Multi-Party Quantum Teleportation with Selective Receiver

Abstract

Quantum teleportation is the protocol of transmitting quantum states without physically moving them. After the proposal for the first scheme of quantum teleportation in 1993, different schemes have been proposed, according to the application required. In this paper, a new quantum teleportation protocol is proposed for an arbitrary number of users. First, an initial entangled stated made of n +1 qubits is distributed between the sender and n users. In each run, the sender decides which user is the receiver of the transmitted state. The receiver can retrieve the sent state with the collaboration of the other users. The applications of this protocol are in quantum networks, cloud computing and also quantum cryptography.

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Appendix: Proving the correctness of the protocol

Appendix: Proving the correctness of the protocol

In this section, we prove that the qubit belongs to Bobn according to the result of Alice’s measurement and the number of even / odd values of “-” in the measurements of other users, corresponds to Table 2. Proof is done by mathematical induction. First we prove the correctness of Table 2 for the k = 2 state (minimum number of receivers for this protocol). Then we show that if the table correctness is valid for k = n − 1, then it is also valid for k = n.

The proof for k = 2

In this section, we rewrite the protocol for dual-receiver mode. Alice first generates the channel used in this protocol as follows.

$$ \left|{\left.G\right\rangle}_{A{B}_1{B}_2}=\right.\frac{1}{\sqrt{2}}\Big(\left|000\right\rangle +\left|\left.111\right\rangle \Big)\right. $$
(13)

She then holds the qubit A, sends the qubit B1 to Bob1 and the qubit B2 to Bob2. The complete state of the system (by considering the state |φA) is as follows:

$$ \left|{\left.\varPsi \right\rangle}_{aA{B}_1{B}_2}=\right.\left|{\left.\varphi \right\rangle}_a\right.\otimes \left|{\left.G\right\rangle}_{A{B}_1{B}_2}=\right.\frac{1}{\sqrt{2}}\left(\alpha \left|0000\right\rangle +\alpha \left|0111\right\rangle +\beta \left|1000\right\rangle +\beta \left|1111\right\rangle \right) $$
(14)

After applying the CNOT operator on qubit a as the control and qubit A as the target and then applying the Hadamard gate on the qubit a, the state of the system changes as follows.

$$ {\displaystyle \begin{array}{l}\left|{\left.\varPsi \right\rangle}_{aA{B}_1{B}_2}=\right.\\ {}\ \frac{1}{2}\left[\alpha \left|0\right.\right.\left.000\right\rangle +\alpha \left|\left.0111\right\rangle +\alpha \left|\left.1000\right\rangle +\alpha \left|\left.1111\right\rangle +\beta \left|0100\right\rangle +\beta \left|0011\right\rangle -\beta \left|1100\right\rangle -\beta \left|1011\right\rangle \right.\right.\right.\Big]\end{array}} $$
(15)

In the third step, Alice measures the two qubits according to computational basis a and A. According to Relation (16), the measurement results and post-measurement states of Bob1 and Bob2 can be seen in Table 4. Alice reveals the result of the measurement on the public channel.

$$ {\left|\Psi \right\rangle}_{\mathrm{aA}{\mathrm{B}}_1{\mathrm{B}}_2}=\left|\left.00\right\rangle \frac{\upalpha \left|00\right\rangle +\upbeta \left|11\right\rangle }{2}+\right.\left|\left.01\right\rangle \frac{\upalpha \left|11\right\rangle +\upbeta \left|00\right\rangle }{2}+\left|\left.10\right\rangle \frac{\upalpha \left|00\right\rangle -\upbeta \left|11\right\rangle }{2}+\right.\left|\left.11\right\rangle \frac{\upalpha \left|11\right\rangle -\upbeta \left|00\right\rangle }{2}\right.\right. $$
(16)
Table 4 Alice measurement

In this step, suppose Bob2 is the receiver of Alice. Bob1 measures its qubit based on the X and declares the result. According to Relations (17) the collapsed stateB2 corresponds to Table 5.

$$ {\displaystyle \begin{array}{c}\left|\left.00\right\rangle +\beta \left|11\right\rangle \right.=\left|\left.+\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle +\beta \left|1\right\rangle \right.}{2}\right)+\right.\left|\left.-\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle -\beta \left|1\right\rangle \right.}{2}\right)\right.\\ {}\alpha \left|\left.11\right\rangle +\beta \left|00\right\rangle \right.=\left|\left.+\right\rangle \left(\frac{\alpha \left|\left.1\right\rangle +\beta \left|0\right\rangle \right.}{2}\right)+\right.\left|\left.-\right\rangle \left(\frac{\beta \left|\left.0\right\rangle -\alpha \left|1\right\rangle \right.}{2}\right)\right.\\ {}\begin{array}{c}\alpha \left|\left.00\right\rangle -\beta \left|11\right\rangle \right.=\left|\left.+\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle -\beta \left|1\right\rangle \right.}{2}\right)+\right.\left|\left.-\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle +\beta \left|1\right\rangle \right.}{2}\right)\right.\\ {}\alpha \left|\left.11\right\rangle -\beta \left|00\right\rangle \right.=\left|\left.+\right\rangle \left(\frac{\alpha \left|\left.1\right\rangle -\beta \left|0\right\rangle \right.}{2}\right)+\right.\left|\left.-\right\rangle \left(\frac{\alpha \left|\left.1\right\rangle +\beta \left|0\right\rangle \right.}{2}\right)\right.\end{array}\end{array}} $$
(17)

According to Table 5, when the measurement result is “+”, it means that num{−} is an even number, and when it is “-”, it means that num{−} is an odd number, the post-measurement state is according to Table 2.

Table 5 User’s measurements

Inductive reasoning

Assume that k = n − 1 mode is valid for the correctness of Table 2. Now we prove that for k = n mode, the correctness of Table 2 is also valid. Proof is provided for the case that the result of Alice’s measurement is “00”. For other results, Alice’s measurement is in the same way.

According to the induction assumption, the α| 0〉⊗(n − 1) + β| 1〉⊗(n − 1) state can be written as follows.

$$ \alpha {\left.|0\right\rangle}^{\otimes \left(n-1\right)}+\beta {\left.|1\right\rangle}^{\otimes \left(n-1\right)}=\sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ even}\left|{\tau}_1\dots {\tau}_{n-1}\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle +\beta \left|1\right\rangle \right.}{2}\right)+\sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ odd}\left|{\tau}_1\dots {\tau}_{n-1}\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle -\beta \left|1\right\rangle \right.}{2}\right) $$
(18)

Now by adding a user, the post-measurement state of Alice according to Table 1 is in the form of relation (19).

$$ \alpha {\left.|0\right\rangle}^{\otimes n}+\beta {\left.|1\right\rangle}^{\otimes n}=\left|0\right\rangle \alpha {\left.|0\right\rangle}^{\otimes \left(n-1\right)}+\left|1\right\rangle \beta {\left.|1\right\rangle}^{\otimes \left(n-1\right)} $$
(19)

According to Relation (18), The Relation (19) can be written as follows:

$$ {\displaystyle \begin{array}{c}\alpha {\left.|0\right\rangle}^{\otimes n}+\beta {\left.|1\right\rangle}^{\otimes n}=\alpha {\left.|0\right\rangle}^{\otimes \left(n-1\right)}\left|0\right\rangle +\beta {\left.|1\right\rangle}^{\otimes \left(n-1\right)}\left|1\right\rangle \kern0.5em =\\ {}\ \sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ even}\left|{\tau}_1\dots {\tau}_{n-1}\right\rangle \left(\frac{\left|0\right\rangle \alpha \left|\left.0\right\rangle +\left|1\right\rangle \beta \left|1\right\rangle \right.}{2}\right)+\sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ odd}\left|{\tau}_1\dots {\tau}_{n-1}\right\rangle \left(\frac{\left|0\right\rangle \alpha \left|\left.0\right\rangle -\left|1\right\rangle \beta \left|1\right\rangle \right.}{2}\right)=\\ {}\begin{array}{c}\ \sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ even}\left|\left.{\tau}_1\dots {\tau}_{n-1}\right\rangle \frac{\left|\left.+\right\rangle +\left|-\right\rangle \right.}{\sqrt{2}}\right.\left(\frac{\alpha \left|0\right\rangle }{2}\right)+\sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ even}\left|\left.{\tau}_1\dots {\tau}_{n-1}\right\rangle \frac{\left|\left.+\right\rangle -\left|-\right\rangle \right.}{\sqrt{2}}\right.\left(\frac{\beta \left|1\right\rangle }{2}\right)+\\ {}\ \sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ odd}\left|\left.{\tau}_1\dots {\tau}_{n-1}\right\rangle \frac{\left|\left.+\right\rangle +\left|-\right\rangle \right.}{\sqrt{2}}\right.\left(\frac{\alpha \left|0\right\rangle }{2}\right)+\sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ odd}\left|\left.{\tau}_1\dots {\tau}_{n-1}\right\rangle \frac{\left|\left.+\right\rangle -\left|-\right\rangle \right.}{\sqrt{2}}\right.\left(\frac{-\beta \left|1\right\rangle }{2}\right)=\\ {}\begin{array}{c}\sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ even}\left|{\tau}_1\dots {\tau}_{n-1}\right\rangle \left|+\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle +\beta \left|1\right\rangle \right.}{2}\right)+\sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ odd}\left|{\tau}_1\dots {\tau}_{n-1}\right\rangle \left|+\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle -\beta \left|1\right\rangle \right.}{2}\right)+\\ {}\begin{array}{c}\ \sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ even}\left|{\tau}_1\dots {\tau}_{n-1}\right\rangle \left|-\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle -\beta \left|1\right\rangle \right.}{2}\right)+\sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ odd}\left|{\tau}_1\dots {\tau}_{n-1}\right\rangle \left|-\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle +\beta \left|1\right\rangle \right.}{2}\right)=\\ {}\sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ even}\left|{\tau}_1\dots {\tau}_n\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle +\beta \left|1\right\rangle \right.}{2}\right)+\sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ odd}\left|{\tau}_1\dots {\tau}_n\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle -\beta \left|1\right\rangle \right.}{2}\right)\end{array}\end{array}\end{array}\end{array}} $$
(20)

Given the equality of Relation (21), which results from Relation (20), the correctness of Table 2 is proved

$$ {\displaystyle \begin{array}{c}\alpha {\left.|0\right\rangle}^{\otimes \left(n-1\right)}+\beta {\left.|1\right\rangle}^{\otimes \left(n-1\right)}=\\ {}\sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ even}\left|{\tau}_1\dots {\tau}_n\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle +\beta \left|1\right\rangle \right.}{2}\right)+\sum \limits_{\tau_i\in \left\{+,-\right\}, num\left\{-\right\} is\ odd}\left|{\tau}_1\dots {\tau}_n\right\rangle \left(\frac{\alpha \left|\left.0\right\rangle -\beta \left|1\right\rangle \right.}{2}\right)\end{array}} $$
(21)

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Bolokian, M., Houshmand, M., Sadeghizadeh, MS. et al. Multi-Party Quantum Teleportation with Selective Receiver. Int J Theor Phys (2021). https://doi.org/10.1007/s10773-020-04702-y

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Keywords

  • Quantum cryptography
  • Teleportation
  • Multi users
  • Entanglement
  • Measurement