期刊菜单

Probabilistic Remote State Preparation of an Arbitrary Two-Qubit State via Two Three-Qubit Partially Entangled States
DOI: 10.12677/MP.2017.76031, PDF, HTML, XML, 下载: 1,770  浏览: 3,820  国家自然科学基金支持

Abstract: In this paper, we presented a novel scheme for probabilistic remote preparation of arbitrary two-qubit state via two three-qubit GHZ states. Any auxiliary particles need not to be introduced in our proposal. The total successful probability is equal to the square of the norm of the minimum amplitude coefficients of the partially entangled channel. The concrete processe of this scheme is presented via some appropriate local unitary operations and measurement basis. When quantum channel is composed of two three-qubit maximally entangled states, the successful probability is equal to 1/4.

1. 引言

2. 概率量子远程态制备协议

$|\psi 〉={c}_{1}|00〉+{c}_{2}|01〉+{c}_{3}|10〉+{c}_{4}|11〉$ (1)

$\begin{array}{l}|{\psi }_{123}〉=a{|000〉}_{123}+b{|111〉}_{123}\\ |{\psi }_{456}〉=d{|000〉}_{456}+e{|111〉}_{456}\end{array}$ (2)

$|{\psi }_{123456}〉=|{\psi }_{123}〉\otimes |{\psi }_{456}〉=\left(a{|000〉}_{123}+b{|111〉}_{123}\text{}\right)\otimes \left(d{|000〉}_{456}+e{|111〉}_{456}\right)$ (3)

Step 1：发送者Alice分别对粒子1与2、粒子4与5执行两粒子幺正矩阵：

$U\left(x,y\right)=\left(\begin{array}{cccc}y/x& -\sqrt{1-{|y/x|}^{2}}& 0& 0\\ \sqrt{1-{|y/x|}^{2}}& y/x& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\text{}\begin{array}{c}x=a,c\\ y=d,e\end{array}$ (4)

$\begin{array}{c}|{\psi }_{123456}^{1}〉=\left[U\left(a,b\right)|{\psi }_{12}^{0}〉\right]\otimes \left[U\left(d,e\right)|{\psi }_{45}^{0}〉\right]\otimes |{\psi }_{36}^{0}〉\\ ={\left(b|000〉+b|101〉+\frac{a}{|a|}\sqrt{{a}^{2}-{|b|}^{2}}|010〉\right)}_{123}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\otimes {\left(e|000〉+e|101〉+\frac{d}{|d|}\sqrt{{d}^{2}-{|e|}^{2}}|010〉\right)}_{456}\end{array}$ (5)

Step 2：为了实现两粒子量子态制备，发送者Alice需要构造一组特殊的测量基：

$\left(\begin{array}{r}\hfill |{\lambda }_{1}〉\\ \hfill |{\lambda }_{2}〉\\ \hfill |{\lambda }_{3}〉\\ \hfill |{\lambda }_{4}〉\end{array}\right)=\left(\begin{array}{rrrr}\hfill {c}_{2}^{*}& \hfill -{c}_{1}^{*}& \hfill {c}_{4}^{*}& \hfill -{c}_{3}^{*}\\ \hfill {c}_{1}& \hfill {c}_{2}& \hfill {c}_{3}& \hfill {c}_{4}\\ \hfill \eta {c}_{2}^{*}& \hfill -\eta {c}_{1}^{*}& \hfill -{\eta }^{-1}{c}_{4}^{*}& \hfill {\eta }^{-1}{c}_{3}^{*}\\ \hfill \eta {c}_{1}& \hfill \eta {c}_{2}& \hfill -{\eta }^{-1}{c}_{3}& \hfill -{\eta }^{-1}{c}_{4}\end{array}\right)\left(\begin{array}{r}\hfill |00〉\\ \hfill |01〉\\ \hfill |10〉\\ \hfill |11〉\end{array}\right)$ (6)

$\begin{array}{c}|{\psi }_{123456}^{1}〉={p}_{1}{|00〉}_{25}\otimes {|{\lambda }_{1}〉}_{14}\otimes {\left({c}_{2}|00〉-{c}_{1}|01〉+{c}_{4}|10〉-{c}_{3}|11〉\right)}_{36}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{p}_{1}{|00〉}_{25}\otimes {|{\lambda }_{2}〉}_{14}\otimes {\left({c}_{1}^{*}|00〉+{c}_{2}^{*}|01〉+{c}_{3}^{*}|10〉+{c}_{4}^{*}|11〉\right)}_{36}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{p}_{1}{|00〉}_{25}\otimes {|{\lambda }_{3}〉}_{14}\otimes {\left(\eta {c}_{2}|00〉-\eta {c}_{1}|01〉-{\eta }^{-1}{c}_{4}|10〉+{\eta }^{-1}{c}_{3}|11〉\right)}_{36}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{p}_{1}{|00〉}_{25}\otimes {|{\lambda }_{4}〉}_{14}\otimes {\left(\eta {c}_{1}^{*}|00〉+\eta {c}_{2}^{*}|01〉-{\eta }^{-1}{c}_{3}^{*}|10〉-{\eta }^{-1}{c}_{4}^{*}|11〉\right)}_{36}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{p}_{2}{|10〉}_{25}\otimes {|010〉}_{13}\otimes {\left(e|00〉+e|11〉\right)}_{46}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{p}_{3}{|01〉}_{25}\otimes {\left(b|00〉+b|11〉\right)}_{13}\otimes {|010〉}_{46}+{p}_{4}{|11〉}_{25}\otimes {|0000〉}_{1346}\end{array}$ (7)

$\begin{array}{ll}{p}_{1}=be\hfill & {p}_{2}=ae\sqrt{1-{|b/a|}^{2}}\hfill \\ {p}_{3}=bd\sqrt{1-{|e/d|}^{2}}\hfill & {p}_{4}=ad\sqrt{\left(1-{|b/a|}^{2}\right)\left(1-{|e/d|}^{2}\right)}\hfill \end{array}$

Step 3：发送者Alice对粒子1与4执行测量 $\left\{|{\lambda }_{i}〉|i=1,2,3,4\right\}$ 。当且仅当测量值等于 ${|{\lambda }_{1}〉}_{14}$ ，且粒子2与5等于 ${|00〉}_{25}$ ，远程态制备可以成功，此时粒子3与6状态可以表示为

${\left({c}_{2}|00〉-{c}_{1}|01〉+{c}_{4}|10〉-{c}_{3}|11〉\right)}_{36}$ (8)

Step 4：随后接受者Bob对粒子进行单粒子幺正矩阵，即可得到原始两粒子量子态，总体成功概率为 ${|{p}_{1}|}^{2}={|be|}^{2}$

3. 总结

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