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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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 31: 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

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 32: 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

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 33: 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

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

Page 34: 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

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

Page 35: 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

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

Page 36: 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

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

Page 37: 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

Dumond diagram: asymmetric Bragg

C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 8 / 18

Page 38: 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

Double Crystal Monochromators

C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 9 / 18

Page 39: 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

Double Crystal Monochromators

C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 9 / 18

Page 40: 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

Double Crystal Monochromators

C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 10 / 18

Page 41: 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

Double Crystal Monochromators

C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 10 / 18

Page 42: 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

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 43: 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

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

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 45: 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

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 46: 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 63: 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

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 64: 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

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

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

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

Page 67: 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

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

Page 68: 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

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

Page 69: 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

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

Page 70: 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

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

Page 71: 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

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

Page 72: 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

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

Page 73: 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

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

Page 74: 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

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

Page 75: 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

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

Page 76: 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

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

Page 77: 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

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 78: 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

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 79: 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

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 80: 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Calculated cross section

C. Segre (IIT) PHYS 570 - Spring 2015 April 02, 2015 18 / 18