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For a potential giving the classical action
S
(
E
)
=
∫
2
E
−
x
2
d
x
S(E) = \int \sqrt{2E - x^2}\, dx
S
(
E
)
=
∫
2
E
−
x
2
d
x
between the turning points
x
=
±
2
E
x = \pm\sqrt{2E}
x
=
±
2
E
, the WKB (Bohr-Sommerfeld) condition is
S
(
E
)
=
(
n
+
1
/
2
)
π
S(E) = (n + 1/2)\pi
S
(
E
)
=
(
n
+
1/2
)
π
. Solving for the
n
=
2
n = 2
n
=
2
level, what energy
E
E
E
satisfies
S
(
E
)
=
(
2
+
1
/
2
)
π
S(E) = (2 + 1/2)\pi
S
(
E
)
=
(
2
+
1/2
)
π
?
≈
2.0
\approx 2.0
≈
2.0
(used
n
n
n
instead of
n
+
1
/
2
n + 1/2
n
+
1/2
)
≈
5.0
\approx 5.0
≈
5.0
(used
2
(
n
+
1
/
2
)
2(n + 1/2)
2
(
n
+
1/2
)
)
≈
2.5
\approx 2.5
≈
2.5
≈
1.5
\approx 1.5
≈
1.5
Check answer