96
5 Quantum Code Constructions
Proof Note first that the inequality n > 1 + (q
2
− 1)q
2 holds.
(a) This follows from the fact that (q
2
+ 1)q
2
≡ q
2
+ 1 mod n.
(b) We prove that each of the cosets C [q 2 +2] , C [2q 2 −1] has exactly two elements. To do
this, assume that q
2
+ j ≡ 1 + jq
2 mod n, where j = 2, . . . , q
2
+ 1. Because
1 + jq
2
< n, we have q
2
+ j = 1 + jq
2 ; hence, j − 1 = ( j − 1)q
2 , which is
a contradiction.
(c) It is clear that coset C [q 2 +1] is disjoint from the other cosets, since it has only one
element. Assume next that C [q 2 +i] = C [q 2 + j] , where 2 ≤ i, j ≤ q
2
− 1, where
i = j, Thus either q
2
+ i ≡ q
2
+ j mod n or q
2
+ i ≡ (q
2
+ j)q
2 mod n,
where 2 ≤ i, j ≤ q
2
− 1. Since 2q
2
+ 1 < q
4
− 1 and 1 + (q
2
− 1)q
2
< q
4
−
1 hold, such inequalities imply that q
2
+ i = q
2
+ j or q
2
+ i = 1 + jq
2 . The
first case implies i = j, a contradiction, and the second implies q
2
|(i − 1), which
is also a contradiction. Therefore, all these cosets are mutually disjoint. The proof
is complete.
In the sequence we use Lemma 5.3.4 to show how to construct quantum codes of
length n = q
4
− 1.
Theorem 5.3.1 Let q ≥ 3 be a prime power and n = q
4
− 1. Then there exists an
[[n, n − 4(q
2
− 2) − 2, d ≥ q
2
]] q quantum error-correcting code.
Proof Let us consider C as the cyclic code generated by the product of the minimal
polynomials
g(x) = M
(q
2 +1)
(x)M
(q
2 +2)
(x) · . . . · M
(q
2 + j)
(x),
where 1 ≤ j ≤ q
2
− 1. We show first that C is Hermitian dual-containing. Seeking
a contradiction, we suppose Z
Z
−q
= ∅. Thus there exist i, j, where 1 ≤ i, j ≤
q
2
− 1 such that C [q 2 + j] = C [−q(q 2 +i)] . Hence, q
2
+ j ≡ −q(q
2
+ i)q
2k , where k =
0 or k = 1. If k = 0, we have q
3
+ qi + q
2
+ j < q
4
− 1, so q
2
+ j = −q
3
− qi,
a contradiction. If k = 1, since gcd(q
2
, n) = 1 and q
4
≡ 1 mod n, we have
q
2
+ j ≡ −q
3
(q
2
+ i) mod n =⇒
q
5
+ q
3 i ≡ −(q
2
+ j) mod n =⇒
q + q
3 i ≡ −(q
2
+ j) mod n,
where 1 ≤ i, j, q
2
− 1.
If i < q then iq
3
+ q + q
2
+ j < q
4
− 1, so q + q
3 i = −(q
2
+ j), a contradiction. On the other hand, if i ≥ q, from the division algorithm we write i = lq + r ,
where r, l are integers such that 0 ≤ r ≤ q − 1. We also have 0 ≤ l ≤ q − 1; hence,
q + q
3 i = q + q
3
(lq + r ) ≡ q + l + q
3 r mod n.
Computing q + l + q
3 r + q
2
+ j we obtain
5 Quantum Code Constructions
Proof Note first that the inequality n > 1 + (q
2
− 1)q
2 holds.
(a) This follows from the fact that (q
2
+ 1)q
2
≡ q
2
+ 1 mod n.
(b) We prove that each of the cosets C [q 2 +2] , C [2q 2 −1] has exactly two elements. To do
this, assume that q
2
+ j ≡ 1 + jq
2 mod n, where j = 2, . . . , q
2
+ 1. Because
1 + jq
2
< n, we have q
2
+ j = 1 + jq
2 ; hence, j − 1 = ( j − 1)q
2 , which is
a contradiction.
(c) It is clear that coset C [q 2 +1] is disjoint from the other cosets, since it has only one
element. Assume next that C [q 2 +i] = C [q 2 + j] , where 2 ≤ i, j ≤ q
2
− 1, where
i = j, Thus either q
2
+ i ≡ q
2
+ j mod n or q
2
+ i ≡ (q
2
+ j)q
2 mod n,
where 2 ≤ i, j ≤ q
2
− 1. Since 2q
2
+ 1 < q
4
− 1 and 1 + (q
2
− 1)q
2
< q
4
−
1 hold, such inequalities imply that q
2
+ i = q
2
+ j or q
2
+ i = 1 + jq
2 . The
first case implies i = j, a contradiction, and the second implies q
2
|(i − 1), which
is also a contradiction. Therefore, all these cosets are mutually disjoint. The proof
is complete.
In the sequence we use Lemma 5.3.4 to show how to construct quantum codes of
length n = q
4
− 1.
Theorem 5.3.1 Let q ≥ 3 be a prime power and n = q
4
− 1. Then there exists an
[[n, n − 4(q
2
− 2) − 2, d ≥ q
2
]] q quantum error-correcting code.
Proof Let us consider C as the cyclic code generated by the product of the minimal
polynomials
g(x) = M
(q
2 +1)
(x)M
(q
2 +2)
(x) · . . . · M
(q
2 + j)
(x),
where 1 ≤ j ≤ q
2
− 1. We show first that C is Hermitian dual-containing. Seeking
a contradiction, we suppose Z
Z
−q
= ∅. Thus there exist i, j, where 1 ≤ i, j ≤
q
2
− 1 such that C [q 2 + j] = C [−q(q 2 +i)] . Hence, q
2
+ j ≡ −q(q
2
+ i)q
2k , where k =
0 or k = 1. If k = 0, we have q
3
+ qi + q
2
+ j < q
4
− 1, so q
2
+ j = −q
3
− qi,
a contradiction. If k = 1, since gcd(q
2
, n) = 1 and q
4
≡ 1 mod n, we have
q
2
+ j ≡ −q
3
(q
2
+ i) mod n =⇒
q
5
+ q
3 i ≡ −(q
2
+ j) mod n =⇒
q + q
3 i ≡ −(q
2
+ j) mod n,
where 1 ≤ i, j, q
2
− 1.
If i < q then iq
3
+ q + q
2
+ j < q
4
− 1, so q + q
3 i = −(q
2
+ j), a contradiction. On the other hand, if i ≥ q, from the division algorithm we write i = lq + r ,
where r, l are integers such that 0 ≤ r ≤ q − 1. We also have 0 ≤ l ≤ q − 1; hence,
q + q
3 i = q + q
3
(lq + r ) ≡ q + l + q
3 r mod n.
Computing q + l + q
3 r + q
2
+ j we obtain