2
$\begingroup$

I am trying to calculate the transition dipole moments between the 1s inner shell and the high-lying states of N2, namely, $<\psi_{1s}|{\bf r}|\psi_{high-lying}>$.

For this, I installed GAMESS (US) and calculated molecular orbitals of N2.
However, I only got HOMO-6 to LUMO+2.
I want LUMO+3, 4, 5... and if possible, 1s.

How can I do this in GAMESS? Or do I need other softwares?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
1
$\begingroup$

Might have found the option you are looking for. GAMESS controls MO printing with the NPREO keyword. If you can specify ranges of orbitals either by their number (e.g. NPREO(1)=1,10 prints the first 10 orbitals) or relative to the HOMO (e.g. NPREO(1)=-3,-5 prints from HOMO-3 to LUMO+4). You can also give 3rd and 4th numbers to list the range of orbital energies to print.

These print options are described on p.23 of the input descriptions document on the Gordon research group page.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you for your comments. I checked that option but the default value gives all the calculated MOs. Actually, I kind of solved the problem myself. By changing the basis from STO to N311, and added NDFUNC, NPFUNC, DIFFS, DIFFSP options in the $BASIS group, I obtained up to LUMO+62. (although I'm very skeptical of the vailidity of those MOs.) $\endgroup$
    – N. Saito
    Commented Dec 19, 2018 at 13:09
  • $\begingroup$ @N.Saito What makes you sceptical of the MOs? It may be worth asking another question to try to figure out whether something is wrong with the MOs. Just be sure to include the level of theory and basis set you are using in that new question. $\endgroup$
    – Tyberius ♦
    Commented Dec 19, 2018 at 13:53
  • $\begingroup$ The energies of the higher LUMO orbitals are too high. (e.g. LUMO+62 has 42.9 a.u. (1167 eV).) Also, these orbitals looks very localized to each nitrogen atom. I will describe the details in another question. $\endgroup$
    – N. Saito
    Commented Dec 19, 2018 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.