A verifiable framework of entanglement-free quantum secret sharing with information-theoretical security

Abstract

Quantum secret sharing (QSS) schemes without entanglement have huge advantages in scalability and are easier to realize as they only require sequential communications of a single quantum system. However, these schemes often come with drawbacks such as exact (n, n) structure, security flaws and absences of effective cheating detections. To address these problems, we propose a verifiable framework by utilizing entanglement-free states to construct (t, n)-QSS schemes. Our work is the heuristic step toward information-theoretical security in entanglement-free QSS, and it sheds light on how to establish effective verification mechanism against cheating. As a result, the proposed framework has a significant importance in constructing QSS schemes for versatile applications in quantum networks due to its intrinsic scalability, flexibility and information-theoretical security.

Appendices

Appendix A: List of abbreviations

In this section, we list the full descriptions for the most frequently used abbreviations in the main text.

See Table 2.

Table 2 List of useful abbreviations

Appendix B: Proof of Eq. (7)

In the paper, with $\omega ={\hbox {e}}^{2\pi i/d}$ we can first prove

$$\begin{aligned} \sum \limits _{j = 0}^{d - 1} {{\omega ^{rj}}} = \left\{ {\begin{array}{*{20}{c}} {d,\quad r = 0\bmod d}\\ {0,\quad r \ne 0\bmod d.} \end{array}} \right. \end{aligned}$$

Proof

We have $\omega ^ { d } = {\hbox {e}} ^ { 2 \pi i } = \cos ( 2 \pi ) + i \sin ( 2 \pi ) = 1$. At first, we suppose $r = 0\bmod d$; thus,

$$\begin{aligned} \sum \limits _{j = 0}^{d - 1} {{\omega ^{rj}}} = \sum \limits _{j = 0}^{d - 1} {{\omega ^{kdj}}} = \sum \limits _{j = 0}^{d - 1} 1 = d(k \in Z). \end{aligned}$$

If $r \ne 0\bmod d$, so ${\omega ^r} \ne 1$. Therefore, by using the sum of geometric series, we can get

$$\begin{aligned} \sum \limits _{j = 0}^{d - 1} {{\omega ^{rj}}} = 1 + {\omega ^r} + {\omega ^{2r}} + \cdots + {\omega ^{(d - 1)r}} = \frac{{1 - {\omega ^{dr}}}}{{1 - {\omega ^r}}} = \frac{{1 - 1}}{{1 - {\omega ^r}}} = 0.\end{aligned}$$

This completes the proof. $\square$

Next, we can define $\left| {{\mu _j}} \right\rangle = \mathrm{QFT}\left| j \right\rangle = \frac{1}{{\sqrt{d} }}\sum \nolimits _{k = 0}^{d - 1} {{\omega ^{jk}}} \left| k \right\rangle ,j=0,1,\ldots ,d-1$. Moreover, consider the generalized Pauli operators $X : = \sum _ { k = 0 } ^ { d - 1 } | k + 1 \rangle \langle k |$ and $Z:=\sum _ { k = 0 } ^ { d - 1 } \omega ^ { k } | k \rangle \langle k |$. After performing these two operators on the state $\left| {{\mu _j}} \right\rangle$, we have

$$\begin{aligned} X\left| {{\mu _j}} \right\rangle = {\omega ^{ - j}}\left| {{\mu _j}} \right\rangle ,Z\left| {{\mu _j}} \right\rangle = \left| {{\mu _{j + 1}}} \right\rangle . \end{aligned}$$

Here, we proof the transformation of the generalized Pauli operator X.

Proof

$$\begin{aligned} \begin{aligned} X\left| {{\mu _j}} \right\rangle&= \frac{1}{{\sqrt{d} }}\sum \limits _{k = 0}^{d - 1} {{\omega ^{jk}}} \left| {k + 1} \right\rangle = {\omega ^{ - j}}\frac{1}{{\sqrt{d} }}\sum \limits _{k = 0}^{d - 1} {{\omega ^{j(k + 1)}}} \left| {k + 1} \right\rangle \\&\mathop = \limits ^{k + 1 = r} {\omega ^{ - j}}\frac{1}{{\sqrt{d} }}\left( \sum \limits _{r = 1}^{d - 1} {{\omega ^{jr}}} \left| r \right\rangle + \left| 0 \right\rangle \right) = {\omega ^{ - j}}\sum \limits _{r = 0}^{d - 1} {{\omega ^{jr}}} \left| r \right\rangle = {\omega ^{ - j}}\left| {{\mu _j}} \right\rangle . \end{aligned} \end{aligned}$$

$\square$

As the definition in the paper, the generalized Pauli operation $U_{m,n}$ is

$$\begin{aligned} {U_{m,n}} = \sum \limits _{k = 0}^{d - 1} {{\omega ^{n \cdot k}}\left| {k + m} \right\rangle } \left\langle k \right| , \end{aligned}$$

where $m,n \in \mathrm{{GF}}(d)$. Moreover, it can be written as ${U_{m,n}}=X^mZ^n$. Because with ${X^m} = \sum \nolimits _{k = 0}^{d - 1} {\left| {k + m} \right\rangle } \left\langle k \right| ,{Z^n} = \sum \nolimits _{k = 0}^{d - 1} {{\omega ^{nk}}\left| k \right\rangle } \left\langle k \right|$, we have

$$\begin{aligned} \begin{aligned} {U_{m,n}}&= \sum \limits _{k = 0}^{d - 1} {{\omega ^{n \cdot k}}\left| {k + m} \right\rangle } \left\langle k \right| \\&=\left( {\sum \nolimits _{k = 0}^{d - 1} {\left| {k + m} \right\rangle } \left\langle k \right| } \right) \left( {\sum \nolimits _{k = 0}^{d - 1} {{\omega ^{nk}}\left| k \right\rangle } \left\langle k \right| } \right) = {X^m}{Z^n}. \end{aligned} \end{aligned}$$

So, Eq. (4) in the paper can be rewritten as

$$\begin{aligned} \begin{aligned} {U_{m,n}}\mathrm{{QFT}}\left| j \right\rangle&= {U_{m,n}}\frac{1}{{\sqrt{d} }}\sum \limits _{k = 0}^{d - 1} {{\omega ^{j \cdot k}}\left| k \right\rangle } \\&= {U_{m,n}}\left| {{\mu _j}} \right\rangle = X^mZ^n\left| {{\mu _j}} \right\rangle = \omega ^{-m(j+n)}\left| {{\mu _{j+n}}} \right\rangle . \end{aligned} \end{aligned}$$

Therefore, we finally give the proof of Eq. (7):

Proof

$$\begin{aligned} \begin{aligned} {\left| \varPsi \right\rangle _m}&=\left( {\prod \limits _{j = 1}^m {{U_{{p_j},{p_j} + {q_j}}}} } \right) {\left| \varPsi \right\rangle _0}=\prod \limits _{j = 1}^m {{X^{{p_j}}}{Z^{{p_j} + {q_j}}}} \left| {{\mu _{{p_0} + {q_0}}}} \right\rangle \\&=\omega ^{-p_1(p_0+q_0+p_1+q_1)}\prod \limits _{j = 2}^m {{X^{{p_j}}}{Z^{{p_j} + {q_j}}}} \left| {{\mu _{{p_0} + {q_0}+{p_1} + {q_1}}}} \right\rangle \\&=\xi _m\left| {{\mu _{\sum \nolimits _{j = 0}^m {({p_j} + {q_j})} }}} \right\rangle \\&=\frac{\xi _m}{{\sqrt{d} }}\sum \limits _{k = 0}^{d - 1} {{\omega ^{\left( {\sum \nolimits _{j = 0}^m {({p_j} + {q_j})} } \right) \cdot k}}} \left| k \right\rangle \\&=\frac{\xi _m}{{\sqrt{d} }}\sum \limits _{k = 0}^{d - 1} {{\omega ^{(\sum \nolimits _{j = 0}^m {{p_j}} \mathrm{{ + }}d - s + \sum \nolimits _{j = 1}^m {{c_j}} ) \cdot k}}\left| k \right\rangle } \\&=\frac{\xi _m}{{\sqrt{d} }}\sum \limits _{k = 0}^{d - 1} {{\omega ^{(\sum \nolimits _{j = 0}^m {{p_j}}+L\cdot d ) \cdot k}}\left| k \right\rangle }, (L\in Z) \end{aligned} \end{aligned}$$

with the overall phase term $\xi _m={\omega ^{ - \sum \nolimits _{a = 1}^m {{p_a}} \left( {\sum \nolimits _{b = 0}^{a} {({p_b} + {q_b})} } \right) }}$. $\square$