6.8 Application: Linear and Circular Dichroism
147
The commutator of the one-electron Hamiltonian with the position operator is given
by:
[H, r]=
p · p
2m
+ V(r)
, r
=−
i
m
p
(6.116)
Here, we used the Heisenberg commutator relation between the conjugate position
and momentum operators: [x,p x ]=i. The magnetic moment matrix element of
the intra-ligand transition with respect to the common origin of the coordinate system is given by:
m A ==ψ A |m|χ A =−
e
2m
ψ A |(R A + r) × p|χ A
=−
e
2m
R A ××ψ A |p|χ A
(6.117)
where it was assumed that the chromophore has no intrinsic magnetic transitionmoment. The momentum matrix element in this equation can now be evaluated with
the help of Eq. (6.116):
ψ A |p|χ A =
im
ψ A |Hr − rH|χ A
=
im
Hψ A |r|χ A −−ψ A |rH|χ A
=
im
(E ψ − E χ )ψ A |r|χ A
= 2πimνψ A |r|χ A
(6.118)
Here, ν is the frequency of the intra-ligand transition. The combination of this result
with Eq. (6.117) yields:
m A = iπν
R A × µ
A
= iπνρμ
0, −
√
2
√
3
,
1
√
3
m B = iπν
R B × µ
B
= iπνρμ
1
√
2
,
1
√
6
,
1
√
3
m C = iπν
R C × µ
C
= iπνρμ
−
1
√
2
,
1
√
6
,
1
√
3
(6.119)
As we indicated the above formalism applies to chromophores that have no intrinsic
magnetic moment.
m
1 A 1 →
1 A 2
=
1
√
3
m
A + m
B + m
C
= iπνρμ
(0, 0, 1)
m
1 A 1 →
1 E ǫ
=
1
√
6
2m
A − m
B − m
C
=−iπνρμ
(0, 1, 0)
m
1 A 1 →
1 E θ
=
1
√
2
−m
B + m
C
=−iπνρμ
(1, 0, 0)
(6.120)