[CP2K-user] [CP2K:18122] fluorescence wavelength by CP2K

[Victor Volkov]

Dear Fr. Hutter:
good evening.
Thank you for the kind explanations.
With best regards,
Victor

On Wed, Nov 30, 2022 at 4:47 PM Jürg Hutter <hutter at chem.uzh.ch> wrote:

> Hi
>
> yes, all your assumptions are correct. At the end of the excited state
> geometry optimization
> you will get the energies
> ground state    Eg
> excitation          Eex          this is the fluorescence energy
> total energy     Eg+Exc   (this was optimized)
>
> Eg+Exc - (ground state energy optimized) is the adiabatic excitation energy
>
> and
>
> excitation energy at ground state optimized   -  Eex is the Stokes shift.
>
> Some other remarks:
>
> these energies are different
> 3) Total Energy [Hartree]
>  -44.8592844650
> 4)  ENERGY| Total FORCE_EVAL ( QS ) energy [a.u.]:   -44.859271658539249
> because your SCF convergence is not tight enough.
> If you use Diagonalization epsscf should be 10E-8 or even smaller.
> But for this system I would use OT as it is much faster and then 1E-6 is
> good enugh.
>
> You are using the CG geometry optimizer. Apparently then you don't get
> detailed output
> on the progress of the optimization. If you use the default optimizers the
> usual
> gradient breakdown will be printed. There is nothing special about TDDFT
> here.
>
> regards
> JH
>
> ________________________________________
> From: cp2k at googlegroups.com <cp2k at googlegroups.com> on behalf of Victor
> Volkov <volkovskr at gmail.com>
> Sent: Sunday, November 27, 2022 1:22 AM
> To: cp2k
> Subject: [CP2K:18094] fluorescence wavelength by CP2K
>
> Dear developers:
> good morning/evening.
> Using pyrimidine, I try to understand how to compute 0'-0 fluorescence
> energy (wavelength).
>
> In gaussian, this relatively well-established series of three calculations:
> 1) B3LYP/6-31+G(d,p) Opt
> 2) B3LYP/6-31+G(d,p) TD=(nstates=xx)
> 3) B3LYP/6-31+G(d,p) Opt TD=(nstates=xx,root=state_of_interest)
> Emission energy is the difference Eex - Eexgr, where
> Eex is the energy of the excited state: print in the string "TD-KS".
> Eexgr is the ground state energy of the excited state geometry: print for
> the string "HF=-".
>
> Here, I attach current CP2K results for pyrimidine structural optimization
> in the first excited state, while accounting the excitation set to include
> three states.
>
> Upon every optimization cycle, the output file reports four energies, for
> example:
> 1) Total energy:
>  -45.00011553431995
>  !--------------------------- Excited State Energy
> ----------------------------!
> 2) Excitation Energy [Hartree]
>  0.1408310694
> 3) Total Energy [Hartree]
>  -44.8592844650
> 4)  ENERGY| Total FORCE_EVAL ( QS ) energy [a.u.]:   -44.859271658539249
>
> Would you comment what are these?
> Why there are three TOTALS?
>
> Next, in the input file I instruct
> &MOTION
>   &GEO_OPT
>     MAX_DR    1.0E-03
>     MAX_FORCE 1.0E-03
>     RMS_DR    1.0E-03
>     RMS_FORCE 1.0E-03
>     OPTIMIZER CG
> but, it seems, the output file does not report on the convergence criteria.
> Does TDDFPT require additional instruction that such data would be printed
> or this is out of the scope of the current package?
>
> Finally, I see that upon the start:
> -                            Excitation analysis
>     -
>
>  -------------------------------------------------------------------------------
>         State             Occupied              Virtual
>  Excitation
>         number             orbital              orbital
>  amplitude
>
>  -------------------------------------------------------------------------------
>              1   3.83221 eV
>                                 15                   16
>  0.995900
>                                 14                   17
> -0.072640
>              2   4.07218 eV
>                                 15                   17
>  0.989483
>                                 14                   16
> -0.140638
>              3   5.17410 eV
>                                 14                   16
> -0.982529
>                                 15                   17
> -0.137943
>                                 15                   18
>  0.080624
>                                 13                   16
> -0.068511
>
>  -------------------------------------------------------------------------------
>
> while later, upon optimization:
> -                            Excitation analysis
>     -
>
>  -------------------------------------------------------------------------------
>         State             Occupied              Virtual
>  Excitation
>         number             orbital              orbital
>  amplitude
>
>  -------------------------------------------------------------------------------
>              1   3.01565 eV
>                                 15                   16
> -0.997677
>              2   3.59856 eV
>                                 15                   17
>  0.989952
>                                 14                   16
> -0.137983
>              3   4.63370 eV
>                                 14                   16
> -0.985780
>                                 15                   17
> -0.135202
>                                 15                   18
> -0.077828
>
>  -------------------------------------------------------------------------------
>
> Since the latter excitation energy  of 3.01565 eV is significantly lower
> than the initial one of 3.83221 eV, should the difference of about 0.8eV
> correspond to the Stokes shift,
> and the 3.01565 eV would be the desired 0'-0 fluorescence energy?
>
> Thank you.
> I hope/wish the users would find my questions valuable.
>
> With best wishes.
> Victor
>
>
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