Chemistry Reference
In-Depth Information
C.3.1 Example: double square well with infinite outer barriers
To illustrate the application of second order perturbation theory we return
to the problem we considered earlier of an infinite square well of width
L
to whose centre is added a barrier of height
V
0
and width
b
(fig. C.2). The
matrix element
ψ
(
0
)
H
|
ψ
(
0
)
|
describing the mixing between the
m
th and
m
n
n
th state is given by
L
/
2
+
b
/
2
V
0
sin
m
sin
n
d
x
2
L
π
x
π
x
ψ
(
0
)
H
|
ψ
(
0
)
|
=
(C.24)
m
n
L
L
L
/
2
−
b
/
2
1
2
This integral can be solved using the identity sin
α
sin
β
=
(
cos
(α
−
β)
−
cos
(α
+
β))
, to give
ψ
(
0
)
H
|
ψ
(
0
)
|
m
n
sin
(
2
L
n
sin
(
2
L
−
)π
/
+
)π
/
2
V
0
π
n
m
b
n
m
b
=
−
(
−
1
)
n
+
m
even
n
−
m
n
+
m
=
0
n
+
m
odd
(C.25)
The dotted (parabolic) lines in fig. C.3 show how the calculated energy
levels vary with barrier height
V
0
in second order perturbation theory.
It can be seen that going to second order gives a useful improvement in
the range of
V
0
values over which perturbation theory applies. You might
then consider extending perturbation theory to third or even higher orders:
however, a law of diminishing returns rapidly sets in, and in practice no
advantage is gained by extending perturbation theory beyond the second
order.
Reference
Schiff, L. I. (1968)
Quantum Mechanics
, 3rd edn, McGraw-Hill, Tokyo.
