3.6 Calderbank–Shor–Steane Construction
39
where + denotes the componentwise addition modulo 2.
Exercise 3.6.1 Show that the relation defined above is an equivalence relation whose
equivalence class of a codeword c ∈ C 1 is the coset c + C 2
Suppose c and c
belong to the disjoint cosets of C 2 ; this implies that there is no
x, x
∈ C 2 such that c + x = c
+ x
, otherwise,
c − c
= x
− x ∈ C 2 =⇒ c + C 2 = c
+ C 2 ,
which is a contradiction. Hence, for distinct c = c
, the corresponding quantum
states |c + C 2 and |c
+ C 2 are orthonormal. Thus, we define the quantum code
CSS(C 1 , C 2 ) to be the vector space spanned by the states |c + C 2 for all c ∈ C 1 .
Since the number of cosets is
|C 1 |
|C 2 |
, the dimension of CSS(C 1 , C 2 ) equals
|C 1 |
|C 2 |
= 2
k 1 −k 2 .
Therefore, the quantum code CSS(C 1 , C 2 ) has parameters [n, k 1 − k 2 ], capable to
correct errors on t qubits.
Let us see how the code works in the error correction. Assume that the initial state
of a quantum system is |c + C 2 . After passing through the channel, the original
state can suffer some kind of error (bit flip and/or phase flip). In the error model it is
assumed that bit flip errors e b are binary vectors of length n (the code length) such
that the component is 1 where the bit flip occurs and 0 otherwise. The phase flip
errors are also binary vectors e p of length n with 1 in the coordinate that a phase flip
occurs and 0 otherwise. Note that both binary vectors cannot have more than t ones.
Adopting this model, we know that the corrupted state is
|c + C 2
channel
− −−− →
1
√
|C 2 |
x∈C 2
(−1)
(c+x)·e p |c + x + e b .
Bit flip Detection. We introduce a sufficient large ancilla system capable of storing
the syndrome for C 1 , which is initially in the all zero state |0. Applying the parity
check matrix H 1 of C 1 to all state |c + x + e b , since H 1 (c + x) = 0 we have
|c + x + e b |0 −→ |c + x + e b |H 1 (c + x + e b ) = |c + x + e b |H 1 e b ,
that is, the error was isolated. Thus, we obtain the state
1
√ |C 2 |
x∈C 2
(−1)
(c+x)·e p |c + x + e b |H 1 e b .
Performing the measurement of the ancilla, we obtain H 1 e b ; discarding the ancilla
we return to the quantum state
1
√ |C 2 |
x∈C 2
(−1)
(c+x)·e p |c + x + e b .
39
where + denotes the componentwise addition modulo 2.
Exercise 3.6.1 Show that the relation defined above is an equivalence relation whose
equivalence class of a codeword c ∈ C 1 is the coset c + C 2
Suppose c and c
belong to the disjoint cosets of C 2 ; this implies that there is no
x, x
∈ C 2 such that c + x = c
+ x
, otherwise,
c − c
= x
− x ∈ C 2 =⇒ c + C 2 = c
+ C 2 ,
which is a contradiction. Hence, for distinct c = c
, the corresponding quantum
states |c + C 2 and |c
+ C 2 are orthonormal. Thus, we define the quantum code
CSS(C 1 , C 2 ) to be the vector space spanned by the states |c + C 2 for all c ∈ C 1 .
Since the number of cosets is
|C 1 |
|C 2 |
, the dimension of CSS(C 1 , C 2 ) equals
|C 1 |
|C 2 |
= 2
k 1 −k 2 .
Therefore, the quantum code CSS(C 1 , C 2 ) has parameters [n, k 1 − k 2 ], capable to
correct errors on t qubits.
Let us see how the code works in the error correction. Assume that the initial state
of a quantum system is |c + C 2 . After passing through the channel, the original
state can suffer some kind of error (bit flip and/or phase flip). In the error model it is
assumed that bit flip errors e b are binary vectors of length n (the code length) such
that the component is 1 where the bit flip occurs and 0 otherwise. The phase flip
errors are also binary vectors e p of length n with 1 in the coordinate that a phase flip
occurs and 0 otherwise. Note that both binary vectors cannot have more than t ones.
Adopting this model, we know that the corrupted state is
|c + C 2
channel
− −−− →
1
√
|C 2 |
x∈C 2
(−1)
(c+x)·e p |c + x + e b .
Bit flip Detection. We introduce a sufficient large ancilla system capable of storing
the syndrome for C 1 , which is initially in the all zero state |0. Applying the parity
check matrix H 1 of C 1 to all state |c + x + e b , since H 1 (c + x) = 0 we have
|c + x + e b |0 −→ |c + x + e b |H 1 (c + x + e b ) = |c + x + e b |H 1 e b ,
that is, the error was isolated. Thus, we obtain the state
1
√ |C 2 |
x∈C 2
(−1)
(c+x)·e p |c + x + e b |H 1 e b .
Performing the measurement of the ancilla, we obtain H 1 e b ; discarding the ancilla
we return to the quantum state
1
√ |C 2 |
x∈C 2
(−1)
(c+x)·e p |c + x + e b .