88
5 Quantum Code Constructions
Quantum codes with parameters [[15, 9, d ≥ 3]] 4 , [[15, 5, d ≥ 4]] 4 , [[24, 18, d
≥ 3]] 5 and [[24, 14, d ≥ 4]] 5 can be constructed by applying Construction I.
5.2.2.3 Construction III
We start by showing Theorem 5.2.9, a particular case of Theorem 5.2.10.
Theorem 5.2.9 Let q ≥ 5 be an odd prime power. Then there exists an [[q
3
− 1, q
3
−21, d ≥ 5]] q quantum code.
Proof It is easy to check that C (
q 3 −1
2 )
contains only one element. Moreover, the cosets
C 1 , C 2 and C 3 are disjoint and each of them has three elements.
We next show that the q-cosets C (
q 3 −1
2 )
, C (
q 3 −1
2 +1)
, C (
q 3 −1
2 +2)
and C (
q 3 −1
2 +3)
are
distinct of the cosets C 1 , C 2 and C 3 . Clearly the coset C (
q 3 −1
2 )
is distinct of such cosets.
Consider the coset C (
q 3 −1
2 +1)
and suppose w.l.o.g. that
q
3 −1
2
+ 1 is even. Then, from
Corollary 5.2.2, C (
q 3 −1
2 +1)
is disjoint from C 1 and C 3 . If (
q
3 −1
2
+ 1)q
i
≡ 2 mod n,
then it follows that
[(q
3
− 1)q
i
+ 2q
i
] ≡ 4 mod n
=⇒ [(q
3
− 1)q
i
+ 2q
i
] ≡ 2q
i
≡ 4 mod n.
Since 2q
i
− 4 < q
3
− 1 and because the equality 2q
i
= 4 is not satisfied, the coset
C (
q 3 −1
2 +1)
is disjoint of C 2 . The other cases are similar to the previous one.
In what follows, we show that each of cosets C (
q 3 −1
2 +1)
, C (
q 3 −1
2 +2)
and C (
q 3 −1
2 +3)
are disjoint among them. From Corollary 5.2.2,
C (
q 3 −1
2 +1)
= C (
q 3 −1
2 +2)
and
C (
q 3 −1
2 +2)
= C (
q 3 −1
2 +3)
.
To show that C (
q 3 −1
2 +1)
is disjoint from C (
q 3 −1
2 +3)
, let us consider the following congruence:
q
3
− 1
2
+ 1
≡
q
3
− 1
2
+ 3
q
i
mod n.
We then obtain
(q
3
− 1 + 2) ≡ [(q
3
− 1)q
i
+ 6q
i
] mod n
=⇒ 6q
i
≡ 2 mod n.