DFT Broken-symmetry for molecules with multiple unpaired electrons
Clicked A Few Times
4:35:27 AM PDT - Thu, May 7th 2015
I'm uncertain how to set up the broken-symmetry input for a molecule with multiple unpaired electrons,
i.e. a Fe2S2(SCH3)4 cluster which has got a high spin configuration on both irons.
The input for the high spin state looks like follows:


# Fe2(III) Fe1(III) high spin (s=5/2 alpha) (s=5/2 beta), S=1 

charge -2

dft

 xc pbe0

 odft

 mult 1 

 convergence nolevelshifting

 direct

 grid fine

 iterations 300

# atoms 1-12 and 13-24 comprise a FeS(SCH3)2 subsystem each

 cdft  1 12 spin  5.0

 cdft 13 24 spin -5.0

 vectors input atomic output "highspin.mos"

end 

# the HOMO is orbital #93 for alpha and beta spins.

Do I have to "shift" all of the highest occupied 5 beta MOs up by one orbital and constrain the
charge of this beta subsystem to +1?
Thanks in advance!

Forum Vet
11:32:54 AM PDT - Thu, May 7th 2015
I am not quite sure how to do this with CDFT
However, you can obtain a broken-symmetry solution using the regulat DFT by means of the fragment guess,
by first converging two separate Fe-containing fragment, one fragment with Fe spin-up (a.k.a. excess of alpha electrons), the second with Fe spin-down (a.k.a. excess of beta electrons).
Here is an example that might be close of what you are looking for.

start fe2s2sch34
#geometry from http://dx.doi.org/10.1038/nchem.2041
geometry big
 Fe 5.22  1.05   -7.95
 S  5.00  0.95   -5.66
 S  4.77  3.18   -8.74
 C  6.00  4.34   -8.17
 H  6.46  4.81   -9.01
 H  5.53  5.08   -7.55
 H  6.74  3.82   -7.60
 C  3.33  1.31   -5.18
 H  2.71  0.46   -5.37
 H  3.30  1.54   -4.13
 H  2.97  2.15   -5.73
 Fe 5.88  -1.05   -9.49
 S  6.10  -0.95  -11.79
 S  6.33  -3.18   -8.71
 C  5.10  -4.34   -9.28
 H  5.56  -5.05   -9.93
 H  4.67  -4.84   -8.44
 H  4.34  -3.81   -9.81
 C  7.77  -1.31  -12.27
 H  7.84  -1.35  -13.34
 H  8.42  -0.54  -11.90
 H  8.06  -2.25  -11.86
 S  3.86  -0.28   -9.06
 S  7.23  0.28   -8.38
end
geometry fe1
 Fe 5.22  1.05   -7.95
 S  5.00  0.95   -5.66
 S  4.77  3.18   -8.74
 C  6.00  4.34   -8.17
 H  6.46  4.81   -9.01
 H  5.53  5.08   -7.55
 H  6.74  3.82   -7.60
 C  3.33  1.31   -5.18
 H  2.71  0.46   -5.37
 H  3.30  1.54   -4.13
 H  2.97  2.15   -5.73
end
geometry fe2
 Fe 5.88  -1.05   -9.49
 S  6.10  -0.95  -11.79
 S  6.33  -3.18   -8.71
 C  5.10  -4.34   -9.28
 H  5.56  -5.05   -9.93
 H  4.67  -4.84   -8.44
 H  4.34  -3.81   -9.81
 C  7.77  -1.31  -12.27
 H  7.84  -1.35  -13.34
 H  8.42  -0.54  -11.90
 H  8.06  -2.25  -11.86
end
geometry s2
 S  3.86  -0.28   -9.06
 S  7.23  0.28   -8.38
end

basis spherical
 * library 6-31g*
end

title "first Fe fragment spin up"

charge -1
set geometry fe1
dft
 mulliken
 mult 6
 xc pbe0
 vectors input atomic output fe1.mos
end

task dft ignore

title "second Fe fragment spin down"

charge -1
set geometry fe2
dft
 mult -6
 vectors input atomic output fe2.mos
end

task dft ignore

title "neutral s2 fragment"

charge 0
set geometry s2
 dft
 odft
 mult 1
 vectors input atomic output s2.mos
end

task dft ignore

title " fe2s2sch34"

charge -2
set geometry big
dft
 mult 1
 vectors input fragment fe1.mos fe2.mos s2.mos output big
 maxiter 99
end

task dft

Clicked A Few Times
12:37:55 AM PDT - Fri, May 8th 2015
Thanks! Seems this kind of [2Fe2S] cluster is a popular problem.
As for the mixed valence state Fe(III)Fe(II) with a total charge of -3:
would the Fe(II) subsystem be defined as charge 0 and mult -5?

Forum Vet
10:28:06 AM PDT - Fri, May 8th 2015
The Fe(II) fragment should have charge -2 and mult -5.
The sum of the charge of the fragments must be equal to the total charge of the aggregate molecule

Forum Vet
1:29:00 PM PDT - Wed, Jul 29th 2015
I have just discovered that the CDFT implementation had a pretty problematic bug since version 6.3.
I have found the fix for it, and the patch for 6.5 is available at

http://nwchemgit.github.io/images/Cdft.patch.gz


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