Chem 3322 homework #2 solutions – out of 32 marks
Problem 1
– 4 marks
Consider the operator
ˆ
A
=
x
d
dx

d
dx
x
(1)
What does this operator do to a function
f
(
x
)? Based on your answer, express this operator
in a simpler form.
Solution:
ˆ
Af
(
x
) =
±
x
d
dx

d
dx
x
²
f
(
x
)
(2)
=
xf
0
(
x
)

d
dx
(
xf
(
x
)) =
xf
0
(
x
)

(
f
(
x
) +
xf
0
(
x
)) =

f
(
x
)
from which we conclude that
ˆ
A
=

1. Namely the operator
A
acts on a function by
multiplying the function by

1.
Problem 2
– 9 marks
Consider a particle in a onedimensional box of length L in its lowest energy (ground)
stationary state. Calculate the probability that the particle is
a – 3 marks)
in the left half of the box
solution:
For the lowest energy state, we have
ψ
(
x
) =
r
2
L
sin
πx
L
(3)
The probability density of this state is thus
2
L
sin
2
πx
L
(4)
You need to integrate this expression. Normally this integral is found in textbooks as
Z
sin
2
x dx
=
1
2
x

1
4
sin 2
x
(5)
Changing variables from
x
to
u
=
πx/L
, we have
Z
b
a
2
L
sin
2
πx
L
dx
=
2
π
Z
πb/L
πa/L
sin
2
u du
(6)
1
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a
= 0 and
b
=
L/
2 we get a probability of 1/2. This makes sense because of symmetry
– see Fig 1.
b – 3 marks)
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