Let’s consider a disordered crystal · Let’sconsider a disordered crystal Long-range order but Unit cell content depends on R uvw Lattice of disordered A and B atoms
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Disordered crystals
Let’s consider a disordered crystal
Long-range order butUnit cell content depends on Ruvw
Lattice of disordered A and B atoms𝐹𝐴 = 1 et 𝐹𝐵 = 2
Bragg’s spot|𝚺(𝒒)|2
1ère zone de Brillouin
Occurrence of Speckles
due to disorder‘‘Speckle pattern’’
𝐴 𝒒 =
𝑢𝑣𝑤
𝐹𝑢𝑣𝑤(𝒒)e−𝑖𝐪∙𝐑𝑢𝑣𝑤 =
𝑛
𝐹𝑛(𝒒)e−𝑖𝐪∙𝐑𝑛
𝐹𝑛 ≠ 𝐹𝑛′
𝐴 𝒒 2
Example of diffuse scattering
Bragg reflexion+
Diffuse scattering
Precession photograph ofreciprocal plane h+k+l=0
Crystal of C60 at 300 K
Cours2.ppt#22. Désordre 1-Effet de la température
General expressionof scattered intensity
Calculation of intensity
Speckle term only visible in coherent diffraction conditions
𝑨∗ 𝒒 𝑨 𝒒 =
𝑛𝑛′
𝐹𝑛∗𝐹𝑛′𝑒
−𝑖𝒒∙(𝒓𝑛′−𝒓𝑛) =
𝑛𝑚
𝐹𝑛∗𝐹𝑛+𝑚 𝑒
−𝑖𝒒∙𝒓𝑚
=
𝑚
𝑁(𝑚)1
𝑁(𝑚)
𝑛
𝐹𝑛∗𝐹𝑛+𝑚 𝑒
−𝑖𝒒∙𝒓𝑚
1
𝑁(𝑚)
𝑛
𝐹𝑛∗𝐹𝑛+𝑚 = 𝐹𝑛
∗𝐹𝑛+𝑚 + Δ𝑚
= Statistical average + fluctuations
𝐼 𝒒 =1
𝑣
𝑚
𝑉 𝒓𝑚 𝐹𝑛∗𝐹𝑛+𝑚 𝑒
−𝑖𝒒∙𝒓𝑚 +1
𝑣
𝑚
𝑉(𝑚)Δ𝑚 𝑒−𝑖𝒒∙𝒓𝑚
𝑉 𝒓𝑚
… 𝑆𝑡𝑎𝑡 = … 𝑡
−𝒓𝑚
Ergodic hypothesis:
How to measure speckle patterns?
1. X-ray beam is coherent enough: 2. Disorder must be static during the time of measurement3. Detector good enough to measure speckles.
- Distance on detector: 𝜆𝑑/𝑎
If these conditiosn are not fulfilled, the speckles are smoothed
Diffuse scattering
𝐷
T
S
Speckles in AuAgZn2F. Livet et al. 𝑎 < 𝜉𝑡 =
𝜆𝐷
𝜎et 𝛿 < 𝜉𝑙 =
1
2
𝜆2
∆𝜆
𝑎
𝑑
𝜆
𝑎
Diffuse scattering
𝐼𝐷(𝒒) :Diffraction
𝐼𝐷𝐷(𝒒) : Diffuse scattering
𝜙𝑛: deviation from average value Fn
𝐹𝑛∗𝐹𝑛+𝑚 = 𝐹𝑛
∗ + 𝛷𝑛∗ 𝐹𝑛+𝑚 +𝛷𝑛+𝑚
= 𝐹 2 + 𝛷𝑛∗𝛷𝑛+𝑚
𝛷𝑛 = 𝐹𝑛 − 𝐹𝑛
𝐼 𝒒 = 𝐹 2
ℎ𝑘𝑙
Σ(𝒒 − 𝑸ℎ𝑘𝑙2
𝑣2+1
𝑣
𝑚
𝑉(𝒓𝑚) 𝛷𝑛∗𝛷𝑛+𝑚 𝑒
−𝑖𝐪∙𝐫𝑚
Cours10_26_10_2011.pptx#1. 4
Disorder
Diffuse scattering:Deviation from perfect order
If very few correlations:
Diffusion scattering ~ N
Diffraction ~ N2 is much stronger
𝐼𝐷𝐷 𝒒 =1
𝑣
𝑚
𝑉(𝒓𝑚) Φ𝑛∗Φ𝑛+𝑚 𝑒
−𝑖𝒒∙𝒓𝑚
lim𝑚→∞
Φ𝑛∗Φ𝑛+𝑚 = 0, and then 𝑉(𝒓𝑚) ≅ 𝑉
𝐼𝐷𝐷 𝒒 = 𝑁
𝑚
Φ𝑛∗Φ𝑛+𝑚 𝑒
−𝑖𝒒∙𝒓𝑚
Two types of disorder
Displacementdisorder
Substitution disorder
Displacement disorder:phonons
N: number of unit cellsM: atomic mass𝒌: wave vector of the phonon modeeak: polarisation of the modeqak: normal coordinates
𝑟𝑛(𝑡) = 𝑟𝑛 + 𝑢𝑛(𝑡)
Harmonic theory• Interaction potential 𝑈, elastic constant 𝐶
Atomic displacements
𝑈 = 𝑈0 +𝐶
2
𝑛
(𝒖𝑛+1 − 𝒖𝑛)2
𝒖𝑛 𝑡 =1
𝑁𝑀
𝛼,𝒌
𝜺𝛼𝒌𝑞𝛼𝒌(𝑡)𝑒𝑖𝒌∙𝒓𝑛
𝑞𝛼𝒌𝑞𝛼−𝒌 =ℏ
2𝜔𝛼(𝒌)coth
ℏ𝜔𝛼(𝒌)
2𝑘𝐵𝑇 𝑘𝐵𝑇≫ℏ𝜔𝛼
𝑘𝐵𝑇
𝜔𝛼2(𝒌)
Phonons
k
𝜔(𝒌)
p/a-p/a
optical
acoustic
Longitudinal
LO
Transverse
TO X 2
LA
TA X 2
10 Thz
Harmonic theoryDebye approx:𝑉𝑘𝐷
3 = 6𝜋2
ℏ𝑘𝐷𝑣𝑠 = 𝑘𝐵𝑇𝐷T at which highest mode excited
Debye-Waller factor
One atom:
Harmonic crystal
Intensity decreased by factor 𝑒−2𝑊
is the Debye-Waller factorRe
Im
Re
N
𝑇 and 𝑞 large ⇒ 𝜃 large
𝐹𝑛(𝐪) = 𝑓 𝑒−𝑖𝐪⋅𝐮𝒏
𝑒−𝑖𝐪⋅𝐮𝒏 = 𝑒−1/2 𝐪⋅𝐮𝒏2= 𝑒−𝑊
𝐼 𝐪 = 𝑁2 𝐹𝑛(𝐪)2 = 𝑁2𝑓2𝑒−2𝑊
𝑒−𝑊
𝜃
Debye-Waller factor 2
Unit cell with n atoms in rj
Isotropic vibrations
Diffraction allows one to measure:
ID e -2W 𝐹𝑛(𝐪) =
𝑗
𝑓𝑗𝑒−𝑊𝑗𝑒−𝑖𝐪⋅𝐫𝑗
𝑊𝑗 =1
2𝐪 ⋅ 𝒖𝑗
2=1
2𝑞2 𝑢𝑗𝑞
2
𝑢𝑗2 = 𝑢𝑥𝑗
2 + 𝑢𝑦𝑗2 + 𝑢𝑧𝑗
2 = 3 𝑢𝑗𝑞2
𝑊𝑗 =1
6
4𝜋 sin 𝜃
𝜆
2
𝑢𝑗2 ≡ 𝐵𝑗,𝑇,𝑖𝑠𝑜
sin 𝜃
𝜆
2
𝐵𝑗,𝑇,𝑖𝑠𝑜 =8𝜋2
3𝑢𝑗2
Calculation of 𝑊
Equipartition theorem
Slow vibrations have large amplitude
𝑊 =𝑞2
𝑁
𝐤
𝑘𝐵𝑇
2𝑀𝜔2(𝐤)
𝑊 =1
2𝑞2 𝑢𝑛
2𝑢𝑛2 =
1
𝑁
𝑛
𝑢𝑛2 =
1
𝑁
𝑘
𝑢𝑘2
1
2𝑀𝜔2(𝒌) 𝑢𝒌
2 =1
2𝑘𝐵𝑇
Einstein model 𝜔 = 𝜔𝑒
∝ 𝑞2𝑘𝐵𝑇
2𝑀𝜔𝑒2
Debye
Debye approximation
Si T >> TD Classical
1
1
0.5
0.5
with
𝑊 =sin𝜃
𝜆
26ℎ2𝑇
𝑀𝑘𝐵𝑇𝐷2
𝐼 ∝ 𝑒−2𝑊 ⇒ ln 𝐼 ∝ −𝑇
𝑊 =sin𝜃
𝜆
26ℎ2
𝑀𝑘𝐵𝑇𝐷
1
4+𝑇
𝑇𝐷Φ(𝑇𝐷𝑇)
élément C Al Cu Mo Ag Pb
TD (K) 3000 390 320 380 226 90
Φ𝑇𝐷𝑇
=𝑇
𝑇𝐷න0
𝑇𝐷𝑇 𝑦
𝑒𝑦 − 1𝑑𝑦
Example 1:Determination of TD
R.M. Nicklow and R.A. Young, Phys. Rev. 152, 591–596 (1966)
Intensity of Al (h00) reflexions
TD ~ 400±5K
Deviation due tozero-point vibrations
http://cornell.mirror.aps.org/abstract/PR/v152/i2/p591_1
Example 2: Lindemann criterion
Solide melts when:
Aluminiumf.c.c.
a=4.04 Å
Melting
𝑢2 = 10% 𝑑1𝑠𝑡 neigh
Example 3:
Simple etals:
Organic compounds:
Anisotropic B:𝑢2 depends
on the directions
𝑃𝐹6
𝐵𝑗,𝑇,𝑖𝑠𝑜 =8𝜋2
3𝑢𝑗2
𝑊𝑗 = 𝐵𝑗,𝑇,𝑖𝑠𝑜sin 𝜃
𝜆
2
𝑢2 = 0,05 − 0,2 Å
𝑢2 = 0,5 Å
Thermal ellipsoids
Influence of dimensionality
Integral is governed by divergence of:
Debye, isotropic
Harmonic theory
𝑊 = 𝑎𝐷𝑞2න𝑘
Influence of dimensionality-2
D=1
𝐿
𝑊 = 𝐴1න𝐿−1
𝑘𝐷 𝑑𝑘
𝑘2= 𝐴1(𝐿 − 𝑘𝐷
−1)𝑊 = 𝐴1න
𝐿−1
𝑘𝐷 𝑑𝑘
𝑘2
𝑊 = 𝐴2න𝐿−1
𝑘𝐷 2𝜋𝑘𝑑𝑘
𝑘2= 2𝜋𝐴2ln(𝐿𝑘𝐷)
𝑊 = 𝐴3න𝐿−1
𝑘𝐷 4𝜋𝑘2𝑑𝑘
𝑘2= 4𝜋𝐴3(𝑘𝐷 − 𝐿
−1)
𝑊 =1
2𝑞2 𝑢2 Si 𝐷 ≤ 2 𝑢2
𝐿→∞∞
D=3
D=2
𝑢2
Influence of dimensionality-3
No long-range order if D 2
If 𝐷 ≤ 2 𝑊𝐿→∞
∞
D=1
D=3
D=2
𝑒−𝑊~𝑒−𝐴1𝐿
𝑒−𝑊~𝐿−2𝜋𝐴2
𝑒−𝑊~exp −2sin2 𝜃
𝜆26ℎ2𝑇
𝑀𝑘𝐵𝑇𝐷2
Thermal Diffuse Scattering
Si 300 K
XRay //
XRay //
Experiment Simulation
M. Holt, Phys. Rev. Lett 83, 3317 (1999)
False colors,Log scale.
𝐼𝐷𝐷 𝒒 = 𝑁
𝑚
Φ𝑛∗Φ𝑛+𝑚 𝑒
−𝑖𝒒∙𝒓𝑚
TDS calculation-1
One atom per cell
Harmonic theory:
At first order,
displacement-displacement correlations
Φ𝑛∗Φ𝑛+𝑚 = 𝐹𝑛
∗𝐹𝑛+𝑚 − 𝐹2
Φ𝑛∗Φ𝑛+𝑚 = 𝑓
2 𝑒−𝑖𝒒∙(𝒖𝑛+𝑚−𝒖𝑛) − 𝑒−2𝑊
𝑒−𝑖𝒒∙(𝒖𝑛+𝑚−𝒖𝑛) = 𝑒−1/2 (𝒒∙(𝒖𝑛+𝑚−𝒖𝑛))2= 𝑒−2𝑊𝑒 (𝒒∙𝒖𝑛+𝑚)(𝒒∙𝒖𝑛)
Φ𝑛∗Φ𝑛+𝑚 = 𝑓
2𝑒−2𝑊 (𝒒 ∙ 𝒖𝑛+𝑚)(𝒒 ∙ 𝒖𝑛)
TDS calculation-3
+k-k
Qhkl Qhkl q
~1/𝑘2
• ~𝑁: diffuse scattering• 𝑘𝐵𝑇: thermal scattering• (𝒒 ∙ 𝒖)2: geometrical factor, (mode selection)• All a modes contribute to the same k
𝐼𝐷𝐷 𝒒 = 𝑸ℎ𝑘𝑙 + 𝒌 = 𝑁𝑓2𝑒−2𝑊𝑘𝐵𝑇
𝛼
(𝒒 ∙ 𝒖𝛼𝒌)2
𝐼𝐷𝐷~(𝒒 ∙ 𝒖𝛼𝒌)2
~1
𝜔𝛼2(𝒌)
Example of TDS
ComparisonX (-)-neutrons(o)
M. Holt, Phys. Rev. Lett 83, 3317 (1999)
Harmonic theory:Born-von Karman model using
constant forcesup to 6th neighbours
Si 300 K
𝐼𝐷𝐷(𝒒) = 𝑁𝑓2𝑒−2𝑊𝑘𝐵𝑇
𝛼
(𝒒 ∙ 𝜺𝛼𝒌)2
𝑀𝜔𝛼2(𝒌)
Substitution disorder
Alloy or solid solution AxB1-x
No information on correlations
• Case of total disorder
Laue scattering:
𝑓 = 𝑥𝑓𝐴 + (1 − 𝑥)𝑓𝐵
𝐼𝐷(𝒒) = 𝑁2 𝑥𝑓𝐴 + 1 − 𝑥 𝑓𝐵
2
𝐼𝐷𝐷 𝒒 = 𝑁 Φ02 = 𝑁 𝑓2 − 𝑓 2
𝐼𝐷𝐷 𝒒 = 𝑁𝑥(1 − 𝑥) 𝑓𝐴 − 𝑓𝐵2
𝐼 𝒒 = 𝑁𝑥(1 − 𝑥)(𝑓𝐴−𝑓𝐵)2
𝑚
(1 −𝑝𝐴 𝑚
𝑥)𝑒−𝑖𝒒∙𝒓𝑚
Correlations
𝑝𝐴(𝑚) A
B
Short-range order:
Conditional probabilities𝑝𝐴(𝑚): probability of having A in 𝒓𝑚 given B at 0𝑝𝐵(𝑚): probability of having B in 𝒓𝑚 given A at 0
AB pairs = BA pairs
Warren-Cowley parameters
⟹ 𝑥𝑝𝐵 𝑚 = (1 − 𝑥)𝑝𝐴(𝑚)
AA ∶ 𝑥 1 − 𝑝𝐵 𝑚 → 𝑓𝐴2
AB ∶ 𝑥𝑝𝐵 𝑚 → 𝑓𝐴𝑓𝐵AB ∶ (1 − 𝑥)𝑝𝐴 𝑚 → 𝑓𝐵𝑓𝐴BB ∶ (1 − 𝑥) 1 − 𝑝𝐴 𝑚 → 𝑓𝐵
2
𝐹𝑛∗𝐹𝑛+𝑚 =
𝑥 1 − 𝑝𝐵 𝑚 𝑓𝐴2 + 𝑥𝑝𝐵 𝑚 𝑓𝐴𝑓𝐵 + 1 − 𝑥 𝑝𝐴 𝑚 𝑓𝐵𝑓𝐴 + (1 − 𝑥) 1 − 𝑝𝐴 𝑚 𝑓𝐵
2
Example
Local order such thatAB pairs are favoredpA(m) A
B
ℎ0 1 2 3
1
1/2
𝑆(𝒒)
Tendency to doubleperiodicity
𝐼 𝒒 = 𝑁𝑥(1 − 𝑥)(𝑓𝐴−𝑓𝐵)2 1 + 2(1 −
𝑝𝐴 1
𝑥) cos(2𝜋ℎ)
𝑝𝐴 1 > 𝑥
Conclusion
• Substitution disorder:
𝐼𝐷𝐷 ~ (𝑓𝐴 − 𝑓𝐵)2
• Visible at small angles• Only electron density contrast at small angles
• Displacement disorder:
𝐼𝐷𝐷 ~ (𝒒. 𝒖)2
• Invisible at small angles𝒒. 𝒖 is too small for interferences to occur
Structural phase transitions Order parameter definition:
𝜂𝒌𝑐𝑒𝑖𝜑: order parameter
𝒌𝑐: critical wave vectorbelongs to the 1st BZ
Displacive Order-desorder
Order parmater 𝑈𝒌𝑐:
Displacement amplitude:
Order parameter :
Site probabilityIsing (pseudo-)spins
𝜂𝑛 = 𝜂𝒌𝑐 cos(𝒌𝑐 ∙ 𝒓𝑛 + 𝜑)
𝑼𝑛 = 𝜺𝒌𝑐𝑈𝒌𝑐 cos(𝒌𝑐 ∙ 𝒓𝑛 + 𝜑)
𝑇𝐶
ExamplesDisplacive transitions:
Order-disorder : • Alloy A0.5B0.5
TC
Critical wave vectors (1/4,0)
• FerroelectricZone center
Not a special point
• Displacive modulation (Peierls)
ab
Zone boundary
𝑇𝐶
𝒌𝐶 = 0 𝒌𝐶 =𝒂∗
2+𝒃∗
2
𝑼𝑛 = 𝜺𝒌𝑐𝑈𝒌𝑐 cos(𝒌𝑐 ∙ 𝒓𝑛 + 𝜑)
𝜺𝒌𝑐 = 𝒂
𝑆𝑛 = 𝑆𝒌𝑐 cos(𝒌𝑐 ∙ 𝒓𝑛 + 𝜑)
𝒌𝐶 =𝒂∗
4
Cours7.ppt#17. BaTiO3Cours7.ppt#17. BaTiO3
Displacive transition
Ordre parameter fluctuations
Susceptibility associated to order parameter
𝑢𝛼𝑘𝑐: composante principale
𝜒(𝒌𝒄) diverges at the transition temperature
𝐼𝐷𝐷 𝒒 = 𝑸ℎ𝑘𝑙 + 𝒌 = 𝑁𝑓2𝑒−2𝑊(𝒒 ∙ 𝜺𝒌)
2 𝑢𝒌𝑢−𝒌
𝑼𝑛 = 𝜺𝒌𝑐𝑈𝒌𝑐 cos(𝒌𝑐 ∙ 𝒓𝑛 + 𝜑)
𝒖𝑛 =1
𝑁
𝛼𝒌
𝜺𝛼𝒌𝑢𝛼𝒌𝒆𝒊𝒌⋅𝒓𝑛
𝒖𝒌 = 𝜒(𝒌)𝒉−𝒌
Fluctuation-dissipation
Example of phonons:
Par le théorème d’équipartition de l’énergie
𝒖𝒌𝒖−𝒌 − 𝒖𝒌 𝒖−𝒌 = 𝑘𝐵𝑇𝜒(𝒌)
1
2𝑀𝜔2(𝒌) 𝑢𝒌
2 =1
2𝑘𝐵𝑇
𝜒 𝒌 =1
𝑀𝜔2(𝒌)
Calculation of the intensity
Fluctuation-dissipation
𝑇 > 𝑇𝑐
𝑇 < 𝑇𝑐
𝐼𝐷 𝒒 = 𝑸ℎ𝑘𝑙 + 𝒌 = 𝑁𝑓2𝑒−2𝑊(𝒒 ∙ 𝜺𝒌)
2(𝑘𝐵𝑇𝜒 𝒌 − 𝑢𝒌 𝑢−𝒌 )
𝐼𝐷𝐷 𝒒 = 𝑁𝑓2𝑒−2𝑊(𝒒 ∙ 𝜺𝒌)
2𝑘𝐵𝑇𝜒 𝒌
𝑢𝒌 = 0
𝐼𝐷𝐷 𝒒 = 𝑁𝑓2𝑒−2𝑊(𝒒 ∙ 𝜺𝒌)
2𝑘𝐵𝑇𝜒 𝒌
+𝑁2𝑓2𝑒−2𝑊(𝒒 ∙ 𝜺𝒌)2 𝑈𝒌𝑐
2𝛿𝐾(𝒌 = ±𝒌𝑐)
𝑢𝒌𝐶 = 𝑁𝑈𝒌𝑐
Qhkl
Ornstein-Zernike
Lorentzian shape
x : correlation length
T>Tc
Qhkl
+𝒌𝑐−𝒌𝑐
T
Critical exponents
Temperature behavior of:
TTc• Susceptibility: 𝜒(𝑘𝑐) ~ 𝑇 −𝑇𝑐
−𝜸
• Correlation lengths: 𝜉 ~ 𝑇 −𝑇𝑐−𝝂 SRO
QLRO
LRO
Example : Order-disorder in AuAgZn2
T< 351.1°C T> 351.1°C
Au/Ag
Zn
Cubicf.c.c.
F. Livet et al. Phys. Rev. B 66, 134108 (2002)
2nd order pahse transition
𝒌𝐶 =𝒂∗
2+𝒃∗
2+𝒄∗
2
FluctuationsDiffuse scattering (1/2,1/2,1/2)
Ising 3Dg =1,24
n = 0,63
h = 0,04
h = 0,03
c(q)~ q-2+h
c~(T-Tc)-g
c-1/g ~(T-Tc)
g =1,242
x ~(T-Tc)-n
x-1/ n~(T-Tc)
n = 0,709
TC+4°C TC+0,13°C
TC+4°C
TC+0,08°C
Example:‘‘Blue bronze’’ K0.3MoO3
b
Octahedra MoO6
Potassium
(Rubidium)
E. Bervas, thesis (1984)
Tp=183 K
c
a
Blue bronze
𝑋𝑌 3𝐷𝛾 = 1,316𝜈 = 0,669𝛽 = 0,346
At T=183 K: appareance of sattelite reflexionsat the critical wave vector:
𝜒~(𝑇 − 𝑇𝑐)−𝛾
𝛾 = 1,33(4) 𝜈 = 0,68(5)
𝛽 = 0,31(5)
𝒌𝐶 = 0,748 𝒃∗ + 0,5 𝒄∗
𝜉~(𝑇 − 𝑇𝑐)−𝜈
𝐼~(𝑇𝑐 − 𝑇)𝛽
Determination of interatomic potential
Ex: Ising model
Within mean field appr.the susceptibility is:
Fit of 𝜒(𝒌) gives the 𝐽′𝑠
𝐻 = −𝑖𝑗𝐽𝑖𝑗 𝜎𝑖𝜎𝑗
𝜒(𝒌) =𝛽
1 + 𝛽𝐽(𝒌)
With,𝐽 𝒌 = 2𝐽𝑎 cos 𝒌 ∙ 𝒂 + 2𝐽𝑏 cos 𝒌 ∙ 𝒃 + 2𝐽𝑐 cos 𝒌 ∙ 𝒄
ExampleBragg
Diffuse scattering
IsotropeJi=Jj
Anisotrope (1D)100xJi=Jj
Local order
Difficultto see
in real space
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