I have been learning some basics of molpro and quantum chemistry, however I am deeply confused now.

As a very simple example, I want to compute the energy of a carbon atom using D2h group

For its ground state, if I give this input:

occ,2,1,1,0,0,0,0,0 closed,1,0,0,0,0,0,0,0 wf,6,4,2

The reason why I used this input is that I want to compute a specific electron configuration which has one alpha electron in the 2px orbital and the other alpha electron in 2py

And my understanding for this input text is that:

2 Ag irreps (for 1s and 2s orbital) and 1 B3u (for 2px) and 1 B2u(for 2py), then the total symetry is B3u*B2u=B3g which coresponds to the number 4 in my WF input, both electrons are spin-up so S=2 for WF input

Is this the correct way to do it? Following this thought I changed my input to

occ,2,0,1,0,1,0,0,0 closed,1,0,0,0,0,0,0,0 wf,6,7,2


occ,2,1,0,0,1,0,0,0 closed,1,0,0,0,0,0,0,0 wf,6,6,2

for the configuration where the 2 aplha electrons are in 2py+2pz and 2px+2pz orbitals.

These 3 inputs actually returned me exactly the same energy which makes sense to me if I think in the way that B1g B2g and B3g should be the same for a single atom because it doesn't matter how I pick my x y and z for a sphere.

However then i tried have those 2 electrons paird in just one orbitals, then there will be 3 possible configurations which are just a pair of electrons in one of the three p orbitals. Here I am confused, beacuse if I think like that there is no difference which axis I should call it as my z axis (as in there is no difference of x y and z axis in the case of an atom) then then the px/pz/pz should give me the same value of energy, which it is the case what molpro tells me, but when we write out the term symbol shouldn't that be like when the electrons are both in px or py orbital then the term is 1D, and when the electrons are both in pz then it is a 1S term? According to the energy level of carbon, 1D and 1S should not be have the same energy, so there is a conflict obviously, what did I miss here? I am pretty sure that I was not using the software in a correct way, but what should be the rigorous way of specifying a electron configuration in molpro?

  • 1
    $\begingroup$ You might instead try posting this at matter modeling SE (I can migrate it if you want). $\endgroup$
    – Buck Thorn
    Commented Jul 27, 2022 at 13:19
  • $\begingroup$ I am not very familiar with molpro, but may I ask why you are using symmetry? Can't you do a MCSCF or something like that without any symmetry? $\endgroup$
    – S R Maiti
    Commented Jul 31, 2022 at 16:45
  • $\begingroup$ Hi, @SRMaiti, I am new to the comp chem topic but the story is that I started with Gaussian trying to compute some atomic excitation energy as practice, but I have a very hard time to understand how to confirm which excited state I am look at, for example, some atoms are neat like Lithium becasue I can excite that single valence electron to 2p, 3s or 3p orbitals, but for a more general atom such as carbon the excited electron configuration is more complex, and I have been told that molpro is very robust of those detail controls. $\endgroup$
    Commented Aug 1, 2022 at 18:23
  • $\begingroup$ Sorry for spliting the comment, but I am reaching the word count limit in the previous one. I have been learning about the idea of Landau-Zener theory and trying to use some software to obtain the potential energy curve for a system that has two ions. From what I have read, the two ions can mutually neutralize themselves at a seperation where corresponds to an avoided crossing of the potential energy curves, and there are commonly multiple PECs because that after neutralization the atoms may take different excited states. So I am trying to learn about the use of molpro for this. Thank you! $\endgroup$
    Commented Aug 1, 2022 at 18:30
  • $\begingroup$ By reading this post: (chemistry.stackexchange.com/questions/105090/…) I am not sure why that a <L^2> means it is a 1D1 sate? Also could you please suggest a systematic way to give the occupancy and where to look at in the output file to determine what is the actual computed excited state? $\endgroup$
    Commented Aug 2, 2022 at 3:35


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