Quantum controlled teleportation with OR-logic-gate-like controllers in noisy environment

Suppose there are four parties: Alice as a sender, Bob as a receiver, Charlie as the first controller, and David as the second controller. To do the teleportation, they should pre-share a four-qubit entangled channel state before doing the teleportation process. We build the channel state by preparing four ∣0〉 states as ∣0000〉₁₂₃₄. Index 1 to 4 represent the 1st to 4th channel's qubit.

We build the four-qubit entanglement by applying Hadamard gates on qubit 1 and 2, followed by a CNOT gate on qubit 4 with qubit 1 as the control qubit, then other CNOT gates on qubit 3 and 4 with qubit 2 as the control qubit. The final channel state is

$$\begin{eqnarray}&&| \psi {\rangle }_{1234}=\displaystyle \frac{1}{2}\left(| 0000\rangle +| 0111\rangle +| 1001\rangle +| 1110\rangle \right).\end{eqnarray} \tag{ 1 }$$

The rank of the reduced density matrix of the state [39] is greater than one, showing that the state is completely entangled. The state also has high persistency of entanglement [40], i.e., we need to perform at least two measurements in order to make the state completely unentangled. If we measure qubit 1 or 4 with computational bases, the remaining qubit collapse into a GHZ state [41], while for qubit 2 and 3, two of the three remaining qubits collapse into a Bell state.

Before the teleportation process, the qubits are distributed to the four parties as ∣ψ〉_ACDB = ∣ψ〉₁₂₃₄. Indexes A, C, D, and B indicate that the corresponding qubit belongs to Alice, Charlie, David, and Bob, respectively. We use the state as the channel because Charlie and David's qubits, i.e., qubits 2 and 3, always have the same state, which means if they were measured in computational bases independently, they would always show the same result. The behavior will generate the OR-logic-gate-like controllers' combination in the teleportation process, described as follows.

Alice, as the sender, has an arbitrary qubit

$$\begin{eqnarray}&&| \phi {\rangle }_{a}={x}_{0}| 0\rangle +{x}_{1}| 1\rangle ,\end{eqnarray} \tag{ 2 }$$

with x₀ and x₁ are complex numbers satisfying normalization condition ${{\rm{\Sigma }}}_{i=0}^{1}| {x}_{i}{| }^{2}=1$. The teleportation circuit scheme is shown in figure 1. Alice wants to teleport her state to Bob with Charlie and David's cooperation. She mixes her state with the channel state as ∣Ψ〉_aACDB = ∣ϕ〉_a ⨂ ∣ψ〉_ACDB , then performs a Bell State Measurement [1] on her two qubits, i.e., a projection measurement on qubits aA in Bell Bases $| {{\rm{\Phi }}}^{\pm }\rangle =\tfrac{1}{\sqrt{2}}(| 00\rangle \pm | 11\rangle )$ and $| {{\rm{\Psi }}}^{\pm }\rangle =\tfrac{1}{\sqrt{2}}(| 01\rangle \pm | 10\rangle )$. Suppose the measurement result is ∣Φ⁺〉, the remaining qubits collapse to

$$\begin{eqnarray}&&| {\rm{\Psi }}^{\prime} {\rangle }_{{CDB}}=\displaystyle \frac{1}{2}\left({x}_{0}| 000\rangle +{x}_{0}| 111\rangle +{x}_{1}| 001\rangle +{x}_{1}| 110\rangle \right).\end{eqnarray} \tag{ 3 }$$

Charlie and David perform computational bases measurement independently. If Charlie's measurement result is ∣0〉, the remaining qubits collapse to ∣Ψ''〉_DB = x₀∣00〉 + x₁∣01〉 = ∣0〉 ⨂ (x₀∣0〉 + x₁∣1〉), indicating that David's result is also ∣0〉. If Charlie's measurement result is ∣1〉, the remaining qubits collapse to $| \tilde{{\rm{\Psi }}}^{\prime\prime} {\rangle }_{{DB}}={x}_{0}| 11\rangle +{x}_{1}| 10\rangle =| 1\rangle \otimes ({x}_{0}| 1\rangle +{x}_{1}| 0\rangle )$, indicating that David's result is also ∣1〉. The states show that David's result is always the same as Charlie's. All possible measurement results and the corresponding receiver's unitary operators are shown in table 1, showing that the controllers' results are correlated with the receiver's operator. However, the controllers always have the same result, which makes the receiver only need one of the controllers' cooperation, and it does not matter which one. Hence, the controllers' behaviour has the properties of the OR logic gate, one of the fundamental gates in communication science. The teleportation is useful if the information to be teleported requires the cooperation of one of two other parties (controllers), and it does not matter which one.

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Table 1. List of Alice, Charlie, and David's measurement results, Bob's qubit state, and Bob's unitary operator.

MA a	MC b	MD c	Bob's $| \phi ^{\prime} {\rangle }_{B}$ state	UB d
∣Φ+〉	∣0〉	∣0〉	x0∣0〉 + x1∣1〉	I
∣Φ+〉	∣1〉	∣1〉	x0∣1〉 + x1∣0〉	σx
∣Φ−〉	∣0〉	∣0〉	x0∣0〉 − x1∣1〉	σz
∣Φ−〉	∣1〉	∣1〉	x0∣1〉 − x1∣0〉	σz σx
∣Ψ+〉	∣0〉	∣0〉	x0∣1〉 + x1∣0〉	σx
∣Ψ+〉	∣1〉	∣1〉	x0∣0〉 + x1∣1〉	I
∣Ψ−〉	∣0〉	∣0〉	x0∣1〉 − x1∣0〉	σz σx
∣Ψ−〉	∣1〉	∣1〉	x0∣0〉 − x1∣1〉	σz

^a Alice's measurement result. ^b Charlie's measurement result. ^c David's measurement result. ^d Bob's Unitary operator.

This subsection's protocol agrees with the previous work by Wang et al [42] who proposed a Hierarchical Quantum Information Splitting which also can be used as a Quantum Controlled Teleportation with Two Controllers, but with different channel state. The advantage of our channel state is that it only requires Hadamard and CNOT gates to build it. Moreover, it can be applied to build a bidirectional protocol as described in the following subsection.

In this subsection, four characters are introduced, Alice and Bob as the senders as well as the receivers, Charlie and David as the first and second controller, respectively. Before the teleportation process, they pre-share a six-qubit entangled state to be used as the teleportation channel. By using the controllers' OR-logic-gate-like behaviour in the previous subsection, we build the state by using equation(1) and preparing another two ∣0〉 qubit states as ∣ψ〉₁₂₃₄ ⨂ ∣00〉₅₆. The additional qubits will be used for the transmission from Bob to Alice in the bidirectional teleportation. We apply a Hadamard gate on qubit 5 and a CNOT gate on qubit 6 with qubit 5 as the control qubit, followed by another CNOT gate on qubit 5 with qubit 3 as the control qubit. The final result is

$$\begin{eqnarray}\begin{array}{rcl}| \psi {\rangle }_{123456} & = & \displaystyle \frac{1}{2\sqrt{2}}\left(| 000000\rangle +| 011110\rangle +| 100100\rangle +| 111010\rangle \right.\\ & & \left.+| 000011\rangle +| 011101\rangle +| 100111\rangle +| 111001\rangle \right).\end{array}\end{eqnarray} \tag{ 4 }$$

All of the reduced density matrix rank of the state is greater than one, indicating that the state is completely entangled [39].

We distribute the channel qubit to the four parties as $| \psi {\rangle }_{{A}_{1}{{CDB}}_{2}{A}_{2}{B}_{1}}=| \psi {\rangle }_{123456}$. Qubit A₁ and A₂ belong to Alice, B₁ and B₂ belong to Bob, C and D belongs to Charlie and David, respectively. Alice and Bob has an arbitrary one-qubit state as

$$\begin{eqnarray}&&| \alpha {\rangle }_{a}={x}_{0}| 0\rangle +{x}_{1}| 1\rangle ,\end{eqnarray} \tag{ 5 }$$

$$\begin{eqnarray}&&| \beta {\rangle }_{b}={y}_{0}| 0\rangle +{y}_{1}| 1\rangle ,\end{eqnarray} \tag{ 6 }$$

respectively, with x₀, x₁, y₀ and y₁ are complex numbers satisfying normalization condition ${{\rm{\Sigma }}}_{i=0}^{1}| {x}_{i}{| }^{2}=1$ and ${{\rm{\Sigma }}}_{i=0}^{1}| {y}_{i}{| }^{2}=1$.

The teleportation circuit scheme is shown in figure 2. Alice and Bob start the teleportation process by mixing their state into the channel state as $| {\rm{\Psi }}{\rangle }_{{{abA}}_{1}{{CDB}}_{2}{A}_{2}{B}_{1}}=| \alpha {\rangle }_{a}\otimes | \beta {\rangle }_{b}\otimes | \psi {\rangle }_{{A}_{1}{{CDB}}_{2}{A}_{2}{B}_{1}}$. Alice and Bob perform Bell State measurement [1] independently on qubit aA₁ and bB₁, respectively. Suppose their measurement results are both ∣Φ⁺〉, the remaining qubits collapse to

$$\begin{eqnarray}\begin{array}{rcl}| {\rm{\Psi }}^{\prime} {\rangle }_{{{CDB}}_{2}{A}_{2}} & = & \displaystyle \frac{1}{2}\left({x}_{0}{y}_{0}| 0000\rangle +{x}_{0}{y}_{0}| 1111\rangle +{x}_{1}{y}_{0}| 0010\rangle +{x}_{1}{y}_{0}| 1101\rangle \right.\\ & & \left.+{x}_{0}{y}_{1}| 0001\rangle +{x}_{0}{y}_{1}| 1110\rangle +{x}_{1}{y}_{1}| 0011\rangle +{x}_{1}{y}_{1}| 1100\rangle \right).\end{array}\end{eqnarray} \tag{ 7 }$$

Charlie and David perform a computational bases measurement independently. Similarly as in Section 2.1, the remaining qubits collapse to

$$\begin{eqnarray}\begin{array}{rcl}| {\rm{\Psi }}^{\prime\prime} {\rangle }_{{B}_{2}{A}_{2}} & = & {x}_{0}{y}_{0}| 00\rangle +{x}_{1}{y}_{0}| 10\rangle +{x}_{0}{y}_{1}| 01\rangle +{x}_{1}{y}_{1}| 11\rangle \\ & = & \left({x}_{0}| 0\rangle +{x}_{1}| 1\rangle \right)\otimes \left({y}_{0}| 0\rangle +{y}_{1}| 1\rangle \right).\end{array}\end{eqnarray} \tag{ 8 }$$

For this case, Alice and Bob perform nothing on their qubit. The teleportation is complete.

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All of the possible measurement result is presented in table 2. The table shows that both receivers cannot rebuild their corresponding initial state without the help of at least one of the controllers. As in the previous subsection, the two controllers' agreement combination mimics the OR logic gate. This result is applied to the cyclic teleportation scheme as described in the next subsection.

Table 2. List of Alice, Bob, and Charlie/David's measurement results, Alice and Bob's qubit state, Alice and Bob's unitary operator.

MA a	MB b	MC c	MD d	$| \phi ^{\prime} {\rangle }_{A}\otimes | \phi ^{\prime} {\rangle }_{B}$ state	UA e	UB f
∣Φ+〉	∣Φ+〉	∣0〉	∣0〉	(x0∣0〉 + x1∣1〉) ⨂ (y0∣0〉 + y1∣1〉)	I	I
∣Φ+〉	∣Φ+〉	∣1〉	∣1〉	(x0∣1〉 + x1∣0〉) ⨂ (y0∣1〉 + y1∣0〉)	σx	σx
∣Φ+〉	∣Φ−〉	∣0〉	∣0〉	(x0∣0〉 + x1∣1〉) ⨂ (y0∣0〉 − y1∣1〉)	I	σz
∣Φ+〉	∣Φ−〉	∣1〉	∣1〉	(x0∣1〉 + x1∣0〉) ⨂ (y0∣1〉 − y1∣0〉)	σx	σz σx
∣Φ+〉	∣Ψ+〉	∣0〉	∣0〉	(x0∣0〉 + x1∣1〉) ⨂ (y0∣1〉 + y1∣0〉)	I	σx
∣Φ+〉	∣Ψ+〉	∣1〉	∣1〉	(x0∣1〉 + x1∣0〉) ⨂ (y0∣0〉 + y1∣1〉)	σx	I
∣Φ+〉	∣Ψ−〉	∣0〉	∣0〉	(x0∣0〉 + x1∣1〉) ⨂ (y0∣1〉 − y1∣0〉)	I	σz σx
∣Φ+〉	∣Ψ−〉	∣1〉	∣1〉	(x0∣1〉 + x1∣0〉) ⨂ (y0∣0〉 − y1∣1〉)	σx	σz
∣Φ−〉	∣Φ+〉	∣0〉	∣0〉	(x0∣0〉 − x1∣1〉) ⨂ (y0∣0〉 + y1∣1〉)	σz	I
∣Φ−〉	∣Φ+〉	∣1〉	∣1〉	(x0∣1〉 − x1∣0〉) ⨂ (y0∣1〉 + y1∣0〉)	σz σx	σx
∣Φ−〉	∣Φ−〉	∣0〉	∣0〉	(x0∣0〉 − x1∣1〉) ⨂ (y0∣0〉 − y1∣1〉)	σz	σz
∣Φ−〉	∣Φ−〉	∣1〉	∣1〉	(x0∣1〉 − x1∣0〉) ⨂ (y0∣1〉 − y1∣0〉)	σz σx	σz σx
∣Φ−〉	∣Ψ+〉	∣0〉	∣0〉	(x0∣0〉 − x1∣1〉) ⨂ (y0∣1〉 + y1∣0〉)	σz	σx
∣Φ−〉	∣Ψ+〉	∣1〉	∣1〉	(x0∣1〉 − x1∣0〉) ⨂ (y0∣0〉 + y1∣1〉)	σz σx	I
∣Φ−〉	∣Ψ−〉	∣0〉	∣0〉	(x0∣0〉 − x1∣1〉) ⨂ (y0∣1〉 − y1∣0〉)	σz	σz σx
∣Φ−〉	∣Ψ−〉	∣1〉	∣1〉	(x0∣1〉 − x1∣0〉) ⨂ (y0∣0〉 − y1∣1〉)	σz σx	σz
∣Ψ+〉	∣Φ+〉	∣0〉	∣0〉	(x0∣1〉 + x1∣0〉) ⨂ (y0∣0〉 + y1∣1〉)	σx	I
∣Ψ+〉	∣Φ+〉	∣1〉	∣1〉	(x0∣0〉 + x1∣1〉) ⨂ (y0∣1〉 + y1∣0〉)	I	σx
∣Ψ+〉	∣Φ−〉	∣0〉	∣0〉	(x0∣1〉 + x1∣0〉) ⨂ (y0∣0〉 − y1∣1〉)	σx	σz
∣Ψ+〉	∣Φ−〉	∣1〉	∣1〉	(x0∣0〉 + x1∣1〉) ⨂ (y0∣1〉 − y1∣0〉)	I	σz σx
∣Ψ+〉	∣Ψ+〉	∣0〉	∣0〉	(x0∣1〉 + x1∣0〉) ⨂ (y0∣1〉 + y1∣0〉)	σx	σx
∣Ψ+〉	∣Ψ+〉	∣1〉	∣1〉	(x0∣0〉 + x1∣1〉) ⨂ (y0∣0〉 + y1∣1〉)	I	I
∣Ψ+〉	∣Ψ−〉	∣0〉	∣0〉	(x0∣1〉 + x1∣0〉) ⨂ (y0∣1〉 − y1∣0〉)	σx	σz σx
∣Ψ+〉	∣Ψ−〉	∣1〉	∣1〉	(x0∣0〉 + x1∣1〉) ⨂ (y0∣0〉 − y1∣1〉)	I	σz
∣Ψ−〉	∣Φ+〉	∣0〉	∣0〉	(x0∣1〉 − x1∣0〉) ⨂ (y0∣0〉 + y1∣1〉)	σz σx	I
∣Ψ−〉	∣Φ+〉	∣1〉	∣1〉	(x0∣0〉 − x1∣1〉) ⨂ (y0∣1〉 + y1∣0〉)	σz	σx
∣Ψ−〉	∣Φ−〉	∣0〉	∣0〉	(x0∣1〉 − x1∣0〉) ⨂ (y0∣0〉 − y1∣1〉)	σz σx	σz
∣Ψ−〉	∣Φ−〉	∣1〉	∣1〉	(x0∣0〉 − x1∣1〉) ⨂ (y0∣1〉 − y1∣0〉)	σz	σz σx
∣Ψ−〉	∣Ψ+〉	∣0〉	∣0〉	(x0∣1〉 − x1∣0〉) ⨂ (y0∣1〉 + y1∣0〉)	σz σx	σx
∣Ψ−〉	∣Ψ+〉	∣1〉	∣1〉	(x0∣0〉 − x1∣1〉) ⨂ (y0∣0〉 + y1∣1〉)	σz	I
∣Ψ−〉	∣Ψ−〉	∣0〉	∣0〉	(x0∣1〉 − x1∣0〉) ⨂ (y0∣1〉 − y1∣0〉)	σz σx	σz σx
∣Ψ−〉	∣Ψ−〉	∣1〉	∣1〉	(x0∣0〉 − x1∣1〉) ⨂ (y0∣0〉 − y1∣1〉)	σz	σz

^a Alice's measurement result. ^b Bob's measurement result. ^c Charlie's measurement result. ^d David's measurement result. ^e Alice's Unitary operator. ^f Bob's Unitary operator.

In this subsection, five characters are introduced. Alice as the first senders (also as the third receiver), Bob as the second senders (also as the first receiver), Evelyn as the third sender (also as the second receiver), Charlie as the first controller, and David as the second controller. Taking the controllers' OR-logic-gate-like behaviour in the previous subsection, we built the channel state by using equation(4) and preparing other two ∣0〉 qubit states as ∣ψ〉₁₂₃₄₅₆ ⨂ ∣00〉₇₈. Qubits 7 and 8 will be used as the third transmission channel in the cyclic scheme. We perfrom a Hadamard gate on qubit 7 followed by a CNOT gate on qubit 8 with qubit 7 as the control qubit. After that, we apply another CNOT gate on qubit 7 with qubit 3 as the control qubit. This resulting a state

$$\begin{eqnarray}\begin{array}{rcl}| \psi {\rangle }_{12345678} & = & \displaystyle \frac{1}{4}\left(| 00000000\rangle +| 01111010\rangle +| 10010000\rangle +| 11101010\rangle \right.\\ & & +| 00001100\rangle +| 01110110\rangle +| 10011100\rangle +| 11100110\rangle \\ & & +| 00000011\rangle +| 01111001\rangle +| 10010011\rangle +| 11101001\rangle \\ & & \left.+| 00001111\rangle +| 01110101\rangle +| 10011111\rangle +| 11100101\rangle \right).\end{array}\end{eqnarray} \tag{ 9 }$$

The channel qubit is distributed to the five parties as $| \psi {\rangle }_{{A}_{1}{{CDB}}_{2}{C}_{2}{B}_{1}{A}_{2}{C}_{1}}=| \psi {\rangle }_{12345678}$. Qubit A₁ and A₂ belong to Alice, B₁ and B₂ belong to Bob, C belongs to Charlie, D belongs to David, E₁ and E₂ belong to Evelyn.

Alice, Bob, and Evelyn, have an arbitrary state

$$\begin{eqnarray}&&| \alpha {\rangle }_{a}={x}_{0}| 0\rangle +{x}_{1}| 1\rangle ,\end{eqnarray} \tag{ 10 }$$

$$\begin{eqnarray}&&| \beta {\rangle }_{b}={y}_{0}| 0\rangle +{y}_{1}| 1\rangle ,\end{eqnarray} \tag{ 11 }$$

$$\begin{eqnarray}&&| \epsilon {\rangle }_{e}={z}_{0}| 0\rangle +{z}_{1}| 1\rangle ,\end{eqnarray} \tag{ 12 }$$

respectively, with x₀, x₁, y₀, y₁, z₀ and z₁ are complex numbers satisfying normalization condition ${{\rm{\Sigma }}}_{i=0}^{1}| {x}_{i}{| }^{2}=1$, ${{\rm{\Sigma }}}_{i=0}^{1}| {y}_{i}{| }^{2}=1$, and ${{\rm{\Sigma }}}_{i=0}^{1}| {z}_{i}{| }^{2}=1$. Alice wants to teleport her state to Bob, Bob to Evelyn, and Evelyn to Alice. They start the teleportation process by mixing their initial state with the channel state as $| {\rm{\Psi }}{\rangle }_{{{abeA}}_{1}{{CDB}}_{2}{E}_{2}{B}_{1}{A}_{2}{E}_{1}}=| \alpha {\rangle }_{a}\otimes | \beta {\rangle }_{b}\otimes | \epsilon {\rangle }_{e}\otimes | \psi {\rangle }_{{A}_{1}{{CDB}}_{2}{E}_{2}{B}_{1}{A}_{2}{E}_{1}}$. The teleportation circuit scheme is shown in figure 3.

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Alice, Bob, and Evelyn perform a Bell State Measurement independently on qubit aA₁, bB₁, and eE₁, respectively. Suppose their measurement results are all ∣Φ⁺〉, the remaining qubits collapse to

$$\begin{eqnarray}\begin{array}{rcl}| \psi {\rangle }_{{{CDB}}_{2}{C}_{2}{A}_{2}} & = & \displaystyle \frac{1}{4}\left({x}_{0}{y}_{0}{z}_{0}| 00000\rangle +{x}_{0}{y}_{0}{z}_{0}| 11111\rangle +{x}_{1}{y}_{0}{z}_{0}| 00100\rangle \right.\\ & & +{x}_{1}{y}_{0}{z}_{0}| 11011\rangle +{x}_{0}{y}_{1}{z}_{0}| 00010\rangle +{x}_{0}{y}_{1}{z}_{0}| 11101\rangle \\ & & +{x}_{1}{y}_{1}{z}_{0}| 00110\rangle +{x}_{1}{y}_{1}{z}_{0}| 11001\rangle +{x}_{0}{y}_{0}{z}_{1}| 00001\rangle \\ & & +{x}_{0}{y}_{0}{z}_{1}| 11110\rangle +{x}_{1}{y}_{0}{z}_{1}| 00101\rangle +{x}_{1}{y}_{0}{z}_{1}| 11010\rangle \\ & & +{x}_{0}{y}_{1}{z}_{1}| 00011\rangle +{x}_{0}{y}_{1}{z}_{1}| 11100\rangle +{x}_{1}{y}_{1}{z}_{1}| 00111\rangle \\ & & \left.+{x}_{1}{y}_{1}{z}_{1}| 11000\rangle \right).\end{array}\end{eqnarray} \tag{ 13 }$$

Similarly, as in the previous subsections, Charlie and David, as the controllers, perform a computational bases measurement independently. Suppose their measurement results are ∣0〉, the remaining qubits can be written as

$$\begin{eqnarray}&&| \psi {\rangle }_{{B}_{2}{C}_{2}{A}_{2}}=\left({x}_{0}| 0\rangle +{x}_{1}| 1\rangle \right)\otimes \left({y}_{0}| 0\rangle +{y}_{1}| 1\rangle \right)\otimes \left({z}_{0}| 0\rangle +{z}_{1}| 1\rangle \right)\end{eqnarray} \tag{ 14 }$$

Alice informs her measurement result to Bob, Bob to Evelyn, and Evelyn to Alice. Charlie and David broadcast their measurement result to the other three parties. Using the information received, for this case, Alice, Bob, and Evelyn perform nothing on their qubit. Equation (14) shows that Alice already receives Evelyn's initial state, as well as Bob and Evelyn receive Alice and Bob's initial state, respectively. The other possible measurement results are evaluated similarly. As in the previous two subsections, each receiver cannot rebuilt their state without the help of at least one of the two controllers, which again, micics the OR logic gate.

The work in [43] proposed a tridirectional quantum teleportation via a twenty-one-qubit state as the quantum channel, with each seven-qubit state facilitating the transmission in each direction. The work shows that any one of the three parties can send their arbitrarily entangled states to the other two simultaneously. This subsection protocol can also be used in another tridirectional scheme, i.e., asymmetric tridirectional controlled teleportation, in which one of the parties can send a single qubit state and a two-qubit state into another two with OR-logic-gate-like controllers via a twenty-four qubit channel state, which each eight-qubit state responsible for the transmission between each two parties. The channel used in each transmission is $| \psi {\rangle }_{{A}_{1}{{CDB}}_{1}{A}_{2}{B}_{2}{A}_{3}{B}_{3}}=| \psi {\rangle }_{12345678}$, where ∣ψ〉_12345678 is from equation (9), in which a single qubit state is transmitted from A₁ to B₁, and a two-qubit state is transmitted from B₂ B₃ to A₂ A₃ with one of C and D's cooperations. The whole tridirectional channel state is $| \psi {\rangle }_{{A}_{1}{{CDB}}_{1}{A}_{2}{B}_{2}{A}_{3}{B}_{3}}\otimes | \psi {\rangle }_{{B}_{1}{{CDE}}_{1}{B}_{2}{E}_{2}{B}_{3}{E}_{3}}\otimes | \psi {\rangle }_{{E}_{1}{{CDA}}_{1}{E}_{2}{A}_{2}{E}_{3}{A}_{3}}$. This channel still uses this subsection's channel state in equation (9), indicating that the OR-logic-gate-like controllers properties are still applied, i.e., the tridirectional transmission can succeed with one of the two controllers' cooperation, and it does not matter which one.

In this subsection, we generalize the previous subsections' result to build a Multi-Way Quantum Controlled Teleportation with N controllers. All of the protocols, i.e., the one-way, bidirectional, and cyclic, has the similar two controllers combination, that is mimicing the OR logic gate. These properties come from the nature of how the channel state is entangled. If we observe the channel used in the three protocols, equations (1), (4) and (9), the controllers' qubit (qubit 2 and 3) always has the same state in each term, which is if we measure one of them in computational bases, the remaining controller's qubit will be in an untangled state (resulting the OR logic gate properties). Then, we can build the one-way protocol with N controllers using a channel state

$$\begin{eqnarray}\begin{array}{rcl}| \pi {\rangle }_{{S}_{1}{C}_{1}{C}_{2}...{C}_{N}{R}_{1}} & = & \displaystyle \frac{1}{2}(\left|0\right.\mathop{\underbrace{00...0}}\limits_{N}\left.0\right\rangle +\left|0\right.\mathop{\underbrace{11...1}}\limits_{N}\left.1\right\rangle +\left|1\right.\mathop{\underbrace{00...0}}\limits_{N}\left.1\right\rangle \\ & & +\left|1\right.\mathop{\underbrace{11...1}}\limits_{N}\left.0\right\rangle ),\end{array}\end{eqnarray} \tag{ 15 }$$

with S₁, R₁, and C_k for k = {1, 2,...,N} is the sender, the receiver, and the k^th controller, respectively. The channel state makes the one-way teleportation will success with only one of the controllers agreement, and it does not matter which one.

We observe the senders and receivers qubit on equations (4) and (9). In the last two subsections, we added the additional sender and receiver by preparing two ∣0〉 states and transform it into a Bell State, i.e., by performing a Hadamard followed by a CNOT gate, then entangling it with the channel state by performing another CNOT gate on one of the qubit in the Bell state with qubit 3 as the control qubit, see sub 2.2 and 2.3. We can use the idea to build a channel state that can be used by multi-senders and multi-receivers. The channel building operation is shown in equation (16).

$$\begin{eqnarray}\begin{array}{rcl}| \psi {\rangle }_{{S}_{1}{C}_{1}{C}_{2}...{C}_{N}{R}_{1}{R}_{2}{S}_{2}...{R}_{M}{S}_{M}} & = & \displaystyle \prod _{i=2}^{N}{\mathrm{CNOT}}_{{C}_{2}i}\displaystyle \prod _{j=2}^{M}{\mathrm{CNOT}}_{{R}_{j}{S}_{j}}{{\rm{H}}}_{{R}_{j}}\\ & & \otimes | \pi {\rangle }_{{S}_{1}{C}_{1}{C}_{2}...{C}_{N}{R}_{1}}| 00...0{\rangle }_{{S}_{2}{R}_{2}...{R}_{M}{S}_{M}}.\end{array}\end{eqnarray} \tag{ 16 }$$

The CNOT${}_{{R}_{j}{S}_{j}}$ and ${{\rm{H}}}_{{R}_{j}}$ gates are entangling the qubit R_j and S_j , the CNOT${}_{{C}_{2}i}$ is entangling the qubit R_j S_j with Qubit S₁ C₁ C₂...C_N R₁. The CNOT${}_{{R}_{j}{S}_{j}}$ and ${{\rm{H}}}_{{R}_{j}}$ gates must be applied first before the CNOT${}_{{C}_{2}i}$ gate. The resulting state can be used as the channel to teleport a single state form sender S_k to receiver R_k with the cooperation of one of the controlles C₁ to C_N , with k = {1, 2, 3,...,M}.

The protocol detail is similar with the previous proposed protocols. Suppose each sender S_k has an arbitrary qubit $| {S}_{k}{\rangle }_{{\mathrm{initial}}_{k}}={a}_{k}| 0\rangle +{b}_{k}| 1\rangle$ with a_k and b_k are complex numbers satisfying normalization conditions ${\left|{a}_{k}\right|}^{2}+{\left|{b}_{k}\right|}^{2}=1$, for k = {1, 2, 3,...,M}. Each sender mixes their qubit into the corresponding channel qubit and performs a Bell State Measurement, i.e., on qubit initial_k and S_k . After that, the controllers perform computational bases measurement. Each sender and the controllers (who agreed to cooperate) send their measurement results to the corresponding receiver, i.e., S_k to R_k and the controllers broadcast to all of the receivers. Each receiver performs an appropriate unitary operation and the teleportaion is finish. The explanation implies that the Multi-Way Quantum Controlled Teleportation with N Controller protocol can be achieved by using a channel state in equation (16), Bell State measurement by the senders, computational bases measurement by the controllers, assisted with classical communication followed by unitary operations. The receivers need only one of the controllers' agreement (and it does not matter which one) to rebuild the corresponding senders' initial state.

As a comparison, the work in [44] evaluates a controlled quantum broadcast protocol in which a receiver can send 2N qubit states into 2N receivers. This subsection protocol can also be modified into the broadcast one. It can be done by setting the ownership of all S_k qubits in equation(16) into only one sender. It modifies the protocol into the broadcast one in which a sender can send M single qubit states into M receivers with the cooperation of N controllers. The modification does not change the controllers' qubits; hence, the OR-logic-gate-like controllers' combination properties remain. Then, even though there are N controllers, the resulting broadcast teleportation only needs one of the controllers' cooperation, and it does not matter which one.

We evaluate the phase-damping noise case first. We choose M = 1 for simplicity. The protocol reduces to one-way scheme as in section 2.1. Equation (19) reduces to ${\xi }^{P}\left(\rho \right)={\sum }_{{m}_{1}}{\sum }_{{m}_{2}}{({E}_{{m}_{1}}^{P})}_{{S}_{1}}{({E}_{{m}_{2}}^{P})}_{{R}_{1}}\rho {({E}_{{m}_{1}}^{\dagger P})}_{{S}_{1}}{({E}_{{m}_{2}}^{\dagger P})}_{{R}_{1}}$, with $\rho =\tfrac{1}{4}\left(| 0000\rangle +| 0111\rangle +| 1001\rangle +| 1110\rangle \right)\left(\langle 0000| +\langle 0111| +\langle 1001| +\langle 1110| \right)$, see equation (1). The equation can be written as

$$\begin{eqnarray}\begin{array}{rcl}{\xi }^{P}(\rho ) & = & \displaystyle \frac{1}{4}{\left(1-{\eta }_{P}\right)}^{2}(| 0000\rangle +| 0111\rangle +| 1001\rangle +| 1110\rangle )(\langle 0000| +\langle 0111| \\ & & +\langle 1001| +\langle 1110| )+\displaystyle \frac{1}{4}(1-{\eta }_{P}){\eta }_{P}[(| 0000\rangle +| 0111\rangle (\langle 0000| )\\ & & +\langle 0111| )+(| 1001\rangle +| 1110\rangle )(\langle 1001| +\langle 1110| )+(| 0000\rangle \\ & & +| 1110\rangle )(\langle 0000| +\langle 1110| )+(| 0111\rangle +| 1001\rangle )(\langle 0111| \\ & & +\langle 1001| )]+\displaystyle \frac{1}{4}{\eta }_{P}^{2}(| 0000\rangle \langle 0000| +| 1110\rangle \langle 1110| \\ & & +| 0111\rangle \langle 0111| +| 1001\rangle \langle 1001| ),\end{array}\end{eqnarray} \tag{ 22 }$$

The trace of equation (22) is always equal to one (we mention it for the discussion in the next section). By using equation (20), after detailed calculations, the output density operator is

$$\begin{eqnarray}\begin{array}{rcl}{\rho }_{{out}}^{P} & = & {\left(1-{\eta }_{P}\right)}^{2}({x}_{0}| 0\rangle +{x}_{1}| 1\rangle )({x}_{0}^{* }\langle 0| +{x}_{1}^{* }\langle 1| )\\ & & +(2{\eta }_{P}-{\eta }_{P}^{2})(| {x}_{0}{| }^{2}| 0\rangle \langle 0| +| {x}_{1}{| }^{2}| 1\rangle \langle 1| ),\end{array}\end{eqnarray} \tag{ 23 }$$

for any sender and controllers' measurement result. From equation (21) the fidelity value is

$$\begin{eqnarray}&&{F}^{P}={\left(1-{\eta }_{P}\right)}^{2}+(2{\eta }_{P}-{\eta }_{P}^{2})(| {x}_{0}{| }^{4}+| {x}_{1}{| }^{4}),\end{eqnarray} \tag{ 24 }$$

which depends on the decoherence rate and initial state's amplitude parameters only. To illustrate it, we plot the fidelity for vaious η_P and ∣x₀∣² value as shown in figure 4(a). Suppose we take the acceptable fidelity value to be equal to 0.8. Then, we add a contour line to divide the greater or lower area than the value. The region outside the contour represents the greater-than case. The figure shows that the protocol works well (the fidelity is always greater than the acceptable value for any initial state) for a relatively low decoherence rate of about 0.2 or less. Besides that, the protocol only has an acceptable fidelity for some particular initial states. However, if ∣x₀∣² = {0, 1}, the fidelity always equals one, even in a noisy environment.

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As an alternative method in the noise evaluation, we also calculate the entropy of the output density operator. Using the Von Neumann entropy formula, $S({\rho }_{{out}}^{r})=-\mathrm{tr}({\rho }_{{out}}^{r}\mathrm{log}({\rho }_{{out}}^{r}))$, we plot the entropy (S) of some initial states and the corresponding fidelity (F) value as shown in figures 4(b)–(d). The plots show that the entropy analysis agreed with the fidelity value, i.e., the lower the F, the higher the S. The higher S means higher information loss, implying lower F, indicating that the teleported state is less similar to the initial state. For some particular cases that satisfy ∣x₀∣² = {0, 1}, the output density operator in equation (23) reduces to a pure state, which has a zero-value entropy, meaning there is no information loss. This also agrees with the fidelity value that is always equal to one.

For the bidirectional, cyclic, or the other M values schemes, the analysis can be done similarly. We evaluate the amplitude-damping noise in the following subsection.

Similarly as in the previous subsections, we choose M = 1. Equation (19) reduces to ${\xi }^{A}\left(\rho \right)={\sum }_{{m}_{1}}{\sum }_{{m}_{2}}{({E}_{{m}_{1}}^{A})}_{{S}_{1}}{({E}_{{m}_{2}}^{A})}_{{R}_{1}}\rho {({E}_{{m}_{1}}^{\dagger A})}_{{S}_{1}}{({E}_{{m}_{2}}^{\dagger A})}_{{R}_{1}}$, or can be written as

$$\begin{eqnarray}\begin{array}{rcl}{\xi }^{A}\left(\rho \right) & = & \displaystyle \frac{1}{4}\left(| 0000\rangle +\sqrt{1-{\eta }_{A}}| 0111\rangle +(1-{\eta }_{A})| 1001\rangle +\sqrt{1-{\eta }_{A}}| 1110\rangle \right)\\ & & \otimes \left(\langle 0000| +\sqrt{1-{\eta }_{A}}\langle 0111| +(1-{\eta }_{A})\langle 1001| +\sqrt{1-{\eta }_{A}}\langle 1110| \right)\\ & & +\displaystyle \frac{1}{4}{\eta }_{A}\left[\left(\sqrt{1-{\eta }_{A}}| 0001\rangle +| 0110\rangle \right)\left(\sqrt{1-{\eta }_{A}}\langle 0001| +\langle 0110| \right)\right.\\ & & +\left.\left(| 0110\rangle +\sqrt{1-{\eta }_{A}}| 1000\rangle \right)\left(\langle 0110| +\sqrt{1-{\eta }_{A}}\langle 1000| \right)\right]\\ & & +\displaystyle \frac{1}{4}{\eta }_{A}^{2}| 0000\rangle \langle 0000| .\end{array}\end{eqnarray} \tag{ 25 }$$

Its trace is always equal to one (again, we mention it for the discussion in the next section). Using equation (20), one can show that the output density operator depends on the sender and controllers' measurement results. Then, we present the operator as a statistical mixture ${\rho }_{{out}}^{A}={\sum }_{i}{P}_{i}{\rho }_{i}$ with ρ_i is the possible output density operator, and P_i is its corresponding probability. After detailed calculations, it can be written as

$$\begin{eqnarray}\begin{array}{rcl}{\rho }_{{out}}^{A} & = & \displaystyle \frac{1}{4}\left(\left({x}_{0}| 0\rangle +(1-{\eta }_{A}){x}_{1}| 1\rangle \right)\left({x}_{0}^{* }\langle 0| +(1-{\eta }_{A}){x}_{1}^{* }\langle 1| \right)\right.\\ & & +\left((1-{\eta }_{A}){x}_{0}| 0\rangle +{x}_{1}| 1\rangle \right)\left((1-{\eta }_{A}){x}_{0}\langle 0| +{x}_{1}\langle 1| \right)\\ & & +2(1-{\eta }_{A})\left({x}_{0}| 0\rangle +{x}_{1}| 1\rangle \right)\left({x}_{0}^{* }\langle 0| +{x}_{1}^{* }\langle 1| \right)\\ & & +2{\eta }_{A}(2-{\eta }_{A})\left(| {x}_{0}{| }^{2}| 1\rangle \langle 1| +| {x}_{1}{| }^{2}| 0\rangle \langle 0| \right)\\ & & \left.+{\eta }_{A}^{2}\left(| {x}_{0}{| }^{2}| 0\rangle \langle 0| +| {x}_{1}{| }^{2}| 1\rangle \langle 1| \right)\right).\end{array}\end{eqnarray} \tag{ 26 }$$

Using equation (21), the mixture produces an average fidelity value equation, i.e.,

$$\begin{eqnarray}\begin{array}{rcl}{F}^{A} & = & \displaystyle \frac{1}{4}\{{\left(| {x}_{0}{| }^{2}+(1-{\eta }_{A})| {x}_{1}{| }^{2}\right)}^{2}+{\left((1-{\eta }_{A})| {x}_{0}{| }^{2}+| {x}_{1}{| }^{2}\right)}^{2}\\ & & +2(1-{\eta }_{A})+4{\eta }_{A}(2-{\eta }_{A})| {x}_{0}{| }^{2}| {x}_{1}{| }^{2}+{\eta }_{A}^{2}\left(| {x}_{0}{| }^{4}+| {x}_{1}{| }^{4}\right)\},\end{array}\end{eqnarray} \tag{ 27 }$$

which only depends on the decoherence rate and the initial state amplitude parameters only. We illustrate the fidelity by plotting it with respect to the parameters as shown in figure 5(a). We also added a contour line to represent acceptable fidelity value (0.8). The area above the contour line has a greater than 0.8 value. We also interested in the entropy evaluation of equation (26). We plot the graph for some initial state to illustrate it, as shown in figures 5(b) and (c), showing that the entropy analysis also agreed with the fidelity analysis, similarly as in the PDN case.

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We compare the fidelity of the ADN and the PDN case by plotting it with respect to the decoherence rate parameter for various initial qubit state amplitude parameter, as shown in figures 6(a)–(c). The comparison shows that the teleportation performs better under the PDN for relatively small (and large) ∣x₀∣² parameter value, i.e., values close to zero or one. However, for ∣x₀∣² = 0.5, it performs better under the ADN for all of the decoherence rate parameter.

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Our noise analysis assumption (the last sentence of the second paragraph of this section) is different from other previous works [29, 34, 35, 38], which will be discussed in the next section.