advances in xafs analysis

Advances in EXAFS Analysis with ifeffit and feff

Matthew Newville

Consortium for Advanced Radiation SourcesUniversity of Chicago

October 15, 2008

XAFS Analysis with IFEFFIT: October 15, 2008

Acknowledgments

Ed Stern John Rehr Bruce Ravel

Univ of Washington, Seattle NIST

Stern Modern description of XAFS (1970 – 1975).

Rehr XAFS theory: feff 3.0 (1990) . . . feff 8.0 (1998).

Ravel atoms, athena, artemis.

XAFS Analysis with IFEFFIT: Introduction October 15, 2008

EXAFS Analysis and Information Theory

The two fundamental equations for EXAFS analysis are in conflict:

The EXAFS Equation to model the data has a sum over paths:

χ(k) =∑

S20 Nj fj(k)e−2Rj/λ(k)e−2k2σ2

kRj2 sin[2kRj + δj(k)]

The Nyquist Formula for the number of independent parameters in the data:

Nidp ≈2∆k∆R

The number of parameters needed to describe g(R) grows exponentially with R.

The number of parameters available grows linearly with R.

EXAFS Analysis and Information Theory

The two fundamental equations for EXAFS analysis are in conflict:

The EXAFS Equation to model the data has a sum over paths:

χ(k) =∑

S20 Nj fj(k)e−2Rj/λ(k)e−2k2σ2

kRj2 sin[2kRj + δj(k)]

The Nyquist Formula for the number of independent parameters in the data:

Nidp ≈2∆k∆R

The number of parameters needed to describe g(R) grows exponentially with R.

The number of parameters available grows linearly with R.

EXAFS Analysis with Limited Information

The basic difficulties with EXAFS Analysis are:

1 The basis functions (Paths) of g(R) grow exponentially with R.

2 The scattering factors f (k), δ(k) are difficult to calculate, and dominate ouruncertainties.

3 There’s not much information in a real measurement: Nidp ≈ 2∆k∆Rπ

Quantitative EXAFS Analysis must conserve the information:

1 reduce the number of Paths to consider (Fourier analysis).

2 reuse or parameterize ab initio calculations of f (k), δ(k) using feff orsomething similar.

3 recycle parameters. Use algebraic constraints to use fewer and moremeaningful independent variables in the modeling.

We parameterize the EXAFS with a physical model built from

Constraints and Restraints.

EXAFS Analysis with Limited Information

The basic difficulties with EXAFS Analysis are:

1 The basis functions (Paths) of g(R) grow exponentially with R.

2 The scattering factors f (k), δ(k) are difficult to calculate, and dominate ouruncertainties.

3 There’s not much information in a real measurement: Nidp ≈ 2∆k∆Rπ

Quantitative EXAFS Analysis must conserve the information:

1 reduce the number of Paths to consider (Fourier analysis).

2 reuse or parameterize ab initio calculations of f (k), δ(k) using feff orsomething similar.

3 recycle parameters. Use algebraic constraints to use fewer and moremeaningful independent variables in the modeling.

We parameterize the EXAFS with a physical model built from

Constraints and Restraints.

Constraints in Analysis: why we need them

ifeffit uses Generalized Variables for constraints:

The Path Parameters R, S20 N, σ2, E0, etc are not directly adjusted in the fit.

Instead, Path Parameters are written as mathematical expressions of theGeneralized Variables.

This makes it easy to share Variables across:

different Path Parameters for a Path

different Paths (including Multiple Scattering) for a Data set.

different data sets (different temperature, edge, polarization . . . )

Constraints enable us to use “Prior Knowledge”to make simple physical models for our data.

Constraints, continued

Examples of Path Parameters written as functions of the Generalized Variables:

shared parameter: E0guess e0 = 1.0

path(1, e0 = e0)

path(2, e0 = e0)

mixed coordination shellset S02 = 0.80

guess x = 0.5

path(1, Amp= S02 * x )

path(2, Amp= S02 * (1-x))

Fit Einstein Temperatureset factor = 24.254337 #= (hbar*c)^2/(2 k_boltz)

# mass and reduced mass in amu

set mass1 = 63.54, mass2 = 63.54

set r_mass = 1/ (1/mass1 + 1/mass2)

# the Einstein Temp will be adjusted in the fit!

guess thetaE = 200

# use for data set 1, T=77

set temp1 = 77

def ss2_path1 = factor*coth(thetaE/(2*temp1))/r_mass )

path(101, sigma2 = ss2_path1 )

set temp2 = 300

Other Examples of Data Modeling with Constraints:

force one R for the same bond for data taken from different edges.

model complex distortions (height of a sorbed atom above a surface).

model non-Gaussian pair distribution function (as GNXAS).

path(1, e0 = e0)

path(2, e0 = e0)

guess x = 0.5

path(2, Amp= S02 * (1-x))

set mass1 = 63.54, mass2 = 63.54

guess thetaE = 200

set temp1 = 77

set temp2 = 300

path(1, e0 = e0)

path(2, e0 = e0)

guess x = 0.5

path(2, Amp= S02 * (1-x))

set mass1 = 63.54, mass2 = 63.54

guess thetaE = 200

set temp1 = 77

set temp2 = 300

path(1, e0 = e0)

path(2, e0 = e0)

guess x = 0.5

path(2, Amp= S02 * (1-x))

set mass1 = 63.54, mass2 = 63.54

guess thetaE = 200

set temp1 = 77

set temp2 = 300

Restraints: Imprecise Prior Knowledge

A Constraint expresses exact prior knowledge (e.g., force 1 E0 for all paths), andreduces the number of variables ~x in the least-squares problem:

χ2 =∑

[χdata

i − χmodeli (~x)

A Restraint expresses imprecise prior knowledge by adding another term to theleast-squares problem:

χ2 =∑

[χdata

i − χmodeli (~x)

[λ0 − λ(~x)

We add “Data” λ0 to the fit, with “Model” λ(~x) and confidence level δλ:

Restraints give us a Bayesian approach to analysis: if we canquantify our prior knowledge, we can include it in the fit.

Restraints can be used for “soft limits” on parameters.

χ2 =∑

[χdata

i − χmodeli (~x)

χ2 =∑

[χdata

i − χmodeli (~x)

[λ0 − λ(~x)

χ2 =∑

[χdata

i − χmodeli (~x)

χ2 =∑

[χdata

i − χmodeli (~x)

[λ0 − λ(~x)

Restraints as “Soft Limits” on Parameters

For some parameters (say, S20 ), there may be an acceptable range of values.

You could use a constraint:

“Hard Wall” Constraint:guess S02Var = 0.9

path(1, S02 = max(0.6, min(1.0, S02Var) ) )

This complicates finding error bars whenthe free variable S02Var goes outside thelimits [0.6, 1.0].

A Restraint makes the fit increasingly worse fit as a variable goes“out-of-bounds”:

Λ(x) =

x − xhi x > xhi

0 xhi ≥ x ≥ xlo

xlo − x xlo ≥ x

Restraint:guess S02Var = 0.9

feffit(..., restraint=penalty(S02Var,0.6,1.0))

Restraints as “Soft Limits” on Parameters

For some parameters (say, S20 ), there may be an acceptable range of values.

You could use a constraint:

“Hard Wall” Constraint:guess S02Var = 0.9

path(1, S02 = max(0.6, min(1.0, S02Var) ) )

This complicates finding error bars whenthe free variable S02Var goes outside thelimits [0.6, 1.0].

A Restraint makes the fit increasingly worse fit as a variable goes“out-of-bounds”:

Λ(x) =

x − xhi x > xhi

0 xhi ≥ x ≥ xlo

xlo − x xlo ≥ x

Restraint:guess S02Var = 0.9

feffit(..., restraint=penalty(S02Var,0.6,1.0))

The Status of EXAFS Analysis (summary so far...)

For analyzing EXAFS data we can use:

Constraints to build complex physical models for our data.

Restraints to include imperfect Prior Knowledge of our system.

Fourier to select which k- and R-components to use in the analysis.

Multiple data sets, multiple k-weights, etc in a single refinement.

so many clever things we can do in the analysis . . .

But our fits are TERRIBLE!

Reduced chi-square, χ2ν = χ2/(N − Nvarys) (for N independent data points)

should be ∼ 1 for a “Good Fit”:

χ2 =1

[χdatai − χmodel

i (~x)]2

We’re lucky if χ2ν is below 50!

What are we doing wrong?

χ2 =1

i (~x)]2

χ2 =1

i (~x)]2

χ2 =1

i (~x)]2

Noise Levels in Data

Is the estimated noise in the data, ε, way off?

Here are typical EXAFS spectra(both transmission, 1 sec/point):

0.2 mM Zn nitrate solution Cu foil Room Temperature

ε ≈ 4.6× 10−4 “Typical Data” ε ≈ 1.6× 10−4 “Good Data”

A simple 1st shell fit to the Cu data gives χ2ν ≈ 240.

For parameter uncertainties, we multiply by√χ2ν ≈ 16, as if ε was 16× larger

than our estimate.

That’s 5× noisier than the Zn data!

Is the estimated noise in the data, ε, way off? Here are typical EXAFS spectra(both transmission, 1 sec/point):

than our estimate.

Room Temperature Cu: Fit Details

Simple fit to 1st shell of Cu foil (300K) with crude 2nd shell and 1st-triangle pathsfor spectral leakage: k = [2, 16] A−1, R = [1.65, 2.6] A, k-weight=2, Nidp = 8.4.Fit results and statistics (Feff 6):

R = 2.543(0.003) A ∆E0 = 4.1(0.5) eVS2

0 = 0.96(0.05) σ2 = 8.5(0.4)× 10−3 A2 εk = 1.6× 10−4

χ2 = 1012 χ2ν = 242 R = 0.0016

R ≈ 0.1% – a very good fit!

Uncertainties include correlations and increase χ2 by χ2ν .

Including C3, χ2ν = 260, and C3 is consistent with 0: just barely insignificant!

R = 2.543(0.003) A ∆E0 = 4.1(0.5) eVS2

0 = 0.96(0.05) σ2 = 8.5(0.4)× 10−3 A2 εk = 1.6× 10−4

χ2 = 1012 χ2ν = 242 R = 0.0016

R = 2.543(0.003) A ∆E0 = 4.1(0.5) eVS2

0 = 0.96(0.05) σ2 = 8.5(0.4)× 10−3 A2 εk = 1.6× 10−4

χ2 = 1012 χ2ν = 242 R = 0.0016

ε: Noise Levels in Data

ε is estimated from the high-R (15 to 25 A) components of the data.

|χ(R)|

The Zn data really shows “white noise” past 10A.

The Zn data is definitely noisier than the Cu data.But you need to look above R = 15 A to be sure of this!

The Cu data has signal above the noise level well past 10A!!Using R = [15, 25] A may overestimate ε for the Cu data.That’s the wrong way!

|χ(R)| log10(|χ(R)|)

ε from high-R spectra?

How good is the estimate of ε from high-R data?

With “mock XAFS data” with a single path at 2.5A + normally-distributed noiselevel ε varying between 10−7 and 1:

Ratio of ε determinedfrom R = [15, 25] A tothat of (normal) randomnoise added to 1st shell χ.

The noise level is predicted well to ∼20% over seven orders of magnitude.

The high-R estimate of ε works for normal, white noise.

Summary (#1)

Our fits are bad: Even for simple Cu metal,

χ2 =1

i (~x)]2 ≈ 1000

We’re lucky if χ2ν = χ2/ν is below 100 in most cases.

Is our estimate of ε off by that much??

No. Even if you invoke “systematic errors” (whatever those are . . . ) or“non-Gaussian noise” (really?), ε is not off by factor of 10.

Conclusion: All our models are bad.

FEFF is not good enough.

Summary (#1)

χ2 =1

i (~x)]2 ≈ 1000

Summary (#1)

χ2 =1

i (~x)]2 ≈ 1000

Summary (#1)

χ2 =1

i (~x)]2 ≈ 1000

Meanwhile . . . femtometer-scale EXAFS?

R. Pettifer et al, Nature 435, pp 78–81 (2005) Measurement of femtometre-scaleatomic displacements by X-ray absorption spectroscopy

Using a differential EXAFS measurement with energy-dispersive EXAFS (ESRFID24), they show EXAFS is sensitive to atomic displacements of ∼ 5 fm.

No kidding: ∆R = 0.00005 A!

OK, they did a clever experiment:

energy dispersive EXAFS gives very consistent energy values.

high flux (to 1013 photons/sec/eV – not uncommonly high).

compared data on a FeCo while switching magnet field.

Why are the rest of us stuck at ∆R & 0.005 A – 100x worse?Has the precision of EXAFS improved in the past 30 years??

Seeing 10 fm distance changes

How easy is it to see a 10 fm shift in distance in “real data” with noise?

Here’s “data” from feff for Cu with σ2 = 0.008 A2 and ε = 3× 10−5 (5x betterthan the earlier Cu data) with R = 2.54780 A and R = 2.54790 A:

∆R = 10 fm shift can be seen!

Any more noise, and a 10 fm shiftprobably could not be seen.

A “feff v. feff” fit for slightly higher noise ε = 1× 10−4 gives:

R = 2.54798(0.00018) A ∆E0 = 0.02(0.03) C3 = 0 (fixed)εk = 1.1× 10−4 S2

0 = 1.001(0.003) σ2 = 8.01(0.02)× 10−3 A2

χ2 = 8.2 χ2ν = 1.4 R = 0.00001

R should be 2.54790A, so we’re within 10 fm, with an uncertainty of 18 fm.

Hey, look: χ2ν ≈ 1 !

R = 2.54798(0.00018) A ∆E0 = 0.02(0.03) C3 = 0 (fixed)εk = 1.1× 10−4 S2

0 = 1.001(0.003) σ2 = 8.01(0.02)× 10−3 A2

χ2 = 8.2 χ2ν = 1.4 R = 0.00001

R = 2.54798(0.00018) A ∆E0 = 0.02(0.03) C3 = 0 (fixed)εk = 1.1× 10−4 S2

0 = 1.001(0.003) σ2 = 8.01(0.02)× 10−3 A2

χ2 = 8.2 χ2ν = 1.4 R = 0.00001

R = 2.54798(0.00018) A ∆E0 = 0.02(0.03) C3 = 0 (fixed)εk = 1.1× 10−4 S2

0 = 1.001(0.003) σ2 = 8.01(0.02)× 10−3 A2

χ2 = 8.2 χ2ν = 1.4 R = 0.00001

A Complication: The Correlation of R and E0

R and E0 are highly correlated.

To see ∆R = 10 fm, we need a precise energy range:Can we see 10 fm in “real data” with noise and an E0 shift of 0.02 eV?

Comparing our previous feff “data”, now with (R = 2.54780 A and E0 = 0 eV)compared to (R = 2.54790 A and E0 = −0.02, 0.00, and0.02 eV):

It is not easy to measure 10 fmwith systematic errors in E atthe level of 0.02 eV!

Finally:an experimental limitation –but one that we might be ableto overcome!

Summary (#2)

Comparing data to data can see ∆R ≈ 5 fm.

Comparing feff to feff can see ∆R ≈ 5 fm.

Comparing data to feff can see ∆R ≈ 500 fm.

Our fits are NOT bad because

we got the statistics wrong.

our data is noisier than we think it is (the opposite, in fact!).

monochromator mis-calibration

background subtraction

any other “Systematic Error”.

Our fits are bad because:

The feff calculations are not even remotely as good as our data.

Summary (#2)

How is feff wrong?

Prime suspect for how feff is wrong:

The photo-electron mean-free-path and loss terms.

With the exchange self-energy Σ and inelastic losses, the photo-electronwavenumber is complex:

p =√

2m(E−Ef−Re[Σ(E)])~2 + i

E x-ray energy

Ef Fermi energy

Σ self-energy

λ mean free path =

√(E/2)

Im[Σ(E)] .

In addition, we can add in an energy broadening term Ei in the analysis:

p = p′ − ip′′ =

√{Re(p)(k)− i

− i Ei

How is feff wrong?

p =√

2m(E−Ef−Re[Σ(E)])~2 + i

E x-ray energy

Ef Fermi energy

Σ self-energy

λ mean free path =

√(E/2)

Im[Σ(E)] .

p = p′ − ip′′ =

√{Re(p)(k)− i

− i Ei

How is feff wrong?

p =√

2m(E−Ef−Re[Σ(E)])~2 + i

E x-ray energy

Ef Fermi energy

Σ self-energy

λ mean free path =

√(E/2)

Im[Σ(E)] .

p = p′ − ip′′ =

√{Re(p)(k)− i

− i Ei

How is feff wrong?

p =√

2m(E−Ef−Re[Σ(E)])~2 + i

E x-ray energy

Ef Fermi energy

Σ self-energy

λ mean free path =

√(E/2)

Im[Σ(E)] .

p = p′ − ip′′ =

√{Re(p)(k)− i

− i Ei

Hope for EXAFS: Improved loss terms

Improving loss terms by going beyond Hedin-Lunqvist (HL) exchange model

HL exchange uses a single plasmon to ac-count for e− – e− interactions in the com-plex self-energy Σ.

Kas, et al. use multiple (10!) excitations.

The dielectric constant Im[ε(ω)−1] is useto set the strength of these excitations.

Effect of Improved Loss Terms

Cu XANES Cu EXAFS

Cu XANES

Cu EXAFS

Cu XANES Cu EXAFS

Room Temperature Cu: feff85 vs. feff6

So, does this help with EXAFS?? Let’s repeat the fit to the 1st shell of roomtemperature Cu data: . . .

Parameter feff6 feff85

R(A) 2.543(0.003) 2.537(0.002)∆E0(eV) 4.1(0.5) 0.6(0.4)

S20 0.96(0.05) 0.86(0.03)σ2(A2) 8.5(0.4)×10−3 7.8(0.3)×10−3

χ2 1012. 584.χ2ν 242. 140.R 1.6× 10−5 0.9× 10−3

factor of 2 improvement in χ2, χ2ν , and R!

slight difference (outside uncertainties) in R and σ2.

C3 is still not significant (consistent with 0, no improvement in χ2ν), but R

appears a little short.

S20 looks too low. . . .

R(A) 2.543(0.003) 2.537(0.002)∆E0(eV) 4.1(0.5) 0.6(0.4)

S20 0.96(0.05) 0.86(0.03)σ2(A2) 8.5(0.4)×10−3 7.8(0.3)×10−3

χ2 1012. 584.χ2ν 242. 140.R 1.6× 10−5 0.9× 10−3

R(A) 2.543(0.003) 2.537(0.002)∆E0(eV) 4.1(0.5) 0.6(0.4)

S20 0.96(0.05) 0.86(0.03)σ2(A2) 8.5(0.4)×10−3 7.8(0.3)×10−3

χ2 1012. 584.χ2ν 242. 140.R 1.6× 10−5 0.9× 10−3

Room Temperature Cu: feff85 vs. feff6 (continued)

With feff85, setting S20 = 1 and instead refining Ei (energy broadening)

improves the fit even more:

Parameter feff6 feff85 (S20 ) feff85 (Ei )

R(A) 2.543(0.003) 2.537(0.002) 2.537(0.002)∆E0(eV) 4.1(0.5) 0.6(0.4) 0.6(0.3)

S20 0.96(0.04) 0.86(0.03) 1.0

Ei (eV) 0.0 0.0 1.1(0.2)σ2(A2) 8.5(0.4)×10−3 7.8(0.3)×10−3 8.2(0.2)×10−3

χ2 1012. 584. 360.χ2ν 242. 140. 86.R 1.6× 10−3 0.9× 10−3 0.6× 10−3

Now almost 3× better χ2, χ2ν , and R than feff6!

need additional 1.0 eV broadening to have no S20 parameter.

Of course, this is one test case . . .

R(A) 2.543(0.003) 2.537(0.002) 2.537(0.002)∆E0(eV) 4.1(0.5) 0.6(0.4) 0.6(0.3)

S20 0.96(0.04) 0.86(0.03) 1.0

Ei (eV) 0.0 0.0 1.1(0.2)σ2(A2) 8.5(0.4)×10−3 7.8(0.3)×10−3 8.2(0.2)×10−3

χ2 1012. 584. 360.χ2ν 242. 140. 86.R 1.6× 10−3 0.9× 10−3 0.6× 10−3

R(A) 2.543(0.003) 2.537(0.002) 2.537(0.002)∆E0(eV) 4.1(0.5) 0.6(0.4) 0.6(0.3)

S20 0.96(0.04) 0.86(0.03) 1.0

Ei (eV) 0.0 0.0 1.1(0.2)σ2(A2) 8.5(0.4)×10−3 7.8(0.3)×10−3 8.2(0.2)×10−3

χ2 1012. 584. 360.χ2ν 242. 140. 86.R 1.6× 10−3 0.9× 10−3 0.6× 10−3

Plots of fit and residuals:

Fit with feff6 Fit with feff85

Both fits are very good, but the feff85 fit clearly has a smaller residual.

Outstanding issues:

Is the shift in R correct??

Why does the residual have a period 2× Re[χ(R)]?

Plots of fit and residuals:

Fit with feff6 Fit with feff85

Both fits are very good, but the feff85 fit clearly has a smaller residual.

Outstanding issues:

Is the shift in R correct??

Why does the residual have a period 2× Re[χ(R)]?

Summary (#3) and Conclusions

We have many tools (algebraic constraints, restraints) for building physicalmodels when refining EXAFS data.

We think we understand most of the fitting statistics and noise in our data

theoretical calculations (feff) are the limiting step in modeling EXAFS data.

improvements in calculations (feff) should:I improve the ultimate precisions in R, N, and Z from real EXAFS data.I allow more robust “automated analysis” of EXAFS data.

improvements in calculations (feff) are being made!

Do we dare think about intelligent, automated analysis methods?

advances in xafs analysis

Documents

fourier analysis

exafs equation

number of paths

number of parameters

krpithe number of parameters

path parameters r

information theorythe

rehr xafs theory