Efficient bit sifting scheme of post-processing in quantum key distribution

Abstract

Bit sifting is an important step in the post-processing of quantum key distribution (QKD). Its function is to sift out the undetected original keys. The communication traffic of bit sifting has essential impact on the net secure key rate of a practical QKD system. In this paper, an efficient bit sifting scheme is presented, of which the core is a lossless source coding algorithm. Both theoretical analysis and experimental results demonstrate that the performance of the scheme is approaching the Shannon limit. The proposed scheme can greatly decrease the communication traffic of the post-processing of a QKD system, which means the proposed scheme can decrease the secure key consumption for classical channel authentication and increase the net secure key rate of the QKD system, as demonstrated by analyzing the improvement on the net secure key rate. Meanwhile, some recommendations on the application of the proposed scheme to some representative practical QKD systems are also provided.

Appendices

Appendix A: Proof of Lemma 1

Proof

For convenience, we denote $p=1-q$, then $p \in \left[ 0.9, 1\right)$ and $q=1-p$. The $\overline{L}\left( z\right)$ in (14) can be rewritten as

$$\begin{aligned} \overline{L}\left( z \right) = \frac{{\left( {1 - p} \right) z}}{{1 - {p^{{2^z} - 1}}}}, \end{aligned}$$

and the partial derivative of the expected codelength $\overline{L}$ with respect to the variable z is given by

$$\begin{aligned} \frac{{\partial \overline{L}}}{{\partial z}} = \frac{{(1 - p)p}}{{{{\left( {p - {p^{{2^z}}}} \right) }^2}}}\left( {p - {p^{{2^z}}} + z{2^z}{p^{{2^z}}}\ln 2\ln p} \right) . \end{aligned}$$

(19)

Due to $p \in [0.9,1)$,

$$\begin{aligned} \frac{{(1 - p)p}}{{{{\left( {p - {p^{{2^z}}}} \right) }^2}}} > 0. \end{aligned}$$

We only need to focus on the sign of

$$\begin{aligned} r\left( z \right) = p - {p^{{2^z}}} + z{2^z}{p^{{2^z}}}\ln 2\ln p. \end{aligned}$$

(20)

The partial derivative of $r\left( z \right)$ with respect to the variable z can be evaluated as

$$\begin{aligned} \frac{{\partial r}}{{\partial z}} = z{2^z}{p^{{2^z}}}{\ln ^2}2\ln p\left( {1 + {2^z}\ln p} \right) . \end{aligned}$$

(21)

Due to $z{2^z}{p^{{2^z}}}{\ln ^2}2\ln p < 0$, the sign of $\frac{{\partial r}}{{\partial z}}$ is determined by the sign of the expression $1 + {2^z}\ln p$. Let $1 + {2^z}\ln p > 0$; then

$$\begin{aligned} z < - {\log _2}\left( { - \ln p} \right) \buildrel \varDelta \over = {z_m}. \end{aligned}$$

So

$$\begin{aligned} \left\{ {\begin{array}{*{20}{l}} {\frac{{\partial r}}{{\partial z}} < 0,}&{}{{{\mathrm{when}}}\;1 \le z{\mathrm{}} < {\mathrm{}}{z_m}}\\ {\frac{{\partial r}}{{\partial z}} = 0,}&{}{{\mathrm{when}}\;z{\mathrm{}} = {\mathrm{}}{z_m}}\\ {\frac{{\partial r}}{{\partial z}} > 0,}&{}{{\mathrm{when}}\;z{\mathrm{}} > {\mathrm{}}{z_m}} \end{array}} \right. . \end{aligned}$$

(22)

Therefore function $r\left( z \right)$ is monotonically decreasing in the domain $z \in \left[ 1, z_m\right)$ and monotonically increasing in domain $\left( z_m, +\infty \right)$ and reaches the global minimum value at the point $z = z_m$, which is

$$\begin{aligned} r\left( {{z_m}} \right) = \frac{{ - 1 + ep + \ln \left( { - \ln p} \right) }}{e}. \end{aligned}$$

(23)

Since

$$\begin{aligned} \frac{{\partial r\left( {{z_m}} \right) }}{{\partial p}} = {\mathrm{}}1 + \frac{1}{{ep\ln p}} < -2.87, \mathrm{}\forall p \in [0.9,1), \end{aligned}$$

function $r\left( z_m\right)$ is monotonically decreasing with respect to the variable p and comes to the maximum value at $p=0.9$, which is about $-0.30$. So

$$\begin{aligned} r\left( {{z_m}} \right) < 0, \forall p \in [0.9,1). \end{aligned}$$

(24)

In addition, as z approaches $+ \infty$, the limit of $r\left( z \right)$ is

$$\begin{aligned} \mathop {\lim }\limits _{\mathrm{z} \rightarrow + \infty } r\left( z \right) = p, \forall p \in [0.9,1). \end{aligned}$$

(25)

Upon (24), (25) and the continuity of $r\left( z \right)$ in the domain $z \in \left[ {z_m, + \infty } \right)$, it can be inferred that there exists at least one ${z_0} \in \left( { z_m, + \infty } \right)$ satisfying that $r\left( {{z_0}} \right) = 0$ according to the intermediate value theorem [23]. Besides, since $r\left( z \right)$ is monotonically increasing function in the range $z \in \left[ {z_m, + \infty } \right)$, the root of $r(z)=0$ is unique, and

$$\begin{aligned} \left\{ {\begin{array}{*{20}{l}} {r\left( z \right) < 0,}&{}{{\mathrm{when}}\;z \in \left[ {{z_m},{z_0}} \right) }\\ {r\left( z \right) = 0,}&{}{\mathrm{when}\; z = z_0}\\ {r\left( z \right) > 0,}&{}{\mathrm{when}\;z \in \left( {{z_0}, + \infty } \right) } \end{array}} \right. \end{aligned}$$

(26)

Since

$$\begin{aligned} {z_m} \ge {\left. { - {{\log }_2}\left( { - \ln p} \right) } \right| _{p = 0.9}} > 3.24, \end{aligned}$$

we have to discuss the sign of $r\left( z \right)$ in the domain $z \in \left[ {1, z_m } \right)$. Now our concern is the sign of

$$\begin{aligned} r\left( 1 \right) = p - {p^2} + 2{p^2}\ln 2\ln p. \end{aligned}$$

(27)

The partial derivative of $r\left( 1 \right)$ with respect to the variable p is given by

$$\begin{aligned} \frac{{\partial r\left( 1 \right) }}{{\partial p}}= 1 - 2p + 2p\ln 2 + 4p\ln 2\ln p, \end{aligned}$$

(28)

and the second partial derivative of $r\left( 1 \right)$ with respect to the variable p is given by

$$\begin{aligned} \frac{{{\partial ^2}r\left( 1 \right) }}{{\partial {p^2}}} = - 2 + 6\ln 2 + 4\ln 2\ln p. \end{aligned}$$

(29)

Obviously in the domain $p \in [0.9,1)$, the function $\frac{{{\partial ^2}r\left( 1 \right) }}{{\partial {p^2}}}$ is monotonically increasing with respect to the variable p and reaches the minimum value at $p = 0.9$, which is about 1.87. Thus the function $\frac{{\partial r\left( 1 \right) }}{{\partial p}}$ with respect to the variable p is also monotonically increasing and also reaches the minimum value at $p = 0.9$, which is about 0.18. Therefore $r\left( 1 \right)$ is a monotonically increasing function with respect to the variable p and reaches the supremum 0 at $p = 1$, so

$$\begin{aligned} r\left( 1 \right) < 0, \forall p \in [0.9,1). \end{aligned}$$

(30)

Since $r\left( z \right)$ is a monotonically decreasing function in the range $z \in \left[ 1, z_m\right)$, we have

$$\begin{aligned} r\left( z \right) \le r\left( 1 \right) < 0, \forall z \in \left[ 1, z_m\right) . \end{aligned}$$

(31)

Combining (26) and (31), we have

$$\begin{aligned} \left\{ {\begin{array}{*{20}{l}} {r\left( z \right) < 0,}&{}{\mathrm{when}\;z \in \left[ {{1},{z_0}} \right) }\\ {r\left( z \right) = 0,}&{}{\mathrm{when}\; z = z_0}\\ {r\left( z \right) > 0,}&{}{\mathrm{when}\;z \in \left( {{z_0}, + \infty } \right) } \end{array}} \right. . \end{aligned}$$

(32)

Since the sign of $\frac{{\partial \overline{L}}}{{\partial z}}$ is the same as the sign of $r\left( z \right)$,

$$\begin{aligned} \left\{ {\begin{array}{*{20}{l}} {\frac{{\partial \overline{L}}}{{\partial z}} < 0,}&{}{\mathrm{when}\;z \in \left[ {1,{z_0}} \right) }\\ {\frac{{\partial \overline{L}}}{{\partial z}} = 0,}&{}{\mathrm{when}\;z = z_0}\\ {\frac{{\partial \overline{L}}}{{\partial z}} > 0,}&{}{\mathrm{when}\;z \in \left( {{z_0}, + \infty } \right) } \end{array}} \right. \end{aligned}$$

where ${z_0} \in \left( { z_m, + \infty } \right)$, i.e.,

$$\begin{aligned} {z_0} \in \left( { - {{\log }_2}\left( { - \ln \left( 1-q\right) } \right) , + \infty } \right) . \end{aligned}$$

On the other hand, since $k \in \mathbb {N}^+$, the minimum of $\overline{L}(k)$ is reached at $k=\left\lfloor {{z_0}} \right\rfloor$ or $k=\left\lceil {{z_0}} \right\rceil$, i.e., the optimal solution of (13) is reached at $k=\left\lfloor {{z_0}} \right\rfloor$ or $k=\left\lceil {{z_0}} \right\rceil$. $\square$

Appendix B: Proof of Theorem 1

Proof

For briefness, we denote $y = - {\log _2}\left( { - \ln \left( {{{1 - q}}} \right) } \right)$. Then Lemma 1 indicates that

$$\begin{aligned} {z_0} > y,\forall q \in \left[ {{{10}^{ - 15}},0.1} \right] . \end{aligned}$$

(33)

At the same time, the value of the partial deviation $\frac{{\partial \overline{L}}}{{\partial z}}$ at $z = y + 3$ is given by

$$\begin{aligned} \frac{{\partial \overline{L}}}{{\partial z}}{|_{z = y + 3}} = \frac{{{e^8}(1 - q)q}}{{{{\left( {1 - {e^8}\left( {1 - q} \right) } \right) }^2}}}A\left( q \right) , \end{aligned}$$

(34)

where

$$\begin{aligned} A\left( q \right) = - 1 - 24\ln 2 + {e^8}(1 - q) + 8\ln \left( { - \ln \left( {1 - q} \right) } \right) . \end{aligned}$$

Since

$$\begin{aligned} \frac{{{e^8}(1 - q)q}}{{{{\left( {1 - {e^8}\left( {1 - q} \right) } \right) }^2}}} > 0, \forall q \in \left[ 10^{-15}, 0.1\right] \end{aligned}$$

(35)

the sign of (34) is same as the sign of $A\left( q \right)$. The partial derivative of $A\left( q \right)$ can be evaluated as

$$\begin{aligned} \frac{{\partial A}}{{\partial q}} = - {e^8} - \frac{8}{{(1 - q)\ln \left( {1 - q} \right) }}, \end{aligned}$$

which is a monotonically decreasing function with respect to the variable q and

$$\begin{aligned} \left\{ {\begin{array}{*{20}{l}} {\frac{{\partial A}}{{\partial q}} > 7.20 \times {{10}^{15}},}&{}{when\;q = {{10}^{ - 15}}}\\ {\frac{{\partial A}}{{\partial q}} < - 2.80 \times {{10}^3},}&{}{when\;q = 0.1} \end{array}} \right. \end{aligned}$$

So the function $A\left( q \right)$ is firstly monotonically increasing and then monotonically decreasing in the range $q \in \left[ 10^{-15}, 0.1\right]$. The minimum must occur at the point $q=10^{-15}$ or $q=0.1$, where ${A\left( {{{10}^{ - 15}}} \right) \approx 2687.85}$ and ${A\left( {0.1} \right) \approx {\mathrm{2647}}{\mathrm{.22}}}$, so

$$\begin{aligned} A\left( q \right) > 0,\forall q \in \left[ 10^{-15}, 0.1\right] . \end{aligned}$$

(36)

Hence, combining (34), (35) and (36),

$$\begin{aligned} \frac{{\partial \overline{L}}}{{\partial z}}{|_{z = y + 3}} > 0,\forall q \in \left[ 10^{-15}, 0.1\right] . \end{aligned}$$

(37)

Making use of (37) and the Lemma 1, it can be concluded that

$$\begin{aligned} {z_0} < y + 3,\forall q \in \left[ {{{10}^{ - 15}},0.1} \right] . \end{aligned}$$

(38)

Combining (33) and (38), we have

$$\begin{aligned} y<{z_0} < y + 3, \forall q \in \left[ {{{10}^{ - 15}},0.1} \right] . \end{aligned}$$

On the other hand, Lemma 1 shows that the optimal solution of (13) is reached at $k=\left\lfloor {{z_0}} \right\rfloor$ or $k=\left\lceil {{z_0}} \right\rceil$, so the optimal parameter $k_\mathrm{opt}$ of (13) is one of the following five values, i.e., $\left\lfloor y \right\rfloor , \left\lfloor y \right\rfloor + 1,\left\lfloor y \right\rfloor + 2,\left\lfloor y \right\rfloor + 3,\left\lceil y \right\rceil + 3$. $\square$

Appendix C: Proof of Theorem 2

Proof

For brevity, let $\left\{ {a \ldots b} \right\} \buildrel \varDelta \over = \left[ {a,b} \right] \cap \mathbb {N}.$

(a) When $n_\mathrm{max} \ge 2^{k_\mathrm{opt}}$, the optimal code alphabet size

$$\begin{aligned} {n_\mathrm{opt}} = {2^{{k_\mathrm{opt}}}} \le {n_\mathrm{max }}, \end{aligned}$$

which is reachable in the domain $\left\{ {2 \ldots {n_\mathrm{max }}} \right\}$. So the optimal solution of (17) is reached at $n=2^{k_\mathrm{opt}}$ in this case.

(b) When $n_\mathrm{max} < 2^{k_\mathrm{opt}}$, the optimal code alphabet size

$$\begin{aligned} {n_\mathrm{opt}} = {2^{{k_\mathrm{opt}}}} > {n_\mathrm{max }}, \end{aligned}$$

which cannot be reachable in the domain $\left\{ {2 \ldots {n_\mathrm{max }}} \right\}$. In the case, the domain $\left\{ {2 \ldots {n_\mathrm{max }}} \right\}$ is divided into $\left\{ {2 \ldots {2^{{k_\mathrm{max }}}}} \right\}$ and $\left\{ {{2^{{k_\mathrm{max }}}} + 1 \ldots {n_\mathrm{max }}} \right\}$, where ${k_\mathrm{max }} = \left\lfloor {{{\log }_2}{n_\mathrm{max }}} \right\rfloor$.

(b.1) Since

$$\begin{aligned} \left\{ {2 \ldots {2^{{k_\mathrm{max }}}}} \right\} = \bigcup \limits _{k = 1}^{{k_\mathrm{max }}} {\left\{ {{2^{k - 1}}+1 \ldots {2^k}} \right\} } \end{aligned}$$

and the minimum of $\overline{L}\left( n\right)$ in the range ${\left\{ {{2^{k - 1}}+1 \ldots {2^k}} \right\} }$ occurs at the point $2^k$ according to (12), $1 \le k \le k_\mathrm{max}$, we just need to consider the minimum of $\overline{L}\left( k\right)$ in the range $\left\{ {1\ldots {k_\mathrm{max }}} \right\}$. According to Lemma 1, it can be inferred that $k_\mathrm{opt} \le \left\lceil {{z_0}} \right\rceil$, then

$$\begin{aligned} {k_\mathrm{max }} = \left\lfloor {{{\log }_2}{n_\mathrm{max }}} \right\rfloor \le {k_\mathrm{opt}} - 1 \le \left\lceil {{z_0}} \right\rceil - 1 < {z_0}. \end{aligned}$$

So $\overline{L}\left( k\right)$ is monotonically decreasing in the range $\left\{ {1 \ldots {k_\mathrm{max }}} \right\}$ and reaches the minimum at the point $k=k_\mathrm{max}$. Therefore, the minimum of $\overline{L}\left( n\right)$ in the range $\left\{ {2 \ldots {2^{{k_\mathrm{max }}}}} \right\}$ occurs at the point $n=2^{k_\mathrm{max}}$, i.e., $n=2^{\left\lfloor {{{\log }_2}{n_\mathrm{max }}} \right\rfloor }$.

(b.2) Since the function $\overline{L}\left( n\right)$ is monotonically decreasing in the range $\big \{ {2^{{k_\mathrm{max }}}} + 1 \ldots {n_\mathrm{max }} \big \}$, the minimum of $\overline{L}\left( n\right)$ in the range occurs at the point $n={n_\mathrm{max }}$.

Combining (b.1) and (b.2), it can be seen that the optimal solution of (17) is reached at $n=2^{\left\lfloor {{{\log }_2}{n_\mathrm{max }}} \right\rfloor }$ or $n=n_\mathrm{max}$ when $n_\mathrm{max} < 2^{k_\mathrm{opt}}$.

Combining the case (a) and the case (b), the theorem is proved. $\square$