Secret sharing based on quantum Fourier transform

Abstract

Secret sharing plays a fundamental role in both secure multi-party computation and modern cryptography. We present a new quantum secret sharing scheme based on quantum Fourier transform. This scheme enjoys the property that each share of a secret is disguised with true randomness, rather than classical pseudorandomness. Moreover, under the only assumption that a top priority for all participants (secret sharers and recovers) is to obtain the right result, our scheme is able to achieve provable security against a computationally unbounded attacker.

Appendix: Proof of Equation (8)

The case that $\zeta =N$ if $K = 0$ mod $N$ is obvious, thus we only focus on the proof of $\zeta =0$ if $K \ne 0$ mod $N$. When $K \ne 0$ mod $N$, we know that

$$\begin{aligned} \zeta =\sum _{j=0}^{N-1} e^{\frac{2\pi ij}{N}K} \end{aligned}$$

(11)

Let $\frac{2\pi K}{N}=t$, then Eq. (11) can be rewritten as

$$\begin{aligned} \zeta =\sum _{j=0}^{N-1} e^{ijt}=\sum _{j=0}^{N-1} \cos {jt}+i\sum _{j=0}^{N-1} \sin {jt} \end{aligned}$$

(12)

It is not hard to verify and prove the following two Lagrange’s trigonometric identities [25]:

$$\begin{aligned} \sum _{n=0}^{N} \cos {n\theta }=-\frac{1}{2}+\frac{\sin {\left(N+\frac{1}{2}\right)\theta }}{2\sin {\frac{1}{2}\theta }} \end{aligned}$$

(13)

and

$$\begin{aligned} \sum _{n=0}^{N} \sin {n\theta }=\frac{\cos {\frac{1}{2}\theta }-\cos {\left(N+\frac{1}{2}\right)\theta }}{2\sin {\frac{1}{2}\theta }} \end{aligned}$$

(14)

Now we consider the first item of the right side of Eq. (12) using (13):

$$\begin{aligned} \sum _{j=0}^{N-1} \cos {jt}&= \sum _{j=1}^{N} \cos {jt}-\cos {Nt}+\cos {0}=\sum _{j=1}^{N} \cos {jt}\nonumber \\&= -\frac{1}{2}+\frac{\sin {\left(N+\frac{1}{2}\right)t}}{2\sin {\frac{1}{2}t}}\nonumber \\&= -\frac{1}{2}+\frac{\sin \left({2\pi K +\frac{\pi K}{N}}\right)}{2\sin {\frac{\pi K}{N}}}\nonumber \\&= -\frac{1}{2}+\frac{\sin {\frac{\pi K}{N}}}{2\sin {\frac{\pi K}{N}}}=0 \end{aligned}$$

(15)

The last step of (15) holds iff $K \ne 0$ mod $N$.

Similarly, we can get that $\sum _{j=0}^{N-1} \sin {jt}=0$ using (14) with $K \ne 0$ mod $N$.

So far, we can safely reach the conclusion that $\zeta =0$ iff $K \ne 0$ mod $N$ and thus Eq. (8) follows.