受控非门(英语:controlled-NOT gate, CNOT)出现在量子线路,是量子版本的逻辑门的一种,牵涉到两个量子比特间的运算。 第一个量子位仅在第二个量子位为1时才会翻转. 数学形式 所涉及的两个量子比特,分别为控制(量子)比特(control qubit)与受控的目标比特(target qubit)。当控制比特是 | 0 ⟩ {\displaystyle |0\rangle } ,则目标比特保持原状态;当控制比特是 | 1 ⟩ {\displaystyle |1\rangle } ,则目标比特翻转,即 | 0 ⟩ {\displaystyle |0\rangle } 成分变为 | 1 ⟩ {\displaystyle |1\rangle } ,而 | 1 ⟩ {\displaystyle |1\rangle } 成分变为 | 0 ⟩ {\displaystyle |0\rangle } 。 写成通式,若c表示控制而t表示目标: ( a | 0 ⟩ + b | 1 ⟩ ) c ⊗ ( α | 0 ⟩ + β | 1 ⟩ ) t {\displaystyle (a|0\rangle +b|1\rangle )_{c}\otimes (\alpha |0\rangle +\beta |1\rangle )_{t}} = a | 0 ⟩ c ⊗ α | 0 ⟩ t + a | 0 ⟩ c ⊗ β | 1 ⟩ t + b | 1 ⟩ c ⊗ α | 0 ⟩ t + b | 1 ⟩ c ⊗ β | 1 ⟩ t {\displaystyle =a|0\rangle _{c}\otimes \alpha |0\rangle _{t}+a|0\rangle _{c}\otimes \beta |1\rangle _{t}+b|1\rangle _{c}\otimes \alpha |0\rangle _{t}+b|1\rangle _{c}\otimes \beta |1\rangle _{t}} = a α | 00 ⟩ c t + a β | 01 ⟩ c t + b α | 10 ⟩ c t + b β | 11 ⟩ c t {\displaystyle =a\alpha |00\rangle _{ct}+a\beta |01\rangle _{ct}+b\alpha |10\rangle _{ct}+b\beta |11\rangle _{ct}} 可以写成张量积的形式,或者拆开来。若经过CNOT的作用: C N O T → a | 0 ⟩ c ⊗ α | 0 ⟩ t + a | 0 ⟩ c ⊗ β | 1 ⟩ t + b | 1 ⟩ c ⊗ α | 1 ⟩ + b | 1 ⟩ c ⊗ β | 0 ⟩ t {\displaystyle {CNOT}\rightarrow a|0\rangle _{c}\otimes \alpha |0\rangle _{t}+a|0\rangle _{c}\otimes \beta |1\rangle _{t}+b|1\rangle _{c}\otimes \alpha |1\rangle +b|1\rangle _{c}\otimes \beta |0\rangle _{t}} = a α | 00 ⟩ c t + a β | 01 ⟩ c t + b α | 11 ⟩ c t + b β | 10 ⟩ c t {\displaystyle =a\alpha |00\rangle _{ct}+a\beta |01\rangle _{ct}+b\alpha |11\rangle _{ct}+b\beta |10\rangle _{ct}} 就一般式子而言不能再写回c和t拆开为张量积的形式 | Ψ ⟩ c ⊗ | Φ ⟩ t {\displaystyle |\Psi \rangle _{c}\otimes |\Phi \rangle _{t}} ,这是量子缠结的来源表征。 若 | 0 ⟩ {\displaystyle |0\rangle } 以 ( 1 0 ) {\displaystyle {\begin{pmatrix}1\\0\end{pmatrix}}} 且 | 1 ⟩ {\displaystyle |1\rangle } 以 ( 0 1 ) {\displaystyle {\begin{pmatrix}0\\1\end{pmatrix}}} 表示,则可将CNOT写为: C N O T = ( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 ) {\displaystyle {CNOT}={\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}}} 操作例子: C N O T | 10 ⟩ = ( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 ) ( 0 0 1 0 ) = ( 0 0 0 1 ) = | 11 ⟩ {\displaystyle {CNOT}|10\rangle ={\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}}{\begin{pmatrix}0\\0\\1\\0\end{pmatrix}}={\begin{pmatrix}0\\0\\0\\1\end{pmatrix}}=|11\rangle } Remove ads与经典逻辑门的对应 CNOT维持|00〉 、|01〉,而将|10〉变|11〉、|11〉变|10〉的特性,相似于古典的异或门(exclusive OR, XOR)维持00、01,将10变11、11变10。 参考文献 Nielsen, Michael A.; Chuang, Isaac L. Quantum Computation and Quantum Information. Cambridge University Press. 2000. ISBN 0-521-63235-8. Monroe, C.; Meekhof, D. & King, B. & Itano, W. & Wineland, D. Demonstration of a Fundamental Quantum Logic Gate. Physical Review Letters. 1995, 75 (25): 4714–4717. Bibcode:1995PhRvL..75.4714M. PMID 10059979. doi:10.1103/PhysRevLett.75.4714. [1](页面存档备份,存于互联网档案馆) 外部连接 Michael Westmoreland: "Isolation and information flow in quantum dynamics" - discussion around the Cnot gate(页面存档备份,存于互联网档案馆) Loading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.Remove ads