t b a b a

Let us take the C1 [7,4,3] Hamming code and its dual as C2. Thus we have C2 ⊂ C1. By
definition the parity check matrix of C2 is equal to the transposed generator matrix of C1:
⎡
⎤
H2 = G T 1 =
⎢ 1
⎢ 0
⎢ 0
⎢
⎣
0
1
0
0
0
1
0
0
0
0
1
1
1
0
1
1 ⎥
1
⎥
0
⎥
⎦
0 0 0 1 1 1 1
An element that obviously must belong to C1 is | 0000000〉. This element belongs to C2 too.
Therefore the first element of C1/C2, which we are going to identify with | 0L〉 is going to
be �
y∈C2 (| 0000000〉 + y). In order to find the second element of C1/C2, which we are going
to identify with | 1L〉 , we need to find a vector in C1 that does not belong to C2. Such a
vector is, for example, | 1111111〉.
Thus, we will give the codewords of the Steane code as:
| 0L〉 = 1 √ 8 [| 0000000〉+ | 1010101〉+ | 0110011〉+ | 1100110〉]
+ | 0001111〉+ | 1011010〉+ | 0111100〉+ | 1101001〉
| 1L〉 = 1 √ 8 [| 1111111〉+ | 0101010〉+ | 1001100〉+ | 0011001〉]
+ | 1110000〉+ | 0100101〉+ | 1000011〉+ | 0010110〉
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