<div dir="ltr"><div>Dear Fr. Hutter: </div><div>good evening.</div><div>Thank you for the kind explanations.</div><div>With best regards,</div><div>Victor </div></div> <div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Wed, Nov 30, 2022 at 4:47 PM Jürg Hutter <<a href="mailto:hutter@chem.uzh.ch">hutter@chem.uzh.ch</a>> wrote: </div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">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: <a href="mailto:cp2k@googlegroups.com" target="_blank">cp2k@googlegroups.com</a> <<a href="mailto:cp2k@googlegroups.com" target="_blank">cp2k@googlegroups.com</a>> on behalf of Victor Volkov <<a href="mailto:volkovskr@gmail.com" target="_blank">volkovskr@gmail.com</a>> 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 -- You received this message because you are subscribed to the Google Groups "cp2k" group. To unsubscribe from this group and stop receiving emails from it, send an email to <a href="mailto:cp2k%2Bunsubscribe@googlegroups.com" target="_blank">cp2k+unsubscribe@googlegroups.com</a><mailto:<a href="mailto:cp2k%2Bunsubscribe@googlegroups.com" target="_blank">cp2k+unsubscribe@googlegroups.com</a>>. 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