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A spin-1/2 particle is prepared in the
S
z
=
+
ℏ
/
2
S_z = +\hbar/2
S
z
=
+
ℏ/2
state
∣
↑
z
⟩
|{\uparrow_z}\rangle
∣
↑
z
⟩
. It is then measured along x, whose eigenstates are
∣
+
x
⟩
=
(
1
/
2
)
(
∣
↑
z
⟩
+
∣
↓
z
⟩
)
|{+x}\rangle = (1/\sqrt{2})(|{\uparrow_z}\rangle + |{\downarrow_z}\rangle)
∣
+
x
⟩
=
(
1/
2
)
(
∣
↑
z
⟩
+
∣
↓
z
⟩)
and
∣
−
x
⟩
=
(
1
/
2
)
(
∣
↑
z
⟩
−
∣
↓
z
⟩
)
|{-x}\rangle = (1/\sqrt{2})(|{\uparrow_z}\rangle - |{\downarrow_z}\rangle)
∣
−
x
⟩
=
(
1/
2
)
(
∣
↑
z
⟩
−
∣
↓
z
⟩)
. What is the probability of obtaining
S
x
=
+
ℏ
/
2
S_x = +\hbar/2
S
x
=
+
ℏ/2
?
🔬 try it before you answer
1/2
1
1
/
2
≈
0.707
1/\sqrt{2} \approx 0.707
1/
2
≈
0.707
1/4
Check answer