How are molecular orbitals constructed from the coefficients and basis set?

Question

When an SCF calculation is done in a QM program, the MO coefficients of the solution are printed into the output. Now, graphical softwares can plot 3D molecular orbitals from these coefficients, presumably by combining the basis set and the coefficients.

This is the first part of the output of an RHF/MIDI calculation on an $\ce{H2O}$ molecule with GAMESS(US):

          ------------
          EIGENVECTORS
          ------------

                      1          2          3          4          5
                  -20.4191    -1.3388    -0.6850    -0.5393    -0.4837
                     A          A          A          A          A   
    1  O  1  S    0.989154  -0.220275  -0.001184   0.079432  -0.000000
    2  O  1  S    0.065271   0.487564   0.002152  -0.177885  -0.000000
    3  O  1  S   -0.034882   0.438551   0.006441  -0.333617  -0.000000
    4  O  1  X    0.001409   0.036986   0.314950   0.210975   0.164841
    5  O  1  Y   -0.000292  -0.007466  -0.121285  -0.044237   0.484575
    6  O  1  Z    0.002563   0.069802  -0.193015   0.375649  -0.035515
    7  O  1  X   -0.002020   0.037929   0.308448   0.249550   0.207751
    8  O  1  Y    0.000419  -0.007817  -0.118676  -0.052406   0.610715
    9  O  1  Z   -0.003659   0.069385  -0.187599   0.443221  -0.044760
   10  H  2  S    0.000324   0.134487   0.264755   0.153642  -0.000000
   11  H  2  S    0.006519   0.005342   0.148311   0.110818  -0.000000
   12  H  3  S    0.000260   0.138002  -0.267972   0.150940  -0.000000
   13  H  3  S    0.006664   0.004777  -0.147661   0.106685  -0.000000

The basis set is:

 O         

      1   S       1           281.8665800    0.069059895330
      1   S       2            42.4160000    0.393159473412
      1   S       3             9.0956200    0.665669054983

      2   S       4            11.4660300   -0.145056191626
      2   S       5             0.8878600    1.044738808830

      3   S       6             0.2788000    1.000000000000

      4   P       7             8.0472400    0.226466081004
      4   P       8             1.6684200    0.868524024343

      5   P       9             0.3725100    1.000000000000

 H         

      6   S      10             4.5018000    0.156285427224
      6   S      11             0.6814440    0.904690578596

      7   S      12             0.1513980    1.000000000000

 H         

      8   S      13             4.5018000    0.156285427224
      8   S      14             0.6814440    0.904690578596

      9   S      15             0.1513980    1.000000000000

At this point, I don't know how to combine the coefficients with the primitives in the basis set. Clearly the symbols on the left side of the MO coefficients indicate S, p_x, p_y, p_z... etc orbitals. So, according to the output, Oxygen should have 3 S, then 3 P and then 3 P orbitals. But the basis set has 6 S and 3 P functions. So I don't know which basis primitives correspond to which AO name (i.e. s, p_x etc.).

Could someone help me understand how the final MOs are constructued from the coefficients and the basis set?

Answer 1

Things are more clear to me when I look at the basis set in CFOUR format, because of the information at the top which tells me how many primitives and how many contractions there are for each type of orbital:

O:MIDI
Huzinaga MIDI

2
0 1
3 2
6 3

281.86658 42.416000 9.0956200 11.46603 0.887860 0.278800 

0.069060  0.000000 0.00000000 
0.393159  0.000000 0.00000000 
0.665669  0.000000 0.00000000 
0.000000 -0.080820 0.00000000 
0.000000  0.582090 0.00000000 
0.000000  0.000000 1.00000000 

8.047240 1.668420 0.372510 

0.124271 0.00000000 
0.476594 0.00000000 
0.000000 1.00000000 

I'm going to use the % symbol to add comments on the numbers at the top, to explain what they mean (but adding comments like this won't work if you actually did this to your GENBAS file in CFOUR):

2   % # of types of orbitals = 2 (S and P).
0 1 % 0 = S, 1 = P, 2 = D, 3 = F, etc.
3 2 % 3 contractions for S, 2 contractions for Px, 2 for Py, 2 for Pz
6 3 % 6 primitives for S,   3 primitives for Px,   3 for Py, 3 for Pz

This explains why the basis set in GAMESS(US) format is showing 6 primitives for S and 3 primitives for P, but in your final eigenvectors, we now consider the contractions:

• 3 for S
• 2 for Px
• 2 for Py
• 2 for Pz

This is exactly what your output shows!