Chemistry Reference
In-Depth Information
ψ
(
1
k
. Finally, we can substitute the results of eq. (C.12b)
into the third equation (C.12c), to determine the second order change,
E
(
2
)
k
wavefunctions,
in the energy levels.
We wish first to consider the form of
ψ
(
1
)
k
ψ
(
0
)
k
(
r
)
, the
change
to
(
r
)
due to
the perturbation
H
. We described in Chapter 1 how any function
f
(
)
r
can
ψ
(
0
)
be written as a linear combination of the complete set of states,
(
r
)
.The
n
ψ
(
0
)
k
, due to the perturbation
H
will involve
change
in the wavefunction,
(
r
)
ψ
(
0
)
k
ψ
(
1
)
k
(
)
(
)
mixing ('adding') other states into
r
, so we expect that
r
can be
written as a linear combination of all the other wavefunctions
ψ
(
1
)
k
ψ
(
0
)
(
r
)
=
a
kn
(
r
)
(C.13)
n
n
=
k
where
a
kn
is the amplitude which the
n
th state contributes to the modi-
fication of the
k
th wavefunction. Substituting eqs (C.12a) and (C.13) into
eq. (C.12b), and rearranging, we obtain
H
0
a
kn
E
(
1
)
k
H
E
(
0
)
k
ψ
(
0
)
ψ
(
0
)
k
−
(
r
)
=
−
(
r
)
(C.14)
n
=
n
k
ψ
(
0
)
E
(
0
)
ψ
(
0
)
which, as
H
0
(
r
)
=
(
r
)
, reduces to
n
n
n
a
kn
E
(
0
)
E
(
1
)
k
H
E
(
0
)
k
ψ
(
0
)
ψ
(
0
)
k
−
(
r
)
=
−
(
r
)
(C.15)
n
n
n
=
k
We can use eq. (C.15) to evaluate the two first order corrections,
E
(
1
)
k
ψ
(
1
)
k
ψ
∗
(
0
)
k
and
(
r
)
. We first multiply both sides of eq. (C.15) by
(
r
)
, and
integrate over all space to find
E
(
1
)
k
:
a
kn
E
(
0
)
E
(
0
)
k
ψ
∗
(
0
)
k
)ψ
(
0
)
−
d
V
(
r
(
r
)
n
n
n
=
k
E
(
1
)
k
H
ψ
∗
(
0
)
k
ψ
(
0
)
k
=
d
V
(
r
)
−
(
r
)
(C.16)
The left hand side of eq.
(C.16)
is identically zero, because the
ψ
(
0
)
k
ψ
(
0
)
wavefunctions
are orthogonal, andwe can rearrange the
right-hand side to give the same result for the first order energy correction
as in the previous section:
(
r
)
and
(
r
)
n
E
(
1
)
k
ψ
∗
(
0
)
k
H
ψ
(
0
)
k
=
d
V
(
r
)
(
r
)
(C.7)
We use the same technique to calculate the coefficients
a
km
from eq. (C.13),
multiplying both sides of eq. (C.15) by
ψ
∗
(
0
)
(
r
)
, and integrating over all
m
