Hi
I am calculating a system with water molecules on the surface. The slab is asymmetric and I used IDIPOL=3 and LDIPOL=T.
I notice that if I use :
1) IDIPOL=3
LDIPOL=T
dipol+quadrupol energy correction -0.060601 eV
E0= -.13517174E+04
2) IDIPOL=3
LDIPOL=F
dipol+quadrupol energy correction 0.345834 eV
E0= -.13515147E+04
I am confused about the total energy and which is to be cosidered correct. Since, with LDIPOL=T I understand that correction to the potential and forces is done and it also goes through the self consistent loop.
I have seen the previous posts about the same issue but the links referred there do not work any more.
I will be thankful if someone can share there idea on this issue.
thank you
Narasimham
USe of LDIPOL
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simha
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USe of LDIPOL
Last edited by simha on Wed Aug 21, 2013 4:58 pm, edited 1 time in total.
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support_vasp
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Re: USe of LDIPOL
Hi,
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