Round-robin differential-phase-shift quantum key distribution with heralded pair-coherent sources

Abstract

Round-robin differential-phase-shift (RRDPS) quantum key distribution (QKD) scheme provides an effective way to overcome the signal disturbance from the transmission process. However, most RRDPS-QKD schemes use weak coherent pulses (WCPs) as the replacement of the perfect single-photon source. Considering the heralded pair-coherent source (HPCS) can efficiently remove the shortcomings of WCPs, we propose a RRDPS-QKD scheme with HPCS in this paper. Both infinite-intensity decoy-state method and practical three-intensity decoy-state method are adopted to discuss the tight bound of the key rate of the proposed scheme. The results show that HPCS is a better candidate for the replacement of the perfect single-photon source, and both the key rate and the transmission distance are greatly increased in comparison with those results with WCPs when the length of the pulse trains is small. Simultaneously, the performance of the proposed scheme using three-intensity decoy states is close to that result using infinite-intensity decoy states when the length of pulse trains is small.

Appendix

1.1 The lower bound of the yield

First, Alice and Bob can estimate the lower bound of the background yield, $Y_{0}^{L}$, by

$$\begin{aligned}&\nu _1^2P_\mathrm{post}(L\nu _{2})Q_{L\nu _{2}}-\nu _2^2P_\mathrm{post}(L\nu _{1})Q_{L\nu _{1}} \nonumber \\&\quad =\left( \nu _1^2-\nu _2^2\right) d_\mathrm{A}Y_0-L^2\nu _1^2\nu _2^2 \sum _{n=2}^{\infty }\left( \frac{(L\nu _1)^{2n-2}-(L\nu _2)^{2n-2}}{(n!)^2} \right. \nonumber \\&\qquad \left. [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^n]Y_n\right) \le \left( \nu _1^2-\nu _2^2\right) d_\mathrm{A}Y_0, \end{aligned}$$

(22)

where the inequality is due to $L^2\nu _1^2\nu _2^2 \sum _{n=2}^{\infty }\Big (\frac{(L\nu _1)^{2n-2}-(L\nu _2)^{2n-2}}{(n!)^2} [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^n]Y_n\Big )\ge 0$. Thus, the lower bound of $Y_{0}$ is given by

$$\begin{aligned} Y_{0}\ge Y_{0}^{L}=\max \left\{ \frac{\nu _{1}^{2}P_\mathrm{post}(L\nu _{2})Q_{L\nu _{2}} -\nu _{2}^{2}P_\mathrm{post}(L\nu _{1})Q_{L\nu _{1}}}{d_\mathrm{A}\left( \nu _{1}^{2}-\nu _{2}^{2}\right) },0 \right\} , \end{aligned}$$

(23)

where the equality sign will hold when $\nu _{2}=0$. By $P_\mathrm{post}(L\nu _{1})Q_{L\nu _{1}}-P_\mathrm{post}(L\nu _{2})Q_{L\nu _{2}}$, Alice and Bob have

$$\begin{aligned}&P_\mathrm{post}(L\nu _{1})Q_{L\nu _{1}}-P_\mathrm{post}(L\nu _{2})Q_{L\nu _{2}} \nonumber \\&\quad =L^2\left( \nu _1^2-\nu _2^2\right) [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})]Y_1+ \sum _{n=2}^{\infty }\left( \frac{(L\nu _1)^{2n}-(L\nu _2)^{2n}}{(n!)^2} \right. \nonumber \\&\qquad \left. \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^n\right] Y_n\right) \nonumber \\&\quad \le L^2\left( \nu _1^2-\nu _2^2\right) [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})]Y_1+ \frac{\nu _1^{4}-\nu _2^{4}}{\mu ^4} \sum _{n=2}^{\infty }\left( \frac{(L\mu )^{2n}}{(n!)^2}\right. \nonumber \\&\qquad \left. \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^n\right] Y_n\right) \nonumber \\&\quad =L^2\left( \nu _1^2-\nu _2^2\right) [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})]Y_1\nonumber \\&\qquad +\, \frac{\nu _1^{4}-\nu _2^{4}}{\mu ^4} \left( P_\mathrm{post}(L\mu )Q_{L\mu }-d_\mathrm{A}Y_0-L^2\mu ^2 \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})\right] Y_1\right) \nonumber \\&\quad \le L^2\left( \nu _1^2-\nu _2^2-\frac{\nu _1^{4}-\nu _2^{4}}{\mu ^2} \right) [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})]Y_1\nonumber \\&\qquad +\,\frac{\nu _1^{4}-\nu _2^{4}}{\mu ^4} \left[ P_\mathrm{post}(L\mu )Q_{L\mu }-d_\mathrm{A}Y_0^L\right] \end{aligned}$$

(24)

Here, the inequality that $a^{i}-b^{i}\le a^{2}-b^{2}$ when $a\ge b\ge 0$, $a+b\le 1$ and $i\ge 2$ is used to prove the first inequality in Eq. 24. The second inequality in Eq. 24 is due to Eq. 23. Then, they can obtain the lower bound of the single-photon pulse trains yield $Y_{1}$,

$$\begin{aligned} Y_{1}\ge Y_{1}^{L}=\frac{P_\mathrm{post}(L\nu _{1})Q_{L\nu _{1}} -P_\mathrm{post}(L\nu _{2})Q_{L\nu _{2}}-\frac{\nu _{1}^{4}-\nu _{2}^{4}}{\mu ^{4}} \left[ P_\mathrm{post}(L\mu )Q_{L\mu }-d_\mathrm{A} Y_{0}^{L}\right] }{L^{2}\left( \nu _{1}^{2}-\nu _{2}^{2}- \frac{\nu _{1}^{4}-\nu _{2}^{4}}{\mu ^{2}}\right) [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})]}. \end{aligned}$$

(25)

Furthermore, Alice and Bob can estimate the lower bound of the two-photon pulse trains yield, $Y_{2}^{L}$, by

$$\begin{aligned}&\left( \nu _2^2-\nu _3^2\right) P_\mathrm{post}(L\nu _{1})Q_{L\nu _{1}}-\left( \nu _1^2-\nu _3^2\right) P_\mathrm{post}(L\nu _{2})Q_{L\nu _{2}}\nonumber \\&\quad +\left( \nu _1^2-\nu _2^2\right) P_\mathrm{post}(L\nu _{3})Q_{L\nu _{3}} \nonumber \\&\qquad =\frac{L^4}{4}\left( \nu _1^2-\nu _2^2\right) \left( \nu _1^2-\nu _3^2\right) \left( \nu _2^2-\nu _3^2\right) \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^2\right] Y_2 \nonumber \\&\qquad \quad + \sum _{n=3}^{\infty }\left( \frac{\left( \nu _2^2-\nu _3^2\right) \left[ (L\nu _1)^{2n}-(L\nu _2)^{2n}\right] +\left( \nu _2^2-\nu _1^2\right) \left[ (L\nu _2)^{2n}-(L\nu _3)^{2n}\right] }{(n!)^2} \right. \nonumber \\&\qquad \quad \left. [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^n]Y_n\right) \nonumber \\&\qquad \le \frac{L^4}{4}\left( \nu _1^2-\nu _2^2\right) \left( \nu _1^2-\nu _3^2\right) \left( \nu _2^2-\nu _3^2\right) \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^2\right] Y_2 \nonumber \\&\qquad \quad + \left( \left( \nu _2^{2}-\nu _3^{2}\right) \frac{\nu _1^{6}-\nu _2^{6}}{\mu ^{6}}+ \left( \nu _2^{2}-\nu _1^{2}\right) \frac{\nu _2^{6}-\nu _3^{6}}{\mu ^{6}}\right) \nonumber \\&\qquad \quad \sum _{n=3}^{\infty }\left( \frac{(L\mu )^{2n}}{(n!)^2}\left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^n\right] Y_n\right) \nonumber \\&\qquad =\frac{L^4}{4}\left( \nu _1^2-\nu _2^2\right) \left( \nu _1^2-\nu _3^2\right) \left( \nu _2^2-\nu _3^2\right) \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^2\right] Y_2 \nonumber \\&\qquad \quad +\,\frac{\left( \nu _1^{2}-\nu _2^{2}\right) \left( \nu _2^{2}-\nu _3^{2}\right) \left( \nu _1^{2}-\nu _3^{2}\right) \left( \nu _1^{2}+\nu _2^{2}+\nu _3^{2}\right) }{\mu ^{6}} \bigg (P_\mathrm{post}(L\mu )Q_{L\mu }-d_\mathrm{A}Y_0 \nonumber \\&\qquad \quad -\,L^2\mu ^2 [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})]Y_1- \frac{L^4\mu ^4}{4} \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^2\right] Y_2\bigg ) \nonumber \\&\qquad \le \frac{L^4}{4}\left( \nu _1^2-\nu _2^2\right) \left( \nu _1^2-\nu _3^2\right) \left( \nu _2^2-\nu _3^2\right) \left( 1-\frac{\nu _1^{2}+\nu _2^{2}+\nu _3^{2}}{\mu ^2}\right) \nonumber \\&\qquad \quad \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^2\right] Y_2 + \frac{\left( \nu _1^{2}-\nu _2^{2}\right) \left( \nu _2^{2}-\nu _3^{2}\right) \left( \nu _1^{2}-\nu _3^{2}\right) \left( \nu _1^{2}+\nu _2^{2}+\nu _3^{2}\right) }{\mu ^{6}} \nonumber \\&\qquad \quad \left( P_\mathrm{post}(L\mu )Q_{L\mu }-d_\mathrm{A}Y_0^L - L^2\mu ^2 \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})\right] Y_1^L\right) \end{aligned}$$

(26)

Here, the inequality that $a^{i}-b^{i}\le a^{3}-b^{3}$ when $a\ge b\ge 0$, $a+b\le 1$ and $i\ge 3$ is used to prove the first inequality in Eq. 26. The second inequality in Eq. 26 is due to Eqs. 25 and 23. The lower bound of $Y_{2}$ is

$$\begin{aligned}&Y_{2}\ge Y_{2}^{L}\nonumber \\&\quad =4\frac{\left( \nu _{2}^{2}-\nu _{3}^{2}\right) P_\mathrm{post}(L\nu _{1})Q_{L\nu _{1}} -\left( \nu _{1}^{2}-\nu _{3}^{2}\right) P_\mathrm{post}(L\nu _{2})Q_{L\nu _{2}} +\left( \nu _{1}^{2}-\nu _{2}^{2}\right) P_\mathrm{post}(L\nu _{3})Q_{L\nu _{3}}}{L^4\left( \nu _{1}^{2}-\nu _{2}^{2}\right) \left( \nu _{1}^{2}-\nu _{3}^{2}\right) \left( \nu _{2}^{2}-\nu _{3}^{2}\right) \left( 1-\frac{\nu _{1}^{2}+\nu _{2}^{2}+\nu _{3}^{2}}{\mu ^{2}}\right) \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^2\right] } \nonumber \\&\qquad -\,4\frac{\frac{\nu _{1}^{2}+\nu _{2}^{2}+\nu _{3}^{2}}{L^4\mu ^{6}} \big (P_\mathrm{post}(L\mu )Q_{L\mu }-d_\mathrm{A} Y_{0}^{L}-(L\mu )^{2}[1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})]Y_{1}^{L}\big )}{\left( 1-\frac{\nu _{1}^{2}+\nu _{2}^{2}+\nu _{3}^{2}}{\mu ^{2}}\right) \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^2\right] }. \end{aligned}$$

(27)

1.2 The upper bound of the error rate

Alice and Bob can estimate the upper bound of the error rate, $e_{1}^{U}$ and $e_{2}^{U}$. According to the following inequality,

$$\begin{aligned}&P_\mathrm{post}(L\nu _{1})E_{L\nu _{1}}Q_{L\nu _{1}}-P_\mathrm{post}(L\nu _{2})E_{L\nu _{2}}Q_{L\nu _{2}} \nonumber \\&\quad =L^2\left( \nu _1^2-\nu _2^2\right) [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})]e_{1}Y_1\nonumber \\&\qquad +\sum _{n=2}^{\infty }\left( \frac{(L\nu _1)^{2n}-(L\nu _2)^{2n}}{(n!)^2} \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^n\right] e_{n}Y_n\right) \nonumber \\&\quad \ge L^2\left( \nu _1^2-\nu _2^2\right) [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})]e_{1}Y_1, \end{aligned}$$

(28)

where the inequality is due to $\sum _{n=2}^{\infty }\left( \frac{(L\nu _1)^{2n}-(L\nu _2)^{2n}}{(n!)^2} [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^n]e_{n}Y_n\right) \ge 0$, the upper bound of $e_{1}$ is given by

$$\begin{aligned} e_{1}\le e_{1}^{U}=\frac{P_\mathrm{post}(L\nu _{1})E_{L\nu _{1}}Q_{L\nu _{1}} -P_\mathrm{post}(L\nu _{2})E_{L\nu _{2}}Q_{L\nu _{2}}}{L^{2}\left( \nu _{1}^{2}-\nu _{2}^{2}\right) [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})]Y_{1}^{L}}. \end{aligned}$$

(29)

Then, Alice and Bob can estimate the upper bound of the two-photon pulse trains error rate, $e_{2}^{U}$, by

$$\begin{aligned}&\left( \nu _2^2-\nu _3^2\right) P_\mathrm{post}(L\nu _{1})E_{L\nu _{1}}Q_{L\nu _{1}} -\left( \nu _1^2-\nu _3^2\right) P_\mathrm{post}(L\nu _{2})E_{L\nu _{2}}Q_{L\nu _{2}}\nonumber \\&\quad +\left( \nu _1^2-\nu _2^2\right) P_\mathrm{post}(L\nu _{3})E_{L\nu _{3}}Q_{L\nu _{3}} \nonumber \\&\qquad =\frac{L^4}{4}\left( \nu _1^2-\nu _2^2\right) \left( \nu _1^2-\nu _3^2\right) \left( \nu _2^2-\nu _3^2\right) \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^2\right] e_2Y_2 \nonumber \\&\qquad \quad + \sum _{n=3}^{\infty }\left( \frac{\left( \nu _2^2-\nu _3^2\right) \left[ (L\nu _1)^{2n}-(L\nu _2)^{2n}\right] +\left( \nu _1^2-\nu _2^2\right) \left[ (L\nu _3)^{2n}-(L\nu _2)^{2n}\right] }{(n!)^2} \right. \nonumber \\&\qquad \quad \left[ 1\left. -(1-d_\mathrm{A})(1-\eta _\mathrm{A})^n\right] e_nY_n\right) \nonumber \\&\qquad \ge \frac{L^4}{4}\left( \nu _1^2-\nu _2^2\right) \left( \nu _1^2-\nu _3^2\right) \left( \nu _2^2-\nu _3^2\right) \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^2\right] e_2Y_2 \end{aligned}$$

(30)

where the inequality is due to $\sum _{n=3}^{\infty }\big (\frac{(\nu _2^2-\nu _3^2) [(L\nu _1)^{2n}-(L\nu _2)^{2n}] +(\nu _1^2-\nu _2^2)[(L\nu _3)^{2n}-(L\nu _2)^{2n}]}{(n!)^2} [1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^n]e_nY_n\big )\ge 0$, and the upper bound of $e_{2}$ is given by

$$\begin{aligned}&e_{2}\le e_{2}^{U}\\&\quad =4\frac{\left( \nu _{2}^{2}-\nu _{3}^{2}\right) P_\mathrm{post}(L\nu _{1})Q_{L\nu _{1}} -\left( \nu _{1}^{2}-\nu _{3}^{2}\right) P_\mathrm{post}(L\nu _{2})Q_{L\nu _{2}} +\left( \nu _{1}^{2}-\nu _{2}^{2}\right) P_\mathrm{post}(L\nu _{3})Q_{L\nu _{3}}}{L^4\left( \nu _{1}^{2}-\nu _{2}^{2}\right) \left( \nu _{1}^{2}-\nu _{3}^{2}\right) \left( \nu _{2}^{2}-\nu _{3}^{2}\right) \left[ 1-(1-d_\mathrm{A})(1-\eta _\mathrm{A})^2\right] Y_{2}^{L}}. \end{aligned}$$