[Ifeffit] Re: EXAFS-Question concerning \delta(k)

[Matt Newville]

Hi,

Bruce summarized Frank's questions about phase shifts as:

> If I understand your question, there are two parts.  (1) Why is
> the shift due to the phase correction always negative (i.e.
> always to a smaller value in FT[chi(R)])?  And (2) Why is the
> shift always in the neighborhood of 1/2 angstrom?

I think there is a simple way to picture why the phase shift is
negative in terms of the quantum mechanical 'scattering from a
potential' problem -- the one where you usually end up talking
about tunneling through a barrier.  I'll give it a try:
                          
          ____               ~~~~~~~> ____
  ~~~~~~>|    |   E<V0               |    |  E>V0
       __|    |__                  __|    |__


Here we're in the E>V0 regime (on the left).  We normally think of
the photoelectron as having energy E= hbar^2k^2/2m, where k is the
wavenumber for the 'free' photoelectron (that is, having escaped
the central atom and far from the scattering atom).  But to
participate in the EXAFS, it does leave the absorbing atom and
scatter from the neighbor.  When it is near the strong potentials
of the central and scattering atom, the wavenumber would be better
written like
  k = sqrt[ 2m * (E-V0)] / hbar 

as for the barrier above (where V0 is the potential energy).  
That is, k is lower (the electron wavelength is longer) in the
presence of a strong potential: kinetic energy being traded for
potential energy.

To describe the EXAFS, we use only k with V0=0, which overestimate
the average kinetic energy of the photoelectron. That means we get
fewer periods of oscillations for a given R than 2*k*R would
suggest, which leads to the negative direction of the phase shift.

In this simplified picture, I think the magnitude of the shift in
the peak of |chi(R)| being around 0.5Ang but slightly variable can
be seen as some measure of the typical average potential sizes of
atoms.  I'm not sure I can give a good feel for why that ends up
being 0.5Ang.  Then again, I'm not sure that really answers why
it's even close to linear in k.

Perhaps someone can correct any mistakes or fill in any of the
many rough spots here....

--Matt