wave function jii, as modified by the interaction Hamiltonian H I . Going forward, we
will assume that the Hamiltonian is for a dipole transition, hence the operator looks
like position r
!
. Furthermore, the core electron wavefunction is so localized near the
nucleus that it is essentially a δ function at the origin. If we then write the scattered
wavefunction in a more conventional form as ψ s , the matrix element just becomes
amplitude of the scattered wave function at the origin:
χ E
ð Þ /j Δf jH I ji
h
i j /
Z
ψ s r
!
Á r
! Á δ r
!
d r
! / ψ s 0
ð Þ
ð6:9Þ
Since the scattered wave function has a phase as well as amplitude, the EXAFS
signal can be positive or negative, depending on whether there is constructive or
destructive interference between outgoing and backscattered waves.
6.3.2 The Scattered Wave function
From the above, we now see that in order to calculate the EXAFS, our task becomes
the evaluation of the scattered wavefunction at the origin of the X-ray absorbing
atom. At this point, we will skip the math and use Fig. 6.7 to illustrate the result.
So far, we have been describing X-ray absorption in terms of the X-ray energy E,
but for EXAFS calculations, we really care about the photoelectron kinetic energy
E kin and associated photoelectron wavelength λ and wavenumber k. Above the
absorption edge E 0 , absorption of an X-ray with energy E ¼ hν (where h is Planck’s
constant) yields an outgoing photoelectron with kinetic energy E kin :
E kin ¼ ΔE ¼ E À E 0
ð
Þ
ð6:10Þ
Using E kin ¼
1
2 m e v
2 and p ¼ m e v, where m e , v, and p are the electron mass,
velocity, and momentum, leads to the momentum p being given by:
p ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2m e E kin
p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2m e E À E 0
ð
Þ
p
ð6:11Þ
From the DeBroglie equation, we know that the wavelength of an electron
depends on its momentum:
λ ¼
h
p
ð6:12Þ
The photoelectron wavelength λ is thus a function of photon energy and is given
by:
λ ¼
h
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2m e E À E 0
ð
Þ
p
ð6:13Þ
6.3 Essential Physics of EXAFS
141
will assume that the Hamiltonian is for a dipole transition, hence the operator looks
like position r
!
. Furthermore, the core electron wavefunction is so localized near the
nucleus that it is essentially a δ function at the origin. If we then write the scattered
wavefunction in a more conventional form as ψ s , the matrix element just becomes
amplitude of the scattered wave function at the origin:
χ E
ð Þ /j Δf jH I ji
h
i j /
Z
ψ s r
!
Á r
! Á δ r
!
d r
! / ψ s 0
ð Þ
ð6:9Þ
Since the scattered wave function has a phase as well as amplitude, the EXAFS
signal can be positive or negative, depending on whether there is constructive or
destructive interference between outgoing and backscattered waves.
6.3.2 The Scattered Wave function
From the above, we now see that in order to calculate the EXAFS, our task becomes
the evaluation of the scattered wavefunction at the origin of the X-ray absorbing
atom. At this point, we will skip the math and use Fig. 6.7 to illustrate the result.
So far, we have been describing X-ray absorption in terms of the X-ray energy E,
but for EXAFS calculations, we really care about the photoelectron kinetic energy
E kin and associated photoelectron wavelength λ and wavenumber k. Above the
absorption edge E 0 , absorption of an X-ray with energy E ¼ hν (where h is Planck’s
constant) yields an outgoing photoelectron with kinetic energy E kin :
E kin ¼ ΔE ¼ E À E 0
ð
Þ
ð6:10Þ
Using E kin ¼
1
2 m e v
2 and p ¼ m e v, where m e , v, and p are the electron mass,
velocity, and momentum, leads to the momentum p being given by:
p ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2m e E kin
p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2m e E À E 0
ð
Þ
p
ð6:11Þ
From the DeBroglie equation, we know that the wavelength of an electron
depends on its momentum:
λ ¼
h
p
ð6:12Þ
The photoelectron wavelength λ is thus a function of photon energy and is given
by:
λ ¼
h
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2m e E À E 0
ð
Þ
p
ð6:13Þ
6.3 Essential Physics of EXAFS
141