PHY4211 Quantum Mechanics I Fall Term of 2005
Homework Set II
(A Possible Set of Solution, October 09, 2005)
Question 1:
(a): Normalization is easy to take as:
1/ 4
2
a
A
π
⎛
⎞
=
⎜
⎟
⎝
⎠
(1)
(b): Using the propagator of a free particle it could be shown that:
(
)
(
)
{
}
1/4
2
2
2
,
exp
exp
2
2
m
m
a
x t
dx
x
x
ax
i t
i t
π
π
⎧
⎫⎛
⎞
′
′
′
Ψ
=
−
−
−
⎨
⎬
⎜
⎟
⎩
⎭⎝
⎠
∫
=
=
(2)
This integral, of course, could be calculated out explicitly:
(
)
(
)
{
}
1/ 4
2
2
1/ 4
2
2
,
exp
exp
2
2
2
1
exp
2
2
1
1
a
m
m
x t
dx
x
x
ax
i t
i t
a
ax
i at
i at
m
m
π
π
π
⎛
⎞
⎧
⎫
′
′
′
Ψ
=
−
−
−
⎨
⎬
⎜
⎟
⎝
⎠
⎩
⎭
⎧
⎫
⎪
⎪
−
⎛
⎞
=
⎨
⎬
⎜
⎟
⎝
⎠
⎪
⎪
+
+
⎩
⎭
∫
=
=
=
=
(3)
And the corresponding
(
)
2
,
x t
Ψ
is:
(
)
1/2
2
2
2
2
2
1
2
,
exp
2
2
1
1
a
ax
x t
at
at
m
m
π
⎧
⎫
⎪
⎪
−
⎪
⎪
⎛
⎞
Ψ
=
⎨
⎬
⎜
⎟
⎝
⎠
⎛
⎞
⎪
⎪
⎛
⎞
+
+
⎜
⎟
⎜
⎟
⎪
⎪
⎝
⎠
⎩
⎭
⎝
⎠
=
=
(4)
(c): the Sketching is ignored here, instead I will give some qualitative explanation.
To consider this problem simply, let us do some transformation:
2
2
1
a
a
a
at
m
→
=
⎛
⎞
+
⎜
⎟
⎝
⎠
±
=
Then equation (4) changes into
(
)
{
}
1/2
2
2
2
,
exp
2
a
x t
ax
π
⎛
⎞
Ψ
=
−
⎜
⎟
⎝
⎠
±
±
(5)
Comparing equation (5) with the
(
)
2
,0
x
Ψ
obtained from equation (3):
(
)
{
}
1/2
2
2
2
,0
exp
2
a
x
ax
π
⎛
⎞
Ψ
=
−
⎜
⎟
⎝
⎠
(6)
you will find that the only change is
a
a
→
±
.

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So after a very large time interval,
a
a
<<
±
and the wave function remains Gaussian type. Since
that
a
±
becomes very small, however, at the final stage the Gaussian wave function has been largely
dispersed.
(d): since at time
t
=0 we have
0
x
=
and
(
)
2
1/ 4
x
a
=
for Gaussian wave function, then it is
direct to generalize this result to
t
>0 cases:
( )
(
)
( )
(
)
2
2
2
2
,
0
1
,
4
x t
dx
x t
x
x
t
dx
x t
x
a
=
Ψ
=
=
Ψ
=
∫
∫
±
(7)
However, the calculation of
( )
p t
and
( )
2
p
t
cannot be generalized in such a way, so I will do
it as following:
For
( )
p t
, you see from the below equation:
( )
(
)
(
)
,
,
p t
dx
x t
x t
i
x
∗
∂
=
Ψ
Ψ
∂
∫
=
(8)
that, the integrand is an odd function, and as a result, the above integral must be equal to 0.
For
( )
2
p
t
, let us assume that:
2
1
a
i at
m
ω
=
+
=
,
then we have
aa
ω ω
∗
=
±
and
2
a
a
ω
ω
∗
+
=
.
( )
(
)
(
)
(
)
{
}
(
)
(
)
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
,
,
,
4
2
4
,
2
,
1
2
2
2
p
t
dx
x t
x t
x
dx
x t
x
dxx
x t
dx
x t
a
a
a
a
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
∗
∗
∂
= −
Ψ
Ψ
∂
= −
Ψ
−
= −
Ψ
+
Ψ
= −
+
⎛
⎞
=
−
⎜
⎟
⎝
⎠
⎛
⎞
=
−
⎜
⎟
⎝
⎠
=
∫
∫
∫
∫
=
=
=
=
=
=
±
=
±
=
=
(9)
So from equation (7), (8) and (9) we have:
2
2
2
2
2
2
2
2
2
1
4
4
2
x
p
m
a t
x
x
am
a
p
p
a
σ
σ
+
=
−
=
=
=
−
=
=
±
=
(10)