Deterministic controlled bidirectional remote state preparation

One of the most specific and also most important concepts in quantum information sciences is quantum entanglement [1]. With the aid of it, a number of classically impossible tasks can be done quantumly (i.e. in a way based on the postulates of quantum mechanics), such as quantum cryptography [2], quantum dense coding [3], quantum information splitting [4] and so on. In 1993, Bennett et al [5] first suggested a quantum method called quantum teleportation to securely and faithfully teleport an unknown quantum state of a single qubit from a sender to a distant receiver by using an Einstein–Podolsky–Rosen pair [6] as the shared quantum channel. Soon after that event, related works appeared widely and achieved great developments in both theoretical [7] and experimental [8] domains. In 2001, Huelga et al [9] proposed a modified scheme of quantum teleportation, called quantum remote control, by using bidirectional quantum teleportation method to implement an arbitrary unitary operation upon a distant quantum system. Since then, the method of bidirectional quantum teleportation has been used in many related quantum remote control protocols [10–12]. In a quantum teleportation protocol, one party, Alice, can transmit any unknown quantum state to another party, Bob. In the case where Alice knows her state, the task can be completed by a simpler method called remote state preparation [13, 14]. The principal concern of remote state preparation is to study whether the required entanglement and classical communication cost can be traded off with respect to quantum teleportation. As we know, in original versions of quantum teleportation and remote state preparation the entanglement is shared only among two parties (Alice and Bob). As an obvious extension, many modified versions have been established using various different non-local resources which are shared among more than two parties to perform global multiparty tasks. For example, Karlsson and Bourennane [15] first proposed a scheme, called controlled quantum teleportation, making use of the so-called three-qubit Greenberger–Horne–Zeilinger state [16] shared among three parties (Alice, Bob and a controller Charlie) as the quantum channel. Afterwards, entangled states other than the Greenberger–Horne–Zeilinger one have also been employed to perform controlled quantum teleportation [17–19]. The merit of controlled quantum teleportation is that Alice and Bob are unable to complete the task without permission of the controller Charlie. Very recently, bidirectional quantum teleportation has been extended to include quantum control [20–22]. Such protocols have been referred to as controlled bidirectional quantum teleportation. In these protocols, by using multi-qubit entangled states, Alice and Bob are able to simultaneously teleport an unknown quantum state to each other under the control of Charlie.

In this work, we devise a protocol called controlled bidirectional remote state preparation which has not been dealt with so far. For the non-local quantum resource shared beforehand among the participants, we use the five-qubit linear cluster state, i.e. the same entangled state as was used in [20, 21]. In order to achieve unit success probability, we follow the adaptive measurement strategy similar to that in [14]. In section 2, we set up the task and briefly mention the type of cluster states we shall use to fulfill the task. Section 3 presents in detail our protocol which consists of four sequential steps. Finally, we summarize our work in section 4.

Suppose that Alice has a quantum state

$$\begin{eqnarray} &&\left| {\psi _{\rm{A}} } \right\rangle = a\left| 0 \right\rangle + b{\rm{e}}^{{\rm{i}}\phi } \left| 1 \right\rangle \end{eqnarray} \tag{ 1 }$$

with $a,b = \sqrt {1 - a^2 }$ and ϕ being real numbers known to her, while Bob has another quantum state

$$\begin{eqnarray} &&\left| {\psi _{\rm{B}} } \right\rangle = x\left| 0 \right\rangle + y{\rm{e}}^{{\rm{i}}\varphi } \left| 1 \right\rangle \end{eqnarray} \tag{ 2 }$$

with $x,y = \sqrt {1 - x^2 }$ and ϕ being real numbers known to him. Alice wishes to remotely prepare |ψ_A 〉 at Bob's lab, Bob wishes to remotely prepare |ψ_B 〉 at Alice's lab, and these are required to be done securely under quantum control of a single third party, Charlie, who need not know any information about |ψ_A 〉 and |ψ_B 〉 but plays a decisive role upon the task completion.

Trivially, of course, this task could be done awkwardly by two independent probabilistic controlled remote state preparation protocols, each consumes one Greenberger–Horne–Zeilinger trio as the quantum channel and one controller, involving in total a non-local resource of six qubits and two separate controllers. Our aim here is to design a single protocol that allows Alice and Bob to exchange their own states simultaneously and deterministically using only a non-local resource of five qubits and only one common controller.

The five qubits we shall use as the original non-local resource in our protocol belong to a special class of multi-particle entangled states, the cluster states, which were introduced in 2001 by Briegel and Raussendorf [23], with the primary purpose of carrying out a new paradigm of quantum computation based on measurements but not on conventional circuits of unitary operations. Cluster states quickly attracted much attention and various schemes for generating them via different physical systems have been proposed [24–27]. A linear cluster state |C_N 〉 _{12 ... N} of N qubits can be represented compactly in the form

$$\begin{eqnarray} &&\left| {C_N } \right\rangle _{12 \ldots N} = \frac{1}{{\sqrt {2^N } }}\mathop \otimes \limits_{n = 1}^N \left( {\left| 0 \right\rangle _n + \left| 1 \right\rangle _n \sigma _z^{(n + 1)} } \right), \end{eqnarray} \tag{ 3 }$$

where σ_z⁽ⁿ⁾ is the Z-Pauli matrix acting on qubit n (σ_z⁽ⁿ⁾ |j〉 _n = (− 1)^j |j〉 _n, $\forall j \in \left\{ {0,1} \right\}$, with the convention σ_z^{(N +1)} ≡ 1). For N = 5,

$$\begin{eqnarray} \left| {C_5 } \right\rangle _{12345} &=& \textstyle\frac{1}{2}\left( \left| {00000} \right\rangle + \left| {00111} \right\rangle + \left| {11010} \right\rangle \right. \nonumber\\ &&\left.+ \left| {11101} \right\rangle \right)_{12345}, \end{eqnarray} \tag{ 4 }$$

which is the state we shall use as the shared non-local resource in our protocol to be described in detail in section 3.

For our convenience we rename the qubits in the state |C₅ 〉 ₁₂₃₄₅ given in equation (4) as

$$\begin{eqnarray} \left| Q \right\rangle _{{\rm{A}}_{\rm{1}} {\rm{A}}_{\rm{2}} {\rm{B}}_{\rm{1}} {\rm{B}}_{\rm{2}} {\rm{C}}} &=& \textstyle\frac{1}{2}\left( \left| {00000} \right\rangle + \left| {01011} \right\rangle + \left| {10101} \right\rangle \right. \nonumber\\ &&\left. + \left| {11110} \right\rangle \right)_{{\rm{A}}_{\rm{1}} {\rm{A}}_{\rm{2}} {\rm{B}}_{\rm{1}} {\rm{B}}_{\rm{2}} {\rm{C}}} \; \end{eqnarray} \tag{ 5 }$$

of which, intuitively, both Alice and Bob should hold two qubits and the remaining qubit is held by Charlie. However, the distribution of qubits among the three parties is subtle: a wrong distribution will lead to a failure. After examining all the possibilities we have come up with the following. While qubit C should always be distributed to Charlie, qubits A₁, A₂, B₁ and B₂ can be distributed as follows: either (i) A₁, A₂ to Alice and B₁, B₂ to Bob or (ii) B₁, B₂ to Alice and A₁, A₂ to Bob. Other distributions, say, A₁, B₁ to Alice and A₂, B₂ to Bob, etc do not work. Here we shall work with the qubits' distribution (i) which is sketched in figure 1. Our protocol is composed of four sequential steps.

Zoom In Zoom Out Reset image size

In the first step, Alice and Bob independently perform the following actions. Alice (Bob) first takes an ancillary qubit $|0\rangle _{{\rm{A}}^\prime_1 } ({|0\rangle _{{\rm{B}}^\prime_2 } } )$, then performs a ${\rm{CNOT}}_{{\rm{A}}_1 {\rm{A}}^\prime _1 } ( {{\rm{CNOT}}_{{\rm{B}}_{\rm{2}} {\rm{B}}^\prime _2 } })$ on qubits A₁ and A' ₁. (B₂ and B' ₂), where CNOT_XY is the controlled-NOT gate acting on two qubits X and Y as CNOT_XY |i〉 _X |j〉 _Y = |i〉 _X |i ⊕ j〉 _Y with ⊕ an addition mod 2. After those actions, qubits A' ₁, B'₂ become entangled with those in state (5), i.e. $\left| Q \right\rangle |j\rangle _{{\rm{A}}_{\rm{1}} {\rm{A}}_{\rm{2}} {\rm{B}}_{\rm{1}} {\rm{B}}_{\rm{2}} {\rm{C}}} \to | {Q^\prime } \rangle _{{\rm{A}}_{\rm{1}} {\rm{A}}^\prime _{\rm{1}} {\rm{A}}_{\rm{2}} {\rm{B}}_{\rm{1}} {\rm{B}}_{\rm{2}} {\rm{B}}^\prime _{\rm{2}} {\rm{C}}}$:

$$\begin{eqnarray} \left| {Q^\prime } \right\rangle _{{\rm{A}}_{\rm{1}} {\rm{A}}^\prime _{\rm{1}} {\rm{A}}_{\rm{2}} {\rm{B}}_{\rm{1}} {\rm{B}}_{\rm{2}} {\rm{B}}^\prime _{\rm{2}} {\rm{C}}} &=& \textstyle \frac{1}{2}\left( \left| {0000000} \right\rangle + \left| {0010111} \right\rangle + \left| {1101001} \right\rangle \right. \nonumber\\ &&\left. + \left| {1111110} \right\rangle \right)_{{\rm{A}}_{\rm{1}} {\rm{A}}^\prime _{\rm{1}} {\rm{A}}_{\rm{2}} {\rm{B}}_{\rm{1}} {\rm{B}}_{\rm{2}} {\rm{B}}^\prime _{\rm{2}} {\rm{C}}} \;. \end{eqnarray} \tag{ 6 }$$

Note that, though |Q' 〉 _{A1 A' 1 A2 B1 B2 B' 2 C} is a seven-qubit entangled state, the actual non-local resource costs just five qubits because the entanglement of qubits A' ₁ and B' ₂ with those in state (5) is made locally.

In the second step, Alice first measures qubit A₁ in the basis $\left\{ {|\upsilon _k \rangle _{{\rm{A}}_1 } ,k = \left\{ {0,1} \right\}} \right\}$ determined by {a, b} as

$$\begin{eqnarray} &&\left( {\begin{array}{@{}c@{}} {\left| {\upsilon _0 } \right\rangle _{{\rm{A}}_1 } } \\ {\left| {\upsilon _1 } \right\rangle _{{\rm{A}}_1 } } \\\end{array}} \right) = \left( {\begin{array}{@{}cc@{}} a & b \\ b & { - a} \\\end{array}} \right)\,\left( {\begin{array}{@{}c@{}} {\left| 0 \right\rangle _{{\rm{A}}_1 } } \\ {\left| 1 \right\rangle _{{\rm{A}}_1 } } \\\end{array}} \right) . \end{eqnarray} \tag{ 7 }$$

Then, if the outcome is k (i.e. |υ_k 〉 _A1 is found), she measures qubit A' ₁ in the basis $\left\{ {|\upsilon ^\prime _{k^\prime } \rangle _{{\rm{A}}^\prime _1 } ,k^\prime = \left\{ {0,1} \right\}} \right\}$ determined by {k,φ} as

$$\begin{eqnarray} &&\left( {\begin{array}{@{}c@{}} {\left| {\upsilon ^\prime _0 } \right\rangle _{{\rm{A}}^\prime _1 } } \\ {\left| {\upsilon ^\prime _1 } \right\rangle _{{\rm{A}}^\prime _1 } } \\\end{array}} \right) = \frac{1}{{\sqrt 2 }}\left( {\begin{array}{@{}cc@{}} 1 & {{\rm{e}}^{ - {\rm{i}}\phi } } \\ {{\rm{e}}^{2{\rm{i}}k\phi } } & { - {\rm{e}}^{{\rm{i}}(2k - 1)\phi } } \\\end{array}} \right)\,\left( {\begin{array}{@{}c@{}} {\left| 0 \right\rangle _{{\rm{A}}^\prime _1 } } \\ {\left| 1 \right\rangle _{{\rm{A}}^\prime _1 } } \\\end{array}} \right).\quad \end{eqnarray} \tag{ 8 }$$

As for Bob, he first measures qubit B₂ in the basis $\left\{ {|u_l \rangle _{{\rm{B}}_2 } ,l = \left\{ {0,1} \right\}} \right\}$ determined by {x, y} as

$$\begin{eqnarray} &&\left( {\begin{array}{@{}c@{}} {\left| {u_0 } \right\rangle _{{\rm{B}}_2 } } \\ {\left| {u_1 } \right\rangle _{{\rm{B}}_2 } } \\\end{array}} \right) = \left( {\begin{array}{@{}cc@{}} x & y \\ y & { - x} \\\end{array}} \right)\,\left( {\begin{array}{@{}c@{}} {\left| 0 \right\rangle _{{\rm{B}}_2 } } \\ {\left| 1 \right\rangle _{{\rm{B}}_2 } } \\\end{array}} \right) \cdot \end{eqnarray} \tag{ 9 }$$

Then, if the outcome is l (i.e. |u_l 〉 _B2 is found), he measures qubit B' ₂ in the basis $\left\{ {|u'_{l'} \rangle _{{\rm{B}}^\prime_2 } ,l' = \left\{ {0,1} \right\}} \right\}$ determined by {l, ϕ} as

$$\begin{eqnarray} &&\left( {\begin{array}{@{}c@{}} {\left| {u^\prime _0 } \right\rangle _{{\rm{B}}^\prime _2 } } \\ {\left| {u^\prime _1 } \right\rangle _{{\rm{B}}^\prime _2 } } \\\end{array}} \right) = \frac{1}{{\sqrt 2 }}\left( {\begin{array}{@{}cc@{}} 1 & {{\rm{e}}^{ - {\rm{i}}\varphi } } \\ {{\rm{e}}^{2{\rm{i}}l\varphi } } & { - {\rm{e}}^{{\rm{i}}(2l - 1)\varphi } } \\\end{array}} \right)\,\left( {\begin{array}{@{}c@{}} {\left| 0 \right\rangle _{{\rm{B}}^\prime _2 } } \\ {\left| 1 \right\rangle _{{\rm{B}}^\prime _2 } } \\\end{array}} \right).\nonumber\\ \end{eqnarray} \tag{ 10 }$$

In terms of the basic states |υ_k 〉 _A1, |υ' _k'〉 _A'1 , |u_l 〉 _B2 and |u' _l' 〉 _{B' 2}, we can reexpress state (6) as follows:

$$\begin{eqnarray} &&\left| {Q^\prime } \right\rangle _{{\rm{A}}_{\rm{1}} {\rm{A}}^\prime _{\rm{1}} {\rm{A}}_{\rm{2}} {\rm{B}}_{\rm{1}} {\rm{B}}_{\rm{2}} {\rm{B}}^\prime _{\rm{2}} {\rm{C}}}\nonumber\\&&\quad = \frac{1}{4}\sum\limits_{k,k^\prime ,l,l^\prime = 0}^1 {\left| {\upsilon _k } \right\rangle _{{\rm{A}}_1 } \left| {\upsilon ^\prime _{k^\prime } } \right\rangle _{{\rm{A}}^\prime _1 } \left| {u_l } \right\rangle _{{\rm{B}}_2 } \left| {u^\prime _l } \right\rangle _{{\rm{B}}_2 } \left| {D_{klk^\prime l^\prime } } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } } \;,\nonumber\\ \end{eqnarray} \tag{ 11 }$$

where

$$\begin{eqnarray} \left| {D_{0000} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &=& \left( ax\left| {000} \right\rangle + ay{\rm{e}}^{{\rm{i}}\varphi } \left| {110} \right\rangle + bx{\rm{e}}^{{\rm{i}}\phi } \left| {101} \right\rangle \right.\nonumber\\ &&\left. + by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } \;, \end{eqnarray} \tag{ 12 }$$

$$\begin{eqnarray} \left| {D_{0001} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &=& \left( ax\left| {000} \right\rangle - ay{\rm{e}}^{{\rm{i}}\varphi } \left| {110} \right\rangle + bx{\rm{e}}^{{\rm{i}}\phi } \left| {101} \right\rangle \right.\nonumber\\ &&\left. - by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } \;, \end{eqnarray} \tag{ 13 }$$

$$\begin{eqnarray} \left| {D_{0010} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &=& \left( ax\left| {000} \right\rangle + ay{\rm{e}}^{{\rm{i}}\varphi } \left| {110} \right\rangle - bx{\rm{e}}^{{\rm{i}}\phi } \left| {101} \right\rangle \right.\nonumber\\ &&\left. - by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } , \end{eqnarray} \tag{ 14 }$$

$$\begin{eqnarray} \left| {D_{0011} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &=& \left(ax\left| {000} \right\rangle - ay{\rm{e}}^{{\rm{i}}\varphi } \left| {110} \right\rangle - bx{\rm{e}}^{{\rm{i}}\phi } \left| {101} \right\rangle \right.\nonumber\\ &&\left. + by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } , \end{eqnarray} \tag{ 15 }$$

$$\begin{eqnarray} \left| {D_{0100} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &=& \left( ay{\rm{e}}^{{\rm{i}}\varphi } \left| {000} \right\rangle - ax\left| {110} \right\rangle + by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {101} \right\rangle \right.\nonumber\\ &&\left. - bx{\rm{e}}^{{\rm{i}}\phi } \left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } , \end{eqnarray} \tag{ 16 }$$

$$\begin{eqnarray} \left| {D_{0101} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_{\rm{1}} } &=& \left( - ay{\rm{e}}^{{\rm{i}}\varphi } \left| {000} \right\rangle - ax\left| {110} \right\rangle + by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {101} \right\rangle \right.\nonumber\\ &&\left. + bx{\rm{e}}^{{\rm{i}}\phi } \left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_{\rm{1}} } \;, \end{eqnarray} \tag{ 17 }$$

$$\begin{eqnarray} \left| {D_{0110} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &=& \left( ay{\rm{e}}^{{\rm{i}}\varphi } \left| {000} \right\rangle - ax\left| {110} \right\rangle - by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {101} \right\rangle \right.\nonumber\\ &&\left. + bx{\rm{e}}^{{\rm{i}}\phi } \left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } , \end{eqnarray} \tag{ 18 }$$

$$\begin{eqnarray} \left| {D_{0111} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &=& \left(- ay{\rm{e}}^{{\rm{i}}\varphi } \left| {000} \right\rangle - ax\left| {110} \right\rangle - by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {101} \right\rangle \right.\nonumber\\ &&\left. - bx{\rm{e}}^{{\rm{i}}\phi } \left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } , \end{eqnarray} \tag{ 19 }$$

$$\begin{eqnarray} \left| {D_{1000} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &=& \left(bx{\rm{e}}^{{\rm{i}}\phi } \left| {000} \right\rangle + by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {110} \right\rangle - ax\left| {101} \right\rangle \right.\nonumber\\ &&\left. - ay{\rm{e}}^{{\rm{i}}\varphi } \left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } \;, \end{eqnarray} \tag{ 20 }$$

$$\begin{eqnarray} \left| {D_{1001} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &=& \left(bx{\rm{e}}^{{\rm{i}}\phi } \left| {000} \right\rangle - by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {110} \right\rangle - ax\left| {101} \right\rangle \right.\nonumber\\ &&\left. + ay{\rm{e}}^{{\rm{i}}\varphi } \left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } \;, \end{eqnarray} \tag{ 21 }$$

$$\begin{eqnarray} \left| {D_{1010} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &=& \left( - bx{\rm{e}}^{{\rm{i}}\phi } \left| {000} \right\rangle - by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {110} \right\rangle - ax\left| {101} \right\rangle \right.\nonumber\\ &&\left. - ay{\rm{e}}^{{\rm{i}}\varphi } \left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } \;, \end{eqnarray} \tag{ 22 }$$

$$\begin{eqnarray} \left| {D_{1011} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &=& \left( - bx{\rm{e}}^{{\rm{i}}\phi } \left| {000} \right\rangle + by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {110} \right\rangle - ax\left| {101} \right\rangle \right.\nonumber\\ &&\left. + ay{\rm{e}}^{{\rm{i}}\varphi } \left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } , \end{eqnarray} \tag{ 23 }$$

$$\begin{eqnarray} \left| {D_{1100} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_{\rm{1}} } &=& \left( by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {000} \right\rangle - bx{\rm{e}}^{{\rm{i}}\phi } \left| {110} \right\rangle - ay{\rm{e}}^{{\rm{i}}\varphi } \left| {101} \right\rangle \right.\nonumber\\ &&\left. + ax\left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } \;, \end{eqnarray} \tag{ 24 }$$

$$\begin{eqnarray} \left| {D_{1101} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &\!=\!&\left(- by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {000} \right\rangle \!-\! bx{\rm{e}}^{{\rm{i}}\phi } \left| {110} \right\rangle \!+\! ay{\rm{e}}^{{\rm{i}}\varphi } \left| {101} \right\rangle \right.\nonumber\\ &&\left. + ax\left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_{\rm{1}} } , \end{eqnarray} \tag{ 25 }$$

$$\begin{eqnarray} \left| {D_{1110} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } &\!=\!& \left(- by{\rm{e}}^{{\rm{i}}(\phi \!+\! \varphi )} \left| {000} \right\rangle \!+\! bx{\rm{e}}^{{\rm{i}}\phi } \left| {110} \right\rangle \!-\! ay{\rm{e}}^{{\rm{i}}\varphi } \left| {101} \right\rangle \right.\nonumber\\ &&\left. + ax\left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } \end{eqnarray} \tag{ 26 }$$

and

$$\begin{eqnarray} \left| {D_{1111} } \right\rangle _{{\rm{CA}}_{\rm{2}} {\rm{B}}_{\rm{1}} } &=& \left(by{\rm{e}}^{{\rm{i}}(\phi + \varphi )} \left| {000} \right\rangle + bx{\rm{e}}^{{\rm{i}}\phi } \left| {110} \right\rangle + ay{\rm{e}}^{{\rm{i}}\varphi } \left| {101} \right\rangle \right.\nonumber\\ &&\left. + ax\left| {011} \right\rangle \right)_{{\rm{CA}}_{\rm{2}} {\rm{B}}_1 } \;. \end{eqnarray} \tag{ 27 }$$

After their measurements, Alice and Bob should let each other know the outcomes kk' and ll'. What is interesting is that it is not necessary for them to send secret massages. Instead, they just need to broadcast their outcomes via any public media since these outcomes in fact mean nothing to any outside parties.

It is worth noting that in the second step Alice and Bob used the adaptive measurement strategy. Namely, the choice of bases for measuring qubits A' ₁ and B' ₂ depends essentially on the outcomes of the prior measurements on qubits A₁ and B₂, respectively. Clearly from equation (11), if the outcomes of Alice's and Bob's measurements are klk' l', then the three unmeasured qubits C, A₂ and B₁ are projected onto the state |D_klk'l'〉CA₂B₁ (see equations (12)–(27)) with an equal probability of 1/16. Note also that at this stage Alice and Bob are still unable to complete the task without Charlie's participation since their qubits A₂ and B₁ are still entangled with qubit C. The role of the controller Charlie will be seen in the next step.

In the third step, Charlie measures qubit C in the basis

$$\begin{eqnarray} &&\left( {\begin{array}{@{}c@{}} {\left| - \right\rangle _{\rm{C}} } \\ {\left| + \right\rangle _{\rm{C}} } \\\end{array}} \right) = \frac{1}{{\sqrt 2 }}\left( {\begin{array}{@{}cc@{}} 1 & { - 1} \\ 1 & 1 \\\end{array}} \right)\,\left( {\begin{array}{@{}c@{}} {\left| 0 \right\rangle _{\rm{C}} } \\ {\left| 1 \right\rangle _{\rm{C}} } \\\end{array}} \right), \end{eqnarray} \tag{ 28 }$$

then publicly announces the outcome m = 1 if qubit C is found in the state | + 〉 _C and m = 0 if it is in the state | − 〉 _C. Depending on the outcomes klk'l'm of all the measurements described above, the state of qubits A₂ and B₁, with an equal probability of 1/32, collapses into |Ψ _{klk' l' m} 〉 _{A2 B1}, which have the following explicit forms:

$$\begin{eqnarray} \left| {\Psi _{00000} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } &=& \left| {\Psi _{00111} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=& \left( {x\left| 0 \right\rangle + y{\rm{e}}^{{\rm{i}}\varphi } \left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {a\left| 0 \right\rangle + b{\rm{e}}^{{\rm{i}}\phi } \left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 29 }$$

$$\begin{eqnarray} \left| {\Psi _{00001} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } &=& \left| {\Psi _{00110} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=& \left( {x\left| 0 \right\rangle - y{\rm{e}}^{{\rm{i}}\varphi } \left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {a\left| 0 \right\rangle - b{\rm{e}}^{{\rm{i}}\phi } \left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 30 }$$

$$\begin{eqnarray} \left| {\Psi _{00010} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } & =& \left| {\Psi _{00101} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=& \left( {x\left| 0 \right\rangle - y{\rm{e}}^{{\rm{i}}\varphi } \left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {a\left| 0 \right\rangle + b{\rm{e}}^{{\rm{i}}\phi } \left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 31 }$$

$$\begin{eqnarray} \left| {\Psi _{00100} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_{\rm{1}} } &=& \left| {\Psi _{00011} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=& \left( {x\left| 0 \right\rangle + y{\rm{e}}^{{\rm{i}}\varphi } \left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {a\left| 0 \right\rangle - b{\rm{e}}^{{\rm{i}}\phi } \left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 32 }$$

$$\begin{eqnarray} \left| {\Psi _{01000} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_{\rm{1}} } &=& \left| {\Psi _{01011} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=& \left( {y{\rm{e}}^{{\rm{i}}\varphi } \left| 0 \right\rangle - x\left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {a\left| 0 \right\rangle + b{\rm{e}}^{{\rm{i}}\phi } \left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 33 }$$

$$\begin{eqnarray} \left| {\Psi _{01100} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } &=& \left| {\Psi _{01111} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 }\nonumber\\ &=& \left( {y{\rm{e}}^{{\rm{i}}\varphi } \left| 0 \right\rangle - x\left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {a\left| 0 \right\rangle - b{\rm{e}}^{{\rm{i}}\phi } \left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 34 }$$

$$\begin{eqnarray} \left| {\Psi _{01001} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } &=& \left| {\Psi _{01010} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=&\left( {y{\rm{e}}^{{\rm{i}}\varphi } \left| 0 \right\rangle + x\left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {a\left| 0 \right\rangle - b{\rm{e}}^{{\rm{i}}\phi } \left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 35 }$$

$$\begin{eqnarray} \left| {\Psi _{01101} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } & =& \left| {\Psi _{01110} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=& \left( {y{\rm{e}}^{{\rm{i}}\varphi } \left| 0 \right\rangle + x\left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {a\left| 0 \right\rangle + b{\rm{e}}^{{\rm{i}}\phi } \left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 36 }$$

$$\begin{eqnarray} \left| {\Psi _{10000} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } & = & \left| {\Psi _{10111} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_{\rm{1}} } \nonumber\\ &=& \left( {x\left| 0 \right\rangle + y{\rm{e}}^{{\rm{i}}\varphi } \left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {b{\rm{e}}^{{\rm{i}}\phi } \left| 0 \right\rangle - a\left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 37 }$$

$$\begin{eqnarray} \left| {\Psi _{10001} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } &=& \left| {\Psi _{10110} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=& \left( {x\left| 0 \right\rangle - y{\rm{e}}^{{\rm{i}}\varphi } \left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {b{\rm{e}}^{{\rm{i}}\phi } \left| 0 \right\rangle + a\left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 38 }$$

$$\begin{eqnarray} \left| {\Psi _{10100} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } &=& \left| {\Psi _{10011} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_{\rm{1}} }\nonumber\\ &=& \left( {x\left| 0 \right\rangle + y{\rm{e}}^{{\rm{i}}\varphi } \left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {b{\rm{e}}^{{\rm{i}}\phi } \left| 0 \right\rangle + a\left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 39 }$$

$$\begin{eqnarray} \left| {\Psi _{10010} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } & =& \left| {\Psi _{10101} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=& \left( {x\left| 0 \right\rangle - y{\rm{e}}^{{\rm{i}}\varphi } \left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {b{\rm{e}}^{{\rm{i}}\phi } \left| 0 \right\rangle - a\left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 40 }$$

$$\begin{eqnarray} \left| {\Psi _{11000} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } &=& \left| {\Psi _{11111} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=& \left( {y{\rm{e}}^{{\rm{i}}\varphi } \left| 0 \right\rangle - x\left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {b{\rm{e}}^{{\rm{i}}\phi } \left| 0 \right\rangle - a\left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 41 }$$

$$\begin{eqnarray} \left| {\Psi _{11001} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } &=& \left| {\Psi _{11110} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=& \left( {y{\rm{e}}^{{\rm{i}}\varphi } \left| 0 \right\rangle + x\left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {b{\rm{e}}^{{\rm{i}}\phi } \left| 0 \right\rangle + a\left| 1 \right\rangle } \right)_{{\rm{B}}_1 } ,\nonumber\\ \end{eqnarray} \tag{ 42 }$$

$$\begin{eqnarray} \left| {\Psi _{11011} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } &=& \left| {\Psi _{11100} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=& \left( {y{\rm{e}}^{{\rm{i}}\varphi } \left| 0 \right\rangle - x\left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {b{\rm{e}}^{{\rm{i}}\phi } \left| 0 \right\rangle + a\left| 1 \right\rangle } \right)_{{\rm{B}}_1 }\nonumber\\ \end{eqnarray} \tag{ 43 }$$

and

$$\begin{eqnarray} \left| {\Psi _{11010} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } &=& \left| {\Psi _{11101} } \right\rangle _{{\rm{A}}_{\rm{2}} {\rm{B}}_1 } \nonumber\\ &=& \left( {y{\rm{e}}^{{\rm{i}}\varphi } \left| 0 \right\rangle + x\left| 1 \right\rangle } \right)_{{\rm{A}}_2 } \otimes \left( {b{\rm{e}}^{{\rm{i}}\phi } \left| 0 \right\rangle - a\left| 1 \right\rangle } \right)_{{\rm{B}}_1 } .\nonumber\\ \end{eqnarray} \tag{ 44 }$$

As is evident from equations (29)–(44), for any possible collection klk' l' m of outcomes, |Ψ _{klk' l' m} 〉 _{A2 B1}turns out to be a product state, but it is not yet readily in the desired form |ψ_A 〉 _A2⊗ |ψ_B 〉 _B1. So, a final step, the fourth step, is needed for Alice and Bob to locally reconstruct the target state.

In the fourth step, Alice (Bob) should apply a proper unitary operator $R_{ll^\prime m}^{\rm{A}} \left( {R_{kk^\prime m}^{\rm{B}} } \right)$, if it exists, on qubit A₂(B₁) to transform its state to |ψ_A 〉_A2(|ψ_B 〉 _B1). That is, to be successful, Alice (Bob) needs to know not only the outcome of Bob's (Alice's) measurement in the second step, but also the outcome of Charlie's measurement in the third step, certifying the controller's role in our protocol. Should Charlie, by some important reasons, decline to carry out the measurement or to disclose the measurement outcome, the task remains unfulfilled. Carefully analyzing the data in equations (29)–(44), we have, for any possible outcomes klk' l' m , come up with the general formulae for R_{ll' m}^A and R_{kk' m}^B as

$$\begin{eqnarray} &&R_{ll^\prime m}^{\rm{A}} = \sigma _x^l \sigma _z^{l \oplus l' \oplus m} \end{eqnarray} \tag{ 45 }$$

and

$$\begin{eqnarray} &&R_{kk^\prime m}^{\rm{B}} = \sigma _x^k \sigma _z^{k \oplus k^\prime \oplus m} , \end{eqnarray} \tag{ 46 }$$

where σ_x is the X-Pauli matrix (σ_x |j〉 = |j ⊕ 1〉; j  = { 0,1}). Since Alice and Bob are always able to reconstruct the desired state by the operators R_{ll' m}^A and R_{kk' m}^B defined above, our controlled bidirectional remote state preparation protocol is deterministic, i.e. the success probability is 1.

To summarize, we have put forward an idea of how two distant parties (Alice and Bob) can simultaneously exchange their quantum states securely, deterministically and under the same control (by Charlie) using only local operations and classical communication. Since each party knows his/her own state, we call our protocol deterministic controlled bidirectional remote state preparation. The local operations involved in the main steps (i.e. from the second step) are simple single-qubit von Neumann measurements. The original quantum channel (5) whose qubits should be a priori distributed through space among the participants consists only of five qubits which are here assumed to be provided off-line in the linear cluster state. The actually working quantum channel (6), though it consumes two more auxiliary qubits and two controlled-NOT gates, can be made locally in the preliminary step (i.e. the first step). The reason for extending the original to the working quantum channel is to make room for adaptive measurements as described in the second step, thanks to which unit success probability is achieved. The classical message each of Alice and Bob has to broadcast costs 2 bits, whereas that of Charlie costs just 1 bit, resulting in the total classical communication of 5 bits. Taken altogether, the presented protocol is feasible within the reach of current technologies.