# 不可克隆原理

## 证明

${\displaystyle |\phi \rangle }$${\displaystyle |\psi \rangle }$ 是两个任意的量子状态，我们要把这两个状态拷贝到另一个与他们完全无关的状态${\displaystyle |k\rangle }$上。我们用一个幺正算符${\displaystyle U}$来描述这个过程。则这个拷贝算符必须具备以下性质：

${\displaystyle U(|\phi \rangle \otimes |k\rangle )=|\phi \rangle \otimes |\phi \rangle }$
${\displaystyle U(|\psi \rangle \otimes |k\rangle )=|\psi \rangle \otimes |\psi \rangle }$

${\displaystyle \langle U(\phi \otimes k)|U(\psi \otimes k)\rangle =\langle \phi \otimes \phi |\psi \otimes \psi \rangle }$
${\displaystyle \langle U(\phi \otimes k)|U(\psi \otimes k)\rangle =\langle \phi \otimes k|\psi \otimes k\rangle }$

${\displaystyle \langle \phi \otimes \phi |\psi \otimes \psi \rangle =\langle \phi \otimes k|\psi \otimes k\rangle ,}$
${\displaystyle {\to }}$
${\displaystyle \langle \phi |\psi \rangle \langle \phi |\psi \rangle =\langle \phi |\psi \rangle \langle k|k\rangle \,.}$

${\displaystyle \langle \phi |\psi \rangle ^{2}=\langle \phi |\psi \rangle .}$

## 举例

${\displaystyle |\psi \rangle }$${\displaystyle |0\rangle }$ 作为输入：

${\displaystyle |\psi \rangle |0\rangle ={\bigg (}a|0\rangle +b|1\rangle {\bigg )}|0\rangle =a|00\rangle +b|10\rangle }$

${\displaystyle a|00\rangle +b|11\rangle }$