Quote:Acorreya Dec 10th 12:52 amWow! Thanks! it did work! I'm clueless of the magic numbers 63 and 58, though.. What should I do to make a bigger crystal? think I have a lot to learn!
I have extracted the magic number 63 and 58 after analyzing the following ouput (generated without the 63-58 swap line)
HOMO = -0.047524
LUMO = -0.047524
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Warning - the HOMO and LUMO are degenerate and you are using symmetry.
This can lead to non-variational energies and poor convergence.
Modify the initial guess, or use an open-shell wavefunction, or turn
off symmetry.
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Symmetry analysis of molecular orbitals - initial
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Numbering of irreducible representations:
1 a1 2 a2 3 e 4 t1 5 t2
Orbital symmetries:
1 a1 2 t2 3 t2 4 t2 5 a1
6 t2 7 t2 8 t2 9 a1 10 t2
11 t2 12 t2 13 t2 14 t2 15 t2
16 t1 17 t1 18 t1 19 e 20 e
21 a1 22 t2 23 t2 24 t2 25 a1
26 t2 27 t2 28 t2 29 a1 30 t2
31 t2 32 t2 33 e 34 e 35 t2
36 t2 37 t2 38 t1 39 t1 40 t1
41 a1 42 t2 43 t2 44 t2 45 a1
46 t2 47 t2 48 t2 49 e 50 e
51 t1 52 t1 53 t1 54 t2 55 t2
56 t2 57 a1 58 t2 59 t2 60 t2
61 e 62 e 63 a1 64 e 65 e
66 t2 67 t2 68 t2
Starting SCF solution at 3.5s
Things to notice in the output:
1) The Warning line tells you that the HOMO and LUMO molecular orbitals are degenerate because of symmetry
2) Next, you need to browse through the "Orbital symmetries" analysis. You have 58 filled orbitals, therefore you need to find -- among the unoccupied orbitals (i.e. the ones number 59 and higher) -- an orbital that avoids the symmetry breaking (i.e. one electron being shared between HOMO and LUMO of the same symmetry). 63 is the first orbital -- with symmetry a1 -- that fits the description
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