Wilson-Amos-Handy method for calculation of NMR shifts
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Bert [QUOTE=Ohlincha Nov 16th 5:07 am]Bert,I don't see much difference between the original 6.1 and my modified 6.1 code. The number of iterations are the same, and while there's a small change in the computed shift, it's small relative what should be obtained if the WIlson-Amos-Handy method has been successfully implemented (isotropic shifts -- C: 4.7 ppm and O: -44 ppm; see first post for references) Here's my test case title "co nmr"
echo
geometry
c 0 0 0
o 0 0 1.13
end
basis
* library "dzp (dunning)"
end
property
shielding
end
dft
direct
grid fine
mult 1
xc HFexch 0.05 slater 0.95 becke88 nonlocal 0.72 vwn_5 1 perdew91 0.81
end
task dft property
Running it using my modified binary (v6.1) I get: 600 iter nsub residual time 601 ---- ------ -------- --------- 602 1 3 1.70D-01 4.9 603 2 6 1.70D-03 5.6 604 3 9 2.93D-05 6.3 For C I get 621 isotropic = -2.8388$ 622 anisotropy = 413.3050$ and for O I get 650 isotropic = -64.1336$ 651 anisotropy = 711.1241$ and with the original binary I see 600 iter nsub residual time$ 601 ---- ------ -------- ---------$ 602 1 3 1.69D-01 5.0$ 603 2 6 1.84D-03 5.8$ 604 3 9 2.47D-05 6.5$ For C I get 621 isotropic = 5.1483$ 622 anisotropy = 401.3242$ and for O I get 650 isotropic = -60.6964$ 651 anisotropy = 705.9682$ when I do cat *|grep "xfac =" I can't find any unedited examples -- i.e. I've already changed them all to 0 and have commented out "if (use_theory.eq.'dft') xfac = bgj_kfac()" I guess the most plausible explanation would be that there's more to these NMR calculations than it at first appears to be -- but then again, as you point out there should be a change in the number of iterations and there isn't.
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