Reference-Frame-Independent Quantum Key Distribution in Uplink and Downlink Free-Space Channel

Abstract

The reference-frame-independent quantum key distribution (RFI-QKD) allows the authorized users to share secrete keys without active alignment of the reference frames, which is beneficial for the implementation over free-space channel. However, the performance of free-space RFI-QKD could notably degrade due to the influence of atmospheric conditions. In this paper, we investigate the transmission attenuation of free-space channel in uplink and downlink scenario, respectively. Furthermore, we also simulate the relationships between the diffraction-caused attenuation and the apertures of sending and receiving optics. Then, the key generation rate of RFI-QKD and BB84 protocol are compared in different links with reference frame deviations. Simulation results show that the transmission attenuation due to diffraction can be reduced by choosing appropriate optical apertures, and better performance of free-space RFI-QKD can be achieved in downlink configuration.

Appendix

Suppose that two single-photon detectors are used at the receiver’s side, the total gain and QBER of λ intensity states in ζ basis can be given as

$$ \begin{array}{@{}rcl@{}} {Q}_{\zeta_{A}\zeta_{B}}^{\lambda}&=&\sum\limits_{n=0}^{\infty} e^{-\lambda}\frac{{\lambda}^{n}}{n!}{Y}_{\zeta_{A}\zeta_{B}}^{n}, \end{array} $$

(17)

$$ \begin{array}{@{}rcl@{}} {Q}_{\zeta_{A}\zeta_{B}}^{\lambda}{E}_{\zeta_{A}\zeta_{B}}^{\lambda}&=&\sum\limits_{n=0}^{\infty} e^{-\lambda}\frac{{\lambda}^{n}}{n!}{Y}_{\zeta_{A}\zeta_{B}}^{n}{e}_{\zeta_{A}\zeta_{B}}^{n}, \end{array} $$

(18)

according to the binomial principle, the successful probability of detecting Alice’s state in Bob’s basis of n-photon pulse is expressed as

$$ \begin{array}{@{}rcl@{}} {P}_{\zeta_{A}\zeta_{B}}^{n} = \sum\limits_{n=0}^{\infty} e^{-\lambda}\frac{{\lambda}^{n}}{n!}\sum\limits_{m=0}^{n}{{C}_{m}^{n}}\eta^{m}(1-\eta)^{n-m}(|\langle \zeta_{A}|\zeta_{B}\rangle|^{2})^{m}P(m), \end{array} $$

(19)

P(m) means the probability of valid detection events. When m = 0, $P(m)=P_{d}-{{P}_{d}^{2}}\approx P_{d}$, when m > 0, P(m) = 1 − P_d. P_d is the dark counts of a single-photon detector. The above calculations are also applied to the decoy state (λ = ν) and vacuum state (λ = 0). It can be acquired intuitively that the gain and error rate of vacuum states are Q⁰ = 2P_d(1 − P_d) and $Q^{0}E^{0}=\frac {1}{2}Q^{0}$.

For non-vacuum states, the total gain of intensity λ is half of the total probability of all kinds of detection events and the quantum bit error is half of the wrong detection events probability. For example,

$$ \begin{array}{@{}rcl@{}} {Q}_{Z_{A}Z_{B}}^{\mu} &=& \frac{1}{2}\left( {P}_{{Z_{A}^{0}}{Z}_{B}^{0}}^{\mu}+{P}_{{Z}_{A}^{0}{Z}_{B}^{1}}^{\mu}+{P}_{{{Z}_{A}^{1}}{{Z}_{B}^{0}}}^{\mu}+P_{{Z}_{A}^{1}{Z}_{B}^{1}}^{\mu}\right) \end{array} $$

(20)

$$ \begin{array}{@{}rcl@{}} {Q}_{Z_{A}Z_{B}}^{\mu}{E}_{Z_{A}Z_{B}}^{\mu} &=& \frac{1}{2}\left( {P}_{{Z}_{A}^{0}{Z}_{B}^{1}}^{\mu}+{P}_{{Z}_{A}^{1}{Z}_{B}^{0}}^{\mu}\right) \end{array} $$

(21)

In order to get the final key rate we also need to evaluate the upperbound of single-photon error rate and the lowerbound of its gain, which are given as

$$ \begin{array}{@{}rcl@{}} {Y}_{Z_{A}Z_{B}}^{1,L} &=& \frac{\mu}{\mu\nu-\nu^{2}}\left( e^{\nu}{Q}_{Z_{A}Z_{B}}^{\nu}-e^{\mu}\frac{\nu^{2}}{\mu^{2}}{Q}_{Z_{A}Z_{B}}^{\mu}-\frac{\mu^{2}-\nu^{2}}{\mu^{2}}Q^{0}\right) \end{array} $$

(22)

$$ \begin{array}{@{}rcl@{}} {e}_{Z_{A}Z_{B}}^{1,U} &=& \min\left\{\frac{1}{2},\frac{e^{\nu}Q_{Z_{A}Z_{B}}^{\nu}{E}_{Z_{A}Z_{B}}^{\nu}-Q^{0}E^{0}}{\nu {Y}_{Z_{A}Z_{B}}^{1,L}}\right\} \end{array} $$

(23)

The formula also apply to the other basis combination to calculate (16).