Dear Soumen,

for closed shell systems, the percentage contribution of a configuration to the resulting excited state TD-DFT wavefunction is generally obtained as 100 x C x C x 2, where C is the coefficient printed by Gaussian. However, this does not work for open shell systems using the unrestricted formalism, TD-UDFT. This is probably because the TD-UDFT results are biorthonormal and Gaussian does not give normalized coefficients in this case. As a consequence, the coefficients are sometimes are larger than unity.

Yopurs, Jens >--

try Gaussum or S-Wizard program for extracting data from Gaussian output file.

I hope this could be helpful

CCL: Summary TD-DFT

http://www.ccl.net/chemistry/resources/messages/2002/03/28.001-dir/

If your aim is to figure out the important transitions, lower excitation energy and higher oscillator strength might help. Moreover, G09 prints the output in decreasing order of transition prob.

100 x twice the square of the CI coefficient (printed by G09 after the MO pair). G09 will print pairs of MO with CI coefficients above a threshold, so the sum over contributions from pairs of MO printer will be < 1.

This is part of my output,

Excited State 3: 6.013-A 0.3900 eV 3179.48 nm f=0.1045 =8.790

291A -> 297A 0.18396

294A -> 297A 0.10363

296A -> 297A 1.09475

296A 297A coefficient is more than 1.Even square of the coefficient multiplied by 2 many time giving percentage more than 100 for a single excitation. What is the solution for these problems?

Ghosh//

Actually, i don't know what kind of calculation do you perform. But i can tell you one thing in your result. You must check the spin state in your calculation. The S2 = 8.790

is Quartet or more than something .. If you send me the output and input, i can give a comment. Also i need the information about spin state and charge.

usually means spin contamination. Since TDDFT uses only unrestricted methods , you'll always have contamination, because you are dealing with unpaired electrons. It shows actually, weather your results are reliable or not. If u have high spin contamination , your TDDFT results are not reliable.From quantum mechanics, =s(s+1), where s-overall spin. For example for triplet we have two unpaired electrons, so s=1 and s(s+1)=2, thus, =2 is normal contamination for triplet. They are also reliable in the range of +10% and -10% from 2.

Dear Alexander,

Spin for my system is S=5/2, so should be 8.75. So, the printed results is within the error range you mentioned.

There is more to this. The mixing of states is a problem if the MOs are mixed too. So try (localized?) natural transition orbitals for interpretation. They base on Natrual Orbitals and are more localized. In this regard even a highly mixed transition may reduce to a "single" excitation between two NOs. (see Martin, R. L. J. Chem. Phys. 2003, 118, 4775−4777, example of application see: dx.doi.org/10.1021/om300710x )

Hi Soumen,

Please take a look at the procedure published in the supporting information (page S14) of the following paper for proper procedure.

http://pubs.acs.org/doi/abs/10.1021/ja303592q?mi=11ve3e8&af=R&pageSize=20&searchText=aromatic

Carlos ///

Thanks a lot for the reference.

You are welcome.

hello.... I'm attaching my output here .I need to know a similar thing! I can get excitation energy and oscillator strength direct from gaussian03 output. they are giving the orbitals and their coefficient also for each transition. but i want to calculate the %homo-lumo or % MLCT associated to a particular transition, say S0-S1. how to do it? please help. and also tell me how to calculate charge density differnce for a specific transition such as S0-S1?part of my output is here.

Excitation energies and oscillator strengths:

Excited State 1: Triplet-A 4.8202 eV 257.22 nm f=0.0000

19 -> 22 0.26242

20 -> 21 0.12370

20 -> 22 0.65422

This state for optimization and/or second-order correction.

Copying the excited state density for this state as the 1-particle RhoCI density.

Excited State 2: Singlet-A 5.2395 eV 236.63 nm f=0.0003

19 -> 22 0.17272

20 -> 21 0.16191

20 -> 22 0.65079

Excited State 3: Triplet-A 5.3249 eV 232.84 nm f=0.0000

20 -> 21 0.67143

20 -> 22 -0.11723

20 -> 23 0.15922

Dear Soumen and Rituparna, both of you can use some software to calculate the percentage contribution of a particular excitation from TDDFT. You can use SWizard or VMOdes for this purpose.

are they free? and how to calculate the electron density difference S0-S1

through gaussview?

Hi sudipta, can you tell me about how to get difference in electron density for a particular transition such as S0-S1 through gaussview

?

you may plot the electron density of molecules in any particular state using Gaussian/GaussView.

U mean any output file for that particular transition?

output file related to optimized molecular structure in state S0, S1, ...

Rituparna, I think you got the answer from Mohammad Alam's post. And the software are freely available.

alam, one file i gave that is ground state S0. another one is with cis=(nstate=1, singlet) or (cis=nstate=1, triplet) should i subtract both the cubes generated from these files and get the difference densities?

I usually don;t optimise excited state..just take ground state geometry and do single point

In total electron density plot of both case (ground and 1st excited state), you can read the value of electron density at different sites of molecules. I dont know how to subtract both cubes.

Dear Soumen,

for closed shell systems, the percentage contribution of a configuration to the resulting excited state TD-DFT wavefunction is generally obtained as 100 x C x C x 2, where C is the coefficient printed by Gaussian. However, this does not work for open shell systems using the unrestricted formalism, TD-UDFT. This is probably because the TD-UDFT results are biorthonormal and Gaussian does not give normalized coefficients in this case. As a consequence, the coefficients are sometimes are larger than unity.

Yopurs, Jens >--

In response to the comment of Soumen Ghosh:

Did you notice this?

296A -> 297A 1.09475

296A

yes, this part is quite common for my system. I think this indicates that electron can easily move between these two orbitals. Am I right?

Then, why TD-DFT and not a wavefunction method that includes static correlation?

My aim was to see how well TD-DFT is working for these kind of systems. Definitely any WFT would be the best way to treat this.

I had the same problem and in my opinion (although I am not 100% sure), the solution lies in the fact, that in a TDDFT framework, the norm is not defined exactly as in a CIS expansion, where the sum over all coefficients squared gives one (or 0.5 in gaussian), since there are also deexcitations, which are not present in the CIS. In fact the sum over all the coefficients of the excitations squared MINUS the sum over all the coefficients of the deexcitations squared is normalized to one: (O,O)=XX-YY=1, see Annu. Rev. Phys. Chem. 2012. 63:287–323 page 291. That means that the coefficient of a certain excitation has to be bigger than one, if it dominates the state and if the corresponding deexcitation has a significant contribution (as in the example discussed above).

Yes, Christof is right. What is usually implemented are the so-called Casida equations (random-phase approximation - RPA):

/ A B \ / X\ = w / 1 0 \ /X\

\ B* A*/ \Y/ ~~ \ 0 -1 / \Y/

where A and B are matrices, X and Y are vectors of coefficients for excitations (X) and deexcitations (Y). If we forget about the coupling part B, we obtain a simpler CIS-like, or Tamm-Dancoff approximation

A X = w X

(~~ are just put to keep space - the RG editor deletes empty spaces:-)

Excited State 1: Singlet-B2 2.7809 eV 445.84 nm f=2.6135 =0.000

146 ->151 -0.10868

147 ->150 0.23132

148 ->149 0.64554

This is in my excited state case. I could able to calculate percentage of contribution for each transition. Among these three transitions, which one i have to consider???

Dear Mr. Ghosh,

Why don't you use the Swizard software to do this?

Your major contribution to that band is HOMO (148) to LUMO (149) transition, and the contribution is: 2x0.64554x0.64554.

Ha, just realized that some questions are posted, and years later, comes the answer. :). But anyway, it may be helpful for others.

Soumen and Rituparna you can use gausssum for calculating percentage contribution its very useful .. please try it

http://gausssum.sourceforge.net/

HOW ABOUT natural transition orbital？

“If it is a closed shell system then square of the coefficient multiplied by 2*100 is the percentage contribution. But for open shell system you have to normalize the coefficient .In that case square of the coefficient divided by sum of the square of the all coefficient for that excited state multiplied by 100 is the percentage contribution.”

 above treatment is obviously wrong!!!

Did you check this link?

http://nbo6.chem.wisc.edu/tut_cmo.htm