100
5 Quantum Code Constructions
s + j + q(s + iq
2h
)
≤
q
2m
− 1
q 2 − 1
+ j + q
q
2m
− 1
q 2 − 1
+ iq
2m−3
≤
q
2m
− 1
q − 1
+ (q
2
− 1)(1 + q
2m−3
).
It is easy to see that
q
2m
− 1
q − 1
+ (q
2
− 1)(1 + q
2m−3
) < q
2m
− 1.
Since s + j = −qs − iq
2h+1 does not hold, we have a contradiction.
If h = m − 1, we will verify the equivalence s + j ≡ −q(s + i)q
2m−2 mod n:
s + j ≡ −q(s + i)q
2m−2 mod n
=⇒ j (q
2
− 1) ≡ −iq
2m−1
(q
2
− 1) mod n
=⇒ ( j + iq
2m−1
)(q
2
− 1) ≡ 0 mod n.
Applying the algorithm of division for i and q we have i = aq + r , where 0 ≤ r < q.
Because 1 ≤ i ≤ q
2
− 1 we also have 0 ≤ a < q; hence,
( j + iq
2m−1
)(q
2
− 1)
≡ [j + (aq + r )q
2m−1
](q
2
− 1)
≡ ( j + a)(q
2
− 1) + r (q
2
− 1)q
2m−1
≡ ( j + a)(q
2
− 1) + rq − rq
2m−1
≡ 0 mod n
=⇒ rq
2m−1
− rq − ( j + a)(q
2
− 1) ≡ 0 mod n.
If r = 0, it follows that ( j + a)(q
2
− 1) < q
2m
− 1, so ( j + a)(q
2
− 1) ≡ 0 mod n.
If r > 0, then 0 < rq
2m−1
− rq − ( j + a)(q
2
− 1) < q
2m
− 1, which is a contradiction.
The cases C [s+ j] = C [−q(s−i)] , C [s− j] = C [−q(s+i)] and C [s− j] = C [−q(s−i)] are
analogous to the previous one, so the proof is omitted. Therefore, C is Hermitian
dual-containing, as required.
Theorem 5.3.2 given in the following is the main result of this subsection.
Theorem 5.3.2 Let q ≥ 4 be a prime power and n = q
2m
− 1, where m = ord n (q
2
) ≥
3. Then there exists an [[n, n − 4m(q
2
− 1) − 2, d ≥ q
2
+ 2]] q quantum errorcorrecting code.
Proof Let C be the cyclic code generated by
M
(s)
(x)M
(s+1)
(x) · . . . · M
(s+q
2 −1)
(x)M
(s−1)
(x) · . . . · M
(s−q
2 +1)
(x).