[Pw_forum] DFT+U

[paresh rout]

Thank you sir for the reply . I was just confused how to use DFT+Ueff .
Still I have some query  If I follow the first method for DFT+Ueff  (U-J)
then isn't just like DFT+U method which will just use a small U=Ueff
(Ueff=U-J) instead of large original U ? Then how will the code
differentiate  DFT+U and DFT+Ueff ?

Kind Regards
Paresh

On Wed, Sep 30, 2015 at 10:28 PM, Matteo Cococcioni <matteo at umn.edu> wrote:

>
> dear Paresh,
>
> for 2) the code does what is explained in PRB 84, 115108 (2011), which is
> not Ueff = U-J
>
> if you need DFT+Ueff just use 1) with U = Ueff.
>
> Best,
>
> Matteo
>
> On Sat, Sep 26, 2015 at 8:58 AM, paresh rout <paresh.rout88 at gmail.com>
> wrote:
>
>> Dear all,
>> I want to confirm DFT+U , DFT+Ueff and DFT+U+J method that quantum
>> espresso uses . So far I understand
>> 1- For simply DFT+U  calculation I have to use
>>     lda_plus_u=.true.
>>     lda_plus_u_kind=0
>>     Hubbard_U(i)=U value
>> 2-For DFT+Ueff    (where Ueff=U-J)
>>     lda_plus_u=.true.
>>     lda_plus_u_kind=0
>>     Hubbard_U(i)=U value
>>     Hubbard_J0(i)=J values
>> 3-For DFT+U+J method
>>    lda_plus_u=.true.
>>    lda_plus_u_kind=1
>>    Hubbard_U(i)=U value
>>    Hubbard_J(i,ityp)=J values
>> If the above ways are right then for DFT+Ueff case
>> does the code will do Ueff=U-J ?
>> Any help would be highly appreciated .
>>
>> Kind Regards
>> Paresh Chandra Rout
>> Research Scholar
>> Indian Institute of Science Education and Research Bhopal
>>
>>
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>
>
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