)\ket{ee}+U_{22}(t)U_{33}(t)\ket{gg})_{1,8},$

is obtained, where its concurrence is calculated as below:

$C_{3}(t)=\frac{2\left|U^{*}_{32}(t)U^{*}_{23}(t)U_{22}(t)U_{33}(t)\right|}{\left|U_{32}(t)U_{23}(t)\right|^{2}+\left|U_{22}(t)U_{33}(t)\right|^{2}}.$
(36)

Notice that $C_{3}(t)=C_{2}(t)=C_{1}(t)$.
On the other hand, the entangled state for atoms $(1,4,5,8)$ after performing
the interaction according to the Hamiltonian (18) with initial state
$\ket{\Psi^{\prime}(t)}_{1,4}\otimes\ket{\Psi(t)}_{5,8}$ can be obtained as

$\displaystyle\ket{\phi^{\prime\prime}}_{1,4,5,8}$
$\displaystyle=\frac{1}{\sqrt{\left|U_{32}(t)\right|^{2}+\left|U_{33}(t)\right|^{2}}\sqrt{\left|U_{22}(t)\right|^{2}+\left|U_{23}(t)\right|^{2}}}$
(37)
$\displaystyle\left\\{U_{33}(t)U_{23}(t)\left[\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1)\ket{eegg}\right.\right.$
$\displaystyle+\left.\left.\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1)\ket{egeg}\right]\right.$
$\displaystyle+\left.U_{32}(t)U_{22}(t)\left[\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1)\ket{gege}\right.\right.$
$\displaystyle+\left.\left.\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1)\ket{ggee}\right]\right.$
$\displaystyle+\left.U_{23}(t)U_{32}(t)\ket{egge}\right.$
$\displaystyle+\left.e^{-2i\lambda^{\prime}(\tau-t)}U_{22}(t)U_{33}(t)\ket{geeg}\right\\}_{1,4,5,8}.$

? If, the output of measurement on the state (37) reads as $\ket{eg}_{4,5}$
and $\ket{ge}_{4,5}$, then the entangled states for atoms $(1,8)$ are achieved
as $\ket{\phi}^{3}_{1,8}$ and $\ket{\phi^{\prime}}^{3}_{1,8}$ where
$\ket{\phi}^{3}_{1,8}=\ket{\phi}^{2}_{1,8}$ and
$\ket{\phi^{\prime}}^{3}_{1,8}=\ket{\phi^{\prime}}^{2}_{1,8}$ in Eqs. (2.2)
and (2.2) with the concurrences as $C^{\prime}_{3}(t)=C^{\prime}_{2}(t)$ and
$C^{\prime\prime}_{3}(t)=C^{\prime\prime}_{2}(t)$ in Eqs. (32) and (34),
respectively (notice that
$C^{\prime}_{3}(t)=C^{\prime\prime}_{3}(t)=C^{\prime}_{2}(t)=C^{\prime\prime}_{2}(t)$).

###  2.4 The case
$\ket{\Psi^{\prime}(t)}_{1,4}\otimes\ket{\Psi^{\prime}(t)}_{5,8}$

For this case, after a BSM with
$\ket{B^{\prime}}_{4,5}=\frac{1}{\sqrt{2}}(\ket{eg}+\ket{ge})_{4,5}$ on state
$\ket{\Psi^{\prime}(t)}_{1,4}\otimes\ket{\Psi^{\prime}(t)}_{5,8}$, the
entangled state for atoms (1,8) is achieved as

$\ket{\gamma}^{4}_{1,8}=\frac{1}{\sqrt{\left|U_{23}(t)\right|^{4}+\left|U_{22}(t)\right|^{4}}}(U^{2}_{23}(t)\ket{eg}+U^{2}_{22}(t)\ket{ge})_{1,8},$
(38)

with the concurrence

$C_{4}(t)=\frac{2\left|U^{*2}_{23}(t)U^{2}_{22}(t)\right|}{\left|U_{23}(t)\right|^{4}+\left|U_{22}(t)\right|^{4}}.$
(39)

Notice that $C_{4}(t)=C_{3}(t)=C_{2}(t)=C_{1}(t)$.
Also, using of BSM performed with
$\ket{B}_{4,5}=\frac{1}{\sqrt{2}}(\ket{ee}+\ket{gg})_{4,5}$ on state
$\ket{\Psi^{\prime}(t)}_{1,4}\otimes\ket{\Psi^{\prime}(t)}_{5,8}$, the Bell
state (14) is obtained for atoms (1,8) with the success probability

$S^{\prime}_{\ket{B}_{1,8}}(t)=\frac{\left|U_{23}(t)U_{22}(t)\right|^{2}}{(\left|U_{23}(t)\right|^{2}+\left|U_{22}(t)\right|^{2})^{2}}.$
(40)

Notice that
$S^{\prime}_{\ket{B}_{1,8}}(t)=S_{\ket{B}_{1,8}}(t)=S_{\ket{B^{\prime}}_{1,8}}(t)$.
Moreover, by performing the interaction according to the Hamiltonian (18)
between atoms (4,5) with initial state
$\ket{\Psi^{\prime}(t)}_{1,4}\otimes\ket{\Psi^{\prime}(t)}_{5,8}$, the
following entangled state of atoms (1,4,5,8) is obtained as below:

$\displaystyle\ket{\phi^{\prime\prime\prime}}_{1,4,5,8}$
$\displaystyle=\frac{1}{\left|U_{23}(t)\right|^{2}+\left|U_{22}(t)\right|^{2}}$
(41)
$\displaystyle\left\\{U^{2}_{23}(t)\left[\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1)\ket{eegg}\right.\right.$
$\displaystyle+\left.\left.\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1)\ket{egeg}\right]\right.$
$\displaystyle+\left.U^{2}_{22}(t)\left[\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1)\ket{gege}\right.\right.$
$\displaystyle+\left.\left.\frac{1}{2}(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1)\ket{ggee}\right]\right.$
$\displaystyle+\left.U_{22}(t)U_{23}(t)\left(\ket{egge}\right.\right.$
$\displaystyle+\left.\left.e^{-2i\lambda^{\prime}(\tau-t)}\ket{geeg}\right)\right\\}_{1,4,5,8}.$

? Now, via measuring $\ket{eg}_{4,5}$ and $\ket{ge}_{4,5}$ on the obtained
state in (41), the atoms (1,8) are respectively converted to the states,

$\displaystyle\ket{\phi}^{4}_{1,8}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{\left|a_{4}(t,\tau)\right|^{2}+\left|b_{4}(t,\tau)\right|^{2}}}\left(a_{4}(t,\tau)\ket{eg}\right.$
$\displaystyle+$ $\displaystyle\left.b_{4}(t,\tau)\ket{ge}\right)_{1,8},$
$\displaystyle a_{4}(t,\tau)$ $\displaystyle=$ $\displaystyle
U^{2}_{23}(t)(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1),$
$\displaystyle b_{4}(t,\tau)$ $\displaystyle=$ $\displaystyle
U^{2}_{22}(t)(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1),$

with the concurrence

$C^{\prime}_{4}(t,\tau)=\frac{2\left|a^{*}_{4}(t,\tau)b_{4}(t,\tau)\right|}{\left|a_{4}(t,\tau)\right|^{2}+\left|b_{4}(t,\tau)\right|^{2}},$
(43)

and

$\displaystyle\ket{\phi^{\prime}}^{4}_{1,8}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{\left|a^{\prime}_{4}(t,\tau)\right|^{2}+\left|b^{\prime}_{4}(t,\tau)\right|^{2}}}\left(a^{\prime}_{4}(t,\tau)\ket{eg}\right.$
$\displaystyle+$
$\displaystyle\left.b^{\prime}_{4}(t,\tau)\ket{ge}\right)_{1,8},$
$\displaystyle a^{\prime}_{4}(t,\tau)$ $\displaystyle=$ $\displaystyle
U^{2}_{23}(t)(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}+1),$
$\displaystyle b^{\prime}_{4}(t,\tau)$ $\displaystyle=$ $\displaystyle
U^{2}_{22}(t)(e^{2i\lambda^{\prime}t}e^{-2i\lambda^{\prime}\tau}-1),$

with the concurrence

$C^{\prime\prime}_{4}(t,\tau)=\frac{2\left|a^{\prime*}_{4}(t,\tau)b^{\prime}_{4}(t,\tau)\right|}{\left|a^{\prime}_{4}(t,\tau)\right|^{2}+\left|b^{\prime}_{4}(t,\tau)\right|^{2}}.$
(45)

Notice that $C^{\prime}_{4}(t,\tau)=C^{\prime\prime}_{1}(t,\tau)$ and
$C^{\prime\prime}_{4}(t,\tau)=C^{\prime}_{1}(t,\tau)$.

##  3 3\. Results and discussion

(a) (b)

Figure 2:  The effect of coupling coefficient on the time evolution of: (a)
concurrence $C_{1}(t)=C_{2}(t)=C_{3}(t)=C_{4}(t)$ (b) success probability
$S_{\ket{B}_{1,8}}(t)=S_{\ket{B^{\prime}}_{1,8}}(t)=S^{\prime}_{\ket{B}_{1,8}}(t)$,
for symmetric condition $g=g_{2}=g_{3}$ (solid red line) and asymmetric
condition $g_{2}=g$, $g_{3}=3g$ (dashed black line) with $\Delta_{2}=2g$ and
$\Delta_{3}=3g$ (BSM method).

(a) (b)

Figure 3:  The effect of detuning on the time evolution of: (a) concurrence
$C_{1}(t)=C_{2}(t)=C_{3}(t)=C_{4}(t)$ (b) success probability
$S_{\ket{B}_{1,8}}(t)=S_{\ket{B^{\prime}}_{1,8}}(t)=S^{\prime}_{\ket{B}_{1,8}}(t)$,
for $\Delta_{3}=3g$ (solid red line) and $\Delta_{3}=10g$ (dashed black line)
with $\Delta_{2}=2g$ for asymmetric condition $g_{2}=g$, $g_{3}=3g$ (BSM
method).

(a) (b)

(c)

Figure 4:  The effect of detuning on the time evolution of concurrence: (a)
$C^{\prime}_{1}(t,\tau)=C^{\prime\prime}_{4}(t,\tau)$ (b)
$C^{\prime}_{4}(t,\tau)=C^{\prime\prime}_{1}(t,\tau)$ (c)
$C^{\prime}_{2}(t,\tau)=C^{\prime\prime}_{2}(t,\tau)=C^{\prime}_{3}(t,\tau)=C^{\prime\prime}_{3}(t,\tau)$,
for $\Delta_{3}=3g$ (solid red line) and $\Delta_{3}=20g$ (dashed black line)
with $\delta=\Delta_{2}=2g$, $g_{2}=g$, $g_{3}=5g$ and $gt=10$ (QED method).

(a) (b)

(c)

Figure 5:  The effect of interaction time on the time evolution of
concurrence: (a) $C^{\prime}_{1}(t,\tau)=C^{\prime\prime}_{4}(t,\tau)$ (b)
$C^{\prime}_{4}(t,\tau)=C^{\prime\prime}_{1}(t,\tau)$ (c)
$C^{\prime}_{2}(t,\tau)=C^{\prime\prime}_{2}(t,\tau)=C^{\prime}_{3}(t,\tau)=C^{\prime\prime}_{3}(t,\tau)$,
for $gt=2$ (solid red line) and $gt=10$ (dashed black line) with
$\delta=\Delta_{2}=2g$, $\Delta_{3}=3g$, $g_{2}=g$ and $g_{3}=5g$ (QED
method).

(a) (b)

(c)

Figure 6:  The effect of coupling coefficient on the time evolution of
concurrence: (a) $C^{\prime}_{1}(t,\tau)=C^{\prime\prime}_{4}(t,\tau)$ (b)
$C^{\prime}_{4}(t,\tau)=C^{\prime\prime}_{1}(t,\tau)$ (c)
$C^{\prime}_{2}(t,\tau)=C^{\prime\prime}_{2}(t,\tau)=C^{\prime}_{3}(t,\tau)=C^{\prime\prime}_{3}(t,\tau)$,
for $g_{3}=g$ (solid red line) and $g_{3}=5g$ (dashed black line) with
$\delta=\Delta_{2}=2g$, $\Delta_{3}=10g$, $g_{2}=g$ and $gt=10$ (QED method).

In this section we present our numerical results and investigate about the
degree of entanglement of the entangled states and success probabilities have
been calculated in previous section. We have shown the time evolution of
concurrence and success probability of entangled states for atoms $(1,8)$
produced by BSM method, where we have considered the effects of atom-field
coupling coefficient and detuning on concurrence and success probability in
figures 2 and 3, respectively. In figure 2(a) concurrence has been reached to
its maximum value in some intervals of time, in symmetric condition, however,
in asymmetric condition the maximum of concurrence and its time period have
been decreased. The success probability in figure 2(b) has been reached to
$0.25$ in some intervals of time for symmetric condition and in asymmetric
condition this value and its time period have been decreased.
In figure 3 we have considered the effect of detuning on concurrence and
success probability. In figures 3(a) and 3(b) the maximum of concurrence and
success probability have been considerably decreased by increasing the
detuning. Also, the time period of both quantities for $\Delta_{3}=3g$ is
twice of the case $\Delta_{3}=10g$.
The effect of detuning on the concurrence of entangled atoms $(1,8)$ produced
by QED method has been considered in figure 4. In figures 4(a) and 4(b) the
death of entanglement has been occurred in large time intervals by increasing
the detuning in asymmetric condition and the entangled state of atoms $(1,8)$
has been converted to atomic Bell state for decreased detuning in more times.
Varying detuning has no observable effect on concurrence in figure 4(c); in
addition, the death of entanglement has been occurred only for some instants
of time.
We have considered the effect of interaction time on the concurrence in figure
5. The entanglement attenuation has been occurred in large time intervals for
decreased interaction time in figures 5(a) and 5(b). In figure 5(c) we can see
similar evolution of concurrence for the two cases, but, maxima and minima of
concurrence have been relocated.
The effects of symmetric and asymmetric conditions have been considered in
figure 6. The maxima of concurrence have been occurred in more times for
asymmetric condition in figures 6(a) and 6(b). Also, the death of entanglement
has been occurred for some finite time intervals and then it suddenly revives
in symmetric condition. In figure 6(c) it is shown that the evolution of
concurrence is independent of coupling coefficient.

##  4 4\. Summary and conclusions

In this paper we considered the quantum repeater protocol using mixed single-
and two-mode Tavis-Cummings models. We investigated eight two-level atoms
prepared in four entangled pairs (1,2), (3,4), (5,6) and (7,8). By performing
the interaction between atoms (2,3) and (6,7) in two-mode cavities, the
entangled atoms (1,4) and (5,8) were produced. Then, by using a BSM or
performing the interaction between atoms (4,5) in a single-mode cavity (QED
method), the atoms (1,8) were converted to an entangled state. The atomic
states $(1,8)$ are converted to atomic Bell states by choosing suitable Bell
states in BSM method. The effect of coupling coefficient and detuning have
been investigated on the concurrence and success probability of entangled
state of atoms (1,8) produced by BSM method. In asymmetric condition the
maxima of concurrence and success probability and their time periods have been
decreased by increasing the detuning in BSM method. Also, we considered the
effects of detuning, interaction time and coupling coefficient on entanglement
produced by QED method. We observed the destructive effects of increased
detuning, i.e., the maxima of concurrence were occurred in less times and the
death of entanglement was happened in large time interval when the detuning is
increased. The effect of decreasing the interaction time $gt$ on the
entanglement has been discussed, i.e., the entanglement attenuation was
happened in large intervals of time by decreasing the interaction time. The
death of entanglement in symmetric condition has been observed in large time
interval for atoms (1,8) which are produced by QED method.
Acknowledgement: The authors would like to thank the referees for their
valuable comments which improved the paper.

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