[Ifeffit] "FEFF+IFEFFIT" and "cumulant expansion+ratio method" approaches
Leandro Langie Araujo
leandrolangie at gmail.com
Thu Jul 20 18:54:39 CDT 2006
Hi Matt,
thanks a lot for the reply.
I actually don't want to and don't have to convert my Ifeffit results to
any kind of effective pair distribution function. I understand that it
is not a trivial conversion and some approximations would be necessary.
I just needed to be sure that by using Ifeffit I get the "real"
cumulants, instead of the so-called "effective" ones, and no
"post-analysis" corrections are necessary to the Ifeffit results.
Knowing that is enough for me to go on and compare my results to some
others obtained by different methods. And the results and talking about
are the thermal evolution of the cumulants for crystalline Ge and Ge
nanocrystals.
Cheers,
Leandro
Matt Newville wrote:
> Hi Leandro,
>
> Sorry for the delay. XAFS 13 was last week, and I'm still catching up!!
>
> Ifeffit reports (as well as it can) cumulants of the "real" distribution
> function, including corrections for the mean-free-path, and the
> 1/R^2 term in the EXAFS equation.
>
> There should be no need to convert back to any sort of 'effective
> pair distribution' as might be found be "other methods", including
> log-ratio methods, etc. The burden should be on those methods to
> correct their effective distributions to the real ones.
>
> So, my immediate reaction would be to not try to do this. Assuming
> that you still want to, I think it's not a trivial conversion because
> a) lambda is k-dependent,
> b) Feffit and Ifeffit use p = k+ Sigma(k) + i/lambda(k) as
> the conjugate variable to R, and so the expansion
> coefficient for the cumulant expansion. Taking lambda(k)
> into account is part of the problem -- so is the Sigma(k).
>
> That additional self-energy term includes an estimate of the difference
> between E_fermi (where k=0) and E_0 (where p=0). You can see
> this in the feff.dat files (the column k and Re[p] are slightly
> different).
> (And it should be that Feff8 is better at this than Feff6, but the jury's
> still out for how big that effect is). In ifeffit, you can see this with
> > path(1, feff0001.dat)
> > get_path(1, group=f1, do_arrays)
> > plot f1.k, f1.rep
>
> But that's all to say that the conversion is not that easy.
>
> Perhaps you could explain what the problem you're trying to solve
> and why you might want the cumulants of an effective potential?
>
> --Matt
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