量子隐形传态(Quantum teleportation)

• 原文: https://zhuanlan.zhihu.com/p/91069177
• 发布日期: 2019-11-09
• 分类: 量子力学基础与量子信息

O. 基本设定:

✦ qubit: 即俩态量子系统, 通常用电子的自旋作为例子, 而态空间基底通常记为 $\left\{|0⟩,|1⟩\right\}$

✦ 四个Bell 态: 即贝尔基, 是两 qubit 空间的一组完备基矢, 同时是系统的四个最大纠缠态.

它们分别被记为 \left\{ \begin{align} & \left| {\Psi }^{\pm } \right\rangle =\frac{1}{\sqrt{2}\left( \left| 01 \right\rangle \pm \left| 10 \right\rangle \right) \\ & \left| {\Phi }^{\pm } \right\rangle =\frac{1}{\sqrt{2}\left( \left| 00 \right\rangle \pm \left| 11 \right\rangle \right) \\ \end{align} \right. 其中 $|xy⟩=|x⟩|y⟩=|x⟩\otimes |y⟩$.

✦ 三个泡利算符的作用: \left\{ \begin{align} & {\sigma }_{z}\left| 0 \right\rangle =\left| 0 \right\rangle \ \ \ \ \ \ {\sigma }_{z}\left| 1 \right\rangle =-\left| 1 \right\rangle \\ & {\sigma }_{y}\left| 0 \right\rangle =i\left| 1 \right\rangle \ \ \ \ {\sigma }_{y}\left| 1 \right\rangle =-i\left| 0 \right\rangle \\ & {\sigma }_{x}\left| 0 \right\rangle =\left| 1 \right\rangle \ \ \ \ \ \ {\sigma }_{x}\left| 1 \right\rangle =\left| 0 \right\rangle \\ \end{align} \right. 这里设置 \left\{ \begin{align} & \left| 0 \right\rangle =\left| \uparrow \right\rangle \\ & \left| 1 \right\rangle =\left| \downarrow \right\rangle \\ \end{align} \right.

I. 背景:

假设Alice 手上有俩qubit , 分别记作 $a$ 与 $c$ , 而Bob 手上的一个记作 $b$ .

已知 $a$ 与 $b$ 处于最大纠缠态(即 Bell 态之一), 即 {\left| {\Psi }^{-} \right\rangle }_{ab}=\frac{1}{\sqrt{2}\left( {\left| 0 \right\rangle }_{a}{\left| 1 \right\rangle }_{b}-{\left| 1 \right\rangle }_{a}{\left| 0 \right\rangle }_{b} \right) , 而系统 $c$ 的态为 ${|\phi ⟩}_c=\alpha {|0⟩}_c+\beta {|1⟩}_c$ , 现在我们的目的就是利用 $a$ 与 $b$ 的纠缠态将$c$的态转移到$b$ 上面去.

我们接下来利用的是一对处于纠缠态的粒子, 所以纠缠实际上是一种资源. 量子隐形传态的目的是不传输粒子本身而是将其量子态传到另一个粒子上.

II. 操作步骤:

1. 先将三个系统看作一个总系统:

\begin{align} & {\left| \psi \right\rangle }_{abc}={\left| {\Psi }^{-} \right\rangle }_{ab}\otimes {\left| \varphi \right\rangle }_{c}=\frac{1}{\sqrt{2}\left( {\left| 0 \right\rangle }_{a}{\left| 1 \right\rangle }_{b}-{\left| 1 \right\rangle }_{a}{\left| 0 \right\rangle }_{b} \right)\left( \alpha {\left| 0 \right\rangle }_{c}+\beta {\left| 1 \right\rangle }_{c} \right) \\ & \Rightarrow {\left| \psi \right\rangle }_{abc}=\frac{1}{\sqrt{2}\left( \alpha {\left| 0 \right\rangle }_{a}{\left| 0 \right\rangle }_{c}{\left| 1 \right\rangle }_{b}+\beta {\left| 0 \right\rangle }_{a}{\left| 1 \right\rangle }_{c}{\left| 1 \right\rangle }_{b}-\alpha {\left| 1 \right\rangle }_{a}{\left| 0 \right\rangle }_{c}{\left| 0 \right\rangle }_{b}-\beta {\left| 1 \right\rangle }_{a}{\left| 1 \right\rangle }_{c}{\left| 0 \right\rangle }_{b} \right) \\ \end{align}

2. 接下来用四个Bell 态$\left\{|{\Phi }^{\pm }⟩,|{\Psi }^{\pm }⟩\right\}$去展开a,c 构成的复合系统:

\left\{ \begin{align} & {\left| 0 \right\rangle }_{a}{\left| 0 \right\rangle }_{c}=\frac{1}{\sqrt{2}\left( {\left| {\Phi }^{+} \right\rangle }_{ac}+{\left| {\Phi }^{-} \right\rangle }_{ac} \right) \\ & {\left| 1 \right\rangle }_{a}{\left| 1 \right\rangle }_{c}=\frac{1}{\sqrt{2}\left( {\left| {\Phi }^{+} \right\rangle }_{ac}-{\left| {\Phi }^{-} \right\rangle }_{ac} \right) \\ \end{align} \right. \left\{ \begin{align} & {\left| 0 \right\rangle }_{a}{\left| 1 \right\rangle }_{c}=\frac{1}{\sqrt{2}\left( {\left| {\Psi }^{+} \right\rangle }_{ac}+{\left| {\Psi }^{-} \right\rangle }_{ac} \right) \\ & {\left| 1 \right\rangle }_{a}{\left| 0 \right\rangle }_{c}=\frac{1}{\sqrt{2}\left( {\left| {\Psi }^{+} \right\rangle }_{ac}-{\left| {\Psi }^{-} \right\rangle }_{ac} \right) \\ \end{align} \right.

将上述形式代入总系统得到

{\left| \psi \right\rangle }_{abc}=\frac{1}{2}\left[ \begin{align} & \alpha \left( {\left| {\Phi }^{+} \right\rangle }_{ac}+{\left| {\Phi }^{-} \right\rangle }_{ac} \right){\left| 1 \right\rangle }_{b}+\beta \left( {\left| {\Psi }^{+} \right\rangle }_{ac}+{\left| {\Psi }^{-} \right\rangle }_{ac} \right){\left| 1 \right\rangle }_{b} \\ & -\alpha \left( {\left| {\Psi }^{+} \right\rangle }_{ac}-{\left| {\Psi }^{-} \right\rangle }_{ac} \right){\left| 0 \right\rangle }_{b}-\beta \left( {\left| {\Phi }^{+} \right\rangle }_{ac}-{\left| {\Phi }^{-} \right\rangle }_{ac} \right){\left| 0 \right\rangle }_{b} \\ \end{align} \right]

整理一下得到

{\left| \psi \right\rangle }_{abc}=\frac{1}{2}\left[ \begin{align} & \left( \alpha {\left| 0 \right\rangle }_{b}+\beta {\left| 1 \right\rangle }_{b} \right){\left| {\Psi }^{-} \right\rangle }_{ac}+\left( -\alpha {\left| 0 \right\rangle }_{b}+\beta {\left| 1 \right\rangle }_{b} \right){\left| {\Psi }^{+} \right\rangle }_{ac} \\ & +\left( -\beta {\left| 0 \right\rangle }_{b}+\alpha {\left| 1 \right\rangle }_{b} \right){\left| {\Phi }^{+} \right\rangle }_{ac}+\left( \beta {\left| 0 \right\rangle }_{b}+\alpha {\left| 1 \right\rangle }_{b} \right){\left| {\Phi }^{-} \right\rangle }_{ac} \\ \end{align} \right]

3. 接下来只需Alice 去测量a,c 构成的复合系统处于哪个Bell 态, 而测量有四个可能结果:

每个可能结果都将给出一个 $b$ 态: \left\{ \begin{align} & {\left| \phi \right\rangle }_{ac}={\left| {\Psi }^{-} \right\rangle }_{ac}\Rightarrow {\left| {\varphi }_{1} \right\rangle }_{b}=\alpha {\left| 0 \right\rangle }_{b}+\beta {\left| 1 \right\rangle }_{b} \\ & {\left| \phi \right\rangle }_{ac}={\left| {\Psi }^{+} \right\rangle }_{ac}\Rightarrow {\left| {\varphi }_{2} \right\rangle }_{b}=-\alpha {\left| 0 \right\rangle }_{b}+\beta {\left| 1 \right\rangle }_{b} \\ & {\left| \phi \right\rangle }_{ac}={\left| {\Phi }^{-} \right\rangle }_{ac}\Rightarrow {\left| {\varphi }_{3} \right\rangle }_{b}=-\beta {\left| 0 \right\rangle }_{b}+\alpha {\left| 1 \right\rangle }_{b} \\ & {\left| \phi \right\rangle }_{ac}={\left| {\Phi }^{+} \right\rangle }_{ac}\Rightarrow {\left| {\varphi }_{4} \right\rangle }_{b}=\beta {\left| 0 \right\rangle }_{b}+\alpha {\left| 1 \right\rangle }_{b} \\ \end{align} \right.

若 ${|\varphi ⟩}_{ac}={|{\Psi }^{-}⟩}_{ac}$

则 $b$ 就已经处于 $c$ 一开始的态 ${|{\phi }_1⟩}_b=\alpha {|0⟩}_b+\beta {|1⟩}_b={|\phi ⟩}_b$了.

若 ${|\varphi ⟩}_{ac}={|{\Psi }^{+}⟩}_{ac}$

则 Bob 需对 $b$ 进行一个 ${\sigma }_z$ 操作, 因为${-|\phi ⟩}_b={\sigma }_z\left(-\alpha {|0⟩}_b+\beta {|1⟩}_b\right)$.

若 ${|\varphi ⟩}_{ac}={|{\Phi }^{-}⟩}_{ac}$

则 Bob 需对 $b$ 进行一个 ${\sigma }_y$ 操作, 因为 $-i{|\phi ⟩}_b={\sigma }_y\left(-\beta {|0⟩}_b+\alpha {|1⟩}_b\right)$.

若 ${|\varphi ⟩}_{ac}={|{\Phi }^{+}⟩}_{ac}$

则Bob 需对 $b$ 进行一个 ${\sigma }_x$ 操作, 因为 ${|\phi ⟩}_b={\sigma }_x\left(\beta {|0⟩}_b+\alpha {|1⟩}_b\right)$.

4. 现在Alice 把测量的结果告诉Bob, 而Bob 只需要按照上面的步骤去进行操作即可. 这样一来我们的b 就取得了c 系统的态. 由于微观粒子不可分辨, 这么一来, 就颇有些借尸还魂的味道.

这里强调一下, 第4步是涉及经典通讯的, 所以没破坏局域性. 主要是担心某些网友看完又搞出个什么大新闻.

值得思考的地方: 你认为经典系统能做到这一点吗?

[额外内容] 纠缠交换(entanglement swapping ):

1. 概念:

纠缠交换就是说我们能将两对分别处于最大纠缠态的粒子的纠缠对象进行交换.

例如粒子 $1,2$ 处于Bell 态 {\left| {\Phi }^{+} \right\rangle }_{12}=\frac{1}{\sqrt{2}\left( {\left| 00 \right\rangle }_{12}\text{+}{\left| 11 \right\rangle }_{12} \right)

而粒子 $3,4$ 处于另一个Bell 态 {\left| {\Psi }^{\text{+} \right\rangle }_{\text{34}=\frac{1}{\sqrt{2}\left( {\left| 01 \right\rangle }_{\text{34}\text{+}{\left| 10 \right\rangle }_{\text{34} \right)

我们现在可以通过对 $2,3$ 进行操作使 $1,4$ 系统处于纠缠态, 这就是纠缠交换.

2. 处理方法:

方法就是对 $2,3$ 进行Bell 基测量, 先写出四个粒子的总系统态矢量:

\begin{align} & \left| \psi \right\rangle \text{=}{\left| {\Phi }^{+} \right\rangle }_{12}\otimes {\left| {\Psi }^{\text{+} \right\rangle }_{\text{34} \\ & \ \ \ \ \ =\frac{\text{1}{\text{2}\left( \left| 0001 \right\rangle +\left| 0010 \right\rangle +\left| 1101 \right\rangle +\left| 1110 \right\rangle \right) \\ & \ \ \ \ \ =\frac{1}{2}\left( {\left| 00 \right\rangle }_{23}{\left| 01 \right\rangle }_{14}+{\left| 01 \right\rangle }_{23}{\left| 00 \right\rangle }_{14}+{\left| 10 \right\rangle }_{23}{\left| 11 \right\rangle }_{14}+{\left| 11 \right\rangle }_{23}{\left| 10 \right\rangle }_{14} \right) \\ \end{align}

用用四个Bell 态展开 $2,3$ 系统:

\left| \psi \right\rangle =\frac{1}{2\sqrt{2}\left[ \begin{align} & \left( {\left| {\Phi }^{+} \right\rangle }_{23}+{\left| {\Phi }^{-} \right\rangle }_{23} \right){\left| 01 \right\rangle }_{14}+\left( {\left| {\Psi }^{+} \right\rangle }_{23}+{\left| {\Psi }^{-} \right\rangle }_{23} \right){\left| 00 \right\rangle }_{14} \\ & +\left( {\left| {\Psi }^{+} \right\rangle }_{23}-{\left| {\Psi }^{-} \right\rangle }_{23} \right){\left| 11 \right\rangle }_{14}+\left( {\left| {\Phi }^{+} \right\rangle }_{23}-{\left| {\Phi }^{-} \right\rangle }_{23} \right){\left| 10 \right\rangle }_{14} \\ \end{align} \right] \ \ \ \ \ \ =\frac{1}{2\sqrt{2}\left[ \begin{align} & {\left| {\Psi }^{+} \right\rangle }_{23}\left( {\left| 00 \right\rangle }_{14}+{\left| 11 \right\rangle }_{14} \right)+{\left| {\Psi }^{-} \right\rangle }_{23}\left( {\left| 00 \right\rangle }_{14}-{\left| 11 \right\rangle }_{14} \right) \\ & +{\left| {\Phi }^{+} \right\rangle }_{23}\left( {\left| 01 \right\rangle }_{14}+{\left| 10 \right\rangle }_{14} \right)+{\left| {\Phi }^{-} \right\rangle }_{23}\left( {\left| 01 \right\rangle }_{14}-{\left| 10 \right\rangle }_{14} \right) \\ \end{align} \right] $=\frac{1}{2}\left[{|{\Psi }^{+}⟩}_{23}{|{\Phi }^{+}⟩}_{14}+{|{\Psi }^{-}⟩}_{23}{|{\Phi }^{-}⟩}_{14}+{|{\Phi }^{+}⟩}_{23}{|{\Psi }^{+}⟩}_{14}+{|{\Phi }^{-}⟩}_{23}{|{\Psi }^{-}⟩}_{14}\right]$

接下来就是对 $2,3$ 进行Bell 基测量, 通过 $2,3$ 系统的结果可以确定 $1,4$ 的结果.

如果希望1,4 处于更 specific 的结果, 可以通过相应的幺正操作( ${\sigma }_x,{\sigma }_y,{\sigma }_z$ )来达到.