[Pw_forum] XSpectra hybrid functional

Giuseppe Mattioli giuseppe.mattioli at ism.cnr.it
Wed Oct 24 15:00:32 CEST 2012

rather meaningful, perhaps by chance...:-)

I can only suppose part of the reason: the screening of core holes in molecules is strongly affected 
by electronic correlation problems, more than in the case of semiconductors or insulators. For 
example, if you try to estimate XPS lines of molecules by using GGA and a full core hole tecnique 
you find horribly large differences with respect to measurements. I've found instead errors as large 
as 0.1 eV on XPS lines by using B3LYP and full core holes. I've obtained some interesting density 
plots of the above screening in pyrimidines, which have been published here.

J. Phys. Chem. A 2009, 113, 13593–13600

In this case, nothing is "empirical" in the calculation. At the same time, I've calculated quite 
good NEXAFS spectra by using the B3LYP charge density and a half core hole technique, as reported in 
the attached image. If I do not use a core hole, thus neglecting final state effects, I obtain poor 
spectra. It is, admittedly, a more "empirical" use of the code, not really justified. We (chemists 
using QE...) would like to have fully justified quantum spectroscopy tools, including also 
turbo_lanczos, with EXX kernels, but we rely on you... In the meanwhile, ci arrangiamo.

All the best


On Wednesday 24 October 2012 14:06:05 Matteo Calandra wrote:
> Le 24/10/12 12:00, pw_forum-request at pwscf.org a écrit :
> >It is possible indeed, in the sense that the XSPECTRA code does not stop
> >complaining if you perform the calculation starting from an electronic
> >density calculated by using hybrid functionals (or DFT+U). But the
> >XSPECTRA code has not a calculation kernel including EXX terms, as just
> >pointed out
> >by Matteo Calandra. Thus:
> Not exactly. In the case of DFT+U, the lanczos is performed, as it must
> be, with the
> U term included in the hamiltonian. So the Hamiltonian H used in
> XSpectra to perform
> H|psi> is the same used to generate the charge density.
> This is of course what must be done. So everything is consistent with
> DFT+U.
> If you take the charge density calculated with hybrids and calculate the
> H|psi>
> without hybrid I do not think that the result is very meaningful...
> M.

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   Giuseppe Mattioli                            
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