Lamda diagnostic

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kaushikhatua
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Affiliation: IIEST India
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Lamda diagnostic

Post by kaushikhatua » 04 Feb 2016, 07:56

Dear all
In some tddft single residue calculation I got the excited state's as well the lamda diagnostic. In the original paper 'i' resemble occupied orbital and 'a' or represent virtual orbital. However I don't get the right orbital pairs of electronic transition. I have attached the image taken from my computer. Say I=189 and a=86 but molecule has only 40 electrons. Any clarification would be appreciable.

taylor
Posts: 589
Joined: 15 Oct 2013, 05:37
First name(s): Peter
Middle name(s): Robert
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Affiliation: Tianjin University
Country: China

Re: Lamda diagnostic

Post by taylor » 04 Feb 2016, 08:33

Actually, you haven't attached anything, or at least nothing that shows up on your forum posting.

Do not post screen scrapes (I assume that's what you mean by "image from your computer"). Upload your output file and then anyone who wants to assist can try to do so.

Best regards
Pete

kaushikhatua
Posts: 35
Joined: 10 Jan 2014, 07:05
First name(s): kaushik
Last name(s): hatua
Affiliation: IIEST India
Country: India

Re: Lamda diagnostic

Post by kaushikhatua » 04 Feb 2016, 10:04

Extremely sorry for that. Please find the attachment.
Attachments
B-B7-TDA.out
(85.38 KiB) Downloaded 206 times

taylor
Posts: 589
Joined: 15 Oct 2013, 05:37
First name(s): Peter
Middle name(s): Robert
Last name(s): Taylor
Affiliation: Tianjin University
Country: China

Re: Lamda diagnostic

Post by taylor » 04 Feb 2016, 11:31

This is straightforward. The numbering used for MOs is continuous across symmetries. That is, all MOs (occupied followed by virtuals) in the first symmetry, then all MOs (again, occupied followed by virtuals) in the second symmetry, etc. To identify the I index 189, you need to subtract the number of orbitals in all preceding symmetries. The number of MOs in each symmetry is printed in the output and (assuming no MOs are deleted because of linear dependence, or user input) is the same as the number of basis functions listed in the output from the integral calculation.

Several further points. You are running 2013.4 and should upgrade to the latest version. We cannot support arbitrarily many versions and you will find it increasingly difficult to get support here. You are running 64-bit integers, why? This is not recommended unless you are looking to access more than 16GB (2 "gigawords": the 32-bit Fortran limit) of memory per process. You are allowing the program to decide symmetry and thus orientation --- can you be certain that you are obtaining the irreps that you think you are? Note that the labels A1, B1, etc., that the program generates are not guaranteed to be consistent with the Mulliken recommendation (A1 and B2 symmetric with respect to the plane of the molecule for planar C2v), although in this case that's what you obtain. You would be much better advised to specify the symmetry yourself. There are forum postings on this and it is documented in detail in the manual.

More philosophically, the basis set you are using is best regarded as sub-standard (I was tempted to use my usual term "rubbish"). While it is augmented with polarization sets and a set of diffuse functions, the notion that this can give quantitative results for anything is, I am afraid, fantasy. And perhaps more importantly, it is even more of a fantasy that twenty roots of the response equations are physically meaningful. I have posted on this so many times on the forum that I refuse to type it all in again. But calculating twenty roots using linear response with almost any basis set and method is simply wasting computer time. At least until you can establish that there are no Rydberg-like states (which your basis is not capable of describing, even with a diffuse set), nor multiply-excited states in the real-world excited-state manifold. The fact that you are running a dianion makes all of this even worse, I would guess, because the diffuse functions are already needed to describe even the ground state.

Best regards
Pete

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