Preliminary Test, (
Answers
)
Question 1
st
:
Method 1
st
:
It could be noticed that in the previous studies we got familiar with another simpler
example of 1D rigid box problem: you have a 1D rigid box whose two boundaries are
located at x=0 and x=a, then you know the wave function of the particle confined in
this box is (in which n are any positive integers):
(
29
2
sin
0
0
0
n
n x
x
a
x
a
a
x
or x
a
π
ψ
<
<
÷
=
<
.
Now the question is similar, differing from the above case only in that the rigid box
now is located at
x
L
= ±
. So you can use the results of the above case directly, with
some mathematical transformation:
Set
y
x
L
=
+
, and we have
[
]
0,2
y
L
∈
as a new variable, and thus this question is
changed into follows: a particle is confined in a rigid box located at
[
]
0,2
y
L
∈
. So
simply using the above result you know that the wave function in our question can be
written as:
(
29
2
sin
0
2
2
2
0
0
2
n
n y
y
L
y
L
L
y
or y
L
<
<
÷
=
<
.
Changing this result back into x we obtain:
(
29
(
29
2
sin
0
2
2
2
0
0
2
n
n
x
L
x
L
L
y
L
L
L
x
L
or x
L
L
+
<
+
<
÷

=
+
<
+
.
This is, of course, equivalent to the following and our final result:
(
29
(
29
1
sin
2
0
n
n
x
L
L
x
L
x
L
L
x
L or x
L
+

<
<
÷
=
.
Method 2
nd
:
You may also use the ordinary method as you do in any other cases, that is, solving
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 Spring '11
 CFLO
 mechanics, Uncertainty Principle, wave function, dx, sin kL

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