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Multiple choice
A particle of energy
E
=
1
E = 1
E
=
1
hits a square barrier of height
V
0
=
2
V_0 = 2
V
0
=
2
and width
L
=
1
L = 1
L
=
1
(units with
ℏ
=
m
=
1
\hbar = m = 1
ℏ
=
m
=
1
). In the WKB approximation with
∣
p
∣
=
2
(
V
0
−
E
)
|p| = \sqrt{2(V_0-E)}
∣
p
∣
=
2
(
V
0
−
E
)
constant inside the barrier, the transmission is
T
=
e
−
2
γ
T = e^{-2\gamma}
T
=
e
−
2
γ
with
γ
=
L
∣
p
∣
\gamma = L\,|p|
γ
=
L
∣
p
∣
. What is
T
T
T
?
≈
0.135
\approx 0.135
≈
0.135
(used
∣
p
∣
=
2
V
0
|p| = \sqrt{2 V_0}
∣
p
∣
=
2
V
0
)
≈
0.243
\approx 0.243
≈
0.243
(forgot the factor of 2:
e
−
γ
e^{-\gamma}
e
−
γ
)
≈
0.0591
\approx 0.0591
≈
0.0591
≈
0.943
\approx 0.943
≈
0.943
(wrong sign in exponent)
Check answer