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

[John J. Rehr]

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