The Feasible Hyper-encoding Measurement-device-independent Deterministic Secure Quantum Communication Protocol

Abstract

Deterministic secure quantum communication (DSQC) can directly transmit ciphertext through the quantum channels. Measurement-device-independent (MDI) DSQC can resist all possible attacks on the imperfect measurement devices. Comparing with conventional DSQC protocols, MDI-DSQC has relatively low ciphertext transmission rate. In this paper, we propose a hyper-encoding MDI-DSQC protocol. The hyper-encoding can effectively increase single photon’s capacity and thus increase the ciphertext transmission rate of MDI-DSQC. Our protocol adopts the high-efficient linear-optical partial hyperentangled Bell state analysis, which is feasible under current experimental condition. Our MDI-DSQC protocol is secure in theory. We provide the numerical simulation of its ciphertext transmission rate. Our hyper-encoding MDI-DSQC protocol may have potential application in the quantum communication field.

Appendix A: Detailed calculation of the yield and QBER

In this section, we provide the detailed calculation of the yield $Y_{ZP}^{1,1}$ and $Y_{XP}^{1,1}$, and the error rate $e_{ZP}^{1,1}$ and $e_{XP}^{1,1}$ of our MDI-DSQC protocol with the HBSA in Ref. [71]. The HBSA system consists of linear-optical elements such as BSs, PBSs, and quarter wave plates (QWPs). Each of the input photons needs to pass through a QWP, which causes $|H\rangle \rightarrow \frac{1}{\sqrt{2}}(|H\rangle + |V\rangle )$ and $|V\rangle \rightarrow \frac{1}{\sqrt{2}}(|H\rangle - |V\rangle )$. The HBSA system contains eight single photon detectors with the background rate of $Y_{0}$ and the detection efficiency of $\eta _{d}$ in total. Charlie could divide the 16 hyperentangled states into 14 groups based on the detector responses and the photons’ time-bin features. We suppose the total misalignment error rates in the polarization and momentum DOFs as $e_{dP}$ and $e_{dM}$, respectively. Besides, we consider the distance from Charlie to Alice $(L_{ac})$ is the same as that from Charlie to Bob $(L_{bc})$. The channel transmission efficiencies of Alice’s and Bob’s channels can be calculated as $t_{a}=10^{-\alpha {{L}_{ac}}/10}$ and $t_{b}=10^{-\alpha L_{bc}/10}$ ($\alpha =0.2$ dB/km for a standard optical fiber).

Firstly, we consider the calculation of the yield in the polarization DOF, which can be divided into two groups based on the basis selection. In the first group, Alice and Bob choose Z basis in the polarization DOF, and the partial HBSA protocol would not cause any error, so that all the detector clicks are valid. There are four specific cases based on the encoded bits. In the first case, Alice and Bob encode 00. There are four possible detection cases as

$$\begin{aligned} |H\rangle _{A}|H\rangle _{B}\rightarrow & {} \frac{1}{\sqrt{2}}(|H\rangle + |V\rangle ) \otimes \frac{1}{\sqrt{2}}(|H\rangle + |V\rangle )=\frac{1}{2}(|HH\rangle + |HV\rangle + |VH\rangle + |VV\rangle ), \nonumber \\ |H\rangle _{A}|0\rangle _{B}\rightarrow & {} \frac{1}{\sqrt{2}}(|H\rangle + |V\rangle ) \otimes |0\rangle = \frac{1}{\sqrt{2}}(|H\rangle |0\rangle + |V\rangle |0\rangle ), \nonumber \\ |0\rangle _{A}|H\rangle _{B}\rightarrow & {} |0\rangle \otimes \frac{1}{\sqrt{2}}(|H\rangle + |V\rangle ) = \frac{1}{\sqrt{2}}(|H\rangle |0\rangle + |V\rangle |0\rangle ), \nonumber \\ |0\rangle _{A}|0\rangle _{B}\rightarrow & {} |0\rangle |0\rangle , \end{aligned}$$

(A1)

where the state $|0\rangle$ denotes that the transmitted photon is lost during the transmission process. In this case, there are four responses of the yield of the signal states as

$$\begin{aligned} Y_{ZP,HH(00)}^{1,1}= & {} (1-Y_{0})^{6}\left[ \frac{1}{4}t_{a}t_{b}\eta _{d}^{2} + \frac{1}{2}t_{a}\eta _{d}(1-t_{b}\eta _{d})Y_{0} \right. \nonumber \\{} & {} \left. +\frac{1}{2}(1-t_{a}\eta _{d})t_{b}\eta _{d}Y_{0} + (1-t_{a}\eta _{d})(1-t_{b}\eta _{d})Y_{0}^{2}\right] , \nonumber \\ Y_{ZP,HV(00)}^{1,1}= & {} (1-Y_{0})^{6}\left[ \frac{1}{4}t_{a}t_{b}\eta _{d}^{2} + t_{a}\eta _{d}(1-t_{b}\eta _{d})Y_{0} \right. \nonumber \\{} & {} \left. +(1-t_{a}\eta _{d})t_{b}\eta _{d}Y_{0} + (1-t_{a}\eta _{d})(1-t_{b}\eta _{d})Y_{0}^{2}\right] , \nonumber \\ Y_{ZP,VH(00)}^{1,1}= & {} (1-Y_{0})^{6}\left[ \frac{1}{4}t_{a}t_{b}\eta _{d}^{2} + t_{a}\eta _{d}(1-t_{b}\eta _{d})Y_{0} \right. \nonumber \\{} & {} \left. +(1-t_{a}\eta _{d})t_{b}\eta _{d}Y_{0} + (1-t_{a}\eta _{d})(1-t_{b}\eta _{d})Y_{0}^{2}\right] , \nonumber \\ Y_{ZP,VV(00)}^{1,1}= & {} (1-Y_{0})^{6}\left[ \frac{1}{4}t_{a}t_{b}\eta _{d}^{2} + \frac{1}{2}t_{a}\eta _{d}(1-t_{b}\eta _{d})Y_{0} \right. \nonumber \\{} & {} \left. +\frac{1}{2}(1-t_{a}\eta _{d})t_{b}\eta _{d}Y_{0} + (1-t_{a}\eta _{d})(1-t_{b}\eta _{d})Y_{0}^{2}\right] . \end{aligned}$$

(A2)

In this way, we can obtain

$$\begin{aligned} Y_{ZP,00}^{1,1}= & {} Y_{ZP,HH(00)}^{1,1}+Y_{ZP,HV(00)}^{1,1} +Y_{ZP,VH(00)}^{1,1}+Y_{ZP,VV(00)}^{1,1}\nonumber \\= & {} (1-Y_{0})^{6}[t_{a}t_{b}\eta _{d}^{2} + 3t_{a}\eta _{d}(1-t_{b}\eta _{d})Y_{0} + 3(1-t_{a}\eta _{d})t_{b}\eta _{d}Y_{0}\nonumber \\{} & {} +4(1-t_{a}\eta _{d})(1-t_{b}\eta _{d})Y_{0}^{2}]. \end{aligned}$$

(A3)

In the second case, when Alice and Bob encode 01, there are four responses of the yield of the signal states as

$$\begin{aligned} Y_{ZP,HH(01)}^{1,1}= & {} (1-Y_{0})^{6}\left[ \frac{1}{4}t_{a}t_{b}\eta _{d}^{2} + \frac{1}{2}t_{a}\eta _{d}(1-t_{b}\eta _{d})Y_{0} \right. \nonumber \\{} & {} \left. +\frac{1}{2}(1-t_{a}\eta _{d})t_{b}\eta _{d}Y_{0} + (1-t_{a}\eta _{d})(1-t_{b}\eta _{d})Y_{0}^{2}\right] , \nonumber \\ Y_{ZP,HV(01)}^{1,1}= & {} (1-Y_{0})^{6}\left[ \frac{1}{4}t_{a}t_{b}\eta _{d}^{2} + t_{a}\eta _{d}(1-t_{b}\eta _{d})Y_{0} \right. \nonumber \\{} & {} \left. +(1-t_{a}\eta _{d})t_{b}\eta _{d}Y_{0} + (1-t_{a}\eta _{d})(1-t_{b}\eta _{d})Y_{0}^{2}\right] , \nonumber \\ Y_{ZP,VH(01)}^{1,1}= & {} (1-Y_{0})^{6}\left[ \frac{1}{4}t_{a}t_{b}\eta _{d}^{2} + t_{a}\eta _{d}(1-t_{b}\eta _{d})Y_{0} \right. \nonumber \\{} & {} \left. +(1-t_{a}\eta _{d})t_{b}\eta _{d}Y_{0} + (1-t_{a}\eta _{d})(1-t_{b}\eta _{d})Y_{0}^{2}\right] , \nonumber \\ Y_{ZP,VV(01)}^{1,1}= & {} (1-Y_{0})^{6}\left[ \frac{1}{4}t_{a}t_{b}\eta _{d}^{2} + \frac{1}{2}t_{a}\eta _{d}(1-t_{b}\eta _{d})Y_{0} \right. \nonumber \\{} & {} \left. +\frac{1}{2}(1-t_{a}\eta _{d})t_{b}\eta _{d}Y_{0} + (1-t_{a}\eta _{d})(1-t_{b}\eta _{d})Y_{0}^{2}\right] . \end{aligned}$$

(A4)

In this case, we can obtain

$$\begin{aligned} Y_{ZP,01}^{1,1}= & {} Y_{ZP,HH(01)}^{1,1}+Y_{ZP,HV(01)}^{1,1} +Y_{ZP,VH(01)}^{1,1}+Y_{ZP,VV(01)}^{1,1}\nonumber \\= & {} (1-Y_{0})^{6}[t_{a}t_{b}\eta _{d}^{2} + 3t_{a}\eta _{d}(1-t_{b}\eta _{d})Y_{0} + 3(1-t_{a}\eta _{d})t_{b}\eta _{d}Y_{0} \nonumber \\{} & {} +4(1-t_{a}\eta _{d})(1-t_{b}\eta _{d})Y_{0}^{2}]. \end{aligned}$$

(A5)

Similarly, we can also obtain

$$\begin{aligned} Y_{ZP,10}^{1,1}= & {} Y_{ZP,11}^{1,1} =(1-Y_{0})^{6}[t_{a}t_{b}\eta _{d}^{2} + 3t_{a}\eta _{d}(1-t_{b}\eta _{d})Y_{0} + 3(1-t_{a}\eta _{d})t_{b}\eta _{d}Y_{0} \nonumber \\{} & {} +4(1-t_{a}\eta _{d})(1-t_{b}\eta _{d})Y_{0}^{2}]. \end{aligned}$$

(A6)

Finally, we can calculate the yield $Y_{ZP}^{1,1}$ and the error rate $e_{ZP}^{1,1}$ as

$$\begin{aligned} Y_{ZP}^{1,1}= & {} \frac{1}{4}(Y_{ZP,00}^{1,1} +Y_{ZP,01}^{1,1}+Y_{ZP,10}^{1,1}+Y_{ZP,11}^{1,1})\nonumber \\= & {} (1-Y_{0})^{6}[t_{a}t_{b}\eta _{d}^{2} + 3t_{a}\eta _{d}(1-t_{b}\eta _{d})Y_{0} \nonumber \\{} & {} +3(1-t_{a}\eta _{d})t_{b}\eta _{d}Y_{0} + 4(1-t_{a}\eta _{d})(1-t_{b}\eta _{d})Y_{0}^{2}]. \end{aligned}$$

(A7)

$$\begin{aligned} e_{ZP}^{1,1}= & {} \frac{1}{2}-\frac{t_{a}t_{b}\eta _{d}^{2}(1-e_{dP})^{2}(1-Y_{0})^{6}}{2Y_{ZP}^{1,1}}. \end{aligned}$$

(A8)

In the second group, when Alice and Bob choose X basis in the polarization DOF, the partial HBSA protocol cannot distinguish states $|\phi _{P}^{\pm }\rangle$ when the quantum state in the momentum DOF is $|\psi _{M}^{\pm }\rangle$. As a result, the partial HBSA may cause error in the polarization DOF with the probability of $\frac{1}{4}$. For ensuring the correctness of the encoded ciphertext, Alice and Bob have to discard the encoded bits in the polarization DOF with the probability of $\frac{1}{4}$. In this way, we can obtain the yield $Y_{XP}^{1,1}$ and the error rate $e_{XP}^{1,1}$ as

$$\begin{aligned} Y_{XP}^{1,1}= & {} \frac{3}{4}Y_{ZP}^{1,1}=(1-Y_{0})^{6}\left[ \frac{3}{4}t_{a}t_{b}\eta _{d}^{2}+\frac{9}{4}Y_{0}(t_{a}\eta _{d} \right. \nonumber \\{} & {} \left. +t_{b}\eta _{d}-2t_{a}t_{b}\eta _{d}^{2})+3Y_{0}^{2}(1-t_{a}\eta _{d})(1-t_{b}\eta _{d})\right] , \end{aligned}$$

(A9)

$$\begin{aligned} e_{XP}^{1,1}= & {} \frac{1}{2}-\frac{t_{a}t_{b}\eta _{d}^{2}(1-e_{dP})^{2}(1-Y_{0})^{6}}{2Y_{ZP}^{1,1}}. \end{aligned}$$

(A10)

On the other hand, in the momentum DOF, the partial HBSA can completely distinguish 4 Bell states, so that the adoption of the partial HBSA cannot case any error in the momentum DOF. In this way, we can deduce the yield $Y_{Z(X)M}^{1,1}$ and the error rate $e_{Z(X)M}^{1,1}$ in the momentum DOF as

$$\begin{aligned} Y_{ZM}^{1,1}= & {} Y_{XM}^{1,1}=Y_{ZP}^{1,1}, \nonumber \\ e_{ZM}^{1,1}= & {} e_{XM}^{1,1}=\frac{1}{2}-\frac{t_{a}t_{b} \eta _{d}^{2}(1-e_{dM})^{2}(1-Y_{0})^{6}}{2Y_{ZM}^{1,1}}. \end{aligned}$$

(A11)