Example: Charge transfer integrals with FDE: water dimer

Download ElectronTransfer_FDE_H2O.run

Expert Option

The electron transfer calculation of a water dimer radical cation in this example is aimed at:

  • calculate site energies and couplings of FDE-derived charge-localized states
  • calculate the charge-transfer excitation energy from a two-state model that includes the two charge-localized states involved in the calculation
  • the output also includes a rough evaluation of the error introduced by the density fitting on the site energies and coupling

First the isolated neutral fragments are obtained. Symmetry NOSYM is used. Next in the first FDE calculation the localized state D+A is calculated, which means that the first water molecule has charge +1, and the second water molecule is neutral. The resulting TAPE21 files must be renamed to fragA1.t21 and fragA2.t21. In the second FDE calculation the localized state DA+ is calculated, now the second water molecule has charge +1, and the first water molecule is neutral. The resulting TAPE21 files must be renamed to fragB1.t21 and fragB2.t21. The FDE freeze and thaw cycle is done manually, and a spin-unrestricted calculation is performed.

The electron transfer calculation is next. The files fragA1.t21, fragA2.t21, fragB1.t21, and fragB2.t21 must exist and must have these names. The program must be execute in serial mode. Hybrids are not supported. An integration parameter of 6 is needed for accuracy.

$ADFBIN/adf -n 1 << eor
Title ElectronTransfer calculation
 rho1 t21.iso.rho1
 rho2 t21.iso.rho2
O         0.0000000000        0.0000000000      0.0000000000   f=rho1
H        -0.9358409558         .2646136961      0.0000000000   f=rho1
H        -0.0304663436       -0.9828924420      0.0000000000   f=rho1
O         0.0000000000       -2.9053396088      0.0000000000   f=rho2
H        -0.4092227596       -3.3374838250     -0.7701260000   f=rho2
H        -0.4092227596       -3.3374838250      0.7701260000   f=rho2
  GGA PW91
  iterations 0
NumericalQuality good
 numfrag 2

The output of this example is discussed here.

============  Electron Transfer RESULTS ===================

Electronic Coupling =         0.000000 eV
Electronic Coupling =        -0.000006 cm-1
H11-H22             =        -1.396836 eV
Excitation Energy   =         1.396836 eV
Overlap             =         0.000000
H11 H22 H12 =  -152.443044906236  -152.391712133030  -151.743951196449 Eh
S11 S22 S12 =     0.981761438554     0.980941502465    -0.000000000038

=========== END Electron Transfer RESULTS ================

Due to symmetry, the overlap is almost diagonal (Overlap = 0.00), thus the transition density is evaluated with one less electron as explained in Ref. [353] in the ADF Manual.

The electronic coupling between the state with a positive charge localized on one water molecule and another with the charge localized on the other water molecule is given by “Electronic Coupling” and is reported in eV and cm^-1.

“H11-H22” is the difference of the site energies in eV. Values of the site energies are given by the first two values of “H11 H22 H12” in atomic units.

“Excitation Energy” reports the value of the transfer excitation energy as calculated by diagonalization of the 2X2 generalized eigenvalue problem in the basis of the charge-localized states, see Refs. [352,353] in the ADF Manual.

“S11 S22 S12” are the values of the non-normalized overlaps.