396
Chapter 39
(b)
The probability of finding the particle is zero where
2
0,
ψ
=
which occurs where sin
0
kx
=
and
2
/
,
0, 1, 2,
kx
x
n
n
πλ π
==
=
…
/ 2,
0, 1, 2,
so
0, / 2, , 3 / 2,
xn
n
x
λλ
λ
=
……
EVALUATE:
The situation is analogous to a standing wave, with the probability analogous to the square of the
amplitude of the standing wave.
39.30.
sin
,
ψω
t
∗∗
Ψ=
so
22
**
2
2
sin
sin
ψψ
ω
t
t
Ψ=ΨΨ=
=
.
2
Ψ
is not timeindependent, so
Ψ
is not the
wavefunction for a stationary state.
39.31.
IDENTIFY:
To describe a real situation, a wave function must be normalizable.
SET UP:


2
dV
is the probability that the particle is found in volume
dV
. Since the particle must be
somewhere
,
must have the property that
∫


2
dV
= 1 when the integral is taken over all space.
EXECUTE: (a)
In one dimension, as we have here, the integral discussed above is of the form
2
(
)

1
xd
x
∞
−∞
=
∫
.
(b)
Using the result from part (a), we have
()
2
2
2
2
ax
ax
ax
e
ed
x
x
a
∞
∞∞
=
∞
∫∫
. Hence this wave function cannot
be normalized and therefore cannot be a valid wave function.
(c)
We only need to integrate this wave function of 0 to
∞
because it is zero for
x
< 0. For normalization we have
2
1

dx
∞
=
∫
=
2
2
00
0
bx
bx
bx
Ae
A
Ae
dx
A e
dx
bb
∞
−
−−
=
−
, which gives
2
1
2
A
b
=
, so
2
Ab
=
.
EVALUATE:
If
b
were positive, the given wave function could not be normalized, so it would not be allowable.
39.32.
(a)
The uncertainty in the particle position is proportional to the width of
ψ
x
, and is inversely proportional to
α
. This can be seen by either plotting the function for different values of
, finding the expectation
value
2
x
ψ
xdx
=
∫
for the normalized wave function or by finding the full width at halfmaximum. The
particle’s uncertainty in position decreases with increasing
. The dependence of the expectation value
2
x
⟨⟩
on
may be found by considering
2
2
2
2
x
x
xe
dx
x
x
∞
−
−∞
∞
−
−∞
=
∫
∫
=
2
2
1
ln
2
x
x
∞
−
−∞
⎡
⎤
∂
−
⎢
⎥
∂
⎣
⎦
∫
2
11
1
ln
,
24
2
u
u
αα
∞
−
−∞
⎡
⎤
∂
=−
=
⎢⎥
∂
⎣⎦
∫
where the substitution
ux
=
has been made.
(b)
Since the uncertainty in position decreases, the uncertainty in momentum must increase.
39.33.
*
(,)
a
n
d (,)
xi
y
y
fxy
f xy
y
y
⎛⎞
⎛
⎞
−+
⎜⎟
⎜
⎟
+−
⎝⎠
⎝
⎠
2
*
1.
y xi
y
ff
f
y
⇒
⋅
=
39.34.
The same.
2
*
(,,)
(,,)(,,)
ψ
xyz
ψ
ψ
=
2
*
(
,
,
)
(
(
,
,
))
(
(
,
,
ii
i
ψ
xyze
ψ
ψ
φφ
φ
=
*
.
ψ
ψ
=
The complex conjugate means convert all
i
’s to–
i
’s and viceversa.
ee
−
⋅=
39.35.
IDENTIFY:
To describe a real situation, a wave function must be normalizable.
SET UP:


2
dV
is the probability that the particle is found in volume
dV
. Since the particle must be
somewhere
,
must have the property that
∫


2
dV
= 1 when the integral is taken over all space.
EXECUTE: (a)
For normalization of the onedimensional wave function, we have
2

dx
∞
=
∫
=
(
)
2 2
bx
bx
bx
bx
Ae
dx
Ae
dx
A e
dx
A e
dx
+=
+
∫
∫
.
0