Chemistry Reference
In-Depth Information
quartet, the singlet states cannot contribute, and we need to couple the triplet to
a
2
T
1
u
state, resulting from a
(t
1
u
)
1
configuration. The orbital part of the triplet
is obtained from the
T
1
×
T
1
=
T
1
coupling table in Appendix
F
:
√
2
−
z(
1
)y(
2
)
1
|
T
1
g
x
=
y(
1
)z(
2
)
+
√
2
x(
1
)z(
2
)
z(
1
)x(
2
)
1
|
T
1
g
y
=
−
√
2
−
y(
1
)x(
2
)
1
|
T
1
g
z
=
x(
1
)y(
2
)
+
The coupling with the third electron can yield
A
1
u
,
E
u
,
T
1
u
, and
T
2
u
states. Our
results is based on the
A
1
u
product. This yields
√
3
|
1
A
1
u
=
T
1
g
x
|
x(
3
)
+|
T
1
g
y
|
y(
3
)
+|
T
1
g
z
|
z(
3
)
x(
1
)y
1
)
1
)
x(
2
)y
2
)
2
)
x(
3
)y
3
)
3
)
1
√
6
=−
This should be multiplied by the product of the three
α
-spins,
α
1
α
2
α
3
, to obtain
the
4
A
1
u
ground state of the
(t
1
u
)
3
configuration.
6.2 The JT problem is determined by the symmetrized direct product of
T
1
u
.Aswe
have seen in the previous problem, this product contains
A
1
g
+
E
g
+
T
2
g
. Since
A
1
g
modes do not break the symmetry, the JT problem is of type
T
1
×
t
2
)
.In
the linear problem only two force elements are required. The distortion matrix
is thus as follows:
(e
+
⎛
⎞
⎛
⎞
Q
θ
0
0
0
−
Q
ζ
−
Q
η
⎝
⎠
+
⎝
⎠
F
E
√
6
F
T
√
2
H
=
0
Q
θ
0
−
Q
ζ
0
−
Q
ξ
00
−
2
Q
θ
−
Q
η
0
ξ
0
6.3 The magnetic dipole operator transforms as
T
1
g
, while the direct square of
e
g
irreps yields
A
1
g
+
E
g
. Since the operator irrep is not contained in the
product space, the selection rules will not allow a dipole matrix element be-
tween
e
g
orbitals.
6.4 We first draw a simple diagram representing the
R
-conformation. The point
group is
C
2
. The twofold-axis is oriented along the
y
-direction, and the centers
of the two chromophores are placed on the positive and negative
x
-axes. The
dipole moments are then oriented as
A
2
g
+
μ
0
,
cos
α
sin
α
2
μ
1
=
2
,
−
μ
0
,
cos
α
2
,
sin
α
μ
2
=
2
