1 Introduction

Entangled states are one of the most important resources in quantum information processing tasks, which has been widely applied in various fields such as quantum teleportation [1,2,3], remote state preparation [4,5,6], photonic quantum interface [7,8,9], quantum information splitting [10,11,12], quantum key distribution [13,14,15], bidirectional controlled quantum teleportation [16,17,18], quantum nonlinear optics [19,20,21], quantum secret sharing [22], quantum communication [23,24,25] and quantum dense coding [26,27,28]. Recently, a scheme for cyclic quantum teleportation of three arbitrary single-qubit states was proposed by using a six-qubit entangled state [29], and it is shown that this scheme can realize perfect quantum teleportation in quantum information networks with N(N ≥ 3) observers in different directions.

In this work, we demonstrate that a seven-qubit entangled state can be used to realize the perfect cyclic controlled teleportation for three arbitrary single-qubit states. In our scheme, Alice can teleport her single-qubit state of qubit a to Bob, Bob can teleport his single-qubit state of qubit b to Charlie and Charlie can also teleport his single-qubit state of qubit c to Alice via the control of the supervisor David. The senders perform a Bell-state measurement (BSM) respectively, and the success probability of this cyclic controlled teleportation scheme is 100%.

2 Cyclic Controlled Teleportation

Let us consider there are three observers Alice, Bob and Charlie, and each of them has an arbitrary single qubit a, b, c in arbitrary single-qubit state, which are given by

$$ \vert \psi \rangle_{a} =a_{0} \vert 0\rangle +a_{1} \vert 1\rangle , $$
(1)
$$ \vert \psi \rangle_{b} =b_{0} \vert 0\rangle +b_{1} \vert 1\rangle , $$
(2)
$$ \vert \psi \rangle_{c} =c_{0} \vert 0\rangle +c_{1} \vert 1\rangle , $$
(3)

where a0,a1,b0,b1,c0,c1 are complex numbers and satisfy that |a0|2 + |a1|2 = 1, |b0|2 + |b1|2 = 1, |c0|2 + |c1|2 = 1. Now Alice wants to teleport her single-qubit state of qubit a to Bob, Bob wants to teleport his single-qubit state of qubit b to Charlie and Charlie wants to teleport his single-qubit state of qubit c to Alice. Suppose that Alice, Bob, Charlie and David share a seven-qubit entangled state, which is given by

$$\begin{array}{@{}rcl@{}} \vert \varphi \rangle_{1234567} &=&\frac{1}{2\sqrt 2 }(\vert 0001110\rangle +\vert 0011100\rangle +\vert 0101010\rangle +\vert 0111000\rangle \\ &&+\vert 1000111\rangle +\vert 1010101\rangle +\vert 1100011\rangle +\vert 1110001\rangle )_{1234567}, \end{array} $$
(4)

where the qubits 1 and 6 belong to Alice, the qubits 2 and 4 belong to Bob, the qubits 3 and 5 belong to Charlie, the qubit 7 belongs to David, respectively. Qubits a,b,cand 1, 2, 3, 4, 5, 6, 7 are in a pure product state, which is expressed as

$$ \vert \xi \rangle_{abc1234567} =\vert \psi \rangle_{a} \otimes \vert \psi \rangle_{b} \otimes \vert \psi \rangle_{c} \otimes \vert \varphi \rangle_{1234567} . $$
(5)

In order to realize the cyclic controlled teleportation, Alice must apply a complete measurement of BSM on her qubits a and 1, and the measurement result of BSM is given by

$$ \vert {\Phi}^{+}\rangle_{a1} =\frac{1}{\sqrt 2 }(\vert 00\rangle +\vert 11\rangle )_{a1} , $$
(6)
$$ \vert {\Phi}^{-}\rangle_{a1} =\frac{1}{\sqrt 2 }(\vert 00\rangle -\vert 11\rangle )_{a1} , $$
(7)
$$ \vert {\Psi}^{+}\rangle_{a1} =\frac{1}{\sqrt 2 }(\vert 01\rangle +\vert 10\rangle )_{a1} , $$
(8)
$$ \vert {\Psi}^{-}\rangle_{a1} =\frac{1}{\sqrt 2 }(\vert 01\rangle -\vert 10\rangle )_{a1} . $$
(9)

If Alice’s BSM result is |Φ+〉a1, the other qubits b,c and 2, 3, 4, 5, 6, 7 are collapsed into the product state

$$\begin{array}{@{}rcl@{}} \vert \mu \rangle_{bc234567} &=&(b_{0} \vert 0\rangle +b_{1} \vert 1\rangle )_{b} \otimes (c_{0} \vert 0\rangle +c_{1} \vert 1\rangle )_{c} \\ &&\otimes \frac{1}{2}(a_{0} \vert 001110\rangle +a_{0} \vert 011100\rangle +a_{0} \vert 101010\rangle +a_{0} \vert 111000\rangle \\ &&+a_{1} \vert 000111\rangle +a_{1} \vert 010101\rangle +a_{1} \vert 100011\rangle +a_{1} \vert 110001\rangle )_{234567} . \end{array} $$
(10)

Subsequently, Bob performs a BSM on his own qubits b and 2, and one has

$$ \vert {\Phi}^{+}\rangle_{b2} =\frac{1}{\sqrt 2 }(\vert 00\rangle +\vert 11\rangle )_{b2} , $$
(11)
$$ \vert {\Phi}^{-}\rangle_{b2} =\frac{1}{\sqrt 2 }(\vert 00\rangle -\vert 11\rangle )_{b2} , $$
(12)
$$ \vert {\Psi}^{+}\rangle_{b2} =\frac{1}{\sqrt 2 }(\vert 01\rangle +\vert 10\rangle )_{b2} , $$
(13)
$$ \vert {\Psi}^{-}\rangle_{b2} =\frac{1}{\sqrt 2 }(\vert 01\rangle -\vert 10\rangle )_{b2} . $$
(14)

If Bob’s BSM result is |Φ+〉b2, the other qubits c and 3, 4, 5, 6, 7 are collapsed into the following product state

$$\begin{array}{@{}rcl@{}} \vert \zeta \rangle_{c34567} &=&(c_{0} \vert 0\rangle +c_{1} \vert 1\rangle )_{c} \otimes \frac{1}{\sqrt 2 }(a_{0} b_{0} \vert 01110\rangle +a_{0} b_{0} \vert 11100\rangle +a_{0} b_{1} \vert 01010\rangle \\ &&+a_{0} b_{1} \vert 11000\rangle +a_{1} b_{0} \vert 00111\rangle +a_{1} b_{0} \vert 10101\rangle +a_{1} b_{1} \vert 00011\rangle +a_{1} b_{1} \vert 10001\rangle )_{34567} . \end{array} $$
(15)

Thirdly, Charlie perform a BSM on his own qubits c and 3, and

$$ \vert {\Phi}^{+}\rangle_{c3} =\frac{1}{\sqrt 2 }(\vert 00\rangle +\vert 11\rangle )_{c3} , $$
(16)
$$ \vert {\Phi}^{-}\rangle_{c3} =\frac{1}{\sqrt 2 }(\vert 00\rangle -\vert 11\rangle )_{c3} , $$
(17)
$$ \vert {\Psi}^{+}\rangle_{c3} =\frac{1}{\sqrt 2 }(\vert 01\rangle +\vert 10\rangle )_{c3} , $$
(18)
$$ \vert {\Psi}^{-}\rangle_{c3} =\frac{1}{\sqrt 2 }(\vert 01\rangle -\vert 10\rangle )_{c3} . $$
(19)

If Charlie’s BSM result is |Φ+〉c3, then the other qubits 4, 5, 6 and 7 are collapsed into the following entangled state

$$\begin{array}{@{}rcl@{}} \vert \upsilon \rangle_{4567} &=&(a_{0} b_{0} c_{0} \vert 1110\rangle +a_{0} b_{0} c_{1} \vert 1100\rangle +a_{0} b_{1} c_{0} \vert 1010\rangle +a_{0} b_{1} c_{1} \vert 1000\rangle \\ &&+a_{1} b_{0} c_{0} \vert 0111\rangle +a_{1} b_{0} c_{1} \vert 0101\rangle +a_{1} b_{1} c_{0} \vert 0011\rangle +a_{1} b_{1} c_{1} \vert 0001\rangle )_{4567} . \end{array} $$
(20)

Finally, a single-qubit measurement is performed by David with measurement base \(\left \{ {\frac {\sqrt 2 }{2}(\vert 0\rangle +\vert 1\rangle )_{7} ,\frac {\sqrt 2 }{2}(\vert 0\rangle -\vert 1\rangle )_{7} } \right \}\). If David’s single-qubit measured result is \(\frac {\sqrt 2 }{2}(\vert 0\rangle +\vert 1\rangle )_{7} \), then the qubits 4, 5 and 6 will collapse into the following product state

$$\begin{array}{@{}rcl@{}} \vert \chi \rangle_{456} &=&\left( a_{0} b_{0} c_{0} \vert 111\rangle +a_{0} b_{0} c_{1} \vert 110\rangle +a_{0} b_{1} c_{0} \vert 101\rangle +a_{0} b_{1} c_{1} \vert 100\rangle \right.\\ &&\left.+a_{1} b_{0} c_{0} \vert 011\rangle +a_{1} b_{0} c_{1} \vert 010\rangle +a_{1} b_{1} c_{0} \vert 001\rangle +a_{1} b_{1} c_{1}\vert 000\rangle \right)_{456}\\ &=&(a_{0} \vert 1\rangle +a_{1} \vert 0\rangle )_{4} \otimes (b_{0} \vert 1\rangle +b_{1} \vert 0\rangle )_{5} \otimes (c_{0} \vert 1\rangle +c_{1} \vert 0\rangle )_{6} . \end{array} $$
(21)

Then Alice, Bob, Charlie perform a local unitary operation \({\sigma _{4}^{x}} \otimes {\sigma _{5}^{x}} \otimes {\sigma _{6}^{x}} \) on qubits 4, 5 and 6, thus the desired state can be reconstructed. Therefore, the cyclic controlled teleportation is successfully realized.

For other 127 measurement results, Alice, Bob, Charlie can perform an appropriate unitary operation on their own qubit according to the BSM results and the single-qubit measurement result, the cyclic controlled teleportation is realized easily.

3 Conclusions

In summary, we have demonstrated that a seven-qubit entangled state can be used to realize the cyclic controlled teleportation. In this work, Alice can teleport her single-qubit state of qubit a to Bob, Bob can teleport his single-qubit state of qubit b to Charlie and Charlie can also teleport his single-qubit state of qubit c to Alice via the control of the supervisor David. Alice, Bob and Charlie can operate an appropriate unitary operation to obtain the desired state. In addition, Alice, Bob and Charlie cannot obtain the original state without the help of David.