Dynamical Diffraction Theorylehre.ikz-berlin.de/physhu/...presentation_english.pdf · V. Dynamical...

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V. Dynamical Diffraction Theory

Dynamical Diffraction Theory

Fraunhofer Approximation

∫⋅−ρ= rQi

ikR

00rad e)r(dVR

ePEr)Q,R(E

vvvv

λ

Θπ=

sin4Qr

k'kQvvr

−=k

k’Q

2Θ

Repetition: Kinematical Theory

S(Q): Structure factor G(Q): Lattice factor

)Q(G)Q(SPR

rE

ee)Q(fR

rPEEE

00

Atoms

G

n,n,n

)ananan(Qi

S

k

1j

rQij

00jcrystal

321

332211j

vv

4444 34444 21444 3444 21

v vvvvvv

××=

=××== ∑ ∑∑ ++−

=

−

dVk

r

R

r-R

Observer

2Θ

k’

(5.1)

(5.2)

Bragg condition: 2dhkl sinθ = λ or Q = G

Introduction: Limitations of Kinematical Theory

When (why at all?) do we need a different and better theory

What simplifications have been made in the framework of kinematical theory?

k

k’No Refraction

No Weakening

Single Scattering Event


k

k’No Refraction

No Weakening

Single Scattering Event

• Refraction of the X-ray waves at the surface (or at other interfaces)

• Weakening through

� Absorption� Extinction (if Bragg condition is fulfilled)

• The Bragg reflected wave can be again reflected at other parallel net-planes

� Multiple Reflections

• The Bragg reflected wave interferes with the incoming wave � Creation of a “Wave Field”

Introduction: Limitations of Kinematical Theory


Introduction: History

• Loss of intensity of incoming wave � Gain of intensity of diffracted wave

• Gain of intensity of incoming wave � Loss of intensity of diffracted wave

� “Dynamical Process“

• Description through new theory (Darwin 1914, Ewald 1917)

� Modification of kinematical theory is not sufficient.� Solution of Maxwell’s equation inside and outside crystal � Boundary conditions for E, D and k at the interface

Charles Galton Darwin (1887 – 1962) Paul Peter Ewald (1888 – 1985)


• Dynamical theory is most relevant for perfect crystals relevant

• This is why dynamical scattering was initially not able to describe the observed intensities.

• However, kinematical theory – which was developed much later – was successful

• Nowadays, along with the presence of highly perfect single crystals - the dynamical theory is well established

Dennoch hat sie (die einfache Theorie) bei der Strukturbestimmung

an Kristallen erstaunlich gute Dienste getan aus einem

absonderlichen Grund: die Kristalle, welche die Natur darbietet,

sind in ihrem Raumgitter so fehlerhaft, dass sie – so möchte man

fast sagen – keine bessere Theorie verdienen

M. v. Laue, Röntgenstrahlinterferenzen,

Akademische Verlagsanstalt, Frankfurt/Main (1960)



• Dynamical theory is relevant for description of diffraction on highly perfect crystals, only!

• Quartz, Calcite (“Kalkspat”): Natural appearance with high perfection

• Nowadays

� Perfect artificial crystals (Silicon, Germanium, ..) � Leibniz-Institute for Crystal Growth

� Epitaxial layers with very smooth interfaces and (near) perfect crystal structure

� Anomal Absorption (Borrmann, 1941)

� Interference double-refraction (Borrmann, 1955)

� Curvature of X-ray trajectories (Hildebrandt, 1959)



Gerhard Borrmann

(1908 – 2006)

Zweikristalltopographie in Transmissionsgeometrie

QuelleKollimatorkristall

Probe

Fo

t op

l atte

durchgehender

Strahl

Air

Crystal

Transmitted Beam Bragg Reflected Wave

Strong Refraction Effect Interference Double Refraction

Wire

X-ray sourceCollimator Crystal

Sample

Transmitted Beam

An Example Why We Need a ‘New Theory”

Photo Plate

How does Dynamical Theory Work?

Wave Optics

• Describes interaction between X-rays and a ‘medium’

• We ask: which (plane) waves are allowed inside and outside of a medium?

• For example: Length of wave vector is different inside medium as compared to vacuum (or air)

• Description of transition from vacuum to crystal via boundary conditions

Dynamical Theory

• Same procedure than above

• Solution of Maxwell’s equations (better: wave equations) inside and outside of a crystal

• Taking into account: boundary conditions


Refraction



Boundary conditions at the interface

�are valid for the transmitted and the diffracted wave

�Parallel components of the wave vectors are continuous across the interface

�Length of wave vector is changing (owing to different refractive index)

�Normal components of the wave vectors are thus different at both sides of the interface

However: The solutions are more complicated than in Optics

since

Planes waves as solutions for the transmitted and reflected beam are not sufficient !



The solutions are usually more complicated than in traditional Optics

since

Planes waves as solutions for the transmitted and reflected waves are not suitable !

)trk(i0eE)t,r(E ω−⋅=

vvvvv

constant amplitude � constant intensity

This interplay cannot be described by a single plane wave !

Incoming wave supplies diffracted wave and vice versa

� We need at least two plane waves with constant amplitude (no absorption)



Incoming wave supplies diffracted wave and vice versa

How can we fulfill this requirement ?

� A spatial change of intensity can be achieved by a slight difference of the length of the wave vectors (not frequency!) of these plane waves

� This is only possible for the normal components of these plane waves (since parallel components have to be continuous across the interface)

� Two waves with slightly different wave vectors undergo a beating (“Schwebung”)

� This beating occurs for both the transmitted as well as for the diffracted beam

� It turns out that transmitted and diffracted waves show exactly opposite beating, i.e. the dominance of either transmitted or diffracted wave is oscillating

We call this modulation “Pendellösung Oscillations”

Derivation of Wave Equation

Maxwell Equations: (5.3)tD

jHrottB

Erot∂

∂+=

∂

∂−=

vvv

vv

(5.4)With the dielectric Polarization P we can write for the dielectric displacement field D PED 0

vvr+ε=

For j = 0 (no free electric current) and with the Ansatz

(5.5)ti

0

ti0

e)r(H)t,r(H

e)r(D)t,r(Dω−

ω−

=

=vvvv

vvvv

(5.6)we find

HitH

DitD

vv

vv

ω−=∂

∂

ω−=∂

∂

So let us start from the very beginning! We will derive the wave equation used in dynamical theory

Five fields (E,B,H,D,j) involved! We need a wave equation for E or D only !




and with (5.3): DitD

Hrotv

vv

ω−=∂

∂= (5.7)

Since div rot H = 0: (5.8)0Ddiv =v

(5.9)For high frequencies we can approximate:

HB 0

wvµ=

We now apply the rotation operator to Eq. (5.3)

(5.10)

Dct

DiHroti

tH

rot)PD(rotrotErotrot

2

2)6.5(

00

)3.5(

00

)6.5(

00

)3.5()4.5(

0

vv

vvvv

ω=

∂

∂µωε=µωε=

=

∂

∂µε−=−=ε



We use of the operator relationship DD)Ddiv(gradDrotrotvvvv

∆−=∆−=

(5.11)ProtrotDc

)PE(rotrotDrotrotD 2

2)10.5(

0

)4.5( vvvvvv−

ω−=+ε−=−=∆and end up with

DEP 00

vvvχε≈χε=Finally, we can approximate:

and (5.11) reads as 0Dc

)D(rotrotD 2

2

=ω

+χ+∆vvv

Wave equation for the dielectric displacement field D

(5.12)

Homework: Derivation of (5.13)

0Dc

)D(rotrotD 2

2

=ω

+χ+∆vvv

Wave equation for the dielectric displacement field D

(5.12)

We can derive an analogous equation for the electric field E:

0E)1(c

Edivgradc

E 2

2

2

2

=χ+ω

+⋅ω

+∆vvv

(5.13)



The 2nd term of this equation leads to a coupling (intermixing) of the components (direction of

polarization) of E. This coupling is very small since we can approximately write:

( ) 0Ddiv1

Ddiv

1D

divEdiv0000 1

1

0

11 ==

χ+≈

χ+=

χ+εεε

vvv

v

0E)1(c

Edivgradc

E 2

2

2

2

=χ+ω

+⋅ω

+∆vvv

0E)1(c

E 2

2

=χ+ω

+∆vv

This equation is valid for each component Ei !

Remark 1



)r(E)r(V̂)r(E)K( 2 vvvv=+∆

)2

cK(KgraddivV̂ 2

λ

π=

ω=χ−=with

Scattering Potential Operator

)r(r4)Re(KK)r(VV̂ e022 vv

ρπ−=χ−≈χ−==

If we use the approximation div E = 0 (see remark 1) and if we neglect absorption (use only real part of dielectric polarizability) we obtain:



0E)1(c

Edivgradc

E 2

2

2

2

=χ+ω

+⋅ω

+∆vvv

Remark 2

Interaction with Matter

Interaction with matter (crystal) � Dielectric Dynamic Polarizability χ(r,ω)

• scalar• complex (� imaginary part describes absorption)• depends on r (� periodic electron density)• dynamic (depends on X-ray energy)

∑γω+ω−ωε

ρ−=ωχ

j j2j

2

j

0

2e

i

g

m

e)(),(

rr

(see lecture 3)

( ) )r(K

r4)r(

m

e)r(Re e2

0e20

2 vvvρ

π−=ρ

ωε−=χ

Neglecting energy dependence (No Hönl corrections)

(5.14)

)2

cK(

λ

π=

ω=


Relationship between χ(r,ω) and structure factor Sg

∑∑ χ+χ=χ=χg

rgiig

rg

g

rgig e)i(e)r(

v

vvvv

v

vvv

v

g20

2

g SVm

evv

ωε=χ

See Lecture 2

(5.15)

gS v gv

: Structure factor for the reciprocal lattice vector V: Volume of unit cell

(5.16)

Interaction with Matter


Denotation of Physical Quantities

Quantities outside of the crystal � Capital Letters (e.g. K, D, E)

Quantities inside the crystal � Small Letters (e.g.. k, d, e)

These quantities are related to each other

(K, D, E) (k,d,e)

via requirement of continuity at the interface

n1

n2

k

k

kr

kt

k||

t

r

Example:Parallel component of the wave vector is continuous

)z,xt(kK tt gg== (5.17)

Most Simple Case: (i) Bragg condition is not fulfilled(ii) Only transmitted beam inside crystal

)r(e)i(e)r( 0g

rgiig

rg

g

rgig

vv

v

vvvv

v

vvv χ=χ≈χ+χ=χ=χ ∑∑ (5.18)

Insert this into (5.12)

{{

0dc

)1(ddc

)d)ddiv(grad(d

dc

drotrotddc

)d(rotrotd

2K

2

2

02

2

00

2

2

02

2

=ω

+χ−∆=ω

+∆−χ+∆=

=ω

+χ+∆=ω

+χ+∆

vvvvvv

vvvvvv

Solution of Wave Equation: One-Beam-Case

� One-beam-case

We can treat this case as if the X-ray wave does not ‘feel’ the presence of a periodic lattice

0Dc

)D(rotrotD 2

2

=ω

+χ+∆vvv And use small letters for

solutions inside crystal


Since χ0 << 1: 0

0 11

1χ+

≈χ−

and we obtain: 0dkd 20 =+∆vr

220

220 Kn)1(Kk =χ+=

211n 00 χ+≈χ+=

(5.19)

(5.20)

Refractive Index



This is the stationary Helmholtz equation which we know from wave optics

Refractive Index

211n 00 χ+≈χ+=

∑+−

=j j

22j

j

0

2e

i

g

m

e

γωωωε

ρωχ

)(),(

rr

0)Re( 0 <χ

β+δ−= i1n (5.21)

Material δ [10−5] β [10−7] β / δ αC

Si 0,7563 1,748 0,023 0,22°GaAs 1,441 4,087 0,028 0,31°Ge 1,444 4,200 0,029 0,31°

Materials parameter δ and β(Cu-Kα1 Radiation)

� n is a complex, scalar number

� Real part describes refraction

� Re(n) < 1

� Imaginary part describes absorption


0dkd 20 =+∆vr

0DKD 2 =+∆vv

Vacuum Inside Medium (Crystal)

rKi0 eD)r(D

vvvvv=

rki0

0ed)r(dvvvvv

=

Possible solutions(plane waves)

PP00(r)

KK00(r)

AA(t)

PPee=P00(t)

AA(r)

kk00

(r)

kk00

(t)

KKee == K

00(r)

α

xx

yy

Medium 2 (Substrat)

Medium 1

Multitude of possible solutions for k vector

Sphere with radius K (Vacuum) k0 (Medium)

Dispersion Surface



PP00(r)

KK00(r)

AA(t)

PPee=P00(t)

AA(r)

kk00

(r)

kk00

(t)

KKee == K

00(r)

α

xx

yy

Medium 2 (Substrat)

Medium 1

Dispersion Surface+

Tangential Condition Excitation Points



Horizontal components of the wave vectors: these are all identical for all three waves

Normal components of the wave vectors: they are different for all three waves

α=−== sinKKKK )r(y0

)t(y0ey α: angle of incidence

022

ex02

y sinKK)1(Kk χ+α=−χ+=

Vacuum

Medium



0sinKK)1(Kk 022

ex02

y =χ+α=−χ+=Medium (α‘=0)

( ) e0

0cr

2Re ρπ

λ=δ=χ−≈α

Critical Angle of Total External Reflection

!

Cu Kα (λ = 1.541 Å )

Material δ [10−5] β [10−7] β / δ αC

Si 0,7563 1,748 0,023 0,22°GaAs 1,441 4,087 0,028 0,31°Ge 1,444 4,200 0,029 0,31°



)r()t()r()t( ddDD +=+

( ) ( ))r()t()t(y

)r()t()t(y ddkDDK −=−

Horizontal components of D and H are continuous at the interface between two media

0d )r( =Media are semi infinite:

yy

yy)t(

)r(

kK

kK

D

Dr

+

−==

yy

y)t(

)t(

kK

K2

D

dt

+==

(for s-Polarization)

Fresnel reflection coefficient transmission coefficient

Comment: The Fresnel coefficients are different for p- and s-Polarization, however, these differences vanish for small angles of incidence.

(5.22)



2rR =2tT =

For the reflected and transmitted intensities we can thus write:

0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,610-5

10-4

10-3

10-2

10-1

100

101

αcR

efle

ktiv

ität

α (Grad)

Calculation:

• Semi infinite Silicon

• λ = 1.54 Å (αc = 0.22°)

• For a more detailed discussion:See Lecture 8



If we fulfill Bragg condition:

� beyond the transmitted (refracted) beam also other ‘beams’ show up inside the crystal

� the (periodic) spatial dependence of the dielectric polarizability has to be taken into account

� We use the same ansatz as applied for solution of Schrödinger equation in solid state physics:

0Dc

)D(rotrotD 2

2

=ω

+χ+∆vvv

)r(v

χ=χwith

∑∑ χ+χ=χ=χg

rgiig

rg

g

rgig e)i(e)r(

v

vvvv

v

vvv

v

∑=g

rgig

rki e)k(de)r(dv

vvv

vv vvv(5.23)Bloch waves

)r(ue)r(d krki vvv

vvv

=

Bloch function, Expansion of d(r) into plane waves

(5.24)


Solution of Wave Equation: Two- and Multi-Beam-Case

)r(uee)k(de)r(d krki

g

rgig

rki vvvvv

vv

v

vvv

vv

== ∑

Remarks: (i) Bloch functions exhibit periodicity of the 3D lattice

(ii) Wave fields do not exhibit periodicity of the 3D lattice

(iii) Intensities exhibit periodicity of the 3D lattice

)r(dvv

2)r(dvv

)r(ukv

v

)k0kk(gkk 0g

vvvvvvvv =+=+= (5.25)Let us define:

Insertion of Bloch waves (5.23) and expansions of χ(r) (5.18) into the wave equation (5.12):

{ } 0ddk

Kk

'ggg'g'ggg02

g

22g

=⋅χ−⋅

χ−

−∑≠

−vvvvvvv

v

v vvv

v

(5.26)

{ }g'gd vvv

: Component of perpendicular to (Direction of polarization!)'gdvv

gk vv

222 cK ω=

(5.23)

Home work: Derivation of Equation (5.26)



{ } 0ddk

Kk

'ggg'g'ggg02

g

22g

=⋅χ−⋅

χ−

−∑≠

−vvvvvvv

v

v vvv

v

(5.26)

Basic equations of dynamical theory

� Infinite set of algebraic equations instead of a differential equation

� This implies the presence of an infinite number of partial plane waves inside the crystal



Let us assume now: Just two ‘strong’ waves are present inside crystal

� transmitted (refracted) and Bragg reflected wave

� with amplitudes and

� with wave vectors and

� intensity is periodic

0dv

gdvv

0kv

gkk 0gvvv

v +=

( ) ( ) rgsinddIm2rgcosddRe2IIededI *go

*gog0

2rki

grki

0go

vvvvvvvvv vvvv

+++=+= (5.27)

Pendellösung Oscillations

The periodic wave field (intensity!) inside the crystal can be written as superposition between two plane waves, which are maximal

• directly on the net-planes (here, the electron density is elevated)

and

• between adjacent net-planes (here, the electron density is reduced)

( ) ( ) rgsinddIm2rgcosddRe2IIededI *go

*gog0

2rki

grki

0go

vvvvvvvvv vvvv

+++=+= (5.27)

The two partial waves experience different electron densities (better: dielectric polarizabilities) and thus exhibit a slightly different length of the wavevector (owing to continuity requirement only in vertical direction)

‘Beating‘ of Intensities � Pendellösung oscillations


Pendellösung =

� interference of two plane waves

� for either the transmitted or the reflected waves

� the intensity oscillates (“pendelt”) between transmitted and reflected waves


Pendellösung Oscillations

Surface

Transmitted Wave Reflected Wave

( ) ( )π

ππππσσ θ≈××==

j

jiBijj]j[ii]j[i d

dd2cosdssdunddd

vvvvv

π-Polarization: Component of d within plane of incidence σ-Polarization: Component of d perpendicular to plane of incidence

Solution of wave equation for two-beam-case:

Angle between and remains constant in the investigated angular range

gk vv

0kvθB : Bragg angle

: Direction vectors of the wave vectors jkv

jsv

0dk

KkPd

0Pddk

Kk

g02g

22g

0g

gg0020

220

=⋅

χ−

−+χ−

=χ−⋅

χ−

−−

vv

vv

(5.28)

(5.29)

−

π

σ

θ= für

2cos

1P

BWith Polarization (5.30)

(5.26)Two

Beams


Solution of Wave Equation: Two-Beam-Case

gv

−χgvχ

0dk

KkPd

0Pddk

Kk

g02g

22g

0g

gg0020

220

=⋅

χ−

−+χ−

=χ−⋅

χ−

−−

vv

vv

• Homogenous linear system of equations• It contains the terms and • Bragg reflected wave can be reflected back into the transmitted wave• Incoming wave supplies diffracted wave and vice versa (� Pendellösung)

(5.29)

0Pk

Kk

k

Kk 2gg02

g

22g

020

220 =χχ−

χ−

−⋅

χ−

−−vv

This (secular) equation describes the shape of the dispersion surface

(5.30)

Nontrivial solution of the system of equations: det (..) = 0!



0ii 1Kk: χ+−=ξ (i = 0,g) (5.31)

Distance of the origins of k0 and kg from the circles with radiusaround the reciprocal lattice points 0 and g, respectively

00 1Kk χ+=

00 2111 χ+≈χ+ (5.32) Further approximations:

22g

20 Kkk ≈≈ (5.33)

For the denominator of (5.30)

gg22

g0 PK41

vv−χχ=ξξ (5.34)

Hyperbola Equation around Lorentz point M



• Shift of Laue point L towards Lorentz point M (caused by refraction)

• Removement of degeneration of point M (Breaking up of dispersion surface into two branches)

• σ and π-Polarization have to be treated independently (� four branches)



Laue Point

Lorentz Point

Ewald sphere inside crystal

Ewald sphere in vacuum

• Two branches with slightly different vertical component of the wave vector � leads to beating

Dispersion Surface

gg22

g0 PK41

vv−χχ=ξξ

• Relative Splitting depends on strength of interaction• Strength of interactions depends on kind of “radiation” (X-rays, electrons, neutrons)• X-rays, 1Å, Silicon � Splitting ≈

gvχ

10-6


Darwin Width

Consequences: Symmetric (111)-Reflection,

270 µm thick Silicon Crystal

• finite width of Bragg reflection (Darwin width)

• pronounced Pendellösung oscillations, in particular for Laue case• shift of Bragg peak (for Bragg case) relative to vacuum Bragg angle

B0

gggD sin

1θγ

γ⋅χχ=θ∆ −

v

vv

00

gg

cos

cos

ψ=γ

ψ=γ vv

(5.35)


B0

gggD sin

1θγ

γ⋅χχ=θ∆ −

v

vv

Darwin width depends on:

( )( )Bg

Bg

0

g

cos

cos

θ+ψ

θ−ψ=

γ

γ=γ

v

• the Fourier components χg und χ-g

� Strength of interaction� Kind of radiation (X-rays, electrons, neutrons)� Kind of Bragg reflection (h k l)

• Wave length of X-rays

• Typical values for X-rays: E = 10 keV, Silicon, ∆θD ≅ 1 .. 5 arcsec

• Scattering geometryAsymmetry parameter


Darwin Width

RemarkBragg reflection g can be also reached via inversion of beam directions. There are however differences in the Darwin widths:

• Steep incidence and glancing exit:� reduced angular acceptance by the factor |γ|1/2

� enlarged divergence at the exit side by the factor |γ|-1/2

• Glancing incidence and steep exit: � enlarged angular acceptance by the factor |γ|-1/2

� reduced divergence at the exit side by the factor |γ|1/2

( )( )Bg

Bg

0

g

cos

cos

θ+ψ

θ−ψ=

γ

γ=γ

v

γ ≡ “asymmetry” parameter

B0

gggD sin

1θγ

γ⋅χχ=θ∆ −

v

vv


Darwin Width

These conditions play a central role in designing a high-resolution X-ray experiment

“Switching on” Absorption“ � Darwin curve becomes asymmetrical

Reason:• The two wave fields exhibit maximum intensity between and on the interatomic planes,

respectively. • When rocking the crystal the relative intensity between both is changed. • At the right side of the Darwin curve field two (here absorption is stronger) is dominant

Darwin Curve


Dynamische Beugungstheorie Dynamische Beugungstheorie

gg

g0

P vv

v

−χχ

γγλ=Λ

Strong reflectivity (close to one) when Bragg condition is fulfilled

Strongly reduced penetration depth of the x-rays

“Extinction”

Si 440 Reflection

λ = 1.54 A

ω = 18°

Extinction length

Extinction

Pen

etra

tion

Dep

th (µ

m)

gg

g0

P vv

v

−χχ

γγλ=Λ

Extinction Length

χ

λ∝Λ 2λ∝χ

∑+−

=j j

22j

j

0

2e

i

g

m

e

γωωωε

ρωχ

)(),(

rr

λ∝Λ

1

Dielectric Polarizability


Extinction Length

Dynamical Theory of X-Ray Diffraction

• Dispersion surfaces describe surfaces of identical energy in k-space.

• Splitting of Dispersion surfaces at Lorentz Point � Expansion of two partial waves whichdiffer in the normal component of the wave vectors.

Calculation of the electronic structure in solid materials

• Completely analogue treatment

• Result of theory: Function E(k)

• Energy band gap at the border of the Brillouin Zone depends on the strength of interaction between the electrons with the solid material.

• At the border of the Brillouin-Zone (i.e. k = ± π/a) Bragg reflection of the electrons take place.

• We can – alternatively - state that, Bragg reflection at the border of the Brillouin zone leads to the appearance of a band gap.

Analogy to Electron Band Structure


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