# [Pw_forum] Hubbard forces without Hubbard U

Dmitry Novoselov dnovoselov at gmail.com
Mon Aug 18 08:08:20 CEST 2014

Dear Prof. Sclauzero,

thank you for your exact answer. You are absolutely right. I found a
corresponding part of the code.
My problem has been solved!

2014-08-15 15:37 GMT+06:00 Sclauzero Gabriele <
gabriele.sclauzero at mat.ethz.ch>:

>
> Dear Dmitry,
>
>     I am sure there is nothing puzzling in there. When lda_plus_u is true,
> but all Hubbard_U’s are zero, the Hubbard component of the forces are still
> computed and added to the total force, but such component will be zero (try
> verbosity=‘high’ to get them printed on output). In the subroutine
> force_hub you can find the line:
>
>  forceh(ipol,alpha) = forceh(ipol,alpha) -  v%ns(m2,m1,is,na) *
> dns(m1,m2,is,na)
>
> which corresponds exactly to the formula you wrote here below. The point
> is that the Hubbard potential
> v%ns  =\frac{U}{2} \sum_{Imm’} (\delta_{mm’} -2n^{I}_{m’m})
> depends linearly on Hubbard_U, so all its components will be zero if U=0.
>
>
> HTH
>
> GS
>
>
>
> > Dear Prof. Cococcioni,
> >
> > I use the QE 5.1 and It seems that the Hubbard forces are calculated
> well. But I'm a little puzzled by the next situation.
> > An analytical expression for the Hubbard forces can be written as:
> >
> > F^{U}_{\alpha i} = - \frac{U}{2} \sum_{Imm'} (\delta_{mm'}
> -2n^{I}_{m'm}) \frac{\partial n^{I}_{mm'}}{\partial \tau_{\alpha i}}.
> >
> > Here alpha is a number of displacement atom in i-direction and U is a
> Hubbard correction.
> > But I have not found the procedure of multiplication by the Hubbard U
> correction in the subroutines where Hubbard forces are calculated. Below
> the appropriate lines of the corresponding subroutines are listed:
> >
> > SUBROUTINE forces()
> >   ...
> >   CALL force_hub(forceh)
> >   ...
> >   force(ipol,na) = force(ipol,na) + forcenl(ipol,na) + forceion(ipol,na)
> + forcelc(ipol,na) + forcecc(ipol,na) + forcescc(ipol,na) + forceh(ipol,na)
> >   ...
> > END SUBROUTINE forces
> >
> >
> > SUBROUTINE force_hub(forceh)
> >   ...
> >   forceh(ipol,alpha) = forceh(ipol,alpha) - v%ns(m2,m1,is,na) *
> dns(m1,m2,is,na)
> >   ...
> > END SUBROUTINE force_hub
> >
> > --
> >
> > Best regards,
> > Dr. Dmitry Novoselov
> >
> > Institute for Metal Physics,
> > Yekaterinburg, Russia
> >
> >
> > 2014-08-14 19:43 GMT+06:00 Matteo Cococcioni <matteo at umn.edu>:
> >
> > Dear Dmitry,
> >
> > it should be working. what version of the code are you using?
> >
> > Matteo
> >
> >
> > On Wed, Aug 13, 2014 at 10:47 AM, Dmitry Novoselov <dnovoselov at gmail.com>
> wrote:
> > Dear all,
> >
> > It seems that in the part of the code where the Hubbard forces is
> calculated (forces and force_hub subroutines), the Hubbard U correction
> does not taken into account. Is it ok?
> >
> > Thank you!
> >
> > --
> >
> > Best regards,
> > Dr. Dmitry Novoselov
> >
> > Institute for Metal Physics,
> > Yekaterinburg, Russia
> >
> >
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> >
> >
> > --
> >
> > Best regards,
> > Dr. Dmitry Novoselov
> >
> > Institute for Metal Physics,
> > Yekaterinburg, Russia
> >
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>
>
> Dr. Gabriele Sclauzero
> Materials Theory - ETHZ
> ETH Zurich, HIT G 43.2
> Wolfgang-Pauli-Str. 27
> 8093 Zürich, Switzerland
>
> Phone +41 44 633 94 10
> Fax +41 44 633 14 59
> gabriele.sclauzero at mat.ethz.ch
> www.theory.mat.ethz.ch
>
>
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--

*Best regards,*

*Dr. Dmitry Novoselov Institute for Metal Physics,*
*Yekaterinburg, Russia*
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