[Ifeffit] "FEFF+IFEFFIT" and "cumulant expansion+ratio method" approaches
Matt Newville
newville at cars.uchicago.edu
Wed Jul 19 23:02:44 CDT 2006
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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