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Discrete-modulation measurement-device-independent continuous-variable quantum key distribution with a quantum scissor: exact non-Gaussian calculation

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Abstract

In this paper, we study non-Gaussian discrete-modulated measurement-device-independent continuous-variable quantum key distribution protocol equipped with a proposed quantum scissor at the receiver side. Our suggested scissor truncates all multiphoton number states with four or more photons and amplifies remaining photon number states in a probabilistic way. Using exact non-Gaussian calculation, we find that quantum scissor meliorates the fidelity and entanglement between two legitimate parties Alice and Bob, at long distances. Therefore, quantum scissor enhances the continuous-variable quantum key distribution protocol range. Examination of the system for different values of the excess noise reveals that this improvement disappears at high noisy channels.

© 2022 Optica Publishing Group under the terms of the Optica Open Access Publishing Agreement

1. Introduction

Quantum key distribution (QKD) that addresses sharing secret key between at least two legitimate parties is a remarkable application of quantum technology [1]. This subject is first proposed by Bennett and Brassard, 1984, and Ekert, 1991, independently [2,3]. The security of shared key is unconditional which is based on uncertainty relation between mutually unbiased observables and no-cloning theorem [4]. In original QKD protocols, the information was modulated on discrete characteristics of single photon such as polarization. This kind of protocols is called discrete-variable quantum key distribution (DV-QKD) protocols, which among them BB84, B92, E91, SARG04 can be mentioned [2,3,5,6]. Continuous variable quantum key distribution (CV-QKD) protocols are another kind of protocols which are based on modulation of information on continuous characteristics of light, such as quadratures of light [2,713]. CV-QKD protocols use low cost homodyne and/or heterodyne detection, to measure light quadratures, that make these protocols compatible with current technology [14,15]. However, these protocols have reference phase adjustment challenge between legitimate parties [16]. Short secure distance and complicated post processing are two drawbacks of CV-QKD protocols. Moreover, this kind of protocol is sensitive to excess noise and suffers from it which restrict this protocol’s secure length rapidly [17]. To overcome CV-QKD limitations some research is done.

Most of CV-QKD protocols usually use the Gaussian modulation scheme to share secret key between two legitimate parties. Discrete modulation (DM) is another scheme that has attracted much attention, recently [18]. DM is utilized to improve the performance of reconciliation efficiency and increase the resilience of the protocol to the excess noise [1922]. Moreover, DM significantly simplifies the modulation scheme as well as the key extraction task [23].

Another important issue of QKD protocols is imperfect devices, which results in security loopholes. Attack on detectors is a type of attacks against imperfect devices, which is crucial in CV QKD protocols. To overcome the security loophole of detector side channel attack, measurement-device-independent (MDI) version of the protocol is innovated [2431]. In MDI CV-QKD protocols, both legitimate users (Alice and Bob) are senders and communicate by connecting to an untrusted third party (Charlie).

Entanglement distillation using non-Gaussian operators such as photon catalysis, photon addition, photon subtraction, photon displacement and so on, is another procedure that increases the secure length of the protocol and improve the key rate in longer distances [3235]. Moreover, noiseless linear amplifiers (NLAs) are suggested to boost the secure key rate and length of the protocol [36,37]. Quit recently, considerable attention has been paid to a Quantum scissor (QS) which is one of the realistic structure that amplifies the signal in a probabilistic way [38,39]. Enhancement of the secrete key rate at large distances using a QS is confirmed previously, which can be as a result of the improvement of the signal to noise ratio [21,40].

QS is a non-Gaussian operation, and hence degaussifies the Guassianity of the channel. In [21] and [40], it has been demonstrated that QS is useful to enhance the performance of Gaussian and non-Gaussian modulated CV-QKD protocols, respectively. In both of these studies, non-Gaussian behavior of the QS is emphasized, and mutual information is exactly calculated, using the conditional distribution of the QS output. In [41], authors introduced a QS to improve the efficiency of the MDI version of DM CV-QKD protocol. Their results show that the QS can enhance the secure transmission distance. However in this study, Gaussian calculation was adopted. A key limitation of this research is that the QS is a non-Gaussian operation, and hence using of Gaussian calculation for investigation of a QS is just an approximation. To the authors’ best knowledge, exact non-Gaussian investigation of the MDI version of CV-QKD protocol with non-Gaussian discrete modulation equipped with a QS is missing. Therefore, it is necessary to adopt exact method to peruse this problem.

In this paper, we use exact non-Gaussian calculation to investigate the impact of QS on the enhancement of DM-MDI CV-QKD protocol secure range. The results demonstrate that how QS can enhance the secure distance of CV-QKD systems. Our approach could also be useful for precise examination of other non-Gaussian operations such as photon catalysis, photon addition and photon subtraction. This paper is organized in four sections. The structure of the protocol and theoretical consideration are described in Section 2. In section 3, we present the security analysis of the protocol and discuss about numerical results. Finally, section 4 summarizes the results and draws conclusions.

2. Theoretical consideration

2.1 Protocol description

In this section, we present the basic notions of a DM-MDI CV-QKD protocol equipped with a quantum scissor at the receiver side. The EB scheme of this protocol is shown in Fig. 1. In the following, we briefly describe this protocol:

 figure: Fig. 1.

Fig. 1. EB scheme of MDI CV-QKD protocol with a quantum scissor. The quantum channel is modeled by a beam splitter ($BS_1$) which mixes Alice’s mode $A'_2$ and Eve’s thermal mode $E'_2$. TMSV: two-mode squeezed vacuum state, QPSK: quadrature phase shifted keying state. $X_C$, $P_D$: Charlie measurement results. Hom: homodyne detection. Het: heterodyne detection.

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  • Step 1: Alice generates a two mode entangled state ${|{\psi }\rangle }_{A_1,A'_2}$, (see Eq. (1)) and sends mode $A'_2$ through the quantum channel to an untrusted third party Charlie.
  • Step 2: Bob generates a TMSV state ${|{\psi }\rangle }_{B'_1 B_2}=\sqrt {1-|\lambda _B|^{2}}\sum _{n=0}^{\infty }{\lambda _B^{n} {|{n}\rangle }_{B'_1} {|{n}\rangle }_{B_2}}$. He sends mode $B_2$ to Charlie which is on the Bob’s side and retains mode $B'_1$. Therefore, Bob needs quantum memories to hold mode $B'_1$ till he receives Charlie’s outcome for displacement and apply the QS.
  • Step 3: Two modes $A_2$ (which is Alice’s mode after passing through quantum channel) and $B_2$ are combined at the 50:50 beam splitter $BS_2$ with two output modes C and D. Charlie performs homodyne detection on two EPR type operators $\hat {X}_C=(\hat {X}_{A_2}-\hat {X}_{B_2})/\sqrt {2}$ and $\hat {P}_D=(\hat {P}_{A_2}+\hat {P}_{B_2})/\sqrt {2}$ [27,42,43]. He announces his measurement results ($\bar {X}_C$ and $\bar {P}_D$) publicly. We will discuss about this strategy later.
  • Step 4: Bob displaces his mode $B'_1$ by displacement operator $\hat {D}(\Delta _X,\Delta _P)$, according to the Charlie’s measured values. If Bob’s state ${|{\psi }\rangle }_{B'_1 B_2}$ was maximally entangled state (which means $\lambda _B \rightarrow 1$), Charlie’s measurement, completely teleports mode $A_2$ to mode $B_1$, if $\Delta _X=\sqrt {2}\bar {X}_C$ and $\Delta _P=-\sqrt {2}\bar {P}_D$ are chosen [43]. However, since $\lambda _B < 1$, the values of $\Delta _X$ and $\Delta _P$ should be chosen to maximize teleportation fidelity. In this paper, we simply select the values of $\Delta _X= g\bar {X}_C$ and $\Delta _P=-g\bar {P}_D$, which may not be the best choice, specially when a QS is used (see the next step). Here $g$ is the gain of displacement [27]. Optimization of the teleportation fidelity to maximize the protocol range, in the presence of QS, is under investigation by our group and results will be reported in the near future.
  • Step 5: In this protocol, in order to improve the secure distance, Bob is equipped with a QS before his homodyne measurement (in Fig. 1, $B_1$ and $B_1^{s}$ are respectively input and output modes of the scissor). He announces when the scissor is successful, so Alice and Bob accept data and will do post-processing, otherwise discard data.
In this paper, we assume that Eavesdropper (Eve) could entangle herself with the state that Alice sends through the fiber ($A'_2$), and could also purify the channel [27]. Therefore, Eve generates a two mode squeezed vacuum (TMSV) state ${|{\psi }\rangle }_{E_1 E'_2}=\sqrt {1-|\lambda _E|^{2}}\sum _{n=0}^{\infty }{\lambda _E^{n} {|{n}\rangle }_{E_1} {|{n}\rangle }_{E'_2}}$, and couples mode $E'_2$ to the quantum channel by means of the beam-splitter $BS_1$. Given that $\rho _{E'_2}=tr_{E_1}({|{\psi }\rangle }_{E_1 E'_2}{\langle {\psi }|})$ is a thermal state (see Eq. (3) for more details), the beam splitter $BS_1$ can also play the role of thermal noise and loss of the channel (which depend on the excess noise $\epsilon _{excess}=2 \lambda _E^{2}/(1-\lambda _E^{2})$ and $BS_1$ transmittance coefficient $t_c$). For simplicity of calculation, sources and detectors are assumed to be perfect. However, we can approximately consider their loss and noise by adjusting the parameters $t_c$ and $\lambda _E$.

It is important to note that, the security analysis of our protocol is more challenging than the Gaussian ones. Therefore, the optimal attack by an eavesdropper could be non-Gaussian. However, we can choose system parameters such that discrete modulated input state remains close to the Gaussian modulation, and therefore, find a appropriate estimation of the secret key rate by assuming Gaussian attack. We will discuss this issue in Section 3.

2.2 Modulation

In this paper, we choose quadrature phase shifted keying (QPSK), which is a special kind of discrete modulation methods [17]. In the EB scheme of QPSK method, Alice prepares the non-Gaussian two-mode entangled state [17]

$${|{\psi}\rangle}_{A_1 A'_2}= \sum_{k = 0}^{3} {\sqrt {{\Lambda _k}} {{\left| {{\varphi _k}} \right\rangle }_{{A_1}}}{{\left| {{\varphi _k}} \right\rangle }_{A'_2}}}$$
where
$$\begin{array}{l} {\Lambda _k} = {e^{ - {{\left| \alpha \right|}^{2}}}}\sum\limits_{n = 0}^{\infty} {\frac{{{{\left| \alpha \right|}^{2(4n + k)}}}}{{(4n + k)!}}} , \\ \left| {{\varphi _k}} \right\rangle = \frac{{{e^{ - \frac{{{{\left| \alpha \right|}^{2}}}}{2}}}}}{{\sqrt {{\Lambda _k}} }}\sum\limits_{n = 0}^{\infty} {{{( - 1)}^{n}}\frac{{{\alpha ^{4n + k}}}}{{\sqrt {(4n + k)!} }}} \left| {4n + k} \right\rangle . \end{array}$$

In the number state basis, this state can be rewritten as

$${|{\psi}\rangle}_{A_1 A'_2}=\sum_{n_1,n_2}{\gamma(n_1,n_2) {|{n_1}\rangle}_{A_1} {|{n_2}\rangle}_{A'_2}},$$
where
$$\gamma ({n_1},{n_2}) = \frac{{{e^{ - {{\left| \alpha \right|}^{2}}}}{{(\alpha {e^{i\frac{\pi }{4}}})}^{{n_1} + {n_2}}}\delta (\tilde{n}_1,\tilde{n}_2)}}{\sqrt {{\Lambda _{\tilde{n}_1}{e^{i\pi \tilde{n}_1}}} {n_1}!{n_2}! }} ,$$
and $\tilde {n}_j$ is the remainder of $n_j$ divided by 4, i.e. $\tilde {n}_j={n_j} \bmod 4$ ($j=1,2$), and $\delta (i,j)$ is the Kronecker delta. Alice retains mode $A_1$ and sends mode $A'_2$ through the quantum channel to Charlie. By partial tracing over state $A_1$, one can obtain the density operator of mode $A'_2$ as ${\rho _{A'_2}} = \frac {1}{4}\sum _{k = 0}^{3} {{{\left | {{\alpha e^{i(2k+1)\pi /4}}} \right \rangle }_{A{'_2}}}} \left \langle {{\alpha e^{i(2k+1)\pi /4}}} \right |,$ which means Alice sends a mixture of coherent states $\Big \{{|{\alpha e^{\frac {i\pi }{4}}}\rangle },{|{\alpha e^{\frac {3i\pi }{4}}}\rangle },{|{\alpha e^{\frac {5i\pi }{4}}}\rangle },{|{\alpha e^{\frac {7i\pi }{4}}}\rangle } \Big \}$, with equal probability, through quantum channel to Bob [17]. In this paper, we do our calculations in the Schrödinger picture and number state basis, therefore Eq. (2) is used.

2.3 Thermal-loss channel

We assume that thermal-loss channel can be modeled by a beam splitter ($BS_1$), which mixes Alice’s mode $A'_2$ and Eve’s thermal state [21]

$$\rho _{E'_2} = \sum_{n = 0}^{\infty} { {(1 - {{\left| {{\lambda _E}} \right|}^{2}})} {\left| {{\lambda _E}} \right|^{2n}}{\left| n \right\rangle}_{E{'_2}} \left\langle n \right|}.$$

We can uniquely define the channel with loss and excess noise. For channel loss of 0.2 dB/km (for standard optical fibre), the transmittance coefficient of $BS_1$ depends on the channel length $L$ (in km unit) as $t_c=10^{-0.01L}$. In this paper, we quantify the excess noise as $\epsilon _{excess} = \frac {{2 {\lambda _E}^{2}}}{{1 - {\lambda _E}^{2}}}$. It is important to note that to obtain the actual value of excess noise, the value of $\epsilon _{excess}$ should be multiplied by the reflectance of $BS_1$ (i.e. $1-|t_c|^{2}$) which depends on the channel length $L$.

The modes $A'_2$ and $E'_2$ are the input modes of $BS_1$, and $A_2$ and $E_2$ are its output modes. Using beam splitter relations $\hat {a}_{E_2}^{\dagger }=t_c \hat {a}_{E'_2}^{\dagger }+i r_c \hat {a}_{A'_2}^{\dagger }$ and $\hat {a}_{A_2}^{\dagger }=t_c \hat {a}_{A'_2}^{\dagger }+i r_c \hat {a}_{E'_2}^{\dagger }$ where $\hat {a}_j^{\dagger }$ is the creation operator of $j$-th mode, it is straightforward to show that this beam splitter convert the input state

$${\left| \psi \right\rangle _{{A_1}{A'_2}{E_1}{E'_2}}}={\left| \psi \right\rangle _{{A_1}{A'_2}}} \otimes{\left| \psi \right\rangle _{{E_1}{E'_2}}}$$
to the output state
$${|{\psi}\rangle}_{A_1 A_2 E_1 E_2}=\sum _{\substack{n_1,n_2 \\ m_1,m_2}}{\xi_{m_1,m_2}^{n_1,n_2}{|{n_1}\rangle}_{A_1} {|{n_2}\rangle}_{A_2} {|{m_1}\rangle}_{E_1} {|{m_2}\rangle}_{E_2}},$$
in which $\xi _{m_1,m_2}^{n_1,n_2}$ is zero for $n_2+m_2<m_1$, otherwise it is equal
$$\xi_{m_1,m_2}^{n_1,n_2}=\sqrt{1-|\lambda_E|^{2}}\lambda_E^{m_1}\gamma(n_1,n_2+m_2-m_1)\mathcal{K}(m_1,m_2,n_2),$$
where
$$\mathcal{K}(m_1,m_2,n_2)=\sum _{\ell=0}^{n_2}\frac{\sqrt{n_2!m_1!m_2!(n_2+m_2-m_1)!}}{\ell !(n_2-\ell)!(m_1-\ell)!(m_2-m_1+\ell)!} (ir_c)^{m_2-m_1+2\ell} \; t_c^{n_2+m_1-2\ell},$$
and $r_c=\sqrt {1-t_c^{2}}$ is the reflectance coefficient of $BS_1$.

2.4 Entanglement in the middle

Untrusted third party Charlie receives Bob’s mode $B_2$ and Alice’s mode $A_2$ (after passing through the quantum channel of length $L_{AC}=L$). It has been proved that, in reverse reconciliation MDI CV-QKD protocols, maximal protocol length can be attained when Charlie be alongside Bob ($L{_{BC}}=0$) [27,41]. Therefore, we choose this strategy and consider a completely asymmetric MDI. According to Fig. 1, these modes interfere at 50:50 beam splitter $BS_2$ and Charlie measures two EPR type operators $\hat {X}_C=(\hat {X}_{A_2}-\hat {X}_{B_2})/\sqrt {2}$ and $\hat {P}_D=(\hat {P}_{A_2}+\hat {P}_{B_2})/\sqrt {2}$ by dual homodyne detection. If Charlie’s measurement results in $\bar {X}_C$ and $\bar {P}_D$ values, Alice $A_2$ and Bob $B_2$ modes will collapse to common eigen state of $\hat {X}_C$ and $\hat {P}_D$ operators, i.e. ${|{\phi _{(\bar {X}_C,\bar {P}_D)}}\rangle }_{A_2 B_2}= \frac {1}{\sqrt {\pi }} \int\limits _{ - \infty }^{\infty } {dx{e^{ - i\sqrt {2}\bar {P}_D x} } {|{x}\rangle }_{A_2}{|{x-\sqrt {2}\bar {X}_C}\rangle }_{B_2}}$ (see Appendix A).

As shown in Fig. 1, after Charlie’s measurement, Bob applies displacement operator $\hat {D}(\Delta _X,\Delta _P)=e^{i( \Delta _P \hat {X}_{B'_1}-\Delta _X \hat {P}_{B'_1})}$ on his mode $B'_1$. According to Appendix B, Charlie’s measurement and Bob’s displacement will transform the input state

$${|{\psi}\rangle}_{A_1 A_2 E_1 E_2}{|{\psi}\rangle}_{B'_1 B_2}= \sqrt{1-|\lambda_B|^{2}} \sum_{\substack{n_1,n_2,\ell\\m_1,m_2}}{\lambda_B^{\ell}\,\xi_{m_1,m_2}^{n_1,n_2}{|{n_1}\rangle}_{A_1}{|{n_2}\rangle}_{A_2}{|{\ell}\rangle}_{B'_1}{|{\ell}\rangle}_{B_2}{|{m_1}\rangle}_{E_1}{|{m_2}\rangle}_{E_2}}$$
to the final state ${|{\psi _{(\bar {X}_C,\bar {P}_D,\Delta _X,\Delta _P)}}\rangle }_{{A_1 B_1 E_1 E_2}}\otimes {|{\phi _{(\bar {X}_C,\bar {P}_D)}}\rangle }_{{A_2 B_2}}$ with
$${|{\psi_{(\bar{X}_C,\bar{P}_D,\Delta_X,\Delta_P)}}\rangle}_{A_1 B_1 E_1 E_2}=\sum _{\substack{n_A,n_B \\ m_1,m_2}}{\Theta_{m_1,m_2}^{n_A,n_B}{(\bar{X}_C,\bar{P}_D,\Delta_X,\Delta_P)}} {|{n_A}\rangle}_{A_1} {|{n_B}\rangle}_{B_1} {|{m_1}\rangle}_{E_1} {|{m_2}\rangle}_{E_2},$$
where $\Theta _{m_1,m_2}^{n_A,n_B}{(\bar {X}_C,\bar {P}_D,\Delta _X,\Delta _P)}$ is given by Eq. (26) of the Appendix B.

For maximally entangled continuous EPR state ${|{\psi }\rangle }_{B'_1 B_2}$ (i.e. $\lambda _B \rightarrow 1$) and displacement values $\Delta _X=\sqrt {2}\bar {X}_C$, $\Delta _P=-\sqrt {2}\bar {P}_D$, it can be shown that $\Theta _{m_1,m_2}^{n_1,n_2}{(\bar {X}_C,\bar {P}_D,\Delta _X,\Delta _P)}=\xi _{m_1,m_2}^{n_1,n_2}$ (apart from the values of $\bar {X}_C$ and $\bar {P}_D$), and therefore, according to Eqs. (4) and (6), state of Alice’s mode $A_2$ (before Charlie’s measurement) will completely teleported to Bob’s mode $B_1$ [43]. However, for actual situation of $\lambda _B<1$ teleportation fidelity is less than one and the displacement values $\Delta _X=\sqrt {2}\bar {X}_C$ and $\Delta _P=-\sqrt {2}\bar {P}_D$ are not the optimal ones. In this paper we select $\Delta _X= g \bar {X}_C$ and $\Delta _P=- g\bar {P}_D$ where

$$g=\sqrt{\frac{2(3\lambda_B^{2}-1)}{3-\lambda_B^{2}}}$$
is the gain of displacement [27]. Optimization of displacement operator to achieve maximum teleportation fidelity, specially in the presence of QS, is out of the scope of this paper.

In order to complete our discussion about Charlie, we need to compute the joint probability $p(\bar {X}_C,\bar {P}_D)$ of measurement outcomes $\bar {X}_C$ and $\bar {P}_D$. If $\rho _{{A_2}{B_2}}$ represents the density operator of modes $A_2 B_2$ before Charlie’s measurement (which is the partial trace of the state of Eq. (5) over $A_1 B'_1 E_1 E_2$ modes), according to completeness relation of Eq. (18) (of Appendix A) it is obvious that

$$p(\bar{X}_C,\bar{P}_D)={{\langle{\phi_{(\bar{X}_C,\bar{P}_D)}}|}\rho_{{A_2}{B_2}}{|{\phi_{(\bar{X}_C,\bar{P}_D)}}\rangle}}.$$

Using above equation, after straightforward calculations (see Appendix C), one can obtain

$$\begin{aligned} p(\bar{X}_C,\bar{P}_D)= & \frac{(1-|\lambda_B|^{2})}{\pi}e^{-\frac{1}{2}({4\bar{X}_C^{2}+\bar{P}_D^{2}})} \\ \times & \sum_{\substack{n_1,n_2,n'_2 \\ m_1,m_2,\ell}}{\frac{|\lambda_B|^{2\ell}\xi_{m_1,m_2}^{n_1,n_2}(\xi_{m_1,m_2}^{n_1,n'_2})^{*}}{2^{\ell}\ell!\sqrt{2^{n_2+n'_2}n_2!n'_2!}}\mathcal{F}_{\ell,n_2}\left(2\bar{X}_C-i\bar{P}_D\right)\mathcal{F}_{n'_2,\ell}\left(2\bar{X}_C-i\bar{P}_D\right)} \end{aligned}$$
where $\mathcal {F}_{\ell,m}(z)$ is given by Eq. (22) of Appendix B.

Gaussian factor of Eq. (7) indicates that $p(\bar {X}_C,\bar {P}_D)$ has a normal-like distribution profile and decreases at larger values of $(|\bar {X}_C|,|\bar {P}_D|)$. Moreover, the standard deviation of this Gaussian function in the $\bar {P}_D$ direction is twice the $\bar {X}_C$ direction which shows its iso-density loci in the $\bar {X}_C-\bar {P}_D$ plane are ellipses. We will use Eq. (7), for calculating average secure key rate of the protocol.

2.5 Quantum scissor

Now, Alice and Bob can share the same secret key, using their $A_1$ and $B_1$ modes. However, thermal-loss channel set a limitation on the maximum distance that they can be apart. QS is a type of quantum amplifier that can increase the protocol range [21,40]. Here, we will study the role of QS on the maximum distance at which secret key can securely be exchanged. This subject has been examined before based on Gaussian CV-QKD. However, QS is a non-Gaussian operation [38], and investigating it using the Gaussian approach is just an approximation. In this paper, we exactly investigate the impact of QS on secret key rate of MDI CV-QKD protocol.

Bob is equipped with a QS whose setup is shown in Fig. 2. This system truncates Bob’s $B_1$ state, leaves only its vacuum ${|{0}\rangle }$, one ${|{1}\rangle }$, two ${|{2}\rangle }$ and three ${|{3}\rangle }$ photon components. This scissor consists of two beam splitters (with transmission and reflection coefficients $t_j$ and $r_j=\sqrt {1-t_j^{2}}$ ($j=1,2$)), two photon number resolved detectors (PNRD) $D_1$ and $D_2$, a single-photon source and finally a source of two-photon Fock state ${|{2}\rangle }$.

 figure: Fig. 2.

Fig. 2. Quantum scissor setup. This scissor consists of two beam splitters with transmission coefficients $t_1$ and $t_2$, two photon number resolved detectors (PNRD) $D_1$ and $D_2$, a source of single-photon ${|{1}\rangle }$ and a source of two-photon Fock states ${|{2}\rangle }$. The scissor is successful if detectors $D{_1}$ and $D{_2}$ count one and two photons, respectively.

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Alice and Bob accept data when detectors $D{_1}$ and $D{_2}$ count one and two photons, respectively. This happen with probability of $P_{ss}$ which is called success probability of scissor. In this situation, the input state of scissor, i.e. ${|{\psi _{(\bar {X}_C,\bar {P}_D,\Delta _X,\Delta _P)}}\rangle }_{{A_1 B_1 E_1 E_2}}$ of Eq. (6) is converted to

$${|{\psi^{s}_{(\bar{X}_C,\bar{P}_D)}}\rangle}_{A_1 B_1^{s} E_1 E_2}=\frac{1}{\sqrt{P_{ss}}}\sum _{\substack{n_A,n_B \\ m_1,m_2}}c_s(n_B)\Theta_{m_1,m_2}^{n_A,n_B} {(\bar{X}_C,\bar{P}_D,\Delta_X,\Delta_P)} {|{n_A}\rangle}_{A_1} {|{n_B}\rangle}_{B_1^{s}} {|{m_1}\rangle}_{E_1} {|{m_2}\rangle}_{E_2},$$
where according to Appendix D, $c_s(n_B)$ is zero for $n_B>3$, and
$$\begin{aligned} c_s(0)= & -3 i r_1 r_2^{2} t_1^{2} t_2, \\ c_s(1)= & -i r_2 t_1 \left(t_1^{2}-2 r_1^{2}\right) \left(r_2^{2}-2 t_2^{2}\right), \\ c_s(2)= & -i r_1 t_2 \left(r_1^{2}-2 t_1^{2}\right) \left(t_2^{2}-2 r_2^{2}\right), \\ c_s(3)= & -3 i r_1^{2} r_2 t_1 t_2^{2}\;. \end{aligned}$$

We remark that success probability $P_{ss}$ can be evaluated as the normalization factor of Eq. (8) and is given by

$$P_{ss}= \sum _{\substack{n_A,n_B \\ m_1,m_2}} |c_s(n_B)\Theta_{m_1,m_2}^{n_A,n_B} {(\bar{X}_C,\bar{P}_D,\Delta_X,\Delta_P)}|^{2}.$$

In the next section, we will use this formula to evaluate $P_{ss}$ numerically. The transmittance coefficients should be chosen such that optimize $P_{ss}$ and entanglement between two modes $A_1$ and $B_1^{s}$, and accordingly improve the performance of the protocol.

Eq. (8) can also be used to investigate the non-Gaussian behavior of our proposed QS. To this end, same as [40], we can evaluate the distribution $f_{x_B}( x_B )$ of homodyne measurement results on x-quadrature of Bob’s $B_1^{s}$ mode (in the same way, one can calculate the distribution function for p-quadrature of Bob’s $B_1^{s}$ mode). From Eq. (8), by partial tracing over Alice and Eve’s modes, one can easily find

$$\rho_{_{B_1^{s}}}= \sum _{n_B=0}^{3}{ \sum _{n_B'=0}^{3}{\sigma_{n_B,n_B'} {|{n_B}\rangle} {\langle{n_B'}|}}}$$
where,
$$\sigma_{n_B,n_B'}=\frac{c_s(n_B)c_s^{*}(n_B')}{P_{ss}} \sum _{n_A,m_1,m_2}{\Theta_{m_1,m_2}^{n_A,n_B} {(\bar{X}_C,\bar{P}_D,\Delta_X,\Delta_P)}(\Theta_{m_1,m_2}^{n_A,n_B'} {(\bar{X}_C,\bar{P}_D,\Delta_X,\Delta_P)})^{*}}.$$

It is important to note that, our proposed QS retains photon number states ${|{0}\rangle }$, ${|{1}\rangle }$, ${|{2}\rangle }$ and ${|{3}\rangle }$, and therefore, the post-selected state $\rho _{_{B_1^{s}}}$ contains these photon number states. In this case, using Eq. (19) of Appendix B, the probability distribution $f_{x_B}( x_B )$, conditional on the success of the QS, can be calculated as

$$f_{x_B}( x_B )={\langle{x_B}|}\rho_{_{B_1^{s}}}{|{x_B}\rangle}=\sqrt{\frac{2}{\pi}} e^{{-}2 x_B^{2}} \sum _{n=0}^{3}{ \sum _{m=0}^{3}{\frac{\sigma_{n,m}}{\sqrt{2^{n+m} n! m!}} H_n(\sqrt{2}x_B) H_m(\sqrt{2}x_B) }}.$$

Above equation can be rewritten as

$$f_{x_B}( x_B )=\frac{1}{3\sqrt{2\pi}}e^{{-}2 x_B^{2}} \left(\nu_0 + \sum_{n=1}^{6}{\nu_n x_B^{n}} \right),$$
with $\nu _0=6 \sigma _{0,0}-3\sqrt {2} \sigma _{0,2}-3\sqrt {2} \sigma _{2,0}+3\sigma _{2,2}$, $\nu _1=12 \sigma _{0,1}-6 \sqrt {6} \sigma _{0,3}+12 \sigma _{1,0}-6 \sqrt {2} \sigma _{1,2}-6 \sqrt {2} \sigma _{2,1}+6 \sqrt {3} \sigma _{2,3}-6 \sqrt {6} \sigma _{3,0}+6 \sqrt {3} \sigma _{3,2}$, $\nu _2=12 \sqrt {2} \sigma _{0,2}+24 \sigma _{1,1}-12 \sqrt {6} \sigma _{1,3}+12 \sqrt {2} \sigma _{2,0}-24 \sigma _{2,2}-12 \sqrt {6} \sigma _{3,1}+36 \sigma _{3,3}$, $\nu _3=8 \sqrt {6} \sigma _{0,3}+24 \sqrt {2} \sigma _{1,2}+24 \sqrt {2} \sigma _{2,1}-32 \sqrt {3} \sigma _{2,3}+8 \sqrt {6} \sigma _{3,0}-32 \sqrt {3} \sigma _{3,2}$, $\nu _4=16 \sqrt {6} \sigma _{1,3}+48 \sigma _{2,2}+16 \sqrt {6} \sigma _{3,1}-96 \sigma _{3,3}$, $\nu _5=32 \sqrt {3} \sigma _{2,3}+32 \sqrt {3} \sigma _{3,2}$, and $\nu _6=64 \sigma _{3,3}$.

The first term of Eq. (13) is a Gaussian term, while the other five terms are non-Gaussian ones. This equation evidently shows that our QS results in a non-Gaussian distribution. Therefore, Gaussian calculation is not proper for our system and leads to incorrect results [44].

3. Security analysis and numerical results

In this section we derive an equation for the secure key rate of the protocol. This will enable us to investigate the impact of QS on the secret key rate, and compare the results with the case that QS is not present. As previously mentioned, we consider a thermal loss channel which can be modeled by a beam splitter with a thermal state at one of it’s inputs. This can effectively describe a Gaussian attack by an eavesdropper [21]. As stated before, the security analysis of non-Gaussian discrete-modulated MDI-CV-QKD protocols are more challenging than the Gaussian ones. Moreover, our protocol uses a QS which is also non-Gaussian. Therefore, the Gaussian attack is not truly the optimal one. In [45], it is proven that Gaussian collective attacks maximize the Holevo quantity of fixed covariance matrix of the shared state, and therefore give the lower bound on the secret key rate for any protocols (including non-Gaussian) and collective attacks (including non-Gaussian ones). However, this theorem requires some conditions to be satisfied. Among these requirements, the condition that the input signal to be a Gaussian modulated is not satisfied in our protocol (Our protocol uses a discrete modulated input signal which is a non-Gaussian modulated state). However, according to [21,46], if the input state of Eq. (1) satisfies the condition $\alpha \le 0.5$, to good approximation, we can consider this input signal to be Gaussian. Here, we choose $\alpha =0.5$ (see Table 1), and therefore Gaussian attacks will not be too far from eavesdropper optimal attack. We assume that Eve could purify the channel [27]. She employs an entangling cloner, by generating a TMSV state ${|{\psi }\rangle }_{E_1 E'_2}$ and coupling mode $E_2$ with Alice’s mode $A'_2$, while retaining mode $E_1$. In reverse reconciliation protocol, secure key rate under collective attack is given by [47,48]

$$K = \beta S(B:A) - S(B:E),$$
where $\beta =0.95$, $S(B:A)$ and $S(B:E)$ are, respectively, the reconciliation efficiency, the mutual information between Alice and Bob, and the mutual information between Eve and Bob. By relating mutual entropy to conditional entropy using $S(B:A)=S(A)-S(A|B)$ and $S(B:E)=S(E)-S(E|B)$ [49], and considering this fact that $S(E)=S(AB)$ (which is based on the assumption that Eve could purify the channel and therefore Eqs. (6) and (8) are pure states) one finds,
$$K = \beta S(A) - S(AB)+(1-\beta)S(A|B).$$

In above equation, the third term is small compare to the first two terms ($1-\beta =0.05$). Therefore, for simplicity of calculation we drop this term, and rewrite key rate as

$$K(\bar{X}_C,\bar{P}_D) \simeq \beta S\left(\rho_A(\bar{X}_C,\bar{P}_D)\right) - S\left(\rho_{AB}(\bar{X}_C,\bar{P}_D)\right),$$
where $S(\rho )=-tr(\rho log(\rho ))$ is the von Neumann entropy of a density matrix $\rho$. In the presence of QS Eq. (8), and in the absence of it Eq. (6), can be used to calculate $\rho _{AB}(\bar {X}_C,\bar {P}_D)$ (by partial tracing over the Eve’s modes $E_1$ and $E_2$) and $\rho _A(\bar {X}_C,\bar {P}_D)$ (which is the partial trace of $\rho _{AB}(\bar {X}_C,\bar {P}_D)$ over Bob’s mode $B_1^{s}$ or $B_1$).

Tables Icon

Table 1. Numerical values of the parameters

In this protocol, when Bob use a QS before his homodyne measurement, the key rate of Eq. (14) should be multiplied by $P_{ss}$, the success probability of scissor. In this case, the information are calculated for the post-selected data when QS is successful (i.e. we use Eq. (8) instead of Eq. (6)). As we will see later, this lead to the reduction of key rate at short distances. However, at longer distance, QS improves the performance of the protocol.

As we state before, for the actual situation of $\lambda _B<1$, teleportation fidelity of mode $A_2$ to mode $B_1$ is less than one. Moreover, in the presence of QS, the fidelity of states $A_2$ and $B_1^{s}$ is strongly affected by $(\bar {X}_C,\bar {P}_D)$. Therefore, the secret key rate depends on the outcomes of Charlie’s measurement ($\bar {X}_C$ and $\bar {P}_D$), as emphasized in Eq. (14). Figure 3 shows the secret key rate of Eq. (14) versus channel length $L$, for some different values of $\bar {X}_C$ and $\bar {P}_D$, in the absence and presence of QS. Table 1 lists numerical values of parameters used in this figure (and in most of the future figures). Apart form the role of scissor in the enhancement of the protocol range, it is obvious that, in the absence of QS, key rate is approximately equal for different values of $(\bar {X}_C,\bar {P}_D)$. However, when Bob use a QS, secret key rate decreases as $\bar {X}_C$ or $\bar {P}_D$ increases and becomes negative for large values of $|\bar {X}_C|$ and $|\bar {P}_D|$. Figure 4 shows the probability of measurement outcomes, Eq. (7), at protocol length L=22.5km. This figure indicates that $p(\bar {X}_C,\bar {P}_D)$ decreases at larger values of $(|\bar {X}_C|,|\bar {P}_D|)$. Therefore, greater values of $(|\bar {X}_C|,|\bar {P}_D|)$, which diminish the key rate are less likely to occur. Therefore, we can adopt another post selection process. At any desired length $L$, we restrict protocol to the measurement outcomes

$$\mathcal{M}=\bigg \{ (\bar{X}_C,\bar{P}_D) \; \bigg| \; |\bar{X}_C|<\bar{X}_{C}^{max}, \; |\bar{P}_D|<\bar{P}_{D}^{max}, \; K(\bar{X}_C,\bar{P}_D)\ge 0 \bigg \},$$
where $K(\bar {X}_C,\bar {P}_D)$ is given in Eq. (14), and call this as a successful measurement. In this case, the information are calculated for the post-selected data when Charlie’s measurement is successful. The probability of successful measurement is simply
$$P_{sm}=\int_{{(\bar{X}_C,\bar{P}_D) \in \mathcal{M}}}{d\bar{X}_C d\bar{P}_D \;p(\bar{X}_C,\bar{P}_D)}.$$

Note that because the key rate $K(\bar {X}_C,\bar {P}_D)$ depends on distance $L$, the set $\mathcal {M}$ and therefore the success probability $P_{sm}$ will also change with channel length $L$.

 figure: Fig. 3.

Fig. 3. Secure key rate per pulse (a) in the absence of quantum scissor, (b) in the presence of quantum scissor.

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 figure: Fig. 4.

Fig. 4. Probability of Charlie’s measurement outcomes at channel length $L=22.5km$.

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In this scenario, since measurement results are nondeterministic, the average value of secure key rate should be evaluated. In this stage, one can consider two strategies.

Strategy 1: Charlie announces his measurement results publicly. In this case, the average key rate is calculated as

$$K_{avg}^{(1)}=P_{ss}\int_{{(\bar{X}_C,\bar{P}_D) \in \mathcal{M}}}{d\bar{X}_C d\bar{P}_D \;p(\bar{X}_C,\bar{P}_D)K(\bar{X}_C,\bar{P}_D)},$$
where we replace $P_{ss}$ with one, in the absence of QS.

Strategy 2: Charlie does not announce measurement results. He only inform whether his results are in the acceptable range (successful measurement) or not. In such circumstances, the average key rate can be expresses as

$$K_{avg}^{(2)}= P_{ss}\,P_{sm}\;(\beta S\left(\tilde{\rho}_A\right) - S\left(\tilde{\rho}_{AB}\right)),$$
where
$$\tilde{\rho}_j=\frac{1}{P_{sm}}\int\limits_{{(\bar{X}_C,\bar{P}_D) \in \mathcal{M}}}{d\bar{X}_C d\bar{P}_D \;p(\bar{X}_C,\bar{P}_D)\;\rho_j(\bar{X}_C,\bar{P}_D)}$$
for $j=A,AB$. Again, if no scissor is used by Bob, $P_{ss}$ should be replaced with one.

Figure 5 compares the average key rate (both in the presence and absence of QS) of two mentioned strategies, for $\bar {X}_{C}^{max}=1.125$ and $\bar {P}_{D}^{max}=2.5$. From this figure it can be seen that the first strategy gives much better result than the second one. Therefore, it is highly suggested that Charlie announces his measurement results publicly. This result shows that although the von Neumann entropy is concave [49], the key rate behaves as a convex function. Figure 6 is same as Fig. 5, but with $\bar {X}_{C}^{max}=2.5$ and $\bar {P}_{D}^{max}=5.0$. These two figures indicate that, $K_{avg}^{(1)}$ (for the first strategy) is increased slightly at larger values of $\left (\bar {X}_{C}^{max},\bar {P}_{D}^{max}\right )$. In fact, Eq. (15) clearly shows the enhancement of $K_{avg}^{(1)}$ by increasing of the integration limits, as long as $K(\bar {X}_C,\bar {P}_D)$ is non-negative. The minimum values of $(\bar {X}_C,\bar {P}_D)$ in which $K(\bar {X}_C,\bar {P}_D)$ become negative depends on the channel length and reduce with its increment. Therefore, if Charlie chooses the first strategy and announce his results publicly, the best choice of $\left (\bar {X}_{C}^{max},\bar {P}_{D}^{max}\right )$, for any desired distance between Alice and Bob, is the largest area in $X_C-P_D$ plane in which $K(\bar {X}_C,\bar {P}_D)$ in non-negative. Figure 7 shows non-negative region of $K(\bar {X}_C,\bar {P}_D)$ in the presence of QS. In this figure, at each channel length $L$, the horizontal cross section of the surface indicates the boundary of the largest region in which $K(\bar {X}_C,\bar {P}_D)$ in non-negative (white contours in Fig. 7).

 figure: Fig. 5.

Fig. 5. Average secure key rate per pulse in the presence and absence of quantum scissor, in (a) strategy 1 (b) strategy 2, for $\bar {X}_{C}^{max}=1.125$ and $\bar {P}_{D}^{max}=2.5$.

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 figure: Fig. 6.

Fig. 6. Average secure key rate per pulse in the presence and absence of quantum scissor, in (a) strategy 1 (b) strategy 2, for $\bar {X}_{C}^{max}=2.5$ and $\bar {P}_{D}^{max}=5.0$.

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 figure: Fig. 7.

Fig. 7. Non-negative region of $K(\bar {X}_C,\bar {P}_D)$. At each length $L$, the horizontal cross section of this surface determine the boundary of non-negative region of $K(\bar {X}_C,\bar {P}_D)$ (white contours).

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According to Fig. 5 and 6, $K_{avg}^{(2)}$ is very sensitive to the choice of $\bar {X}_{C}^{max}$ and $\bar {P}_{D}^{max}$, and its range is decreased in Fig. 6 compared to Fig. 5. Moreover, from Eq. (16) one can find that $K_{avg}^{(2)}$ would be negligible for very small values of $\bar {X}_{C}^{max}$ and $\bar {P}_{D}^{max}$ (in which $P_{sm}$ is approximately zero). Therefore, according to mean value theorem [50], there exists at least one point $\left (\bar {X}_{C}^{max},\bar {P}_{D}^{max}\right )$ such that $K_{avg}^{(2)}$ takes its maximum value. Because of the better results of the first strategy, we will not deal with finding this extremum point which depends on the channel length.

Hereafter, we choose the first strategy and assume that Charlie announces his measurement outcomes publicly. According to Fig. 6, QS reduces the secure key rate at lower values of the protocol length. However,QS enhances the key rate at larger protocol length, and improves the protocol range. These results are in good agreement with the previous studies [21,40]. Therefore, QS is useful for extending the secure range of the CV-QKD protocols.

In order to understand why QS shows different behavior in short and long channel lengths, in Fig. 8, we have plotted the average fidelity of Alice ($A_1$) and Bob ($B_1^{s}$ if scissor is used, otherwise $B_1$) modes

$$\int\limits_{{(\bar{X}_C,\bar{P}_D) \in \mathcal{M}}}{d\bar{X}_C d\bar{P}_D \;p(\bar{X}_C,\bar{P}_D) \underline{F} \left(\rho_A(\bar{X}_C,\bar{P}_D),\rho_B(\bar{X}_C,\bar{P}_D)\right)}$$
as a function of channel length $L$, where [49]
$$\underline{F} \left(\rho_A,\rho_B\right)=\left(Tr \sqrt{\sqrt{\rho_A}\rho_B\sqrt{\rho_A}}\right)^{2}.$$

According to Fig. 6(a), the average secure key rate per pulse becomes negative after about 50km and 115km, in the absence and presence of QS, respectively. Therefore, in the absence of QS, for $L>50km$, the is no any secure key between Alice and Bob. Given that the purpose of Fig. 8 is to explain the effect of average fidelity on the secure key rate, only the regions with positive secure key rate is shown in this figure. By increasing the channel length, attenuation results in the lower contribution of multiphoton Fock states in Bob’s $B_1$ mode. Therefore, the fidelity of Alice $A_1$ and Bob $B_1$ modes decreases (Blue line diagram in Fig. 8). Moreover, the thermal noise of channel reduces the correlation between these two modes causes higher values of von Neumann entropy $S\left (\rho _{AB}\right )$, and subsequently lower secure Key rate.

 figure: Fig. 8.

Fig. 8. Average Fidelity of Alice and Bob modes as a function of channel length.

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QS of Bob’s side, at short values of $L$, by complete removing of higher order number states (according to Eq. (8), all multiphoton number states with four or more photons) causes the reduction of Alice $A_1$ and Bob $B_1^{s}$ mode’s fidelity. However, at higher values of channel length, scissor acts as an amplifier for ${|{n_B}\rangle }_{B_1^{s}}$ with $n_B \le 3$ and reduces the effect of channel loss on these states; hence gradually increases the fidelity of $A_1$ and $B_1^{s}$ modes. These results can be clearly seen in Fig. 8. According to Eq. (9), scissor coefficient $c_s(n_B)$ depends on the transmission coefficients $t_1$ and $t_2$. Consequently, these transmission coefficients determine how the amplification process occurs, and therefore, their proper selection has a very large effect on the improvement of the protocol range by QS. In this paper, we examined different values of transmission coefficients, and found that $t_1=0.4$ and $t_2=0.9$ are appropriate values (see Table 1).

In addition, QS, by removing the higher order multiphoton states and reducing the impact of thermal noise, slows down the speed of correlation reduction during the channel length $L$. Figure 9, which shows the variation of average logarithmic negativity (as a measure of quantum entanglement)

$$\int\limits_{{(\bar{X}_C,\bar{P}_D) \in \mathcal{M}}}{d\bar{X}_C d\bar{P}_D \;p(\bar{X}_C,\bar{P}_D) \underline{E_N} \left(\rho_{AB}(\bar{X}_C,\bar{P}_D)\right)}$$
between Alice ($A_1$) and Bob ($B_1^{s}$ in the presence, or, $B_1$ in the absence of QS) modes, confirms these results. Here $\underline {E_N} \left (\rho _{AB}\right )=\log _2(||\rho _{AB}^{T_A}||_1)$, where $T_A$ denotes the partial transpose operation and $||A||_1=Tr\left (\sqrt {\rho ^{\dagger } \rho }\right )$ is the trace norm [35]. Again, it is important to emphasize that, sane as Fig. 8, only the regions with the positive secure key rate is shown in Fig. 9.

 figure: Fig. 9.

Fig. 9. Average logarithmic negativity of Alice and Bob modes as a function of channel length.

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In order to complete our analysis, we evaluate the average secure key rate as a function of channel length $L$ for different values of excess noise. The results are summarized in Fig. 10. As expected we find that excess noise, by reduction of the correlation between Alice and Bob modes, shortens the secure range of the protocol, both in the presence and absence of QS. Figure 10 indicates that QS only enhances the secure range of protocols for small values of the excess noise. However, this utility disappears when we deal with a high noisy channel ($\epsilon _{excess}=0.22$). Therefore, QS is not a solution for extending the protocol range at high noisy channels.

 figure: Fig. 10.

Fig. 10. Average secure key rate per pulse in the presence and absence of QS for different values of excess noise.

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4. Conclusion

In this paper we have investigated the impact of QS on the extending of DM-MDI CV-QKD protocol secure range. Given that the QS is intrinsically a non-Gaussian operation, it is essential to use exact method instead of Gaussian approximation, which is adopted in this paper. Our numerical results clearly show that QS reduces secure key rate at short distances. Therefore, QS is inappropriate for short channel lengths. However, at longer distances, QS improves the protocol range. Accordingly, it can be concluded that QS is a proper choice for enhancement of CV-QKD protocol range.

Comparison of the results for two cases of the presence and absence of QS, shows that the QS (which acts as an amplifier) diminishes the impact of channel loss at long distances, and therefore increase the fidelity of teleportation. Moreover, we have calculated the quantum entanglement between two parties Alice and Bob as a function of the channel length in the presence and absence of QS, and find that QS moderates the rate of correlation reduction during the channel length.

In summary, we conclude that QS ameliorates the fidelity and entanglement between two parties Alice and Bob at long distances, and as a result improves the CV-QKD protocol range. In addition, our investigation for different values of excess noise, shows that this betterment disappears when we deal with a high noisy channel.

Appendix A. Common eigen states of EPR type operators

The commutator of two EPR type operators $\hat {X}_C=(\hat {X}_{A}-\hat {X}_{B})/\sqrt {2}$ and $\hat {P}_D=(\hat {P}_{A}+\hat {P}_{B})/\sqrt {2}$ are zero. Therefore, these operators have common eigen states which are maximally entangled [42]. Here, we show that the continuous variable EPR state

$${|{\phi_{(\bar{X}_C,\bar{P}_D)}}\rangle}_{A B}= \frac{1}{\sqrt{\pi}} \int\limits_{ - \infty }^{\infty} {dx{e^{ - i\sqrt{2}\bar{P}_D x} } {|{x}\rangle}_{A}{|{x-\sqrt{2}\bar{X}_C}\rangle}_{B}}$$
is the common eigen state of $\hat {X}_C$ and $\hat {P}_D$, with eigenvalues $\bar {X}_C$ and $\bar {P}_D$ [43]. Here, ${|{x}\rangle }_{j}$ is the eigen state of $\hat {X}_j$, i.e. $\hat {X}_j {|{x}\rangle }_{j}=x_j {|{x}\rangle }_{j}$ (j=A,B). By applying $\hat {X}_C$ on this state we have,
$$\begin{aligned} & \hat{X}_C{|{\phi_{(\bar{X}_C,\bar{P}_D)}}\rangle}_{A B} \\ & =\frac{1}{\sqrt{2\pi}} \int\limits_{ - \infty }^{\infty} {dx{e^{ - i\sqrt{2}\bar{P}_D x} }\left(\hat{X}_{A}-\hat{X}_{B} \right) {|{x}\rangle}_{A}{|{x-\sqrt{2}\bar{X}_C}\rangle}_{B}} \\ & =\frac{1}{\sqrt{2\pi}} \int\limits_{ - \infty }^{\infty} {dx{e^{ - i\sqrt{2}\bar{P}_D x} }\left(x-(x-\sqrt{2}\bar{X}_C) \right) {|{x}\rangle}_{A}{|{x-\sqrt{2}\bar{X}_C}\rangle}_{B}} \\ & =\bar{X}_C{|{\phi_{(\bar{X}_C,\bar{P}_D)}}\rangle}_{A B}. \end{aligned}$$
which shows that the state of Eq. (17) is the eigenstate of $\hat {X}_C$ with eigenvalue $\bar {X}_C$. Now, we apply $\hat {P}_D$ on this state; we obtain
$$\begin{aligned} & \hat{P}_D{|{\phi_{(\bar{X}_C,\bar{P}_D)}}\rangle}_{A B} \\ & =\frac{1}{\sqrt{2\pi}} \int\limits_{ - \infty }^{\infty} {dx{e^{ - i\sqrt{2}\bar{P}_D x} }\left(\hat{P}_{A}+\hat{P}_{B} \right) {|{x}\rangle}_{A}{|{x-\sqrt{2}\bar{X}_C}\rangle}_{B}} \\ & =\frac{-i}{\sqrt{2\pi}} \int\limits_{ - \infty }^{\infty} {dx{e^{ - i\sqrt{2}\bar{P}_D x} }\left(\frac{\partial}{\partial x}{|{x}\rangle}_{A}{|{x-\sqrt{2}\bar{X}_C}\rangle}_{B}+{|{x}\rangle}_{A}\frac{\partial}{\partial x}{|{x-\sqrt{2}\bar{X}_C}\rangle}_{B} \right) } \\ & =\frac{i}{\sqrt{2\pi}} \int\limits_{ - \infty }^{\infty} dx \left[ {|{x}\rangle}_{A} \frac{\partial}{\partial x}\left({e^{ - i\sqrt{2}\bar{P}_D x} }{|{x-\sqrt{2}\bar{X}_C}\rangle}_{B}\right) \right. \\ & \left. -{e^{ - i\sqrt{2}\bar{P}_D x} }{|{x}\rangle}_{A}\frac{\partial}{\partial x}{|{x-\sqrt{2}\bar{X}_C}\rangle}_{B} \right] \\ & =\bar{P}_D{|{\phi_{(\bar{X}_C,\bar{P}_D)}}\rangle}_{A B}, \end{aligned}$$
where in the fourth line integration by part is used for the first term. This equation reveals that ${|{\phi _{(\bar {X}_C,\bar {P}_D)}}\rangle }_{A B}$ is the eigenstate of $\hat {P}_D$ with eigenvalue $\bar {P}_D$.

Since $\hat {X}_C$ and $\hat {P}_D$ are Hermitian operators, their eigenstates form a complete orthogonal set. However, these states are non-normalizable. It is easy to show that the factor $\frac {1}{\sqrt {\pi }}$ of Eq. (17) is chosen such that the completeness relation

$$\int\limits_{ - \infty }^{\infty} \int\limits_{ - \infty }^{\infty} d\bar{X}_C d\bar{P}_D {|{\phi_{(\bar{X}_C,\bar{P}_D)}}\rangle}_{A B} {\langle{\phi_{(\bar{X}_C,\bar{P}_D)}}|} = I_A\otimes I_B$$
holds. Here, $I_A$ and $I_B$ are the identity operators on the Hilbert spaces of modes $A$ and $B$, respectively.

Appendix B. Teleportation

In this section, we derive Eq. (6) of the paper. Equation (5) is the state of system before Charlie’s measurement. After Charlie’s measurement (with measurement outcomes $\bar {X}_C$ and $\bar {P}_D$), Alice $A_2$ and Bob $B_2$ modes will collapse to common eigen state of $\hat {X}_C$ and $\hat {P}_D$ operators, i.e. ${|{\phi _{(\bar {X}_C,\bar {P}_D)}}\rangle }_{A_2 B_2}.$ In this case, the state of $A_1 B'_1 E_1 E_2$ modes become

$$\begin{aligned} & {|{\psi_{(\bar{X}_C,\bar{P}_D)}}\rangle}_{A_1 B'_1 E_1 E_2}={_{A_2 B_2} {\langle{\phi_{(\bar{X}_C,\bar{P}_D)}}|}} . \left({|{\psi}\rangle}_{A_1 A_2 E_1 E_2}\otimes {|{\psi}\rangle}_{B'_1 B_2} \right) \\ & =\sqrt{\frac{1-|\lambda_B|^{2}}{\pi}} \sum_{\substack{n_1,n_2,\ell\\m_1,m_2}} \left[{\lambda_B^{\ell}\,\xi_{m_1,m_2}^{n_1,n_2}{|{n_1}\rangle}_{A_1} {|{\ell}\rangle}_{B'_1} {|{m_1}\rangle}_{E_1}{|{m_2}\rangle}_{E_2}} \int\limits_{ - \infty }^{\infty} {dx{e^{ i\sqrt{2}\bar{P}_D x} } _{A_2}{\langle{x}|}{{n_2}\rangle}_{A_2\; B_2} {{\langle{x-\sqrt{2}\bar{X}_C}}{|{\ell}\rangle}_{B_2}}}\right] . \end{aligned}$$

Using the relation

$${\langle{x}}{|{n}\rangle}=\frac{1}{\sqrt{2^{n} n!}}\left(\frac{2}{\pi}\right)^{1/4}e^{{-}x^{2}}H_n\left(\sqrt{2}x \right),$$
where $H_n(x)$ is the Hermite function [51], we obtain
$$\begin{aligned} & {|{\psi_{(\bar{X}_C,\bar{P}_D)}}\rangle}_{A_1 B'_1 E_1 E_2}=\sqrt{\frac{2(1-|\lambda_B|^{2})}{\pi^{2}}}e^{{-}2\bar{X}_C^{2}} \\ & \times \sum_{\substack{n_1,n_2,\ell\\m_1,m_2}} \left[{\frac{\lambda_B^{\ell}}{\sqrt{2^{\ell+n_2}\ell!n_2!}}\xi_{m_1,m_2}^{n_1,n_2}{|{n_1}\rangle}_{A_1} {|{\ell}\rangle}_{B'_1} {|{m_1}\rangle}_{E_1}{|{m_2}\rangle}_{E_2}}\right. \\ & \left. \times \int\limits_{ - \infty }^{\infty} {dx{e^{ {-}2x^{2}+(i\sqrt{2}\bar{P}_D+2\sqrt{2}\bar{X}_C) x} } H_{n_2}(\sqrt{2}x) H_{\ell}(\sqrt{2}x-2\bar{X}_C)} \right] \\ & =\sqrt{\frac{1-|\lambda_B|^{2}}{\pi^{2}}}e^{i\bar{P}_D\bar{X}_C-\frac{1}{4}(4\bar{X}_C^{2}+\bar{P}_D^{2})} \\ & \times \sum_{\substack{n_1,n_2,\ell\\m_1,m_2}} \left[{\frac{\lambda_B^{\ell}}{\sqrt{2^{\ell+n_2}\ell!n_2!}}\xi_{m_1,m_2}^{n_1,n_2}{|{n_1}\rangle}_{A_1} {|{\ell}\rangle}_{B'_1} {|{m_1}\rangle}_{E_1}{|{m_2}\rangle}_{E_2}}\right. \\ & \left. \times \int\limits_{ - \infty }^{\infty} {dx\; {e^{-\frac{1}{4}\left(2x-i\bar{P}_D \right)^{2} }} H_{\ell}(x-\bar{X}_C) H_{n_2}(x+\bar{X}_C)} \right] . \end{aligned}$$

In order to calculate the last integral of Eq. (20), we evaluate $\int _{ - \infty }^{\infty }dx\;\mathcal {G}(t_1,x- x_0)\mathcal {G}(t_2,x+ x_0) e^{-\frac {1}{4}(2x-ip_0)^{2}}$ for the Hermite generating function [51]

$$\mathcal{G}(t,x)=e^{{-}t^{2}+2tx}=\sum_{n=0}^{\infty}{\frac{t^{n}}{n!}H_n(x )}.$$

After straightforward calculation, and comparing the coefficient of $t_1^{\ell }t_2^{m}$ in both sides of equation, one can find,

$$\begin{aligned} & \int\limits_{ - \infty }^{\infty}{dx\;e^{-\frac{1}{4}(2x-ip_0)^{2}}H_{\ell}(x-x_0)H_m(x+x_0)} = ({-}1)^{\ell}\sqrt{\pi} \\ & \times (2x_0-ip_0)^{\ell}(2x_0+ip_0)^{m} \sum _{r=0}^{m}{\frac{({-}1)^{r} \ell ! m!}{r! (\ell -r)! (m-r)!}\left( \frac{2}{4x_0^{2}+p_0^{2}} \right)^{r}}. \end{aligned}$$

The above equation can be simplified, and the final result is

$$\int\limits_{ - \infty }^{\infty}{dx\;e^{-\frac{1}{4}(2x-ip_0)^{2}}H_{\ell}(x-x_0)H_m(x+x_0)} =({-}1)^{\ell} \sqrt{\pi} \mathcal{F}_{\ell, m}\left(2x_0-ip_0 \right),$$
where
$$\mathcal{F}_{\ell,m}(z)= \sum _{r=0}^{m}{\frac{({-}2)^{r} \ell ! m!}{r! (\ell -r)! (m-r)!} z^{\ell-r}(z^{*})^{m-r}} =z^{\ell}(z^{*})^{m}\left(\frac{2}{|z|^{2}}\right)^{\ell}\mathcal{U}\left(-\ell,1-\ell+m,\frac{|z|^{2}}{2}\right),$$
and
$$\mathcal{U}(a,b,x)=\frac{1}{(a-1)!}\int\limits_{0}^{\infty}{dt e^{{-}x t}t^{a-1}(1+t)^{b-a-1}}$$
is confluent Hypergeometric function of the second kind [51], $z^{*}$ is the complex conjugate of $z$ and $|z|=\sqrt {z^{*} z}$ is the absolute value of $z$.

Combining Eqs. (20) and (21), we have,

$$\begin{aligned} & {|{\psi_{(\bar{X}_C,\bar{P}_D)}}\rangle}_{A_1 B'_1 E_1 E_2}=\sqrt{\frac{1-|\lambda_B|^{2}}{\pi}}e^{i\bar{P}_D\bar{X}_C-\frac{1}{4}(4\bar{X}_C^{2}+\bar{P}_D^{2})} \\ & \times \sum_{\substack{n_1,n_2,\ell\\m_1,m_2}} {\frac{(-\lambda_B)^{\ell}\xi_{m_1,m_2}^{n_1,n_2}}{\sqrt{2^{\ell+n_2}\ell!n_2!}} \mathcal{F}_{\ell, n_2}\left(2\bar{X}_C-i\bar{P}_D \right) {|{n_1}\rangle}_{A_1} {|{\ell}\rangle}_{B'_1} {|{m_1}\rangle}_{E_1}{|{m_2}\rangle}_{E_2}} \end{aligned}$$

After Charlie’s measurement, Bob applies displacement operator $\hat {D}(\Delta _X,\Delta _P)=e^{i( \Delta _P \hat {X}_{B'_1}-\Delta _X \hat {P}_{B'_1})}$ on his mode $B'_1$

$${|{\psi_{(\bar{X}_C,\bar{P}_D,\Delta_X,\Delta_P)}}\rangle}_{A_1 B_1 E_1 E_2}=\hat{D}(\Delta_X,\Delta_P){|{\psi_{(\bar{X}_C,\bar{P}_D)}}\rangle}_{A_1 B'_1 E_1 E_2}.$$

In order to simplify above equation, first we calculate the action of displacement operator on the number state ${|{\ell }\rangle }$,

$$\begin{aligned} & \hat{D}(\Delta_X,\Delta_P){|{\ell}\rangle} =e^{i( \Delta_P \hat{X} -\Delta_X \hat{P} )}\int\limits_{ - \infty }^{\infty}{dx{|{x}\rangle}{\langle{x}}{|{\ell}\rangle}} \\ & =e^{\frac{i\Delta_X \Delta_P}{2}}\int\limits_{ - \infty }^{\infty}{dxe^{ix\Delta_P}{|{x+\Delta_X}\rangle}{\langle{x}}{|{\ell}\rangle}} =\int\limits_{ - \infty }^{\infty}{dxe^{ix\Delta_P}{|{x+\frac{\Delta_X}{2}}\rangle}{\langle{x-\frac{\Delta_X}{2}}}{|{\ell}\rangle}} \\ & =\sum _{m=0}^{\infty}{|{m}\rangle}\int\limits_{ - \infty }^{\infty}{dxe^{ix\Delta_P}{\langle{m}}{|{x+\frac{\Delta_X}{2}}\rangle}{\langle{x-\frac{\Delta_X}{2}}}{|{\ell}\rangle}}. \\ & =\frac{e^{-\Delta_X^{2}/2}}{\sqrt{\pi \; 2^{\ell} \ell !}}\sum _{m=0}^{\infty}\frac{{|{m}\rangle}}{\sqrt{2^{m} m!}}\int\limits_{ - \infty }^{\infty}{dxe^{{-}x^{2}+ix\frac{\Delta_P}{\sqrt{2}}}H_{\ell}(x-\frac{\Delta_X}{\sqrt{2}})H_m(x+\frac{\Delta_X}{\sqrt{2}})}, \end{aligned}$$
where in the last line, Eq. (19) is applied. Again, using Eq. (21), we have,
$$\hat{D}(\Delta_X,\Delta_P){|{\ell}\rangle}=e^{-\frac{4\Delta_X^{2}+\Delta_P^{2}}{8}} \sum _{m=0}^{\infty}\frac{({-}1)^{\ell}{|{m}\rangle}}{\sqrt{2^{\ell+m}\ell ! m!}} \mathcal{F}_{\ell, m}\left( \frac{2 \Delta_X-i\Delta_P}{\sqrt{2}} \right).$$

Starting from Eq. (24), using Eqs. (23) and (25), on can simply finds Eq. (6) of the paper, where

$$\Theta_{m_1,m_2}^{n_A,n_B}{(\bar{X}_C,\bar{P}_D,\Delta_X,\Delta_P)}=\sum_{n=0}^{\infty}{\xi_{m_1,m_2}^{n_A,n} \Gamma_{n,n_B}(\bar{X}_C,\bar{P}_D,\Delta_X,\Delta_P)},$$
and
$$\begin{aligned} & \Gamma_{n,m}(\bar{X}_C,\bar{P}_D,\Delta_X,\Delta_P)=\sqrt{\frac{1-|\lambda_B|^{2}}{\pi\; 2^{n+m} n! m!}} \;e^{-\frac{4\bar{X}_C^{2}+\bar{P}_D^{2}}{4}-\frac{4\Delta_X^{2}+\Delta_P^{2}}{8}+i\bar{P}_D\bar{X}_C } \\ & \times\sum _{\ell=0}^{\infty}{\frac{\lambda_B^{\ell}}{2^{\ell}\ell !} }\mathcal{F}_{\ell,n}\left(2\bar{X}_C-i\bar{P}_D\right)\mathcal{F}_{\ell,m}\left(\frac{2 \Delta_X-i\Delta_P}{\sqrt{2}}\right). \end{aligned}$$

Appendix C. Probability of Charlie’s measurement outcome

In this section, we compute the joint probability $p(\bar {X}_C,\bar {P}_D)$ of measurement outcomes $\bar {X}_C$ and $\bar {P}_D$. At the first step, we evaluate the state of Alice $A_2$ and Bob $B_2$ modes before Charlie’s measurement. By partial tracing of Eq. (5) of the paper over $A_1B'_1E_1E_2$ modes, it is straightforward to find,

$$\begin{aligned} \rho_{{A_2}{B_2}}=(1-|\lambda_B|^{2})\sum_{\substack{n_1,n_2,n'_2 \\ m_1,m_2,\ell}} |\lambda_B|^{2\ell}\xi_{m_1,m_2}^{n_1,n_2}(\xi_{m_1,m_2}^{n_1,n'_2})^{*} {|{n_2}\rangle}_{A_2}{\langle{n'_2}|} \otimes {|{\ell}\rangle}_{B_2}{\langle{\ell}|}. \end{aligned}$$

If Charlie’s measurement results in $\bar {X}_C$ and $\bar {P}_D$ values, Alice $A_2$ and Bob $B_2$ modes will collapse to common eigenstate of $\hat {X}_C$ and $\hat {P}_D$ operators, i.e. ${|{\phi _{(\bar {X}_C,\bar {P}_D)}}\rangle }_{A_2 B_2}$ of Eq. (17). The probability of this collapse is given by

$$p(\bar{X}_C,\bar{P}_D)=\;_{_{A_2 B_2}}{{\langle{\phi_{(\bar{X}_C,\bar{P}_D)}}|}\rho_{{A_2}{B_2}}{|{\phi_{(\bar{X}_C,\bar{P}_D)}}\rangle}_{A_2 B_2}}.$$

Using Eqs. (17) and (28), we have

$$\begin{aligned} p(\bar{X}_C,\bar{P}_D) & =\frac{1-|\lambda_B|^{2}}{\pi} \sum_{\substack{n_1,n_2,n'_2 \\ m_1,m_2,\ell}} |\lambda_B|^{2\ell}\xi_{m_1,m_2}^{n_1,n_2}(\xi_{m_1,m_2}^{n_1,n'_2})^{*} \\ & \times \int\limits_{-\infty}^{\infty}{dx\;e^{i\sqrt{2}\bar{P}_D x} {\langle{x}}{|{n_2}\rangle}{\langle{x-\sqrt{2}\bar{X}_C}}{|{\ell}\rangle} } \int\limits_{-\infty}^{\infty}{dy\;e^{{-}i\sqrt{2}\bar{P}_D y} {\langle{n'_2}}{|{y}\rangle}{\langle{\ell}}{|{y-\sqrt{2}\bar{X}_C}\rangle} } \\ & =\frac{2(1-|\lambda_B|^{2})}{\pi^{2}} \sum_{\substack{n_1,n_2,n'_2 \\ m_1,m_2,\ell}} \frac{|\lambda_B|^{2\ell}\xi_{m_1,m_2}^{n_1,n_2}(\xi_{m_1,m_2}^{n_1,n'_2})^{*}}{2^{\ell}\ell!\sqrt{2^{n_2+n'_2}n_2!n'_2!}} \\ & \times \frac{e^{i\bar{P}_D\bar{X}_C-\frac{1}{4}(4\bar{X}_C^{2}+\bar{P}_D^{2})}}{\sqrt{2}} \int\limits_{ - \infty }^{\infty} {dx\; {e^{-\frac{1}{4}\left(2x-i\bar{P}_D \right)^{2} }} H_{\ell}(x-\bar{X}_C) H_{n_2}(x+\bar{X}_C) } \\ & \times \frac{e^{{-}i\bar{P}_D\bar{X}_C-\frac{1}{4}(4\bar{X}_C^{2}+\bar{P}_D^{2})}}{\sqrt{2}} \int\limits_{ - \infty }^{\infty} {dx\; {e^{-\frac{1}{4}\left(2y+i\bar{P}_D \right)^{2} }} H_{\ell}(y-\bar{X}_C) H_{n'_2}(y+\bar{X}_C) } \end{aligned}$$
where in the last two integrals, Eq. (19) and change of variables $x \rightarrow \frac {x+X_c}{\sqrt {2}}$ and $y \rightarrow \frac {y+X_c}{\sqrt {2}}$ are used. These two integrals are in the form of Eq. (21). Therefore, one can finds
$$\begin{aligned} & p(\bar{X}_C,\bar{P}_D)=\frac{(1-|\lambda_B|^{2})}{\pi}e^{-\frac{1}{2}({4\bar{X}_C^{2}+\bar{P}_D^{2}})} \\ & \sum_{\substack{n_1,n_2,n'_2 \\ m_1,m_2,\ell}}{\frac{|\lambda_B|^{2\ell}\xi_{m_1,m_2}^{n_1,n_2}(\xi_{m_1,m_2}^{n_1,n'_2})^{*}}{2^{\ell}\ell!\sqrt{2^{n_2+n'_2}n_2!n'_2!}}\mathcal{F}_{\ell,n_2}\left(2\bar{X}_C-i\bar{P}_D\right)\mathcal{F}_{\ell,n'_2}\left(2\bar{X}_C+i\bar{P}_D\right)}. \end{aligned}$$

From Eq. (22) it is obvious that $\mathcal {F}_{\ell,m}(z^{*})=\mathcal {F}_{m,\ell }(z)$. Therefore, Eq. (29) is equivalent to Eq. (7) of the paper.

Appendix D. Quantum scissor

In this section we calculate the output state of quantum scissor of Fig. 2 . The initial state of this system is

$${|{\Phi_{in}}\rangle}={|{2}\rangle}_v{|{1}\rangle}_h{|{\psi_{in}}\rangle}_{in}={|{2}\rangle}_v{|{1}\rangle}_h\sum_{n=0}^{\infty}{\eta_n {|{n}\rangle}_{in}},$$
where ${|{\psi _{in}}\rangle }_{in}=\sum _{n=0}^{\infty }{\eta _n {|{n}\rangle }_{in}}$ is an arbitrary input state of the scissor. This state can be rewritten as,
$${|{\Phi_{in}}\rangle}=\sum_{n=0}^{\infty}{\frac{\eta_n}{\sqrt{2\;n!}} (\hat{a}_v^{{\dagger}})^{2}\hat{a}_h^{{\dagger}}(\hat{a}_{in}^{{\dagger}})^{n}{|{0}\rangle}_v{|{0}\rangle}_h{|{0}\rangle}_{in}}.$$

According to Fig. 2, we have

$$\begin{aligned} & \hat{a}_{in}^{{\dagger}}=ir_2\hat{a}_{1}^{{\dagger}}+t_2\hat{a}_{2}^{{\dagger}}, \;\;\;\;\;\;\;\;\;\; \hat{a}_{m}^{{\dagger}}=t_2\hat{a}_{1}^{{\dagger}}+ir_2\hat{a}_{2}^{{\dagger}} \\ & \hat{a}_{h}^{{\dagger}}=ir_1\hat{a}_{m}^{{\dagger}}+t_1\hat{a}_{out}^{{\dagger}}, \;\;\;\;\;\;\;\; \hat{a}_{v}^{{\dagger}}=t_1\hat{a}_{m}^{{\dagger}}+ir_1\hat{a}_{out}^{{\dagger}} \end{aligned}$$
which results in
$$\begin{aligned} & \hat{a}_{in}^{{\dagger}}=ir_2\hat{a}_{1}^{{\dagger}}+t_2\hat{a}_{2}^{{\dagger}} \\ & \hat{a}_{h}^{{\dagger}}=ir_1t_2\hat{a}_{1}^{{\dagger}}-r_1r_2\hat{a}_{2}^{{\dagger}}+t_1\hat{a}_{out}^{{\dagger}} \\ & \hat{a}_{v}^{{\dagger}}=t_1t_2\hat{a}_{1}^{{\dagger}}+it_1r_2\hat{a}_{2}^{{\dagger}}+ir_1\hat{a}_{out}^{{\dagger}} \end{aligned}$$

Substituting Eq. (32) in Eq. (31), we obtain

$$\begin{aligned} & {|{\Phi_{out}}\rangle}=\sum_{n=0}^{\infty}\frac{\eta_n}{\sqrt{2\;n!}} (t_1t_2\hat{a}_{1}^{{\dagger}}+it_1r_2\hat{a}_{2}^{{\dagger}}+ir_1\hat{a}_{out}^{{\dagger}})^{2} \\ & \times (ir_1t_2\hat{a}_{1}^{{\dagger}}-r_1r_2\hat{a}_{2}^{{\dagger}}+t_1\hat{a}_{out}^{{\dagger}}) (ir_2\hat{a}_{1}^{{\dagger}}+t_2\hat{a}_{2}^{{\dagger}})^{n}{|{0}\rangle}_1{|{0}\rangle}_2{|{0}\rangle}_{out}, \end{aligned}$$
which can be simplified as
$$\begin{aligned} & {|{\Phi_{out}}\rangle}=\sum_{n=0}^{\infty}\sum_{\ell=0}^{n}\frac{\eta_n \sqrt{n!}}{\sqrt{2}\ell!(n-\ell)!} (t_1t_2\hat{a}_{1}^{{\dagger}}+it_1r_2\hat{a}_{2}^{{\dagger}}+ir_1\hat{a}_{out}^{{\dagger}})^{2} \\ & \times (ir_1t_2\hat{a}_{1}^{{\dagger}}-r_1r_2\hat{a}_{2}^{{\dagger}}+t_1\hat{a}_{out}^{{\dagger}}) (ir_2\hat{a}_{1}^{{\dagger}})^{\ell} (t_2\hat{a}_{2}^{{\dagger}})^{n-\ell}{|{\{\mathbf{0}\}}\rangle} . \end{aligned}$$

Here, the notation ${|{\{\mathbf {0}\}}\rangle }={|{0}\rangle }_1{|{0}\rangle }_2{|{0}\rangle }_{out}$ is used. This is the final state of QS in the operator form. We can simplify this state and convert it to the photon number state representation. In our numerical study, we examined different outcomes for detector $D_1$ and $D_2$, and find that the case in which these detectors measure one and two photons respectively, gives better secure key rate. Therefore, we only interested to the case in which detectors $D_1$ and $D_2$ count one and two photons, respectively. In this special case, the final state of QS is only the terms which contain $\hat {a}_{1}^{\dagger }(\hat {a}_{2}^{\dagger })^{2}$, i.e.,

$$\begin{aligned} & {|{\Phi_{out}\left( D_1=1, D_2=2 \right)}\rangle} \propto \frac{\hat{a}_{1}^{{\dagger}}(\hat{a}_{2}^{{\dagger}})^{2}}{2} \left({-}3i\sqrt{2}r_1 r_2^{2} t_1^{2} t_2 \eta_0 -i\sqrt{2}r_2 t_1 \left(t_1^{2}-2 r_1^{2}\right) \left(r_2^{2}-2 t_2^{2}\right) \eta_1 \hat{a}_{out}^{{\dagger}} \right. \\ & \left. -i r_1 t_2 \left(r_1^{2}-2 t_1^{2}\right) \left(t_2^{2}-2 r_2^{2}\right) \eta_2 (\hat{a}_{out}^{{\dagger}})^{2} -i \sqrt{3} r_1^{2} r_2 t_1 t_2^{2} \eta_3 (\hat{a}_{out}^{{\dagger}})^{3} \right) {|{\{\mathbf{0}\}}\rangle} \end{aligned}$$

Finally using $(\hat {a}^{\dagger })^{n} {|{0}\rangle }=\sqrt {n!}{|{n}\rangle }$, one can easily obtains

$${|{\Phi_{out}\left( D_1=1, D_2=2 \right)}\rangle}={|{1}\rangle}_1{|{2}\rangle}_2{|{\psi_{out}}\rangle}_{out} =\frac{1}{\sqrt{P_{ss}}} {|{1}\rangle}_1{|{2}\rangle}_2\sum_{n=0}^{3}{c_s(n)\eta_n {|{n}\rangle}_{out}},$$
where $c_s(n)$ are given by Eq. (9) of the paper, and normalization factor $P_{ss}$ is the success probability of scissor, which can be obtain as
$$P_{ss}=\sum_{n=0}^{3}{ |c_s(n)\eta_n|^{2} }$$

Disclosures

The authors declare no conflicts of interest.

Data availability

No data were generated or analyzed in the presented research.

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Data availability

No data were generated or analyzed in the presented research.

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Figures (10)

Fig. 1.
Fig. 1. EB scheme of MDI CV-QKD protocol with a quantum scissor. The quantum channel is modeled by a beam splitter ($BS_1$) which mixes Alice’s mode $A'_2$ and Eve’s thermal mode $E'_2$. TMSV: two-mode squeezed vacuum state, QPSK: quadrature phase shifted keying state. $X_C$, $P_D$: Charlie measurement results. Hom: homodyne detection. Het: heterodyne detection.
Fig. 2.
Fig. 2. Quantum scissor setup. This scissor consists of two beam splitters with transmission coefficients $t_1$ and $t_2$, two photon number resolved detectors (PNRD) $D_1$ and $D_2$, a source of single-photon ${|{1}\rangle }$ and a source of two-photon Fock states ${|{2}\rangle }$. The scissor is successful if detectors $D{_1}$ and $D{_2}$ count one and two photons, respectively.
Fig. 3.
Fig. 3. Secure key rate per pulse (a) in the absence of quantum scissor, (b) in the presence of quantum scissor.
Fig. 4.
Fig. 4. Probability of Charlie’s measurement outcomes at channel length $L=22.5km$.
Fig. 5.
Fig. 5. Average secure key rate per pulse in the presence and absence of quantum scissor, in (a) strategy 1 (b) strategy 2, for $\bar {X}_{C}^{max}=1.125$ and $\bar {P}_{D}^{max}=2.5$.
Fig. 6.
Fig. 6. Average secure key rate per pulse in the presence and absence of quantum scissor, in (a) strategy 1 (b) strategy 2, for $\bar {X}_{C}^{max}=2.5$ and $\bar {P}_{D}^{max}=5.0$.
Fig. 7.
Fig. 7. Non-negative region of $K(\bar {X}_C,\bar {P}_D)$. At each length $L$, the horizontal cross section of this surface determine the boundary of non-negative region of $K(\bar {X}_C,\bar {P}_D)$ (white contours).
Fig. 8.
Fig. 8. Average Fidelity of Alice and Bob modes as a function of channel length.
Fig. 9.
Fig. 9. Average logarithmic negativity of Alice and Bob modes as a function of channel length.
Fig. 10.
Fig. 10. Average secure key rate per pulse in the presence and absence of QS for different values of excess noise.

Tables (1)

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Table 1. Numerical values of the parameters

Equations (63)

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| ψ A 1 A 2 = k = 0 3 Λ k | φ k A 1 | φ k A 2
Λ k = e | α | 2 n = 0 | α | 2 ( 4 n + k ) ( 4 n + k ) ! , | φ k = e | α | 2 2 Λ k n = 0 ( 1 ) n α 4 n + k ( 4 n + k ) ! | 4 n + k .
| ψ A 1 A 2 = n 1 , n 2 γ ( n 1 , n 2 ) | n 1 A 1 | n 2 A 2 ,
γ ( n 1 , n 2 ) = e | α | 2 ( α e i π 4 ) n 1 + n 2 δ ( n ~ 1 , n ~ 2 ) Λ n ~ 1 e i π n ~ 1 n 1 ! n 2 ! ,
ρ E 2 = n = 0 ( 1 | λ E | 2 ) | λ E | 2 n | n E 2 n | .
| ψ A 1 A 2 E 1 E 2 = | ψ A 1 A 2 | ψ E 1 E 2
| ψ A 1 A 2 E 1 E 2 = n 1 , n 2 m 1 , m 2 ξ m 1 , m 2 n 1 , n 2 | n 1 A 1 | n 2 A 2 | m 1 E 1 | m 2 E 2 ,
ξ m 1 , m 2 n 1 , n 2 = 1 | λ E | 2 λ E m 1 γ ( n 1 , n 2 + m 2 m 1 ) K ( m 1 , m 2 , n 2 ) ,
K ( m 1 , m 2 , n 2 ) = = 0 n 2 n 2 ! m 1 ! m 2 ! ( n 2 + m 2 m 1 ) ! ! ( n 2 ) ! ( m 1 ) ! ( m 2 m 1 + ) ! ( i r c ) m 2 m 1 + 2 t c n 2 + m 1 2 ,
| ψ A 1 A 2 E 1 E 2 | ψ B 1 B 2 = 1 | λ B | 2 n 1 , n 2 , m 1 , m 2 λ B ξ m 1 , m 2 n 1 , n 2 | n 1 A 1 | n 2 A 2 | B 1 | B 2 | m 1 E 1 | m 2 E 2
| ψ ( X ¯ C , P ¯ D , Δ X , Δ P ) A 1 B 1 E 1 E 2 = n A , n B m 1 , m 2 Θ m 1 , m 2 n A , n B ( X ¯ C , P ¯ D , Δ X , Δ P ) | n A A 1 | n B B 1 | m 1 E 1 | m 2 E 2 ,
g = 2 ( 3 λ B 2 1 ) 3 λ B 2
p ( X ¯ C , P ¯ D ) = ϕ ( X ¯ C , P ¯ D ) | ρ A 2 B 2 | ϕ ( X ¯ C , P ¯ D ) .
p ( X ¯ C , P ¯ D ) = ( 1 | λ B | 2 ) π e 1 2 ( 4 X ¯ C 2 + P ¯ D 2 ) × n 1 , n 2 , n 2 m 1 , m 2 , | λ B | 2 ξ m 1 , m 2 n 1 , n 2 ( ξ m 1 , m 2 n 1 , n 2 ) 2 ! 2 n 2 + n 2 n 2 ! n 2 ! F , n 2 ( 2 X ¯ C i P ¯ D ) F n 2 , ( 2 X ¯ C i P ¯ D )
| ψ ( X ¯ C , P ¯ D ) s A 1 B 1 s E 1 E 2 = 1 P s s n A , n B m 1 , m 2 c s ( n B ) Θ m 1 , m 2 n A , n B ( X ¯ C , P ¯ D , Δ X , Δ P ) | n A A 1 | n B B 1 s | m 1 E 1 | m 2 E 2 ,
c s ( 0 ) = 3 i r 1 r 2 2 t 1 2 t 2 , c s ( 1 ) = i r 2 t 1 ( t 1 2 2 r 1 2 ) ( r 2 2 2 t 2 2 ) , c s ( 2 ) = i r 1 t 2 ( r 1 2 2 t 1 2 ) ( t 2 2 2 r 2 2 ) , c s ( 3 ) = 3 i r 1 2 r 2 t 1 t 2 2 .
P s s = n A , n B m 1 , m 2 | c s ( n B ) Θ m 1 , m 2 n A , n B ( X ¯ C , P ¯ D , Δ X , Δ P ) | 2 .
ρ B 1 s = n B = 0 3 n B = 0 3 σ n B , n B | n B n B |
σ n B , n B = c s ( n B ) c s ( n B ) P s s n A , m 1 , m 2 Θ m 1 , m 2 n A , n B ( X ¯ C , P ¯ D , Δ X , Δ P ) ( Θ m 1 , m 2 n A , n B ( X ¯ C , P ¯ D , Δ X , Δ P ) ) .
f x B ( x B ) = x B | ρ B 1 s | x B = 2 π e 2 x B 2 n = 0 3 m = 0 3 σ n , m 2 n + m n ! m ! H n ( 2 x B ) H m ( 2 x B ) .
f x B ( x B ) = 1 3 2 π e 2 x B 2 ( ν 0 + n = 1 6 ν n x B n ) ,
K = β S ( B : A ) S ( B : E ) ,
K = β S ( A ) S ( A B ) + ( 1 β ) S ( A | B ) .
K ( X ¯ C , P ¯ D ) β S ( ρ A ( X ¯ C , P ¯ D ) ) S ( ρ A B ( X ¯ C , P ¯ D ) ) ,
M = { ( X ¯ C , P ¯ D ) | | X ¯ C | < X ¯ C m a x , | P ¯ D | < P ¯ D m a x , K ( X ¯ C , P ¯ D ) 0 } ,
P s m = ( X ¯ C , P ¯ D ) M d X ¯ C d P ¯ D p ( X ¯ C , P ¯ D ) .
K a v g ( 1 ) = P s s ( X ¯ C , P ¯ D ) M d X ¯ C d P ¯ D p ( X ¯ C , P ¯ D ) K ( X ¯ C , P ¯ D ) ,
K a v g ( 2 ) = P s s P s m ( β S ( ρ ~ A ) S ( ρ ~ A B ) ) ,
ρ ~ j = 1 P s m ( X ¯ C , P ¯ D ) M d X ¯ C d P ¯ D p ( X ¯ C , P ¯ D ) ρ j ( X ¯ C , P ¯ D )
( X ¯ C , P ¯ D ) M d X ¯ C d P ¯ D p ( X ¯ C , P ¯ D ) F _ ( ρ A ( X ¯ C , P ¯ D ) , ρ B ( X ¯ C , P ¯ D ) )
F _ ( ρ A , ρ B ) = ( T r ρ A ρ B ρ A ) 2 .
( X ¯ C , P ¯ D ) M d X ¯ C d P ¯ D p ( X ¯ C , P ¯ D ) E N _ ( ρ A B ( X ¯ C , P ¯ D ) )
| ϕ ( X ¯ C , P ¯ D ) A B = 1 π d x e i 2 P ¯ D x | x A | x 2 X ¯ C B
X ^ C | ϕ ( X ¯ C , P ¯ D ) A B = 1 2 π d x e i 2 P ¯ D x ( X ^ A X ^ B ) | x A | x 2 X ¯ C B = 1 2 π d x e i 2 P ¯ D x ( x ( x 2 X ¯ C ) ) | x A | x 2 X ¯ C B = X ¯ C | ϕ ( X ¯ C , P ¯ D ) A B .
P ^ D | ϕ ( X ¯ C , P ¯ D ) A B = 1 2 π d x e i 2 P ¯ D x ( P ^ A + P ^ B ) | x A | x 2 X ¯ C B = i 2 π d x e i 2 P ¯ D x ( x | x A | x 2 X ¯ C B + | x A x | x 2 X ¯ C B ) = i 2 π d x [ | x A x ( e i 2 P ¯ D x | x 2 X ¯ C B ) e i 2 P ¯ D x | x A x | x 2 X ¯ C B ] = P ¯ D | ϕ ( X ¯ C , P ¯ D ) A B ,
d X ¯ C d P ¯ D | ϕ ( X ¯ C , P ¯ D ) A B ϕ ( X ¯ C , P ¯ D ) | = I A I B
| ψ ( X ¯ C , P ¯ D ) A 1 B 1 E 1 E 2 = A 2 B 2 ϕ ( X ¯ C , P ¯ D ) | . ( | ψ A 1 A 2 E 1 E 2 | ψ B 1 B 2 ) = 1 | λ B | 2 π n 1 , n 2 , m 1 , m 2 [ λ B ξ m 1 , m 2 n 1 , n 2 | n 1 A 1 | B 1 | m 1 E 1 | m 2 E 2 d x e i 2 P ¯ D x A 2 x | n 2 A 2 B 2 x 2 X ¯ C | B 2 ] .
x | n = 1 2 n n ! ( 2 π ) 1 / 4 e x 2 H n ( 2 x ) ,
| ψ ( X ¯ C , P ¯ D ) A 1 B 1 E 1 E 2 = 2 ( 1 | λ B | 2 ) π 2 e 2 X ¯ C 2 × n 1 , n 2 , m 1 , m 2 [ λ B 2 + n 2 ! n 2 ! ξ m 1 , m 2 n 1 , n 2 | n 1 A 1 | B 1 | m 1 E 1 | m 2 E 2 × d x e 2 x 2 + ( i 2 P ¯ D + 2 2 X ¯ C ) x H n 2 ( 2 x ) H ( 2 x 2 X ¯ C ) ] = 1 | λ B | 2 π 2 e i P ¯ D X ¯ C 1 4 ( 4 X ¯ C 2 + P ¯ D 2 ) × n 1 , n 2 , m 1 , m 2 [ λ B 2 + n 2 ! n 2 ! ξ m 1 , m 2 n 1 , n 2 | n 1 A 1 | B 1 | m 1 E 1 | m 2 E 2 × d x e 1 4 ( 2 x i P ¯ D ) 2 H ( x X ¯ C ) H n 2 ( x + X ¯ C ) ] .
G ( t , x ) = e t 2 + 2 t x = n = 0 t n n ! H n ( x ) .
d x e 1 4 ( 2 x i p 0 ) 2 H ( x x 0 ) H m ( x + x 0 ) = ( 1 ) π × ( 2 x 0 i p 0 ) ( 2 x 0 + i p 0 ) m r = 0 m ( 1 ) r ! m ! r ! ( r ) ! ( m r ) ! ( 2 4 x 0 2 + p 0 2 ) r .
d x e 1 4 ( 2 x i p 0 ) 2 H ( x x 0 ) H m ( x + x 0 ) = ( 1 ) π F , m ( 2 x 0 i p 0 ) ,
F , m ( z ) = r = 0 m ( 2 ) r ! m ! r ! ( r ) ! ( m r ) ! z r ( z ) m r = z ( z ) m ( 2 | z | 2 ) U ( , 1 + m , | z | 2 2 ) ,
U ( a , b , x ) = 1 ( a 1 ) ! 0 d t e x t t a 1 ( 1 + t ) b a 1
| ψ ( X ¯ C , P ¯ D ) A 1 B 1 E 1 E 2 = 1 | λ B | 2 π e i P ¯ D X ¯ C 1 4 ( 4 X ¯ C 2 + P ¯ D 2 ) × n 1 , n 2 , m 1 , m 2 ( λ B ) ξ m 1 , m 2 n 1 , n 2 2 + n 2 ! n 2 ! F , n 2 ( 2 X ¯ C i P ¯ D ) | n 1 A 1 | B 1 | m 1 E 1 | m 2 E 2
| ψ ( X ¯ C , P ¯ D , Δ X , Δ P ) A 1 B 1 E 1 E 2 = D ^ ( Δ X , Δ P ) | ψ ( X ¯ C , P ¯ D ) A 1 B 1 E 1 E 2 .
D ^ ( Δ X , Δ P ) | = e i ( Δ P X ^ Δ X P ^ ) d x | x x | = e i Δ X Δ P 2 d x e i x Δ P | x + Δ X x | = d x e i x Δ P | x + Δ X 2 x Δ X 2 | = m = 0 | m d x e i x Δ P m | x + Δ X 2 x Δ X 2 | . = e Δ X 2 / 2 π 2 ! m = 0 | m 2 m m ! d x e x 2 + i x Δ P 2 H ( x Δ X 2 ) H m ( x + Δ X 2 ) ,
D ^ ( Δ X , Δ P ) | = e 4 Δ X 2 + Δ P 2 8 m = 0 ( 1 ) | m 2 + m ! m ! F , m ( 2 Δ X i Δ P 2 ) .
Θ m 1 , m 2 n A , n B ( X ¯ C , P ¯ D , Δ X , Δ P ) = n = 0 ξ m 1 , m 2 n A , n Γ n , n B ( X ¯ C , P ¯ D , Δ X , Δ P ) ,
Γ n , m ( X ¯ C , P ¯ D , Δ X , Δ P ) = 1 | λ B | 2 π 2 n + m n ! m ! e 4 X ¯ C 2 + P ¯ D 2 4 4 Δ X 2 + Δ P 2 8 + i P ¯ D X ¯ C × = 0 λ B 2 ! F , n ( 2 X ¯ C i P ¯ D ) F , m ( 2 Δ X i Δ P 2 ) .
ρ A 2 B 2 = ( 1 | λ B | 2 ) n 1 , n 2 , n 2 m 1 , m 2 , | λ B | 2 ξ m 1 , m 2 n 1 , n 2 ( ξ m 1 , m 2 n 1 , n 2 ) | n 2 A 2 n 2 | | B 2 | .
p ( X ¯ C , P ¯ D ) = A 2 B 2 ϕ ( X ¯ C , P ¯ D ) | ρ A 2 B 2 | ϕ ( X ¯ C , P ¯ D ) A 2 B 2 .
p ( X ¯ C , P ¯ D ) = 1 | λ B | 2 π n 1 , n 2 , n 2 m 1 , m 2 , | λ B | 2 ξ m 1 , m 2 n 1 , n 2 ( ξ m 1 , m 2 n 1 , n 2 ) × d x e i 2 P ¯ D x x | n 2 x 2 X ¯ C | d y e i 2 P ¯ D y n 2 | y | y 2 X ¯ C = 2 ( 1 | λ B | 2 ) π 2 n 1 , n 2 , n 2 m 1 , m 2 , | λ B | 2 ξ m 1 , m 2 n 1 , n 2 ( ξ m 1 , m 2 n 1 , n 2 ) 2 ! 2 n 2 + n 2 n 2 ! n 2 ! × e i P ¯ D X ¯ C 1 4 ( 4 X ¯ C 2 + P ¯ D 2 ) 2 d x e 1 4 ( 2 x i P ¯ D ) 2 H ( x X ¯ C ) H n 2 ( x + X ¯ C ) × e i P ¯ D X ¯ C 1 4 ( 4 X ¯ C 2 + P ¯ D 2 ) 2 d x e 1 4 ( 2 y + i P ¯ D ) 2 H ( y X ¯ C ) H n 2 ( y + X ¯ C )
p ( X ¯ C , P ¯ D ) = ( 1 | λ B | 2 ) π e 1 2 ( 4 X ¯ C 2 + P ¯ D 2 ) n 1 , n 2 , n 2 m 1 , m 2 , | λ B | 2 ξ m 1 , m 2 n 1 , n 2 ( ξ m 1 , m 2 n 1 , n 2 ) 2 ! 2 n 2 + n 2 n 2 ! n 2 ! F , n 2 ( 2 X ¯ C i P ¯ D ) F , n 2 ( 2 X ¯ C + i P ¯ D ) .
| Φ i n = | 2 v | 1 h | ψ i n i n = | 2 v | 1 h n = 0 η n | n i n ,
| Φ i n = n = 0 η n 2 n ! ( a ^ v ) 2 a ^ h ( a ^ i n ) n | 0 v | 0 h | 0 i n .
a ^ i n = i r 2 a ^ 1 + t 2 a ^ 2 , a ^ m = t 2 a ^ 1 + i r 2 a ^ 2 a ^ h = i r 1 a ^ m + t 1 a ^ o u t , a ^ v = t 1 a ^ m + i r 1 a ^ o u t
a ^ i n = i r 2 a ^ 1 + t 2 a ^ 2 a ^ h = i r 1 t 2 a ^ 1 r 1 r 2 a ^ 2 + t 1 a ^ o u t a ^ v = t 1 t 2 a ^ 1 + i t 1 r 2 a ^ 2 + i r 1 a ^ o u t
| Φ o u t = n = 0 η n 2 n ! ( t 1 t 2 a ^ 1 + i t 1 r 2 a ^ 2 + i r 1 a ^ o u t ) 2 × ( i r 1 t 2 a ^ 1 r 1 r 2 a ^ 2 + t 1 a ^ o u t ) ( i r 2 a ^ 1 + t 2 a ^ 2 ) n | 0 1 | 0 2 | 0 o u t ,
| Φ o u t = n = 0 = 0 n η n n ! 2 ! ( n ) ! ( t 1 t 2 a ^ 1 + i t 1 r 2 a ^ 2 + i r 1 a ^ o u t ) 2 × ( i r 1 t 2 a ^ 1 r 1 r 2 a ^ 2 + t 1 a ^ o u t ) ( i r 2 a ^ 1 ) ( t 2 a ^ 2 ) n | { 0 } .
| Φ o u t ( D 1 = 1 , D 2 = 2 ) a ^ 1 ( a ^ 2 ) 2 2 ( 3 i 2 r 1 r 2 2 t 1 2 t 2 η 0 i 2 r 2 t 1 ( t 1 2 2 r 1 2 ) ( r 2 2 2 t 2 2 ) η 1 a ^ o u t i r 1 t 2 ( r 1 2 2 t 1 2 ) ( t 2 2 2 r 2 2 ) η 2 ( a ^ o u t ) 2 i 3 r 1 2 r 2 t 1 t 2 2 η 3 ( a ^ o u t ) 3 ) | { 0 }
| Φ o u t ( D 1 = 1 , D 2 = 2 ) = | 1 1 | 2 2 | ψ o u t o u t = 1 P s s | 1 1 | 2 2 n = 0 3 c s ( n ) η n | n o u t ,
P s s = n = 0 3 | c s ( n ) η n | 2
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