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Re^9: How to check for acceptance of Kasha's rule from TDDFT data...?

Siddheshwar Chopra
sidhusai@gmail.com

Dear Ilya,
Again my thanks to you for this information.

"First, you need to decide which excited states are responsible for fluorescence. If the system was initially excited to S1 then the answer is more or less unambiguous. If it was excited to some higher excited state, make an educated guess whether the Kasha's rule holds. The lower is the gap between Sn and S1, the higher is the probability that it does"

Ilya I use ISTATE=1, that means first singlet excited state. Again I don't know how can I comment on "HOW LOWER IS THE GAP between Sn and S1"?

"As for the emission wavelength, for fluorescence from Sn you need, indeed, to subtract the DFT energy of S0 from the TDDFT energy of Sn."
You mean to say the Optimized S0 state DFT energy and the TDDFT energy of Sn?? I am unsure of this.

"In fact, the result is already printed in the output, see "TDDFT EXCITATION ENERGIES"
Ilya please read the following o/p I get from the Emission calc. by searching for "TDDFT EXCITATION ENERGIES" (only first few lines):

------------------------------------------------------------------------------
TDDFT EXCITATION ENERGIES
STATE       HARTREE        EV      KCAL/MOL       CM-1   NANOMETERS  OSC. STR.
------------------------------------------------------------------------------
1A      0.0170083214    0.4628     10.6729      3732.90    2678.89  0.0044801
1A      0.0607884179    1.6541     38.1453     13341.52     749.54  0.0091386
1A      0.0631347976    1.7180     39.6177     13856.49     721.68  0.0289015
1A      0.0740760156    2.0157     46.4834     16257.81     615.09  0.0035655
1A      0.0798165815    2.1719     50.0857     17517.71     570.85  0.0464085
1A      0.0919979731    2.5034     57.7296     20191.22     495.26  0.1471384
1A      0.0959600579    2.6112     60.2159     21060.80     474.82  0.0897933
1A      0.1034845547    2.8160     64.9375     22712.23     440.29  0.0700157
1A      0.1054237637    2.8687     66.1544     23137.84     432.19  0.1377923
1A      0.1142962652    3.1102     71.7220     25085.13     398.64  0.0004947
1A      0.1156280862    3.1464     72.5577     25377.43     394.05  0.0041491
1A      0.1179649259    3.2100     74.0241     25890.31     386.24  0.0868358
1A      0.1243301305    3.3832     78.0183     27287.31     366.47  0.0759754
1A      0.1307953962    3.5591     82.0754     28706.27     348.36  0.0074417
1A      0.1308869697    3.5616     82.1328     28726.37     348.11  0.0136588
1A      0.1339895192    3.6460     84.0797     29407.30     340.05  0.0305930
1A      0.1361664704    3.7053     85.4458     29885.09     334.62  0.0080487
1A      0.1383482109    3.7646     86.8148     30363.92     329.34  0.0474317
1A      0.1394772519    3.7954     87.5233     30611.72     326.67  0.1485271

It is actually the emission spectra... The first value is what you must be pointing at..Right (which actually corresponds to the S1)?? My problem is that the emission wavelength must come around 620-700nm approximately... I am unable to get this...Is it that higher excited states are more probable (fall down to S1) rapidly? Should I look for their osc. str. and then discuss?
Can you help me in resolving this? If you need any o/p files..I will provide you on your mail id.

Kind Regards,

On Sat Sep 27 '14 10:25pm, Ilya Ioffe wrote
-------------------------------------------
>Dear Siddheshwar,

>First, you need to decide which excited states are responsible for fluorescence. If the system was initially excited to S1 then the answer is more or less unambiguous. If it was excited to some higher excited state, make an educated guess whether the Kasha's rule holds. The lower is the gap between Sn and S1, the higher is the probability that it does, i.e. that Sn rapidly decays into S1 and one observes fluorescence from S1. In fact, there are theoretical approaches to estimate the spontaneous transition rates between the states but that would be much more complex and we'd better skip that.
>As for the emission wavelength, for fluorescence from Sn you need, indeed, to subtract the DFT energy of S0 from the TDDFT energy of Sn. In fact, the result is already printed in the output, see "TDDFT EXCITATION ENERGIES".The HOMO-LUMO gap can be an inappropriate approximation for the S0-S1 transition and you've done TDDFT anyway. Note that TDDFT energies are not ideal either. You are advised to learn about the limitations of TDDFT from the appropriate sources, this discussion board is not a very suitable medium for that.

>Best regards
>
>
>On Sat Sep 27 '14 8:36am, Siddheshwar Chopra wrote
>--------------------------------------------------
>>Dear Ilya,
>>I think I have finally started understanding now. I must really thank you
>>for helping me.
>>So now I understand that the excited states mentioned in the o/p are the same... Sn states.. like we get in abdorption calcs.. The only difference in emission calc. is that we start with optimized S1 geometry here.
>>Now could you help me how to analyse the emission spectra from the EXCITED STATES information? As Pavlo sir mentioned that it is not mandatory for kasha's rule to hold good, please explain how do we discuss with respect to these excited states.
>>Also how to get the value of emission wavelength? Is it just the difference between homo, lumo values? or I need to subtract the energy values of relaxed Sn and S0?

>>Kind Regards,
>>
>>
>>On Fri Sep 26 '14 8:36pm, Ilya Ioffe wrote
>>------------------------------------------
>>>Dear Siddheshwar,

>>>In TDDFT, the reference is always the ground state, irrespective of the state you optimize. In other words, TDDFT makes possible to minimize energy of the excited state with respect to geometric variations but electronically you still optimize the ground state by means of DFT and then compute the excitations from it in the linear response approximation. Accordingly, the output contains data for the transitions between the ground electronic state, S0, and the excited electronic states, S1 to S_NSTATE (same osc. str. for absorption and stimulated emission). There are no vibrational levels, that's a completely different task. For absorption spectra optimize S0, for fluorescence from Sn optimize Sn.
>>>
>>>
>>>
>>>
>>>On Fri Sep 26 '14 4:49pm, Siddheshwar Chopra wrote
>>>--------------------------------------------------
>>>>Dear Ilya,
>>>>Thats really nice of you to explain in such simple words. Ilya I understand the point 1 you said about kasha's rule. I also understood that osc. strs. are reported between S0 and excited states.
>>>>My doubt is that:: Say I have the optimized geometry of S1 now.. Now Iwant the emission spectra. So in this case, the osc. strs. etc. printed in the output file will be for S1 and excited states(S0 vibr. levels) OR will be for S0 and excited states (S1 vibr. levels)??
>>>>I hope you got my confusion well...

>>>>Kind Regards,
>>>>
>>>>
>>>>On Fri Sep 26 '14 2:40pm, Ilya Ioffe wrote
>>>>------------------------------------------
>>>>>Dear Siddheshwar,

>>>>>1) Kasha's rule is of kinetic nature. It suggests that radiationless relaxation from Sn to S1 (with further radiative decay to S0) is normally of higher probability than fluorescence directly from Sn. But nothing requires S1 to have higher oscillator strength.

>>>>>2) ISTATE=x means that you are requesting properties for the excited state no. x and/or that you are optimizing it. The excitation energies and oscillator strengths are always reported between S0 and each of the excited states requested by NSTATE, not between the excited states (you would need a multiconfiguration method rather than TDDFT to compute that). If you'd like to calculate fluorescence energy from a particular relaxed Sx, optimize this very Sx.

>>>>>Best regards,

>>>>>Ilya
>>>>>
>>>>>
>>>>>
>>>>>On Thu Sep 25 '14 2:18pm, Siddheshwar Chopra wrote
>>>>>--------------------------------------------------
>>>>>>Dear Pavlo,

>>>>>>Please have a look at the energies for 3 excited states from absorption and emission runs:
>>>>>>EMISSION CALCULATION:: ISTATE=1, NSTATE=3

>>>>>> EXCITED STATE   1  ENERGY=     -1355.7217994424  S = 0.0  SPACE SYM = A

>>>>>> EXCITED STATE   2  ENERGY=     -1355.6780193460  S = 0.0  SPACE SYM = A

>>>>>> EXCITED STATE   3  ENERGY=     -1355.6756729663  S = 0.0  SPACE SYM = A
>>>>>>
>>>>>>
>>>>>>
>>>>>>UV CALCULATION:::: ISTATE=1, NSTATE=3

>>>>>> EXCITED STATE   1  ENERGY=     -1355.7080939866  S = 0.0  SPACE SYM = A

>>>>>> EXCITED STATE   2  ENERGY=     -1355.6840777343  S = 0.0  SPACE SYM = A

>>>>>> EXCITED STATE   3  ENERGY=     -1355.6782292662  S = 0.0  SPACE SYM = A

>>>>>>
>>>>>> IN ALL THE TRANSITIONS... EXCITED STATE 1 HAS THE LOWEST ENERGY...BE IT ABSORPTION OR EMISSION...So for emission, if the initial state is S1,
>>>>>>then what are these excited states? Ideally speaking in this run, S1 is the ground state. The excited state 1 has the lowest energy (-1355.7217994424). Then it should be the lowest vibrational level of S0.

>>>>>>Kind Regards,

>>>>>>On Thu Sep 25 '14 6:47am, Siddheshwar Chopra wrote
>>>>>>--------------------------------------------------
>>>>>>>Dear Pavlo,
>>>>>>>That is some real relief... Actually the same under study is a doped one. Sir please read my last message and answer them. The picture of Emission process is unclear to me, with regard to the firefly calculations. The ISSTATE=1 is ENSURING that we are finding a transitions from S1 only..Is it correct?
>>>>>>>So now SPECIFICALLY about the emission calculations::: What are these EXCITED STATE 1, 2 and so on (depending on NSTATE vaue)? I mean are these the S0 vibrational levels now (because now we are studying de-excitation)? If yes, then how exactly do I find the emission wavelength? Is the analysis similar to the absorption spectra analysis then? I mean using osc. str. etc.? Please guide me.
>>>>>>>Also if Kasha's rule holds good, then do I get a high osc. str. at 1st excited state itself in the o/p file?

>>>>>>>Regards,

>>>>>>>On Thu Sep 25 '14 1:42am, Pavlo Solntsev wrote
>>>>>>>----------------------------------------------
>>>>>>>>Kasha's rule doesn't always work. For example, if you have redox active group near chromophore.

>>>>>>>>Try to figure out absorbtion--emission situation for a porphyrin.

>>>>>>>>-Pavlo.
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>On Wed Sep 24 '14 9:11pm, Siddheshwar Chopra wrote
>>>>>>>>--------------------------------------------------
>>>>>>>>>Dear users,
>>>>>>>>>Somebody please help me in understanding this concept. I need to know that by setting istate =1 in emission spectra calculation, I am actually setting or calculating the de-excitation from the S1 state. Am I correct? If yes, then the TDDFT o/p gives me a set of transitions possible for every excited state (set by variable nstate).What are these nstates now? Do they hold any significance in S1-->S0 transition? Do we mean that these are the vibrational levels of S0 level? Please explain this in the light of kasha's rule.

>>>>>>>>>Kind Regards,
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>n Tue Sep 23 '14 12:21pm, Siddheshwar Chopra wrote
>>>>>>>>>---------------------------------------------------
>>>>>>>>>>Dear Users,
>>>>>>>>>>Please have a look at the attached file which contains a table describing the contributions/transitions from first four states only for both the absorption and emission calculations. For absorption, I understand that the final state can be any of the excited states which has highest osc. strength. Am I right? But my doubt is regarding kasha's rule... If you see the emission data, the first excited state has a very low oscillator strength of 0.004 (HOMO-->LUMO(97‰ at 2678.9 nm. However the higher states have a better contribution than this state. So please tell me how to analyse this properly. To my knowledge emission wavelength is around 600-730nm.
>>>>>>>>>>I would be grateful for any help.

>>>>>>>>>>Kind Regards,
>>
>>
>>

Mon Sep 29 '14 11:41am