定义：量子通信是稠密编码的逆过程
稠密编码与量子通信的对比：
源发送传递稠密编码 ∣ ϕ + ⟩ \mid\phi^+\rang ∣ϕ+⟩1 qubit2 bits量子通信 ∣ ϕ + ⟩ \mid\phi^+\rang ∣ϕ+⟩2 bits1 qubit定义：量子通信是一个信息协议，Alice通过与Bob共享一个最大纠缠态，传送一个未知的qubit、发送两个经典bits给远处的Bob
假定Alice和Bob在两个不同的地方，现在Alice打算传递一个未知的qubit ∣ ψ ⟩ = α ∣ 0 ⟩ + β ∣ 1 ⟩ |\psi\rang=\alpha|0\rang+\beta|1\rang ∣ψ⟩=α∣0⟩+β∣1⟩给Bob，即：从 ∣ ψ ⟩ A |\psi\rang_A ∣ψ⟩A到 ∣ ψ ⟩ B |\psi\rang_B ∣ψ⟩B
注意：在量子通信之后， ∣ ψ ⟩ A |\psi\rang_A ∣ψ⟩A必须被销毁。否则， ∣ ψ ⟩ A → ∣ ψ ⟩ A ⊗ ∣ ψ ⟩ B |\psi\rang_A\rightarrow|\psi\rang_A\otimes|\psi\rang_B ∣ψ⟩A→∣ψ⟩A⊗∣ψ⟩B将会违背不可克隆定理
Alice和Bob共享一个最大纠缠态 ∣ β 00 ⟩ |\beta_{00}\rang ∣β00⟩，则此时准备的态为： ∣ ψ ⟩ A ⊗ ∣ β 00 ⟩ |\psi\rang_A\otimes|\beta_{00}\rang ∣ψ⟩A⊗∣β00⟩
定理： ∣ ψ ⟩ A ⊗ ∣ β 00 ⟩ A B = 1 2 ∑ i , j = 0 1 ∣ β i j ⟩ A A ⊗ X j Z i ∣ ψ ⟩ B |\psi\rang_A\otimes|\beta_{00}\rang_{AB}=\dfrac{1}{2}\sum\limits_{i,j=0}^1|\beta_{ij}\rang_{AA}\otimes X^jZ^i|\psi\rang_B ∣ψ⟩A⊗∣β00⟩AB=21i,j=0∑1∣βij⟩AA⊗XjZi∣ψ⟩B 证明： ∣ ψ ⟩ = α ∣ 0 ⟩ + β ∣ 1 ⟩ |\psi\rang=\alpha|0\rang+\beta|1\rang ∣ψ⟩=α∣0⟩+β∣1⟩
∣ 00 ⟩ = 1 2 ( ∣ β 00 ⟩ + ∣ β 10 ⟩ ) |00\rang=\dfrac{1}{\sqrt 2}(|\beta_{00}\rang+|\beta_{10}\rang) ∣00⟩=2 1(∣β00⟩+∣β10⟩)
∣ 01 ⟩ = 1 2 ( ∣ β 01 ⟩ + ∣ β 11 ⟩ ) |01\rang=\dfrac{1}{\sqrt 2}(|\beta_{01}\rang+|\beta_{11}\rang) ∣01⟩=2 1(∣β01⟩+∣β11⟩)
∣ 10 ⟩ = 1 2 ( ∣ β 01 ⟩ − ∣ β 11 ⟩ ) |10\rang=\dfrac{1}{\sqrt 2}(|\beta_{01}\rang-|\beta_{11}\rang) ∣10⟩=2 1(∣β01⟩−∣β11⟩)
∣ 11 ⟩ = 1 2 ( ∣ β 00 ⟩ − ∣ β 10 ⟩ ) |11\rang=\dfrac{1}{\sqrt 2}(|\beta_{00}\rang-|\beta_{10}\rang) ∣11⟩=2 1(∣β00⟩−∣β10⟩)
∣ ψ ⟩ A ∣ β 00 ⟩ A B = 1 2 ( α ∣ 0 ⟩ A + β ∣ 1 ⟩ A ) ( ∣ 00 ⟩ A B + ∣ 11 ⟩ A B ) = 1 2 ( α ∣ 000 ⟩ + α ∣ 011 ⟩ + β ∣ 100 ⟩ + β ∣ 111 ⟩ ) A A B = 1 2 α ( ∣ β 00 ⟩ + ∣ β 10 ⟩ ) ∣ 0 ⟩ A A B + 1 2 α ( ∣ β 01 ⟩ + ∣ β 11 ⟩ ) A A ∣ 1 ⟩ B + 1 2 β ( ∣ β 01 ⟩ − ∣ β 11 ⟩ ) A A ∣ 0 ⟩ B + 1 2 β ( ∣ β 00 ⟩ A A − ∣ β 10 ⟩ A A ) ∣ 1 ⟩ B = 1 2 ∣ β 00 ⟩ A A ( α ∣ 0 ⟩ + β ∣ 1 ⟩ ) B + 1 2 ∣ β 10 ⟩ A A ( α ∣ 0 ⟩ − β ∣ 1 ⟩ ) B + 1 2 ∣ β 01 ⟩ A A ( α ∣ 1 ⟩ + β ∣ 0 ⟩ ) B + 1 2 ∣ β 11 ⟩ A A ( α ∣ 1 ⟩ − β ∣ 0 ⟩ ) B = 1 2 ∣ β 00 ⟩ A A ∣ ψ ⟩ B + 1 2 ∣ β 10 ⟩ A A Z ∣ ψ ⟩ B + 1 2 ∣ β 01 ⟩ A A X ∣ ψ ⟩ B + 1 2 ∣ β 11 ⟩ A A X Z ∣ ψ ⟩ B = 1 2 ∑ i , j = 0 1 ∣ β i j ⟩ A A X j Z i ∣ ψ ⟩ B \begin{aligned}|\psi\rang_A|\beta_{00}\rang_{AB}&=\frac{1}{\sqrt 2}(\alpha|0\rang_A+\beta|1\rang_A)(|00\rang_{AB}+|11\rang_{AB})\\ &=\frac{1}{\sqrt 2}(\alpha|000\rang+\alpha|011\rang+\beta|100\rang+\beta|111\rang)_{AAB}\\ &=\frac{1}{2}\alpha(|\beta_{00}\rang+|\beta_{10}\rang)|0\rang_{AAB}+\frac{1}{2}\alpha(|\beta_{01}\rang+|\beta_{11}\rang)_{AA}|1\rang_B\\ &+\frac{1}{2}\beta(|\beta_{01}\rang-|\beta_{11}\rang)_{AA}|0\rang_B+\frac{1}{2}\beta(|\beta_{00}\rang_{AA}-|\beta_{10}\rang_{AA})|1\rang_B\\ &=\frac{1}{2}|\beta_{00}\rang_{AA}(\alpha|0\rang+\beta|1\rang)_B+\frac{1}{2}|\beta_{10}\rang_{AA}(\alpha|0\rang-\beta|1\rang)_B\\ &+\frac{1}{2}|\beta_{01}\rang_{AA}(\alpha|1\rang+\beta|0\rang)_B+\frac{1}{2}|\beta_{11}\rang_{AA}(\alpha|1\rang-\beta|0\rang)_B\\ &=\frac{1}{2}|\beta_{00}\rang_{AA}|\psi\rang_B+\frac{1}{2}|\beta_{10}\rang_{AA}Z|\psi\rang_B\\ &+\frac{1}{2}|\beta_{01}\rang_{AA}X|\psi\rang_B+\frac{1}{2}|\beta_{11}\rang_{AA}XZ|\psi\rang_B\\ &=\frac{1}{2}\sum\limits_{i,j=0}^1|\beta_{ij}\rang_{AA}X^jZ^i|\psi\rang_B\end{aligned} ∣ψ⟩A∣β00⟩AB=2 1(α∣0⟩A+β∣1⟩A)(∣00⟩AB+∣11⟩AB)=2 1(α∣000⟩+α∣011⟩+β∣100⟩+β∣111⟩)AAB=21α(∣β00⟩+∣β10⟩)∣0⟩AAB+21α(∣β01⟩+∣β11⟩)AA∣1⟩B+21β(∣β01⟩−∣β11⟩)AA∣0⟩B+21β(∣β00⟩AA−∣β10⟩AA)∣1⟩B=21∣β00⟩AA(α∣0⟩+β∣1⟩)B+21∣β10⟩AA(α∣0⟩−β∣1⟩)B+21∣β01⟩AA(α∣1⟩+β∣0⟩)B+21∣β11⟩AA(α∣1⟩−β∣0⟩)B=21∣β00⟩AA∣ψ⟩B+21∣β10⟩AAZ∣ψ⟩B+21∣β01⟩AAX∣ψ⟩B+21∣β11⟩AAXZ∣ψ⟩B=21i,j=0∑1∣βij⟩AAXjZi∣ψ⟩B
批注：这里运用了态函数的线性叠加原理
Alice对系统进行Bell测量：
X ⊗ X ∣ β i j ⟩ A A = ( − 1 ) i ∣ β i j ⟩ A A X\otimes X|\beta_{ij}\rang_{AA}=(-1)^i|\beta_{ij}\rang_{AA} X⊗X∣βij⟩AA=(−1)i∣βij⟩AA
Z ⊗ Z ∣ β i j ⟩ A A = ( − 1 ) j ∣ β i j ⟩ A A Z\otimes Z|\beta_{ij}\rang_{AA}=(-1)^j|\beta_{ij}\rang_{AA} Z⊗Z∣βij⟩AA=(−1)j∣βij⟩AA
其中，i和j是两个经典bits，对于不同的态，测量结果如下：
∣ β i j ⟩ \mid\beta_{ij}\rang ∣βij⟩ X ⊗ X X\otimes X X⊗X / ( − 1 ) i (-1)^i (−1)i Z ⊗ Z Z\otimes Z Z⊗Z / ( − 1 ) j (-1)^j (−1)jtwo-bit ∣ β 00 ⟩ \mid\beta_{00}\rang ∣β00⟩11(0,0) ∣ β 01 ⟩ \mid\beta_{01}\rang ∣β01⟩1-1(0,1) ∣ β 10 ⟩ \mid\beta_{10}\rang ∣β10⟩-11(1,0) ∣ β 11 ⟩ \mid\beta_{11}\rang ∣β11⟩-1-1(1,1)在进行Bell测量前，密度矩阵为：
ρ A A B = ∣ Ψ ⟩ A A B ⟨ Ψ ∣ \rho_{AAB}=|\Psi\rang_{AAB}\lang\Psi| ρAAB=∣Ψ⟩AAB⟨Ψ∣
其中： ∣ Ψ ⟩ A A B = ∣ Ψ ⟩ A ∣ β 00 ⟩ A B |\Psi\rang_{AAB}=|\Psi\rang_A|\beta_{00}\rang_{AB} ∣Ψ⟩AAB=∣Ψ⟩A∣β00⟩AB
在进行Bell测量后， ∣ Ψ ⟩ A A B |\Psi\rang_{AAB} ∣Ψ⟩AAB分别有四分之一的概率变为 ∣ Φ i j ⟩ A A B |\Phi_{ij}\rang_{AAB} ∣Φij⟩AAB
∣ Φ i j ⟩ A A B = ∣ β i j ⟩ A A X j Z i ∣ Ψ ⟩ B |\Phi_{ij}\rang_{AAB}=|\beta_{ij}\rang_{AA}X^jZ^i|\Psi\rang_B ∣Φij⟩AAB=∣βij⟩AAXjZi∣Ψ⟩B
混合态：
ρ A A B 1 = ∑ i , j = 0 1 1 4 ∣ Φ i j ⟩ A A B ⟨ Φ i j ∣ \rho_{AAB}^1=\sum\limits_{i,j=0}^1\dfrac{1}{4}|\Phi_{ij}\rang_{AAB}\lang\Phi_{ij}| ρAAB1=i,j=0∑141∣Φij⟩AAB⟨Φij∣
在Bob的体系中，密度矩阵简化：
ρ B 1 = t r A A ρ A A B 1 = ∑ k , l = 0 1 A A ⟨ β k l ∣ ρ A A B 1 ∣ β k l ⟩ A A \rho_B^1=tr_{AA}\rho_{AAB}^1=\sum\limits_{k,l=0}^1\ _{AA}\lang\beta_{kl}|\rho_{AAB}^1|\beta_{kl}\rang_{AA} ρB1=trAAρAAB1=k,l=0∑1 AA⟨βkl∣ρAAB1∣βkl⟩AA
ρ B 1 = 1 4 ∑ k , l = 0 1 X k Z l ∣ ψ ⟩ B ⟨ ψ ∣ Z l X k \rho_B^1=\dfrac{1}{4}\sum\limits_{k,l=0}^1X^kZ^l|\psi\rang_B\lang\psi| Z^lX^k ρB1=41k,l=0∑1XkZl∣ψ⟩B⟨ψ∣ZlXk
∣ ψ ⟩ B = α ∣ 0 ⟩ + β ∣ 1 ⟩ |\psi\rang_B=\alpha|0\rang+\beta|1\rang ∣ψ⟩B=α∣0⟩+β∣1⟩
ρ B 1 = 1 2 ( ∣ 0 ⟩ B ⟨ 0 ∣ + ∣ 1 ⟩ B ⟨ 1 ∣ ) = 1 2 I 2 \rho_B^1=\dfrac{1}{2}(|0\rang_B\lang 0|+|1\rang_B\lang 1|)=\dfrac{1}{2}I_2 ρB1=21(∣0⟩B⟨0∣+∣1⟩B⟨1∣)=21I2
此混合态正好对应于Bloch球的球心
Alice将测量的输出值(i, j)发送给Bob：
∣ Ψ ⟩ A A B → ∣ Φ i j ⟩ A A B = ∣ β i j ⟩ A A ⊗ X j Z i ∣ ψ ⟩ B |\Psi\rang_{AAB}\rightarrow|\Phi_{ij}\rang_{AAB}=|\beta_{ij}\rang_{AA}\otimes X^jZ^i|\psi\rang_B ∣Ψ⟩AAB→∣Φij⟩AAB=∣βij⟩AA⊗XjZi∣ψ⟩B
在Alice测量结束的时候，Bob对自己的qubit的情况是不知道的
根据Alice和Bob之间的协议，Bob对Alice的信息(i, j)进行操作：
( I 2 ⊗ I 2 ⊗ Z i X j ) ( ∣ β i j ⟩ A A ⊗ X j Z i ∣ ψ ⟩ B ) = ∣ β i j ⟩ A A ⊗ ∣ ψ ⟩ B (I_2\otimes I_2\otimes Z^iX^j)(|\beta_{ij}\rang_{AA}\otimes X^jZ^i|\psi\rang_B)=|\beta_{ij}\rang_{AA}\otimes|\psi\rang_B (I2⊗I2⊗ZiXj)(∣βij⟩AA⊗XjZi∣ψ⟩B)=∣βij⟩AA⊗∣ψ⟩B
此时，Alice的原始 ∣ ψ ⟩ |\psi\rang ∣ψ⟩被摧毁，Bob得到了一个完美的 ∣ ψ ⟩ |\psi\rang ∣ψ⟩复制品
量子通信的量子电路如图1所示：
图1 量子通信的量子电路其中：
Alice测量的结果：
测量时，使用了投影测量方式，相应的图像表示分别为：
