1 Introduction

It is an elementary problem to safely and securely transmit quantum states in quantum network communication and quantum distributed computation. One of the most remarkable schemes for the transmission of quantum states is the so-called quantum teleportation, originally proposed by Bennett et al. [1], in which quantum states can be transmitted between remote locations via quantum channel and classical communication. Afterward, Lo [2] investigated how to send quantum information using a prior shared entanglement and the classical communication when the sender knows fully the transmitted state. This communication protocol is called remote state preparation (RSP). Compared with the usual teleportation [1, 3,4,5,6,7,8,9,10,11,12,13], the RSP proposal can be preformed with simpler measurement and using less classical information in some special cases [14,15,16,17,18,19,20].

So far, lots of RSP proposals have been presented, including joint RSP [21,22,23,24,25], controlled RSP [26,27,28], optimal RSP [29] and so on. In the process of joint remote state preparation (JRSP), the senders separately own the partial classical information of quantum state they want to prepare. If and only if all of the senders cooperate with each other, the JRSP scheme can be realized. The main advantage from the RSP protocol of two-party is that each sender cannot have the final prepared state which is very useful for security. Hence, the JRSP proposal has acquired lots of attention recently. The deterministic controlled JRSP scheme [30] was presented to remotely prepare arbitrary single- and two-qubit states using partially entangled quantum channel. Chen et al. [31] propose a scheme to perform joint remote preparation of an arbitrary two-qubit state using a generalized seven-qubit brown state. Luo et al. presented that two senders can jointly prepare three-qubit states [32] and four-qubit \(\chi \)-states [33] to a remote receiver via the shared GHZ states. Meanwhile, some JRSP schemes for special kinds of multi-qubit states have been explored [34,35,36,37,38]. For example, Wang [34] proposed a method to remotely prepare a two-qubit state via three bipartite entanglements, and generalized this method to the multi-qubit GHZ-class state case. Li et al. [35] proposed a scheme for joint remote preparation of multi-qubit equatorial states with unit successful probability. Long et al. [36] took advantages of the positive operator-valued measurement to perform multi-party joint remote preparation of an arbitrary GHZ-class state. Nowadays, some theoretical and experimental schemes of quantum information processing [39,40,41,42] have been investigated to promote the physical realization of remote state preparation.

The purpose of this paper is to present a novel JRSP proposal for arbitrary multi-qubit states from two senders to one receiver in a deterministic manner by using of maximally three-qubit entangled states. The concrete processes for our JRSP proposal are given, and some useful measurement basis are constructed without the introduction of auxiliary particles. The total classical information cost and successful probability regarding this JRSP scheme are calculated, respectively. It should be emphasized that arbitrary multi-qubit states can be remotely prepared via our protocol with the 100% successful probability. This is the most important advantage of this novel JRSP scheme.

The rest of this paper is organized as follows: In Sect. 2, an efficient scheme for remote preparation of an arbitrary n-qubit state is presented with some special measurement basis, of which the solution expressions are shown in detail. The total successful probability of this JRSP scheme can reach up to one, and the classical information cost is equal to 3n bits. In Sect. 3, concrete realization processes for jointly preparing single-qubit and two-qubit states are illustrated to demonstrate explicitly the feasibility of our JRSP scheme. The paper concludes with Sect. 4.

2 Deterministic JRSP of arbitrary multi-qubit states

Suppose that the senders Alice and Charlie want to help the receiver Bob remotely prepare arbitrary n-qubit state

$$\begin{aligned} |\psi ^0_n\rangle =\sum ^{2^n-1}_{x=0}~a_xe^{i\phi _x}|d_n \ldots d_2 d_1 \rangle \qquad \quad d_i\in \{0,1\};~~x=\sum ^n_{i=1}d_i\cdot 2^{i-1}. \end{aligned}$$
(1)

where \(a_x,\phi _x\in {\mathcal {R}}~(x=0,1,\cdots ,2^n-1)\), \(\sum ^{2^n-1}_{x=0}|a_x|^2=1\) and \(\phi _0=0\). Usually, \(a_x\) can be considered as the amplitude factor of quantum state, and \(\phi _x\) is known as the (relative) phase parameter. The information of \(a_x\) and \(\phi _x\) are only available for Alice and Charlie, respectively. Note that x is the decimal form of the binary string \(d_n\ldots d_2d_1\). Quantum channel is composed of n maximally entangled three-qubit states below

$$\begin{aligned} |\varPsi \rangle _{A_kC_kB_k}=\frac{1}{\sqrt{2}}(|000\rangle +|111\rangle )_{A_kC_kB_k} \quad k=1,2\cdots n. \end{aligned}$$
(2)

Without loss of generality, Alice, Charlie and Bob have particles \(A_k\), \(C_k\) and \(B_k\), respectively. The concrete processes for our deterministic JRSP protocol can be elaborated as follows:

Step 1: For the purpose to realize the JRSP, Alice needs to construct the special projective measurements basis \(\{|\varGamma _{m}\rangle ~|m=0,1\cdots 2^n-1\}\), of which the form can be presented as

$$\begin{aligned} \left[ \begin{array}{ccccc} |\varGamma _0\rangle ,&|\varGamma _1\rangle ,&\cdots&|\varGamma _{2^n-2}\rangle ,&|\varGamma _{2^n-1}\rangle \end{array} \right] ^T=U[n] \left( \begin{array}{c} |0\ldots 00\rangle \\ |0\ldots 01\rangle \\ \vdots \\ |1\ldots 10\rangle \\ |1\ldots 11\rangle \\ \end{array}\right) \end{aligned}$$
(3)

here

(4)

Let us begin with a brief statement of the elements of this matrix U[n] (For more details, see Ref. [43]). If \(n=1\), i.e., some single states need to be prepared, U[1] could be presented as \([a_0,a_1;~a_0,-a_1]\) from Eq. (4). When \(n\ge 2\), the (red) elements at the upper left corner of U[n] can be determined by setting \(U[n](i,j)=U[n-1](i,j)\), here \(2\le i,j\le 2^{n-1}\). The element in the i-th row and j-th column of the complex conjugate of the matrix U[n] would be denoted as the parameter U[n](i, j). According to \(\sum ^{2^n}_{j=1}U[\varTheta ^n_n](i,j)\cdot U[\varTheta ^n_n](2^n+1,j)=0~(2\le i\le 2^{n-1})\), one can obtain the (green) parameters at the upper right corner. Subsequently, the other (blue) coefficients from the \((2^{n-1}-1)\)-th row to \(2^{n}\)-th row can be fulfilled in terms of \(\sum ^{2^n}_{i=1}U[n](i,j)\cdot U[n](i,k)=0~(k=1,2^{n-1}+1;~2\le j\le 2^{n}; j\ne k)\). After that, we can obtain all of the elements of the matrix U[n]. Furthermore, it could be find that

$$\begin{aligned} \langle \varGamma _i|\varGamma _j\rangle =\left\{ \begin{array}{c} 0,\quad i\ne j; \\ 1,\quad i= j. \end{array} \right. \quad \quad \quad \quad \quad \sum ^{2^n-1}_{i=1}~|\varGamma _i\rangle \langle \varGamma _i|=I_{2^n} \end{aligned}$$
(5)

where \(I_{2^n}\) is the identical \(2^n\times 2^n\) matrix. Hence, the orthogonal states \(\{|\varGamma _{m}\rangle ~|m=0,1\cdots 2^n-1\}\) could be used as the projective measurement basis. When the result of particles \((A_1A_2\cdots {A_n})\) is \(|\varGamma _{m}\rangle \), particles \((C_1C_2\cdots {C_n}~B_1B_2\cdots {B_n})\) between Charlie and Bob could collapse into \(|\varPhi _i\rangle \), which can be described as follows:

$$\begin{aligned}&|\varPhi _m\rangle _{C_1B_1C_2B_2\cdots {C_nB_n}}\nonumber \\&\quad =[~|\varGamma _m\rangle _{A_1A_2\cdots {A_n}}]^{\dag }~\left[ ~|\varPsi \rangle _{A_1C_1B_1}\otimes |\varPsi \rangle _{A_2C_2B_2}\cdots |\varPsi \rangle _{A_nC_nB_n}\right] \nonumber \\&\quad =\left( \frac{1}{\sqrt{2}} \right) ^n~\sum ^{2^n-1}_{k=0}~U[n](m,k)\cdot |l_nl_n\cdots l_2l_2l_1l_1 \rangle _{C_1B_1C_2B_2\cdots {C_nB_n}} \end{aligned}$$
(6)

It is should be emphasized that \(l_n\cdots l_2l_1\) is the binary string of the number k. Hence, the state \(|l_nl_n\cdots l_2l_2l_1l_1 \rangle \) can be determined as long as k is set. From Eqs. (4) and (6), it could be found that \(|\varPhi _m\rangle \) can be modified to \(|\varTheta \rangle =\sum ^{2^n-1}_{x=0}a_x|l_nl_n \cdots l_2l_2 l_1l_1 \rangle \) by exchanging the relative positions of the factors of \(|\varPhi _m\rangle \), here \(x=\sum ^n_{i=1}l_i\cdot 2^{i-1}\). For instance, \(|\varPhi _{2^{n-1}+1}\rangle \) can be transported into \(|\varTheta \rangle \) using the permutation operation \(S_n^{2^{n-1}+1}=S_n[1,\overline{2^{n-1}+1};~{\overline{2}},2^{n-1}+2;~{\overline{3}},2^{n-1}+3;\cdots ;~\overline{2^{n-1}},2^n]\):

$$\begin{aligned} |\varTheta \rangle\simeq & {} S_n^{2^{n-1}+1}\cdot |\varPhi _{2^{n-1}}\rangle \nonumber \\\simeq & {} S_n[1,\overline{2^{n-1}+1};~{\overline{2}},2^{n-1}+2;~{\overline{3}},2^{n-1}+3;\cdots ;~\overline{2^{n-1}},2^n]\cdot |\varPhi _{2^{n-1}}\rangle \end{aligned}$$
(7)

where \(S_n[i,j]\) (\(S_n[i,{\overline{j}}])\) means the i-th column and j-th column of the identical matrix \(I_{2^n\times 2^n}\) have changed the positions (with a coefficient \(-1\)). Meanwhile, \(S_n[i,j; {\overline{k}},l]\) equals to the product of \(S_n[i,j]\) and \(S_n[ {\overline{k}},l]\).

Step 2: After the measurements \(\{|\varGamma _{m}\rangle ~|m=0,1\cdots 2^n-1\}\), Alice informs Bob and Charlie of her measurement results via classical channel. Then, Charlie constructs the projective measurements \(\{|\varOmega ^m_{k}\rangle ~|k=0,1\cdots 2^n-1\}\), which can be given by

$$\begin{aligned} \left( \begin{array}{l} |\varOmega ^m_0\rangle \\ |\varOmega ^m_1\rangle \\ \cdots \\ |\varOmega ^m_{2^n-2}\rangle \\ |\varOmega ^m_{2^n-1}\rangle \\ \end{array} \right) =\left( \frac{1}{\sqrt{2}}\right) ^n~S_n^{m}\left( \begin{array}{ccc} 1 &{} exp\left( -i\phi _1\right) &{} exp\left( -i\phi _2\right) \\ 1 &{} exp\left( \frac{i\pi \cdot 1\cdot 1}{2^{n-1}}-i\phi _1\right) &{} exp\left( \frac{i\pi \cdot 2\cdot 1}{2^{n-1}}-i\phi _2\right) \\ 1 &{} exp\left( \frac{i\pi \cdot 1\cdot 2}{2^{n-1}}-i\phi _1\right) &{} exp\left( \frac{i\pi \cdot 2\cdot 2}{2^{n-1}}-i\phi _2\right) \\ \vdots &{} \ldots &{} \cdots \\ 1 &{} exp\left( \frac{i\pi \cdot 1\cdot (2^n-1)}{2^{n-1}}-i\phi _1\right) &{} exp\left( \frac{i\pi \cdot 2\cdot (2^n-1)}{2^{n-1}}-i\phi _1\right) \\ \end{array} \right. \quad \quad&\nonumber \\ \left. \begin{array}{cc} \cdots &{} exp\left( -i\phi _{2^n-1}\right) \\ \cdots &{} exp\left( \frac{i\pi \cdot (2^n-1)\cdot 1}{2^{n-1}}-i\phi _{2^n-1}\right) \\ \cdots &{} exp\left( \frac{i\pi \cdot (2^n-1)\cdot 2}{2^{n-1}}-i\phi _{2^n-1}\right) \\ \ddots &{} \vdots \\ \cdots &{} exp\left( \frac{i\pi \cdot (2^n-1)\cdot (2^n-1)}{2^{n-1}}-i\phi _{2^n-1}\right) \\ \end{array} \right) \left( \begin{array}{c} |0\ldots 00\rangle \\ |0\ldots 01\rangle \\ \vdots \\ |1\ldots 10\rangle \\ |1\ldots 11\rangle \\ \end{array} \right) \nonumber \\ \end{aligned}$$
(8)

The unitary transformation \(S_n^m\) is the corresponding permutation operation for the measurement result \(|\varGamma _{m}\rangle \). Thus, one can find that

$$\begin{aligned}&|\varDelta ^m_k\rangle _{B_1B_2\ldots B_n}\nonumber \\&\quad =[~|\varGamma _m\rangle _{A_1A_2\ldots A_n}\otimes |\varOmega ^m_{k}\rangle _{C_1C_2\ldots C_n}~]^{\dag }~|\varPsi \rangle _{A_1C_1B_1}\otimes |\varPsi \rangle _{A_2C_2B_2}\cdots |\varPsi \rangle _{A_nC_nB_n}\nonumber \\&\quad =\frac{1}{2^n}~\sum ^{2^n-1}_{j=0}~U[n](m,j)\cdot exp\left( i\phi _x-\frac{i\pi \cdot x \cdot k}{2^{n-1}}\right) |d_n \cdots d_2d_1 \rangle _{B_1B_2\ldots B_n} \end{aligned}$$
(9)

here \(x=\sum ^n_{i=1}~d_i2^{i-1}\). After the projective measurements \(\{|\varOmega ^m_{k}\rangle ~|k=0,1\cdots 2^n-1\}\), Charlie tells the results to Bob via classical channel. From Eq. (9), we can find that the probability for each outcome \(|\varDelta ^m_k\rangle \) is always equal to \((1/2^n)^2=4^{-n}\). Based on the quantum measurement postulate, we can get that the state of quantum channel after the projective measurements would be known exactly if the measurement outcome is obtained. Furthermore, the states of particles \((B_1,~B_2\cdots B_n)\) after projective measurements are pure, and could be converted into the prepared states by using of some unitary operations.

Step 3: According to the result \(|\varOmega ^m_{k}\rangle \) of Charlie, Bob need to introduce the unitary operation \(U_n^k\)

$$\begin{aligned} U_n^k=\left( \begin{array}{rr} 1 &{}\quad 0 \\ 0 &{}\quad e^{i\pi k} \end{array}\right) \otimes \left( \begin{array}{rr} 1 &{}\quad 0 \\ 0 &{}\quad e^{i\frac{\pi k}{2}} \end{array}\right) \otimes \cdots \otimes \left( \begin{array}{rr} 1 &{} \quad 0 \\ 0 &{}\quad e^{i\frac{\pi k}{2^n}} \end{array}\right) \end{aligned}$$
(10)

For the state in Eq. (9), Bob only need to perform the permutation operation \(S_n^{m}\) and the unitary gate \(U_n^k\) on particles \((B_1,~B_2\cdots B_n)\) to convert the initial multi-qubit state shown as Eq. (1).

$$\begin{aligned} |\psi ^0_n\rangle _{B_1B_2\ldots B_n}\simeq U_n^k\cdot S^m_n~|\varDelta ^m_k\rangle _{B_1B_2\ldots B_n} \end{aligned}$$
(11)

The successful probability and required classical communication cost play important roles in the JRSP schemes. Note that the probability for each measurement outcome \(|\varDelta ^m_k\rangle ~(m,k=0,1\cdots 2^n-1)\) is \((1/2^n)^2=4^{-n}\). So that the total successful probability can be presented as

$$\begin{aligned} P_{total}=\sum ^{2^n-1}_{m=0}\sum ^{2^n-1}_{k=0}~4^{-n}=100\% \end{aligned}$$
(12)

Based on the calculation method of classical information for RSP proposals in Refs. [44, 45], this classical information required in our three-party RSP proposal can be divided into two transmitted processes. One is the classical information \(S_{A-BC}\) (including \(S_{A-C}\) and \(S_{A-B}\)) sent from Alice to Charlie and Bob, and the other is the classical information \(S_{C-B}\) from Charlie to Bob. In order to realize this JRSP proposal, Alice would perform a projective measurement on particles \(A_1A_2\cdots A_n\) and informs them of her measurement result. It should be emphasized that the measurement basis is \(\{|\varGamma _{m}\rangle ~|m=0,1\cdots 2^n-1\}\) shown as Eq. (3). Particles \(A_1A_2\cdots A_n\) after this measurement will collapse into one of the \(2^n\) kinds of outcomes. Meanwhile, it can be obtained that the \(2^n\) kinds of results have the same probability of \(2^{-n}\). The classical information \(S_{A-BC}\) sent from Alice to Charlie and Bob could be presented as

$$\begin{aligned} S_{A-BC}=S_{A-C}+S_{A-B}=-\frac{1}{2^n}\sum ^{2^n-1}_{m=0}\log \frac{1}{2^n}-\frac{1}{2^n}\sum ^{2^n-1}_{m=0}\log \frac{1}{2^n}=2n\quad bits \end{aligned}$$
(13)

According to the measurement result of \(A_1A_2\cdots A_n\), Charlie measures particles \(C_1C_2\cdots C_n\) with the orthogonal states \(\{|\varOmega ^m_{k}\rangle ~|k=0,1\cdots 2^n-1\}\) in Eq. (8), and informs Bob of his outcome. There are \(2^n\) kinds of measurement results with the same probability of \(2^{-n}\). Hence, the classical information cost from Charlie to Bob can be given by

$$\begin{aligned} S_{C-B}=-\frac{1}{2^n}\sum ^{2^n-1}_{k=0}\log \frac{1}{2^n}=n\quad bits \end{aligned}$$
(14)

From the above discussions, we could find that arbitrary n-qubit states can be prepared in a deterministic manner by using our proposal. Meanwhile, the required classical communication cost of this JRSP protocol is equal to

$$\begin{aligned} S_{total}=S_{A-BC}+S_{C-B}=2n+n=3n\quad bits \end{aligned}$$
(15)

It should be noted that n maximally entangled three-qubit states are required for remotely preparing one n-qubit state, whatever the prepared state is entangled or not.

3 Examples of JRSP

To illustrate our scheme for deterministic joint remote state preparation explicitly, we will study how to remotely prepare arbitrary single- and two-qubit states, which are elementary resources for quantum information.

3.1 JRSP of arbitrary single-qubit states

Suppose that the senders Alice and Charlie want to help the receiver Bob remotely prepare the following single-qubit state

$$\begin{aligned} |\psi \rangle =a_0|0\rangle +a_1e^{i\phi _1}|1\rangle \end{aligned}$$
(16)

where \(a_0,~a_1,~\phi _1\) are real, and \(|a_0|^2+|a_1|^2=1\). Quantum channel is composed of the maximally entangled three-qubit states below

$$\begin{aligned} |\varPsi \rangle _{ACB}=\frac{1}{\sqrt{2}}~(|000\rangle +|111\rangle )_{ACB} \end{aligned}$$
(17)

The subscript A, C and B denote that the particles belong to Alice, Charlie and Bob, respectively. The concrete processes can be presented as follows:

Step 1: Due to transmit the initial state given by Eq. (16), the projective measurement \(\{~|\varGamma _m\rangle ~|~m=0,1~\}\) need to be performed by Alice on particle A.

$$\begin{aligned} \left( \begin{array}{c} |\varGamma _0\rangle \\ |\varGamma _1\rangle \end{array}\right) = \left( \begin{array}{cc} a_0 &{}\quad a_1\\ a_1 &{}\quad -a_0 \end{array}\right) \left( \begin{array}{c} |0\rangle \\ |1\rangle \end{array}\right) \end{aligned}$$
(18)

After that, particles C and B would collapse into one of two kinds of outcomes below

$$\begin{aligned}&|\varPhi _0\rangle _{CB}\simeq [~|\varGamma _0\rangle _A~]^\dag ~|\varPsi \rangle _{ACB}=\frac{1}{\sqrt{2}}(a_0|00\rangle +a_1|11\rangle )_{CB} \end{aligned}$$
(19)
$$\begin{aligned}&|\varPhi _1\rangle _{CB}\simeq [~|\varGamma _1\rangle _A~]^\dag ~|\varPsi \rangle _{ACB}=\frac{1}{\sqrt{2}}(a_1|00\rangle -a_0|11\rangle )_{CB} \end{aligned}$$
(20)

Meanwhile, Alice informs Bob and Charlie of her measurement results using classical information. From the analysis of Step 1 in Sect. 2, one can obtain that

$$\begin{aligned} \begin{array}{ll} S_1^0=I_2 &{} \quad S_1^1=S_1[1,-{\overline{2}}]=\sigma _z \\ \end{array} \end{aligned}$$
(21)

Step 2: Charlie need to perform the following projective measurement \(\{~|\varOmega ^m_k\rangle ~|~k=0,1~\}\) on particle C in terms of Alice’s measurement result \(|\varGamma _m\rangle \).

$$\begin{aligned} \left( \begin{array}{c} |\varOmega ^0_0\rangle \\ |\varOmega ^0_1\rangle \\ |\varOmega ^1_0\rangle \\ |\varOmega ^1_1\rangle \end{array}\right) = \frac{1}{\sqrt{2}}\left( \begin{array}{cc} 1 &{} \quad e^{-i\phi _1} \\ 1 &{}\quad -e^{-i\phi _1} \\ -e^{-i\phi _1} &{}\quad 1 \\ e^{-i\phi _1} &{}\quad 1 \\ \end{array}\right) \left( \begin{array}{c} |0\rangle \\ |1\rangle \end{array}\right) \end{aligned}$$
(22)

Meanwhile, Eqs. (17) and (18) would become

$$\begin{aligned}&|\varDelta ^0_0\rangle _B=[~|M_0\rangle _A|N^0_0\rangle _C~]^\dag ~|Q\rangle _{ACB}=\frac{1}{2}(a_0|0\rangle +a_1e^{i\phi _1}|1\rangle )_{B} \end{aligned}$$
(23)
$$\begin{aligned}&|\varDelta ^0_1\rangle _B=[~|M_0\rangle _A|N^0_1\rangle _C~]^\dag ~|Q\rangle _{ACB}=\frac{1}{2}(a_0|0\rangle -a_1e^{i\phi _1}|1\rangle )_{B} \end{aligned}$$
(24)
$$\begin{aligned}&|\varDelta ^1_0\rangle _B=[~|M_1\rangle _A|N^1_0\rangle _C~]^\dag ~|Q\rangle _{ACB}=\frac{1}{2}(-a_1e^{i\phi _1}|0\rangle +a_0|1\rangle )_{B} \end{aligned}$$
(25)
$$\begin{aligned}&|\varDelta ^1_1\rangle _B=[~|M_1\rangle _A|N^1_1\rangle _C~]^\dag ~|Q\rangle _{ACB}=\frac{1}{2}(a_1e^{i\phi _1}|0\rangle +a_0|1\rangle )_{B} \end{aligned}$$
(26)

The measurements results of particle C can be transmitted from Charlie to Bob via classical channel.

Step 3: Based on the outcomes \(\{~m,k=0,1~\}\) of Alice and Charlie, Bob performs the relative unitary operation \(U_2^kS_2^m\) to prepare the original state. Table 1 indicates how to select the unitary transformation for particle B based on the measurements results of particles A and C. The unitary operation \(U_1^k~(k=0,1)\) in Table 1 can be presented as

$$\begin{aligned} U_n^k=\left( \begin{array}{rr} 1 &{} ~~0 \\ 0 &{} ~~e^{i\pi k} \end{array}\right) \end{aligned}$$
(27)
Table 1 The results on particles (A, C) and the unitary gates on particles B
Full size table

here \(\{~\sigma _x,\sigma _y,\sigma _z,I\}\) are the Pauli matrices. From Table 1, it can be found that our scheme for joint remote state preparation of arbitrary single-qubit states can be realized with the 100% successful probability at the cost of 3 bits classical information. The results about successful probability are in agreement with the probabilities of Refs. [21, 30].

3.2 JRSP of arbitrary two-qubit states

The two-qubit states prepared from the senders Alice and Charlie to the receiver Bob can be presented as

$$\begin{aligned} |\psi \rangle =a_0|00\rangle +a_1e^{i\phi _1}|01\rangle +a_2e^{i\phi _2}|10\rangle +a_3e^{i\phi _3}|11\rangle \end{aligned}$$
(28)

where \(a_i~(i=0,1\cdots 3)\) and \(\phi _j~(j=1,2,3)\) are real, and \(\varSigma ^{3}_{x=0}|a_x|^2=1\). Assume that Alice has the information of \(a_x\), and Charlie has \(\phi _y\). The three-qubit GHZ states shared along Alice and Charlie with Bob are presented as

$$\begin{aligned}&|\varPsi \rangle _{A_1C_1B_1}=\frac{1}{\sqrt{2}}(|000\rangle +|111\rangle )_{A_1C_1B_1} \end{aligned}$$
(29)
$$\begin{aligned}&|\varPsi \rangle _{A_2C_2B_2}=\frac{1}{\sqrt{2}}(|000\rangle +|111\rangle )_{A_2C_2B_2} \end{aligned}$$
(30)

Note that Alice, Charlie and Bob have particles \(A_j\), \(C_j\) and \(B_j\) respectively, here \(j=1,2\). Moreover, the detailed processes of our JRSP proposal for arbitrary two-qubit states are elaborated as follows:

Step 1: For the purpose to remotely prepare two-qubit state, Alice needs to perform the projective measurements \(\{~|\varGamma _m\rangle ~|~m=0,1,2,3~\}\) on her particles \(A_1\) and \(A_2\).

$$\begin{aligned} \left( \begin{array}{c} |\varGamma _0\rangle \\ |\varGamma _1\rangle \\ |\varGamma _2\rangle \\ |\varGamma _3\rangle \end{array}\right) = \left( \begin{array}{rrrr} a_0 &{}\quad a_1 &{}\quad a_2 &{}\quad a_3 \\ a_1 &{}\quad -a_0 &{}\quad a_3 &{}\quad -a_2 \\ a_2 &{}\quad -a_3 &{}\quad -a_0 &{}\quad a_1 \\ a_3 &{}\quad a_2 &{}\quad -a_1 &{}\quad -a_0 \end{array}\right) \left( \begin{array}{c} |00\rangle \\ |01\rangle \\ |10\rangle \\ |11\rangle \end{array}\right) \end{aligned}$$
(31)

Thus, quantum channel composed of the three-qubit GHZ states could be rewritten as

$$\begin{aligned}&|\varPsi \rangle _{A_1C_1B_1}\otimes |\varPsi \rangle _{A_2C_2B_2}=\frac{1}{2}\cdot \sum ^{2^n-1}_{j=0}~~{|\varGamma _j\rangle _{A_1A_2}\otimes |\varPhi _j\rangle _{C_1B_1C_2B_2}} \end{aligned}$$
(32)

here

$$\begin{aligned}&|\varPhi _0\rangle =a_0|0000\rangle +a_1|0011\rangle +a_2|1100\rangle +a_3|1111\rangle \nonumber \\&|\varPhi _1\rangle =a_1|0000\rangle -a_0|0011\rangle +a_3|1100\rangle -a_2|1111\rangle \nonumber \\&|\varPhi _2\rangle =a_2|0000\rangle -a_3|0011\rangle -a_0|1100\rangle +a_1|1111\rangle \nonumber \\&|\varPhi _3\rangle =a_3|0000\rangle +a_2|0011\rangle -a_1|1100\rangle -a_0|1111\rangle \end{aligned}$$
(33)

Meanwhile, Alice informs Bob and Charlie of her measurement results using classical information. From the analysis of Step 1 in Sect. 1 and Eq. (29), it can be find that

$$\begin{aligned} \begin{array}{ll} S_2^0=I_4 &{} \quad S_2^1=S_n[1,{\overline{2}};~3,{\overline{4}}]=I_2\otimes i\sigma _y \\ S_2^2=S_n[1,{\overline{3}};~{\overline{2}},4]=i\sigma _y\otimes \sigma _z &{} \quad S_2^3=S_n[1,{\overline{4}};~2,{\overline{3}}]=i\sigma _y \otimes \sigma _x \\ \end{array} \end{aligned}$$
(34)

Step 2: Charlie need to perform the following projective measurement \(\{~|\varOmega ^m_k\rangle ~|~k=0,1~\}\) on particles \(C_1\) and \(C_2\) in terms of Alice’s measurement result \(|\varGamma _m\rangle \).

$$\begin{aligned} \left( \begin{array}{l} |\varOmega ^m_0\rangle \\ |\varOmega ^m_1\rangle \\ |\varOmega ^m_2\rangle \\ |\varOmega ^m_3\rangle \\ \end{array} \right) =\frac{1}{2}~S_2^{m} \left( \begin{array}{rrrr} 1 &{}\quad e^{-i\phi _1} &{} \quad e^{-i\phi _2} &{} \quad e^{-i\phi _3} \\ 1 &{} \quad ie^{-i\phi _1} &{}\quad -e^{-i\phi _2} &{}\quad -ie^{-i\phi _3} \\ 1 &{}\quad -e^{-i\phi _1} &{}\quad e^{-i\phi _2} &{}\quad -e^{-i\phi _3} \\ 1 &{}\quad -ie^{-i\phi _1} &{}\quad -e^{-i\phi _2} &{}\quad ie^{-i\phi _3} \\ \end{array} \right) \left( \begin{array}{c} |00\rangle \\ |01\rangle \\ |10\rangle \\ |11\rangle \\ \end{array} \right) \end{aligned}$$
(35)

Thus, one can find that

$$\begin{aligned}&|\varDelta ^0_0\rangle _{B_1B_2}=1/4\cdot (a_0|00\rangle +a_1e^{i\phi _1}|01\rangle +a_2e^{i\phi _2}|10\rangle +a_3e^{i\phi _3}|11\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^0_1\rangle _{B_1B_2}=1/4\cdot (a_0|00\rangle +a_1ie^{i\phi _1}|01\rangle -a_2e^{i\phi _2}|10\rangle -a_3ie^{i\phi _3}|11\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^0_2\rangle _{B_1B_2}=1/4\cdot (a_0|00\rangle -a_1e^{i\phi _1}|01\rangle +a_2e^{i\phi _2}|10\rangle -a_3e^{i\phi _3}|11\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^0_3\rangle _{B_1B_2}=1/4\cdot (a_0|00\rangle -a_1ie^{i\phi _1}|01\rangle -a_2e^{i\phi _2}|10\rangle +ia_3e^{i\phi _3}|11\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^1_0\rangle _{B_1B_2}=1/4\cdot (-a_0|01\rangle +a_1e^{i\phi _1}|00\rangle -a_2e^{i\phi _2}|11\rangle -a_3e^{i\phi _3}|10\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^1_1\rangle _{B_1B_2}=1/4\cdot (-a_0|01\rangle +a_1ie^{i\phi _1}|00\rangle +a_2e^{i\phi _2}|11\rangle +a_3ie^{i\phi _3}|10\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^1_2\rangle _{B_1B_2}=1/4\cdot (-a_0|01\rangle -a_1e^{i\phi _1}|00\rangle -a_2e^{i\phi _2}|11\rangle +a_3e^{i\phi _3}|10\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^1_3\rangle _{B_1B_2}=1/4\cdot (-a_0|01\rangle -a_1ie^{i\phi _1}|00\rangle +a_2e^{i\phi _2}|11\rangle -ia_3e^{i\phi _3}|10\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^2_0\rangle _{B_1B_2}=1/4\cdot (-a_0|10\rangle +a_1e^{i\phi _1}|11\rangle +a_2e^{i\phi _2}|00\rangle -a_3e^{i\phi _3}|01\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^2_1\rangle _{B_1B_2}=1/4\cdot (-a_0|10\rangle +a_1ie^{i\phi _1}|11\rangle -a_2e^{i\phi _2}|00\rangle +a_3ie^{i\phi _3}|01\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^2_2\rangle _{B_1B_2}=1/4\cdot (-a_0|10\rangle -a_1e^{i\phi _1}|11\rangle +a_2e^{i\phi _2}|00\rangle +a_3e^{i\phi _3}|01\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^2_3\rangle _{B_1B_2}=1/4\cdot (-a_0|10\rangle -a_1ie^{i\phi _1}|11\rangle -a_2e^{i\phi _2}|00\rangle -ia_3e^{i\phi _3}|01\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^3_0\rangle _{B_1B_2}=1/4\cdot (-a_0|11\rangle -a_1e^{i\phi _1}|10\rangle +a_2e^{i\phi _2}|01\rangle +a_3e^{i\phi _3}|00\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^3_1\rangle _{B_1B_2}=1/4\cdot (-a_0|11\rangle -a_1ie^{i\phi _1}|10\rangle -a_2e^{i\phi _2}|01\rangle -a_3ie^{i\phi _3}|00\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^3_2\rangle _{B_1B_2}=1/4\cdot (-a_0|11\rangle +a_1e^{i\phi _1}|10\rangle +a_2e^{i\phi _2}|01\rangle -a_3e^{i\phi _3}|00\rangle )_{B_1B_2}\nonumber \\&|\varDelta ^3_3\rangle _{B_1B_2}=1/4\cdot (-a_0|11\rangle +a_1ie^{i\phi _1}|10\rangle -a_2e^{i\phi _2}|01\rangle +ia_3e^{i\phi _3}|00\rangle )_{B_1B_2}\nonumber \\ \end{aligned}$$
(36)

Meanwhile, Table 2 shows the relationship between the measurement results \(\{~m,k~|~m,k=0,1,2,3~\}\) with the unitary transformation \(U_2^k\cdot S_2^m \) on particles \(B_1\) and \(B_2\).

Table 2 The results \(\{m,k\}\) and the unitary gates \(U_2^k\cdot S_2^m\) on particles (\(B_1,B_2\))
Full size table

here, \(\{~S_2^m~|m=0,1,2,3~\}\) are given by Eq. (30), and \(\{~U_2^k~|k=0,1,2,3~\}\) can be presented as

$$\begin{aligned} \begin{array}{ll} U_2^0=I_4 &{} \quad U_2^1=\sigma _z\otimes \left( \begin{array}{cc} 1 &{}\quad 0 \\ 0 &{}\quad i \end{array}\right) \\ U_2^2=I_2\otimes \sigma _z &{} \quad U_2^3=\sigma _z\otimes \left( \begin{array}{cc} 1 &{}\quad 0 \\ 0 &{}\quad -i \end{array}\right) \\ \end{array} \end{aligned}$$
(37)

From the above discussions, we can obtain that arbitrary two-qubit states can be transmitted deterministically by using our scheme. It should be emphasized that the successful probability is equal to one. In this JRSP scheme, Alice needs 4 bits classical information to tell the measurement outcomes to Charlie and Bob. Meanwhile, 2 bits classical communication cost is required for Charlie to inform Bob of his measurement results. Thus, the total classical information is equal to 6 bits. We would like to point out that an efficient three-party RSP scheme for entangled two-qubit states from a sender to either of two receivers is presented by Dai et al. [44]. It is shown that total classical communication costs of such a RSP scheme in a general case and two particular cases via the maximally entangled channel are 2.5 bits and 5 bits, respectively. The classical information costs are less than our RSP protocol for two-qubit states. Actually, the classical message required in RSP proposals is primarily determined by the initial condition, communication task, entanglement sources, implement steps and so on. Compared with our novel protocol, the sender of former scheme has the whole information of quantum state they want to prepare, while the senders of current JRSP scheme separately have partial information. The communication task in Ref. [44] is to prepare entangled two-qubit states from one sender to either of two receiver, and arbitrary two-qubit states from two senders to one receiver can be prepared via our RSP protocol, whatever the prepared state is entangled or not. Quantum channel in Ref. [44] is the combination of a non-maximally entangled two-qubit state and a partially entangled three-qubit state, and two maximally entangled three-qubit states are required for remotely preparing one two-qubit state in this novel proposal. The implement steps of the former scheme include one single-qubit measurement and one two-qubit measurement, only the two-qubit measurement is relative with the information of the prepared state, and two kinds of two-qubit quantum measurements, both of which are corresponding to the prepared two-qubit state, would be performed by the two senders in our scheme, respectively. Additionally, the projective measurements of previous scheme are mutually independent, meanwhile they are simpler than ones of our JRSP scheme. These features are useful for the physical realization. The new method in this paper and the former proposal complement each other.

4 Discussion and conclusions

In summary, we put forward a general proposal to deterministically prepare arbitrary multi-qubit states by using of maximally three-qubit entangled states. Two special kinds of mutually orthogonal measurement basis, of which the analytical expressions are presented in the form of iterative process, are constructed without the introduction of auxiliary particles. Furthermore, the realization procedures of this novel protocol are elaborated in detail, and the total classical communication cost required in our JRSP scheme is also calculated. In contrast to previous methods, the significant advantage is that the successful probability of this JRSP proposal for arbitrary multi-qubit states can reach up to 100%. Frankly speaking, it is a prerequisite of this advantage that quantum channel is composed of maximally three-qubit entangled state in our scheme. Actually, there are several protocols presented for RSP via various quantum entanglement channels. Further research will focus on the schemes for preparing arbitrary multi-qubit states by using partially entangled states.