Chem 120A
Free Particle and a Particle Meeting a Potential Barrier
02/22/06
Spring 2006
Lecture 15
READING:
Engel: Chapters 4 and 5
In the last few lectures, we have been dealing with particle in a potential well. Now we will look at the
the solutions to the Schrodinger equation for a free particle. That is a particle which is not conﬁned by any
potential ( V=0 everywhere). The Schrodinger equation for a particle in 1D is:

¯
h
2
2
m
d
2
dx
2
Ψ
(
x
) =
E
Ψ
(
x
)
(1)
The general solution to this 2ndorder differential equation is:
Ψ
(
x
) =
A
1
e
ikx
+
A
2
e

ikx
(2)
where
k
2
=
2
mE
¯
h
2
There are several interesting comparisons between the free particle and the particle in a box. First of all, the
energy eigenvalue for the free particle solution is
E
=
¯
h
2
k
2
2
m
, so the energies are not quantized. Without ap
plying constraints (boundary conditions) on
Ψ
, the energy can take on continuous values. Secondly, there is
no zeropoint energy (which means that the lowest energy is E = 0). Finally, we can think of the free particle
as a particle in a box of length
a
, where
a
→
∞
Remember that the uncertainty in momentum is inversely
propotinal to
a
as
σ
p
=
n
π
¯
h
a
. Thus,
σ
p
→
0 as
a
→
0. On the otherhand, the uncertainty in position is directly
proportional to
a
, thus
σ
x
→
∞
as
a
→
∞
. For a free particle, the momentum is completely deﬁned, while
the position is completely uncertain.
a)For the bound solutions when
E
<
0, k will be a complex number. For convience, we can set
k
=
i
κ
to get
rid of the complexity (remember
i
2
=

1), thus
Ψ
(
x
) =
A
1
e

κ
x
+
A
2
e
κ
x
(3)
You can easily show that since the bound solutions are real, the probability current,
∂
j
(
x
,
t
)
∂
x
is zero, where
j
(
x
,
t
) =
i
¯
h
2
m
[
Ψ
*
(
x
,
t
)
∂
Ψ
(
x
,
t
)
∂
x

∂
Ψ
*
(
x
,
t
)
∂
x
Ψ
(
x
,
t
)]
(4)
b) Now we want to look at the probability current for the unbound solutions,
E
>
0. We know that the
Chem 120A, Spring 2006, Lecture 15
1