Today’s Outline - November 14, 2016csrri.iit.edu/~segre/phys570/16F/lecture_22.pdf · Today’s Outline - November 14, 2016 Kramers-Kronig relations Friedel’s Law Bijvoet (Bay-voot)

Today’s Outline - November 14, 2016

• Kramers-Kronig relations

• Friedel’s Law

• Bijvoet (Bay-voot) Pairs

• MAD Phasing

• Quantum Origin of Resonant Scattering

• Imaging

Homework Assignment #7:Chapter 7: 2,3,9,10,11due Monday, November 28, 2016

Final Exam, Wednesday, December 7, 2016, Stuart Building 2132 sessions: 09:00-12:00; 13:00-17:00; (this may change) 16:00-18:00

Provide me with the paper you intend to present and a preferred sessionfor the examSend me your presentation in Powerpoint or PDF format before beforeyour session

C. Segre (IIT) PHYS 570 - Fall 2016 November 14, 2016 1 / 19



• Friedel’s Law


• MAD Phasing


• Imaging







• Friedel’s Law


• MAD Phasing


• Imaging







• Friedel’s Law


• MAD Phasing


• Imaging







• Friedel’s Law


• MAD Phasing


• Imaging







• Friedel’s Law


• MAD Phasing


• Imaging







• Friedel’s Law


• MAD Phasing


• Imaging







• Friedel’s Law


• MAD Phasing


• Imaging







• Friedel’s Law


• MAD Phasing


• Imaging







• Friedel’s Law


• MAD Phasing


• Imaging



Provide me with the paper you intend to present and a preferred sessionfor the exam

Send me your presentation in Powerpoint or PDF format before beforeyour session




• Friedel’s Law


• MAD Phasing


• Imaging





Refractive index

Scattering and refraction are alternative ways to approach the phenomenonof x-ray interaction with matter.

Thus the resonant response we have seenin scattering must be manifested in the index of refraction as well.

the electric field from the x-rays, ~E (t), induces a polar-ization response, ~P(t), in thematerial, whereχ = (ε/ε0 − 1)is the electric susceptibility

given an electron density ρand using the displacementfunction for the electrons inthe forced oscillator model

index of refraction can thusbe computed

~P(t) = ε0χ~E (t) = (ε− ε0)~E (t) = −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0 =

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2=

ε

ε0


Refractive index

Scattering and refraction are alternative ways to approach the phenomenonof x-ray interaction with matter. Thus the resonant response we have seenin scattering must be manifested in the index of refraction as well.




~P(t) = ε0χ~E (t) = (ε− ε0)~E (t) = −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0 =

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2=

ε

ε0


Refractive index


the electric field from the x-rays, ~E (t), induces a polar-ization response, ~P(t), in thematerial,

whereχ = (ε/ε0 − 1)is the electric susceptibility



~P(t) = ε0χ~E (t) = (ε− ε0)~E (t) = −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0 =

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2=

ε

ε0


Refractive index






~P(t) = ε0χ~E (t)

= (ε− ε0)~E (t) = −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0 =

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2=

ε

ε0


Refractive index






~P(t) = ε0χ~E (t) = (ε− ε0)~E (t)

= −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0 =

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2=

ε

ε0


Refractive index





~P(t) = ε0χ~E (t) = (ε− ε0)~E (t)

= −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0 =

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2=

ε

ε0


Refractive index





~P(t) = ε0χ~E (t) = (ε− ε0)~E (t)

= −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0 =

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2=

ε

ε0


Refractive index





~P(t) = ε0χ~E (t) = (ε− ε0)~E (t) = −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0 =

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2=

ε

ε0


Refractive index





~P(t) = ε0χ~E (t) = (ε− ε0)~E (t) = −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0 =

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2=

ε

ε0


Refractive index





~P(t) = ε0χ~E (t) = (ε− ε0)~E (t) = −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0

=

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2=

ε

ε0


Refractive index





~P(t) = ε0χ~E (t) = (ε− ε0)~E (t) = −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0 =

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2=

ε

ε0


Refractive index





~P(t) = ε0χ~E (t) = (ε− ε0)~E (t) = −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0 =

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2

=ε

ε0


Refractive index





~P(t) = ε0χ~E (t) = (ε− ε0)~E (t) = −eρx(t)

= −eρ(− e

m

) E0e−iωt

(ω2s − ω2 − iωΓ)

P(t)

E (t)= ε− ε0 =

(e2ρ

m

)1

(ω2s − ω2 − iωΓ)

n2 =c2

v2=

ε

ε0


Refractive index

n2 = 1 +

(e2ρ

ε0m

)1

(ω2s − ω2 − iωΓ)

= 1 +

(e2ρ

ε0m

)ω2s − ω2

(ω2s − ω2)2 + (ωΓ)2

+ iωΓ

(ω2s − ω2)2 + (ωΓ)2

-0.5

0

0.5

1

1.5

2

2.5

3

0.1 1 10

Re

[n2]

ω/ωs

0

0.5

1

1.5

2

2.5

3

0.1 1 10

Im[n

2]

ω/ωs


Refractive index

n2 = 1 +

(e2ρ

ε0m

)1

(ω2s − ω2 − iωΓ)

= 1 +

(e2ρ

ε0m

)ω2s − ω2

(ω2s − ω2)2 + (ωΓ)2

+ iωΓ

(ω2s − ω2)2 + (ωΓ)2

-0.5

0

0.5

1

1.5

2

2.5

3

0.1 1 10

Re

[n2]

ω/ωs

0

0.5

1

1.5

2

2.5

3

0.1 1 10

Im[n

2]

ω/ωs


Refractive index

n2 = 1 +

(e2ρ

ε0m

)1

(ω2s − ω2 − iωΓ)

= 1 +

(e2ρ

ε0m

)ω2s − ω2

(ω2s − ω2)2 + (ωΓ)2

+ iωΓ

(ω2s − ω2)2 + (ωΓ)2

-0.5

0

0.5

1

1.5

2

2.5

3

0.1 1 10

Re

[n2]

ω/ωs

0

0.5

1

1.5

2

2.5

3

0.1 1 10

Im[n

2]

ω/ωs


Absorption cross-section

Electrons in atoms are bound andtherefore have resonant effects dueto the binding forces

the imaginary part of the resonantscattering for an electron bound toan atom shows a frequency depen-dence with a peak at ω ≈ ωs

this single oscillator model, how-ever, does not reproduce theobserved absorption cross-sectionjump at an absorption edge

f ′′s (ω) =ω2sωΓ

(ω2 − ω2s )2 + (ωΓ)2

r20

0

0.2

0.4

0.6

0.8

1

1.2

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

,s

ω/ωs

σa,s(ω) = 4πr0cω2s Γ

(ω2 − ω2s )2 + (ωΓ)2







(ω2 − ω2s )2 + (ωΓ)2

r20

0

0.2

0.4

0.6

0.8

1

1.2

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

,s

ω/ωs


(ω2 − ω2s )2 + (ωΓ)2







(ω2 − ω2s )2 + (ωΓ)2

r20

0

0.2

0.4

0.6

0.8

1

1.2

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

,s

ω/ωs


(ω2 − ω2s )2 + (ωΓ)2







(ω2 − ω2s )2 + (ωΓ)2

r20

0

0.2

0.4

0.6

0.8

1

1.2

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

,s

ω/ωs


(ω2 − ω2s )2 + (ωΓ)2







(ω2 − ω2s )2 + (ωΓ)2

r20

0

0.2

0.4

0.6

0.8

1

1.2

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

,s

ω/ωs


(ω2 − ω2s )2 + (ωΓ)2







(ω2 − ω2s )2 + (ωΓ)2

r20

0

0.2

0.4

0.6

0.8

1

1.2

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

,s

ω/ωs


(ω2 − ω2s )2 + (ωΓ)2


Multi-oscillator model

The damping constant, Γis gener-ally much less than the resonantfrequency, ωs

thus the single oscillator is essen-tially a delta function

in a real atom, exceeding the ab-sorption edge allows the electronto be excited into a continuum ofstates which can be approximatedby a sum of resonant oscillatorswith frequency distribution g(ωs)

σa(ω) = 2π2r0c∑s

g(ωs)δ(ω − ωs)

a similar effect is seen in the reso-nant scattering term f ′(w)


(ω2 − ω2s )2 + (ωΓ)2

≈4πr0cπ

2δ(ω − ωs)

0

0.5

1

1.5

2

2.5

3

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

ω/ωs










(ω2 − ω2s )2 + (ωΓ)2

≈4πr0cπ

2δ(ω − ωs)

0

0.5

1

1.5

2

2.5

3

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

ω/ωs










(ω2 − ω2s )2 + (ωΓ)2

≈4πr0cπ

2δ(ω − ωs)

0

0.5

1

1.5

2

2.5

3

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

ω/ωs










(ω2 − ω2s )2 + (ωΓ)2

≈4πr0cπ

2δ(ω − ωs)

0

0.5

1

1.5

2

2.5

3

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

ω/ωs










(ω2 − ω2s )2 + (ωΓ)2

≈4πr0cπ

2δ(ω − ωs)

0

0.5

1

1.5

2

2.5

3

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

ω/ωs










(ω2 − ω2s )2 + (ωΓ)2

≈4πr0cπ

2δ(ω − ωs)

0

0.5

1

1.5

2

2.5

3

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

ω/ωs










(ω2 − ω2s )2 + (ωΓ)2

≈4πr0cπ

2δ(ω − ωs)

0

0.5

1

1.5

2

2.5

3

0.6 0.8 1 1.2 1.4 1.6 1.8 2

σa

ω/ωs










(ω2 − ω2s )2 + (ωΓ)2

≈4πr0cπ

2δ(ω − ωs)

-0.1

-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.6 0.8 1 1.2 1.4 1.6 1.8 2

f’

ω/ωs


Kramers-Kronig relations

It is generally not a good idea to rely on theoretical calculations of thedispersion corrections.

However, if it is possible to obtain the experimentalabsorption cross-section, σa, the resonant scattering can be computed.

first compute f ′′(ω) from themeasured absorption cross-section

f ′′(ω) = −(

ω

4πr0c

)σa(ω)

then use the Kramers-Kronig relations which connect the resonant term tothe absorptive term and where all the integrals are “principal value”integrals

f ′(ω) =1

πP∫ +∞

−∞

f ′′(ω′)

(ω′ − ω)dω′ =

2

πP∫ +∞

0

ω′f ′′(ω′)

(ω′2 − ω2)dω′

f ′′(ω) = − 1

πP∫ +∞

−∞

f ′(ω′)

(ω′ − ω)dω′ = −2ω

πP∫ +∞

0

f ′(ω′)

(ω′2 − ω2)dω′



It is generally not a good idea to rely on theoretical calculations of thedispersion corrections. However, if it is possible to obtain the experimentalabsorption cross-section, σa, the resonant scattering can be computed.


f ′′(ω) = −(

ω

4πr0c

)σa(ω)


f ′(ω) =1

πP∫ +∞

−∞

f ′′(ω′)

(ω′ − ω)dω′ =

2

πP∫ +∞

0

ω′f ′′(ω′)

(ω′2 − ω2)dω′

f ′′(ω) = − 1

πP∫ +∞

−∞

f ′(ω′)

(ω′ − ω)dω′ = −2ω

πP∫ +∞

0

f ′(ω′)

(ω′2 − ω2)dω′





f ′′(ω) = −(

ω

4πr0c

)σa(ω)


f ′(ω) =1

πP∫ +∞

−∞

f ′′(ω′)

(ω′ − ω)dω′ =

2

πP∫ +∞

0

ω′f ′′(ω′)

(ω′2 − ω2)dω′

f ′′(ω) = − 1

πP∫ +∞

−∞

f ′(ω′)

(ω′ − ω)dω′ = −2ω

πP∫ +∞

0

f ′(ω′)

(ω′2 − ω2)dω′





f ′′(ω) = −(

ω

4πr0c

)σa(ω)


f ′(ω) =1

πP∫ +∞

−∞

f ′′(ω′)

(ω′ − ω)dω′ =

2

πP∫ +∞

0

ω′f ′′(ω′)

(ω′2 − ω2)dω′

f ′′(ω) = − 1

πP∫ +∞

−∞

f ′(ω′)

(ω′ − ω)dω′ = −2ω

πP∫ +∞

0

f ′(ω′)

(ω′2 − ω2)dω′





f ′′(ω) = −(

ω

4πr0c

)σa(ω)

then use the Kramers-Kronig relations which connect the resonant term tothe absorptive term

and where all the integrals are “principal value”integrals

f ′(ω) =1

πP∫ +∞

−∞

f ′′(ω′)

(ω′ − ω)dω′ =

2

πP∫ +∞

0

ω′f ′′(ω′)

(ω′2 − ω2)dω′

f ′′(ω) = − 1

πP∫ +∞

−∞

f ′(ω′)

(ω′ − ω)dω′ = −2ω

πP∫ +∞

0

f ′(ω′)

(ω′2 − ω2)dω′





f ′′(ω) = −(

ω

4πr0c

)σa(ω)



f ′(ω) =1

πP∫ +∞

−∞

f ′′(ω′)

(ω′ − ω)dω′

=2

πP∫ +∞

0

ω′f ′′(ω′)

(ω′2 − ω2)dω′

f ′′(ω) = − 1

πP∫ +∞

−∞

f ′(ω′)

(ω′ − ω)dω′ = −2ω

πP∫ +∞

0

f ′(ω′)

(ω′2 − ω2)dω′





f ′′(ω) = −(

ω

4πr0c

)σa(ω)



f ′(ω) =1

πP∫ +∞

−∞

f ′′(ω′)

(ω′ − ω)dω′

=2

πP∫ +∞

0

ω′f ′′(ω′)

(ω′2 − ω2)dω′

f ′′(ω) = − 1

πP∫ +∞

−∞

f ′(ω′)

(ω′ − ω)dω′

= −2ω

πP∫ +∞

0

f ′(ω′)

(ω′2 − ω2)dω′





f ′′(ω) = −(

ω

4πr0c

)σa(ω)



f ′(ω) =1

πP∫ +∞

−∞

f ′′(ω′)

(ω′ − ω)dω′ =

2

πP∫ +∞

0

ω′f ′′(ω′)

(ω′2 − ω2)dω′

f ′′(ω) = − 1

πP∫ +∞

−∞

f ′(ω′)

(ω′ − ω)dω′

= −2ω

πP∫ +∞

0

f ′(ω′)

(ω′2 − ω2)dω′





f ′′(ω) = −(

ω

4πr0c

)σa(ω)



f ′(ω) =1

πP∫ +∞

−∞

f ′′(ω′)

(ω′ − ω)dω′ =

2

πP∫ +∞

0

ω′f ′′(ω′)

(ω′2 − ω2)dω′

f ′′(ω) = − 1

πP∫ +∞

−∞

f ′(ω′)

(ω′ − ω)dω′ = −2ω

πP∫ +∞

0

f ′(ω′)

(ω′2 − ω2)dω′





f ′′(ω) = −(

ω

4πr0c

)σa(ω)


f ′(ω) =1

πP∫ +∞

−∞

f ′′(ω′)

(ω′ − ω)dω′ =

2

πP∫ +∞

0

ω′f ′′(ω′)

(ω′2 − ω2)dω′

f ′′(ω) = − 1

πP∫ +∞

−∞

f ′(ω′)

(ω′ − ω)dω′ = −2ω

πP∫ +∞

0

f ′(ω′)

(ω′2 − ω2)dω′


Calculated cross-sections


Calculated cross-sections


Scattering from two unlike atoms

Two unlike atoms with scatter-ing factors f1 and f2 are orientedby a vector pointing from thelarger to the smaller.

Consider two cases, with thescattering vector Q in the samedirection as the orientation vec-tor and opposite to the orienta-tion vector.

Now compute the scattered in-tensity in each case, assumingscattering factors are purely real.




Consider two cases, with thescattering vector Q

in the samedirection as the orientation vec-tor and opposite to the orienta-tion vector.





Consider two cases, with thescattering vector Q in the samedirection as the orientation vec-tor

and opposite to the orienta-tion vector.













Friedel’s Law

A(+Q) = f1 + f2e+iQx

I (+Q) = (f1 + f2e+iQx)(f1 + f2e

−iQx)

= f 21 + f 22 + 2f1f2 cos(Qx)

I (+Q) = I (−Q) Friedel′s Law

A(−Q) = f1 + f2e−iQx

I (−Q) = (f1 + f2e−iQx)(f1 + f2e

+iQx)

= f 21 + f 22 + 2f1f2 cos(Qx)


Friedel’s Law


I (+Q) = (f1 + f2e+iQx)(f1 + f2e

−iQx)

= f 21 + f 22 + 2f1f2 cos(Qx)



I (−Q) = (f1 + f2e−iQx)(f1 + f2e

+iQx)

= f 21 + f 22 + 2f1f2 cos(Qx)


Friedel’s Law


I (+Q) = (f1 + f2e+iQx)(f1 + f2e

−iQx)

= f 21 + f 22 + 2f1f2 cos(Qx)



I (−Q) = (f1 + f2e−iQx)(f1 + f2e

+iQx)

= f 21 + f 22 + 2f1f2 cos(Qx)


Friedel’s Law


I (+Q) = (f1 + f2e+iQx)(f1 + f2e

−iQx)

= f 21 + f 22 + 2f1f2 cos(Qx)



I (−Q) = (f1 + f2e−iQx)(f1 + f2e

+iQx)

= f 21 + f 22 + 2f1f2 cos(Qx)


Friedel’s Law


I (+Q) = (f1 + f2e+iQx)(f1 + f2e

−iQx)

= f 21 + f 22 + 2f1f2 cos(Qx)



I (−Q) = (f1 + f2e−iQx)(f1 + f2e

+iQx)

= f 21 + f 22 + 2f1f2 cos(Qx)


Breakdown of Friedel’s Law

If the scattering factor has resonant terms which are not negligible, wehave to include them in the computation

fj = f 0j + f ′j + if ′′j

= rjeiφj

j = 1, 2 rj = |fj |A(Q) = r1e

iφ1 + r2eiφ2e iQx

I (Q) = (r1eiφ1 + r2e

iφ2e iQx)(r1e−iφ1 + r2e

−iφ2e−iQx)

= r21 + r22 + r1r2eiφ1e−iφ2e−iQx + r1r2e

−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))

I (Q) = |f1|2 + |f2|2 + 2r1r2 cos(Qx + φ1 − φ2) 6= I (−Q)

Thus, Friedel’s Law breaks down unless there is a center of symmetry:

F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)

= [2r1 cos(Qx1)]e−iφ1 + [2r2 cos(Qx2)]e−iφ2

I (Q) = 4|f1|2 + 4|f2|2 + 8|f1||f2| cos(Qx1) cos(Qx2) cos(φ2 − φ1)




fj = f 0j + f ′j + if ′′j

= rjeiφj

j = 1, 2

rj = |fj |A(Q) = r1e

iφ1 + r2eiφ2e iQx

I (Q) = (r1eiφ1 + r2e


−iφ2e−iQx)


−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))



F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)






fj = f 0j + f ′j + if ′′j = rjeiφj j = 1, 2 rj = |fj |

A(Q) = r1eiφ1 + r2e

iφ2e iQx

I (Q) = (r1eiφ1 + r2e


−iφ2e−iQx)


−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))



F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)








iφ2e iQx

I (Q) = (r1eiφ1 + r2e


−iφ2e−iQx)


−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))



F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)








iφ2e iQx

I (Q) = (r1eiφ1 + r2e


−iφ2e−iQx)


−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))



F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)








iφ2e iQx

I (Q) = (r1eiφ1 + r2e


−iφ2e−iQx)


−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))



F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)








iφ2e iQx

I (Q) = (r1eiφ1 + r2e


−iφ2e−iQx)


−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))



F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)








iφ2e iQx

I (Q) = (r1eiφ1 + r2e


−iφ2e−iQx)


−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))

I (Q) = |f1|2 + |f2|2 + 2r1r2 cos(Qx + φ1 − φ2)

6= I (−Q)


F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)








iφ2e iQx

I (Q) = (r1eiφ1 + r2e


−iφ2e−iQx)


−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))



F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)








iφ2e iQx

I (Q) = (r1eiφ1 + r2e


−iφ2e−iQx)


−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))



F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)








iφ2e iQx

I (Q) = (r1eiφ1 + r2e


−iφ2e−iQx)


−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))



F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)








iφ2e iQx

I (Q) = (r1eiφ1 + r2e


−iφ2e−iQx)


−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))



F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)








iφ2e iQx

I (Q) = (r1eiφ1 + r2e


−iφ2e−iQx)


−iφ1e iφ2e iQx

= |f1|2 + |f2|2 + r1r2(e−(Qx+φ1−φ2) + e+(Qx+φ1−φ2))



F = r1e−i(φ1+Qx1) + r1e

−i(φ1−Qx1) + r2e−i(φ2+Qx2) + r2e

−i(φ2−Qx2)




Argand diagram

This can all be described graphically using an Argand diagram:

no resonant terms

including resonant terms


Argand diagram

This can all be described graphically using an Argand diagram:

no resonant terms including resonant terms


ZnS example

The ZnS structure is not cen-trosymmetric and when viewedalong the 〈111〉 direction, it showsalternating stacked planes of Znand S atoms.

Scattering from opposite faces of asingle crystal of ZnS gives a dif-ferent scattering factor and onecan deduce the terminating surfaceatom.


ZnS example

The ZnS structure is not cen-trosymmetric and when viewedalong the 〈111〉 direction, it showsalternating stacked planes of Znand S atoms.

Scattering from opposite faces of asingle crystal of ZnS gives a dif-ferent scattering factor and onecan deduce the terminating surfaceatom.


Bijvoet pairs - chiral molecules

Consider a tetrahedral molecule of carbon with four different species ateach corner, oriented so the lightest is projected to the origin.


Bijvoet pairs - chiral molecules

Consider a tetrahedral molecule of carbon with four different species ateach corner, oriented so the lightest is projected to the origin.


Atomic scattering factors

Each of the three atoms not at the origin has a scattering factor for ~Q asshown


Left handed scattering factor

FS = |fs |+ |fm|e−iφme iφ + |fl |e−iφl e−iφ


Left handed scattering factor

FS = |fs |+ |fm|e−iφme iφ + |fl |e−iφl e−iφ


Right handed scattering factor

FR = |fs |+ |fm|e−iφme−iφ + |fl |e−iφl e iφ


Right handed scattering factor

FR = |fs |+ |fm|e−iφme−iφ + |fl |e−iφl e iφ


Scattering factor comparison

It is thus possible to tell the difference in handedness of chiral moleculesimply by x-ray scattering

s

∣∣∣|fs |+ |fm|e−iφme iφ + |fl |e−iφl e−iφ∣∣∣2 6= ∣∣∣|fs |+ |fm|e−iφme−iφ + |fl |e−iφl e iφ

∣∣∣2


Scattering factor comparison

It is thus possible to tell the difference in handedness of chiral moleculesimply by x-ray scattering

s

∣∣∣|fs |+ |fm|e−iφme iφ + |fl |e−iφl e−iφ∣∣∣2 6= ∣∣∣|fs |+ |fm|e−iφme−iφ + |fl |e−iφl e iφ

∣∣∣2C. Segre (IIT) PHYS 570 - Fall 2016 November 14, 2016 17 / 19

MAD phasing


MAD phasing


MAD phasing


MAD phasing


Comparison of matrix elements

Absorption

e~A · ~pm

Thomson scattering

e2A2

2m

Resonant scattering

(e~A · ~pm

)2



Absorption

e~A · ~pm

Thomson scattering

e2A2

2m

Resonant scattering

(e~A · ~pm

)2



Absorption

e~A · ~pm

Thomson scattering

e2A2

2m

Resonant scattering

(e~A · ~pm

)2


Documents

Today’s Outline - November 14, 2016csrri.iit.edu/~segre/phys570/16F/lecture_22.pdf · Today’s Outline - November 14, 2016 Kramers-Kronig relations Friedel’s Law Bijvoet (Bay-voot)