Pavlo Solntsev
pavlo.solntsev@gmail.com
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.
Pavel.
$SYSTEM MWORDS=50 $END
$CONTRL RUNTYP=energy dfttyp=bp86 $END
$CONTRL SCFTYP=UHF $END
$CONTRL MAXIT=100 $END
$CONTRL ICHARG=-2 MULT=1 $END
$BASIS GBASIS=n31 ngauss=6 $END
$BASIS ndfunc=1 $END
$guess mix=.t. $end
$DATA
[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
$end