A novel dynamic quantum secret sharing in high-dimensional quantum system

Abstract

This paper proposed a novel dynamic quantum secret sharing protocol in high-dimensional quantum system. Via transmitting the particles circularly and local unitary operations, the dealer and all agents can share the multi-particle entangled GHZ state in high-dimensional quantum system. The proposed protocol allows a dealer to share the predetermined dits without directly distributing any piece of shares to agents. To recover the secrets, dealer and all agents only need to perform the single-particle measurement operation and then implement the simple modular arithmetic operation according to their corresponding measurement results. Besides, in the proposed protocol, only a little fraction of entangled GHZ states are employed for eavesdropping check instead of lots of decoy (or detecting) particles used in previous protocols. Since the protocol is designed in the high-dimensional quantum system, it makes our protocol have higher resource capacity and better security in detecting the illegal eavesdropper than former protocols designed in two-dimensional quantum system. Security analyses indicate that the proposed protocol is immune to general attacks of intercept-and-resend, entangle-and-measure, collusion, and revoked a dishonest agent. Furthermore, the efficiency of proposed DQSS protocol in noise environment is deduced through quantum fidelity. The obtained results indicate that the efficiency of proposed protocol decreases with the increase in channel noise parameter and it has a quite different efficiency in four types of noise, i.e., dit-flip, d-phase-flip, amplitude-damping and depolarizing.

Appendix A

Let us assume that Alice first prepares the generalized Bell state $\left| {\psi \left( {0,0} \right)} \right\rangle = \frac{1}{\sqrt d }\sum\limits_{m = 0}^{d - 1} {\left| {m,m} \right\rangle_{12} }$, and transmits the second particle among all agents (Bob₁, Bob₂, …, Bob_n) and herself. After each agent’s local unitary operation RS_c, it would form a (n + 2)-particle generalized GHZ state $\left| {\psi \left( {0,0,0, \cdots ,0} \right)} \right\rangle_{1,2,3, \cdots ,n + 2}$, and the whole operation can be expressed as:

$$ \left| {\psi \left( {0,0,0, \cdots ,0} \right)} \right\rangle_{1,2,3, \cdots ,n + 2} = \left[ {\prod\nolimits_{j = 3}^{n} {RS_{C} \left( {2,j} \right)} } \right]\left| {\psi \left( {0,0} \right)} \right\rangle_{12} \left| 0 \right\rangle^{ \otimes n} = \frac{1}{\sqrt d }\sum\limits_{m = 0}^{d - 1} {\left| {m,m,m, \cdots ,m} \right\rangle }_{1,2,3, \cdots ,n + 2} $$

(46)

where $RS_{C} \left( {2,j} \right)$ represents the implementation of local unitary operation RS_C on two qudits, and the second qudit and j-th qudit, respectively, act as the control qudit and target qudit.

When Alice performs local unitary operation LS_C on two qudits (i.e., the first qudit and second qudit), where the first qudit acts control qudit and second qudit acts the target qudit, the above (n + 2)-qudit GHZ state would transform into following forms:

$$ \left| {\psi^{ * } } \right\rangle = \frac{1}{\sqrt d }\sum\limits_{m = 0}^{d - 1} {\left| {m,0,m, \cdots ,m} \right\rangle_{1,2,3, \cdots ,n + 2} } $$

(47)

Now, it is easy to find that second qudit in above quantum state $\left| {\psi^{ * } } \right\rangle$ (i.e., particles in sequence S_T) being quantum state $\left| 0 \right\rangle$ has disentangled with other (n + 1) qudits. Thus, when Alice measures the second qudit in Z-basis, the remaining (n + 1) qudits transform into following GHZ state:

$$ \left| {\psi \left( {0,0, \cdots ,0} \right)} \right\rangle = \frac{1}{\sqrt d }\sum\limits_{m = 0}^{d - 1} {\left| {m,m, \cdots ,m} \right\rangle_{1,2,3, \cdots ,n + 1} } $$

(48)

Hence, when the generalized Hadamard transform H_d is performed on each qudit in state $\left| {\psi \left( {0,0, \cdots ,0} \right)} \right\rangle_{1,2,3, \cdots ,n + 1}$, Eq. (48) would transform into following form:

$$ \begin{gathered} \, \left| {\Psi^{ * } \left( {0,0, \cdots ,0} \right)} \right\rangle = H_{d}^{{ \otimes \left( {n{ + }1} \right)}} \left| {\psi \left( {0,0, \cdots ,0} \right)} \right\rangle \hfill \\ \quad= \frac{1}{\sqrt d }\sum\limits_{m = 0}^{d - 1} {\left( {\frac{1}{\sqrt d }\sum\limits_{{s_{H} = 0}}^{d - 1} {w^{{ms_{H} }} } \left| {s_{H} } \right\rangle \times \frac{1}{\sqrt d }\sum\limits_{{s_{1} = 0}}^{d - 1} {w^{{ms_{1} }} } \left| {s_{1} } \right\rangle \times } \right.} \left. { \cdots \times \frac{1}{\sqrt d }\sum\limits_{{s_{n} = 0}}^{d - 1} {w^{{ms_{n} }} \left| {s_{n} } \right\rangle } } \right) \hfill \\ \quad= \frac{1}{{\sqrt {d^{n + 2} } }}\sum\limits_{{s_{H} ,s_{1} , \cdots ,s_{n} = 0}}^{d - 1} {\sum\limits_{m = 0}^{d - 1} {w^{{m\left( {s_{H} + s_{1} + \cdots + s_{n} } \right)}} \left| {s_{H} } \right\rangle \left| {s_{1} } \right\rangle } \cdots \left| {s_{n} } \right\rangle } \hfill \\ \quad= \frac{1}{{\sqrt {d^{n} } }}\sum\limits_{{\begin{array}{*{20}c} {s_{H} ,s_{1} , \cdots ,s_{n} = 0} \\ {s_{H} + s_{1} + \cdots + s_{n} \equiv 0\left( {mod \, d} \right)} \\ \end{array} }}^{d - 1} {\left| {s_{H} } \right\rangle \left| {s_{1} } \right\rangle \cdots \left| {s_{n} } \right\rangle } \hfill \\ \end{gathered} $$

(49)

The last equation in Eq. (49) can be further proved as follows:

Let us denote that $S = \sum\nolimits_{m = 0}^{d - 1} {w^{km} }$ and accordingly denotes $k = s_{H} + s_{1} + \cdots + s_{n}$, where $w = e^{{{{2\pi i} \mathord{\left/ {\vphantom {{2\pi i} d}} \right. \kern-\nulldelimiterspace} d}}}$, then we can calculate the sum S as:

If $k \ne 0 \, \left( {mod \, d} \right)$, then $w = e^{{{{2\pi ik} \mathord{\left/ {\vphantom {{2\pi ik} d}} \right. \kern-\nulldelimiterspace} d}}} \ne 1$. Hence, the sum S is calculated as:

$$ S = \sum\limits_{m = 0}^{d - 1} {w^{mk} } = \sum\limits_{m = 0}^{d - 1} {e^{{\frac{2\pi ik}{d}m}} } = \frac{{1 * \left( {1 - e^{{\frac{2\pi ik}{d}d}} } \right)}}{{1 - e^{{\frac{2\pi ik}{d}}} }} = \frac{{1 * \left( {1 - e^{2\pi ik} } \right)}}{{1 - e^{{\frac{2\pi ik}{d}}} }} = 0 $$

(50)

In other case, if $k \equiv 0 \, \left( {mod \, d} \right)$, then we have $e^{{{{2\pi ikm} \mathord{\left/ {\vphantom {{2\pi ikm} d}} \right. \kern-\nulldelimiterspace} d}}} = e^{2\pi im} = 1$ for any $m \in \left\{ {0, \, d - 1} \right\}$. Thus, the sum S is calculated as:

$$ S = \sum\limits_{m = 0}^{d - 1} {w^{mk} } = \sum\limits_{m = 0}^{d - 1} {e^{{\frac{2\pi ik}{d}m}} } = \sum\limits_{m = 0}^{d - 1} {e^{2\pi im} } = d $$

(51)

In summary, we have:

$$ S = \sum\limits_{m = 0}^{d - 1} {w^{{m\left( {s_{H} + s_{1} + \cdots + s_{n} } \right)}} } = \left\{ \begin{gathered} 0,s_{H} + s_{1} + \cdots + s_{n} \ne 0\left( {\bmod \;d} \right) \hfill \\ d,s_{H} + s_{1} + \cdots + s_{n} \equiv 0\left( {\bmod \;d} \right) \hfill \\ \end{gathered} \right. $$

(52)

Hence, based on Eq. (52), we can verify the last equation of above-mentioned Eq. (49). And accordingly, the correctness of previous mentioned Eq. (11) is proved.