Handbook of Solvents - George Wypych - ChemTech - Ventech!

658 Mati Karelson
For the solution of equation [11.1.54], the molecular wavefunction can be presented as a
proper spin-projected antisymmetrized product of molecular (or atomic) orbitals 27
|ψ� = OA[
φ, , φ ]
� [11.1.56]
s 1 n
Recalling that the dipole moment operator is a one-electron operator (as are all electric moment
operators), the following orbital equations are obtained
with
or
() φ() ε φ()
fk k = k
[11.1.57]
i i i
() k = () k − () k
f f0
Γψμ� tot ψμ�
el
[11.1.58]
() k = () k − Γψμ ψμ () k
f f
0
1
2
� � [11.1.59]
tot el
where f 0(k) is the usual Fock operator for the isolated molecule, ε i is the molecular orbital
energy for |φ i > and �μ el (k) is the electronic part of the dipole moment operator. Both equations
are solved iteratively, using the usual SCF procedure and the expectation value of the
total dipole moment from the previous SCF cycle.
A scheme for the treatment of the solvent effects on the electronic absorption spectra
in solution had been proposed in the framework of the electrostatic SCRF model and quantum
chemical configuration interaction (CI) method. 27 Within this approach, the absorption
of the light by chromophoric molecules was considered as an instantaneous process. Therefore,
during the photon absorption no change in the solvent orientational polarization was
expected. Only the electronic polarization of solvent would respond to the changed electron
density of the solute molecule in its excited (Franck-Condon) state. Consequently, the solvent
orientation for the excited state remains the same as it was for the ground state, the solvent
electronic polarization, however, must reflect the excited state dipole and other electric
moments of the molecule. Considering the SCRF Hamiltonian
� � 0
H= H + Γφμ� φμ�
tot
it is possible to write for the state |ψ I � the following expression
0 0
HII = ψI H + ψμψ μψI = ψI H ψI − ψ0μψ0 ψI μψI
[11.1.60]
� Γ � � � Γ � � [11.1.61]
that is the zeroth order estimate of the energy of the state |ψI �. Then, for a single excitation,
I→ A,
the excitation energy is given as follows