PHY4221 Quantum Mechanic I (Fall term of 2004)
Problem Set No. 3
(Answers)
Question 1:
(a): You could see from the timedepended Schrodinger equation (1) that, although the potential
is changed suddenly, the wave function should be continuous as a function of time
t
.
2
2
2
2
i
V
t
m
x
ψ
ψ
ψ
∂
∂
= 
+
∂
∂
h
h
(1)
As a result, immediately after the motion of the wall of this potential well, the wave function of
this system stays undisturbed.
For our two systems, we have been familiar with their eigenfunctions:
(
29
(
29
2
2
2
2
2
sin
1, 2, 3,
2
n
n
n
x
n
x
E
n
L
L
mL
π
π
ψ
=
↔
=
=
×××
h
(2)
(
29
(
29
2
2
2
2
1
sin
1, 2, 3,
2
8
n
n
n
x
n
x
E
n
L
L
mL
π
π
ψ
′
′
=
↔
=
=
×××
h
(3)
Now let us analyse the condition we are facing:
At first, the system is in its stable state (and particle in ground state); then suddenly we move the
right wall of this potential well rightward (in an assumed very large velocity so that we do not
consider the process of moving it). Once the wall is moved,
the system changes into another, even
though its wave function has not been changed as well
; as a result, immediately after the motion of
the wall, we would better do our consideration using the quantities of the new system.
Before the change of the potential wall, the particle is in its ground state simply given by:
(
29
[
]
2
2
2
2
sin
0,
2
before
before
x
x
E
x
L
L
L
mL
π
π
ψ
=
↔
=
∈
h
(4)
The eigenfunctions in the new system is given in equation (3) and the realistic wave function is
given by equation (4); so we could expand equation (4) using equation (3)’s as:
(
29
(
29
(
29
2
1
sin
sin
2
immediately after
before
n
n
n
n
n
x
n
x
x
x
c
x
c
L
L
L
L
π
π
ψ
ψ
ψ
′
=
=
=
=
∑
∑
(5)
Then it is easy to obtain the coefficients:
(
29
(
29
(
29
2
0
4
2
sin
2
4
2
1
sin
sin
2
2
2
2
L
n
before
n
n
n
x
n
x
c
dx
x
x
dx
L
L
L
L
for n
π
π
π
π
ψ
ψ


′
=
=
×
×
=
=
∫
∫
(6)
So obviously the most probable result is when
n
=2, that is to say, the measurement will give the
following result most probably:
2
2
2
2
2
1
2
2
2
n
n
E
p
c
mL
π
′
=
↔
=
→
=
=
h
(7)
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(b):
Obviously, the next most probable result is given by:
2
2
2
1
2
2
32
1
8
9
n
n
E
p
c
mL
π
π
′
=
↔
=
→
=
=
h
(8)
(c):
if you use the above result to calculate the energy expectation value, you will find yourself
confronted with an infinite series and thus the calculation becomes difficult to carry out. Thus you
should try another method.
In fact, because the wave function is not influenced at all, we conclude that the motion of the
potential wall doesn’t ever influence the particle, and thus the expectation value of the energy of
the particle does not change. So we have this expectation value as:
2
2
2
2
immediately after
before
before
E
E
E
mL
π
=
=
=
h
(9)
In equation (9) the 2
nd
‘=’ is due to the fact that before changing the potential the particle is in its
ground state, not a mixed state.
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 CFLO
 Schrodinger Equation, Fundamental physics concepts, wave function

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