Conclusive multiparty quantum state sharing in amplitude-damping channel

Abstract

In this paper, two conclusive multiparty quantum state sharing protocols in amplitude damping channel are proposed that, respectively share an arbitrary unknown single-qubit state and single-qutrit state. To achieve this aim, the detailed processes of sharing pure entangled quantum states as quantum channel in amplitude damping channel via entanglement compensation is proposed first. Then, based on the pure entangled quantum state shared among dealer and all agents, the dealer’s secret quantum information, i.e., single-qubit state (or single-qutrit state) is split in such a way that it can be probabilistically reconstructed through introducing an auxiliary qubit (or qutrit) and performing appropriate operations provided that all the receivers collaborate together. Through the analysis, we have found that whether in single-qubit or single-qutrit state sharing protocols, the successful probability of receiver recovering the secret quantum information is only determined by the small one among the absolute values of the coefficients characterizing the quantum channel. In addition, through the analysis, it proves that the conclusive multiparty single-qutrit state sharing has higher efficiency than single-qubit state sharing under the same amplitude damping strength.

Appendices

Appendix 1: Proof of Theorem 1

Theorem 1

On condition that Alice prepares two-qubit maximal entangled Bell state $\left| {B_{0,0} } \right\rangle = {1 \mathord{\left/ {\vphantom {1 {\sqrt 2 }}} \right. \kern-0pt} {\sqrt 2 }}\left( {\left| {00} \right\rangle + \left| {11} \right\rangle } \right)_{{{\text{A}}_{1} {\text{A}}_{2} }}$, when she measures qubit A₂ in Z-basis and obtains the result $\left| 0 \right\rangle$ in Step 1.4 of Sect. 3.1, Alice shares three-qubit pure entangled state $\left| \Phi \right\rangle_{{{\text{A}}_{{1}} {\text{BC}}}} = {1 \mathord{\left/ {\vphantom {1 {\sqrt {1 + \left( {1 - D} \right)^{3} } }}} \right. \kern-0pt} {\sqrt {1 + \left( {1 - D} \right)^{3} } }}\left( {\left| {000} \right\rangle + \sqrt {\left( {1 - D} \right)^{3} } \left| {111} \right\rangle } \right)_{{{\text{A}}_{{1}} {\text{BC}}}}$ with Bob and Charlie.

Proof 1

When Alice sends the qubit A₂ to Bob via amplitude damping channel, the density matrix of two-qubit A₁ and A₂ can be expressed as:

$$ \begin{gathered} \rho_{b} = \varepsilon \left( \rho \right) = \sum\nolimits_{i = 0}^{1} {\left( {I \otimes K_{i} } \right) * \rho * \left( {I \otimes K_{i} } \right)^{\dag } } \hfill \\ \quad = \frac{1}{2}\left| {00} \right\rangle \left\langle {00} \right| + \frac{1}{2}\sqrt {1 - D} \left| {00} \right\rangle \left\langle {11} \right| + \frac{1}{2}\sqrt {1 - D} \left| {11} \right\rangle \left\langle {00} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{2}D\left| {10} \right\rangle \left\langle {10} \right| + \frac{1}{2}\left( {1 - D} \right)\left| {11} \right\rangle \left\langle {11} \right| \hfill \\ \end{gathered} $$

(A1)

where $\rho = \left| {B_{0,0} } \right\rangle_{{{\text{A}}_{1} {\text{A}}_{2} }} \left\langle {B_{0,0} } \right|$, and $\rho_{b}$ is the density matrix of qubits (A₁, A₂) when Bob receives the qubit A₂.

When Bob performs the CNOT operation on qubit A₂ and ancillary qubit $\left| 0 \right\rangle_{B}$, the density matrix of three qubits (A₁, A₂, B) is written as:

$$ \begin{aligned} \rho_{b}^{ * } = & \frac{1}{2}\left| {000} \right\rangle \left\langle {000} \right| + \frac{1}{2}\sqrt {1 - D} \left| {000} \right\rangle \left\langle {111} \right| + \frac{1}{2}\sqrt {1 - D} \left| {111} \right\rangle \left\langle {000} \right| \\ & + \frac{1}{2}D\left| {100} \right\rangle \left\langle {100} \right| + \frac{1}{2}\left( {1 - D} \right)\left| {111} \right\rangle \left\langle {111} \right| \\ \end{aligned} $$

(A2)

Then, Bob further sends qubit A₂ to Charlie via amplitude damping channel. When Charlie receives the qubit A₂, the density matrix of three qubits (A₁, A₂, B) evolves into:

$$ \begin{gathered} \rho_{c} = \varepsilon \left( {\rho_{b}^{ * } } \right) = \sum\nolimits_{i = 0}^{1} {\left( {I \otimes K_{i} \otimes I} \right) * \rho_{b}^{ * } * \left( {I \otimes K_{i} \otimes I} \right)^{\dag } } \hfill \\ \quad \,\, = \frac{1}{2}\left| {000} \right\rangle \left\langle {000} \right| + \frac{1}{2}\left( {1 - D} \right)\left| {000} \right\rangle \left\langle {111} \right| + \frac{1}{2}\left( {1 - D} \right)\left| {111} \right\rangle \left\langle {000} \right| + \frac{1}{2}D\left| {100} \right\rangle \left\langle {100} \right| \hfill \\ \quad \,\,\,\,\, + \frac{1}{2}D\left( {1 - D} \right)\left| {101} \right\rangle \left\langle {101} \right| + \frac{1}{2}\left( {1 - D} \right)^{2} \left| {111} \right\rangle \left\langle {111} \right| \hfill \\ \end{gathered} $$

(A3)

When Charlie performs the CNOT operation on qubit A₂ and ancillary qubit $\left| 0 \right\rangle_{C}$, the density matrix of four qubit (A₁, A₂, B, C) is written as:

$$ \begin{aligned} \rho_{c}^{ * } = & \frac{1}{2}\left| {0000} \right\rangle \left\langle {0000} \right| + \frac{1}{2}\left( {1 - D} \right)\left| {0000} \right\rangle \left\langle {1111} \right| \\ & + \frac{1}{2}\left( {1 - D} \right)\left| {1111} \right\rangle \left\langle {0000} \right| + \frac{1}{2}D\left| {1000} \right\rangle \left\langle {1000} \right| \\ & + \frac{1}{2}D\left( {1 - D} \right)\left| {1010} \right\rangle \left\langle {1010} \right| + \frac{1}{2}\left( {1 - D} \right)^{2} \left| {1111} \right\rangle \left\langle {1111} \right| \\ \end{aligned} $$

(A4)

Then, Charlie sends qubit A₂ to Alice via amplitude damping channel. When Alice receives the qubit A₂, the density matrix of four qubit (A₁, A₂, B, C) evolves into:

$$ \begin{gathered} \rho_{a} = \varepsilon \left( {\rho_{c}^{ * } } \right) = \sum\nolimits_{i = 0}^{1} {\left( {I \otimes K_{i} \otimes I \otimes I} \right) * \rho_{c}^{ * } * \left( {I \otimes K_{i} \otimes I \otimes I} \right)^{\dag } } \hfill \\ \quad \,\, = \frac{1}{2}\left| {0000} \right\rangle \left\langle {0000} \right| + \frac{1}{2}\sqrt {\left( {1 - D} \right)^{3} } \left| {0000} \right\rangle \left\langle {1111} \right| \hfill \\ \quad \,\,\,\,\, + \frac{1}{2}\sqrt {\left( {1 - D} \right)^{3} } \left| {1111} \right\rangle \left\langle {0000} \right| + \frac{1}{2}D\left| {1000} \right\rangle \left\langle {1000} \right| \hfill \\ \quad \,\,\,\,\, + \frac{1}{2}D\left( {1 - D} \right)^{2} \left| {1010} \right\rangle \left\langle {1010} \right| + \frac{1}{2}\left( {1 - D} \right)^{3} \left| {1111} \right\rangle \left\langle {1111} \right| \hfill \\ \end{gathered} $$

(A5)

Final, Alice first performs the CNOT operation on two qubits (A₁, A₂), the density matrix of four qubit (A₁, A₂, B, C) evolves into:

$$ \begin{aligned} \rho_{a}^{ * } = & \frac{1}{2}\left| {0000} \right\rangle \left\langle {0000} \right| + \frac{1}{2}\sqrt {\left( {1 - D} \right)^{3} } \left| {0000} \right\rangle \left\langle {1011} \right| \\ & + \frac{1}{2}\sqrt {\left( {1 - D} \right)^{3} } \left| {1011} \right\rangle \left\langle {0000} \right| + \frac{1}{2}D\left| {1100} \right\rangle \left\langle {1100} \right| \\ & + \frac{1}{2}D\left( {1 - D} \right)^{2} \left| {1110} \right\rangle \left\langle {1110} \right| + \frac{1}{2}\left( {1 - D} \right)^{3} \left| {1011} \right\rangle \left\langle {1011} \right| \\ \end{aligned} $$

(A6)

Hence, when Alice measures qubit A₂ in Z-basis $\left\{ {\left| 0 \right\rangle ,\left| 1 \right\rangle } \right\}$, it would disentangle the qubit A₂ with other three qubits (A₁, B, C). If she obtains the result $\left| 0 \right\rangle$, the density matrix of three qubits (A₁, B, C) is collapsed to:

$$ \begin{gathered} \rho_{{{\text{A}}_{{1}} {\text{BC}}}} = \frac{1}{2}\left| {000} \right\rangle \left\langle {000} \right| + \frac{1}{2}\sqrt {\left( {1 - D} \right)^{3} } \left| {000} \right\rangle \left\langle {111} \right| \hfill \\ \quad \quad \quad + \frac{1}{2}\sqrt {\left( {1 - D} \right)^{3} } \left| {111} \right\rangle \left\langle {000} \right| + \frac{1}{2}\left( {1 - D} \right)^{3} \left| {111} \right\rangle \left\langle {111} \right| \hfill \\ \quad \quad \,\, = \frac{1}{\sqrt 2 }\left( {\left| {000} \right\rangle + \sqrt {\left( {1 - D} \right)^{3} } \left| {111} \right\rangle } \right)\frac{1}{\sqrt 2 }\left( {\left\langle {000} \right| + \sqrt {\left( {1 - D} \right)^{3} } \left\langle {111} \right|} \right) \hfill \\ \end{gathered} $$

(A7)

After renormalization of above Eq. (A7), Alice shares the pure entangled state with two agents Bob and Charlie written as:

$$ \left| \Phi \right\rangle_{{{\text{A}}_{{1}} {\text{BC}}}} = {1 \mathord{\left/ {\vphantom {1 {\sqrt {1 + \left( {1 - D} \right)^{3} } }}} \right. \kern-0pt} {\sqrt {1 + \left( {1 - D} \right)^{3} } }}\left( {\left| {000} \right\rangle + \sqrt {\left( {1 - D} \right)^{3} } \left| {111} \right\rangle } \right)_{{{\text{A}}_{{1}} {\text{BC}}}} $$

(A8)

Appendix 2: Proof of Theorem 2

Theorem 2

On condition that Alice prepares the generalized Bell state $\left| {GB_{0,0} } \right\rangle = {1 \mathord{\left/ {\vphantom {1 {\sqrt 3 }}} \right. \kern-0pt} {\sqrt 3 }}\left( {\left| {00} \right\rangle + \left| {11} \right\rangle + \left| {22} \right\rangle } \right)_{{{\text{A}}_{1} {\text{A}}_{2} }}$, when she measures qutrit A₂ in Z-basis and obtains the result $\left| 0 \right\rangle$ in Step 1.4 of Sect. 4.1, Alice shares the pure entangled state $\left| \Psi \right\rangle_{{{\text{A}}_{{1}} {\text{BC}}}} = {1 \mathord{\left/ {\vphantom {1 {\sqrt {1 + 2\left( {1 - D} \right)^{3} } }}} \right. \kern-0pt} {\sqrt {1 + 2\left( {1 - D} \right)^{3} } }}\left( {\left| {000} \right\rangle + \sqrt {\left( {1 - D} \right)^{3} } \left| {111} \right\rangle + \sqrt {\left( {1 - D} \right)^{3} } \left| {222} \right\rangle } \right)_{{{\text{A}}_{{1}} {\text{BC}}}}$ with Bob and Charlie.

Proof 2

When Alice sends the qutrit A₂ to Bob via amplitude damping channel, the density matrix of qutrits (A₁, A₂) can be expressed as:

$$ \begin{gathered} \rho_{b} = \varepsilon \left( \rho \right) = \sum\nolimits_{i = 0}^{2} {\left( {I \otimes F_{i} } \right) * \rho * \left( {I \otimes F_{i} } \right)^{\dag } } \hfill \\ \quad \, = \frac{1}{3}\left| {00} \right\rangle \left\langle {00} \right| + \frac{1}{3}\sqrt {1 - D} \left| {00} \right\rangle \left\langle {11} \right| + \frac{1}{3}\sqrt {1 - D} \left| {00} \right\rangle \left\langle {22} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{3}D\left| {10} \right\rangle \left\langle {10} \right| + \frac{1}{3}\sqrt {1 - D} \left| {11} \right\rangle \left\langle {00} \right| + \frac{1}{3}\left( {1 - D} \right)\left| {11} \right\rangle \left\langle {11} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{3}\left( {1 - D} \right)\left| {11} \right\rangle \left\langle {22} \right| + \frac{1}{3}D\left| {20} \right\rangle \left\langle {20} \right| + \frac{1}{3}\sqrt {1 - D} \left| {22} \right\rangle \left\langle {00} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{3}\left( {1 - D} \right)\left| {22} \right\rangle \left\langle {11} \right| + \frac{1}{3}\left( {1 - D} \right)\left| {22} \right\rangle \left\langle {22} \right| \hfill \\ \end{gathered} $$

(A9)

where $\rho = \left| {GB_{0,0} } \right\rangle_{{{\text{A}}_{1} {\text{A}}_{2} }} \left\langle {GB_{0,0} } \right|$, and $\rho_{b}$ is the density matrix of qutrits (A₁, A₂) when Bob receives the qutrit A₂.

When Bob performs the GCNOT operation on qutrit A₂ and ancillary qutrit $\left| 0 \right\rangle_{{\text{B}}}$, the density matrix of three qutrits (A₁, A₂, B) is written as:

$$ \begin{aligned} \rho_{b}^{ * } = & \frac{1}{3}\left| {000} \right\rangle \left\langle {000} \right| + \frac{1}{3}\sqrt {1 - D} \left| {000} \right\rangle \left\langle {111} \right| \\ & + \frac{1}{3}\sqrt {1 - D} \left| {000} \right\rangle \left\langle {222} \right| + \frac{1}{3}D\left| {100} \right\rangle \left\langle {100} \right| \\ & + \frac{1}{3}\sqrt {1 - D} \left| {111} \right\rangle \left\langle {000} \right| + \frac{1}{3}\left( {1 - D} \right)\left| {111} \right\rangle \left\langle {111} \right| \\ & + \frac{1}{3}\left( {1 - D} \right)\left| {111} \right\rangle \left\langle {222} \right| + \frac{1}{3}D\left| {200} \right\rangle \left\langle {200} \right| \\ & + \frac{1}{3}\sqrt {1 - D} \left| {222} \right\rangle \left\langle {000} \right| + \frac{1}{3}\left( {1 - D} \right)\left| {222} \right\rangle \left\langle {111} \right| \\ & + \frac{1}{3}\left( {1 - D} \right)\left| {222} \right\rangle \left\langle {222} \right| \\ \end{aligned} $$

(A10)

Then, Bob sends qutrit A₂ to Charlie via amplitude damping channel. When Charlie receives the qutrit A₂, the density matrix of three qutrits (A₁, A₂, B) evolves into:

$$ \begin{gathered} \rho_{c} { = }\varepsilon \left( {\rho_{b}^{ * } } \right){ = }\sum\nolimits_{i = 0}^{2} {\left( {I \otimes F_{i} \otimes I} \right) * \rho_{b}^{ * } * \left( {I \otimes F_{i} \otimes I} \right)^{\dag } } \hfill \\ \quad = \frac{1}{3}\left| {000} \right\rangle \left\langle {000} \right| + \frac{1}{3}\left( {1 - D} \right)\left| {000} \right\rangle \left\langle {111} \right| + \frac{1}{3}\left( {1 - D} \right)\left| {000} \right\rangle \left\langle {222} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{3}D\left| {100} \right\rangle \left\langle {100} \right| + \frac{1}{3}D\left( {1 - D} \right)\left| {101} \right\rangle \left\langle {101} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{3}\left( {1 - D} \right)\left| {111} \right\rangle \left\langle {000} \right| + \frac{1}{3}\left( {1 - D} \right)^{2} \left| {111} \right\rangle \left\langle {111} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{3}\left( {1 - D} \right)^{2} \left| {111} \right\rangle \left\langle {222} \right| + \frac{1}{3}D\left| {200} \right\rangle \left\langle {200} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{3}D\left( {1 - D} \right)\left| {202} \right\rangle \left\langle {202} \right| + \frac{1}{3}\left( {1 - D} \right)\left| {222} \right\rangle \left\langle {000} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{3}\left( {1 - D} \right)^{2} \left| {222} \right\rangle \left\langle {111} \right| + \frac{1}{3}\left( {1 - D} \right)^{2} \left| {222} \right\rangle \left\langle {222} \right| \hfill \\ \end{gathered} $$

(A11)

When Charlie performs the GCNOT operation on qutrit A₂ and ancillary qutrit $\left| 0 \right\rangle_{C}$, the density matrix of four qutrits (A₁, A₂, B, C) is written as:

$$ \begin{aligned} \rho_{c}^{ * } = & \frac{1}{3}\left| {0000} \right\rangle \left\langle {0000} \right| + \frac{1}{3}\left( {1 - D} \right)\left| {0000} \right\rangle \left\langle {1111} \right| \\ & + \frac{1}{3}\left( {1 - D} \right)\left| {0000} \right\rangle \left\langle {2222} \right| + \frac{1}{3}D\left| {1000} \right\rangle \left\langle {1000} \right| \\ & + \frac{1}{3}D\left( {1 - D} \right)\left| {1010} \right\rangle \left\langle {1010} \right| + \frac{1}{3}\left( {1 - D} \right)\left| {1111} \right\rangle \left\langle {0000} \right| \\ & + \frac{1}{3}\left( {1 - D} \right)^{2} \left| {1111} \right\rangle \left\langle {1111} \right| \\ & + \frac{1}{3}\left( {1 - D} \right)^{2} \left| {1111} \right\rangle \left\langle {2222} \right| + \frac{1}{3}D\left| {2000} \right\rangle \left\langle {2000} \right| \\ & + \frac{1}{3}D\left( {1 - D} \right)\left| {2020} \right\rangle \left\langle {2020} \right| \\ & + \frac{1}{3}\left( {1 - D} \right)\left| {2222} \right\rangle \left\langle {0000} \right| + \frac{1}{3}\left( {1 - D} \right)^{2} \left| {2222} \right\rangle \left\langle {1111} \right| \\ & + \frac{1}{3}\left( {1 - D} \right)^{2} \left| {2222} \right\rangle \left\langle {2222} \right| \\ \end{aligned} $$

(A12)

Then, Charlie sends qutrit A₂ to Alice via amplitude damping channel. When Alice receives the qutrit A₂, the density matrix of four qutrit (A₁, A₂, B, C) evolves into:

$$ \begin{gathered} \rho_{a} = \varepsilon \left( {\rho_{c}^{ * } } \right) = \sum\nolimits_{i = 0}^{2} {\left( {I \otimes F_{i} \otimes I \otimes I} \right) * \rho_{c}^{ * } * \left( {I \otimes F_{i} \otimes I \otimes I} \right)^{\dag } } \hfill \\ \quad \, = \frac{1}{3}\left| {0000} \right\rangle \left\langle {0000} \right| + \frac{1}{3}\sqrt {\left( {1 - D} \right)^{3} } \left| {0000} \right\rangle \left\langle {1111} \right| + \frac{1}{3}\sqrt {\left( {1 - D} \right)^{3} } \left| {0000} \right\rangle \left\langle {2222} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{3}D\left| {1000} \right\rangle \left\langle {1000} \right| + \frac{1}{3}D\left( {1 - D} \right)\left| {1010} \right\rangle \left\langle {1010} \right| + \frac{1}{3}D\left( {1 - D} \right)^{2} \left| {1011} \right\rangle \left\langle {1011} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{3}\sqrt {\left( {1 - D} \right)^{3} } \left| {1111} \right\rangle \left\langle {0000} \right| + \frac{1}{3}\left( {1 - D} \right)^{3} \left| {1111} \right\rangle \left\langle {1111} \right| + \frac{1}{3}\left( {1 - D} \right)^{3} \left| {1111} \right\rangle \left\langle {2222} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{3}D\left| {2000} \right\rangle \left\langle {2000} \right| + \frac{1}{3}D\left( {1 - D} \right)\left| {2020} \right\rangle \left\langle {2020} \right| + \frac{1}{3}D\left( {1 - D} \right)^{2} \left| {2022} \right\rangle \left\langle {2022} \right| \hfill \\ \quad \,\,\,\, + \frac{1}{3}\sqrt {\left( {1 - D} \right)^{3} } \left| {2222} \right\rangle \left\langle {0000} \right| + \frac{1}{3}\left( {1 - D} \right)^{3} \left| {2222} \right\rangle \left\langle {1111} \right| + \frac{1}{3}\left( {1 - D} \right)^{3} \left| {2222} \right\rangle \left\langle {2222} \right| \hfill \\ \end{gathered} $$

(A13)

Final, Alice first performs the IGCNOT operation on two qutrits (A₁, A₂), the density matrix of four qutrit (A₁, A₂, B, C) evolves into:

$$ \begin{aligned} \rho_{a}^{ * } = & \frac{1}{3}\left| {0000} \right\rangle \left\langle {0000} \right| + \frac{1}{3}\sqrt {\left( {1 - D} \right)^{3} } \left| {0000} \right\rangle \left\langle {1011} \right| + \frac{1}{3}\sqrt {\left( {1 - D} \right)^{3} } \left| {0000} \right\rangle \left\langle {2022} \right| \\ & + \frac{1}{3}D\left| {1100} \right\rangle \left\langle {1100} \right| + \frac{1}{3}D\left( {1 - D} \right)\left| {1110} \right\rangle \left\langle {1110} \right| + \frac{1}{3}D\left( {1 - D} \right)^{2} \left| {1111} \right\rangle \left\langle {1111} \right| \\ & + \frac{1}{3}\sqrt {\left( {1 - D} \right)^{3} } \left| {1011} \right\rangle \left\langle {0000} \right| + \frac{1}{3}\left( {1 - D} \right)^{3} \left| {1011} \right\rangle \left\langle {1011} \right| + \frac{1}{3}\left( {1 - D} \right)^{3} \left| {1011} \right\rangle \left\langle {2022} \right| \\ & + \frac{1}{3}D\left| {2200} \right\rangle \left\langle {2200} \right| + \frac{1}{3}D\left( {1 - D} \right)\left| {2220} \right\rangle \left\langle {2220} \right| + \frac{1}{3}D\left( {1 - D} \right)^{2} \left| {2222} \right\rangle \left\langle {2222} \right| \\ & + \frac{1}{3}\sqrt {\left( {1 - D} \right)^{3} } \left| {2022} \right\rangle \left\langle {0000} \right| + \frac{1}{3}\left( {1 - D} \right)^{3} \left| {2022} \right\rangle \left\langle {1011} \right| + \frac{1}{3}\left( {1 - D} \right)^{3} \left| {2022} \right\rangle \left\langle {2022} \right| \\ \end{aligned} $$

(A14)

Hence, when Alice measures qutrit A₂ in Z-basis $\left\{ {\left| 0 \right\rangle ,\left| 1 \right\rangle ,\left| 2 \right\rangle } \right\}$, it would disentangle the qutrit A₂ with other three qutrits (A₁, B, C). If she obtains the result $\left| 0 \right\rangle$, the density matrix of three qutrits (A₁, B, C) can be expressed as:

$$ \begin{gathered} \rho_{{{\text{A}}_{{1}} {\text{BC}}}} = \frac{1}{3}\left| {000} \right\rangle \left\langle {000} \right| + \frac{1}{3}\sqrt {\left( {1 - D} \right)^{3} } \left| {000} \right\rangle \left\langle {111} \right| + \frac{1}{3}\sqrt {\left( {1 - D} \right)^{3} } \left| {000} \right\rangle \left\langle {222} \right| \hfill \\ \quad \quad \,\,\,\, + \frac{1}{3}\sqrt {\left( {1 - D} \right)^{3} } \left| {111} \right\rangle \left\langle {000} \right| + \frac{1}{3}\left( {1 - D} \right)^{3} \left| {111} \right\rangle \left\langle {111} \right| + \frac{1}{3}\left( {1 - D} \right)^{3} \left| {111} \right\rangle \left\langle {222} \right| \hfill \\ \quad \quad \,\,\,\, + \frac{1}{3}\sqrt {\left( {1 - D} \right)^{3} } \left| {222} \right\rangle \left\langle {000} \right| + \frac{1}{3}\left( {1 - D} \right)^{3} \left| {222} \right\rangle \left\langle {111} \right| + \frac{1}{3}\left( {1 - D} \right)^{3} \left| {222} \right\rangle \left\langle {222} \right| \hfill \\ \quad \quad \,{ = }\frac{1}{\sqrt 3 }\left( {\left| {000} \right\rangle + \sqrt {\left( {1 - D} \right)^{3} } \left| {111} \right\rangle + \sqrt {\left( {1 - D} \right)^{3} } \left| {222} \right\rangle } \right)\frac{1}{\sqrt 3 }\left( {\left\langle {000} \right| + \sqrt {\left( {1 - D} \right)^{3} } \left\langle {111} \right| + \sqrt {\left( {1 - D} \right)^{3} } \left\langle {222} \right|} \right) \hfill \\ \end{gathered} $$

(A15)

After renormalization of above Eq. (A15), Alice now shares the pure entangled state with two agents Bob and Charlie written as:

$$ \left| \Psi \right\rangle_{{{\text{A}}_{{1}} {\text{BC}}}} = {1 \mathord{\left/ {\vphantom {1 {\sqrt {1 + 2\left( {1 - D} \right)^{3} } }}} \right. \kern-0pt} {\sqrt {1 + 2\left( {1 - D} \right)^{3} } }}\left( {\left| {000} \right\rangle + \sqrt {\left( {1 - D} \right)^{3} } \left| {111} \right\rangle + \sqrt {\left( {1 - D} \right)^{3} } \left| {222} \right\rangle } \right)_{{{\text{A}}_{{1}} {\text{BC}}}} $$

(A16)