# New entanglement-assisted quantum MDS codes with length n=q^2+1/5

The entanglement-assisted stabilizer formalism can transform arbitrary classical linear codes into entanglement-assisted quantum error correcting codes (EAQECCs). In this work, we construct some new entanglement-assisted quantum MDS (EAQMDS) codes with length n=q^2+1/5 from cyclic codes. Compared with all the previously known parameters with the same length, all of them have flexible parameters and larger minimum distance.

## Authors

• 10 publications
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03/12/2018

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## 1 Introduction

In recent years, quantum computation has become a hot research issue. As in the case of classical digital communications, quantum error-correcting codes (QECCs) play an important role in quantum information processing and quantum computation [1]-[2]. There are many good works about QECCs [3]-[9].

In kinds of construction methods of QECCs, the CSS construction is most frequently used. It establishes a relationship between classical error correcting codes and QECCs, but it needs classical codes to be dual-containing or self-orthogonal which is not easy to satisfy all the time. This problem is solved after Brun et al. proposed EAQECCs which allows non-dual-containing classical codes to construct QECCs if the sender and receiver shared entanglement bits in advance [10]. This has inspired more and more scholars to focus on constructing EAQECCs [11]-[23].

As we all know, many EAQMDS codes with a small number of entangled states have been constructed. In [24], Lu et al. constructed new EAQMDS codes of length with larger minimum distance and consumed four entanglement bits. Besides, Chen et al. constructed EAQMDS codes by using constacyclic codes with length and consumed four entanglement bits in [25]. Actually, the larger the minimum distance of EAQECCs are, the more the entanglement bits will be employed. However, it is not an easy task to analyze the accurate parameters if the value of is too large or flexible. Recently, some scholars have obtained great progress. In [26], Qian and Zhang constructed some new EAQMDS codes with length and some new entanglement-assisted quantum almost MDS codes with length . Besides, Wang et al. obtained series of EAQECCs with flexible parameters of length by using -cyclotomic coset modulo [27]. And almost all of those known results about EAQECCs with the same length are some special cases of their.

Inspired by the above work, we consider to use cyclic codes to construct EAQECCs of length , with flexible parameters naturally. Based on cyclic codes, we construct four classes of EAQMDS codes with the following parameters:

(1) where

is an odd prime power and

.

(2) where is an odd prime power and .

(3) , where , is a positive integer and .

(4) , where , is a positive integer and .

The main organization of this paper is as follows. In Sect.2, some basic background and results about cyclic codes and EAQECCs are reviewed. In Sect.3, we construct four classes of optimal EAQMDS codes with length . Sect.4 concludes the paper.

## 2 Preliminaries

In this section, we will review some relevant concepts on cyclic codes and EAQECCs. For further and detailed information on cyclic codes can be found in [6, 28], EAQECCs please see [4, 10, 16, 26, 27, 29].

### 2.1 Review of Cyclic Codes

For given a positive integer and prime number , let and be the finite field of elements. A -dimensional subspace of the

-dimensional vector space

is a linear code of length over and this linear code is denoted by . If an linear code can detect errors but not errors, it is an linear code . For any , the conjugation of is denoted by and the conjugation transpose of an matrix entries in is an matrix .

The Hermitian inner product of the vectors and in is

 ⟨μ,υ⟩h=n−1∑i=0¯¯¯¯¯uivi=uq0v0+uq1v1+⋯+uqn−1vn−1.

The Hermitian dual code of is defined as

 C⊥h={μ∈Fnq2 |⟨μ,υ⟩h=0 for all υ∈C}.

If , then is called a Hermitian self-orthogonal code. If , then is called a Hermitian dual-containing code.

A linear code of length over is said to be cyclic if for any codeword implies its cyclic shift . For a cyclic code , each codeword is customarily represented in its polynomial form: , and the is in turn identified with the set of all polynomial representations of its codewords. Then, a -ary cyclic code of length is an ideal of and can be generated by a monic polynomial factors of , i.e. and . The is called the generator polynomial of .

Note that has no repeated root over if and only if gcd. Let denote a primitive -th root of unity in some extension field of . Hence, . For , the -cyclotomic coset modulo containing is defined by the set

 Ci={i,iq2,iq4,...,iq2(mi−1)},

where is the smallest positive integer such that . Each corresponds to an irreducible divisor of over . The defining set of a cyclic code of length is the set .

Let be an cyclic code over with defining set . Obviously must be a union of some -cyclotomic coset modulo and . It is clear to see that the Hermitian dual code has a defining set . Note that . Then, contains its Hermitian dual code if and only if from Lemma 2.2 in [6]. The following is a lower bound for cyclic codes:

###### Proposition

[28]( The BCH Bound for Cyclic Codes) Let be a -ary cyclic code of length with defining set . If contains consecutive elements, then the minimum distance of is at least .

### 2.2 Review of EAQECCs

Let be a nonnegative integer. Via pairs of maximally entanglement states, an EAQECC encodes

information qubits into

qubits, and it can correct up to errors which act on qubits, where is called minimum distance of the EAQECC.

A parity check of an linear code with respect to Hermitian inner product is an matrix whose rows constitute a basis for , then has an parity check matrix .

Brun et al. established a bound on the parameters of an EAQECC.

###### Proposition

[10, 16] Assume that is an entanglement-assisted quantum code with parameters . If , then satisfies the entanglement-assisted Singleton bound . If satisfies the equality for , then it is called an entanglement-assisted quantum MDS code.

According to literature [4, 10, 29], EAQECCs can be constructed from arbitrary linear codes over , which is given by the following proposition.

###### Proposition

If is an classical code and is its parity check matrix over , then there exist entanglement-assisted quantum codes with parameters , where .

Although it is possible to construct an EAQECCs from any classical linear code over , it is not easy to calculate the parameter of ebits . However, can be easily determined for some special classes of linear codes. In [4] defining the decomposition of the defining set of cyclic codes was initially introduced.

###### Definition

[4] Let be a -ary cyclic code of length with defining set . Assume that and , where . Then, is called a decomposition of the defining set of .

###### Lemma

Let be a cyclic code with length n over , where . Suppose that is the defining set of the cyclic code and is a decomposition of . Then, the number of entangled states required is .

## 3 New EAQMDS codes of length q2+15

In this section, we use cyclic codes of length to construct some new EAQMDS codes. Let , where is an odd prime power, it is clear that that the -cyclotomic cosets modulo are

 C0={0},C1={1,−1}={1,n−1},C2={2,−2}={2,n−2},... ,Cq2+110={q2+110}.

Now, we first give a useful lemma that will be used in later constructions.

###### Lemma

Let and be an odd prime power.
1) When , then , where and  , if , or , if .

2) When , then , where , and if .

###### Proof

1) Note that for and if , or if .

Since . This gives that .

2) The proof is similar to case 1), so it is omitted here.

From Lemma 3, we can also obtain , where the range of , and is listed below:

1) When , we have and if , or if .

2) When , we have , and if  .

Based on the discussions above, we can give the first construction as follows.

Case I

In order to determine the number of entangled states , we give the following Lemma for preparation.

###### Lemma

Let and be an odd prime power. For a positive integer , let

 Z1=⋃m≤i1≤q+25−m,0≤s≤m−2Csq+i1⋃q−35+m≤i2≤2q+45−m,0≤s≤m−2Csq+i2⋃2q−15+m≤i3≤3q+15−m,0≤s≤m−2Csq+i3
 ⋃q−35+m≤j1≤2q+45−m,1≤t≤m−1Ctq−j1⋃m−1≤j2≤q+25−m,1≤t≤m−1Ctq−j2.

Then .

###### Proof

For a positive integer with , let

 Z1=⋃m≤i1≤q+25−m,0≤s≤m−2Csq+i1⋃q−35+m≤i2≤2q+45−m,0≤s≤m−2Csq+i2⋃2q−15+m≤i3≤3q+15−m,0≤s≤m−2Csq+i3
 ⋃q−35+m≤j1≤2q+45−m,1≤t≤m−1Ctq−j1⋃m−1≤j2≤q+25−m,1≤t≤m−1Ctq−j2.

Then by Lemma 3, we have

 −qZ1=⋃m≤i1≤q+25−m,0≤s≤m−2Ci1q−s⋃q−35+m≤i2≤2q+45−m,0≤s≤m−2Ci2q−s⋃2q−15+m≤i3≤3q+15−m,0≤s≤m−2Ci3q−s
 ⋃q−35+m≤j1≤2q+45−m,1≤t≤m−1Cj1q+t⋃m−1≤j2≤q+25−m,1≤t≤m−1Cj2q+t.

When it follows that

 sq+i1≤(m−2)q+q+25−m, mq−m+2≤i1q−s.

When it follows that

 sq+i2≤(m−2)q+2q+45−m, (q−35+m)q−m+2≤i2q−s.

When it follows that

 sq+i3≤(m−2)q+3q+15−m, (2q−15+m)q−m+2≤i3q−s.

When it follows that

 tq−j1≤(m−1)q−q−35−m, (q−35+m)q+1≤j1q+t.

When it follows that

 tq−j2≤(m−1)q−m+1, (m−1)q+1≤j2q+t.

It is easy to check that

 sq+i1
 sq+i2
 sq+i3
 tq−j1
 tq−j2

For the range of and , note that , the subscripts of is the smallest number in the set. Then . The desired results follows.

Example 3.1  Let and Then, According to Lemma 3, we can obtain

 Z1=⋃2≤i1≤3,s=0Csq+i1⋃6≤i2≤8,s=0Csq+i2⋃11≤i3≤12,s=0Csq+i3⋃6≤j1≤8,t=1Ctq−j1⋃1≤j2≤3,t=1Ctq−j2=C2⋃C3⋃C6⋃C7⋃C8⋃C11⋃C12⋃C15⋃C16⋃C17⋃C20⋃C21   ⋃C22.

It is easy to check that .

Example 3.2  Let and Then, respectively.
1) Let according to Lemma 3, we can obtain

 Z1=⋃2≤i1≤7,s=0Csq+i1⋃10≤i2≤16,s=0Csq+i2⋃19≤i3≤24,s=0Csq+i3⋃10≤j1≤16,t=1Ctq−j1⋃1≤j2≤7,t=1Ctq−j2=C2⋃C3⋃…⋃C7⋃C10⋃…⋃C16⋃C19⋃…⋃C24⋃C27⋃…   ⋃C33⋃C36⋃…⋃C42.

It is easy to check that
2) Let according to Lemma 3, we can obtain

 Z1=⋃3≤i1≤6,0≤s≤1Csq+i1⋃11≤i2≤15,0≤s≤1Csq+i2⋃20≤i3≤23,0≤s≤1Csq+i3⋃2≤j1≤6,1≤t≤2Ctq−j1⋃11≤j2≤15,1≤t≤2Ctq−j2.

It is easy to check that .
3) Let , according to Lemma 3, we can obtain

 Z1=⋃4≤i1≤5,0≤s≤2Csq+i1⋃12≤i2≤14,0≤s≤2Csq+i2⋃21≤i3≤22,0≤s≤2Csq+i3⋃3≤j1≤5,1≤t≤3Ctq−j1⋃12≤j2≤14,1≤t≤3Ctq−j2.

It is easy to check that .

Based on Lemma 3, we can obtain the number of entangled states in the following theorem.

###### Theorem

Let and be an odd prime power. For a positive integer m with , let be a cyclic code with defining set given as follows

 Z=C0⋃C1⋃…⋃C(m−1)q.

Then .

###### Proof

Let

 Z1=⋃m≤i1≤q+25−m,0≤s≤m−2Csq+i1⋃q−35+m≤i2≤2q+45−m,0≤s≤m−2Csq+i2⋃2q−15+m≤i3≤3q+15−m,0≤s≤m−2Csq+i3
 ⋃q−35+m≤j1≤2q+45−m,1≤t≤m−1Ctq−j1⋃m−1≤j2≤q+25−m,1≤t≤m−1Ctq−j2.

and

 Z′1=⋃0≤i1≤m−1,0≤s≤m−2Csq+i1⋃q+25−m+1≤i2≤q−35+m−1,0≤s≤m−2Csq+i2⋃2q+45−m+1≤i3≤2q−15+m−1,0≤s≤m−2Csq+i3
 ⋃0≤j1≤m−2,1≤t≤m−1Ctq−j1⋃q+25−m+1≤j2≤q−35+m−1,1≤t≤m−1Ctq−j2⋃2q+45−m+1≤j3≤2q−15+m−1,1≤t≤m−1Ctq−j3.

From Lemma 3, we have

 −qZ′1=⋃0≤i1≤m−1,0≤s≤m−2Ci1q−s⋃q+25−m+1≤i2≤q−35+m−1,0≤s≤m−2Ci2q−s⋃2q+45−m+1≤i3≤2q−15+m−1,0≤s≤m−2Ci3q−s
 ⋃0≤j1≤m−2,1≤t≤m−1Cj1q+t⋃q+25−m+1≤j2≤q−35+m−1,1≤t≤m−1Cj2q+t⋃2q+45−m+1≤j3≤2q−15+m−1,1≤t≤m−1Cj3q+t.

It is easy to check that . From the definitions of , and , we have . Then from the definition of ,

 Z2=Z⋂(−qZ)=(Z1⋃Z′1)⋂(−qZ1⋃−qZ′1)=(Z1⋂−qZ1)⋃(Z1⋂−qZ′1)⋃(Z′1⋂−qZ1)⋃(Z′1⋂−qZ′1)=Z′1.

Therefore, .