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A constant perturbation switched on for time
t
t
t
gives transition probability
P
=
V
f
i
2
ℏ
2
⋅
4
sin
2
(
w
f
i
t
/
2
)
w
f
i
2
P = \frac{V_{fi}^2}{\hbar^2} \cdot \frac{4 \sin^2(w_{fi} t/2)}{w_{fi}^2}
P
=
ℏ
2
V
f
i
2
⋅
w
f
i
2
4
s
i
n
2
(
w
f
i
t
/2
)
. With
ℏ
=
1
\hbar = 1
ℏ
=
1
,
V
f
i
=
0.3
V_{fi} = 0.3
V
f
i
=
0.3
,
w
f
i
=
1.5
w_{fi} = 1.5
w
f
i
=
1.5
, and
t
=
2
t = 2
t
=
2
, what is
P
P
P
?
≈ 0.3989 (took the square root, i.e. the amplitude)
≈ 0.0398 (forgot the factor of
4
4
4
)
≈ 0.1592
≈ 0.6368 (used
sin
2
(
w
f
i
t
)
\sin^2(w_{fi} t)
sin
2
(
w
f
i
t
)
instead of
sin
2
(
w
f
i
t
/
2
)
\sin^2(w_{fi} t/2)
sin
2
(
w
f
i
t
/2
)
)
Check answer