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5 Quantum Code Constructions
which is a contradiction. Moreover, it is easy to see that the last equivalence does
not hold, a contradiction.
Therefore, C is Euclidean dual-containing code, as required. The proof is complete.
We next recall Corollary 5.1.2 shown in [62].
Corollary 5.3.3 Assume that we have an [N 0 , K 0 ] linear code L which contains its
Euclidean dual, L
⊥
≤ L, and which can be enlarged to an [N 0 , K
0 ] linear code L
,
where K
0 ≥ K 0 + 2. Then there exists a quantum code with parameters [[N 0 , K 0 +
K
0 − N 0 , d ≥ min{d,
q+1
q
d
}]], where d = w(L\L
⊥ ) and d
= w(L
\L
⊥ ).
We are now ready to state the main results of this subsection.
Theorem 5.3.3 Let q ≥ 4 be a prime power and n = q
m
− 1, where m = ord n (q) ≥
3. Then there exists an [[n, n − m(2c − 1) − 2, d ≥ c + 2]] q quantum code, for all
1 ≤ c ≤ q − 2.
Proof Let C be the cyclic code generated by
M
(s)
(x)M
(s+1)
(x) · . . . · M
(s+i)
(x)M
(s−1)
(x) · . . . · M
(s− j)
(x),
where 1 ≤ i + j = c ≤ q − 2. It is easy to see that C is an [n, n − mc − 1, d ≥
c + 2] q code, where 1 ≤ c ≤ q − 2. Moreover, from Lemma 5.1.3, C is Euclidean
dual-containing.
Let C
be the code generated by
M
(s)
(x)M
(s+1)
(x) · . . . · M
(s+i)
(x)M
(s−1)
(x) · . . . · M
(s− j+1)
(x).
We know that C
is an enlargement of C and has parameters [n, n − m(c − 1) −
1, d
≥ c + 1] q . Applying Corollary 5.1.2 to C and C
we obtain an [[n, n − m(2c −
1) − 2, d ≥ c + 2]] q code, as required. The proof is complete.
Theorem 5.3.4 Let q ≥ 4 be a prime power and n = q
m
− 1, where m = ord n (q) ≥
3. Then there exist quantum codes with parameters
• [[n, n − m(2q − 3) − 2, d ≥ q + 1]] q ;
• [[n, n − m(2q − 1) − 1, d ≥ q + 3]] q ;
• [[n, n − m(2c − 4) − 2, d ≥ c + 2]] q ;
• [[n, n − m(4q − 8) − 2, d ≥ 2q]] q ;
• [[n, n − m(4q − 5) − 2, d ≥ 2q + 2]] q , where q + 1 < c < 2q − 2.
Exercise 5.3.5 Show Theorem 5.3.4.