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Let
∣
v
⟩
=
(
2
+
i
)
∣
0
⟩
+
(
1
−
2
i
)
∣
1
⟩
|v\rangle = (2+i)|0\rangle + (1-2i)|1\rangle
∣
v
⟩
=
(
2
+
i
)
∣0
⟩
+
(
1
−
2
i
)
∣1
⟩
. What is the squared norm
⟨
v
∣
v
⟩
=
∣
2
+
i
∣
2
+
∣
1
−
2
i
∣
2
\langle v|v\rangle = |2+i|^2 + |1-2i|^2
⟨
v
∣
v
⟩
=
∣2
+
i
∣
2
+
∣1
−
2
i
∣
2
?
10
\sqrt{10}
10
≈ 3.16
0 (the cross terms cancel)
6 (using real parts only:
2
2
+
1
2
+
(
−
1
)
2^2 + 1^2 + (-1)
2
2
+
1
2
+
(
−
1
)
)
10
Check answer