[Pw_forum] compatibility between pseudopotential choice and input_dft token

Giuseppe Mattioli giuseppe.mattioli at ism.cnr.it
Mon Apr 8 17:01:33 CEST 2013


Dear Cristian
As usual, what is not suggested should be tested...:-) AFAIK you should do not too much harm if you use a PBE pseudopotential in a input_dft='vdwdf' 
calculation. If you are able to generate a rPBE pseudopotential, you may like to check differences on the same systems used in the related papers 
(graphite, ...).
Yours
G. 

On Monday 08 April 2013 15:55:42 Cristian Degli Esposti Boschi wrote:
>  >> Or a GGA (say C.pbe-mt_fhi.UPF in my case)
>  > 
>  > All the vdW corrections (semiempirical DFTD2 included, london=.true.,
>  > in &system) are well tested with a given kind of xc functional. This
>  > is often discussed in the related papers. If you are not certain of >
>  > the functional recipes, look into the funct.f90 file. But this is not
>  > a hard limit. You may build your own functional, including vdW. I've
>  > tested, with PBE pseudopotentials, an hybrid+vdW functional time ago
>  > ('sla+pw+hse+vdw1'), with good results on molecule-surface
>  > interactions.
> 
> Thanks for the answers; now my question can be more specific:
> suppose I want to use a pseudopotential built constructed for PBE
> (say C.pbe-mt_fhi.UPF in my case). If I set input_dft = "vdw-df",
> that means "sla+pw+rpb+vdw1",
> as far as I can see the gradient
> correction on exchange is revPBE and not PBE (third index is 4), and
> also there is no gradient correction on correlation (fourth index is 0).
> Is this compatible with a PBE pseudopotential file that, if used without
> vdw-df specification, would report an exchange-correlation scheme
> "sla+pw+pbx+pbc" (1 4 3 4 0)?
> 
> Or, should I set something like "sla+pw+pbx+pbc+vdw1" in order
> to be consistent with the PBE pseudopotential file?
> 
> The same question may hold for Cooper's variant (c09x instead of rpb above)
> 
> The question is on how the code works internally; at the beginning
> of the output there is a warning "Any further DFT definition will be
> discarded", and I wonder what it means exactly...
> 
> Thanks again for your time. Cristian


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   Giuseppe Mattioli                            
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