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An infinite-square-well ground state gives probability density
∣
ψ
(
x
)
∣
2
=
2
sin
2
(
π
x
)
|\psi(x)|^2 = 2\sin^2(\pi x)
∣
ψ
(
x
)
∣
2
=
2
sin
2
(
π
x
)
on
0
≤
x
≤
1
0 \le x \le 1
0
≤
x
≤
1
. What is the probability of finding the particle in the interval
0
≤
x
≤
1
/
3
0 \le x \le 1/3
0
≤
x
≤
1/3
? Use
∫
2
sin
2
(
π
x
)
d
x
=
x
−
sin
(
2
π
x
)
/
(
2
π
)
\int 2\sin^2(\pi x) \, dx = x - \sin(2\pi x)/(2\pi)
∫
2
sin
2
(
π
x
)
d
x
=
x
−
sin
(
2
π
x
)
/
(
2
π
)
.
🔬 try it before you answer
2
/
3
≈
0.667
2/3 \approx 0.667
2/3
≈
0.667
1
/
3
−
3
/
(
4
π
)
≈
0.196
1/3 - \sqrt{3}/(4\pi) \approx 0.196
1/3
−
3
/
(
4
π
)
≈
0.196
1
/
3
≈
0.333
1/3 \approx 0.333
1/3
≈
0.333
1
/
3
+
3
/
(
4
π
)
≈
0.471
1/3 + \sqrt{3}/(4\pi) \approx 0.471
1/3
+
3
/
(
4
π
)
≈
0.471
Check answer