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Broken-symmetry HF and DFT.

Pavlo Solntsev

Dear FF users.

Could you share your experience with broken-symmetry calculation using HF and DFT? I have an example below. If i use DFT i have Sz=0, S^2 = 0 for broken-symmetry singlet. If i use HF i have Sz=0 S^2 = 1.8. Which is reasonable. Does it mean, mix=.t. works only for HF? I know, we can remove switch symmetry completely off (nosym=1), and in this case i got S^2 = 0.8 for DFT which is also acceptable. I confused why we can use full orbital symmetry within HF, but not for DFT.


$CONTRL RUNTYP=energy dfttyp=bp86  $END
$BASIS GBASIS=n31 ngauss=6  $END
$BASIS ndfunc=1 $END
$guess mix=.t. $end
[Cu2F6]^-2 BS singlet
Dnh 2

CU         29.0   1.5298305586   0.0000000000   0.0000000000
F           9.0   2.7561841734   1.3744839332   0.0000000000
O           8.0   0.0000000000   1.1934572146   0.0000000000
H           1.0   0.0000000000   2.1349088145   0.0000000000

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