[Ifeffit] lambda in parameter Ei

[Matt Newville]

Dear Arkadisz,

The ei parameter is a refinement to the imaginary part of
Feff's complex photoelectron wavenumber, p, so that

    lambda(k) = 1 / Im[p(k)].

When Ei = 0, lamba(k) is easy to get as it is the lambda(k)
straight from Feff.    When Ei is not 0,  ifeffit does not
currently generate a 'refined lambda' -- it could.  Also sadly,
Ifeffit does not currently support complex math, so that
calculating it yourself is somewhat harder than it should be.


Here are some hints for how to do this. For a particular path,
you could do this:

  > path(1, feff0001.dat)
 > get_path(1, prefix=p1, do_arrays)
 > plot p1.k, p1.lambda

With get_path(N, prefix=p1, do_arrays), arrays named p1.k,
p1. amp,  p1.phase, p1.caps, p1.rep, and p1.lambda are
generated.  'rep' means 'The real part of p', and lamba is as
above.   All these are k-dependent functions.  Also note that
Re[p] is not exactly k, as the energy origins differ by a few eV.

For non-zero Ei, the complex wavenumber p becomes
  p^2 =    ( Re[p] - i / lambda ) ^2  - i * EI * 2m/(hbar*hbar)

(where m = electron mass). Mapping that onto non-complex
math is left as an exercise for the reader ;), but you'd still
want to calculate lamba(k) as 1/ Im[p].

Hope that helps.  If not, let me know,

--Matt