Re: k

["John J. Rehr" <jjr@leonardo.phys.washington.edu>]

Hi Konstantin,

> Q1: What is the physical meaning of the XAFS k? Is it just a convenient
> something for subsequent Fourier analysis?
> Q2: This artificial "k" (or "p") is included somehow into FEFF calculations.
> Does this  imitation of real k affect the calculations? In which step
> mostly? Extrinsic losses are incorporated into FEFF - as k- or E- dependent?

  I fully agree with Matt Newville's discussion of the difference between k and p
but thought I would add a couple of technical points about the difference:

 FEFF retained the "experimental definition" of k = sqrt (E-E_0) where E_0 is the
chemical potential (mu) or Fermi energy, for two reasons:

1) the definition is consistent with the traditional definition of the XAFS Fourier
  Transform with respect to k.

2) the physical momentum p is not accessible experimentally since there is no way
to tell what the interstitial potential and self-energy are from EXAFS measurements.
Thus as Matt notes the connection between p and k requires knowledge of the complex
self energy Sigma(E,p).  In the interstitial region,

   p^2 + Sigma(E,p) + Vint = E
   k_F^2 + Sigma(mu,k_F) + Vint = mu

so the explicit relation between k and p is:

k^2=E-mu=p^2-k_F^2 +Sigma(E,p)-Sigma(mu,k_F).

The definitions of f_eff(k) and other variables are renormalized to be consistent
with how one defines these quantities experimentally.

FEFF evaluates the self-energy corrections using the Hedin Lundqvist self energy
in the local density approximation at the interstitial density r_s. For other
details see the section of the FEFF doc on Variables in EXAFS and XANES formulae.

 Cheers,
  John