![Page 1: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/1.jpg)
Today’s Outline - April 02, 2015
• PHYS 570 days at 10-ID
• Characteristics of the Darwin Curve
• Asymmetric Reflections
• Dumond Diagrams
• Monochromators
• Photoelectric Absorption
• Cross-Section of an Isolated Atom
Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 14, 2015
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 1 / 18
![Page 2: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/2.jpg)
Today’s Outline - April 02, 2015
• PHYS 570 days at 10-ID
• Characteristics of the Darwin Curve
• Asymmetric Reflections
• Dumond Diagrams
• Monochromators
• Photoelectric Absorption
• Cross-Section of an Isolated Atom
Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 14, 2015
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 1 / 18
![Page 3: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/3.jpg)
Today’s Outline - April 02, 2015
• PHYS 570 days at 10-ID
• Characteristics of the Darwin Curve
• Asymmetric Reflections
• Dumond Diagrams
• Monochromators
• Photoelectric Absorption
• Cross-Section of an Isolated Atom
Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 14, 2015
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 1 / 18
![Page 4: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/4.jpg)
Today’s Outline - April 02, 2015
• PHYS 570 days at 10-ID
• Characteristics of the Darwin Curve
• Asymmetric Reflections
• Dumond Diagrams
• Monochromators
• Photoelectric Absorption
• Cross-Section of an Isolated Atom
Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 14, 2015
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 1 / 18
![Page 5: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/5.jpg)
Today’s Outline - April 02, 2015
• PHYS 570 days at 10-ID
• Characteristics of the Darwin Curve
• Asymmetric Reflections
• Dumond Diagrams
• Monochromators
• Photoelectric Absorption
• Cross-Section of an Isolated Atom
Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 14, 2015
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 1 / 18
![Page 6: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/6.jpg)
Today’s Outline - April 02, 2015
• PHYS 570 days at 10-ID
• Characteristics of the Darwin Curve
• Asymmetric Reflections
• Dumond Diagrams
• Monochromators
• Photoelectric Absorption
• Cross-Section of an Isolated Atom
Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 14, 2015
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 1 / 18
![Page 7: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/7.jpg)
Today’s Outline - April 02, 2015
• PHYS 570 days at 10-ID
• Characteristics of the Darwin Curve
• Asymmetric Reflections
• Dumond Diagrams
• Monochromators
• Photoelectric Absorption
• Cross-Section of an Isolated Atom
Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 14, 2015
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 1 / 18
![Page 8: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/8.jpg)
Today’s Outline - April 02, 2015
• PHYS 570 days at 10-ID
• Characteristics of the Darwin Curve
• Asymmetric Reflections
• Dumond Diagrams
• Monochromators
• Photoelectric Absorption
• Cross-Section of an Isolated Atom
Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 14, 2015
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 1 / 18
![Page 9: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/9.jpg)
Today’s Outline - April 02, 2015
• PHYS 570 days at 10-ID
• Characteristics of the Darwin Curve
• Asymmetric Reflections
• Dumond Diagrams
• Monochromators
• Photoelectric Absorption
• Cross-Section of an Isolated Atom
Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 14, 2015
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 1 / 18
![Page 10: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/10.jpg)
PHYS 570 days at 10-ID
1 April 10, 2015, 09:00 – 16:00
2 April 24, 2015, 09:00 – 16:00
3 Activities
• Absolute flux measurement• Reflectivity measurement• EXAFS measurement• Rocking curve measurement (possibly)
4 Make sure your badge is ready
5 Leave plenty of time to get the badge
6 Let me know when you plan to come!
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 2 / 18
![Page 11: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/11.jpg)
PHYS 570 days at 10-ID
1 April 10, 2015, 09:00 – 16:00
2 April 24, 2015, 09:00 – 16:00
3 Activities• Absolute flux measurement• Reflectivity measurement• EXAFS measurement• Rocking curve measurement (possibly)
4 Make sure your badge is ready
5 Leave plenty of time to get the badge
6 Let me know when you plan to come!
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 2 / 18
![Page 12: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/12.jpg)
PHYS 570 days at 10-ID
1 April 10, 2015, 09:00 – 16:00
2 April 24, 2015, 09:00 – 16:00
3 Activities• Absolute flux measurement• Reflectivity measurement• EXAFS measurement• Rocking curve measurement (possibly)
4 Make sure your badge is ready
5 Leave plenty of time to get the badge
6 Let me know when you plan to come!
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 2 / 18
![Page 13: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/13.jpg)
PHYS 570 days at 10-ID
1 April 10, 2015, 09:00 – 16:00
2 April 24, 2015, 09:00 – 16:00
3 Activities• Absolute flux measurement• Reflectivity measurement• EXAFS measurement• Rocking curve measurement (possibly)
4 Make sure your badge is ready
5 Leave plenty of time to get the badge
6 Let me know when you plan to come!
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 2 / 18
![Page 14: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/14.jpg)
PHYS 570 days at 10-ID
1 April 10, 2015, 09:00 – 16:00
2 April 24, 2015, 09:00 – 16:00
3 Activities• Absolute flux measurement• Reflectivity measurement• EXAFS measurement• Rocking curve measurement (possibly)
4 Make sure your badge is ready
5 Leave plenty of time to get the badge
6 Let me know when you plan to come!
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 2 / 18
![Page 15: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/15.jpg)
Darwin widths
the quantities below the widths are f 0(Q), f ′, and f ′′ (forλ = 1.5405 A). For an angular width, multiply times tan θand for π polarization, multiply by cos(2θ).
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 3 / 18
![Page 16: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/16.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi
=1√b
(ζD tan θ)bHe =√b(ζD tan θ)
= δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 17: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/17.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi
=1√b
(ζD tan θ)bHe =√b(ζD tan θ)
= δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 18: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/18.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi
=1√b
(ζD tan θ)bHe =√b(ζD tan θ)
= δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 19: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/19.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi
=1√b
(ζD tan θ)bHe =√b(ζD tan θ)
= δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 20: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/20.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi
=1√b
(ζD tan θ)bHe =√b(ζD tan θ)
= δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 21: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/21.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi
=1√b
(ζD tan θ)bHe =√b(ζD tan θ)
= δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 22: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/22.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi
=1√b
(ζD tan θ)bHe =√b(ζD tan θ)
= δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 23: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/23.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi
=1√b
(ζD tan θ)bHe =√b(ζD tan θ)
= δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 24: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/24.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi
=1√b
(ζD tan θ)bHe =√b(ζD tan θ)
= δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 25: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/25.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi
=1√b
(ζD tan θ)bHe =√b(ζD tan θ) = δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 26: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/26.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi =1√b
(ζD tan θ)bHe
=√b(ζD tan θ) = δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 27: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/27.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi =1√b
(ζD tan θ)bHe =√b(ζD tan θ)
= δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 28: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/28.jpg)
Asymmetric Geometry
In general the diffracting planes are not precisely aligned with the surfaceof the crystal.
Parameterized by the asymmetryangle 0 < α < θBragg
This leads to a beam compression
b =sin θisin θe
=sin(θ + α)
sin(θ − α)
He =Hi
b
according to Liouville’s theorem,the divergence of the beam mustalso change
δθe =√b(ζD tan θ)
δθi =1√b
(ζD tan θ)
δθiHi =1√b
(ζD tan θ)bHe =√b(ζD tan θ) = δθeHe
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 4 / 18
![Page 29: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/29.jpg)
Rocking Curve Measurements
The measured “rocking” curve from a two crystal system is a convolutionof the Darwin curves of both crystals.
When the two crystals have amatched asymmetry, we get a triangle. When one asymmetry is muchhigher, then we can measure the Darwin curve of a single crystal.
output divergence on left, input divergence on right
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 5 / 18
![Page 30: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/30.jpg)
Rocking Curve Measurements
The measured “rocking” curve from a two crystal system is a convolutionof the Darwin curves of both crystals. When the two crystals have amatched asymmetry, we get a triangle.
When one asymmetry is muchhigher, then we can measure the Darwin curve of a single crystal.
output divergence on left, input divergence on right
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 5 / 18
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Rocking Curve Measurements
The measured “rocking” curve from a two crystal system is a convolutionof the Darwin curves of both crystals. When the two crystals have amatched asymmetry, we get a triangle.
When one asymmetry is muchhigher, then we can measure the Darwin curve of a single crystal.
output divergence on left, input divergence on right
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 5 / 18
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Rocking Curve Measurements
The measured “rocking” curve from a two crystal system is a convolutionof the Darwin curves of both crystals. When the two crystals have amatched asymmetry, we get a triangle. When one asymmetry is muchhigher, then we can measure the Darwin curve of a single crystal.
output divergence on left, input divergence on right
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 5 / 18
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Dumond diagram: no Darwin width
Transfer function of an optical element parameterized by angle andwavelength.
Here Darwin width is ignored.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 6 / 18
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Dumond diagram: no Darwin width
Transfer function of an optical element parameterized by angle andwavelength. Here Darwin width is ignored.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 6 / 18
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Dumond diagram: symmetric Bragg
Including the Darwin width, we have a bandpass in wavelength.
If inputbeam is perfectly collimated, so is output (vertical black line).
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 7 / 18
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Dumond diagram: symmetric Bragg
Including the Darwin width, we have a bandpass in wavelength. If inputbeam is perfectly collimated, so is output (vertical black line).
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 7 / 18
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Dumond diagram: asymmetric Bragg
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 8 / 18
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Double Crystal Monochromators
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 9 / 18
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Double Crystal Monochromators
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 9 / 18
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Double Crystal Monochromators
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 10 / 18
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Double Crystal Monochromators
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 10 / 18
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Total Cross Section
The total cross-section forphoton “absorption” in-cludes elastic (or coher-ent) scattering, Compton(inelastic) scattering, andphotoelectric absorption.
Characteristic absorptionjumps depend on the ele-ment
These quantities vary significantly over many decades but can easily puton an equal footing.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 11 / 18
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Total Cross Section
The total cross-section forphoton “absorption” in-cludes elastic (or coher-ent) scattering, Compton(inelastic) scattering, andphotoelectric absorption.
Characteristic absorptionjumps depend on the ele-ment
These quantities vary significantly over many decades but can easily puton an equal footing.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 11 / 18
![Page 44: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/44.jpg)
Total Cross Section
The total cross-section forphoton “absorption” in-cludes elastic (or coher-ent) scattering, Compton(inelastic) scattering, andphotoelectric absorption.
Characteristic absorptionjumps depend on the ele-ment
These quantities vary significantly over many decades but can easily puton an equal footing.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 11 / 18
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Scaled Absorption
T =I
Io= e−µz
µ =ρmNA
Mσa
σa ∼Z 4
E 3
scale σa for different ele-ments by E 3/Z 4 and plottogether
remarkably, all values lie on a common curve above the K edge andbetween the L and K edges and below the L edge
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 12 / 18
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Scaled Absorption
T =I
Io= e−µz
µ =ρmNA
Mσa
σa ∼Z 4
E 3
scale σa for different ele-ments by E 3/Z 4 and plottogether
remarkably, all values lie on a common curve above the K edge andbetween the L and K edges and below the L edge
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 12 / 18
![Page 47: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/47.jpg)
Scaled Absorption
T =I
Io= e−µz
µ =ρmNA
Mσa
σa ∼Z 4
E 3
scale σa for different ele-ments by E 3/Z 4 and plottogether
remarkably, all values lie on a common curve above the K edge andbetween the L and K edges and below the L edge
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 12 / 18
![Page 48: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/48.jpg)
Scaled Absorption
T =I
Io= e−µz
µ =ρmNA
Mσa
σa ∼Z 4
E 3
scale σa for different ele-ments by E 3/Z 4 and plottogether
remarkably, all values lie on a common curve above the K edge andbetween the L and K edges and below the L edge
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 12 / 18
![Page 49: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/49.jpg)
Scaled Absorption
T =I
Io= e−µz
µ =ρmNA
Mσa
σa ∼Z 4
E 3
scale σa for different ele-ments by E 3/Z 4 and plottogether
remarkably, all values lie on a common curve above the K edge andbetween the L and K edges and below the L edge
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 12 / 18
![Page 50: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/50.jpg)
Scaled Absorption
T =I
Io= e−µz
µ =ρmNA
Mσa
σa ∼Z 4
E 3
scale σa for different ele-ments by E 3/Z 4 and plottogether
remarkably, all values lie on a common curve above the K edge andbetween the L and K edges and below the L edge
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 12 / 18
![Page 51: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/51.jpg)
Calculation of σa
From first-order perturbation theory, the absorption cross section is givenby
σa =2π
~cV 2
4π3
∫|Mif |2δ(Ef − Ei )q2 sin θdqdθdϕ
where the matrix element Mif be-tween the initial, 〈i |, and final, |f 〉,states is given by
The interaction Hamiltonian is ex-pressed in terms of the electromag-netic vector potential
Mif = 〈i |HI |f 〉
HI =e~p · ~Am
+e2A2
2m
~A = ε
√~
2ε0Vω
[ake
i~k·~r + a†ke−i~k·~r
]The first term gives absorption while the second produces Thomsonscattering so we take only the first into consideration now.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 13 / 18
![Page 52: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/52.jpg)
Calculation of σa
From first-order perturbation theory, the absorption cross section is givenby
σa =2π
~cV 2
4π3
∫|Mif |2δ(Ef − Ei )q2 sin θdqdθdϕ
where the matrix element Mif be-tween the initial, 〈i |, and final, |f 〉,states is given by
The interaction Hamiltonian is ex-pressed in terms of the electromag-netic vector potential
Mif = 〈i |HI |f 〉
HI =e~p · ~Am
+e2A2
2m
~A = ε
√~
2ε0Vω
[ake
i~k·~r + a†ke−i~k·~r
]The first term gives absorption while the second produces Thomsonscattering so we take only the first into consideration now.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 13 / 18
![Page 53: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/53.jpg)
Calculation of σa
From first-order perturbation theory, the absorption cross section is givenby
σa =2π
~cV 2
4π3
∫|Mif |2δ(Ef − Ei )q2 sin θdqdθdϕ
where the matrix element Mif be-tween the initial, 〈i |, and final, |f 〉,states is given by
The interaction Hamiltonian is ex-pressed in terms of the electromag-netic vector potential
Mif = 〈i |HI |f 〉
HI =e~p · ~Am
+e2A2
2m
~A = ε
√~
2ε0Vω
[ake
i~k·~r + a†ke−i~k·~r
]The first term gives absorption while the second produces Thomsonscattering so we take only the first into consideration now.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 13 / 18
![Page 54: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/54.jpg)
Calculation of σa
From first-order perturbation theory, the absorption cross section is givenby
σa =2π
~cV 2
4π3
∫|Mif |2δ(Ef − Ei )q2 sin θdqdθdϕ
where the matrix element Mif be-tween the initial, 〈i |, and final, |f 〉,states is given by
The interaction Hamiltonian is ex-pressed in terms of the electromag-netic vector potential
Mif = 〈i |HI |f 〉
HI =e~p · ~Am
+e2A2
2m
~A = ε
√~
2ε0Vω
[ake
i~k·~r + a†ke−i~k·~r
]The first term gives absorption while the second produces Thomsonscattering so we take only the first into consideration now.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 13 / 18
![Page 55: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/55.jpg)
Calculation of σa
From first-order perturbation theory, the absorption cross section is givenby
σa =2π
~cV 2
4π3
∫|Mif |2δ(Ef − Ei )q2 sin θdqdθdϕ
where the matrix element Mif be-tween the initial, 〈i |, and final, |f 〉,states is given by
The interaction Hamiltonian is ex-pressed in terms of the electromag-netic vector potential
Mif = 〈i |HI |f 〉
HI =e~p · ~Am
+e2A2
2m
~A = ε
√~
2ε0Vω
[ake
i~k·~r + a†ke−i~k·~r
]The first term gives absorption while the second produces Thomsonscattering so we take only the first into consideration now.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 13 / 18
![Page 56: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/56.jpg)
Calculation of σa
From first-order perturbation theory, the absorption cross section is givenby
σa =2π
~cV 2
4π3
∫|Mif |2δ(Ef − Ei )q2 sin θdqdθdϕ
where the matrix element Mif be-tween the initial, 〈i |, and final, |f 〉,states is given by
The interaction Hamiltonian is ex-pressed in terms of the electromag-netic vector potential
Mif = 〈i |HI |f 〉
HI =e~p · ~Am
+e2A2
2m
~A = ε
√~
2ε0Vω
[ake
i~k·~r + a†ke−i~k·~r
]
The first term gives absorption while the second produces Thomsonscattering so we take only the first into consideration now.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 13 / 18
![Page 57: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/57.jpg)
Calculation of σa
From first-order perturbation theory, the absorption cross section is givenby
σa =2π
~cV 2
4π3
∫|Mif |2δ(Ef − Ei )q2 sin θdqdθdϕ
where the matrix element Mif be-tween the initial, 〈i |, and final, |f 〉,states is given by
The interaction Hamiltonian is ex-pressed in terms of the electromag-netic vector potential
Mif = 〈i |HI |f 〉
HI =e~p · ~Am
+e2A2
2m
~A = ε
√~
2ε0Vω
[ake
i~k·~r + a†ke−i~k·~r
]The first term gives absorption
while the second produces Thomsonscattering so we take only the first into consideration now.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 13 / 18
![Page 58: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/58.jpg)
Calculation of σa
From first-order perturbation theory, the absorption cross section is givenby
σa =2π
~cV 2
4π3
∫|Mif |2δ(Ef − Ei )q2 sin θdqdθdϕ
where the matrix element Mif be-tween the initial, 〈i |, and final, |f 〉,states is given by
The interaction Hamiltonian is ex-pressed in terms of the electromag-netic vector potential
Mif = 〈i |HI |f 〉
HI =e~p · ~Am
+e2A2
2m
~A = ε
√~
2ε0Vω
[ake
i~k·~r + a†ke−i~k·~r
]The first term gives absorption while the second produces Thomsonscattering so we take only the first into consideration now.
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 13 / 18
![Page 59: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/59.jpg)
Free electron approximation
In order to evaluate the Mif matrix element we define the initial and finalstates
the initial state has a photon and a K electron(no free electron)
similarly, the final state has no photon and anejected free electron (ignoring the core holeand charged ion)
|i〉 = |1〉γ |0〉e
〈f | = e〈1|γ〈0|
Thus
Mif =e
m
√~
2ε0Vω
[e〈1|γ〈0|(~p · ε)ae i
~k·~r + (~p · ε)a†e−i~k·~r |1〉γ |0〉e
]The calculation is simplified if the interaction Hamiltonian is applied to theleft since the final state has only a free electron and no photon
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 14 / 18
![Page 60: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/60.jpg)
Free electron approximation
In order to evaluate the Mif matrix element we define the initial and finalstates
the initial state has a photon and a K electron(no free electron)
similarly, the final state has no photon and anejected free electron (ignoring the core holeand charged ion)
|i〉 = |1〉γ |0〉e
〈f | = e〈1|γ〈0|
Thus
Mif =e
m
√~
2ε0Vω
[e〈1|γ〈0|(~p · ε)ae i
~k·~r + (~p · ε)a†e−i~k·~r |1〉γ |0〉e
]The calculation is simplified if the interaction Hamiltonian is applied to theleft since the final state has only a free electron and no photon
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 14 / 18
![Page 61: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/61.jpg)
Free electron approximation
In order to evaluate the Mif matrix element we define the initial and finalstates
the initial state has a photon and a K electron(no free electron)
similarly, the final state has no photon and anejected free electron (ignoring the core holeand charged ion)
|i〉 = |1〉γ |0〉e
〈f | = e〈1|γ〈0|
Thus
Mif =e
m
√~
2ε0Vω
[e〈1|γ〈0|(~p · ε)ae i
~k·~r + (~p · ε)a†e−i~k·~r |1〉γ |0〉e
]The calculation is simplified if the interaction Hamiltonian is applied to theleft since the final state has only a free electron and no photon
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 14 / 18
![Page 62: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/62.jpg)
Free electron approximation
In order to evaluate the Mif matrix element we define the initial and finalstates
the initial state has a photon and a K electron(no free electron)
similarly, the final state has no photon and anejected free electron (ignoring the core holeand charged ion)
|i〉 = |1〉γ |0〉e
〈f | = e〈1|γ〈0|
Thus
Mif =e
m
√~
2ε0Vω
[e〈1|γ〈0|(~p · ε)ae i
~k·~r + (~p · ε)a†e−i~k·~r |1〉γ |0〉e
]The calculation is simplified if the interaction Hamiltonian is applied to theleft since the final state has only a free electron and no photon
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 14 / 18
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Free electron approximation
In order to evaluate the Mif matrix element we define the initial and finalstates
the initial state has a photon and a K electron(no free electron)
similarly, the final state has no photon and anejected free electron (ignoring the core holeand charged ion)
|i〉 = |1〉γ |0〉e
〈f | = e〈1|γ〈0|
Thus
Mif =e
m
√~
2ε0Vω
[e〈1|γ〈0|(~p · ε)ae i
~k·~r + (~p · ε)a†e−i~k·~r |1〉γ |0〉e
]The calculation is simplified if the interaction Hamiltonian is applied to theleft since the final state has only a free electron and no photon
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 14 / 18
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Free electron approximation
In order to evaluate the Mif matrix element we define the initial and finalstates
the initial state has a photon and a K electron(no free electron)
similarly, the final state has no photon and anejected free electron (ignoring the core holeand charged ion)
|i〉 = |1〉γ |0〉e
〈f | = e〈1|γ〈0|
Thus
Mif =e
m
√~
2ε0Vω
[e〈1|γ〈0|(~p · ε)ae i
~k·~r + (~p · ε)a†e−i~k·~r |1〉γ |0〉e
]
The calculation is simplified if the interaction Hamiltonian is applied to theleft since the final state has only a free electron and no photon
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 14 / 18
![Page 65: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/65.jpg)
Free electron approximation
In order to evaluate the Mif matrix element we define the initial and finalstates
the initial state has a photon and a K electron(no free electron)
similarly, the final state has no photon and anejected free electron (ignoring the core holeand charged ion)
|i〉 = |1〉γ |0〉e
〈f | = e〈1|γ〈0|
Thus
Mif =e
m
√~
2ε0Vω
[e〈1|γ〈0|(~p · ε)ae i
~k·~r + (~p · ε)a†e−i~k·~r |1〉γ |0〉e
]The calculation is simplified if the interaction Hamiltonian is applied to theleft since the final state has only a free electron and no photon
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 14 / 18
![Page 66: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/66.jpg)
Free electron approximation
e〈1|~p = (~~q)e〈1|
γ〈n|a = (√n + 1)γ〈n + 1|a
γ〈n|a† = (√n)γ〈n − 1|a
The free electron state is an eigen-function of the electron momentumoperator
The annihilation operator appliedto the left creates a photon whilethe creation operator annihilates aphoton when applied to the left.
e〈1|γ〈0|(~p · ε)a = ~(~q · ε)e〈1|γ〈1|
e〈1|γ〈0|(~p · ε)a† = 0
Mif =e
m
√~
2ε0Vω
[~(~q · ε)e〈1|γ〈1|e i
~k·~r |1〉γ |0〉e + 0]
=e~m
√~
2ε0Vω(~q · ε)e〈1|e i
~k·~r |0〉e =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 15 / 18
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Free electron approximation
e〈1|~p = (~~q)e〈1|
γ〈n|a = (√n + 1)γ〈n + 1|a
γ〈n|a† = (√n)γ〈n − 1|a
The free electron state is an eigen-function of the electron momentumoperator
The annihilation operator appliedto the left creates a photon whilethe creation operator annihilates aphoton when applied to the left.
e〈1|γ〈0|(~p · ε)a = ~(~q · ε)e〈1|γ〈1|
e〈1|γ〈0|(~p · ε)a† = 0
Mif =e
m
√~
2ε0Vω
[~(~q · ε)e〈1|γ〈1|e i
~k·~r |1〉γ |0〉e + 0]
=e~m
√~
2ε0Vω(~q · ε)e〈1|e i
~k·~r |0〉e =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 15 / 18
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Free electron approximation
e〈1|~p = (~~q)e〈1|
γ〈n|a = (√n + 1)γ〈n + 1|a
γ〈n|a† = (√n)γ〈n − 1|a
The free electron state is an eigen-function of the electron momentumoperator
The annihilation operator appliedto the left creates a photon
whilethe creation operator annihilates aphoton when applied to the left.
e〈1|γ〈0|(~p · ε)a = ~(~q · ε)e〈1|γ〈1|
e〈1|γ〈0|(~p · ε)a† = 0
Mif =e
m
√~
2ε0Vω
[~(~q · ε)e〈1|γ〈1|e i
~k·~r |1〉γ |0〉e + 0]
=e~m
√~
2ε0Vω(~q · ε)e〈1|e i
~k·~r |0〉e =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 15 / 18
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Free electron approximation
e〈1|~p = (~~q)e〈1|
γ〈n|a = (√n + 1)γ〈n + 1|a
γ〈n|a† = (√n)γ〈n − 1|a
The free electron state is an eigen-function of the electron momentumoperator
The annihilation operator appliedto the left creates a photon
whilethe creation operator annihilates aphoton when applied to the left.
e〈1|γ〈0|(~p · ε)a = ~(~q · ε)e〈1|γ〈1|
e〈1|γ〈0|(~p · ε)a† = 0
Mif =e
m
√~
2ε0Vω
[~(~q · ε)e〈1|γ〈1|e i
~k·~r |1〉γ |0〉e + 0]
=e~m
√~
2ε0Vω(~q · ε)e〈1|e i
~k·~r |0〉e =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 15 / 18
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Free electron approximation
e〈1|~p = (~~q)e〈1|
γ〈n|a = (√n + 1)γ〈n + 1|a
γ〈n|a† = (√n)γ〈n − 1|a
The free electron state is an eigen-function of the electron momentumoperator
The annihilation operator appliedto the left creates a photon whilethe creation operator annihilates aphoton when applied to the left.
e〈1|γ〈0|(~p · ε)a = ~(~q · ε)e〈1|γ〈1|
e〈1|γ〈0|(~p · ε)a† = 0
Mif =e
m
√~
2ε0Vω
[~(~q · ε)e〈1|γ〈1|e i
~k·~r |1〉γ |0〉e + 0]
=e~m
√~
2ε0Vω(~q · ε)e〈1|e i
~k·~r |0〉e =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 15 / 18
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Free electron approximation
e〈1|~p = (~~q)e〈1|
γ〈n|a = (√n + 1)γ〈n + 1|a
γ〈n|a† = (√n)γ〈n − 1|a
The free electron state is an eigen-function of the electron momentumoperator
The annihilation operator appliedto the left creates a photon whilethe creation operator annihilates aphoton when applied to the left.
e〈1|γ〈0|(~p · ε)a = ~(~q · ε)e〈1|γ〈1|
e〈1|γ〈0|(~p · ε)a† = 0
Mif =e
m
√~
2ε0Vω
[~(~q · ε)e〈1|γ〈1|e i
~k·~r |1〉γ |0〉e + 0]
=e~m
√~
2ε0Vω(~q · ε)e〈1|e i
~k·~r |0〉e =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 15 / 18
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Free electron approximation
e〈1|~p = (~~q)e〈1|
γ〈n|a = (√n + 1)γ〈n + 1|a
γ〈n|a† = (√n)γ〈n − 1|a
The free electron state is an eigen-function of the electron momentumoperator
The annihilation operator appliedto the left creates a photon whilethe creation operator annihilates aphoton when applied to the left.
e〈1|γ〈0|(~p · ε)a = ~(~q · ε)e〈1|γ〈1|
e〈1|γ〈0|(~p · ε)a† = 0
Mif =e
m
√~
2ε0Vω
[~(~q · ε)e〈1|γ〈1|e i
~k·~r |1〉γ |0〉e + 0]
=e~m
√~
2ε0Vω(~q · ε)e〈1|e i
~k·~r |0〉e =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 15 / 18
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Free electron approximation
e〈1|~p = (~~q)e〈1|
γ〈n|a = (√n + 1)γ〈n + 1|a
γ〈n|a† = (√n)γ〈n − 1|a
The free electron state is an eigen-function of the electron momentumoperator
The annihilation operator appliedto the left creates a photon whilethe creation operator annihilates aphoton when applied to the left.
e〈1|γ〈0|(~p · ε)a = ~(~q · ε)e〈1|γ〈1|
e〈1|γ〈0|(~p · ε)a† = 0
Mif =e
m
√~
2ε0Vω
[~(~q · ε)e〈1|γ〈1|e i
~k·~r |1〉γ |0〉e + 0]
=e~m
√~
2ε0Vω(~q · ε)e〈1|e i
~k·~r |0〉e =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 15 / 18
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Free electron approximation
e〈1|~p = (~~q)e〈1|
γ〈n|a = (√n + 1)γ〈n + 1|a
γ〈n|a† = (√n)γ〈n − 1|a
The free electron state is an eigen-function of the electron momentumoperator
The annihilation operator appliedto the left creates a photon whilethe creation operator annihilates aphoton when applied to the left.
e〈1|γ〈0|(~p · ε)a = ~(~q · ε)e〈1|γ〈1|
e〈1|γ〈0|(~p · ε)a† = 0
Mif =e
m
√~
2ε0Vω
[~(~q · ε)e〈1|γ〈1|e i
~k·~r |1〉γ |0〉e + 0]
=e~m
√~
2ε0Vω(~q · ε)e〈1|e i
~k·~r |0〉e =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 15 / 18
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Free electron approximation
e〈1|~p = (~~q)e〈1|
γ〈n|a = (√n + 1)γ〈n + 1|a
γ〈n|a† = (√n)γ〈n − 1|a
The free electron state is an eigen-function of the electron momentumoperator
The annihilation operator appliedto the left creates a photon whilethe creation operator annihilates aphoton when applied to the left.
e〈1|γ〈0|(~p · ε)a = ~(~q · ε)e〈1|γ〈1|
e〈1|γ〈0|(~p · ε)a† = 0
Mif =e
m
√~
2ε0Vω
[~(~q · ε)e〈1|γ〈1|e i
~k·~r |1〉γ |0〉e + 0]
=e~m
√~
2ε0Vω(~q · ε)e〈1|e i
~k·~r |0〉e
=e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 15 / 18
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Free electron approximation
e〈1|~p = (~~q)e〈1|
γ〈n|a = (√n + 1)γ〈n + 1|a
γ〈n|a† = (√n)γ〈n − 1|a
The free electron state is an eigen-function of the electron momentumoperator
The annihilation operator appliedto the left creates a photon whilethe creation operator annihilates aphoton when applied to the left.
e〈1|γ〈0|(~p · ε)a = ~(~q · ε)e〈1|γ〈1|
e〈1|γ〈0|(~p · ε)a† = 0
Mif =e
m
√~
2ε0Vω
[~(~q · ε)e〈1|γ〈1|e i
~k·~r |1〉γ |0〉e + 0]
=e~m
√~
2ε0Vω(~q · ε)e〈1|e i
~k·~r |0〉e =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 15 / 18
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Photoelectron integral
Mif =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r
=e~m
√~
2ε0Vω(~q · ε)φ(~Q)
ψi = ψ1s(~r) ψf =
√1
Ve i~q·~r
φ(~Q) =
√1
V
∫e−i~q·~re i
~k·~rψ1s(~r)d~r
=
√1
V
∫ψ1s(~r)e i(
~k−~q)·~rd~r
=
√1
V
∫ψ1s(~r)e i
~Q·~rd~r
The initial electron wavefunction issimply that of a 1s atomic statewhile the final state is approxi-mated as a plane wave
The integral thus becomes
which is the Fourier transform ofthe initial state 1s electron wavefunction
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 16 / 18
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Photoelectron integral
Mif =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r =e~m
√~
2ε0Vω(~q · ε)φ(~Q)
ψi = ψ1s(~r) ψf =
√1
Ve i~q·~r
φ(~Q) =
√1
V
∫e−i~q·~re i
~k·~rψ1s(~r)d~r
=
√1
V
∫ψ1s(~r)e i(
~k−~q)·~rd~r
=
√1
V
∫ψ1s(~r)e i
~Q·~rd~r
The initial electron wavefunction issimply that of a 1s atomic statewhile the final state is approxi-mated as a plane wave
The integral thus becomes
which is the Fourier transform ofthe initial state 1s electron wavefunction
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 16 / 18
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Photoelectron integral
Mif =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r =e~m
√~
2ε0Vω(~q · ε)φ(~Q)
ψi = ψ1s(~r) ψf =
√1
Ve i~q·~r
φ(~Q) =
√1
V
∫e−i~q·~re i
~k·~rψ1s(~r)d~r
=
√1
V
∫ψ1s(~r)e i(
~k−~q)·~rd~r
=
√1
V
∫ψ1s(~r)e i
~Q·~rd~r
The initial electron wavefunction issimply that of a 1s atomic state
while the final state is approxi-mated as a plane wave
The integral thus becomes
which is the Fourier transform ofthe initial state 1s electron wavefunction
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 16 / 18
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Photoelectron integral
Mif =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r =e~m
√~
2ε0Vω(~q · ε)φ(~Q)
ψi = ψ1s(~r)
ψf =
√1
Ve i~q·~r
φ(~Q) =
√1
V
∫e−i~q·~re i
~k·~rψ1s(~r)d~r
=
√1
V
∫ψ1s(~r)e i(
~k−~q)·~rd~r
=
√1
V
∫ψ1s(~r)e i
~Q·~rd~r
The initial electron wavefunction issimply that of a 1s atomic state
while the final state is approxi-mated as a plane wave
The integral thus becomes
which is the Fourier transform ofthe initial state 1s electron wavefunction
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 16 / 18
![Page 81: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/81.jpg)
Photoelectron integral
Mif =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r =e~m
√~
2ε0Vω(~q · ε)φ(~Q)
ψi = ψ1s(~r)
ψf =
√1
Ve i~q·~r
φ(~Q) =
√1
V
∫e−i~q·~re i
~k·~rψ1s(~r)d~r
=
√1
V
∫ψ1s(~r)e i(
~k−~q)·~rd~r
=
√1
V
∫ψ1s(~r)e i
~Q·~rd~r
The initial electron wavefunction issimply that of a 1s atomic statewhile the final state is approxi-mated as a plane wave
The integral thus becomes
which is the Fourier transform ofthe initial state 1s electron wavefunction
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 16 / 18
![Page 82: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/82.jpg)
Photoelectron integral
Mif =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r =e~m
√~
2ε0Vω(~q · ε)φ(~Q)
ψi = ψ1s(~r) ψf =
√1
Ve i~q·~r
φ(~Q) =
√1
V
∫e−i~q·~re i
~k·~rψ1s(~r)d~r
=
√1
V
∫ψ1s(~r)e i(
~k−~q)·~rd~r
=
√1
V
∫ψ1s(~r)e i
~Q·~rd~r
The initial electron wavefunction issimply that of a 1s atomic statewhile the final state is approxi-mated as a plane wave
The integral thus becomes
which is the Fourier transform ofthe initial state 1s electron wavefunction
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 16 / 18
![Page 83: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/83.jpg)
Photoelectron integral
Mif =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r =e~m
√~
2ε0Vω(~q · ε)φ(~Q)
ψi = ψ1s(~r) ψf =
√1
Ve i~q·~r
φ(~Q) =
√1
V
∫e−i~q·~re i
~k·~rψ1s(~r)d~r
=
√1
V
∫ψ1s(~r)e i(
~k−~q)·~rd~r
=
√1
V
∫ψ1s(~r)e i
~Q·~rd~r
The initial electron wavefunction issimply that of a 1s atomic statewhile the final state is approxi-mated as a plane wave
The integral thus becomes
which is the Fourier transform ofthe initial state 1s electron wavefunction
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 16 / 18
![Page 84: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/84.jpg)
Photoelectron integral
Mif =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r =e~m
√~
2ε0Vω(~q · ε)φ(~Q)
ψi = ψ1s(~r) ψf =
√1
Ve i~q·~r
φ(~Q) =
√1
V
∫e−i~q·~re i
~k·~rψ1s(~r)d~r
=
√1
V
∫ψ1s(~r)e i(
~k−~q)·~rd~r
=
√1
V
∫ψ1s(~r)e i
~Q·~rd~r
The initial electron wavefunction issimply that of a 1s atomic statewhile the final state is approxi-mated as a plane wave
The integral thus becomes
which is the Fourier transform ofthe initial state 1s electron wavefunction
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 16 / 18
![Page 85: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/85.jpg)
Photoelectron integral
Mif =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r =e~m
√~
2ε0Vω(~q · ε)φ(~Q)
ψi = ψ1s(~r) ψf =
√1
Ve i~q·~r
φ(~Q) =
√1
V
∫e−i~q·~re i
~k·~rψ1s(~r)d~r
=
√1
V
∫ψ1s(~r)e i(
~k−~q)·~rd~r
=
√1
V
∫ψ1s(~r)e i
~Q·~rd~r
The initial electron wavefunction issimply that of a 1s atomic statewhile the final state is approxi-mated as a plane wave
The integral thus becomes
which is the Fourier transform ofthe initial state 1s electron wavefunction
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 16 / 18
![Page 86: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/86.jpg)
Photoelectron integral
Mif =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r =e~m
√~
2ε0Vω(~q · ε)φ(~Q)
ψi = ψ1s(~r) ψf =
√1
Ve i~q·~r
φ(~Q) =
√1
V
∫e−i~q·~re i
~k·~rψ1s(~r)d~r
=
√1
V
∫ψ1s(~r)e i(
~k−~q)·~rd~r
=
√1
V
∫ψ1s(~r)e i
~Q·~rd~r
The initial electron wavefunction issimply that of a 1s atomic statewhile the final state is approxi-mated as a plane wave
The integral thus becomes
which is the Fourier transform ofthe initial state 1s electron wavefunction
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 16 / 18
![Page 87: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/87.jpg)
Photoelectron integral
Mif =e~m
√~
2ε0Vω(~q · ε)
∫ψ∗f e
i~k·~rψid~r =e~m
√~
2ε0Vω(~q · ε)φ(~Q)
ψi = ψ1s(~r) ψf =
√1
Ve i~q·~r
φ(~Q) =
√1
V
∫e−i~q·~re i
~k·~rψ1s(~r)d~r
=
√1
V
∫ψ1s(~r)e i(
~k−~q)·~rd~r
=
√1
V
∫ψ1s(~r)e i
~Q·~rd~r
The initial electron wavefunction issimply that of a 1s atomic statewhile the final state is approxi-mated as a plane wave
The integral thus becomes
which is the Fourier transform ofthe initial state 1s electron wavefunction
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 16 / 18
![Page 88: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/88.jpg)
Photoelectron cross-section
the overall matrix element squared for a particular photoelectron finaldirection (ϕ, θ) is
|Mif |2 =
(e~m
)2 ~2ε0V 2ω
(q2 sin2 θ cos2 ϕ)φ2(~Q)
and the final cross-section per K electron can now be computed as
σa =2π
~cV 2
4π3
(e~m
)2 ~2ε0V 2ω
I3 =
(e~m
)2 1
4π2ε0cωI3
where the integral I3 is given by
I3 =
∫φ2(~Q)q2 sin2 θ cos2 ϕδ(Ef − Ei )q2 sin θdqdθdφ
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 17 / 18
![Page 89: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/89.jpg)
Photoelectron cross-section
the overall matrix element squared for a particular photoelectron finaldirection (ϕ, θ) is
|Mif |2 =
(e~m
)2 ~2ε0V 2ω
(q2 sin2 θ cos2 ϕ)φ2(~Q)
and the final cross-section per K electron can now be computed as
σa =2π
~cV 2
4π3
(e~m
)2 ~2ε0V 2ω
I3 =
(e~m
)2 1
4π2ε0cωI3
where the integral I3 is given by
I3 =
∫φ2(~Q)q2 sin2 θ cos2 ϕδ(Ef − Ei )q2 sin θdqdθdφ
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 17 / 18
![Page 90: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/90.jpg)
Photoelectron cross-section
the overall matrix element squared for a particular photoelectron finaldirection (ϕ, θ) is
|Mif |2 =
(e~m
)2 ~2ε0V 2ω
(q2 sin2 θ cos2 ϕ)φ2(~Q)
and the final cross-section per K electron can now be computed as
σa =2π
~cV 2
4π3
(e~m
)2 ~2ε0V 2ω
I3 =
(e~m
)2 1
4π2ε0cωI3
where the integral I3 is given by
I3 =
∫φ2(~Q)q2 sin2 θ cos2 ϕδ(Ef − Ei )q2 sin θdqdθdφ
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 17 / 18
![Page 91: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/91.jpg)
Photoelectron cross-section
the overall matrix element squared for a particular photoelectron finaldirection (ϕ, θ) is
|Mif |2 =
(e~m
)2 ~2ε0V 2ω
(q2 sin2 θ cos2 ϕ)φ2(~Q)
and the final cross-section per K electron can now be computed as
σa =2π
~cV 2
4π3
(e~m
)2 ~2ε0V 2ω
I3
=
(e~m
)2 1
4π2ε0cωI3
where the integral I3 is given by
I3 =
∫φ2(~Q)q2 sin2 θ cos2 ϕδ(Ef − Ei )q2 sin θdqdθdφ
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 17 / 18
![Page 92: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/92.jpg)
Photoelectron cross-section
the overall matrix element squared for a particular photoelectron finaldirection (ϕ, θ) is
|Mif |2 =
(e~m
)2 ~2ε0V 2ω
(q2 sin2 θ cos2 ϕ)φ2(~Q)
and the final cross-section per K electron can now be computed as
σa =2π
~cV 2
4π3
(e~m
)2 ~2ε0V 2ω
I3 =
(e~m
)2 1
4π2ε0cωI3
where the integral I3 is given by
I3 =
∫φ2(~Q)q2 sin2 θ cos2 ϕδ(Ef − Ei )q2 sin θdqdθdφ
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 17 / 18
![Page 93: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/93.jpg)
Photoelectron cross-section
the overall matrix element squared for a particular photoelectron finaldirection (ϕ, θ) is
|Mif |2 =
(e~m
)2 ~2ε0V 2ω
(q2 sin2 θ cos2 ϕ)φ2(~Q)
and the final cross-section per K electron can now be computed as
σa =2π
~cV 2
4π3
(e~m
)2 ~2ε0V 2ω
I3 =
(e~m
)2 1
4π2ε0cωI3
where the integral I3 is given by
I3 =
∫φ2(~Q)q2 sin2 θ cos2 ϕδ(Ef − Ei )q2 sin θdqdθdφ
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 17 / 18
![Page 94: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/94.jpg)
Photoelectron cross-section
the overall matrix element squared for a particular photoelectron finaldirection (ϕ, θ) is
|Mif |2 =
(e~m
)2 ~2ε0V 2ω
(q2 sin2 θ cos2 ϕ)φ2(~Q)
and the final cross-section per K electron can now be computed as
σa =2π
~cV 2
4π3
(e~m
)2 ~2ε0V 2ω
I3 =
(e~m
)2 1
4π2ε0cωI3
where the integral I3 is given by
I3 =
∫φ2(~Q)q2 sin2 θ cos2 ϕδ(Ef − Ei )q2 sin θdqdθdφ
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 17 / 18
![Page 95: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/95.jpg)
Calculated cross section
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 18 / 18
![Page 96: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/96.jpg)
Calculated cross section
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 18 / 18
![Page 97: Today’s Outline - April 02, 2015segre/phys570/15S/lecture_18.pdfHomework Assignment #06: Chapter 6: 1,6,7,8,9 due Tuesday, April 14, 2015 C. Segre (IIT) PHYS 570 - Spring 2015 April](https://reader036.vdocuments.us/reader036/viewer/2022071105/5fdfe08fcf21c6201d25fb75/html5/thumbnails/97.jpg)
Calculated cross section
C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 18 / 18