H Solutions to Problems
253
quartet, the singlet states cannot contribute, and we need to couple the triplet to
a 2 T 1u state, resulting from a (t 1u ) 1 configuration. The orbital part of the triplet
is obtained from the T 1 × T 1 = T 1 coupling table in Appendix F:
|T 1g x=
1
√
2
−y(1)z(2) + z(1)y(2)
|T 1g y=
1
√
2
x(1)z(2) − z(1)x(2)
|T 1g z=
1
√
2
−x(1)y(2) + y(1)x(2)
The coupling with the third electron can yield A 1u , E u , T 1u , and T 2u states. Our
results is based on the A 1u product. This yields
A 1u =
1
√
3
|T 1g x|x(3)+|T 1g y|y(3)+|T 1g z|z(3)
=−
1
√
6
x(1)y ( 1)z ( 1)
x(2)y ( 2)z ( 2)
x(3)y ( 3)z ( 3)
This should be multiplied by the product of the three α-spins, α 1 α 2 α 3 , to obtain
the 4 A 1u ground state of the (t 1u ) 3 configuration.
6.2 The JT problem is determined by the symmetrized direct product of T 1u .Aswe
have seen in the previous problem, this product contains A 1g + E g + T 2g . Since
A 1g modes do not break the symmetry, the JT problem is of type T 1 ×(e +t 2 ).In
the linear problem only two force elements are required. The distortion matrix
is thus as follows:
H
′ =
F E
√
6
⎛
⎜
⎜
⎝
Q θ
00
0 Q θ
0
00 −2Q θ
⎞
⎟
⎟
⎠ +
F T
√
2
⎛
⎜
⎜
⎝
0
−Q ζ −Q η
−Q ζ
0
−Q ξ
−Q η
0 ξ
0
⎞
⎟
⎟
⎠
6.3 The magnetic dipole operator transforms as T 1g , while the direct square of e g
irreps yields A 1g + A 2g + E g . Since the operator irrep is not contained in the
product space, the selection rules will not allow a dipole matrix element between 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
µ 1 = μ
0, cos
α
2
, − sin
α
2
µ 2 = μ
0, cos
α
2
, sin
α
2
253
quartet, the singlet states cannot contribute, and we need to couple the triplet to
a 2 T 1u state, resulting from a (t 1u ) 1 configuration. The orbital part of the triplet
is obtained from the T 1 × T 1 = T 1 coupling table in Appendix F:
|T 1g x=
1
√
2
−y(1)z(2) + z(1)y(2)
|T 1g y=
1
√
2
x(1)z(2) − z(1)x(2)
|T 1g z=
1
√
2
−x(1)y(2) + y(1)x(2)
The coupling with the third electron can yield A 1u , E u , T 1u , and T 2u states. Our
results is based on the A 1u product. This yields
A 1u =
1
√
3
|T 1g x|x(3)+|T 1g y|y(3)+|T 1g z|z(3)
=−
1
√
6
x(1)y ( 1)z ( 1)
x(2)y ( 2)z ( 2)
x(3)y ( 3)z ( 3)
This should be multiplied by the product of the three α-spins, α 1 α 2 α 3 , to obtain
the 4 A 1u ground state of the (t 1u ) 3 configuration.
6.2 The JT problem is determined by the symmetrized direct product of T 1u .Aswe
have seen in the previous problem, this product contains A 1g + E g + T 2g . Since
A 1g modes do not break the symmetry, the JT problem is of type T 1 ×(e +t 2 ).In
the linear problem only two force elements are required. The distortion matrix
is thus as follows:
H
′ =
F E
√
6
⎛
⎜
⎜
⎝
Q θ
00
0 Q θ
0
00 −2Q θ
⎞
⎟
⎟
⎠ +
F T
√
2
⎛
⎜
⎜
⎝
0
−Q ζ −Q η
−Q ζ
0
−Q ξ
−Q η
0 ξ
0
⎞
⎟
⎟
⎠
6.3 The magnetic dipole operator transforms as T 1g , while the direct square of e g
irreps yields A 1g + A 2g + E g . Since the operator irrep is not contained in the
product space, the selection rules will not allow a dipole matrix element between 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
µ 1 = μ
0, cos
α
2
, − sin
α
2
µ 2 = μ
0, cos
α
2
, sin
α
2