[Ifeffit] questions concerning paper

Matt Newville newville at cars.uchicago.edu
Thu May 22 10:43:02 CDT 2008


Dear Eckhard,

On Thu, May 22, 2008 at 6:32 AM,  <s440697 at stud.uni-goettingen.de> wrote:
> Hello,
>
> i have some questions concerning the xafs-formula in this paper (attached):
> "Theoretical approaches to x-ray absorption fine structure" (July 2000).
> The author explained on page 625 and 626 the parameters of this formula.
> I think, most of you have read this paper.
>
> I don't understand the meaning of lambda(k), the author wrote "... and
> lambda(k) is the energy-dependent XAFS mean free path (not to be confused
> with the de Broglie wavelength)."
> My question is, the mean free path whereof? The propagating
> photoelektron(wave) or the (back)scattered waves?

Yes (that is both). This lambda(k) is the mean-free-path for a
photo-electron of wavenumber k.
That is, it is the mean distance that a photo-electron will travel
before it scatters inelastically (and incoherently) from electrons (or
phonons, etc) in the material.   These scatterings are from the
"low-energy electrons", and do not include the "back-scattering" from
the highly localized electrons of the cores of neighboring atoms:
those scattering events will be elastic and coherent, and give rise to
the XAFS.

This lamdba(k) is essentially the same mean-free-path for a
photo-electron used in other electron spectroscopies, and has a
classic U-shape when plotted as a function of k.  I say essentially
because we often fold into this term that the finite lifetime of the
core-hole, which also limits how far the photo-electron can travel and
still have an empty space at home to return to.

Odysseus, the photo-electron (an EveryParticle: though we like to
think of it as special, it is really indistinguishable from its
neighbors) leaves home for war, and has many adventures on his way
home which might prevent him from ever getting back.  Meanwhile, there
is a long line of people waiting to take his place at home.

> In the next sentence, what is the overall amplitude factor S_0^2? How is
> it defined?

S_0^2 accounts for the relaxation of ALL THE OTHER electrons in the
absorbing atom.   The other core electrons respond a bit to ripping
the 1s (or other) core electron out of the atom, which makes the atom
that the photo-electron sees on returning slightly different from when
it left.
S_0^2 is simply the overlap of the N-1 electrons in the ground state
atom with the N-1 electrons in the excited atom.  It has to be less
than 1, but is typically close to 1.0.   With this definition, S_0^2
should have no energy dependence, but there are other processes
(multi-electron excitations, for example) that can be considered as
relaxation of the other electrons in the atom that may have energy
dependence.

Odysseus is gone for a very long time: can he even recognize home when
he gets back?  Will they recognize him

> In conjunction with question one i don't understand the expression in the
> next paragraph for the term e^{-2R/lambda}. The author wrote: "The decay
> of the wave (which wave does he mean?) due to the mean free path or finite
> lifetime (what does he mean?, finite lifetime of the emitted and
> propagating photoelektron wave?) [including core-hole lifetime] of the
> photoelectron is captured by ..." Is it right, that the emitted
> photoelektron is detectable outside the sample?

The decay is of the photo-electron in that it may scatter into other
low-energy electron states.

The lifetime is the core-hole lifetime (how long the core-hole will
live before being filled, generally through an emission process such
as fluorescence.  This is typically in the femtosecond range.

The emitted photo-electron CAN escape the sample, though as the
mean-free-path is on the order of 5 to 30Ang, the photo-electrons
would only be from the top surface of the sample.   In fact, one can
(and some do!) measure XAFS using the emitted electrons.  Typically
the emitted electron current is dominated not by the primary
photo-electrons, but by the cascade of electrons the photo-electrons
and emitted Auger electrons cause as they scatter toward the sample
surface.  This is similar to the photo-currents measured with Auger
spectroscopy.


> Another parameter is the phase factor phi = arg f(k) "...reflects the
> quantum-mechanical wavelike nature of the backscattering." Please explain
> me the meaning of this parameter.

I'm not sure I can explain quantum mechanics.  The phi = arg f(k) here
(page 626, Fig 6b)  is the phase-shift of the photo-electron as it
scatters from the neighboring atom.   That is, scattering has a finite
effect on both the amplitude and phase of the photo-electron
wave-function.

The war changes Odysseus both physically and mentally.

> At the bottom of this page the author wrote: "...the Debye-Waller factor,
> which is given to a good approximation by e^{-2 sigma^2 k^2}." In this
> context he refers to figure 8, where k^2 chi(k) vs. k is plotted. I don't
> understand this context because there is no chi in the approximation for
> the Debye-Waller factor. Where comes the chi here?

chi(k)  is proportional to e^(-2 sigma^2 k^2), where sigma^2 is the
mean-square-displacement in the bond length R to the neighboring atom,
say from thermal vibrations.   That is, we average over very many
(10^9 or more) photo-electrons scattering from different neighboring
atoms, and each one may be sampling a different bond length as the
atoms are moving.     The figure (showing k^2 * chi(k) for a sample at
different temperatures) shows
k^2 * chi(k) being dramatically reduced at high k as the temperature
increases, and reduced less substantially at low k.     As the
temperature increases, so does sigma^2 (generally linear with T), and
so there is an exponential decay of chi(k) that is strongly
k-dependent.

The more wars Odysseus sees (or the more violent the wars are), the
less chance he has of making it home.

> I would be very grateful for any help you would offer to me.
> Thank you very much for your help and time.

If that's not enough, perhaps the description in the tutorial
documents at  http://xafs.org/Tutorials  would be helpful.

--Matt



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