Chemistry Reference
In-Depth Information
where
A
q
s
q
0
s
0
¼
d
qq
0
d
ss
0
o
q
s
þ
K
q
s
q
0
s
0
,
B
q
s
q
0
s
0
¼
K
q
s
q
0
s
0
,
X
q
s
¼
P
q
s
,
Y
q
s
¼
P
q
s
, and
ð
dr
ð
dr
0
q
s
ð
r
Þ
f
HXC
ss
0
ð
rr
0
o
Þ
q
0
s
0
ð
r
0
Þ
K
q
s
q
0
s
0
ð
o
Þ¼
½
45
with
ð
dr
q
s
ð
r
Þ
d
d
v
q
s
ð
o
Þ¼
v
ext
ð
r
o
Þ
½
46
When
will be an excitation frequency of the true system. If the
KS orbital are chosen as real and using the fact that
d
v
¼
0, then
o
will be positive
definite, then we can simplify Eq. [44] as the eigenvalue problem in Eq. [47]:
X
ð
A
B
Þ
~
2
q
0
s
0
ð
o
Þ
~
~
a
q
0
s
0
¼
o
a
q
s
½
47
q
s
q
0
s
0
where
~
¼ð
A
B
Þ
1
=
2
1
=
2
ð
A
þ
B
Þð
A
B
Þ
½
48
or
2
p
~
2
q
q
0
s
0
ð
o
Þ¼
o
s
d
qq
0
d
ss
0
þ
o
s
o
K
q
s
q
0
s
0
½
49
q
s
q
q
0
s
0
Oscillator strengths
f
q
may be calculated
191
from the normalized eigenvectors
using
2
2
2
2
3
x
T
S
1=2
y
T
S
1=2
z
T
S
1=2
f
q
s
¼
ðj~
~
a
q
s
j
þj~
~
a
q
s
j
þj~
~
a
q
s
j
Þ
½
50
where
S
qq
0
¼
d
qq
0
d
ss
0
w
q
0
s
0
½
51
Figure 4 shows the results of
exact
DFT calculations for the He atom. On
the left side of the diagram, we consider just transitions from the exact ground-
state KS occupied orbital (1
s
) to unoccupied orbitals. These are
not
the true
excitations of the system, nor are they supposed to be. However, applying
TDDFT linear response theory, using the exact kernel with the exact orbitals,
yields the exact excitation frequencies of the He atom. Spin decomposing pro-
duces both singlet and triplet excitations.
