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Multiple choice
In the single-qubit QPE of this lesson, after phase kickback the counting qubit holds
(
∣
0
⟩
+
e
2
π
i
ϕ
∣
1
⟩
)
/
2
(|0\rangle + e^{2\pi i\phi}|1\rangle)/\sqrt{2}
(
∣0
⟩
+
e
2
π
i
ϕ
∣1
⟩)
/
2
. After the final Hadamard, what are the measurement probabilities?
🔬 try it before you answer
P
(
0
)
=
ϕ
P(0) = \phi
P
(
0
)
=
ϕ
and
P
(
1
)
=
1
−
ϕ
P(1) = 1 - \phi
P
(
1
)
=
1
−
ϕ
P
(
0
)
=
sin
2
(
π
ϕ
)
P(0) = \sin^2(\pi\phi)
P
(
0
)
=
sin
2
(
π
ϕ
)
and
P
(
1
)
=
cos
2
(
π
ϕ
)
P(1) = \cos^2(\pi\phi)
P
(
1
)
=
cos
2
(
π
ϕ
)
P
(
0
)
=
P
(
1
)
=
1
/
2
P(0) = P(1) = 1/2
P
(
0
)
=
P
(
1
)
=
1/2
for every
ϕ
\phi
ϕ
P
(
0
)
=
cos
2
(
π
ϕ
)
P(0) = \cos^2(\pi\phi)
P
(
0
)
=
cos
2
(
π
ϕ
)
and
P
(
1
)
=
sin
2
(
π
ϕ
)
P(1) = \sin^2(\pi\phi)
P
(
1
)
=
sin
2
(
π
ϕ
)
Check answer