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Improving long-distance distribution of entangled coherent state with the method of twin-field quantum key distribution

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Abstract

The twin-field quantum key distribution (TFQKD) protocol has garnered considerable attention in quantum communication because it overcomes the well-known fundamental limit of the secret key rate without quantum repeaters. In this study, we employ this scheme to demonstrate the long-distance distribution of entangled coherent state (ECS), which has not been addressed in the existing literature. We show a scheme for the distribution of ECS with a yield of $\sqrt {\eta }$, where η is the total efficiency of the whole transmission link. Compared to the cat-state based scheme, the success probability for fidelity is enhanced by 1.86 to 11.54 times in our new scheme, where the ECS is |ψECS(α = 2.0)〉 and the fixed fidelity (F) ranges from 0.99 to 0.75. The performance of our scheme in the presence of realistic on-off photon detector has also been investigated. Our work provides the application of TFQKD method toward continuous variable entanglement distribution and we believe that its application to other quantum information processing protocols are worth investigation in the near future.

© 2019 Optical Society of America under the terms of the OSA Open Access Publishing Agreement

1. Introduction

Entangled coherent states (ECSs) are a fundamentally interesting class of quantum states of light with several applications in quantum information processing [1,2]. Since its first appearance in 1967 [3], ECS have attracted considerable research interest for Bell-like inequality violation [4], quantum teleportation tasks [57], quantum metrology [810], and so on. Several schemes for the generation of ECS have been proposed, which are based on nonlinear Kerr evolution [7], electromagnetically induced transparency [11], and cavity quantum electrodynamics system [12,13]. More recently, the optical generation of ECS using the non-local photon subtraction on cat state [14] and the mixture between squeezed vacuum and coherent light has been experimentally demonstrated [15].

Although ECS has been successfully generated experimentally, its distribution between two remote users is still a major challenge. This is because ECS is easily distorted by loss. In this study, we show that twin-field quantum key distribution (TFQKD) can be used to improve the performance of the distribution of ECSs over long distances.

Essentially, the key rate in QKD with weak coherent states is always bounded by $O(\eta ^2)$ [16,17]. The decoy-state technique [1820] facilitates the use of stronger coherent states with average photon number $O(1)$ and the finally the key rate scales as $O(\eta )$ [21]. The recent advancement in TFQKD [22,23] further improves the secret key rates to the order of $O(\sqrt {\eta })$ [24]. The essence for this improvement is that TFQKD can work if only one of the detectors ($D_c$ and $D_d$) registers a click (see Fig. 1 for more details on the allocation of these two detectors). This is in strong contrast to device-independent QKD where both $D_c$ and $D_d$ should register non-zero photons [25].

 figure: Fig. 1.

Fig. 1. (a) Scheme for the direct transmission of ECS via two symmetric sides of a lossy channel. (b) Scheme for distributing ECS with Schrödinger$'$s cat state. (c) Distribution of ECS using TFQKD method. BS1 and BS2 are two beam splitters (BSs) with transmittance $\sqrt {\eta }$, which simulate the channel transmittance during halfway transmission. BS3 and BS4 are two BSs with transmittance $T$. BS5 is a $50:50$ BS for interference. $D_c$ and $D_d$ are two single photon detectors. The trash boxes denote the discarding operation.

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TFQKD has also stimulated many upgraded variants. For example, Wang et al. [26,27] proposed a sending-or-not-sending (SNS) protocol in which Alice and Bob do not perform post selection for the bits in the Z basis of the signal pulses. Hence, the traditional calculation formulas directly apply. It automatically resolves the issue of security proof. Han et al. gave a detailed security proof to the No-Phase-Post-Selection TFQKD in case of finite-key size and intensity fluctuations [28]. F. Grasselli et al. presented a security bound of the TFQKD protocol when the parities of both Alice and Bob used different settings of decoy intensity and when the protocol was performed in asymmetric loss scenarios [29]. More recently, the method of TFQKD protocol has also been extended to coherent state coding [30].

Motivated by such a striking improvement of $\sqrt {\eta }$ in the secure key rate of TFQKD, we are now investigating whether it is possible to harvest this improvement in continuous variable quantum information processing. Typically, in the prototype scheme of TFQKD [24], an entangled state $|\psi _{AC}\rangle =|\psi _{BD}\rangle =\sqrt {q}|00\rangle +\sqrt {1-q}|11\rangle$ is used. Here, we show that by replacing $|\psi _{AC}\rangle =|\psi _{BD}\rangle$ with a hybrid entangled state (Eq.(13) in Sec. 2.2), one can implement a long-distance distribution of ECS.

Moreover, this scheme has the following advantages: (1) It inherits the intrinsic advantages of TFQKD ($\sqrt {\eta }$ scaling of secret key rate) and the high generation rate of ECS. (2) The scheme generates a genuine odd ECS $|\alpha \rangle |-\alpha \rangle -|-\alpha \rangle |\alpha \rangle$, and not a damped or approximated version of $|\sqrt {T}\alpha \rangle |-\sqrt {T}\alpha \rangle - |-\sqrt {T}\alpha \rangle |\sqrt {T}\alpha \rangle$ [14], which has been frequently reported [14,31,32]. (3) This new scheme exhibits a better probability-fidelity tradeoff, with $1.86$ to $11.54$ times improvement in success probability when fixed fidelity $F$ varies from $0.99$ to $0.75$. Thus, the TFQKD method, originally developed for the distribution of keys between remote users, still has an advantage toward the distribution of continuous variable quantum entanglement. The $\sqrt {\eta }$ improvement in the key rate provides a powerful tool to counteract the decoherence in extremely long-distance quantum communications.

2. Schemes for the distribution of ECS

We have considered the following ECS:

$$|\psi_\mathrm{ECS}(\alpha)\rangle=\frac{1}{\sqrt{N_-(\alpha)}}(|\alpha\rangle|-\alpha\rangle-|-\alpha\rangle|\alpha\rangle),$$
where $N_-(\alpha )=2-2e^{-4\alpha ^2}$.

This is an odd cat state, which has been reported earlier [2]. The even and other forms of ECS can be analyzed using the same method. Figure 1 shows three different schemes for long-distance distribution of ECS.

2.1 Direct transmission

Figure 1(a) presents the direct distribution scheme. The ECS $|\psi _\mathrm {ECS}\rangle$ is prepared in the center and it is sent toward the two sides of a lossy channel. We assume that the channel has a total transmission efficiency of $\eta$ and the source is just located at the center of the channel. The transmittance of each halfway channel is $\sqrt {\eta }$. We use two beam splitters (BS1 and BS2) with transmittance $\sqrt {\eta }$ to simulate this two-side lossy channel. As $|\psi _\mathrm {ECS}\rangle$ is a non-Gaussian state, it is quite convenient to express the state evolution in Fock state space. In what follows, we assume that $\alpha$ is a real and positive number $\alpha >0$. The coherent states can be represented by a superposition $|\alpha \rangle =\sum _{n=0}^\infty \frac {e^{-\frac {\alpha ^2}{2}}\alpha ^n}{\sqrt {n!}}|n\rangle$ [33]. Mathematically, the operation of a BS on modes $k$ and $\ell$ can be equivalently represented by an unitary operator [33]

$$U_{k\ell}(t)=\exp[(\mathrm{arctan}\sqrt{(1-t)/t})(\hat{a}_k \hat{a}_\ell^\dagger{-}\hat{a}_k^\dagger \hat{a}_\ell)],$$
where $t$ is the transmittance.

Generally, BS is described by a linear operation and it evolves two coherent states ($|\alpha \rangle$ and $|\beta \rangle$) into two new coherent states

$$U(t)|\alpha\rangle|\beta\rangle=|\alpha\sqrt{t}-\beta\sqrt{1-t}\rangle|\alpha\sqrt{1-t}+\beta\sqrt{t}\rangle.$$
With this evolution, one can investigate the state decay during the direct transmission. Firstly, it is important to note that direct transmission is a deterministic process, i.e., its success probability is $P^{(a)}=1$.

After the BS operation, the four-mode state $ACBD$ can be written as

$$\begin{aligned}&|\psi^{(a)}\rangle=U_{AC}(\sqrt{\eta})\otimes U_{BD}(\sqrt{\eta})|\psi_\mathrm{ECS}(\alpha)\rangle_{AB}|00\rangle_{CD}\\ &=\frac{1}{\sqrt{N_-}}( |\alpha\eta^{\frac{1}{4}},\alpha \sqrt{1-\sqrt{\eta}},-\alpha\eta^{\frac{1}{4}},-\alpha \sqrt{1-\sqrt{\eta}}\rangle_{ACBD}\\ &~~ ~~~- |-\alpha\eta^{\frac{1}{4}},-\alpha \sqrt{1-\sqrt{\eta}} ,\alpha\eta^{\frac{1}{4}},\alpha \sqrt{1-\sqrt{\eta}}\rangle_{ACBD}). \end{aligned}$$
The output state is obtained by taking partial trace of the optical modes $C$ and $D$, i.e.,
$$\begin{aligned}\rho^{(a)}&=\mathrm{Tr}_{CD}|\psi^{(a)}\rangle\langle \psi^{(b)}|\\ &=\frac{e^{{-}4\alpha^2(1-\sqrt{\eta})}N_-(\alpha\eta^{\frac{1}{4}})}{N_-(\alpha)} |\psi_\mathrm{ECS}(\alpha \eta^\frac{1}{4})\rangle_{AB}\langle \psi_\mathrm{ECS}(\alpha \eta^\frac{1}{4})|\\ &~~~~+\frac{1-e^{{-}4\alpha^2(1-\sqrt{\eta})}}{N_-(\alpha)} |\alpha\eta^{\frac{1}{4}},-\alpha\eta^{\frac{1}{4}}\rangle\langle \alpha\eta^{\frac{1}{4}},-\alpha\eta^{\frac{1}{4}}|\\ &~~~~+\frac{1-e^{{-}4\alpha^2(1-\sqrt{\eta})}}{N_-(\alpha)} |-\alpha\eta^{\frac{1}{4}},\alpha\eta^{\frac{1}{4}}\rangle\langle -\alpha\eta^{\frac{1}{4}},\alpha\eta^{\frac{1}{4}}|, \end{aligned}$$
which contains a damped version of ECS: $|\psi _\mathrm {ECS}(\alpha \eta ^{\frac {1}{4}})\rangle$.

2.2 Non-local photon subtraction on cat state

Figure 1(b) shows the most typical scheme for generating ECS [32]. It uses two local Schrödinger$'$s cat states $|\psi _\mathrm {cat}\rangle =\frac {1}{\sqrt {N_c}}(|\alpha \rangle +|-\alpha \rangle )$ ($N_c=2+2e^{-2\alpha ^2}$) and one non-local photon subtraction in the center. The photon subtraction is heralded by a single click in the single photon detector.

First, the coupling between $AC$ and $BD$ generates a four-mode state

$$\begin{aligned}&|\psi^{(b)}\rangle_{ACBD}\\ &= \frac{1}{N_c}(|\sqrt{T}\alpha,\sqrt{1-T}\alpha\rangle_{AC}+|-\sqrt{T}\alpha,-\sqrt{1-T}\alpha\rangle_{AC})\\ &\otimes( |\sqrt{T}\alpha,\sqrt{1-T}\alpha\rangle_{BD}+|-\sqrt{T}\alpha,-\sqrt{1-T}\alpha\rangle_{BD}). \end{aligned}$$
Subsequently, $C$ and $D$ are sent through the halfway lossy channel. The coupling between $C$ and $E$ (as well as between $D$ and $F$) generates a six-mode state
$$\begin{aligned}&|\psi^{(b)}\rangle_{ACEBDF}=\frac{1}{N_c}\times\\ &(|\sqrt{T}\alpha,f_1\alpha,f_2\alpha,\sqrt{T}\alpha,f_1\alpha,f_2\alpha\rangle_{ACEBDF}\\ & +|\sqrt{T}\alpha,f_1\alpha,f_2\alpha,-\sqrt{T}\alpha,-f_1\alpha,-f_2\alpha\rangle_{ACEBDF}\\ & + |-\sqrt{T}\alpha,-f_1\alpha,-f_2\alpha,\sqrt{T}\alpha,f_1\alpha,f_2\alpha\rangle_{ACEBDF}\\ & +|-\sqrt{T}\alpha,-f_1\alpha,-f_2\alpha,-\sqrt{T}\alpha,-f_1\alpha,-f_2\alpha\rangle_{ACEBDF}),\\ \end{aligned}$$
where $f_1=\eta ^{\frac {1}{4}}\sqrt {1-T}$ and $f_2=\sqrt {1-\sqrt {\eta } }\sqrt {1-T}$.

Finally, the optical modes $C$ and $F$ interfere at the $50:50$ BS (BS5). Again using Eq.(3), we obtain

$$\begin{aligned}&|\psi^{(b)' }\rangle_{ACEBDF}=\frac{1}{N_c}\times\\ &(|\sqrt{T}\alpha,0,f_2\alpha,\sqrt{T}\alpha,\sqrt{2}f_1\alpha,f_2\alpha\rangle_{ACEBDF}\\ & +|\sqrt{T}\alpha,\sqrt{2}f_1\alpha,f_2\alpha,-\sqrt{T}\alpha,0,-f_2\alpha\rangle_{ACEBDF}\\ & + |-\sqrt{T}\alpha,-\sqrt{2}f_1\alpha,-f_2\alpha,\sqrt{T}\alpha,0,f_2\alpha\rangle_{ACEBDF}\\ & +|-\sqrt{T}\alpha,0,-f_2\alpha,-\sqrt{T}\alpha,-\sqrt{2}f_1\alpha,-f_2\alpha\rangle_{ACEBDF}).\\ \end{aligned}$$
We consider the case when detectors $(D_c,D_d)$ register $(1,0)$ photons. This corresponds to projection of $C$ and $D$ mode on the Fock state $|1\rangle _C\langle 1|\otimes |0\rangle _D\langle 0|$. The probability of such a projection as well as the normalized output state $\rho _{10}^{(b)}$ follows
$$\begin{aligned} &P_{10}^{(b)}\rho_{10}^{(b)}=\mathrm{Tr}_{CDEF}[|\psi^{(b) '}\rangle_{ACEBDF}\langle \psi^{(b) '}|(|10\rangle_{CD}\langle 10|\otimes I_{EF})].\\ \end{aligned}$$
After a lengthy calculation, we obtain
$$\begin{aligned}&\rho_{10}^{(b)}=\frac{e^{{-}4f_2^2\alpha^2}N_-(\sqrt{T}\alpha)}{2-2e^{4(f_1^2-1)\alpha^2}} |\psi_\mathrm{ECS}(\sqrt{T}\alpha)\rangle\langle \psi_\mathrm{ECS}(\sqrt{T}\alpha)|\\ &~~~~~~~+\frac{1-e^{{-}4f_2^2\alpha^2} }{2-2e^{4(f_1^2-1)\alpha^2}}|\sqrt{T}\alpha,-\sqrt{T}\alpha \rangle\langle \sqrt{T}\alpha,-\sqrt{T}\alpha|\\ &~~~~~~~+\frac{1-e^{{-}4f_2^2\alpha^2} }{2-2e^{4(f_1^2-1)\alpha^2}}|-\sqrt{T}\alpha,\sqrt{T}\alpha \rangle\langle -\sqrt{T}\alpha,\sqrt{T}\alpha|.\\ \end{aligned}$$
The probability $P_{10}^{(b)}$ follows
$$P_{10}^{(b)}=\frac{4f_1^2\alpha^2 e^{{-}2f_1^2\alpha^2}}{N_c^2}(1-e^{4(f_1^2-1)\alpha^2})= \sqrt{\eta}(1-T)\frac{4 \alpha^2 e^{{-}2f_1^2\alpha^2}}{N_c^2}(1-e^{4(f_1^2-1)\alpha^2}).$$

2.3 TFQKD method

Figure 1(c) shows a schematic of the proposed scheme, which is based on TFQKD protocol. Here, we use two identical hybrid entangled states

$$|\psi_h\rangle_{AC}=\frac{1}{ \sqrt{2}} (|\alpha\rangle_{A} |0\rangle_{C}+|-\alpha\rangle_{A}|1\rangle_{C}),$$
$$|\psi_h\rangle_{DB}=\frac{1}{ \sqrt{2}} (|0\rangle_{D}|\alpha\rangle_{B} +|1\rangle_{D}|-\alpha\rangle_{B}).$$
It can be easily seen that the only difference is that we have replaced $|0\rangle _A|1\rangle _A$ with $|\alpha \rangle _A,|-\alpha \rangle _A$ ($|0\rangle ,|1\rangle$ with $|\alpha \rangle _C,|-\alpha \rangle _C$). It will become clear later that such a replacement makes effective use of direct variable quantum information [34].

Then, $C$ and $D$ modes are sent through the two halfway channels each with transmittance $\sqrt {\eta }$. Finally, both beams interfere at a $50:50$ BS. Again, we consider the case when the two single photon detectors $(D_c,D_d)$ register $(1,0)$. The advantage of this new scheme is that the states $|0\rangle$ and $|1\rangle$ suffer the channel loss and the states $A$ and $B$ that contain the information of $\alpha$ remain invariant. This is the reason for obtaining the original ECS as the output state.

The output can be derived in Fock state space. First, we see that after the halfway channel, the state in optical modes $A$ and $C$ (similarly, that in the modes $B$ and $D$) is a mixed state:

$$\begin{aligned}&\rho_{AC} =\rho_{BD} =Tr_E[U_{CE}(\sqrt{\eta})|\psi_{h}\rangle\langle \psi_h |\otimes |0\rangle_E\langle 0|U_{CE}(\sqrt{\eta})^\dagger]\\ &=\frac{1}{2}(|\widetilde{\phi}\rangle\langle\widetilde{\phi}|+(1-\sqrt{\eta})|-\alpha,0\rangle\langle-\alpha,0 |), \end{aligned}$$
with unnormalized state $|\widetilde {\phi }\rangle =|\alpha ,0\rangle +\eta ^{\frac {1}{4}}|-\alpha ,1\rangle .$ To obtain the $|\psi _{ECS}(\alpha )\rangle$ state, we only consider the case when photon detectors $(D_c,D_d)$ register $(1,0)$. The corresponding probability and normalized output state follow:
$$P_{10}^{(c)}\rho_{10}^{(c)}=\mathrm{Tr}_{CD}[U_{CD}(\frac{1}{2})(\rho_{AC}\otimes \rho_{DB})U_{CD}(\frac{1}{2})^\dagger |10\rangle_{CD}\langle 10|].$$
Using straightforward calculations, we obtain
$$\begin{aligned}&\rho_{10}^{(c)}=\frac{N_-}{N_-{+}2-2\sqrt{\eta}}|\psi_\mathrm{ECS}(\alpha)\rangle\langle \psi_\mathrm{ECS}(\alpha)|+\frac{2(1-\sqrt{\eta})}{N_-{+}2(1-\sqrt{\eta})} |-\alpha,-\alpha\rangle\langle -\alpha,-\alpha|, \end{aligned}$$
$$\begin{aligned}&P_{10}^{(c)}=\frac{1}{4} \left(2-e^{{-}4 \alpha ^2}-\sqrt{\eta }\right) \sqrt{\eta }. \end{aligned}$$
It should be noted that the success probability of both the schemes in Figs. 1(b) and 1(c) scales as $\sqrt {\eta }$. However, they are different. It can be easily seen that there is a modulation factor of $1-T$ in $P_{10}^{(b)}$. Moreover, $T$ is chosen as $1-T \ll 1$ such that a higher fidelity with respect to $|\psi _\mathrm {ECS}(\alpha )\rangle$ can be obtained. Typically, $T=0.95$, as is shown in an earlier experiment [14].

Moreover, as mentioned before, the output state $\rho _{10}^{(c)}$ is exactly similar to the ECS in Eq. (1), and not the damped version $|\psi _\mathrm {ECS}(\alpha \eta ^{\frac {1}{4}})\rangle$ or $|\psi _\mathrm {ECS}(\sqrt {T}\alpha )\rangle$.

3. Performance and numerical comparisons of the three schemes

We can use the fidelity $F=\langle \psi _\mathrm {ECS}|\rho |\psi _\mathrm {ECS}\rangle$ ($\rho =\rho ^{(a)}, \rho _{ 10}^{(b)}, \rho _{10}^{(c)}$) to compare the performance of the three schemes. $F$ can be expressed as

$$\begin{aligned}&F^{(a)}=\frac{e^{2 \alpha ^2 \left(1-\eta ^{1/4}\right)^2} \left(1+e^{4 \alpha ^2 \left(1-\sqrt{\eta }\right)}\right) \left(1-e^{4 \alpha ^2 \eta ^{1/4}}\right)^2}{2 \left(1-e^{4 \alpha ^2}\right)^2},\\ &F^{(b)}=\frac{e^{2 \left({-}1+\sqrt{T}\right)^2 \alpha ^2} \left({-}1+e^{4 \sqrt{T} \alpha ^2}\right)^2 \left(1+e^{4 ({-}1+T) \alpha ^2 \left({-}1+\sqrt{\eta }\right)}\right)}{2 \left(1-e^{4 \alpha ^2}\right) \left(1-e^{4 \alpha ^2 \left(1+({-}1+T) \sqrt{\eta }\right)}\right)},\\ &F^{(c)}=\frac{1-e^{4 \alpha ^2}}{1+e^{4 \alpha ^2} \left({-}2+\sqrt{\eta }\right)}. \end{aligned}$$
Figure 2(a) shows the fidelities $F^{(a)}$, $F^{(b)}$, and $F^{(c)}$ as a function of channel efficiency $\eta$. We consider the distribution of a stronger ECS with $\alpha =2.0$. The corresponding success probability is shown in Fig. 2(b). As expected, the fidelity of schemes Figs. 1(b) and 1(c) outperforms that of the direct transmission scheme (Fig. 1(a)). The fidelity of our scheme is almost the same as that of the scheme in Fig. 1(b), but the success probability is improved. Figure 2(c) shows the comparison of the probability-fidelity tradeoff in these three schemes. It is evident that the scheme based on TFQKD method exhibits a much higher success probability for a given fidelity. When $F$ ranges from $0.99$ to $0.75$, the success probability is improved by $1.86$ to $11.54$ times.

 figure: Fig. 2.

Fig. 2. (a) Fidelity and (b) success probability for the distribution of ECS $|\psi _\mathrm {ECS}\rangle$ as a function of channel efficiency. (c) Tradeoff between fidelity and success probability for the schemes in Figs. 1(b) and (c). Curves are plotted according to the analytical results in Sec. 2 and Sec. 3 without any approximations. Other parameters: $\alpha =2.0$ and $T=0.95$.

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4. Advantage of $\sqrt {\eta }$ scaling

It is quite interesting to find out whether our scheme for the distribution of ECS also benefits from the advantage of $\sqrt {\eta }$ scaling of the security key rate , as is shown by most of the papers based on TFQKD [26,3540]. Here, we show that the answer is affirmative.

To give a fair comparison between the three schemes in Fig. 1, we define a quantity yield to quantify how well the target ECS can be obtained. The direct transmission scheme shown in Fig. 1(a) is deterministic (success probability = $1$) and we can define the yield as $Y^{(a)}= \langle \psi _\mathrm {ECS}| \rho ^{(a)} |\psi _\mathrm {ECS}\rangle$. For the scheme in Fig. 1(b), the yield can be defined as $Y^{(b)}=P_{(10)}^{(b)} \langle \psi _\mathrm {ECS}| \rho ^{(b)} | \psi _\mathrm {ECS} \rangle$. For the scheme Fig. 1(c), the yield can be defined as $Y^{(c)}=P_{(10)}^{(c)} \langle \psi _\mathrm {ECS}| \rho ^{(c)} | \psi _\mathrm {ECS} \rangle$.

After lengthy calculations, we obtain

$$Y^{(a)}=c_{a1}\sqrt{\eta}+c_{a2 }\eta+O(\eta^{\frac{3}{2}}),$$
$$Y^{(b)}=c_{b1}\sqrt{\eta}+c_{b2 }\eta+O(\eta^{\frac{3}{2}}),$$
$$Y^{(c)}=\frac{1}{4} \left(1- e^{{-}4 \alpha ^2}\right)\sqrt{\eta}+O(\eta^{\frac{5}{2}}),$$
where
$$\begin{aligned}&c_{a1}=\frac{8 e^{2 \alpha ^2} \left(1+e^{4 \alpha ^2}\right) \alpha ^4}{\left({-}1+e^{4 \alpha ^2}\right)^2},\\ &c_{a2}=\frac{16 e^{2 \alpha ^2} \alpha ^6 \left(3-3 e^{4 \alpha ^2}+2 \alpha ^2+2 e^{4 \alpha ^2} \alpha ^2\right)}{3 \left({-}1+e^{4 \alpha ^2}\right)^2},\\ &c_{b1}=\frac{e^{6 \alpha ^2-4 \sqrt{T} \alpha ^2-2 T \alpha ^2} \left( e^{4 \sqrt{T} \alpha ^2}-1\right)^2 \left(e^{4 \alpha ^2}+e^{4 T \alpha ^2}\right) (1-T) \alpha ^2}{2 \left( e^{4 \alpha ^2}-1\right)^3},\\ &c_{b2}= \frac{e^{6 \alpha ^2-4 \sqrt{T} \alpha ^2-2 T \alpha ^2} \left( e^{4 \sqrt{T} \alpha ^2}-1\right)^2 \left(e^{4 T \alpha ^2}-e^{4 \alpha ^2} \right) (1-T)^2 \alpha ^4}{\left(e^{4 \alpha ^2}-1\right)^3}.\\ \end{aligned}$$
As expected, the yield in scheme Fig. 1(c) scales as $\sqrt {\eta }$. Interestingly, the terms that scale as $\sqrt {\eta }$ can also be found in $Y^{(a)}$ and $Y^{(b)}$. However, it should be noted that $c_{a1}$ decays exponentially as $\alpha \gtrsim \ 1$. In fact, for $\alpha =2.2$ and $T=0.98$, we obtain $c_{a1}=0.01$, $c_{a2}=0.25$, $c_{b1}=0.08$, and $c_{b2}=-0.003$.

A visual comparison of the yields $Y^{(a)}, Y^{(b)}, Y^{(c)}$ as a function of transmission distance is given in Fig. 3. We consider that $\alpha =2.2$, $T=0.98$, and $\eta =10^{-\frac {0.20L}{10}}$, where the channel efficiency is assumed to be $0.20$ dB/km [32]. The direct transmission scheme exhibits reliable performance only for short distances, where its $100\%$ probability of success makes a pronounced contribution. However, this advantage is lost with increasing distance. It is evident that the scheme based on TFQKD outperforms the other two schemes in long-distance transmission. Moreover, the yield of this new scheme is exactly equal to $\frac {1}{4}\sqrt {\eta }$, as shown by the black solid line in Fig. 3. The factor $\frac {1}{4}$ can be attributed to the probability of photon detection $(D_c,D_d)=(1,0)$. $(D_c,D_d)=(0,1)$ can provide another ECS $\frac {1}{\sqrt {N_+}}(|\alpha ,-\alpha \rangle +|-\alpha ,\alpha \rangle )$. Thus, in terms of ECS, the yield of our TFQKD scheme scales as $\frac {1}{2}\sqrt {\eta }=O(\sqrt {\eta })$, similar to most of the TFQKD-based schemes.

 figure: Fig. 3.

Fig. 3. Yield as a function of transmission distance. The channel efficiency is assumed to be $0.20$ dB/km. Green dashed line represents the linear trend $Y=\eta$. Black solid line represents the square root trend $Y=\frac {1}{4}\sqrt {\eta }$. Other parameters: $\alpha =2.2$ and $T=0.98$.

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5. Performance of the schemes for non-ideal detectors

In the previous section, we showed the performance of our scheme using ideal photon detector with $100 \%$ efficiency, which is difficult to achieve in experiments. Here, we alleviate this condition by using on-off photon detectors that are not $100 \%$ efficient. It is quite interesting to investigate how the performance of the schemes (Figs. 1(b) and 1(c)) degrade in case of on-off detectors. The on-off detector can only register two results. ‘Off’ implies that no photons are detected, while ‘on’ implies that one or more photons are detected. In this case, successful distribution of ECS is realized when the output of detectors $D_c$ and $D_d$ is $(D_c,D_d)=(\mathrm {'on'},\mathrm {'off'})$.

In photon number space, the measurement operators corresponding to the detection of ‘on’ and ‘off’ states can be expressed as positive-definite operators $\{ \hat {\Pi }_\mathrm {on}=\sum\limits _{k=1}^\infty [1-(1-\tau )^k] |k\rangle \langle k|$, $\hat {\Pi }_\mathrm {off}=\sum\limits _{k=0}^\infty (1-\tau )^k |k\rangle \langle k| \}$ [41,42]. Here, $0<\tau <1$ denotes the efficiency of the detector. The derivation of state evolution in Sec. 2 can be applied here as well, except the projection operator $|10\rangle _{CD}\langle 10|$ in Eq. (9) and Eq. (15) should be replaced by $\hat {\Pi }_{\mathrm {on},C}\otimes \hat {\Pi }_{\mathrm {off},D}$. The yield in this case can be defined in an analogous way as that in Sec. 4, i.e.,

$$Y_\mathrm{on,off}^{(b)}=P_\mathrm{on,off}^{(b)}F^{(b)}_\mathrm{on,off},Y_\mathrm{on,off}^{(c)}=P_\mathrm{on,off}^{(c)}F^{(c)}_\mathrm{on,off}.$$
The performance of the two schemes can be conveniently evaluated using numerical calculation and truncation of Fock states within a finite-dimensional subspace. In Fig. 4, we again consider $\alpha =2.5$ and truncate quantum state of each optical mode within a subspace spanned by $|0\rangle ,|1\rangle ,\ldots |19\rangle$ for numerical convergence and computational feasibility. We compare the fidelity and success probability of the schemes in Figs. 1(b) and 1(c) for a fixed channel efficiency $\eta =0.9$ (Figs. 4(a) and 4(b)), $\eta =0.5$ (Figs. 4(c) and 4(d)), and for a increasing detection efficiency $\tau$. It is evident that both the schemes provide a fidelity that is almost immune to the efficiency loss in the on-off photon detectors. In fact, this is an advantage of all entanglement-swapping based schemes [34]. Physically, this can be understood by the fact that the detector affects only the detection rate, not the projection to photonic non-vacuum space. However, it should be noted that the on-off detector cannot discriminate the photon number in the corresponding optical mode. Therefore, a detection event $(D_c,D_d)=(\mathrm {'on'},\mathrm {'off'})$ also counts the contribution of $|10\rangle _{CD}\langle 10|$ and $|20\rangle _{CD}\langle 20|$. A pronounced increase in the success probability can be observed in Figs. 4(b) and 4(d). Moreover, the projection with $|20\rangle _{CD}\langle 20|$ contributes to unwanted noise terms $|-\alpha ,-\alpha \rangle \langle -\alpha ,-\alpha |$, causing a decrease in the output fidelity. Even for $\tau =1$, we observe that the fidelity decreases and probability increases compared to that in the scheme based on perfect single photon detection (Fig. 2).

 figure: Fig. 4.

Fig. 4. Fidelity and success probability when an inefficient on-off detector is used. Other parameters: $\alpha =2.5,T=0.98$.

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Figure 5 compares the yield of the schemes for $\tau =0.9$ (Fig. 5(a)) and $\tau =0.6$ (Fig. 5(b)). A yield scaling with $O(\sqrt {\eta })$ can be identified for the case of on-off photon detectors as well. Here, we observe a strong coincidence between the yield of scheme in Fig. 1(c) and the scaling of $\frac {\tau }{4}\sqrt {\eta }$.

 figure: Fig. 5.

Fig. 5. Yield of distribution of $|\psi _\mathrm {ECS}(\alpha )\rangle$ as a function of transmission distance. The channel efficiency is assumed to be $0.20$ dB/km. Green dashed line represents the linear trend $Y=\eta$. Black solid line represents the square root trend $Y=\frac {\tau }{4}\sqrt {\eta }$. Other parameters: $\alpha =2.5$, $T=0.98$, and (a) $\tau =0.90$ (b) $\tau =0.60$.

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6. Conclusions

Based on entanglement swapping, TFQKD provides an excellent method for countering the linear bound in long-distance key distribution. Here, we used this method for the distribution of ECS, which is very important for continuous variable quantum information processing. This method can be extended to the distribution of entangled squeezed state such as $|\xi ,-\xi \rangle -|-\xi ,\xi \rangle$, where $|\xi \rangle =S(\xi )|0\rangle$ is the single-mode squeezed vacuum state.

Compared with the direct transmission protocol in Fig. 1(a), the interference protocols in Figs. 1(b) and 1(c) are complicated and have more challenges to overcome in experimental implementations. For example, (1) the single-photon interference at the central measuring center requires indistinguishable spatial and spectral characteristics. This requires a precise control of the source and the beam splitter. The experiment in Ref. [35] also adds a polarization controller and a feedback phase modulation before and after the input of interference (BS5). (2) The dark count in detectors $D_c$ and $D_d$. Recent dark count of the detector for the infrared single-photon is about 50 Hz [43]. Although the count rate is small, it will have a serious effect because it could finally cause an erroneous heralding when the photon number in the optic pulse of each halfway is comparable with the dark count. This happens in long-distance quantum communication, and a super-conducting single-photon detector with a dark count rate as low as 1 Hz [44] can be used to help further reduce the dark count rate. (3) The channel noise could be another factor to be considered in long-distance fiber-based channels. The fiber coupling efficiency, guided acoustic wave Brillouin scattering (GAWBS) [45], Rayleigh scattering, and Raman scattering are not negligible when $\eta \rightarrow 0$. The fiber coupling efficiency can be considered as an additional transmission line loss and the noise in the scattering can forms a thermal noise in the fiber, as shown in a recent study by Jia et al. [46]. In their experiment, they used a phase-locking technique to further reduce the influence of noise in fiber.

The advantage of $\sqrt {\eta }$ scaling inherited from the TFQKD provides an efficient and promising tool to combat long-distance induced decoherence and is expected to be beneficial for several quantum information processing protocols. This work provides an effective step in such a direction and we hope our theoretical perspective can be validated experimentally in the near future.

Funding

National Natural Science Foundation of China (11574400, 11204379, 11981240356); Beijing Institute of Technology Research Fund Program for Young Scholars.

Acknowledgments

The authors thank Prof. XiongFeng Ma at TsingHua University and Prof. Shuang Wang at University of Science and Technology of China for fruitful discussion in ShanXi University on twin-field quantum key distribution.

Disclosures

The authors declare no conflicts of interest.

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Figures (5)

Fig. 1.
Fig. 1. (a) Scheme for the direct transmission of ECS via two symmetric sides of a lossy channel. (b) Scheme for distributing ECS with Schrödinger$'$s cat state. (c) Distribution of ECS using TFQKD method. BS1 and BS2 are two beam splitters (BSs) with transmittance $\sqrt {\eta }$, which simulate the channel transmittance during halfway transmission. BS3 and BS4 are two BSs with transmittance $T$. BS5 is a $50:50$ BS for interference. $D_c$ and $D_d$ are two single photon detectors. The trash boxes denote the discarding operation.
Fig. 2.
Fig. 2. (a) Fidelity and (b) success probability for the distribution of ECS $|\psi _\mathrm {ECS}\rangle$ as a function of channel efficiency. (c) Tradeoff between fidelity and success probability for the schemes in Figs. 1(b) and (c). Curves are plotted according to the analytical results in Sec. 2 and Sec. 3 without any approximations. Other parameters: $\alpha =2.0$ and $T=0.95$.
Fig. 3.
Fig. 3. Yield as a function of transmission distance. The channel efficiency is assumed to be $0.20$ dB/km. Green dashed line represents the linear trend $Y=\eta$. Black solid line represents the square root trend $Y=\frac {1}{4}\sqrt {\eta }$. Other parameters: $\alpha =2.2$ and $T=0.98$.
Fig. 4.
Fig. 4. Fidelity and success probability when an inefficient on-off detector is used. Other parameters: $\alpha =2.5,T=0.98$.
Fig. 5.
Fig. 5. Yield of distribution of $|\psi _\mathrm {ECS}(\alpha )\rangle$ as a function of transmission distance. The channel efficiency is assumed to be $0.20$ dB/km. Green dashed line represents the linear trend $Y=\eta$. Black solid line represents the square root trend $Y=\frac {\tau }{4}\sqrt {\eta }$. Other parameters: $\alpha =2.5$, $T=0.98$, and (a) $\tau =0.90$ (b) $\tau =0.60$.

Equations (23)

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| ψ E C S ( α ) = 1 N ( α ) ( | α | α | α | α ) ,
U k ( t ) = exp [ ( a r c t a n ( 1 t ) / t ) ( a ^ k a ^ a ^ k a ^ ) ] ,
U ( t ) | α | β = | α t β 1 t | α 1 t + β t .
| ψ ( a ) = U A C ( η ) U B D ( η ) | ψ E C S ( α ) A B | 00 C D = 1 N ( | α η 1 4 , α 1 η , α η 1 4 , α 1 η A C B D           | α η 1 4 , α 1 η , α η 1 4 , α 1 η A C B D ) .
ρ ( a ) = T r C D | ψ ( a ) ψ ( b ) | = e 4 α 2 ( 1 η ) N ( α η 1 4 ) N ( α ) | ψ E C S ( α η 1 4 ) A B ψ E C S ( α η 1 4 ) |         + 1 e 4 α 2 ( 1 η ) N ( α ) | α η 1 4 , α η 1 4 α η 1 4 , α η 1 4 |         + 1 e 4 α 2 ( 1 η ) N ( α ) | α η 1 4 , α η 1 4 α η 1 4 , α η 1 4 | ,
| ψ ( b ) A C B D = 1 N c ( | T α , 1 T α A C + | T α , 1 T α A C ) ( | T α , 1 T α B D + | T α , 1 T α B D ) .
| ψ ( b ) A C E B D F = 1 N c × ( | T α , f 1 α , f 2 α , T α , f 1 α , f 2 α A C E B D F + | T α , f 1 α , f 2 α , T α , f 1 α , f 2 α A C E B D F + | T α , f 1 α , f 2 α , T α , f 1 α , f 2 α A C E B D F + | T α , f 1 α , f 2 α , T α , f 1 α , f 2 α A C E B D F ) ,
| ψ ( b ) A C E B D F = 1 N c × ( | T α , 0 , f 2 α , T α , 2 f 1 α , f 2 α A C E B D F + | T α , 2 f 1 α , f 2 α , T α , 0 , f 2 α A C E B D F + | T α , 2 f 1 α , f 2 α , T α , 0 , f 2 α A C E B D F + | T α , 0 , f 2 α , T α , 2 f 1 α , f 2 α A C E B D F ) .
P 10 ( b ) ρ 10 ( b ) = T r C D E F [ | ψ ( b ) A C E B D F ψ ( b ) | ( | 10 C D 10 | I E F ) ] .
ρ 10 ( b ) = e 4 f 2 2 α 2 N ( T α ) 2 2 e 4 ( f 1 2 1 ) α 2 | ψ E C S ( T α ) ψ E C S ( T α ) |               + 1 e 4 f 2 2 α 2 2 2 e 4 ( f 1 2 1 ) α 2 | T α , T α T α , T α |               + 1 e 4 f 2 2 α 2 2 2 e 4 ( f 1 2 1 ) α 2 | T α , T α T α , T α | .
P 10 ( b ) = 4 f 1 2 α 2 e 2 f 1 2 α 2 N c 2 ( 1 e 4 ( f 1 2 1 ) α 2 ) = η ( 1 T ) 4 α 2 e 2 f 1 2 α 2 N c 2 ( 1 e 4 ( f 1 2 1 ) α 2 ) .
| ψ h A C = 1 2 ( | α A | 0 C + | α A | 1 C ) ,
| ψ h D B = 1 2 ( | 0 D | α B + | 1 D | α B ) .
ρ A C = ρ B D = T r E [ U C E ( η ) | ψ h ψ h | | 0 E 0 | U C E ( η ) ] = 1 2 ( | ϕ ~ ϕ ~ | + ( 1 η ) | α , 0 α , 0 | ) ,
P 10 ( c ) ρ 10 ( c ) = T r C D [ U C D ( 1 2 ) ( ρ A C ρ D B ) U C D ( 1 2 ) | 10 C D 10 | ] .
ρ 10 ( c ) = N N + 2 2 η | ψ E C S ( α ) ψ E C S ( α ) | + 2 ( 1 η ) N + 2 ( 1 η ) | α , α α , α | ,
P 10 ( c ) = 1 4 ( 2 e 4 α 2 η ) η .
F ( a ) = e 2 α 2 ( 1 η 1 / 4 ) 2 ( 1 + e 4 α 2 ( 1 η ) ) ( 1 e 4 α 2 η 1 / 4 ) 2 2 ( 1 e 4 α 2 ) 2 , F ( b ) = e 2 ( 1 + T ) 2 α 2 ( 1 + e 4 T α 2 ) 2 ( 1 + e 4 ( 1 + T ) α 2 ( 1 + η ) ) 2 ( 1 e 4 α 2 ) ( 1 e 4 α 2 ( 1 + ( 1 + T ) η ) ) , F ( c ) = 1 e 4 α 2 1 + e 4 α 2 ( 2 + η ) .
Y ( a ) = c a 1 η + c a 2 η + O ( η 3 2 ) ,
Y ( b ) = c b 1 η + c b 2 η + O ( η 3 2 ) ,
Y ( c ) = 1 4 ( 1 e 4 α 2 ) η + O ( η 5 2 ) ,
c a 1 = 8 e 2 α 2 ( 1 + e 4 α 2 ) α 4 ( 1 + e 4 α 2 ) 2 , c a 2 = 16 e 2 α 2 α 6 ( 3 3 e 4 α 2 + 2 α 2 + 2 e 4 α 2 α 2 ) 3 ( 1 + e 4 α 2 ) 2 , c b 1 = e 6 α 2 4 T α 2 2 T α 2 ( e 4 T α 2 1 ) 2 ( e 4 α 2 + e 4 T α 2 ) ( 1 T ) α 2 2 ( e 4 α 2 1 ) 3 , c b 2 = e 6 α 2 4 T α 2 2 T α 2 ( e 4 T α 2 1 ) 2 ( e 4 T α 2 e 4 α 2 ) ( 1 T ) 2 α 4 ( e 4 α 2 1 ) 3 .
Y o n , o f f ( b ) = P o n , o f f ( b ) F o n , o f f ( b ) , Y o n , o f f ( c ) = P o n , o f f ( c ) F o n , o f f ( c ) .
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