RE: [Ifeffit] Re: EXAFS-Question concerning \delta(k)
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 __| |__ __| |__ 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
Hi Everyone,
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?
This is an interesting result. I don't know of a good quantitative explanation, but I do know that the result involves several considerations. The following is an attempt to explain the sign of the effect: The total phase shift delta in the EXAFS equation chi ~ [f_eff(pi,k)/kR^2] sin (2kR + delta) has two parts: delta = 2 delta_c + arg f_eff(pi,k) where delta_c is the central atom phase shift and f_eff(pi,k) is the phase of the effective curved wave backscattering amplitude. In my experince the central atom phase shift (i.e., the p-wave phase shift for K-edge) dominates. The behavior of the phase shift at high energies can be estimated from the WKB approximation of quantum mechanics for electrons of a given angular momentum l: delta_c = \int_0^Rmt [k(r) - k0(r)] dr Here k(r) = sqrt 2[E-V(r)-V_l(r)] is the local wave number inside the atom. V(r) is the coulomb potential (negative) and V_l = l(l+1)/r^2 is the positive centrifugal barrier, and k0 is the local wave number in the absence of any potential. At high energies the centrifugal terms cancel and one has to linear order in the coulomb potential delta_c = - (1/k) \int_0^Rmt V(r) dr = + const/k where the constant is the strength of the potential. This quantity clearly gets smaller with increasing energy. It is also positive since V(r) must be negative for the coulomb potential of an atom to be attractive and bind electrons. The contribution to the distance shift in the FT from this term is the slope of the phase shift which is thus negative. delta R = d delta_c/dk = - const/k^2 = -const / E The WKB estimate only works at high energies where phase shifts are small, but I think it at least explains the sign of the effect. J. Rehr
participants (2)
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John J. Rehr -
Kelly, Shelly D.