68
5 Quantum Code Constructions
other q-cosets are C 4 = {4, 52, 100}, C 5 = {5, 101, 53}, C 7 = {7, 55, 103}, C 8 =
{8, 104, 56}, C 10 = {10, 58, 106}, C 11 = {11, 107, 59}. Let C be the cyclic code generated by M
(3)
(x)M
(4)
(x)M
(5)
(x)M
(6)
(x)M
(7)
(x) M
(8)
(x)M
(9)
(x) · M
(10)
(x)
M
(11)
(x)M
(12)
(x). It is straightforward to show that C is Hermitian dual-containing
and has parameters [144, 122, d ≥ 11] 7 2 . Thus, applying the Hermitian construction, we obtain an [[144, 100, d ≥ 11]] 7 quantum code. Similarly one can construct quantum codes with parameters [[144, 102, d ≥ 10]] 7 , [[144, 108, d ≥ 9]] 7 ,
[[144, 114, d ≥ 8]] 7 , [[144, 116, d ≥ 7]] 7 , [[144, 122, d ≥ 6]] 7 , [[144, 128, d ≥ 5]] 7 ,
[[144, 130, d ≥ 4]] 7 and [[144, 136, d ≥ 3]] 7 .
Theorem 5.1.8 Suppose that q > 3 is a prime power and n > q
2 is an integer such
that gcd(q
2
, n) = 1. Assume also that (q
2
− 1) | n and m = ord n (q
2
) = 2 hold.
Then there exists a quantum code with parameters [[n, n − 4(r − 2) − 2, d ≥ r ]] q ,
where r satisfies n = r (q
2
− 1).
Proof Let C be the cyclic code generated by
M
(r )
(x)M
(r +1)
(x) · . . . · M
(2r −2)
(x).
We first show that C is Hermitian dual-containing. For this, let us consider the defining
set Z of C consisting of the q
2 -ary cosets given by C [r ] = {r }, C [r +1] = {r + 1, r +
q
2
}, C [r +2] = {r + 2, r + 2q
2
}, . . . , C [2r −2] = {2r − 2, r + (r − 2)q
2
}.
We know that gcd(q, n) = 1 holds. From Lemma 5.1.4, it suffices to show that
Z ∩ Z
−q
= ∅. Seeking a contradiction, we assume that Z ∩ Z
−q
= ∅. Then there
exist i, j, where 0 ≤ i, j ≤ r − 2, such that (r + j)q
l
≡ −q(r + i) mod n, where
l = 0 or l = 2. If l = 0, one has r + j ≡ −q(r + i) mod n, so q(r + i) + r + j ≡
0 mod n. Since both q(r + i) + r + j < n and q(r + i) + r + j = 0 are true, one
has a contradiction. If l = 2, it implies that (r + j)q
2
≡ −q(r + i) mod n and since
gcd(q
2
, n) = 1 and rq
2
≡ r mod n one obtains
(r + j)q
2
≡ −q(r + i) mod n
=⇒ r + jq
2
≡ −q(r + i) mod n
=⇒ (q + 1)r ≡ −q(i + jq) mod n
=⇒ −q(i + jq)(q − 1) ≡ 0 mod n
=⇒ n | q(i + jq)(q − 1)
=⇒ r (q + 1) | q(i + jq).
Since gcd(r, q) = 1 and gcd(q + 1, q) = 1 hold, it follows that r (q + 1) | (i + jq),
which is a contradiction because i + jq < r (q + 1). Thus C is Hermitian dualcontaining.
It is easy to see that these cosets are mutually disjoint. With exception of C [r ] ,
the other q-cosets have two elements. Thus, C has dimension k = n − 2(r − 2) − 1.
By construction, the defining set of C contains the sequence r, r + 1, . . . , 2r − 2, of
r − 1 consecutive integers and, so the minimum distance of C is greater than or equal
5 Quantum Code Constructions
other q-cosets are C 4 = {4, 52, 100}, C 5 = {5, 101, 53}, C 7 = {7, 55, 103}, C 8 =
{8, 104, 56}, C 10 = {10, 58, 106}, C 11 = {11, 107, 59}. Let C be the cyclic code generated by M
(3)
(x)M
(4)
(x)M
(5)
(x)M
(6)
(x)M
(7)
(x) M
(8)
(x)M
(9)
(x) · M
(10)
(x)
M
(11)
(x)M
(12)
(x). It is straightforward to show that C is Hermitian dual-containing
and has parameters [144, 122, d ≥ 11] 7 2 . Thus, applying the Hermitian construction, we obtain an [[144, 100, d ≥ 11]] 7 quantum code. Similarly one can construct quantum codes with parameters [[144, 102, d ≥ 10]] 7 , [[144, 108, d ≥ 9]] 7 ,
[[144, 114, d ≥ 8]] 7 , [[144, 116, d ≥ 7]] 7 , [[144, 122, d ≥ 6]] 7 , [[144, 128, d ≥ 5]] 7 ,
[[144, 130, d ≥ 4]] 7 and [[144, 136, d ≥ 3]] 7 .
Theorem 5.1.8 Suppose that q > 3 is a prime power and n > q
2 is an integer such
that gcd(q
2
, n) = 1. Assume also that (q
2
− 1) | n and m = ord n (q
2
) = 2 hold.
Then there exists a quantum code with parameters [[n, n − 4(r − 2) − 2, d ≥ r ]] q ,
where r satisfies n = r (q
2
− 1).
Proof Let C be the cyclic code generated by
M
(r )
(x)M
(r +1)
(x) · . . . · M
(2r −2)
(x).
We first show that C is Hermitian dual-containing. For this, let us consider the defining
set Z of C consisting of the q
2 -ary cosets given by C [r ] = {r }, C [r +1] = {r + 1, r +
q
2
}, C [r +2] = {r + 2, r + 2q
2
}, . . . , C [2r −2] = {2r − 2, r + (r − 2)q
2
}.
We know that gcd(q, n) = 1 holds. From Lemma 5.1.4, it suffices to show that
Z ∩ Z
−q
= ∅. Seeking a contradiction, we assume that Z ∩ Z
−q
= ∅. Then there
exist i, j, where 0 ≤ i, j ≤ r − 2, such that (r + j)q
l
≡ −q(r + i) mod n, where
l = 0 or l = 2. If l = 0, one has r + j ≡ −q(r + i) mod n, so q(r + i) + r + j ≡
0 mod n. Since both q(r + i) + r + j < n and q(r + i) + r + j = 0 are true, one
has a contradiction. If l = 2, it implies that (r + j)q
2
≡ −q(r + i) mod n and since
gcd(q
2
, n) = 1 and rq
2
≡ r mod n one obtains
(r + j)q
2
≡ −q(r + i) mod n
=⇒ r + jq
2
≡ −q(r + i) mod n
=⇒ (q + 1)r ≡ −q(i + jq) mod n
=⇒ −q(i + jq)(q − 1) ≡ 0 mod n
=⇒ n | q(i + jq)(q − 1)
=⇒ r (q + 1) | q(i + jq).
Since gcd(r, q) = 1 and gcd(q + 1, q) = 1 hold, it follows that r (q + 1) | (i + jq),
which is a contradiction because i + jq < r (q + 1). Thus C is Hermitian dualcontaining.
It is easy to see that these cosets are mutually disjoint. With exception of C [r ] ,
the other q-cosets have two elements. Thus, C has dimension k = n − 2(r − 2) − 1.
By construction, the defining set of C contains the sequence r, r + 1, . . . , 2r − 2, of
r − 1 consecutive integers and, so the minimum distance of C is greater than or equal