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

[Kelly, Shelly D.]

Hi,

> -----Original Message-----
> From: Newville, Matthew G.
> Sent: Wednesday, December 08, 2004 11:27 PM
> To: XAFS Analysis using Ifeffit
> Cc: Frank Haaß
> Subject: Re: [Ifeffit] Re: EXAFS-Question concerning \delta(k)
> 
> 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.
> 

[Kelly, Shelly D.] I think that my explanation for a negative phase shift is just the same, but slightly easier for my brain to comprehend.  Which means that I'll make some mistakes but it works for me..:) 

The photoelectron is negative and it scatters off of the electrons from the surrounding atoms.  Hence the photoelectron is slowed down when it gets close to the other electrons.  Negative and Negative repel, giving rise to fewer periods of oscillations, and a negative phase shift.

As for the magnitude of the shift, lighter elements have a larger shift than the heaver elements.  You can think about that by remembering that lighter elements scatter more at low k.  If you have a photoelectron that scatters with very little energy (moving slowly) it will be slowed down more by the electrons.  Heaver elements scatter more at high k.  So, a photoelectron that scatters with a lot of energy will not get slowed down as much, hence a smaller phase shift.

HTH
Shelly