Deterministic bidirectional controlled remote preparation without information splitting

Abstract

By using a five-qubit entangled state as the quantum channel, we propose an efficient protocol for implementing bidirectional controlled remote preparation of arbitrary single-qubit states with unit success probability. The senders do not need to perform information splitting and additional unitary operations due to the elaborately constructed measurement basis. Furthermore, three-bit classical communication can be saved at the controller’s broadcast channel with the aid of network coding. Then, we consider the effect of five-type noises on the proposed protocol. It is found that the fidelities of the output states are dependent on the coefficients of the prepared state and the noise parameter. Interestingly, the fidelity is irrelevant to the participators’ measurement results except in the amplitude-damping and phase-damping noises.

Appendix

Take the measurement result $|\xi ^1_{00}\rangle |\xi ^2_{10}\rangle |1\rangle$ as an example, we calculate the output state under the bit flip noise in detail.

The effect on the shared channel $|\varOmega '\rangle$ under the bit-flip noise is

$$\begin{aligned} \varepsilon _{bf}(\rho )= & {} \frac{1}{8}\{\lambda ^4\sum \limits _{j=0}^1[(|101\rangle +(-1)^j|010\rangle )^{\otimes 2}|j\rangle ] \sum \limits _{j=0}^1[(\langle 101|+(-1)^j\langle 010|)^{\otimes 2}\langle j|]\\&+(1-\lambda )^4\sum \limits _{j=0}^1[(|000\rangle +(-1)^j|111\rangle )^{\otimes 2}|j\rangle ] \sum \limits _{j=0}^1[(\langle 000|+(-1)^j\langle 111|)^{\otimes 2}\langle j|]\\&+\lambda ^2(1-\lambda )^2\sum \limits _{j=0}^1[(|001\rangle +(-1)^j|110\rangle )(|100\rangle +(-1)^j|011\rangle )|j\rangle \\&\times \sum \limits _{j=0}^1[(\langle 001|+(-1)^j\langle 110|)(\langle 100|+(-1)^j\langle 011|)\langle j|]\\&+\lambda ^2(1-\lambda )^2\sum \limits _{j=0}^1[(|100\rangle +(-1)^j|011\rangle )(|001\rangle +(-1)^j|110\rangle )|j\rangle ] \\&\times \sum \limits _{j=0}^1[(\langle 100|+(-1)^j\langle 011|)(\langle 001|+(-1)^j\langle 110|)\langle j|] \}. \end{aligned}$$

If Alice$_1$’s measurement result is $|\xi ^1_{00}\rangle$, the system of qubits $(2,3,3',4,5)$ becomes

where $g(\lambda )=[\lambda ^2+(1-\lambda )^2]^2$.

If Alice$_2$’s measurement result is $|\xi ^2_{10}\rangle$, the system of qubits (2, 4, 5) becomes

If Charlie’s measurement outcome is $|1\rangle$, the system of qubits (2, 4) becomes

$$\begin{aligned} \rho _3= & {} \mathrm{tr}_{5}\left( \frac{M_C\rho _2M_C^\dag }{\mathrm{tr}(M_C^\dag M_C\rho _2)}\right) \\= & {} \frac{1}{g(\lambda )}\left\{ \lambda ^4(\alpha _1|1\rangle -\beta _1^*|0\rangle )(\alpha _2|0\rangle +\beta _2^*|1\rangle ) (\alpha _1\langle 1|-\beta _1\langle 0|)(\alpha _2\langle 0|+\beta _2\langle 1|)\right. \\&+(1-\lambda )^4(\alpha _1|0\rangle -\beta _1|1\rangle )(\alpha _2|1\rangle +\beta _2|0\rangle ) (\alpha _1\langle 0|-\beta _1^*\langle 1|)(\alpha _2\langle 1|+\beta _2^*\langle 0|)\\&+\lambda ^2(1-\lambda )^2[(\alpha _1|1\rangle -\beta _1|0\rangle )(\alpha _2|1\rangle +\beta _2^*|0\rangle ) (\alpha _1\langle 1|-\beta _1^*\langle 0|)(\alpha _2\langle 1|+\beta _2\langle 0|)\\&\left. +(\alpha _1|0\rangle -\beta _1^*|1\rangle )(\alpha _2|0\rangle +\beta _2|1\rangle ) (\alpha _1\langle 0|-\beta _1\langle 1|)(\alpha _2\langle 0|+\beta _2^*\langle 1|)]\right\} . \end{aligned}$$

According to the measurement result, Alice$_1$ and Alice$_2$ perform recovery unitary operation $U_{A_1A_2}=X_4Z_2$ and get the output state $\rho _{out}^{bf} =U_{A_1A_2}\rho _3U_{A_1A_2}^\dagger$ in Eq. (23).