Entanglement swapping and swapped entanglement

Quantum networks play a very important role in quantum information science and have applications in quantum communication [1], computation [2], metrology [3] and fundamental tests [4]. Quantum networks are comprised of a large number of nodes interconnected by quantum channels. The stationary qubits at the separated nodes develop, store, and manipulate quantum states while the flying qubits constitute quantum channels and can be realized by photons. These channels teleport quantum states between the nodes with high fidelity, allowing the distribution of quantum entanglement across the whole network. Thus, the task of building quantum networks requires the ability to establish quantum entanglement between distant quantum nodes. H. J. Briegel et al. proposed a quantum repeater protocol that enables the transmission of entanglement over long distances [5]. The ability to distribute and manipulate entanglement between distant parties serves as the basis for quantum applications. Entanglement swapping is one such tool that helps us to connect many separable nodes for long-distance communication in a quantum network. Specifically, entanglement swapping is a protocol by which quantum systems that have never interacted in the past can become entangled [6, 7]. The nomenclature “entanglement swapping” describes the transfer of entanglement from a priori entangled systems to a priori separable systems [8]. It is a very useful tool for entanglement purification [9], teleportation [10], and plays an important role in quantum computing and quantum cryptography [11, 12]. We can also use entanglement swapping for the creation of multipartite entangled states from bipartite entanglement [13].

Let us describe the phenomenon of entanglement swapping. Suppose two entangled particles $(A,B)$ are shared between Alice and Bob. Similarly Cara and Danny also share another entangled pair of particles $(C,D)$. Initially there is no entanglement between Alice’s and Danny’s particles $(A,D)$, shown in Fig. (1(a)). If Bob and Cara who are situated in the same laboratory make measurement in a suitable basis on the pair $(B,C)$ and classically communicate the outcome with distant partners then Alice’s and Danny’s particles who are at very large distance become entangled as shone in Fig. (1(b)). This entanglement swapping protocol can be generalized in different ways: by modifying the initial states, or by modifying the measurement performed by Bob and Cara, or by extending the number of parties [13, 14].

An ever-increasing body of literature shows that the entanglement swapping and purification of quantum systems need specific protocols. Smaller changes may bring huge changes to the output state because of the relative sensitivity of the operation and quantum systems. For this reason, several previous studies suggest entanglement swapping of initial states into maximally entangled states. For example, in Ref. [15], the authors provided a scheme of entanglement swapping of initial states into biqubit maximally entangled states when influenced by an amplitude damping channel. The concurrence of the measuring basis for entanglement swapping caused a two-fold entanglement matching effect has been witnessed in Ref. [16]. Recently, Ref. [17] investigated that hyper-entangled states produce deterministic entanglement swapping while considering projections of biqubit systems on symmetric, and iso-entangled states. It is found that biqubit entanglement generated through entanglement swapping, will depart from a Bell-type inequality even for visibilities smaller than 50% [18]. From the above literature, we found that entanglement swapping has been previously considered using different procedures, and various important results have been achieved. This research work constitutes a relatively more generalized entanglement swapping protocol that covers the maximum possible cases of swapping. We consider pure, mixed as well as noisy systems for entanglement swapping. Moreover, we explore the application of the final state as a channel for the teleportation of an unknown qubit. We also investigate the fidelity of teleported qubit with the initial unknown qubit.

The article is organized as follows. In Section 2, we present the mathematical tools that are useful for the present research. The section 3 describes the details of the entanglement swapping scenario for non-noisy systems and the application of swapped entanglement. Next, in Section 4, we demonstrate the entanglement swapping among noisy qubits and teleportation using a noisy quantum channel. Finally, we conclude with a short discussion in Section 5.

In this section, we explain in brief the Schmidt decomposition, entangled quantum states and famous entanglement quantifiers. We shall quantify the swapped entanglement with these quantifiers and compare them in next sections.

We consider two pairs of qubits for entanglement swapping. The entangled qubits $A$ and $B$ make the first pair and the second pair consist of entangled qubits $C$ and $D$ (the qubits’ names are chosen such that they make the initial of Alice, Bob, Cara and Danny, shown in Fig.(1)). In terms of Schmidt decomposition, these subsystems can be written as

$$\begin{split}\ket{\phi}_{AB}&=\sqrt{p_{0}}\ket{00}_{AB}+\sqrt{p_{1}}\ket{11}_{AB},\\ \ket{\phi}_{CD}&=\sqrt{p_{0}^{\prime}}\ket{00}_{CD}+\sqrt{p_{1}^{\prime}}\ket{11}_{CD}\;.\end{split}$$

(7)

The concurrence and negativity of these systems are $\textbf{C}(\ket{\phi}_{AB})=2\sqrt{p_{0}p_{1}}$, $\textbf{C}(\ket{\phi}_{CD})=2\sqrt{p_{0}^{\prime}p_{1}^{\prime}}$ and $\mathcal{N}(\ket{\phi}_{AB})=2\sqrt{p_{0}}\sqrt{p_{1}}$, $\mathcal{N}(\ket{\phi}_{CD})=2\sqrt{p_{0}^{\prime}}\sqrt{p_{1}^{\prime}}$. It means for these bi-dimensional systems concurrence and negativity produce similar results. The initial state of our four-qubit system can be written as

$$\begin{split}\ket{\Phi}&=\ket{\phi}_{AB}\ket{\phi}_{CD}\\ &=\sqrt{p_{0}p_{0}^{\prime}}\ket{0000}_{ABCD}+\sqrt{p_{0}p_{1}^{\prime}}\ket{0011}_{ABCD}+\sqrt{p_{1}p_{0}^{\prime}}\ket{1100}_{ABCD}+\sqrt{p_{1}p_{1}^{\prime}}\ket{1111}_{ABCD}\;.\end{split}$$

(8)

Let us do some simple algebra. We rearrange the qubits in each term so that we can write qubits $A$ and $D$ together and qubits $B$ and $C$ together:

$$\ket{\Phi}=\sqrt{p_{0}p_{0}^{\prime}}\ket{00}_{AD}\ket{00}_{BC}\sqrt{p_{0}p_{1}^{\prime}}\ket{01}_{AD}\ket{01}_{BC}+\sqrt{p_{1}p_{0}^{\prime}}\ket{10}_{AD}\ket{10}_{BC}+\sqrt{p_{1}p_{1}^{\prime}}\ket{11}_{AD}\ket{11}_{BC}\;.$$

(9)

In order to do measurements over $BC$ qubits, we also need to define a set of four orthonormal basis [16]

$$\begin{split}\ket{\tilde{\Phi}_{+}}_{BC}&=\alpha_{0}\ket{00}_{BC}+\beta_{0}\ket{11}_{BC},\\ \ket{\tilde{\Phi}_{-}}_{BC}&=\beta_{0}^{*}\ket{00}_{BC}-\alpha_{0}^{*}\ket{11}_{BC},\\ \ket{\tilde{\Psi}_{+}}_{BC}&=\alpha_{1}\ket{01}_{BC}+\beta_{1}\ket{10}_{BC},\\ \ket{\tilde{\Psi}_{-}}_{BC}&=\beta_{1}^{*}\ket{01}_{BC}-\alpha_{1}^{*}\ket{10}_{BC}\;,\end{split}$$

(10)

where, without loss of generality, we assume the amplitudes $\alpha_{i}$ and $\beta_{i}$ to be real and non-negative numbers and for normalization $\lvert\alpha_{i}\rvert^{2}+\lvert\beta_{i}\rvert^{2}=1$ for $i\in\{0,1\}$. Conversely, we have $\ket{00}_{BC}=\alpha_{0}^{*}\ket{\tilde{\Phi}_{+}}_{BC}+\beta_{0}\ket{\tilde{\Phi}_{-}}_{BC}$ and likewise we can find expressions for $\ket{01}_{BC},\ket{10}_{BC},\ket{11}_{BC}$. Now by using these expressions we can write Eq. (9) as

$$\ket{\Phi}=\sqrt{p_{\tilde{\Phi}_{+}}}\ket{\ddot{\Phi}_{+}}_{AD}\ket{\tilde{\Phi}_{+}}_{BC}+\sqrt{p_{\tilde{\Phi}_{-}}}\ket{\ddot{\Phi}_{-}}_{AD}\ket{\tilde{\Phi}_{-}}_{BC}\\ +\sqrt{p_{\tilde{\Psi}_{+}}}\ket{\ddot{\Psi}_{+}}_{AD}\ket{\tilde{\Psi}_{+}}_{BC}+\sqrt{p_{\tilde{\Psi}_{-}}}\ket{\ddot{\Psi}_{-}}_{AD}\ket{\tilde{\Psi}_{-}}_{BC}\;,$$

(11)

where possible outcome states of the qubits AD can be defined as,

$$\begin{split}\ket{\ddot{\Phi}_{+}}_{AD}=&\left(\sqrt{p_{0}p_{0}^{\prime}}\alpha_{0}^{*}\ket{0}_{A}\ket{0}_{D}+\sqrt{p_{1}p_{1}^{\prime}}\beta_{0}^{*}\ket{1}_{A}\ket{1}_{D}\right)/\sqrt{p_{\tilde{\Phi}_{+}}}\;,\\ \ket{\ddot{\Phi}_{-}}_{AD}=&\left(\sqrt{p_{0}p_{0}^{\prime}}\beta_{0}\ket{0}_{A}\ket{0}_{D}-\sqrt{p_{1}p_{1}^{\prime}}\alpha_{0}\ket{1}_{A}\ket{1}_{D}\right)/\sqrt{p_{\tilde{\Phi}_{-}}}\;,\\ \ket{\ddot{\Psi}_{+}}_{AD}=&\left(\sqrt{p_{0}p_{1}^{\prime}}\alpha_{1}^{*}\ket{0}_{A}\ket{1}_{D}+\sqrt{p_{0}^{\prime}p_{1}}\beta_{1}^{*}\ket{1}_{A}\ket{0}_{D}\right)/\sqrt{p_{\tilde{\Psi}_{+}}}\;,\\ \ket{\ddot{\Psi}_{-}}_{AD}=&\left(\sqrt{p_{0}p_{1}^{\prime}}\beta_{1}\ket{0}_{A}\ket{1}_{D}-\sqrt{p_{0}^{\prime}p_{1}}\alpha_{1}\ket{1}_{A}\ket{0}_{D}\right)/\sqrt{p_{\tilde{\Psi}_{-}}}\;,\end{split}$$

(12)

where the associated probabilities are

$$\begin{split}p_{\tilde{\Phi}_{+}}&=p_{0}p_{0}^{\prime}\lvert\alpha_{0}\rvert^{2}+p_{1}p_{1}^{\prime}\lvert\beta_{0}\rvert^{2}\;,\\ p_{\tilde{\Phi}_{-}}&=p_{0}p_{0}^{\prime}\lvert\beta_{0}\rvert^{2}+p_{1}p_{1}^{\prime}\lvert\alpha_{0}\rvert^{2}\;,\\ p_{\tilde{\Psi}_{+}}&=p_{0}p_{1}^{\prime}\lvert\alpha_{1}\rvert^{2}+p_{0}^{\prime}p_{1}\lvert\beta_{1}\rvert^{2}\;,\\ p_{\tilde{\Psi}_{-}}&=p_{0}p_{1}^{\prime}\lvert\beta_{1}\rvert^{2}+p_{0}^{\prime}p_{1}\lvert\alpha_{1}\rvert^{2}\;.\end{split}$$

(13)

We observe in Eq. (11) that the state of qubits $A$ and $D$ is similar to the state of the basis $BC$. It is clear from equation (12) that after measurements in the basis $BC$, Alice and Danny’s qubits $A$, $D$ which are initially separable, become entangled in one of the four possible forms. We can compute the average concurrence for the final state as

$$\begin{split}\textbf{C}_{av}&=p_{\tilde{\Phi}_{+}}\textbf{C}_{\ddot{\Phi}_{+}}+p_{\tilde{\Phi}_{-}}\textbf{C}_{\ddot{\Phi}_{-}}+p_{\tilde{\Psi}_{+}}\textbf{C}_{\ddot{\Psi}_{+}}+p_{\tilde{\Psi}_{-}}\textbf{C}_{\ddot{\Psi}_{-}}\;,\\ &=4\sqrt{p_{0}p_{0}^{\prime}p_{1}p_{1}^{\prime}}\left(\lvert\alpha_{0}\beta_{0}\rvert+\lvert\alpha_{1}\beta_{1}\rvert\right)\;.\end{split}$$

(14)

The Bell states are maximally entangled biqubit states. The states in Eq. (10) transform into maximally entangled Bell states if we take $\alpha_{i}=\beta_{i}=\frac{1}{\sqrt{2}}$ for $i\in\{0,1\}$. The measurement in maximally entangled Bell basis return maximally entangled $A$, $D$ qubits states with the average concurrence

$$\begin{split}\textbf{C}_{av}&=4\sqrt{p_{0}p_{0}^{\prime}p_{1}p_{1}^{\prime}}\;\\ &=\textbf{C}_{AB}\textbf{C}_{CD}\;.\end{split}$$

(15)

Similarly, the average negativity of the qubits $A$ and $D$ states when the measurement is done in Bell basis takes the form

$$\begin{split}\mathcal{N}_{av}&=4\sqrt{p_{0}p_{0}^{\prime}p_{1}p_{1}^{\prime}}\;\\ &=\mathcal{N}_{AB}\mathcal{N}_{CD}\;.\end{split}$$

(16)

We obtain from Eq. (15) and Eq. (16) that if initial quantum states are maximally entangled and we make measurements in the Bell basis, then average concurrence and average negativity are equivalent. We simply obtain the average concurrence (average negativity) by taking the product of concurrences (negativities) of the initial states. Besides Eq. (14) shows that measurement in non-maximally entangled basis during entanglement swapping degrades the average swapped entanglement.

Now we extend entanglement swapping between two pairs of qubits to three pairs of qubits. We take three pairs of entangled qubits and make a measurement in the GHZ basis and analyze the outcome state. If the three pairs of qubits in Schmidt form are $\ket{\phi}_{AB}=\sqrt{\lambda_{0}}\ket{00}_{AB}+\sqrt{\lambda_{1}}\ket{11}_{AB})$, $\ket{\phi}_{CD}=\sqrt{\mu_{0}}\ket{00}_{CD}+\sqrt{\mu_{1}}\ket{11}_{CD}$ and $\ket{\phi}_{EF}\sqrt{\nu_{0}}\ket{00}_{EF}+\sqrt{\nu_{1}}\ket{00}_{EF}$ then the six-qubit system can be written as

$$\ket{\Phi^{\prime}}=|\phi\rangle_{AB}\otimes\ket{\phi}_{CD}\otimes\ket{\phi}_{EF}.$$

(17)

Let us make measurements on the $B$, $D$, and $F$ qubits. For this purpose, we can define the triqubit GHZ basis as [31]

	$\displaystyle\left|G_{0,1}\right\rangle$	$\displaystyle=\frac{1}{\sqrt{2}}(|000\rangle\pm|111\rangle)\;,$		(18)
	$\displaystyle\left|G_{2,3}\right\rangle$	$\displaystyle=\frac{1}{\sqrt{2}}(|001\rangle\pm|110\rangle)\;,$
	$\displaystyle\left|G_{4,5}\right\rangle$	$\displaystyle=\frac{1}{\sqrt{2}}(|010\rangle\pm|101\rangle)\;,$
	$\displaystyle\left|G_{6,7}\right\rangle$	$\displaystyle=\frac{1}{\sqrt{2}}(|011\rangle\pm|100\rangle)\;.$

Here the $-$ sign applies to states with odd indices. Now we can write Eq. (17) in terms of GHZ basis as

	$\displaystyle\ket{\Phi^{\prime}}$	$\displaystyle=\frac{1}{\sqrt{2}}\left(\sqrt{\lambda_{0}\mu_{0}\nu_{0}}|000\rangle_{ACE}+\sqrt{\lambda_{1}\mu_{1}\nu_{1}}|111\rangle_{ACE}\right)|G_{0}\rangle$		(19)
		$\displaystyle+\frac{1}{\sqrt{2}}\left(\sqrt{\lambda_{0}\mu_{0}\nu_{0}}|000\rangle_{ACE}-\sqrt{\lambda_{1}\mu_{1}\nu_{1}}|111\rangle_{ACE}\right)|G_{1}\rangle$
		$\displaystyle+\frac{1}{\sqrt{2}}\left(\sqrt{\lambda_{0}\mu_{0}\nu_{1}}|001\rangle_{ACE}+\sqrt{\lambda_{1}\mu_{1}\nu_{0}}|110\rangle_{ACE}\right)|G_{2}\rangle$
		$\displaystyle+\frac{1}{\sqrt{2}}\left((\sqrt{\lambda_{0}\mu_{0}\nu_{1}}|001\rangle_{ACE}-\sqrt{\lambda_{1}\mu_{1}\nu_{0}}|110\rangle_{ACE}\right)|G_{3}\rangle$
		$\displaystyle+\frac{1}{\sqrt{2}}\left(\sqrt{\lambda_{0}\mu_{1}\nu_{0}}|010\rangle_{ACE}+\sqrt{\lambda_{1}\mu_{0}\nu_{1}}|101\rangle_{ACE}\right)|G_{4}\rangle$
		$\displaystyle+\frac{1}{\sqrt{2}}\left(\sqrt{\lambda_{0}\mu_{1}\nu_{0}}|010\rangle_{ACE}-\sqrt{\lambda_{1}\mu_{0}\nu_{1}}|101\rangle_{ACE}\right)|G_{5}\rangle$
		$\displaystyle+\frac{1}{\sqrt{2}}\left(\sqrt{\lambda_{0}\mu_{1}\nu_{1}}|011\rangle_{ACE}+\sqrt{\lambda_{1}\mu_{0}\nu_{0}}|100\rangle_{ACE}\right)|G_{6}\rangle$
		$\displaystyle+\frac{1}{\sqrt{2}}\left(\sqrt{\lambda_{0}\mu_{1}\nu_{1}}|011\rangle_{ACE}-\sqrt{\lambda_{1}\mu_{0}\nu_{0}}|100\rangle_{ACE}\right)|G_{7}\rangle\;.$

This equation shows that after measurements on $BDF$ qubits, we gain $ACE$ qubits in any one of the eight possible forms of entangled state. For example, if measurement gives us $|G_{0}\rangle$ then $ACE$ qubits have state

$$\frac{1}{\sqrt{2p_{0}}}\left(\sqrt{\lambda_{0}\mu_{0}\nu_{0}}|000\rangle_{\text{ACE}}+\sqrt{\lambda_{1}\mu_{1}\nu_{1}}|111\rangle_{\text{ACE}}\right)\;.$$

(20)

Here probability of getting $|G_{0}\rangle$ state is $p_{0}=\left(\lambda_{0}\mu_{0}\nu_{0}+\lambda_{1}\mu_{1}\nu_{1}\right)/2$.

Yu and Song [32] showed that any good bipartite entanglement measure $M_{A-B}$ can be extended to multipartite systems by taking bipartite partitions of them. So a tripartite entanglement quantifier can be defined as

$$M_{ABC}^{+}=\frac{1}{3}\left(M_{A-BC}+M_{B-AC}+M_{C-AB}\right).$$

(21)

But $M_{ABC}^{+}$ could be nonzero for pure biseparable states. It can be avoided by using the geometric mean:

$$M_{ABC}^{\times}=\left(M_{A-BC}M_{B-AC}M_{C-AB}\right)^{\frac{1}{3}}\;.$$

(22)

Now by considering this bi-partition for the final triqubit entangled state, we can compute the swapped entanglement in the form of concurrence as

$$\textbf{C}_{ACE}=\left(\textbf{C}_{A-CE}\textbf{C}_{C-AE}\textbf{C}_{E-AC}\right)^{\frac{1}{3}}\;,$$

(23)

where $\textbf{C}_{A-CE}=\sqrt{2\left(1-\operatorname{Tr}\left(\rho_{A}^{2}\right)\right)}$ and similarly we can compute $\textbf{C}_{C-AE}$, $\textbf{C}_{E-AC}$. Here, $\rho_{A}$ is the one-qubit reduced density matrix of the qubit $A$, obtained after tracing out the other qubits. The average concurrence for the final three qubits state now can be written as

$$\textbf{C}_{ACE}^{av}=\textbf{C}_{AB}\textbf{C}_{CD}\textbf{C}_{EF}\;.$$

(24)

It is again equal to the product of the concurrences of the initial three states.

We can compute the the negativity of triqubit state $\rho_{ACE}$ as

$$\mathcal{N}(\rho_{ACE})=\left(\mathcal{N}_{A-CE}\mathcal{N}_{C-AE}\mathcal{N}_{E-AC}\right)^{\frac{1}{3}}\;,$$

(25)

where $\mathcal{N}_{A-CE}=-2\sum_{i}\lambda_{i}\left(\rho_{ACE}^{T_{A}}\right)$, $\lambda_{i}\left(\rho_{ACE}^{T_{A}}\right)$ are the negative eigenvalues of $\rho_{ACE}^{T_{A}}$, partial transpose of $\rho_{ACE}$ with respect to subsystem $A$ is defined as $\left\langle i_{A},j_{CE}\left|\rho_{ACE}^{T_{A}}\right|k_{A},l_{CE}\right\rangle=\left\langle k_{A},j_{CE}|\rho|i_{A},l_{Ce}\right\rangle$ and similarly, we can define $\mathcal{N}_{C-AE}$, $\mathcal{N}_{E-AC}$. The average negativity of the final triqubit state can be written as

$$\mathcal{N}_{ACE}^{av}=\mathcal{N}_{AB}\mathcal{N}_{CD}\mathcal{N}_{EF}\;,$$

(26)

where $\mathcal{N}_{AB},\mathcal{N}_{CD}$ and $\mathcal{N}_{EF}$ are the negativities of the initial three biqubit states.

We have used so far pure quantum systems. These quantum systems are isolated from external environments which comprise a variety of disorders and noises. In reality, quantum systems interact with the environment. One of the important types of noise is called depolarizing noise or white noise. This type of noise takes a quantum state and replaces it with a completely mixed state $\frac{1}{N}\mathbb{I}$, where $N$ is the dimension of the quantum system and $\mathbb{I}$ is identity matrix. Let us consider a biqubit noisy state that is prepared by mixing a pure state with white noise:

	$\displaystyle\rho_{\alpha}$	$\displaystyle=\alpha\rho_{AB}+(1-\alpha)\mathbb{I}_{2}\otimes\mathbb{I}_{2}/4$		(31)
		$\displaystyle=\left(\begin{array}[]{cccc}\frac{1-\alpha}{4}+\alpha p_{0}&0&0&\alpha\sqrt{p_{0}}\sqrt{p_{1}}\\ 0&\frac{1-\alpha}{4}&0&0\\ 0&0&\frac{1-\alpha}{4}&0\\ \alpha\sqrt{p_{0}}\sqrt{p_{1}}&0&0&\frac{1-\alpha}{4}+\alpha p_{1}\\ \end{array}\right)\;,$

where $\rho_{AB}=\ket{\phi}_{AB}\bra{\phi}$ is the density operator of biqubit system $AB$ described in Eq. (7), $\mathbb{I}_{2}$ identity matrix and parameter $\alpha$ called visibility of system $AB$. If we take $p_{0}=p_{1}=1/2$, the Eq. (31) becomes an isotropic state [35] with maximally entangled $\rho_{AB}$. The isotropic states are invariant under all transformations of the form $U\otimes U^{*}$, where the asterisk denotes complex conjugation in a certain basis.

We can also represent a biqubit noisy system in the Bloch form as [20]

$\displaystyle\rho=$

$\displaystyle\frac{1}{4}\mathbb{I}_{2}\otimes\mathbb{I}_{2}+\sum_{\mu=1}^{3}r_{\mu}\frac{\sigma_{\mu}}{\sqrt{2}}\otimes\frac{\mathbb{I}_{2}}{2}+\sum_{\nu=1}^{3}s_{\nu}\frac{\mathbb{I}_{2}}{2}\otimes\frac{\sigma_{\nu}}{\sqrt{2}}$

$\displaystyle+\sum_{\mu=1}^{3}\sum_{\nu=1}^{3}t_{\mu\nu}\frac{\sigma_{\mu}}{\sqrt{2}}\otimes\frac{\sigma_{\nu}}{\sqrt{2}}\;,$

(32)

where $\sigma$ represents Pauli matrices, $\mathbb{I}$ is identity matrix, $r_{\mu}=\operatorname{Tr}\left(\rho\frac{\sigma_{\mu}}{\sqrt{2}}\otimes\frac{\mathbb{I}_{2}}{2}\right)$ and $s_{\nu}=\operatorname{Tr}\left(\rho\frac{\mathbb{I}_{2}}{2}\otimes\frac{\sigma_{\nu}}{\sqrt{2}}\right)$ are Bloch vectors of given two qubits and $t_{\mu\nu}=\operatorname{Tr}\left(\rho\frac{\sigma_{\mu}}{\sqrt{2}}\otimes\frac{\sigma_{\nu}}{\sqrt{2}}\right)$ called a correlation tensor. We can construct Bloch matrix from $\vec{r}$, $\vec{s}$ and $3\times 3$ dimensional correlation matrix $T$ as

$$\tilde{\mathcal{T}}=\left(\begin{array}[]{cc}c&\vec{s}\\ \vec{r}&T\\ \end{array}\right)\;,$$

(33)

where $c$ is a scalar number. The Bloch matrix form of Eq. (31) contains $c=\alpha\sqrt{p_{0}}\sqrt{p_{1}}$, $\vec{r}=\vec{s}=0$ and correlation matrix

$$T=\left(\scriptsize\begin{array}[]{ccc}-\alpha\sqrt{p_{0}}\sqrt{p_{1}}&0&0\\ 0&\frac{\alpha-1}{4}+\frac{1}{2}\left(\frac{1-\alpha}{4}+\alpha p_{0}\right)+\frac{1}{2}\left(\frac{1-\alpha}{4}+\alpha p_{1}\right)&\frac{1-\alpha}{8}+\frac{\alpha-1}{8}+\frac{1}{2}\left(\frac{1-\alpha}{4}+\alpha p_{0}\right)+\frac{1}{2}\left(\frac{\alpha-1}{4}-\alpha p_{1}\right)\\ 0&\frac{1-\alpha}{8}+\frac{\alpha-1}{8}+\frac{1}{2}\left(\frac{1-\alpha}{4}+\alpha p_{0}\right)+\frac{1}{2}\left(\frac{\alpha-1}{4}-\alpha p_{1}\right)&\frac{1-\alpha}{4}+\frac{1}{2}\left(\frac{1-\alpha}{4}+\alpha p_{0}\right)+\frac{1}{2}\left(\frac{1-\alpha}{4}+\alpha p_{1}\right)\\ \end{array}\right)\;.$$

(34)

As the coherence vectors $\vec{r}$, $\vec{s}$ of the subsystems $A$ and $B$ have zero magnitudes that means the state is maximally mixed. According to combo separability criteria [20] if $f(\alpha,p_{0})=\|\tilde{\mathcal{T}}_{\alpha}\|_{KF}-1>0$ then state $\rho_{\alpha}$ is an entangled state. We plotted $f(\alpha,p_{0})$ in Fig. (2) which represents entanglement of the mixed state $\rho_{\alpha}$.

It is clear from the figure that $\rho_{\alpha}$ remains entangled when $0<p_{0}<1$ and the minimum value of $\alpha$ is $1/3$. This state becomes maximally entangled when $p_{0}=0.5$ and $\alpha$ approaches 1.

As $\rho_{\alpha}$ is a X-form mixed-state with nonzero entries only along the diagonal and anti-diagonal so its concurrence is given by [36]

	$\displaystyle\textbf{C}({\rho_{\alpha}})$	$\displaystyle=\max\left[0,2\left(\left|\rho_{\alpha}^{(14)}\right|-\sqrt{\rho_{\alpha}^{(22)}\rho_{\alpha}^{(33)}}\right),2\left(\left|\rho_{\alpha}^{(23)}\right|-\sqrt{\rho_{\alpha}^{(11)}\rho_{\alpha}^{(44)}}\right)\right]$		(35)
		$\displaystyle=\max\left[0,2\left(\alpha\sqrt{p_{0}p_{1}}-\frac{1-\alpha}{4}\right)\right]\;.$

This relation also gives a lower bound for the probability that keeps the $\rho_{\alpha}$ entangled as

$$\alpha>\frac{1}{\left(1+4\sqrt{p_{0}p_{1}}\right)}\;.$$

(36)

If $\rho_{AB}$ is a maximally entangled state then $p_{0}=p_{1}=\frac{1}{2}$, in this case the state remains entangled for $\alpha>\frac{1}{3}$ that we can also observe from Fig (3(a)).

The negativity of $X-$form state $\rho_{\alpha}$ can be computed as

$$\mathcal{N}(\rho_{X})=-2\min\left\{0,r_{+}-\sqrt{r_{-}^{2}+\left(\rho^{(14)}\right)^{2}},u_{+}-\sqrt{u_{-}^{2}+\left(\rho^{(23)}\right)^{2}}\right\}\;,$$

(37)

where $u_{\pm}=\left(\rho^{(11)}\pm\rho^{(44)}\right)/2$, $r_{\pm}=\left(\rho^{(22)}\pm\rho^{(33)}\right)/2$. The Eq. (37) can be reduced to

$$\mathcal{N}(\rho_{\alpha})=-2\min\left\{0,\frac{1}{4}\left(1-\alpha-4\alpha\sqrt{p_{0}p_{1}}\right)\right\}\;,$$

(38)

because $\rho^{(}23)=0$ and $u_{+}-u_{-}=\rho^{(44)}$ is a positive number. For a maximally entangled state, the Eq. (38) also gives $\alpha>\frac{1}{3}$ for entanglement retain and can be observed from Fig. (3(b)).

It is clear from Fig. (3) that the concurrence and negativity produce similar results in the case of biqubit noisy state.

Now we want to explore entanglement swapping between two states of the form given in Eq. (31). For simplicity, we assume that the states are similar and we make standard Bell measurements in order to accomplish the entanglement swapping. Therefore our four-qubit noisy state in terms of Bell basis is

	$\displaystyle\rho_{ABCD}=$	$\displaystyle\frac{\alpha^{2}}{2}\left(p_{0}|00\rangle_{AD}\pm p_{1}|11\rangle_{AD}\right)\left(p_{0}\langle 00|\pm p_{1}\langle 11|\right)|\Phi_{\pm}\rangle_{\text{BC}}\langle\Phi_{\pm}|$		(39)
		$\displaystyle+\frac{\alpha^{2}p_{0}p_{1}}{2}\left(|01\rangle_{AD}\pm|10\rangle_{AD}\right)\left(\langle 01|\pm\langle 10|\right)|\Psi_{\pm}\rangle_{\text{BC}}\langle\Psi_{\pm}|$
		$\displaystyle+\mathbf{d}_{AD}\left(|\Phi_{\pm}\rangle_{\text{BC}}\langle\Phi_{\pm}|+|\Psi_{\pm}\rangle_{\text{BC}}\langle\Psi_{\pm}|\right)\;,$

where

$$\mathbf{d}_{AD}=\frac{\alpha\left(1-\alpha\right)}{8}\left(\left(p_{0}\ket{0}_{A}\bra{0}+p_{1}\ket{1}_{A}\bra{1}\right)\otimes\mathbb{I}_{D}+\mathbb{I}_{A}\otimes\left(p_{0}\ket{0}_{D}\bra{0}+p_{1}\ket{1}_{D}\bra{1}\right)\right)\\ +\frac{(1-\alpha)^{2}}{16}\mathbb{I}_{AD}s\;,$$

(40)

and $\mathbb{I}$ is an identitiy matrix. The measurement of the qubits $B$ and $C$ will give us one of the following four Bell states

	$\displaystyle\left|\Phi_{\pm}\right\rangle_{BC}$	$\displaystyle=\frac{1}{\sqrt{2}}\left(|00\rangle_{BC}\pm|11\rangle_{BC}\right)\;,$		(41)
	$\displaystyle\left|\Psi_{\pm}\right\rangle_{BC}$	$\displaystyle=\frac{1}{\sqrt{2}}\left(|01\rangle_{BC}\pm|10\rangle_{BC}\right)\;.$

If we obtain $\left|\Phi_{\pm}\right\rangle_{BC}$ then the qubits $A$ and $D$ become entangled in the state

$$\rho_{AD}^{\Phi_{\pm}}=\frac{1}{P_{\Phi}}\left(\frac{\alpha^{2}}{2}\left(p_{0}|00\rangle_{AD}\pm p_{1}|11\rangle_{AD}\right)\left(p_{0}\langle 00|\pm p_{1}\langle 11|\right)+\mathbf{d}_{AD}\right)\;,$$

(42)

where $P_{\Phi}=\frac{\alpha^{2}}{2}\left(p_{0}^{2}+p_{1}^{2}\right)+\frac{1-\alpha^{2}}{4}$ is the probability of $\ket{\Phi_{+}}$ and $\ket{\Phi_{-}}$. If measurement gives us $\left|\Psi_{\pm}\right\rangle_{BC}$ then the qubits $A$ and $D$ make entangled state

$$\rho_{AD}^{\Psi_{\pm}}=\frac{1}{P_{\Psi}}\left(\frac{\alpha^{2}p_{0}p_{1}}{2}\left(|01\rangle_{AD}\pm|10\rangle_{AD}\right)\left(\langle 01|\pm\langle 10|\right)+\mathbf{d}_{AD}\right)\;.$$

(43)

Here, $P_{\Psi}=\alpha^{2}p_{0}p_{1}+\frac{1-\alpha^{2}}{4}$ is the probability of $\ket{\Psi_{+}}$ and $\ket{\Psi_{-}}$.

In order to evaluate the transferred entanglement between qubits $A$ and $D$, we first compute the concurrence of all four types of density matrix $\rho_{AD}$. As all density matrices of qubits $A$ and $D$ in Eq. (42) and Eq. (43) have X-form state form so, their concurrence can easily be computed by using Eq. (35). The state $\rho_{AD}^{\Phi_{+}}$ and $\rho_{AD}^{\Phi_{-}}$ have the same amount of concurrence and is given by

$$\textbf{C}\left(\rho_{AD}^{\Phi_{\pm}}\right)=\frac{1}{P_{\phi}}\max\left(0,\alpha^{2}p_{0}p_{1}-\frac{1}{8}\left(1-\alpha^{2}\right)\right)\;.$$

(44)

and similarly, the concurrence of $\rho_{AD}^{\Psi_{\pm}}$ is

$\displaystyle\textbf{C}\left(\rho_{AD}^{\Psi_{\pm}}\right)=\frac{1}{P_{\psi}}\max\left(0,\alpha^{2}p_{0}p_{1}-\frac{1}{8}(1-\alpha)\sqrt{1+2\alpha-3\alpha^{2}+16\alpha^{2}p_{0}p_{1}}\right)\;.$

(45)

Now the average of teleported entanglement in terms of concurrence can be computed as

$$\textbf{C}_{av}=2P_{\Phi}\textbf{C}\left(\rho_{AD}^{\Phi_{+}}\right)+2P_{\Psi}\textbf{C}\left(\rho_{AD}^{\Psi_{+}}\right)\;.$$

(46)

This average concurrence of the final states has been plotted in Fig. (4(a)). Besides Fig. (4(b)) put forward that for mixed states, the product of the concurrences of the initial states (dashed line plots) is an upper bound to the average concurrence of the finally swapped entanglement (solid line plots).