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Today’s Outline - March 27, 2018 C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 1 / 20

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Page 1: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Today’s Outline - March 27, 2018

• Dumond diagrams

• Monochromators

• Photoelectric absorption

• Cross-section of an isolated atom

• X-ray absorption spectroscopy

Homework Assignment #05:Chapter 5: 1, 3, 7, 9, 10due Thursday, March 29, 2018

Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 10, 2018

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 1 / 20

Page 2: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Today’s Outline - March 27, 2018

• Dumond diagrams

• Monochromators

• Photoelectric absorption

• Cross-section of an isolated atom

• X-ray absorption spectroscopy

Homework Assignment #05:Chapter 5: 1, 3, 7, 9, 10due Thursday, March 29, 2018

Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 10, 2018

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 1 / 20

Page 3: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Today’s Outline - March 27, 2018

• Dumond diagrams

• Monochromators

• Photoelectric absorption

• Cross-section of an isolated atom

• X-ray absorption spectroscopy

Homework Assignment #05:Chapter 5: 1, 3, 7, 9, 10due Thursday, March 29, 2018

Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 10, 2018

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 1 / 20

Page 4: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Today’s Outline - March 27, 2018

• Dumond diagrams

• Monochromators

• Photoelectric absorption

• Cross-section of an isolated atom

• X-ray absorption spectroscopy

Homework Assignment #05:Chapter 5: 1, 3, 7, 9, 10due Thursday, March 29, 2018

Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 10, 2018

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 1 / 20

Page 5: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Today’s Outline - March 27, 2018

• Dumond diagrams

• Monochromators

• Photoelectric absorption

• Cross-section of an isolated atom

• X-ray absorption spectroscopy

Homework Assignment #05:Chapter 5: 1, 3, 7, 9, 10due Thursday, March 29, 2018

Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 10, 2018

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 1 / 20

Page 6: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Today’s Outline - March 27, 2018

• Dumond diagrams

• Monochromators

• Photoelectric absorption

• Cross-section of an isolated atom

• X-ray absorption spectroscopy

Homework Assignment #05:Chapter 5: 1, 3, 7, 9, 10due Thursday, March 29, 2018

Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 10, 2018

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 1 / 20

Page 7: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Today’s Outline - March 27, 2018

• Dumond diagrams

• Monochromators

• Photoelectric absorption

• Cross-section of an isolated atom

• X-ray absorption spectroscopy

Homework Assignment #05:Chapter 5: 1, 3, 7, 9, 10due Thursday, March 29, 2018

Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 10, 2018

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 1 / 20

Page 8: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Today’s Outline - March 27, 2018

• Dumond diagrams

• Monochromators

• Photoelectric absorption

• Cross-section of an isolated atom

• X-ray absorption spectroscopy

Homework Assignment #05:Chapter 5: 1, 3, 7, 9, 10due Thursday, March 29, 2018

Homework Assignment #06:Chapter 6: 1,6,7,8,9due Tuesday, April 10, 2018

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 1 / 20

Page 9: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

PHYS 570 day at 10-BM

1. Friday, March 30, 2018, 09:00 – 16:00

2. Activities

• Absolute flux measurement• Rocking curve measurement• SDD setup & measurement• EXAFS measurement

3. Make sure your badge is ready

4. Leave plenty of time to get the badge

5. Let me know when you plan to come!

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 2 / 20

Page 10: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

PHYS 570 day at 10-BM

1. Friday, March 30, 2018, 09:00 – 16:00

2. Activities• Absolute flux measurement• Rocking curve measurement• SDD setup & measurement• EXAFS measurement

3. Make sure your badge is ready

4. Leave plenty of time to get the badge

5. Let me know when you plan to come!

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 2 / 20

Page 11: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

PHYS 570 day at 10-BM

1. Friday, March 30, 2018, 09:00 – 16:00

2. Activities• Absolute flux measurement• Rocking curve measurement• SDD setup & measurement• EXAFS measurement

3. Make sure your badge is ready

4. Leave plenty of time to get the badge

5. Let me know when you plan to come!

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 2 / 20

Page 12: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

PHYS 570 day at 10-BM

1. Friday, March 30, 2018, 09:00 – 16:00

2. Activities• Absolute flux measurement• Rocking curve measurement• SDD setup & measurement• EXAFS measurement

3. Make sure your badge is ready

4. Leave plenty of time to get the badge

5. Let me know when you plan to come!

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 2 / 20

Page 13: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

PHYS 570 day at 10-BM

1. Friday, March 30, 2018, 09:00 – 16:00

2. Activities• Absolute flux measurement• Rocking curve measurement• SDD setup & measurement• EXAFS measurement

3. Make sure your badge is ready

4. Leave plenty of time to get the badge

5. Let me know when you plan to come!

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 2 / 20

Page 14: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Final projects & presentations

In-class student presentations on research topics

• Choose a research article which features a synchrotrontechnique

• Get it approved by instructor first!

• Schedule a 15 minute time on Final Exam Day(tentatively, Monday, April 30, 2018, 10:30-12:30)

Final project - writing a General User Proposal

• Think of a research problem (could be yours) that canbe approached using synchrotron radiation techniques

• Make proposal and get approval from instructor beforestarting

• Must be different techique than your presentation!

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 3 / 20

Page 15: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Final projects & presentations

In-class student presentations on research topics

• Choose a research article which features a synchrotrontechnique

• Get it approved by instructor first!

• Schedule a 15 minute time on Final Exam Day(tentatively, Monday, April 30, 2018, 10:30-12:30)

Final project - writing a General User Proposal

• Think of a research problem (could be yours) that canbe approached using synchrotron radiation techniques

• Make proposal and get approval from instructor beforestarting

• Must be different techique than your presentation!

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 3 / 20

Page 16: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Dumond diagram: no Darwin width

Transfer function of an optical element parametrized by angle andwavelength.

Here Darwin width is ignored.

0 0θi-θB θe-θB

λ

2d

∆θ

cosθΒ ∆θ

sin

θB

for small angular de-viations sin θ is lin-ear with a slope ofcos θB

non-zero diffractedbeam only for pointson the line

a horizontal linetransfers inputto output beamcharacteristics

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 4 / 20

Page 17: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Dumond diagram: no Darwin width

Transfer function of an optical element parametrized by angle andwavelength. Here Darwin width is ignored.

0 0θi-θB θe-θB

λ

2d

∆θ

cosθΒ ∆θ

sin

θB

for small angular de-viations sin θ is lin-ear with a slope ofcos θB

non-zero diffractedbeam only for pointson the line

a horizontal linetransfers inputto output beamcharacteristics

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 4 / 20

Page 18: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Dumond diagram: no Darwin width

Transfer function of an optical element parametrized by angle andwavelength. Here Darwin width is ignored.

0 0θi-θB θe-θB

λ

2d

∆θ

cosθΒ ∆θ

sin

θB

for small angular de-viations sin θ is lin-ear with a slope ofcos θB

non-zero diffractedbeam only for pointson the line

a horizontal linetransfers inputto output beamcharacteristics

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 4 / 20

Page 19: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Dumond diagram: no Darwin width

Transfer function of an optical element parametrized by angle andwavelength. Here Darwin width is ignored.

0 0θi-θB θe-θB

λ

2d

∆θ

cosθΒ ∆θ

sin

θB

for small angular de-viations sin θ is lin-ear with a slope ofcos θB

non-zero diffractedbeam only for pointson the line

a horizontal linetransfers inputto output beamcharacteristics

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 4 / 20

Page 20: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Dumond diagram: no Darwin width

Transfer function of an optical element parametrized by angle andwavelength. Here Darwin width is ignored.

0 0θi-θB θe-θB

λ

2d

∆θ

cosθΒ ∆θ

sin

θB

for small angular de-viations sin θ is lin-ear with a slope ofcos θB

non-zero diffractedbeam only for pointson the line

a horizontal linetransfers inputto output beamcharacteristics

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 4 / 20

Page 21: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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).

0 0θi-θB θe-θB

λ

2d

w0=sinθB ζD

the bandwidth ofa collimated (noangular divergence)beam denoted bythe black line canbe accepted by theinput function ofthe crystal

this input band-width is transferredto a similar outputbandwidth which isalso collimated

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 5 / 20

Page 22: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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).

0 0θi-θB θe-θB

λ

2d

w0=sinθB ζD

the bandwidth ofa collimated (noangular divergence)beam denoted bythe black line canbe accepted by theinput function ofthe crystal

this input band-width is transferredto a similar outputbandwidth which isalso collimated

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 5 / 20

Page 23: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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).

0 0θi-θB θe-θB

λ

2d

w0=sinθB ζD

the bandwidth ofa collimated (noangular divergence)beam denoted bythe black line canbe accepted by theinput function ofthe crystal

this input band-width is transferredto a similar outputbandwidth which isalso collimated

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 5 / 20

Page 24: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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).

0 0θi-θB θe-θB

λ

2d

w0=sinθB ζD

the bandwidth ofa collimated (noangular divergence)beam denoted bythe black line canbe accepted by theinput function ofthe crystal

this input band-width is transferredto a similar outputbandwidth which isalso collimated

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 5 / 20

Page 25: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Dumond diagram: asymmetric Bragg

For an asymmetric crystal, the output beam is no longer collimated butacquires a divergence αe

0 0θi-θB θe-θB

λ

2d w0 b

b

w0

αe

a perfectly colli-mated input beamtransfers to anoutput beam thathas an angulardivergence whichdepends on theasymmetry factor b

this is in additionto a compression (inthis case) of thebeam height (Liou-ville’s theorem!)

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 6 / 20

Page 26: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Dumond diagram: asymmetric Bragg

For an asymmetric crystal, the output beam is no longer collimated butacquires a divergence αe

0 0θi-θB θe-θB

λ

2d w0 b

b

w0

αe

a perfectly colli-mated input beamtransfers to anoutput beam thathas an angulardivergence whichdepends on theasymmetry factor b

this is in additionto a compression (inthis case) of thebeam height (Liou-ville’s theorem!)

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 6 / 20

Page 27: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Dumond diagram: asymmetric Bragg

For an asymmetric crystal, the output beam is no longer collimated butacquires a divergence αe

0 0θi-θB θe-θB

λ

2d w0 b

b

w0

αe

a perfectly colli-mated input beamtransfers to anoutput beam thathas an angulardivergence whichdepends on theasymmetry factor b

this is in additionto a compression (inthis case) of thebeam height (Liou-ville’s theorem!)

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 6 / 20

Page 28: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Double crystal monochromator: Non-dispersive

∆θin

the transfer functions ofthe the two crystalsmatch and full bandwithand divergence is pre-served

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 7 / 20

Page 29: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Double crystal monochromator: Non-dispersive

∆θin

the transfer functions ofthe the two crystalsmatch and full bandwithand divergence is pre-served

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 7 / 20

Page 30: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Double crystal monochromator: Non-dispersive

∆θin

the transfer functions ofthe the two crystalsmatch and full bandwithand divergence is pre-served

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 7 / 20

Page 31: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Double crystal monochromators: Dispersive

∆θin

the transfer functionmatches only in smallband that varies withangle of the secondcrystal

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 8 / 20

Page 32: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Double crystal monochromators: Dispersive

∆θin

the transfer functionmatches only in smallband that varies withangle of the secondcrystal

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 8 / 20

Page 33: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Double crystal monochromators: Dispersive

∆θin

the transfer functionmatches only in smallband that varies withangle of the secondcrystal

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 8 / 20

Page 34: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 2018 March 27, 2018 9 / 20

Page 35: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 2018 March 27, 2018 9 / 20

Page 36: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 2018 March 27, 2018 9 / 20

Page 37: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Scaled absorption

T =I

I0= 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 2018 March 27, 2018 10 / 20

Page 38: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Scaled absorption

T =I

I0= 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 2018 March 27, 2018 10 / 20

Page 39: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Scaled absorption

T =I

I0= 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 2018 March 27, 2018 10 / 20

Page 40: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Scaled absorption

T =I

I0= 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 2018 March 27, 2018 10 / 20

Page 41: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Scaled absorption

T =I

I0= 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 2018 March 27, 2018 10 / 20

Page 42: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Scaled absorption

T =I

I0= 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 2018 March 27, 2018 10 / 20

Page 43: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 = ε̂

√~

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 2018 March 27, 2018 11 / 20

Page 44: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 = ε̂

√~

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 2018 March 27, 2018 11 / 20

Page 45: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 = ε̂

√~

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 2018 March 27, 2018 11 / 20

Page 46: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 = ε̂

√~

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 2018 March 27, 2018 11 / 20

Page 47: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 = ε̂

√~

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 2018 March 27, 2018 11 / 20

Page 48: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 = ε̂

√~

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 2018 March 27, 2018 11 / 20

Page 49: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 = ε̂

√~

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 2018 March 27, 2018 11 / 20

Page 50: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 = ε̂

√~

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 2018 March 27, 2018 11 / 20

Page 51: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)ae i

~k·~r + (~p · ε̂)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 2018 March 27, 2018 12 / 20

Page 52: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)ae i

~k·~r + (~p · ε̂)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 2018 March 27, 2018 12 / 20

Page 53: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)ae i

~k·~r + (~p · ε̂)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 2018 March 27, 2018 12 / 20

Page 54: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)ae i

~k·~r + (~p · ε̂)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 2018 March 27, 2018 12 / 20

Page 55: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)ae i

~k·~r + (~p · ε̂)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 2018 March 27, 2018 12 / 20

Page 56: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)ae i

~k·~r + (~p · ε̂)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 2018 March 27, 2018 12 / 20

Page 57: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)ae i

~k·~r + (~p · ε̂)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 2018 March 27, 2018 12 / 20

Page 58: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)a = ~(~q · ε̂)e〈1|γ〈1|

e〈1|γ〈0|(~p · ε̂)a† = 0

Mif =e

m

√~

2ε0Vω

[~(~q · ε̂)e〈1|γ〈1|e i

~k·~r |1〉γ |0〉e + 0]

=e~m

√~

2ε0Vω(~q · ε̂)e〈1|e i

~k·~r |0〉e =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 13 / 20

Page 59: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)a = ~(~q · ε̂)e〈1|γ〈1|

e〈1|γ〈0|(~p · ε̂)a† = 0

Mif =e

m

√~

2ε0Vω

[~(~q · ε̂)e〈1|γ〈1|e i

~k·~r |1〉γ |0〉e + 0]

=e~m

√~

2ε0Vω(~q · ε̂)e〈1|e i

~k·~r |0〉e =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 13 / 20

Page 60: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)a = ~(~q · ε̂)e〈1|γ〈1|

e〈1|γ〈0|(~p · ε̂)a† = 0

Mif =e

m

√~

2ε0Vω

[~(~q · ε̂)e〈1|γ〈1|e i

~k·~r |1〉γ |0〉e + 0]

=e~m

√~

2ε0Vω(~q · ε̂)e〈1|e i

~k·~r |0〉e =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 13 / 20

Page 61: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)a = ~(~q · ε̂)e〈1|γ〈1|

e〈1|γ〈0|(~p · ε̂)a† = 0

Mif =e

m

√~

2ε0Vω

[~(~q · ε̂)e〈1|γ〈1|e i

~k·~r |1〉γ |0〉e + 0]

=e~m

√~

2ε0Vω(~q · ε̂)e〈1|e i

~k·~r |0〉e =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 13 / 20

Page 62: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)a = ~(~q · ε̂)e〈1|γ〈1|

e〈1|γ〈0|(~p · ε̂)a† = 0

Mif =e

m

√~

2ε0Vω

[~(~q · ε̂)e〈1|γ〈1|e i

~k·~r |1〉γ |0〉e + 0]

=e~m

√~

2ε0Vω(~q · ε̂)e〈1|e i

~k·~r |0〉e =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 13 / 20

Page 63: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)a = ~(~q · ε̂)e〈1|γ〈1|

e〈1|γ〈0|(~p · ε̂)a† = 0

Mif =e

m

√~

2ε0Vω

[~(~q · ε̂)e〈1|γ〈1|e i

~k·~r |1〉γ |0〉e + 0]

=e~m

√~

2ε0Vω(~q · ε̂)e〈1|e i

~k·~r |0〉e =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 13 / 20

Page 64: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)a = ~(~q · ε̂)e〈1|γ〈1|

e〈1|γ〈0|(~p · ε̂)a† = 0

Mif =e

m

√~

2ε0Vω

[~(~q · ε̂)e〈1|γ〈1|e i

~k·~r |1〉γ |0〉e + 0]

=e~m

√~

2ε0Vω(~q · ε̂)e〈1|e i

~k·~r |0〉e =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 13 / 20

Page 65: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)a = ~(~q · ε̂)e〈1|γ〈1|

e〈1|γ〈0|(~p · ε̂)a† = 0

Mif =e

m

√~

2ε0Vω

[~(~q · ε̂)e〈1|γ〈1|e i

~k·~r |1〉γ |0〉e + 0]

=e~m

√~

2ε0Vω(~q · ε̂)e〈1|e i

~k·~r |0〉e =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 13 / 20

Page 66: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)a = ~(~q · ε̂)e〈1|γ〈1|

e〈1|γ〈0|(~p · ε̂)a† = 0

Mif =e

m

√~

2ε0Vω

[~(~q · ε̂)e〈1|γ〈1|e i

~k·~r |1〉γ |0〉e + 0]

=e~m

√~

2ε0Vω(~q · ε̂)e〈1|e i

~k·~r |0〉e =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 13 / 20

Page 67: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)a = ~(~q · ε̂)e〈1|γ〈1|

e〈1|γ〈0|(~p · ε̂)a† = 0

Mif =e

m

√~

2ε0Vω

[~(~q · ε̂)e〈1|γ〈1|e i

~k·~r |1〉γ |0〉e + 0]

=e~m

√~

2ε0Vω(~q · ε̂)e〈1|e i

~k·~r |0〉e

=e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 13 / 20

Page 68: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 · ε̂)a = ~(~q · ε̂)e〈1|γ〈1|

e〈1|γ〈0|(~p · ε̂)a† = 0

Mif =e

m

√~

2ε0Vω

[~(~q · ε̂)e〈1|γ〈1|e i

~k·~r |1〉γ |0〉e + 0]

=e~m

√~

2ε0Vω(~q · ε̂)e〈1|e i

~k·~r |0〉e =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 13 / 20

Page 69: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Photoelectron integral

Mif =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r

=e~m

√~

2ε0Vω(~q · ε̂)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 2018 March 27, 2018 14 / 20

Page 70: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Photoelectron integral

Mif =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r =e~m

√~

2ε0Vω(~q · ε̂)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 2018 March 27, 2018 14 / 20

Page 71: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Photoelectron integral

Mif =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r =e~m

√~

2ε0Vω(~q · ε̂)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 2018 March 27, 2018 14 / 20

Page 72: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Photoelectron integral

Mif =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r =e~m

√~

2ε0Vω(~q · ε̂)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 2018 March 27, 2018 14 / 20

Page 73: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Photoelectron integral

Mif =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r =e~m

√~

2ε0Vω(~q · ε̂)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 2018 March 27, 2018 14 / 20

Page 74: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Photoelectron integral

Mif =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r =e~m

√~

2ε0Vω(~q · ε̂)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 2018 March 27, 2018 14 / 20

Page 75: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Photoelectron integral

Mif =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r =e~m

√~

2ε0Vω(~q · ε̂)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 2018 March 27, 2018 14 / 20

Page 76: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Photoelectron integral

Mif =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r =e~m

√~

2ε0Vω(~q · ε̂)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 2018 March 27, 2018 14 / 20

Page 77: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Photoelectron integral

Mif =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r =e~m

√~

2ε0Vω(~q · ε̂)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 2018 March 27, 2018 14 / 20

Page 78: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Photoelectron integral

Mif =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r =e~m

√~

2ε0Vω(~q · ε̂)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 2018 March 27, 2018 14 / 20

Page 79: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Photoelectron integral

Mif =e~m

√~

2ε0Vω(~q · ε̂)

∫ψ∗f e

i~k·~rψid~r =e~m

√~

2ε0Vω(~q · ε̂)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 2018 March 27, 2018 14 / 20

Page 80: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 2018 March 27, 2018 15 / 20

Page 81: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 2018 March 27, 2018 15 / 20

Page 82: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 2018 March 27, 2018 15 / 20

Page 83: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 2018 March 27, 2018 15 / 20

Page 84: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 2018 March 27, 2018 15 / 20

Page 85: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 2018 March 27, 2018 15 / 20

Page 86: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

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 2018 March 27, 2018 15 / 20

Page 87: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Calculated cross section

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 16 / 20

Page 88: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Calculated cross section

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 16 / 20

Page 89: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

Calculated cross section

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 16 / 20

Page 90: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

What is XAFS?

X-ray Absorption Fine-Structure (XAFS) is the modulation of the x-rayabsorption coefficient at energies near and above an x-ray absorption edge.XAFS is also referred to as X-ray Absorption Spectroscopy (XAS) and isbroken into 2 regimes:

XANES X-ray Absorption Near-Edge SpectroscopyEXAFS Extended X-ray Absorption Fine-Structure

which contain related, but slightly different information about an element’slocal coordination and chemical state.

EXAFS

XANES

E (eV)

µ

(

E

)

77007600750074007300720071007000

2.0

1.5

1.0

0.5

0.0

Fe K-edge XAFS for FeO

XAFS Characteristics:

• local atomic coordination

• chemical / oxidation state

• applies to any element

• works at low concentrations

• minimal sample requirements

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 17 / 20

Page 91: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

The x-ray absorption process

An x-ray is absorbed by anatom when the energy of thex-ray is transferred to a core-level electron (K, L, or Mshell).

The atom is in an excitedstate with an empty elec-tronic level: a core hole.

Any excess energy fromthe x-ray is given to anejected photoelectron

, whichexpands as a spherical wave,reaches the neighboring elec-tron clouds, and scattersback to the core hole, cre-ating interference patternscalled XANES and EXAFS.

x−ray

L

K

M

En

erg

y

photo−electron

Continuum

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 18 / 20

Page 92: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

The x-ray absorption process

An x-ray is absorbed by anatom when the energy of thex-ray is transferred to a core-level electron (K, L, or Mshell).

The atom is in an excitedstate with an empty elec-tronic level: a core hole.

Any excess energy fromthe x-ray is given to anejected photoelectron

, whichexpands as a spherical wave,reaches the neighboring elec-tron clouds, and scattersback to the core hole, cre-ating interference patternscalled XANES and EXAFS.

x−ray

L

K

M

En

erg

y

photo−electron

Continuum

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 18 / 20

Page 93: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

The x-ray absorption process

An x-ray is absorbed by anatom when the energy of thex-ray is transferred to a core-level electron (K, L, or Mshell).

The atom is in an excitedstate with an empty elec-tronic level: a core hole.

Any excess energy fromthe x-ray is given to anejected photoelectron

, whichexpands as a spherical wave,reaches the neighboring elec-tron clouds, and scattersback to the core hole, cre-ating interference patternscalled XANES and EXAFS.

x−ray

L

K

M

En

erg

y

photo−electron

Continuum

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 18 / 20

Page 94: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

The x-ray absorption process

An x-ray is absorbed by anatom when the energy of thex-ray is transferred to a core-level electron (K, L, or Mshell).

The atom is in an excitedstate with an empty elec-tronic level: a core hole.

Any excess energy fromthe x-ray is given to anejected photoelectron, whichexpands as a spherical wave

,reaches the neighboring elec-tron clouds, and scattersback to the core hole, cre-ating interference patternscalled XANES and EXAFS.

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 18 / 20

Page 95: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

The x-ray absorption process

An x-ray is absorbed by anatom when the energy of thex-ray is transferred to a core-level electron (K, L, or Mshell).

The atom is in an excitedstate with an empty elec-tronic level: a core hole.

Any excess energy fromthe x-ray is given to anejected photoelectron, whichexpands as a spherical wave

,reaches the neighboring elec-tron clouds, and scattersback to the core hole, cre-ating interference patternscalled XANES and EXAFS.

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 18 / 20

Page 96: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

The x-ray absorption process

An x-ray is absorbed by anatom when the energy of thex-ray is transferred to a core-level electron (K, L, or Mshell).

The atom is in an excitedstate with an empty elec-tronic level: a core hole.

Any excess energy fromthe x-ray is given to anejected photoelectron, whichexpands as a spherical wave,reaches the neighboring elec-tron clouds

, and scattersback to the core hole, cre-ating interference patternscalled XANES and EXAFS.

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 18 / 20

Page 97: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

The x-ray absorption process

An x-ray is absorbed by anatom when the energy of thex-ray is transferred to a core-level electron (K, L, or Mshell).

The atom is in an excitedstate with an empty elec-tronic level: a core hole.

Any excess energy fromthe x-ray is given to anejected photoelectron, whichexpands as a spherical wave,reaches the neighboring elec-tron clouds, and scattersback to the core hole

, cre-ating interference patternscalled XANES and EXAFS.

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 18 / 20

Page 98: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

The x-ray absorption process

An x-ray is absorbed by anatom when the energy of thex-ray is transferred to a core-level electron (K, L, or Mshell).

The atom is in an excitedstate with an empty elec-tronic level: a core hole.

Any excess energy fromthe x-ray is given to anejected photoelectron, whichexpands as a spherical wave,reaches the neighboring elec-tron clouds, and scattersback to the core hole

, cre-ating interference patternscalled XANES and EXAFS.

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 18 / 20

Page 99: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

The x-ray absorption process

An x-ray is absorbed by anatom when the energy of thex-ray is transferred to a core-level electron (K, L, or Mshell).

The atom is in an excitedstate with an empty elec-tronic level: a core hole.

Any excess energy fromthe x-ray is given to anejected photoelectron, whichexpands as a spherical wave,reaches the neighboring elec-tron clouds, and scattersback to the core hole, cre-ating interference patternscalled XANES and EXAFS.

11500 12000 12500

E(eV)

-2

-1.5

-1

-0.5

0

0.5

ln(I

o/I

)

EXAFS

XA

NE

S /

NE

XA

FS

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 18 / 20

Page 100: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

EXAFS data extraction

normalize by fitting pre-edgeand post-edge trends

remove “smooth” µ0 back-ground

convert to photoelectron mo-mentum space

k =2π

hc

√E − E0

weight by appropriate powerof k to obtain “good” enve-lope which clearly shows thatEXAFS is a sum of oscilla-tions with varying frequen-cies and phases

Fourier transform to get realspace EXAFS

11500 12000 12500

E(eV)

-2

-1.5

-1

-0.5

0

0.5

ln(I

o/I

)

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 19 / 20

Page 101: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

EXAFS data extraction

normalize by fitting pre-edgeand post-edge trends

remove “smooth” µ0 back-ground

convert to photoelectron mo-mentum space

k =2π

hc

√E − E0

weight by appropriate powerof k to obtain “good” enve-lope which clearly shows thatEXAFS is a sum of oscilla-tions with varying frequen-cies and phases

Fourier transform to get realspace EXAFS

11500 12000 12500

E(eV)

0

0.5

1

ln(I

o/I

)

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 19 / 20

Page 102: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

EXAFS data extraction

normalize by fitting pre-edgeand post-edge trends

remove “smooth” µ0 back-ground

convert to photoelectron mo-mentum space

k =2π

hc

√E − E0

weight by appropriate powerof k to obtain “good” enve-lope which clearly shows thatEXAFS is a sum of oscilla-tions with varying frequen-cies and phases

Fourier transform to get realspace EXAFS

11500 12000 12500

E(eV)

0

0.5

1

ln(I

o/I

)

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 19 / 20

Page 103: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

EXAFS data extraction

normalize by fitting pre-edgeand post-edge trends

remove “smooth” µ0 back-ground

convert to photoelectron mo-mentum space

k =2π

hc

√E − E0

weight by appropriate powerof k to obtain “good” enve-lope which clearly shows thatEXAFS is a sum of oscilla-tions with varying frequen-cies and phases

Fourier transform to get realspace EXAFS

0 5 10 15

k(Å-1

)

-0.2

-0.1

0

0.1

0.2

χ

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 19 / 20

Page 104: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

EXAFS data extraction

normalize by fitting pre-edgeand post-edge trends

remove “smooth” µ0 back-ground

convert to photoelectron mo-mentum space

k =2π

hc

√E − E0

weight by appropriate powerof k to obtain “good” enve-lope which clearly shows thatEXAFS is a sum of oscilla-tions with varying frequen-cies and phases

Fourier transform to get realspace EXAFS

0 5 10 15

k(Å-1

)

-0.2

-0.1

0

0.1

0.2

χ

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 19 / 20

Page 105: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

EXAFS data extraction

normalize by fitting pre-edgeand post-edge trends

remove “smooth” µ0 back-ground

convert to photoelectron mo-mentum space

k =2π

hc

√E − E0

weight by appropriate powerof k to obtain “good” enve-lope which clearly shows thatEXAFS is a sum of oscilla-tions with varying frequen-cies and phases

Fourier transform to get realspace EXAFS

0 2 4 6

R(Å)

0

5

10

15

20

25

30

χ(R

)

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 19 / 20

Page 106: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

XANES edge shifts and pre-edge peaks

5460 5470 5480 5490 5500

E(eV)

0

0.2

0.4

0.6

0.8

1

1.2

ln(I

o/I)

V metal

V2O

3

V2O

5

LiVOPO4

The shift of the edge positioncan be used to determine thevalence state.

The heights and positions ofpre-edge peaks can also be re-liably used to determine ionicratios for many atomic species.

XANES can be used as a fin-gerprint of phases and XANESanalysis can be as simple asmaking linear combinations of“known” spectra to get com-position.

Modern codes, such as FEFF9,are able to accurately computeXANES features.

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 20 / 20

Page 107: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

XANES edge shifts and pre-edge peaks

5460 5470 5480 5490 5500

E(eV)

0

0.2

0.4

0.6

0.8

1

1.2

ln(I

o/I)

V metal

V2O

3

V2O

5

LiVOPO4

The shift of the edge positioncan be used to determine thevalence state.

The heights and positions ofpre-edge peaks can also be re-liably used to determine ionicratios for many atomic species.

XANES can be used as a fin-gerprint of phases and XANESanalysis can be as simple asmaking linear combinations of“known” spectra to get com-position.

Modern codes, such as FEFF9,are able to accurately computeXANES features.

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 20 / 20

Page 108: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

XANES edge shifts and pre-edge peaks

5460 5470 5480 5490 5500

E(eV)

0

0.2

0.4

0.6

0.8

1

1.2

ln(I

o/I)

V metal

V2O

3

V2O

5

LiVOPO4

The shift of the edge positioncan be used to determine thevalence state.

The heights and positions ofpre-edge peaks can also be re-liably used to determine ionicratios for many atomic species.

XANES can be used as a fin-gerprint of phases and XANESanalysis can be as simple asmaking linear combinations of“known” spectra to get com-position.

Modern codes, such as FEFF9,are able to accurately computeXANES features.

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 20 / 20

Page 109: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

XANES edge shifts and pre-edge peaks

5460 5470 5480 5490 5500

E(eV)

0

0.2

0.4

0.6

0.8

1

1.2

ln(I

o/I)

V metal

V2O

3

V2O

5

LiVOPO4

The shift of the edge positioncan be used to determine thevalence state.

The heights and positions ofpre-edge peaks can also be re-liably used to determine ionicratios for many atomic species.

XANES can be used as a fin-gerprint of phases and XANESanalysis can be as simple asmaking linear combinations of“known” spectra to get com-position.

Modern codes, such as FEFF9,are able to accurately computeXANES features.

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 20 / 20

Page 110: Today’s Outline - March 27, 2018 - IITphys.iit.edu/~segre/phys570/18S/lecture_19.pdfHomework Assignment #05: Chapter 5: 1, 3, 7, 9, 10 due Thursday, March 29, 2018 Homework Assignment

XANES edge shifts and pre-edge peaks

5460 5470 5480 5490 5500

E(eV)

0

0.2

0.4

0.6

0.8

1

1.2

ln(I

o/I)

V metal

V2O

3

V2O

5

LiVOPO4

The shift of the edge positioncan be used to determine thevalence state.

The heights and positions ofpre-edge peaks can also be re-liably used to determine ionicratios for many atomic species.

XANES can be used as a fin-gerprint of phases and XANESanalysis can be as simple asmaking linear combinations of“known” spectra to get com-position.

Modern codes, such as FEFF9,are able to accurately computeXANES features.

C. Segre (IIT) PHYS 570 - Spring 2018 March 27, 2018 20 / 20