9
$\begingroup$

I am trying to calculate electronically excited states for the CO molecule, with the aug-cc-pVTZ basis set, and a CASSCF calculation. The documentation describes how I select the number of orbitals in the closed and active spaces for each symmetry (in the C2v point group). But I can't find information about the criteria that is used for selecting which orbitals will be the active ones in each symmetry.

Using the command orbprint,12 I can tell that orbitals are not selected simply in increasing energy order (since some unoccupied orbitals have lower energy than the occupied ones).

How does Molpro pick out orbitals for the active space? And, can I control this process if I am not happy with some particular selection?

A condensed version of my input:

***,CASSCF test

MEMORY,2,G

GEOMETRY = {
  C
  O, C, qco
}

basis = aug-cc-pVTZ

dq = 0.125
DO qco = 1, 10, dq
  IF (qco.EQ.1.5) dq = 0.0625
  IF (qco.EQ.3) dq = 0.125
  IF (qco.EQ.4) dq = 0.25
  IF (qco.EQ.5) dq = 0.5
  IF (qco.EQ.6) dq = 1

  CASSCF
    orbprint,12
    closed,2,0,0,0
    occ,6,3,3,2
    wf,14,1,0
      state,5
  
  ! Storing distance, energy, and tr-dip-mom. in memory [..]

END DO

! Saving all computed values to disk [..]

---

For nuclear distance 2.75, these are eight orbitals in each symmetry. Taken from the output file.

Orbital  Occupation       Energy
 1.1        2.00000    -20.72392
 2.1        2.00000    -11.41840
 3.1        1.99027     -1.38533
 4.1        1.93048     -0.81010
 5.1        1.55554     -0.48414
 6.1        0.07730      0.43260
 7.1       -0.00000      0.06413
 8.1       -0.00000      0.08645
 1.2        1.73314     -0.55388
 2.2        0.54582     -0.05416
 3.2        0.01415      0.95790
 4.2       -0.00000      0.09084
 5.2       -0.00000      0.25216
 6.2       -0.00000      0.48245
 7.2       -0.00000      0.57317
 8.2       -0.00000      0.77930
 1.3        1.55073     -0.50443
 2.3        0.39516     -0.01699
 3.3        0.19777      0.09396
 4.3       -0.00000      0.17954
 5.3       -0.00000      0.37646
 6.3       -0.00000      0.50901
 7.3       -0.00000      0.77746
 8.3       -0.00000      1.00082
 1.4        0.00610      1.10176
 2.4        0.00353      2.22594
 3.4       -0.00000      0.47199
 4.4       -0.00000      1.09344
 5.4       -0.00000      1.21291
 6.4       -0.00000      2.37566
 7.4       -0.00000      3.10900
 8.4       -0.00000      3.49887
$\endgroup$
1
  • $\begingroup$ Let's continue this conversation in chat. $\endgroup$ May 11 at 16:28
3
$\begingroup$

You fixed the occupation within symmetries by the command "occ,6,3,3,2" and asked the program to make CASSCF for the lowest 5 singlet states of symmetry 1 ($A_1$). And indeed, as you see, the first 6, 3, 3, and 2 orbitals in symmetries 1, 2, 3, 4, respectively, have nonzero occupation numbers, which sum up to 6+8=14, the number of electrons of $\ce{CO}$. These orbitals are ordered by energy, but within symmetries, so the criterion is the orbital energy.

There are many options in the CASSCF Molpro program and it is difficult to advise without knowing what you do not like in your results. You can, e.g., run a HF calculation, use these orbitals as starting orbitals and freeze some orbitals in the subsequent CASSCF calculation. Since you have a linear molecule, you probably would like to select orbitals with "extra" symmetries (i.e. differentiate between orbitals, which are e.g. in the same $A_1$ representation in $C_{2v}$, but in reality they are in different representation in $C_{\infty v}$, look for the EXTRA keyword in the manual). You can also restrict CSFs (so you have RAS, not CAS), by the RESTRICT keyword, e.g. to single and double excitations. Another option is to play with the MERGE command and to prepare initial orbitals from atomic orbitals (there is an example for a $\ce{NO}$ molecule in the Molpro manual, i.e. something very similar to your molecule).

$\endgroup$
4
  • 1
    $\begingroup$ Welcome to the site and thank you for a great first answer. We hope to see more of you around. $\endgroup$
    – Tyberius
    May 18 at 17:14
  • $\begingroup$ Thanks @tkorona! But as far as I can tell it is not true that orbitals are ordered by energy within each symmetry. For instance, orbital 6.1 is occupied and has energy 0.43260, while orbital 7.1 is un-occupied but with the lower energy 0.06413. (See the table above that I took from my output file.) $\endgroup$ May 19 at 6:55
  • $\begingroup$ @NikeDattani The first digit (before te dot) is an enumeration of orbitals, and the digit after the dot is the symmetry. Which is one of the four irreducible representations in C2v, where #.1 = A1, #.2 = B1, #.3 = B2, #.4 = A2. $\endgroup$ May 25 at 14:21
  • $\begingroup$ @stack-delay Sorry, I was thinking about D2h which has 8 irreps. Can you please post your output file here in a folder called 4956 (since 4956 is the number in this question's URL)? Let's see more details about the HF orbitals for example. $\endgroup$ May 25 at 15:29
0
$\begingroup$

I believe my question is based on a misconception, so this is an answer according to my current understanding. Please, if you find additional misconceptions, comment below and I will try to update accordingly. Okay, let's go!

When you run a CASSCF calculation, note that the coefficients which are building up the orbitals are optimised at the same time as the CI coefficients. So there are no set of predefined orbitals which the calculations picks from (when creating the CI expansion).

Instead, what's happening is that the orbitals are optimised such that they are the best set available—for the chosen basis—to describe the "electron correlation" (i.e. the difference between the product wave-function of the Slater determinant, and the true eigenstate wave-function).

This procedure does not imply that the "selected" orbitals must be the lowest energy ones. What makes the determination is instead how useful their shapes are for describing the electron correlation. Though it can be said that very high energy orbitals are probably not useful.

$\endgroup$
1
  • $\begingroup$ This might be better off as an addendum to the question, rather than as a separate post in the answers section. $\endgroup$ Jun 13 at 3:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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