# How do I visualise excited state MOs?

I did an excited state calculation (TD-SCF)in Gaussian and wanted to see the MOs of the first excited state. But when I opened the checkpoint file in GaussView5 and clicked edit -> MOs, I could only see the configuration and MOs of the ground state. How do I see the excited state MOs in GaussView or other visualisers that can be downloaded for free?

• Please let it open, as in my eyes there need some misunderstandings be cleared. Apr 19 '18 at 21:19
• I have overruled the community decision of closing; if any of the close-voters are unhappy about that, please open a question on Chemistry Meta. To the original poster I do however strongly recommend to add more context to the question. In any case it is a good idea to specify the versions of the programs you are using. It would also be helpful to know what route section you use. (cc @ph13) Apr 20 '18 at 7:34
• Hm. I am not a Gaussian user but this is almost certainly because one does not need the time-dependent density to calculate the time-dependent energy. Adding density=current should do it. Apr 20 '18 at 21:56
• @pentavalentcarbon That keyword doesn't work. Apr 22 '18 at 21:03

What you see, when you open your checkpoint file, is all you can see. I assume you use TDDFT and with this method you can't obtain other orbitals than the ones from the ground state. Why is this so?

Because in LR-TDDFT you calculate your ground state and apply a perturbation to it. (You can look up on the behind theory in good books like Introduction to Computational Chemistry from Frank Jensen.) The result from this calculation is presented from Gaussian like this:

Excited State   1:      Singlet-A      1.0662 eV 1162.87 nm  f=0.0007  <S**2>=0.000
110 ->113         0.60155  (0.72)
110 ->115        -0.23139  (0.11)
110 ->116         0.22569  (0.10)
Excited State   2:      Singlet-A      1.2549 eV  988.02 nm  f=0.0002  <S**2>=0.000
109 ->113        -0.39288  (0.31)
110 ->114         0.51726  (0.54)
Excited State   3:      Singlet-A      1.2929 eV  958.99 nm  f=0.0027  <S**2>=0.000
109 ->113         0.50172  (0.50)
110 ->114         0.29522  (0.17)
(The weights in spaces are added by me.)


These are the configurations of three excited states. The first excited state, for example, consists mainly (72%) from an excitation from MO 110 (HOMO) to MO 113 (LUMO+2). There are some contributions with weights of about 10% into the MOs 115 and 116. (On how to get the weights, have a look at an answer by me on this site here.)

You ask, how you can see the excited state in GaussView. You can't ... directly. And this question is also a little bit tricky, because what is it that you want to see? Here are some options:

You could

• calculate the electron density of the excited state TD(Root=1) Density(Current) and subtract the electron density from the ground state. Then this difference shows you the "hole" where now is less density and the "hole" where there is more density.

• calculate Charge Density Differences (CDD) with Multiwfn or Electronic Difference Density Maps (EDDM) with GaussSum which show you the same but a little bit easier as you do not need an additional calculation because they are based on the excitations and the CI coefficients.

• calculate the transition density for the excited state TD(Root=1) Transition=1

• calculate Natural Transition Orbitals with, e.g., Gaussian or Multiwfn, for very diffuse excitated states as NTOs reduces/optimizes the number of contributing excitations for each root with Density=(Transition=1) Pop=(Minimal,NTO,SaveNTO)

Just have a look on the manual for TDDFT, Density, FAQ and Pop.

• my "Pop"-Link is wrong, I'll change it when I extend my answer soon Apr 22 '18 at 10:36
• What I want to see is the MO of the excited state. In your example, in the first excited state, an electron transfers from MO 110 to MO 113, so I hope to see MO 110 and MO 113. Of course I can see the ground state version of them in GaussView, but I can't see the first excited state version of the two MOs. Apr 22 '18 at 20:45
• That is what I said, you will not get different orbitals. If you optimize your geometry for this excited state, you'll get the orbitals you want. But still you would have to look at the "ground state" orbitals in the geometry of the optimized excited state. Apr 22 '18 at 22:31
• Ok thanks. I'm not trying to visualise it any more. By the way, I don't think the weights you added are useful. We just look at the coefficients and they're helpful enough. Apr 23 '18 at 18:26