1. Introduction
Quantum entanglement is a pivotal concept in quantum information technology on which most of the applications of quantum information theory are dependent. With the rapid development of quantum technology in recent times, quantum communication theory is becoming more attractive to physicists, mathematicians and engineers throughout the globe. The use of quantum entanglement has been widely explored in the different fields of quantum communication such as quantum teleportation (QT) [1, 2], remote state preparation (RSP) [3, 4], joint remote state preparation (JRSP) [5, 6], quantum secure dense coding [7, 8], quantum key distribution [9, 10] and quantum secret sharing [11, 12].
QT protocol was introduced by Bennett et al [1] in 1993 through which an arbitrary single-qubit quantum state was transferred to a distant party using a pre-shared maximally entangled state as a quantum channel with the help of some classical communication. Teleportation protocols can be classified into various types based on their nature. These include bidirectional quantum teleportation [13, 14], controlled bidirectional quantum teleportation [15], probabilistic teleportation [16], cyclic quantum teleportation [17], multi-hop quantum teleportation [18], conference teleportation (CT) [19], etc. They are designed to serve different purposes in the science of quantum communication. The requirements of entangled resources also vary with these variants of teleportation.
RSP protocols constitute another important class of protocols similar to QT, first introduced by Pati [20] in 2000. Its central feature lies in the knowledge of the quantum state to be transmitted. In RSP protocols, the sender knows the quantum state that the sender wants to prepare on the receiver's end whereas in QT the quantum state is arbitrary at least within a specified class. Following the work of Pati [20], several RSP protocols have been published [21–24].
Furthermore, for security and confidentiality reasons, it may be undesirable for a single party to access all the information about the quantum state to be transmitted. In addition, there are situations where the complete information of the state is not available to a single party due to some technical constraints. Instead it is distributed between two parties. In these cases, a new protocol called JRSP was introduced by Xia et al [25] in 2007 in which two parties in possession of partial information cooperate to prepare the state at the receiver's end. Several variants of JRSP intended for different purposes have appeared in works such as [26–30].
In a protocol for performing multiple tasks of quantum communication, it may be the case that the importance is not the same for all the tasks. There may be different levels of importance attached to them and therefore it may be desirable to treat them differently within the framework of the same integrated protocol. The above consideration has produced the concept of hierarchical quantum communication protocol.
Hierarchy was introduced in quantum information splitting (QIS) by Gottesman [31] who introduced a scheme where the participants were endowed with different levels of access to the split information. QIS protocols with different hierarchical concepts between the participants were discussed by many authors such as Wang et al [32], Shukla et al [33], Ma et al [34], etc. Hierarchical teleportation was discussed by Bich and An [35] in which teleportation under different levels of control was considered. Hierarchy in the context of RSP protocol was discussed by Ma et al [36]. JRSP protocol for the preparation of a single-qubit state with hierarchy of power amongst three receivers was discussed by Chen et al in [37]. Hierarchical JRSP in a noisy environment using a five-qubit cluster state has been discussed by Shukla et al in [38]. Furthermore, a bidirectional teleportation scheme with different levels of control creating hierarchy has been proposed by Cao et al [39] while a hierarchically controlled asymmetric bidirectional teleportation scheme for a single- and a two-qubit state was discussed by Yang [40]. Several more recent works discussed both teleportation and RSP protocols with parties possessing different levels of powers for the purpose of producing the states intended for transfer at their respective ends for which some references are [41–45].
In the present study, we consider one party, Alice, who intends to create a two-qubit state at the site of Bob while the parties Bob and Eve together intend to produce a three-qubit state at the site of Alice. The two-qubit state is fully known to Alice while the information of the three-qubit state is divided between Bob and Eve. The protocol is a bidirectional protocol being a hybrid combination of an RSP and a JRSP scheme. There are two controllers, Charlie and David. The RSP intended by Alice is controlled by Charlie only while the accomplishment of JRSP requires the action of both the controllers, Charlie and David. The protocol is represented through a quantum circuit and is run on an IBM quantum platform using IBM's Qiskit package. Furthermore, we consider the effect of noise on our protocol. Since entanglement is fragile in nature, it is very much affected whenever there is a scope of interaction with the environment. In particular, this occurs when the qubits constituting parts of the quantum channel we used are distributed amongst different parties. The quantum noise is represented through Kraus operators. As particular cases, we consider amplitude-damping, bit-flip and phase-flip noise and study their effect on our protocol by analysis of fidelity, which is identically of unit value when there is no noise.
2. Main protocol
Let us consider a scenario where Alice aims to remotely prepare a known two-qubit state ∣φ⟩ in Bob's laboratory and at the same time, Bob and Eve concurrently seek to jointly prepare a known entangled state ∣ψ⟩ in Alice's laboratory. Here, the states ∣φ⟩ and ∣ψ⟩ are defined as follows:
$\begin{eqnarray}| \phi \rangle =a(| 00\rangle +| 11\rangle )+b(| 01\rangle +| 10\rangle ),\end{eqnarray}$
$\begin{eqnarray}| \psi \rangle ={x}_{0}| 000\rangle +{y}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 111\rangle ,\end{eqnarray}$
where, the coefficients a, b, x0 and y0 are all assumed to be real and satisfy the normalization conditions a2 + b2 = 1 and ${x}_{0}^{2}+{y}_{0}^{2}=1$. In addition, it is assumed that θ adheres to the relation 0 ≤ θ < 2π. Specifically, Alice possesses knowledge of coefficients a and b, whereas Bob holds information regarding coefficients x0 and y0, and only Eve possesses knowledge of θ.In our proposed protocol, the state that Bob and Eve intend to jointly prepare at Alice's site is more important, which is overseen by controllers Charlie and David. In contrast, the state that Alice aims to remotely prepare at Bob's site is less important and is only controlled by Charlie.
To accomplish the task defined above, all the involved parties, namely, Alice, Bob, Charlie, David and Eve share a ten-qubit entangled state, which is given by:
$\begin{eqnarray}\begin{array}{rcl}| {\rm{\Xi }}\rangle & = & \displaystyle \frac{1}{2\sqrt{2}}\left[| 0000111000\rangle +| 0110111000\rangle \right.\\ & & -\ | 0001000111\rangle -| 0111000111\rangle -| 1011000011\rangle \\ & & -\ | 1101000011\rangle +| 1010111100\rangle \\ & & {\left.+\ | 1100111100\rangle \right]}_{{A}_{1}{B}_{1}{B}_{3}{B}_{2}{A}_{2}{A}_{3}{A}_{4}{CDE}}.\end{array}\end{eqnarray}$
Qubits A1, A2, A3 and A4 belong to Alice, qubits B1, B2 and B3 are in the hands of Bob, Charlie possesses qubit C, David owns qubit D and Eve has qubit E.In the following, the proposed protocol is described step by step.
Step 1: Alice performs a single-qubit projective measurement on her qubit A1 in the basis {∣ν1⟩, ∣ν2⟩}, which is given as:
$\begin{eqnarray}\begin{array}{rcl}| {\nu }^{1}{\rangle }_{{A}_{1}} & = & \sqrt{2}(a| 0\rangle +b| 1\rangle ),\\ | {\nu }^{2}{\rangle }_{{A}_{1}} & = & \sqrt{2}(b| 0\rangle -a| 1\rangle ).\end{array}\end{eqnarray}$
Since Alice possesses knowledge of a and b, she can make a choice of basis as above. After the measurement, Alice publicly announces her measurement results via a classical channel.Step 2: At the same time, Bob executes a projective measurement on his qubit B2 in the basis {∣ξ1⟩, ∣ξ2⟩}, which is defined as:
$\begin{eqnarray}\begin{array}{rcl}| {\xi }^{1}\rangle & = & {x}_{0}| 0\rangle +{y}_{0}| 1\rangle ,\\ | {\xi }^{2}\rangle & = & {y}_{0}| 0\rangle -{x}_{0}| 1\rangle .\end{array}\end{eqnarray}$
Bob's choice is possible since he has knowledge of x0 and y0. After the measurement, Bob publicly announces his measurement results via a classical channel.Step 3: Eve performs a projective measurement on her particle E in the basis $\{| {\tau }_{1}^{j}\rangle ,| {\tau }_{2}^{j}\rangle \}$, which is dependent on Bob's measurement results as follows. If Bob's measurement yields ∣ξ1⟩, then Eve selects $\{| {\tau }_{1}^{1}\rangle ,| {\tau }_{2}^{1}\rangle \}$ as her basis. Otherwise, she opts for $\{| {\tau }_{1}^{2}\rangle ,| {\tau }_{2}^{2}\rangle \}$. The measurement basis $\{| {\tau }_{1}^{j}\rangle ,| {\tau }_{2}^{j}\rangle \}$ is defined as:
$\begin{eqnarray}\left(\begin{array}{c}| {\tau }_{1}^{j}\rangle \\ | {\tau }_{2}^{j}\rangle \end{array}\right)=\displaystyle \frac{1}{\sqrt{2}}{G}_{j}\left(\begin{array}{c}| 0\rangle \\ | 1\rangle \end{array}\right),\,j=1,2,\end{eqnarray}$
where $\begin{eqnarray}{G}_{1}=\left(\begin{array}{cc}1 & {{\rm{e}}}^{-{\rm{i}}\theta }\\ 1 & -{{\rm{e}}}^{-{\rm{i}}\theta }\end{array}\right)\rm{and}{G}_{2}=\left(\begin{array}{cc}{{\rm{e}}}^{-{\rm{i}}\theta } & 1\\ {{\rm{e}}}^{-{\rm{i}}\theta } & -1\end{array}\right).\end{eqnarray}$
This choice is possible since Eve has knowledge of the parameter θ.Step 4: There are two controllers, Charlie and David. If they are satisfied with all the measurement results from Alice, Bob and Eve, they measure their respective single-qubit and share their outcome with the respective parties. The creation of state ∣φ⟩ given in (1 ) in Bob's hands, needs assistance from the controller, Charlie. However, the joint creation of the state ∣ψ⟩ given in (2 ), at Alice's place, needs assistance from both the controllers, Charlie and David.
Both the controllers, Charlie and David, measure their respective qubits C and D in the basis given by:
$\begin{eqnarray}| {\eta }^{1}\rangle =\displaystyle \frac{1}{\sqrt{2}}(| 0\rangle +| 1\rangle ),\,| {\eta }^{2}\rangle =\displaystyle \frac{1}{\sqrt{2}}(| 0\rangle -| 1\rangle ),\end{eqnarray}$
and Charlie shares his measurement results with Alice and Bob through a classical channel, while David shares his measurement results with Alice via a classical channel.Step 5: Finally, Alice and Bob apply a unitary operation to reconstruct the desired respective states. According to the classical information about the measurement results from Bob, Eve, Charlie and David, Alice needs to apply a unitary operator on her qubits A2,A3 and A4 to reconstruct the state ∣ψ⟩ given in (2 ). At the same time, to correspond to the classical messages from Alice and Charlie, Bob needs to apply a unitary operation on his qubits B1 and B2 to reconstruct the state ∣φ⟩ given in (1 ).
This is the end of the protocol.
Now, we illustrate the protocol in detail.
Equation (3 ) can be rewritten as:4 ), $| {\xi }^{j}\rangle ^{\prime} s$ are given in equation (5 ) and $| {\tau }_{k}^{j}\rangle ^{\prime} s$ are given in equations (6 ) and (7 ). In addition, $| {M}_{{ijk}}\rangle ^{\prime} s$ are given by:9 ), it can be concluded that after Steps 1, 2 and 3 are performed, if Alice's, Bob's and Eve's measurement outcomes are $| {\nu }^{i}{\rangle }_{{A}_{1}}$, $| {\xi }^{j}{\rangle }_{{B}_{2}}$ and $| {\tau }_{k}^{j}{\rangle }_{E}$, respectively, then the state of the remaining qubits collapses into the state $| {M}_{{ijk}}{\rangle }_{{B}_{1}{B}_{3}{A}_{2}{A}_{3}{A}_{4}{CD}}$ with probability $\tfrac{1}{8}$. Suppose Alice's, Bob's and Eve's measurement results are $| {\nu }^{1}{\rangle }_{{A}_{1}}$, $| {\xi }^{1}{\rangle }_{{B}_{2}}$ and $| {\tau }_{1}^{1}{\rangle }_{E}$, respectively, then the reduced state of the remaining qubits B1, B3 A2, A3, A4, C and D is:11 ), it can be seen that irrespective of Charlie's measurement results, the state of qubits B1 and B3 is decoupled from the rest of the qubits. Thus, Bob can reconstruct the desired state ∣φ⟩ with the help of the controller, Charlie; there is no need for assistance from controller David. However, at this point, Alice cannot reconstruct the state ∣ψ⟩ because David's qubit D remains entangled with qubits A2, A3 and A4.
$\begin{eqnarray}| {\rm{\Xi }}\rangle =\displaystyle \frac{1}{2\sqrt{2}}\displaystyle \sum _{i,j,k=1}^{2}| {\nu }^{i}{\rangle }_{{A}_{1}}| {\xi }^{j}{\rangle }_{{B}_{2}}| {\tau }_{k}^{j}{\rangle }_{E}| {M}_{{ijk}}{\rangle }_{{B}_{1}{B}_{3}{A}_{2}{A}_{3}{A}_{4}{CD}},\end{eqnarray}$
where $| {\nu }^{i}\rangle ^{\prime} s$ are given in equation ( $\begin{eqnarray*}\begin{array}{rcl}| {M}_{111}\rangle & = & ({{ax}}_{0}| 0011100\rangle +{{ax}}_{0}| 1111100\rangle \\ & & -{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0000011\rangle -{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1100011\rangle \\ & & -{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0100001\rangle -{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1000001\rangle \\ & & +{{bx}}_{0}| 0111110\rangle +{{bx}}_{0}| 1011110\rangle ),\\ | {M}_{112}\rangle & = & ({{ax}}_{0}| 0011100\rangle +{{ax}}_{0}| 1111100\rangle \\ & & +{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0000011\rangle \\ & & +{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1100011\rangle \\ & & +{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0100001\rangle +{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1000001\rangle \\ & & +{{bx}}_{0}| 0111110\rangle +{{bx}}_{0}| 1011110\rangle ),\\ | {M}_{121}\rangle & = & ({{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0011100\rangle +{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1111100\rangle \\ & & +{{ax}}_{0}| 0000011\rangle +{{ax}}_{0}| 1100011\rangle \\ & & +{{bx}}_{0}| 0100001\rangle +{{bx}}_{0}| 1000001\rangle \\ & & +{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0111110\rangle \\ & & +{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1011110\rangle ),\\ | {M}_{122}\rangle & = & ({{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0011100\rangle +{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1111100\rangle \\ & & -{{ax}}_{0}| 0000011\rangle -{{ax}}_{0}| 1100011\rangle \\ & & -{{bx}}_{0}| 0100001\rangle -{{bx}}_{0}| 1000001\rangle \\ & & +{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0111110\rangle \\ & & +{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1011110\rangle ),\\ | {M}_{211}\rangle & = & ({{bx}}_{0}| 0011100\rangle +{{bx}}_{0}| 1111100\rangle \\ & & -{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0000011\rangle \\ & & -{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1100011\rangle \\ & & +{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0100001\rangle +{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1000001\rangle \\ & & -{{ax}}_{0}| 0111110\rangle -{{ax}}_{0}| 1011110\rangle ),\\ | {M}_{212}\rangle & = & ({{bx}}_{0}| 0011100\rangle +{{bx}}_{0}| 1111100\rangle +{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0000011\rangle \\ & & +{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1100011\rangle \\ & & -{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0100001\rangle -{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1000001\rangle \\ & & -{{ax}}_{0}| 0111110\rangle -{{ax}}_{0}| 1011110\rangle ),\\ | {M}_{221}\rangle & = & ({{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0011100\rangle +{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1111100\rangle \\ & & +{{bx}}_{0}| 0000011\rangle +{{bx}}_{0}| 1100011\rangle \\ & & -{{ax}}_{0}| 0100001\rangle -{{ax}}_{0}| 1000001\rangle \\ & & -{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0111110\rangle \\ & & -{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1011110\rangle ),\\ | {M}_{222}\rangle & = & ({{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0011100\rangle +{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1111100\rangle \\ & & -{{bx}}_{0}| 0000011\rangle -{{bx}}_{0}| 1100011\rangle \\ & & +{{ax}}_{0}| 0100001\rangle +{{ax}}_{0}| 1000001\rangle \\ & & -{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0111110\rangle \\ & & -{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1011110\rangle ).\end{array}\end{eqnarray*}$
From the above equation ( $\begin{eqnarray}\begin{array}{rcl}| {M}_{111}\rangle & = & \left({{ax}}_{0}| 0011100\rangle +{{ax}}_{0}| 1111100\rangle -{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0000011\rangle \right.\\ & & -{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1100011\rangle \\ & & -{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0100001\rangle -{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 1000001\rangle \\ & & {\left.+{{bx}}_{0}| 0111110\rangle +{{bx}}_{0}| 1011110\rangle \right)}_{{B}_{1}{B}_{3}{A}_{2}{A}_{3}{A}_{4}{CD}}.\end{array}\end{eqnarray}$
The above state ∣M111⟩ can be written as: $\begin{eqnarray}\begin{array}{rcl}| {M}_{111}\rangle & = & \displaystyle \frac{1}{\sqrt{2}}| {\eta }^{1}{\rangle }_{C}\otimes \left[a(| 00\rangle +| 11\rangle )\right.\\ & & {\left.+b(| 01\rangle +| 10\rangle )\right]}_{{B}_{1}{B}_{3}}\\ & & \otimes {\left({x}_{0}| 1110\rangle -{y}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0001\rangle \right)}_{{A}_{2}{A}_{3}{A}_{4}D}\\ & & +\displaystyle \frac{1}{\sqrt{2}}| {\eta }^{2}{\rangle }_{C}\otimes \left[a(| 00\rangle -| 11\rangle )\right.\\ & & {\left.+b(| 01\rangle +| 10\rangle )\right]}_{{B}_{1}{B}_{3}}\\ & & \otimes {\left({x}_{0}| 1110\rangle +{y}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0001\rangle \right)}_{{A}_{2}{A}_{3}{A}_{4}D}.\end{array}\end{eqnarray}$
Controller Charlie measures his qubit C in the measurement basis {∣η1⟩, ∣η2⟩} described in Step 4. There are two possible outcomes in Charlie's measurement. From equation (If Charlie's measurement result is ∣η1⟩C, then the state of qubits B1 and B3 is a(∣00⟩ + ∣11⟩) + b(∣01⟩ + ∣10⟩), which is Bob's desired state. At the same time, the joint state of qubits A2, A3, A4 and D is given by:
$\begin{eqnarray}| q\rangle ={\left({x}_{0}| 1110\rangle -{y}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 0001\rangle \right)}_{{A}_{2}{A}_{3}{A}_{4}D}.\end{eqnarray}$
The above state ∣q⟩ can be written as: $\begin{eqnarray}\begin{array}{rcl}| q\rangle & = & \displaystyle \frac{1}{\sqrt{2}}| {\eta }^{1}{\rangle }_{D}\otimes {\left({x}_{0}| 111\rangle -{y}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 000\rangle \right)}_{{A}_{2}{A}_{3}{A}_{4}}\\ & & +\displaystyle \frac{1}{\sqrt{2}}| {\eta }^{2}{\rangle }_{D}\otimes {\left({x}_{0}| 111\rangle +{y}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 000\rangle \right)}_{{A}_{2}{A}_{3}{A}_{4}}.\end{array}\end{eqnarray}$
Now, the controller, David, makes a measurement on his qubit D in the basis described in Step 4. David will obtain two possible outcomes, ∣η1⟩D and ∣η2⟩D, with equal probability. David shares his measurement outcomes with Alice and then, based on that, Alice applies a unitary operator on his qubits A2, A3 and A4 to reconstruct the desired state ∣ψ⟩.If David's measurement result is ∣η1⟩D, then the reduced state of qubits A2, A3 and A4 is (x0∣111⟩ − y0eiθ∣000⟩), which is not the desired state. Thus, Alice needs to apply ${\left({\sigma }_{z}{\sigma }_{x}\right)}_{{A}_{2}}\otimes {\left({\sigma }_{x}\right)}_{{A}_{3}}\otimes {\left({\sigma }_{x}\right)}_{{A}_{4}}$ as a unitary operator, then the state will become the desired state ∣ψ⟩.
If David's measurement outcome is ∣η2⟩D, then the reduced state of qubits A2, A3 and A4 is (x0∣111⟩ + y0eiθ∣000⟩), which is also not the desired state. Thus, Alice needs to apply ${\left({\sigma }_{x}\right)}_{{A}_{2}}\otimes {\left({\sigma }_{x}\right)}_{{A}_{3}}\otimes {\left({\sigma }_{x}\right)}_{{A}_{4}}$ as a unitary operator, then the state will become the desired state ∣ψ⟩.
Table 1. Reduced state of Bob's qubits B1 and B3 and unitary operators for Bob's execution corresponding to the measurement results of Alice and Charlie. |
| Alice's measurement result (∣νi⟩) | Charlie's measurement result (∣ηl⟩) | State of qubits B1 and B3 | Bob's operation ${\sigma }_{{B}_{1}{B}_{3}}^{{il}}$ |
|---|---|---|---|
| ∣ν1⟩ | ∣η1⟩C | $\{a(| 00\rangle +| 11\rangle )+b(| 01\rangle +| 10\rangle )\}{}_{{B}_{1}{B}_{3}}$ | I ⨂ I |
| ∣ν1⟩ | ∣η2⟩C | $\{a(| 00\rangle +| 11\rangle )-b(| 01\rangle +| 10\rangle )\}{}_{{B}_{1}{B}_{3}}$ | σz ⨂ σz |
| ∣ν2⟩ | ∣η1⟩C | $\{b(| 00\rangle +| 11\rangle )-a(| 01\rangle +| 10\rangle )\}{}_{{B}_{1}{B}_{3}}$ | σz ⨂ σxσz |
| ∣ν2⟩ | ∣η2⟩C | $\{b(| 00\rangle +| 11\rangle )+a(| 01\rangle +| 10\rangle )\}{}_{{B}_{1}{B}_{3}}$ | I ⨂ σx |
Table 2. Reduced state of Alice's qubits A2, A3 and A4 and unitary operators for execution by Alice corresponding to the measurement results of Bob, Eve, Charlie and David. |
| Bob's and Eve's measurement result $(| {\xi }^{j}\rangle ,| {\tau }_{k}^{j}\rangle )$ | Charlie's measurement result (∣ηl⟩) | David's measurement result (∣ηm⟩) | State of qubits A2, A3 and A4 | Alice's operation ${\sigma }_{{A}_{2}{A}_{3}{A}_{4}}^{{jklm}}$ |
|---|---|---|---|---|
| (∣ξ1⟩, $| {\tau }_{1}^{1}\rangle $) | ∣η1⟩C | ∣η1⟩D | x0∣111⟩ − y0eiθ∣000⟩ | σzσx ⨂ σx ⨂ σx |
| ∣η2⟩D | x0∣111⟩ + y0eiθ∣000⟩ | σx ⨂ σx ⨂ σx | ||
| | ||||
| (∣ξ1⟩, $| {\tau }_{1}^{1}\rangle $) | ∣η2⟩C | ∣η1⟩D | x0∣111⟩ + y0eiθ∣000⟩ | σx ⨂ σx ⨂ σx |
| ∣η2⟩D | x0∣111⟩ − y0eiθ∣000⟩ | σzσx ⨂ σx ⨂ σx | ||
| | ||||
| (∣ξ1⟩, $| {\tau }_{1}^{2}\rangle $) | ∣η1⟩C | ∣η1⟩D | x0∣111⟩ + y0eiθ∣000⟩ | σx ⨂ σx ⨂ σx |
| ∣η2⟩D | x0∣111⟩ − y0eiθ∣000⟩ | σzσx ⨂ σx ⨂ σx | ||
| | ||||
| (∣ξ1⟩, $| {\tau }_{1}^{2}\rangle $) | ∣η2⟩C | ∣η1⟩D | x0∣111⟩ − y0eiθ∣000⟩ | σzσx ⨂ σx ⨂ σx |
| ∣η2⟩D | x0∣111⟩ + y0eiθ∣000⟩ | σx ⨂ σx ⨂ σx | ||
| | ||||
| (∣ξ2⟩, $| {\tau }_{2}^{1}\rangle $) | ∣η1⟩C | ∣η1⟩D | x0∣000⟩ + y0eiθ∣111⟩ | I ⨂ I ⨂ I |
| ∣η2⟩D | −x0∣000⟩ + y0eiθ∣111⟩ | σxσzσx ⨂ I ⨂ I | ||
| | ||||
| (∣ξ2⟩, $| {\tau }_{2}^{1}\rangle $) | ∣η2⟩C | ∣η1⟩D | −x0∣000⟩ + y0eiθ∣111⟩ | σxσzσx ⨂ I ⨂ I |
| ∣η2⟩D | x0∣000⟩ + y0eiθ∣111⟩ | I ⨂ I ⨂ I | ||
| | ||||
| (∣ξ2⟩, $| {\tau }_{2}^{2}\rangle $) | ∣η1⟩C | ∣η1⟩D | −x0∣000⟩ + y0eiθ∣111⟩ | σxσzσx ⨂ I ⨂ I |
| ∣η2⟩D | x0∣000⟩ + y0eiθ∣111⟩ | I ⨂ I ⨂ I | ||
| | ||||
| (∣ξ2⟩, $| {\tau }_{2}^{2}\rangle $) | ∣η2⟩C | ∣η1⟩D | x0∣000⟩ + y0eiθ∣111⟩ | I ⨂ I ⨂ I |
| ∣η2⟩D | −x0∣000⟩ + y0eiθ∣111⟩ | σxσzσx ⨂ I ⨂ I | ||
3. Generation of the entangled channel
The generation of the quantum channel is important because, without the quantum channel, it is impossible to perform quantum information transfer. The generation of the ten-qubit quantum channel used in our protocol can be done by taking the initial state of all the ten qubits as ∣0⟩ and applying different quantum gates on the qubits. The steps are the following:
| • | Step 1: Consider the initial state of the ten qubits as: $\begin{eqnarray}\begin{array}{rcl}| {{\rm{\Xi }}}_{0}\rangle & = & | 0{\rangle }_{{q}_{0}}\otimes | 0{\rangle }_{{q}_{1}}\otimes | 0{\rangle }_{{q}_{2}}\otimes | 0{\rangle }_{{q}_{3}}\otimes | 0{\rangle }_{{q}_{4}}\\ & & \otimes | 0{\rangle }_{{q}_{5}}\otimes | 0{\rangle }_{{q}_{6}}\otimes | 0{\rangle }_{{q}_{7}}\otimes | 0{\rangle }_{{q}_{8}}\otimes | 0{\rangle }_{{q}_{9}}.\end{array}\end{eqnarray}$ |
| • | Step 2: A Hadamard gate (H) is applied to each of the three qubits q0, q1 and q4. An X-gate is applied to qubit q3, then the state ∣Ξ0⟩ is transformed into the following state: $\begin{eqnarray}\begin{array}{rcl}| {{\rm{\Xi }}}_{1}\rangle & = & \displaystyle \frac{1}{2\sqrt{2}}{\left(| 0\rangle +| 1\rangle \right)}_{{q}_{0}}\otimes {\left(| 0\rangle +| 1\rangle \right)}_{{q}_{1}}\\ & & \otimes | 0{\rangle }_{{q}_{2}}\otimes | 0{\rangle }_{{q}_{3}}\otimes {\left(| 0\rangle +| 1\rangle \right)}_{{q}_{4}}\\ & & \otimes | 0{\rangle }_{{q}_{5}}\otimes | 0{\rangle }_{{q}_{6}}\otimes | 0{\rangle }_{{q}_{7}}\otimes | 0{\rangle }_{{q}_{8}}\otimes | 0{\rangle }_{{q}_{9}}.\end{array}\end{eqnarray}$ |
| • | Step 3: Four controlled-not (CNOT) gate operations are performed on the pairs of qubits (q1, q2), (q4, q3), (q4, q5) and (q4, q6), where q1 and q4 serve as control qubits and q2, q3, q5 and q6 are the target qubits. Then, the above state ∣Ξ1⟩ becomes: $\begin{eqnarray}\begin{array}{rcl}| {{\rm{\Xi }}}_{2}\rangle & = & \displaystyle \frac{1}{2\sqrt{2}}(| 0001000000\rangle +| 0000111000\rangle \\ & & +| 0111000000\rangle +| 0110111000\rangle \\ & & +| 1001000000\rangle +| 1000111000\rangle \\ & & +| 1111000000\rangle +| 1110111000\rangle ).\end{array}\end{eqnarray}$ |
| • | Step 4: A Z-gate is applied to qubit q3. Two CNOT gates are applied on the qubit pairs (q0, q2) and (q0, q7), where qubit q0 acts as a controlled qubit and qubits q2 and q7 are target qubits. Then, the state ∣Ξ2⟩ is transformed into: $\begin{eqnarray}\begin{array}{rcl}| {{\rm{\Xi }}}_{3}\rangle & = & \displaystyle \frac{1}{2\sqrt{2}}(-| 0001000000\rangle +| 0000111000\rangle \\ & & -| 0111000000\rangle +| 0110111000\rangle \\ & & -| 1011000100\rangle +| 1010111100\rangle \\ & & -| 1101000100\rangle +| 1100111100\rangle ).\end{array}\end{eqnarray}$ |
| • | Step 5: Three CNOT gates are executed on the pairs of qubits (q3, q7), (q3, q8) and (q3, q9) with qubit q3 acting as the control qubit and qubits q7, q8 and q9 as the target qubits. Then, the quantum state ∣Ξ3⟩ becomes: $\begin{eqnarray}\begin{array}{rcl}| {{\rm{\Xi }}}_{4}\rangle & = & \displaystyle \frac{1}{2\sqrt{2}}(-| 0001000111\rangle +| 0000111000\rangle \\ & & -| 0111000111\rangle +| 0110111000\rangle \\ & & -| 1011000011\rangle +| 1010111100\rangle \\ & & -| 1101000011\rangle +| 1100111100\rangle ),\end{array}\end{eqnarray}$ which is used as a quantum channel ∣Ξ⟩ in this paper. The complete circuit diagram for generating the entangled channel ( |
Figure 1. Circuit diagram for the generation of the entangled channel. |
4. Protocol under noisy environments
Quantum noise is an unavoidable phenomenon in the practical scenario. When the qubits are distributed after the generation of the quantum entangled state to the involved parties, quantum noise comes into play. Suppose the quantum channel used in our protocol is prepared by Alice in her laboratory and she keeps particles A1, A2, A3 and A4 for herself and distributes particles B1, B2 and B3 to Bob, particle C to Charlie, D to David and E to Eve through quantum channels. Only particles B1, B2, B3, C, D and E are affected by noise. We have used ‘Wolfram Mathematica' software for the calculation of noise and its effects on the protocol.
The quantum channel (3 ) can be rewritten as:
$\begin{eqnarray}\begin{array}{rcl}| {\rm{\Xi }}\rangle & = & \displaystyle \frac{1}{2\sqrt{2}}\left[| 0111000000\rangle +| 0111110000\rangle \right.\\ & & -| 0000001111\rangle \\ & & -| 0000111111\rangle -| 1000011011\rangle \\ & & -| 1000101011\rangle +| 1111010100\rangle \\ & & {\left.+| 1111100100\rangle \right]}_{{A}_{1}{A}_{2}{A}_{3}{A}_{4}{B}_{1}{B}_{3}{B}_{2}{CDE}}.\end{array}\end{eqnarray}$
The evolution of the quantum channel under the influence of quantum noise can be depicted through the transformation: $\begin{eqnarray}T(\rho )=\displaystyle \sum _{i,j,k,l,m,n}{K}_{{ijklmn}}| {\rm{\Xi }}{\rangle }_{{A}_{1}{A}_{2}{A}_{3}{A}_{4}{B}_{1}{B}_{3}{B}_{2}{CDE}}\langle {\rm{\Xi }}| {K}_{{ijklmn}}^{\dagger },\end{eqnarray}$
where $\,{\sum }_{i,j,k,l,m,n}({K}_{{ijklmn}}^{\dagger }{K}_{{ijklmn}})=I$, ${K}_{{ijklmn}}={I}_{{A}_{1}}\otimes {I}_{{A}_{2}}\,\otimes {I}_{{A}_{3}}\otimes {I}_{{A}_{4}}\otimes {X}_{i}\otimes {X}_{j}\otimes {X}_{k}\otimes {X}_{l}\otimes {X}_{m}\otimes {X}_{n}$ and ${X}_{i}^{{\prime} }s$ are the Kraus operators corresponding to different noises. All the parties do their respective jobs described in the previous section. Suppose that Alice's, Bob's, Eve's, Charlie's and David's measurement outcomes are $| {\nu }^{i}{\rangle }_{{A}_{1}}$, $| {\xi }^{j}{\rangle }_{{B}_{2}}$, $| {\tau }_{k}^{j}{\rangle }_{E}$, ∣ηl⟩C and ∣ηm⟩D, respectively. Then, the output state of the protocol can be expressed as: $\begin{eqnarray}{\rho }_{{ijklm}}^{{\rm{out}}}={{\rm{Tr}}}_{{A}_{1}{B}_{2}{ECD}}\left(\displaystyle \frac{{U}_{{ijklm}}[T(\rho )]{U}_{{ijklm}}^{\dagger }}{{\rm{tr}}({U}_{{ijklm}}^{\dagger }{U}_{{ijklm}}[T(\rho )])}\right),\end{eqnarray}$
where the partial trace ${{\rm{Tr}}}_{{A}_{1}{B}_{2}{ECD}}$ is performed over qubits (A1, B2, E, C, D) and Uijklm is defined as follows: $\begin{eqnarray}\begin{array}{rcl}{U}_{{ijklm}} & = & \{{I}_{{A}_{1}}\otimes {\sigma }_{{B}_{1}{B}_{3}}^{{il}}\otimes {I}_{{B}_{2}}\otimes {\sigma }_{{A}_{2}{A}_{3}{A}_{4}}^{{jklm}}\\ & & \otimes {I}_{C}\otimes {I}_{D}\otimes {I}_{E}\}\\ & & \{{I}_{{A}_{1}}\otimes {I}_{{B}_{1}}\otimes {I}_{{B}_{3}}\otimes {I}_{{B}_{2}}\otimes {I}_{{A}_{2}}\otimes {I}_{{A}_{3}}\\ & & \otimes {I}_{{A}_{4}}\otimes {I}_{C}\otimes | {\eta }^{m}{\rangle }_{D}\langle {\eta }^{m}| \otimes {I}_{E}\}\\ & & \{{I}_{{A}_{1}}\otimes {I}_{{B}_{1}}\otimes {I}_{{B}_{3}}\otimes {I}_{{B}_{2}}\otimes {I}_{{A}_{2}}\otimes {I}_{{A}_{3}}\\ & & \otimes {I}_{{A}_{4}}\otimes | {\eta }^{l}{\rangle }_{C}\langle {\eta }^{l}| \otimes {I}_{D}\otimes {I}_{E}\}\\ & & \{{I}_{{A}_{1}}\otimes {I}_{{B}_{1}}\otimes {I}_{{B}_{3}}\otimes {I}_{{B}_{2}}\otimes {I}_{{A}_{2}}\otimes {I}_{{A}_{3}}\otimes {I}_{{A}_{4}}\\ & & \otimes {I}_{C}\otimes {I}_{D}\otimes | {\tau }_{k}^{j}{\rangle }_{E}\langle {\tau }_{k}^{j}| \}\\ & & \{{I}_{{A}_{1}}\otimes {I}_{{B}_{1}}\otimes {I}_{{B}_{3}}\otimes | {\xi }^{j}{\rangle }_{{B}_{2}}\langle {\xi }^{j}| \otimes {I}_{{A}_{2}}\\ & & \otimes {I}_{{A}_{3}}\otimes {I}_{{A}_{4}}\otimes {I}_{C}\otimes {I}_{D}\otimes {I}_{E}\}\\ & & \{| {\nu }^{i}{\rangle }_{{A}_{1}}\langle {\nu }^{i}| \otimes {I}_{{B}_{1}}\otimes {I}_{{B}_{3}}\otimes {I}_{{B}_{2}}\otimes {I}_{{A}_{2}}\\ & & \otimes {I}_{{A}_{3}}\otimes {I}_{{A}_{4}}\otimes {I}_{C}\otimes {I}_{D}\otimes {I}_{E}\},\end{array}\end{eqnarray}$
where i, j, k, l, m ∈ {1, 2}. Unitary operators ${\sigma }_{{B}_{1}{B}_{3}}^{{il}}$ and ${\sigma }_{{A}_{2}{A}_{3}{A}_{4}}^{{jklm}}$ are given in table 1 and table 2, respectively.The influence of noisy channels on the quantum protocol can be measured by fidelity. When Alice's, Bob's, Eve's, Charlie's and David's measurement outcomes are $| {\nu }^{i}{\rangle }_{{A}_{1}}$, $| {\xi }^{j}{\rangle }_{{B}_{2}}$, $| {\tau }_{k}^{j}{\rangle }_{E}$, ∣ηl⟩C and ∣ηm⟩D, respectively, fidelity of the protocol is given by:
$\begin{eqnarray}{F}_{{ijkl}}=\langle {\rm{\Psi }}| {\rho }_{{ijklm}}^{{\rm{out}}}| {\rm{\Psi }}\rangle ,\end{eqnarray}$
where ∣$\Psi$⟩ represents the ideal output state. In the present protocol, the ideal output state is: $\begin{eqnarray}\begin{array}{rcl}| {\rm{\Psi }}\rangle & = & {\left({x}_{0}| 000\rangle +{y}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 111\rangle \right)}_{{A}_{2}{A}_{3}{A}_{4}}\\ & & \otimes {\left(a(| 00\rangle +| 11\rangle )+b(| 01\rangle +| 10\rangle )\right)}_{{B}_{1}{B}_{3}}\\ & = & \left[{{ax}}_{0}| 00000\rangle +{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 11100\rangle \right.\\ & & +{{ax}}_{0}| 00011\rangle +{{ay}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 11111\rangle \\ & & +{{bx}}_{0}| 00001\rangle +{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 11101\rangle \\ & & {\left.+{{bx}}_{0}| 00010\rangle +{{by}}_{0}{{\rm{e}}}^{{\rm{i}}\theta }| 11110\rangle \right]}_{{A}_{2}{A}_{3}{A}_{4}{B}_{1}{B}_{3}}.\end{array}\end{eqnarray}$
Since final output state ${\rho }_{{ijklm}}^{{\rm{out}}}$ is obtained by applying unitary operation by the receivers, after the measurement is performed by all the parties on respective qubits, the final output states ${\rho }_{{ijklm}}^{{\rm{out}}}$ for i, j, k, l, m ∈ {1, 2} are the same, and the total fidelity of the protocol is defined as: $\begin{eqnarray}F=\displaystyle \frac{1}{32}\displaystyle \sum _{i,j,k,l,m}{F}_{{ijkl}}={F}_{11111}.\end{eqnarray}$
In the following, we analyze cases of specific types of noises applying the general case discussed above.4.1. Amplitude-damping noisy environment
Amplitude-damping (AD) noise typically occurs when a quantum system interacts with its surrounding environment, causing the system to lose energy and transition to a lower-energy state. In the case of a two-level quantum system (a qubit), the Kraus operators for amplitude-damping noise are typically represented by a set of operators denoted as:
$\begin{eqnarray}{X}_{0}=\left(\begin{array}{cc}1 & 0\\ 0 & \sqrt{1-q}\end{array}\right)\rm{and}{X}_{1}=\left(\begin{array}{cc}0 & \sqrt{q}\\ 0 & 0\end{array}\right),\end{eqnarray}$
where q is the noise intensity rate in an AD noisy environment.The quantum channel is transformed to TAD(ρ) according to equation (20 ). Then, all the parties do their measurement, as in the illustration. Suppose Alice's, Bob's, Eve's, Charlie's and David's measurement outcomes are $| {\nu }^{1}{\rangle }_{{A}_{1}}$, $| {\xi }^{1}{\rangle }_{{B}_{2}}$, $| {\tau }_{1}^{1}{\rangle }_{E}$, ∣η1⟩C and ∣η1⟩D, respectively. Then, the output state according to formula (21 ) of the protocol can be represented as:
$\begin{eqnarray}{\rho }_{{\rm{AD}}}^{{\rm{out}}}={\rho }_{11111}^{{\rm{out}}}=\displaystyle \sum _{i=1}^{64}| {R}_{i}{\rangle }_{{A}_{2}{A}_{3}{A}_{4}{B}_{1}{B}_{3}}\langle {R}_{i}| ,\end{eqnarray}$
where the expressions of ∣Ri⟩ (for i = 1 to 64) are given in appendix A, table A1.According to formulas (23 ) and (25 ), the fidelity is calculated as:
$\begin{eqnarray}\begin{array}{rcl}{F}^{{\rm{AD}}} & = & \displaystyle \frac{1}{2}\left[2+q\{-2+{b}^{2}(4-6q)+8{b}^{4}(-1+q)+q\}\right.\\ & & +{{qy}}_{0}^{2}\{(-3+2q)(2+(-2+q)q)\\ & & -2{b}^{2}(-2+q)(4-9q+6{q}^{2})\\ & & +8{b}^{4}(-1+q)(4+q(-7+2q))\\ & & -2(-1+q)(2-2q+(-1+2{b}^{2})\\ & & \left.\times (4{b}^{2}(-2+q)(-1+q)-{q}^{2})){y}_{0}^{2}\}\right].\end{array}\end{eqnarray}$
The fidelity FAD is dependent on the AD noise parameter q, on Alice's known state amplitude b and on Bob's known parameter y0. The variations of the fidelity are given in figure 2.
Figure 2. (a) Variation of fidelity FAD with ∣y0∣2 and ∣b∣2 when q = 0.5. (b) Variation of fidelity FAD with ∣b∣2 and q when ∣y0∣2 = 0.5. (c) Variation of fidelity FAD with ∣y0∣2 and q when ∣b∣2 = 0.25. (d) Variation of fidelity FAD with q when ∣b∣2 = 0.25 and ∣y0∣2 = 0.125, 0.25, 0.5, 1. (e) Variation of fidelity FAD with q when ∣y0∣2 = 0.5 and ∣b∣2 = 0.125, 0.25, 0.35, 0.45. |
4.2. Bit-flip noisy environment
In a noisy environment susceptible to bit-flip (BF) errors, these errors occur with a probability s $({\rm{where}}\,0\leqslant s\leqslant 1)$, and the state remains unchanged with a probability of (1 − s). Here, s is the noise parameter. The Kraus operators associated with bit-flip noise are:20 ), which is a mixed state. Then, all the parties measure their qubits according to the protocol. After the measurement is complete, suppose Alice's, Bob's, Eve's, Charlie's and David's measurement outcomes are $| {\nu }^{1}{\rangle }_{{A}_{1}}$, $| {\xi }^{1}{\rangle }_{{B}_{2}}$, $| {\tau }_{1}^{1}{\rangle }_{E}$, ∣η1⟩C and ∣η1⟩D, respectively. The output state according to formula (21 ) of the protocol can be represented as:
$\begin{eqnarray}{X}_{0}=\left(\begin{array}{cc}\sqrt{1-s} & 0\\ 0 & \sqrt{1-s}\end{array}\right)\rm{and}{X}_{1}=\left(\begin{array}{cc}0 & \sqrt{s}\\ \sqrt{s} & 0\end{array}\right).\end{eqnarray}$
The quantum channel is transformed according to formula ( $\begin{eqnarray}{\rho }_{{\rm{BF}}}^{{\rm{out}}}={\rho }_{11111}^{{\rm{out}}}=\displaystyle \sum _{i=1}^{64}| {H}_{i}{\rangle }_{{A}_{2}{A}_{3}{A}_{4}{B}_{1}{B}_{3}}\langle {H}_{i}| ,\end{eqnarray}$
where the expressions of ∣Hi⟩ (for i = 1 to 64) are given in appendix B, table B1.According to formulas (23 ) and (25 ), the fidelity is calculated as:
$\begin{eqnarray}\begin{array}{rcl}{F}^{{\rm{BF}}} & = & -\left(2{\left(1-4{b}^{2}\right)}^{2}(s-1)s+1\right)\\ & & \times \left(4s\left({y}_{0}^{2}-1\right){y}_{0}^{2}{\cos }^{2}(\theta )+s-1\right).\end{array}\end{eqnarray}$
We observe that the fidelity FBF of the protocol is influenced by the noise parameter s as well as the coefficients of the input states. Figure 3 illustrates how the fidelity varies with different parameters.
Figure 3. (a) Variation of fidelity FBF with ∣y0∣2 and ∣b∣2 when s = 0.5. (b) Variation of fidelity FBF with ∣b∣2 and s when ∣y0∣2 = 0.5. (c) Variation of fidelity FBF with ∣y0∣2 and s when ∣b∣2 = 0.25. (d) Variation of fidelity FBF with s when ∣b∣2 = 0.25 and ∣y0∣2 = 0.25, 1, 0.5, 0.125. (e) Variation of fidelity FBF with s when ∣y0∣2 = 0.5 and ∣b∣2 = 0.25, 0.5, 0.45, 0.125. |
4.3. Phase-flip noisy environment
In this environment filled with noise, the relative phase undergoes a flip with a probability of p $({\rm{where}}\,0\leqslant p\leqslant 1)$ and remains unchanged with a probability of (1 − p). Here, p is the noise parameter. The Kraus operators associated with phase-flip (PF) noise are:20 ), the quantum channel is transformed. If Alice's, Bob's, Eve's, Charlie's and David's measurement outcomes are $| {\nu }^{1}{\rangle }_{{A}_{1}}$, $| {\xi }^{1}{\rangle }_{{B}_{2}}$, $| {\tau }_{1}^{1}{\rangle }_{E}$, ∣η1⟩C and ∣η1⟩D, respectively, then the output state according to formula (21 ) of the protocol can be represented as:
$\begin{eqnarray}{X}_{0}=\left(\begin{array}{cc}\sqrt{1-p} & 0\\ 0 & \sqrt{1-p}\end{array}\right)\rm{and}{X}_{1}=\left(\begin{array}{cc}\sqrt{p} & 0\\ 0 & -\sqrt{p}\end{array}\right).\end{eqnarray}$
According to formula ( $\begin{eqnarray}{\rho }_{{\rm{PF}}}^{{\rm{out}}}={\rho }_{11111}^{{\rm{out}}}=\displaystyle \sum _{i=1}^{64}| {G}_{i}{\rangle }_{{A}_{2}{A}_{3}{A}_{4}{B}_{1}{B}_{3}}\langle {G}_{i}| ,\end{eqnarray}$
where the expressions of ∣Gi⟩ (for i = 1 to 64) are given in appendix C, table C1.According to formulas (23 ) and (25 ), the fidelity is calculated as:
$\begin{eqnarray}\begin{array}{rcl}{F}^{{\rm{PF}}} & = & -16(p-1)p({y}_{0}^{2}-1){y}_{0}^{2}\\ & & \times (4{b}^{4}(4(p-1)p(2(p-1)p+1)+1)\\ & & +{b}^{2}(8p(p(-2(p-2)p-3)+1)-2)\\ & & +{\left(2(p-1)p+1\right)}^{2})-2\\ & & \times \left(8{b}^{4}-4{b}^{2}-1\right)(p-1)p+1.\end{array}\end{eqnarray}$
It can be seen that fidelity FPF of the protocol depends on the noise parameter p and on the coefficients of the input states. Variations of the fidelity with different parameters are presented in figure 4.
Figure 4. (a) Variation of fidelity FPF with ∣y0∣2 and ∣b∣2 when p = 0.5. (b) Variation of fidelity FPF with ∣b∣2 and p when ∣y0∣2 = 0.5. (c) Variation of fidelity FPF with ∣y0∣2 and p when ∣b∣2 = 0.25. (d) Variation of fidelity FPF with p when ∣b∣2 = 0.25 and ∣y0∣2 = 0.125, 0.25, 0.5, 1. (e) Variation of fidelity FPF with p when ∣y0∣2 = 0.5 and ∣b∣2 = 0.125, 0.25, 0.35, 0.45. |
5. Execution of the protocol on the IBM quantum platform
In order to test the feasibility of our proposed scheme, we initiated a simulation experiment using IBM's Qiskit Aer (1.0) on the IBM quantum platform. This involved constructing a quantum circuit comprising ten qubits and ten classical bits, which we subsequently executed utilizing the ‘Aer Simulator'. Following the execution, we captured and analyzed the resultant output states of both Alice and Bob. The entire process, along with its associated circuit diagram, is depicted in figure 5.
Figure 5. Quantum circuit for the protocol. Symbols utilized are conventional and the steps performed herein are described in section 5. |
The comprehensive description of the entire process of the quantum circuit diagram is described part by part in the following. In figure 5, qubits q0, q1, q2, q3, q4, q5, q6, q7, q8 and q9 represent qubits A1, B1, B3, B2, A2, A3, A4, C, D and E, respectively.
Part I involves the preparation of an entangled channel.
Part II is for the projective measurement made by Alice. Here, quantum gate U1 is employed on qubit q0 for the basis defined in equation (4 ), where U1 is defined as:
$\begin{eqnarray*}{U}_{1}=\left(\begin{array}{cc}\sqrt{2}a & \sqrt{2}b\\ \sqrt{2}b & -\sqrt{2}a\end{array}\right).\end{eqnarray*}$
Part III of the circuit involves Bob and Eve conducting a projection measurement on qubits B2 and E. Quantum gate U2 utilized by Bob is represented by the matrix: $\begin{eqnarray*}{U}_{2}=\left(\begin{array}{cc}{x}_{0} & {y}_{0}\\ {y}_{0} & -{x}_{0}\end{array}\right).\end{eqnarray*}$
Based on the Bob's measurement outcome, Eve performs either gate U3 or U4. Quantum gates U3 and U4 are given as: $\begin{eqnarray*}{U}_{3}=\displaystyle \frac{1}{\sqrt{2}}\left(\begin{array}{cc}1 & {{\rm{e}}}^{{\rm{i}}\theta }\\ 1 & -{{\rm{e}}}^{{\rm{i}}\theta }\end{array}\right),\,\,\,\,{U}_{4}=\displaystyle \frac{1}{\sqrt{2}}\left(\begin{array}{cc}{{\rm{e}}}^{{\rm{i}}\theta } & 1\\ {{\rm{e}}}^{{\rm{i}}\theta } & -1\end{array}\right).\end{eqnarray*}$
Part IV of the circuit illustrates the circuit employed by Charlie and David to perform the measurement using the basis {∣η1⟩, ∣η2⟩}.Part V of the circuit involves the unitary operation executed by Bob based on the measurement outcomes of Alice and Charlie.
Part VI of the circuit comprises the unitary operation carried out by Alice based on the measurement outcomes of Bob, Eve, Charlie and David.
Parts VII and VIII of the circuit represent the measurement verification conducted after the protocol is completed.
Experiment 1: Suppose Alice knows the information about the state $| \phi \rangle =\tfrac{1}{\sqrt{3}}(| 00\rangle +| 11\rangle )+\tfrac{1}{\sqrt{6}}(| 01\rangle +| 10\rangle )$ and Bob and Eve jointly know the state $| {\rm{\Psi }}\rangle =\tfrac{1}{\sqrt{2}}(| 000\rangle +| 111\rangle )$, that is the values ${x}_{0}=\tfrac{1}{\sqrt{2}},\,{y}_{0}=\tfrac{1}{\sqrt{2}}$ are known to Bob only while Eve knows only the value of θ, which is θ = 0. Then, we run the quantum circuit by taking $a=\tfrac{1}{\sqrt{3}},b=\tfrac{1}{\sqrt{6}}$, ${x}_{0}=\tfrac{1}{\sqrt{2}}$, ${y}_{0}=\tfrac{1}{\sqrt{2}}$ and θ = 0. The output state of qubits (q1, q2) and (q4, q5, q6) after the experiment are given in figure 6. From figure 6, it can be seen that our protocol is feasible in a quantum simulation platform.
Figure 6. Histogram of the experimental results for experiment 1. |
Experiment 2: If Alice possesses knowledge of the state $| \phi \rangle =\tfrac{1}{2}(| 00\rangle +| 11\rangle )+\tfrac{1}{2}(| 01\rangle +| 10\rangle )$, and Bob and Eve are jointly aware of the state $| {\rm{\Psi }}\rangle =\sqrt{\tfrac{2}{3}}| 000\rangle +\sqrt{\tfrac{1}{3}}| 111\rangle $ in that Bob has knowledge that ${x}_{0}=\sqrt{\tfrac{2}{3}}$, ${y}_{0}=\sqrt{\tfrac{1}{3}}$ while θ = 0 is known only to Eve. We execute the quantum circuit with parameters set as $a=\tfrac{1}{2}$, $b=\tfrac{1}{2}$, ${x}_{0}=\sqrt{\tfrac{2}{3}}$, ${y}_{0}=\sqrt{\tfrac{1}{3}}$, and θ = 0. The resulting output states of qubits (q1, q2) and (q4, q5, q6) after the experiment are depicted in figure 7. It is evident from figure 7 that our protocol demonstrates feasibility within a quantum simulation framework.
Figure 7. Histogram of the experimental results for experiment 2. |
6. Efficiency of the protocol
In this section, we explore the efficiency of our protocol. For quantum communication protocols, such as teleportation, RSP and JRSP, efficiency is defined as $\eta =\tfrac{{q}_{a}}{{q}_{b}+{q}_{c}}$. Here, qa represents the number of qubits to be teleported or remotely prepared, qb indicates the number of resource qubits used as a quantum channel and qc accounts for the classical bits required for the entire protocol, which is the classical communication cost.
In our proposed protocol, we have qa = 5, qb = 10 and qc = 5. Consequently, the efficiency of our protocol is given by:
$\begin{eqnarray*}\eta =\displaystyle \frac{5}{10+5}\approx 33.33 \% .\end{eqnarray*}$
In the following, we present a comparison of the performance of our protocol with some similar works. This comparison is with respect to efficiency η given above.Consideration of the above table shows that our protocol performs better in terms of efficiency η amongst those considered here.
7. Discussion and conclusion
Hierarchical quantum communication protocols are needed to satisfy the practical demands arising out of the fact that often in real-life situations we need to differentiate between tasks on the basis of their importance. In this paper, with a view to such a situation, we consider the different levels of control of the supervisors on different parties. What is important is the choice of an appropriate quantum channel, which, in our case, is a ten-qubit entangled state. Although the creation and use of multipartite entanglement is difficult, its involvement is an unavoidable feature in this situation. Another difficulty that may arise due to the fragile nature of entanglement, is that it makes them prone to noise. We have considered quantum noise modeled through Kraus operators. The particular cases of amplitude-damping, bit-flip, phase-flip noises have been considered. This list is not exhaustive, as other kinds of noises that have not been considered may also be present. There are some interesting features in the fidelity analysis. For instance, the states $| \phi \rangle =\tfrac{1}{2}(| 00\rangle +| 11\rangle )+\tfrac{1}{2}(| 01\rangle +| 10\rangle )$ and $| \psi \rangle =\tfrac{1}{\sqrt{2}}| 000\rangle +\tfrac{1}{\sqrt{2}}| 111\rangle $ remain unaffected by bit-flip noise, as can be seen in figures 3(d) and (e). In all the cases of fidelity analysis, it can be seen that the fidelity tends to unit value as the noise parameter tends to zero.
We also note that the efficiency of the protocol is better than many other similar protocols, as shown in table 3.
Table 3. Comparison of the efficiency of our present protocol with other works. |
| References | Type | qa | qb | Classical cost (qc) | Noise analysis | Efficiency |
|---|---|---|---|---|---|---|
| [46] | BRSP | (1 + 2) | 10 | 6-bit | No | ≈18.75% |
| [47] | HBQCP | (1 + 1) | 5 | 3-bit | No | ≈25% |
| [48] | HBQCP | (1 + 1) | 7 | 6-bit | No | ≈15.38% |
| [49] | HBQCP | (2 + 1) | 8 | 7-bit | No | ≈20% |
| [50] | HBQCP | (1 + 1) | 5 | 4-bit | Yes | ≈22.22% |
| Our Protocol | HBQCP | (2 + 3) | 10 | 5-bit | Yes | ≈33.33% |
Appendix A. Expressions of ∣Ri⟩
Table A1. Values of ∣Ri⟩. |
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Appendix B. Expressions of ∣Hi⟩
Table B1. Values of ∣Hi⟩. |
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Appendix C. Expressions of ∣Gi⟩
Table C1. Values of ∣Gi⟩. |
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