Wang2005-公式推导

Wang X B. Beating the photon-number-splitting attack in practical quantum cryptography[J]. Physical review letters, 2005, 94(23): 230503.

主要内容：

• 提出诱骗态协议（1强度信号$\mu$ + 1强度诱骗$\mu^\prime$ + 1真空诱骗，$\mu^{\prime}>\mu$）；
• 得到了一个比Hwang2003更紧凑的$\Delta$估值；
• 对诱骗态协议首次进行有限码长分析。

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1 (1)

命题：
如果$\eta<(1-e^{-\mu}-\mu e^{-\mu})/\mu$，则Eve能通过PNS攻击获得Bob所有sifted key信息。

证明：
通过Hwang2003$(1)$式密钥生成条件可直接导出。

2 (5)

命题：
$(5)$式是$|0\rangle\langle0|$、$|1\rangle\langle1|$、$\rho_{c}$、$\rho_{d}$的凸组合

证明：
即要证明$d=1-e^{-\mu^{\prime}}-\mu^{\prime}e^{-\mu^{\prime}}-c\frac{\mu^{\prime{2}}e^{-\mu^{\prime}}}{\mu^{2}e^{-\mu}}\geq0$.

因为$P_{n}(\mu^{\prime})/P_2(\mu^{\prime})>P_n(\mu)/P_2(\mu)$,

所以：

$$\begin{aligned} \frac{1-e^{-\mu^{\prime}}-\mu^{\prime}e^{-\mu^{\prime}}}{c}-\frac{\mu^{\prime{2}}e^{-\mu^{\prime}}}{\mu^{2}e^{-\mu}} &= \frac{1-e^{-\mu^{\prime}}-\mu^{\prime}e^{-\mu^{\prime}}}{1-e^{-\mu}-\mu e^{-\mu}}-\frac{\mu^{\prime{2}}e^{-\mu^{\prime}}}{\mu^{2}e^{-\mu}}\\ &= \frac{\sum_{n=2}^\infty P_n(\mu^{\prime})}{\sum_{n=2}^\infty P_n(\mu)}-\frac{ P_2(\mu^{\prime})}{ P_2(\mu)} \\ &\geq \frac{\sum_{n=2}^\infty P_n(\mu) \frac{ P_2(\mu^{\prime})}{ P_2(\mu)}}{\sum_{n=2}^\infty P_n(\mu)}-\frac{ P_2(\mu^{\prime})}{ P_2(\mu)} \\ &= 0 \end{aligned}$$

3 (13)

由$(11)$式可得：

$$s_{1}=\frac{S_{\mu}-e^{-\mu}s_{0}-cs_{c}}{\mu e^{-\mu}}$$

带入$(9)$可得：

$$\begin{aligned} cs_{c} &\leq \frac{\mu^{2}e^{-\mu}}{\mu^{\prime2}e^{-\mu^{\prime}}}(S_{\mu^{\prime}}-e^{-\mu^{\prime}}s_{0}-\mu^{\prime}e^{-\mu^{\prime}}s_{1}) \\ &= \frac{\mu^{2}e^{-\mu}}{\mu^{\prime2}e^{-\mu^{\prime}}}[S_{\mu^{\prime}}-e^{-\mu^{\prime}}s_{0}-\frac{\mu^{\prime}e^{-\mu^{\prime}}}{\mu e^{-\mu}}(S_{\mu}-e^{-\mu}s_{0}-cs_{c})] \end{aligned}$$

即：

$$\begin{aligned} (1-\frac{\mu}{\mu^{\prime}})cs_{c} &\leq \frac{\mu^{2}e^{-\mu}}{\mu^{\prime2}e^{-\mu^{\prime}}}[S_{\mu^{\prime}}-e^{-\mu^{\prime}}s_{0}-\frac{\mu^{\prime}e^{-\mu^{\prime}}}{\mu e^{-\mu}}(S_{\mu}-e^{-\mu}s_{0})] \\ &= \frac{\mu^{2}e^{-\mu}}{\mu^{\prime2}e^{-\mu^{\prime}}}S_{\mu^{\prime}}-\frac{\mu}{\mu^{\prime}}S_{\mu}+\frac{\mu^{2}e^{-\mu}}{\mu^{\prime2}e^{-\mu^{\prime}}}(\frac{\mu^{\prime}e^{-\mu^{\prime}}}{\mu e^{-\mu}}e^{-\mu}-e^{-\mu^{\prime}})s_0 \end{aligned}$$

两边同乘$\frac{\mu^{\prime}}{\mu}$，得：

$$\begin{aligned} (\frac{\mu^{\prime}}{\mu}-1)cs_c &\leq \frac{\mu e^{-\mu}}{\mu^{\prime}e^{-\mu^{\prime}}}S_{\mu^{\prime}}-S_{\mu}+(e^{-\mu}-\frac{\mu e^{-\mu}}{\mu^{\prime}})s_0 \end{aligned}$$

将之带入$(7)$式稍加整理即可得式$(13)$.

4 (14)

命题：
假如$\eta\ll1$，则Bob端多光子成分占比是$1-e^{-\mu}$.

证明：($\eta\mu-0-\eta\mu e^{-\mu}$)/$\eta\mu=1-e^{-\mu}$

5 (15)

由$1-\Delta-\frac{s_0e^{-\mu}}{S_\mu}=\frac{s_1\mu e^{-\mu}}{S_\mu}$可得：

$$\begin{aligned} s_1=\frac{S_\mu}{\mu e^{-\mu}}(1-\Delta-\frac{s_0e^{-\mu}}{S_\mu}) \end{aligned}$$

由$(8)$式可得：

$$\begin{aligned} \Delta^{\prime} &= \frac{c\frac{\mu^{\prime2}e^{-\mu^{\prime}}}{\mu^2e^{-\mu}}s_c+ds_d}{S_{\mu^{\prime}}} \\ &= \frac{S_{\mu^{\prime}}-e^{-\mu^{\prime}}s_{0}-\mu^{\prime}e^{-\mu^{\prime}}s_{1}}{S_{\mu^{\prime}}} \\ &= \frac{S_{\mu^{\prime}}-e^{-\mu^{\prime}}s_{0}-\mu^{\prime}e^{-\mu^{\prime}}\frac{S_\mu}{\mu e^{-\mu}}(1-\Delta-\frac{s_0e^{-\mu}}{S_\mu})}{S_{\mu^{\prime}}} \\ &= 1-\frac{e^{-\mu^{\prime}}s_0}{s_{\mu^{\prime}}}-e^{\mu-\mu^{\prime}}\frac{S_\mu\mu^{\prime}}{S_{\mu^{\prime}}\mu}(1-\Delta-\frac{s_0e^{-\mu}}{S_\mu}) \end{aligned}$$

6 (17) 推导不符？

由$(16)$式得：

$$\begin{aligned} c(1-r_c)s_c \leq\frac{\mu^2e^{-\mu}}{\mu^{\prime2}e^{-\mu^{\prime}}} \times(S_{\mu^{\prime}}-\mu^{\prime}e^{-\mu^{\prime}}(1-r_1)s_1-e^{-\mu^{\prime}}(1+r_0)s_0) \end{aligned}$$

即：

$$\begin{aligned} (1-r_c)\Delta &\leq\frac{\mu^2e^{-\mu}}{\mu^{\prime2}e^{-\mu^{\prime}}} \frac{S_{\mu^{\prime}}-\mu^{\prime}e^{-\mu^{\prime}}(1-r_1)s_1-e^{-\mu^{\prime}}(1+r_0)s_0}{S_{\mu}} \\ &= \frac{\mu^2e^{-\mu}}{\mu^{\prime2}e^{-\mu^{\prime}}} \frac{S_\mu^{\prime}}{S_\mu}-\frac{\mu^2e^{-\mu}}{\mu^{\prime}}\frac{(1-r_1)s_1}{S_\mu}-\frac{\mu^2e^{-\mu}}{\mu^{\prime2}}\frac{(1+r_0)s_0}{S_\mu} \end{aligned}$$

两边同乘$\frac{\mu^{\prime2}}{\mu}e^{\mu}$，得：

$$\begin{aligned} \frac{\mu^{\prime2}}{\mu}e^{\mu}(1-r_c)\Delta &\leq \mu e^{\mu^{\prime}} \frac{S_\mu^{\prime}}{S_\mu}-\mu^{\prime}\mu\frac{(1-r_1)s_1}{S_\mu}-\mu \frac{(1+r_0)s_0}{S_\mu} \end{aligned}$$

所以：

$$\begin{aligned} \mu^{\prime}e^{\mu}\left[(1-r_c)\frac{\mu^{\prime}}{\mu}-1\right]\Delta &\leq \mu e^{\mu^{\prime}} \frac{S_\mu^{\prime}}{S_\mu}-\mu^{\prime}\mu\frac{(1-r_1)s_1}{S_\mu}-\mu \frac{(1+r_0)s_0}{S_\mu}-\mu^{\prime}e^{\mu}\Delta \\ &= \mu e^{\mu^{\prime}} \frac{S_\mu^{\prime}}{S_\mu}-\mu^{\prime}\mu\frac{(1-r_1)s_1}{S_\mu} \\ &-\mu \frac{(1+r_0)s_0}{S_\mu}-\mu^{\prime}e^{\mu}(1-\frac{s_0e^{-\mu}}{S_\mu}-\frac{s_1\mu e^{-\mu}}{S_\mu}) \\ &= \mu e^{\mu^{\prime}} \frac{S_\mu^{\prime}}{S_\mu}-\mu^{\prime}\mu\frac{(1-r_1)s_1}{S_\mu} \\ &-\mu \frac{(1+r_0)s_0}{S_\mu}-\mu^{\prime}e^{\mu}+\frac{\mu^{\prime}s_0}{S_\mu}+\frac{s_1\mu\mu^{\prime} }{S_\mu} \\ &= \mu e^{\mu^{\prime}} \frac{S_\mu^{\prime}}{S_\mu}+\mu^{\prime}\mu \frac{r_1s_1}{S_\mu}-\mu\frac{r_0s_0}{S_\mu}+(\mu^{\prime}-\mu)\frac{s_0}{S_\mu}-\mu^{\prime}e^\mu \end{aligned}$$

7 (17) 推导不符？

命题：
Given $N_1 + N_2$ copies of state ρ, suppose the counting rate for $N_1$ randomly chosen states is $s_{ρ}$ and the counting rate for the remaining states is $s’_{ρ}$, the probability that $s_{ρ} - s’_{ρ} > δ_{ρ}$ is less than exp(-$\frac{1}{4}\delta_{ρ}^{2}N_{0}/s_{ρ}$) and $N_{0} = Min(N_{1},N_{2}).$

证明：
设$N_1 + N_2$个ρ的平均计数率为$\hat{s}_{ρ}$，则有$N_1s_{ρ} + N_2s’_{ρ}=(N_1 + N_2)\hat{s}_{ρ}$，所以$s_{ρ} - s’_{ρ} > δ_{ρ}$的概率即$s_{ρ} - \hat{s}_{ρ} > \frac{N_2}{N_1+N_2}δ_{ρ}$的概率。由于$s_{ρ}$和$\hat{s}_{ρ}$都在$(0,1)$区间，所以直接带入$Hoeffding$ $inequality$得:

$$P[s_{ρ} - \hat{s}_{ρ} > \frac{N_2}{N_1+N_2}δ_{ρ}] \leq e^{-2\frac{(\frac{N_2}{N_1+N_2})^2N_1(N_1+N_2)}{N_2+1}}$$