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A qubit prepared in
∣
+
⟩
=
(
∣
0
⟩
+
∣
1
⟩
)
/
2
|+\rangle = (|0\rangle+|1\rangle)/\sqrt{2}
∣
+
⟩
=
(
∣0
⟩
+
∣1
⟩)
/
2
passes through a dephasing channel with
p
=
0.3
p = 0.3
p
=
0.3
, giving
ρ
\rho
ρ
with populations
0.5
0.5
0.5
and off-diagonals
0.5
(
1
−
2
p
)
0.5(1-2p)
0.5
(
1
−
2
p
)
. The fidelity with
∣
+
⟩
|+\rangle
∣
+
⟩
is
F
=
⟨
+
∣
ρ
∣
+
⟩
=
0.5
(
1
+
2
⋅
0.5
(
1
−
2
p
)
)
=
1
−
p
F = \langle +|\rho|+\rangle = 0.5(1 + 2 \cdot 0.5(1-2p)) = 1 - p
F
=
⟨
+
∣
ρ
∣
+
⟩
=
0.5
(
1
+
2
⋅
0.5
(
1
−
2
p
))
=
1
−
p
. What is
F
F
F
?
🔬 try it before you answer
0.7
0.3
0.5
0.85
Check answer