Cyclic codes and some new entanglement-assisted quantum MDS codes

Abstract

Entanglement-assisted quantum error correcting codes (EAQECCs) play a significant role in protecting quantum information from decoherence and quantum noise. In this work, we construct six families of new EAQECCs of lengths $n=(q^2+1)/a$, $n=q^2+1$ and $n=(q^2+1)/2$ from cyclic codes, where $a=m^2+1$ ($m\ge 1$ is odd) and q is an odd prime power with the form of $a|(q+m)$ or $a|(q-m)$. Moreover, those EAQECCs are entanglement-assisted quantum maximum distance separable (EAQMDS) codes when $d\le (n+2)/2$. In particular, the length of EAQECCs we studied is more general and the method of selecting defining set is different from others. Compared with all the previously known results, the EAQECCs in this work have flexible parameters and larger minimum distance. All of these EAQECCs are new in the sense that their parameters are not covered by the quantum codes available in the literature.

A Appendix: The proof of Lemma 3.3

Proof

For a positive integer $\alpha$ with $1 \le \alpha \le k$, let

$$\begin{aligned}&T_{1}= \bigcup _{\begin{array}{c} s+(m+t)k+h+\alpha \le v \le s+ (m+t+2)k+(h-1)-\alpha ,\\ if~ v\le s+mk,~0 \le u\le \alpha ,~else~ 0 \le u\le \alpha -1\\ -m\le t\le (2m-1)m~and~t~is~odd \end{array}}C_{uq+v} \\&when~ -m\le t \le -1,~h=1. \\&when~ 1\le t\le 2m-1,~h=2. \\&when~ 2m+1 \le t \le 4m-1,~h=3. \\&\dots \dots \\&when~ (2m-2)m+1 \le t \le (2m-1)m,~h=m+1. \end{aligned}$$

Then by Lemma 3.2, we have

$$\begin{aligned}&-qT_{1}= \bigcup _{\begin{array}{c} s+(m+t)k+h+\alpha \le v \le s+ (m+t+2)k+(h-1)-\alpha ,\\ if~ v\le s+mk,~0 \le u\le \alpha ,~else~ 0 \le u\le \alpha -1\\ -m\le t\le (2m-1)m~and~t~is~odd \end{array}}C_{vq-u} \\&when~ -m\le t \le -1,~h=1. \\&when~ 1\le t\le 2m-1,~h=2. \\&when~ 2m+1 \le t \le 4m-1,~h=3. \\&\dots \dots \\&when~ (2m-2)m+1 \le t \le (2m-1)m,~h=m+1. \end{aligned}$$

When $-m\le t_1 \le -1,~h_1=1$, then $s+1+\alpha \le v_1\le s+mk$ and $0\le u_1\le \alpha$, it follows that

$$\begin{aligned} u_1q+v_1\le \alpha q+s+mk,~(s+1+\alpha )q-\alpha \le v_1q-u_1. \end{aligned}$$

When $-m\le t_2 \le -1,~h_2=1$, then $s+mk+1 \le v_2\le s+(m+1)k-\alpha$ and $0\le u_2\le \alpha -1$, it follows that

$$\begin{aligned} u_2q+v_2\le (\alpha -1)q+s+(m+1)k-\alpha ,~(s+mk+1)q-\alpha +1 \le v_2q-u_2. \end{aligned}$$

When $1\le t_3 \le 2m-1,~h_3=2$, then $s+(m+1)k+2+\alpha \le v_3\le s+(3m+1)k-\alpha +1$ and $0\le u_3\le \alpha -1$, it follows that

$$\begin{aligned}&u_3q+v_3\le (\alpha -1)q+s+(3m+1)k-\alpha +1,~[s+(m+1)k+2+\alpha ]q\\&-\alpha +1 \le v_3q-u_3. \end{aligned}$$

When $2m+1\le t_4 \le 4m-1,~h_4=3$, then $s+(3m+1)k+3+\alpha \le v_4\le s+(5m+1)k+2-\alpha$ and $0\le u_4\le \alpha -1$, it follows that

$$\begin{aligned}&u_4q+v_4\le (\alpha -1)q+s+(5m+1)k+2-\alpha ,~[s+(3m+1)k+3+\alpha ]q\\&-\alpha +1 \le v_4q-u_4. \dots \dots \end{aligned}$$

When $2(m-1)m+1\le t_{m+2} \le (2m-1)m,~h_{m+2}=m+1$, then $s+[(2m-1)m+1]k+m+1+\alpha \le v_{m+2}\le s+2(m^2+1)k+m-\alpha$ and $0\le u_{m+2}\le \alpha -1$, it follows that

$$\begin{aligned}&u_{m+2}q+v_{m+2}\le (\alpha -1)q+s+2(m^2+1)k+m-\alpha , \\&[s+(2m^2-m+1)k+m+1+\alpha ]q-\alpha +1 \le v_{m+2}q-u_{m+2}. \end{aligned}$$

It is easy to check that

$u_1q+v_1<v_1q-u_1,~u_1q+v_1<v_2q-u_2,~\ldots ~,~u_1q+v_1<v_{m+2}q+u_{m+2}.$ $u_2q+v_2<v_1q-u_1,~u_2q+v_2<v_2q-u_2,~\ldots ~,~u_2q+v_2<v_{m+2}q+u_{m+2}.$

$$\begin{aligned} \dots \dots \end{aligned}$$

$u_{m+2}q+v_{m+2}<v_1q-u_1,~\ldots ~,~u_{m+2}q+v_{m+2}<v_{m+2}q+u_{m+2}.$

For the range of $~v_1,~v_2,~\ldots ~,~v_{m+2}$ and $~u_1,~u_2,~\ldots ~,~u_{m+2}$, note that $u_iq+v_i\le (m+q)k~(i=1,2,\ldots ,m+2)$, the subscripts of $C_{u_i+v_i}$ is the biggest number in the set. Then $T_1\cap -qT_1=\emptyset$. The desired results follows. $\square$