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Multiple choice
The quantum Singleton bound for an
[
[
n
,
k
,
d
]
]
[[n,k,d]]
[[
n
,
k
,
d
]]
code states
k
≤
n
−
2
(
d
−
1
)
k \le n - 2(d-1)
k
≤
n
−
2
(
d
−
1
)
, equivalently
n
−
k
≥
2
(
d
−
1
)
n - k \ge 2(d-1)
n
−
k
≥
2
(
d
−
1
)
. Why does each unit of distance beyond 1 cost TWO qubits of redundancy, rather than one as in the classical Singleton bound?
🔬 try it before you answer
Because the codespace is always two-dimensional regardless of
k
k
k
Because quantum errors come in two independent flavours, X and Z, that must both be corrected
Because the no-cloning theorem requires storing two physical copies of each logical qubit
Because every qubit must be measured twice to read out its syndrome
Check answer