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Peter Hertel

Overview Normal matter Natural units Atomic units

Length and

Electric and Peter Hertel magnetic fields

More University of Osnabr¨uck,Germany examples Lecture presented at APS, Nankai University, China

http://www.home.uni-osnabrueck.de/phertel

October/November 2011 • Normal matter • Atomic units • SI to AU conversion

Natural units Overview Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • Atomic units • SI to AU conversion

Natural units Overview Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields • Normal matter More examples • SI to AU conversion

Natural units Overview Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields • Normal matter More examples • Atomic units Natural units Overview Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields • Normal matter More examples • Atomic units • SI to AU conversion • normal matter is governed by electrostatic and non-relativistic • heavy nuclei are surrounded by neutral , molecules, liquids and solids • Planck’s constant ~ • unit charge e • m

• 4π0 from ’s law • typical values for normal matter are reasonable numbers • products of powers of these constants

Natural units Normal matter Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • heavy nuclei are surrounded by electrons neutral atoms, molecules, liquids and solids • Planck’s constant ~ • unit charge e • electron mass m

• 4π0 from Coulomb’s law • typical values for normal matter are reasonable numbers • times products of powers of these constants

Natural units Normal matter Peter Hertel

Overview

Normal matter Atomic units • normal matter is governed by electrostatic forces and Length and non-relativistic quantum mechanics energy

Electric and magnetic fields

More examples • Planck’s constant ~ • unit charge e • electron mass m

• 4π0 from Coulomb’s law • typical values for normal matter are reasonable numbers • times products of powers of these constants

Natural units Normal matter Peter Hertel

Overview

Normal matter Atomic units • normal matter is governed by electrostatic forces and Length and non-relativistic quantum mechanics energy Electric and • heavy nuclei are surrounded by electrons neutral atoms, magnetic fields molecules, liquids and solids

More examples • unit charge e • electron mass m

• 4π0 from Coulomb’s law • typical values for normal matter are reasonable numbers • times products of powers of these constants

Natural units Normal matter Peter Hertel

Overview

Normal matter Atomic units • normal matter is governed by electrostatic forces and Length and non-relativistic quantum mechanics energy Electric and • heavy nuclei are surrounded by electrons neutral atoms, magnetic fields molecules, liquids and solids More • examples Planck’s constant ~ • electron mass m

• 4π0 from Coulomb’s law • typical values for normal matter are reasonable numbers • times products of powers of these constants

Natural units Normal matter Peter Hertel

Overview

Normal matter Atomic units • normal matter is governed by electrostatic forces and Length and non-relativistic quantum mechanics energy Electric and • heavy nuclei are surrounded by electrons neutral atoms, magnetic fields molecules, liquids and solids More • examples Planck’s constant ~ • unit charge e • 4π0 from Coulomb’s law • typical values for normal matter are reasonable numbers • times products of powers of these constants

Natural units Normal matter Peter Hertel

Overview

Normal matter Atomic units • normal matter is governed by electrostatic forces and Length and non-relativistic quantum mechanics energy Electric and • heavy nuclei are surrounded by electrons neutral atoms, magnetic fields molecules, liquids and solids More • examples Planck’s constant ~ • unit charge e • electron mass m • typical values for normal matter are reasonable numbers • times products of powers of these constants

Natural units Normal matter Peter Hertel

Overview

Normal matter Atomic units • normal matter is governed by electrostatic forces and Length and non-relativistic quantum mechanics energy Electric and • heavy nuclei are surrounded by electrons neutral atoms, magnetic fields molecules, liquids and solids More • examples Planck’s constant ~ • unit charge e • electron mass m

• 4π0 from Coulomb’s law • times products of powers of these constants

Natural units Normal matter Peter Hertel

Overview

Normal matter Atomic units • normal matter is governed by electrostatic forces and Length and non-relativistic quantum mechanics energy Electric and • heavy nuclei are surrounded by electrons neutral atoms, magnetic fields molecules, liquids and solids More • examples Planck’s constant ~ • unit charge e • electron mass m

• 4π0 from Coulomb’s law • typical values for normal matter are reasonable numbers Natural units Normal matter Peter Hertel

Overview

Normal matter Atomic units • normal matter is governed by electrostatic forces and Length and non-relativistic quantum mechanics energy Electric and • heavy nuclei are surrounded by electrons neutral atoms, magnetic fields molecules, liquids and solids More • examples Planck’s constant ~ • unit charge e • electron mass m

• 4π0 from Coulomb’s law • typical values for normal matter are reasonable numbers • times products of powers of these constants • today, the Syst`emeinternational d’unit´es (SI) is in common use, in particular in science and engineering • also called MKSA • MKSA = meter (m), (kg), (s) and (A) • other SI units are derived from it, like (V) −34 2 −1 • ~ = 1.054572 ×10 kg m s • e = 1.602177 ×10−19 A s • m = 9.10938 ×10−31 kg −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A

Natural units Constants of nature Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • also called MKSA • MKSA = meter (m), kilogram (kg), second (s) and ampere (A) • other SI units are derived from it, like volt (V) −34 2 −1 • ~ = 1.054572 ×10 kg m s • e = 1.602177 ×10−19 A s • m = 9.10938 ×10−31 kg −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A

Natural units Constants of nature Peter Hertel

Overview

Normal matter Atomic units • today, the Syst`emeinternational d’unit´es (SI) is in Length and energy common use, in particular in science and engineering

Electric and magnetic fields

More examples • MKSA = meter (m), kilogram (kg), second (s) and ampere (A) • other SI units are derived from it, like volt (V) −34 2 −1 • ~ = 1.054572 ×10 kg m s • e = 1.602177 ×10−19 A s • m = 9.10938 ×10−31 kg −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A

Natural units Constants of nature Peter Hertel

Overview

Normal matter Atomic units • today, the Syst`emeinternational d’unit´es (SI) is in Length and energy common use, in particular in science and engineering Electric and • magnetic also called MKSA fields

More examples • other SI units are derived from it, like volt (V) −34 2 −1 • ~ = 1.054572 ×10 kg m s • e = 1.602177 ×10−19 A s • m = 9.10938 ×10−31 kg −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A

Natural units Constants of nature Peter Hertel

Overview

Normal matter Atomic units • today, the Syst`emeinternational d’unit´es (SI) is in Length and energy common use, in particular in science and engineering Electric and • magnetic also called MKSA fields • MKSA = meter (m), kilogram (kg), second (s) and More examples ampere (A) −34 2 −1 • ~ = 1.054572 ×10 kg m s • e = 1.602177 ×10−19 A s • m = 9.10938 ×10−31 kg −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A

Natural units Constants of nature Peter Hertel

Overview

Normal matter Atomic units • today, the Syst`emeinternational d’unit´es (SI) is in Length and energy common use, in particular in science and engineering Electric and • magnetic also called MKSA fields • MKSA = meter (m), kilogram (kg), second (s) and More examples ampere (A) • other SI units are derived from it, like volt (V) • e = 1.602177 ×10−19 A s • m = 9.10938 ×10−31 kg −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A

Natural units Constants of nature Peter Hertel

Overview

Normal matter Atomic units • today, the Syst`emeinternational d’unit´es (SI) is in Length and energy common use, in particular in science and engineering Electric and • magnetic also called MKSA fields • MKSA = meter (m), kilogram (kg), second (s) and More examples ampere (A) • other SI units are derived from it, like volt (V) −34 2 −1 • ~ = 1.054572 ×10 kg m s • m = 9.10938 ×10−31 kg −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A

Natural units Constants of nature Peter Hertel

Overview

Normal matter Atomic units • today, the Syst`emeinternational d’unit´es (SI) is in Length and energy common use, in particular in science and engineering Electric and • magnetic also called MKSA fields • MKSA = meter (m), kilogram (kg), second (s) and More examples ampere (A) • other SI units are derived from it, like volt (V) −34 2 −1 • ~ = 1.054572 ×10 kg m s • e = 1.602177 ×10−19 A s −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A

Natural units Constants of nature Peter Hertel

Overview

Normal matter Atomic units • today, the Syst`emeinternational d’unit´es (SI) is in Length and energy common use, in particular in science and engineering Electric and • magnetic also called MKSA fields • MKSA = meter (m), kilogram (kg), second (s) and More examples ampere (A) • other SI units are derived from it, like volt (V) −34 2 −1 • ~ = 1.054572 ×10 kg m s • e = 1.602177 ×10−19 A s • m = 9.10938 ×10−31 kg Natural units Constants of nature Peter Hertel

Overview

Normal matter Atomic units • today, the Syst`emeinternational d’unit´es (SI) is in Length and energy common use, in particular in science and engineering Electric and • magnetic also called MKSA fields • MKSA = meter (m), kilogram (kg), second (s) and More examples ampere (A) • other SI units are derived from it, like volt (V) −34 2 −1 • ~ = 1.054572 ×10 kg m s • e = 1.602177 ×10−19 A s • m = 9.10938 ×10−31 kg −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A −34 2 −1 • ~ = 1.054572 ×10 kg m s • e = 1.602177 ×10−19 A s • m = 9.10938 ×10−31 kg −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A • the powers of SI units m kg s A ~ 2 1 -1 0 e 0 0 1 1 m 0 1 0 0 4π0 -3 -1 4 2

Natural units Powers of MKSA Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • e = 1.602177 ×10−19 A s • m = 9.10938 ×10−31 kg −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A • the powers of SI units m kg s A ~ 2 1 -1 0 e 0 0 1 1 m 0 1 0 0 4π0 -3 -1 4 2

Natural units Powers of MKSA Peter Hertel

Overview

Normal matter Atomic units −34 2 −1 • ~ = 1.054572 ×10 kg m s Length and energy

Electric and magnetic fields

More examples • m = 9.10938 ×10−31 kg −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A • the powers of SI units m kg s A ~ 2 1 -1 0 e 0 0 1 1 m 0 1 0 0 4π0 -3 -1 4 2

Natural units Powers of MKSA Peter Hertel

Overview

Normal matter Atomic units −34 2 −1 • ~ = 1.054572 ×10 kg m s Length and energy • e = 1.602177 ×10−19 A s Electric and magnetic fields

More examples −10 −1 −3 2 • 4π0 = 1.112650 ×10 kg m A • the powers of SI units m kg s A ~ 2 1 -1 0 e 0 0 1 1 m 0 1 0 0 4π0 -3 -1 4 2

Natural units Powers of MKSA Peter Hertel

Overview

Normal matter Atomic units −34 2 −1 • ~ = 1.054572 ×10 kg m s Length and energy • e = 1.602177 ×10−19 A s Electric and −31 magnetic • m = 9.10938 ×10 kg fields

More examples • the powers of SI units m kg s A ~ 2 1 -1 0 e 0 0 1 1 m 0 1 0 0 4π0 -3 -1 4 2

Natural units Powers of MKSA Peter Hertel

Overview

Normal matter Atomic units −34 2 −1 • ~ = 1.054572 ×10 kg m s Length and energy • e = 1.602177 ×10−19 A s Electric and −31 magnetic • m = 9.10938 ×10 kg fields • 4π = 1.112650 ×10−10 kg−1 m−3 A2 More 0 examples Natural units Powers of MKSA Peter Hertel

Overview

Normal matter Atomic units −34 2 −1 • ~ = 1.054572 ×10 kg m s Length and energy • e = 1.602177 ×10−19 A s Electric and −31 magnetic • m = 9.10938 ×10 kg fields • 4π = 1.112650 ×10−10 kg−1 m−3 A2 More 0 examples • the powers of SI units m kg s A ~ 2 1 -1 0 e 0 0 1 1 m 0 1 0 0 4π0 -3 -1 4 2 • the former 4 × 4 matrix can be inverted • SI units can be expressed as a product of powers of the involved constants of nature

~ e m 4π0 m 2 -2 -1 1 kg 0 0 1 0 s 3 -4 -1 2 A -3 5 1 -2 3 −4 −1 2 • read: s = number multiplied by ~ e m (4π0) etc. • find out number and exponents for arbitrary SI unit

Natural units Powers of natural constants Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • SI units can be expressed as a product of powers of the involved constants of nature

~ e m 4π0 m 2 -2 -1 1 kg 0 0 1 0 s 3 -4 -1 2 A -3 5 1 -2 3 −4 −1 2 • read: s = number multiplied by ~ e m (4π0) etc. • find out number and exponents for arbitrary SI unit

Natural units Powers of natural constants Peter Hertel

Overview

Normal matter Atomic units • the former 4 × 4 matrix can be inverted Length and energy

Electric and magnetic fields

More examples 3 −4 −1 2 • read: s = number multiplied by ~ e m (4π0) etc. • find out number and exponents for arbitrary SI unit

Natural units Powers of natural constants Peter Hertel

Overview

Normal matter Atomic units • the former 4 × 4 matrix can be inverted Length and energy • SI units can be expressed as a product of powers of the Electric and magnetic involved constants of nature fields e m 4π0 More ~ examples m 2 -2 -1 1 kg 0 0 1 0 s 3 -4 -1 2 A -3 5 1 -2 • find out number and exponents for arbitrary SI unit

Natural units Powers of natural constants Peter Hertel

Overview

Normal matter Atomic units • the former 4 × 4 matrix can be inverted Length and energy • SI units can be expressed as a product of powers of the Electric and magnetic involved constants of nature fields e m 4π0 More ~ examples m 2 -2 -1 1 kg 0 0 1 0 s 3 -4 -1 2 A -3 5 1 -2 3 −4 −1 2 • read: s = number multiplied by ~ e m (4π0) etc. Natural units Powers of natural constants Peter Hertel

Overview

Normal matter Atomic units • the former 4 × 4 matrix can be inverted Length and energy • SI units can be expressed as a product of powers of the Electric and magnetic involved constants of nature fields e m 4π0 More ~ examples m 2 -2 -1 1 kg 0 0 1 0 s 3 -4 -1 2 A -3 5 1 -2 3 −4 −1 2 • read: s = number multiplied by ~ e m (4π0) etc. • find out number and exponents for arbitrary SI unit Natural units

Peter Hertel function [value,]=atomic_unit(si) Overview val=[1.05457e-34,1.60218e-19,9.10938e-31,4*pi*8.85419E-12]; Normal matter dim=[2 1 -1 0; 0 0 1 1; 0 1 0 0; -3 -1 4 2]; Atomic units mid=round(inv(dim)); Length and energy power=si*mid; Electric and value=prod(val.^power); magnetic fields

More examples

>> length = [0 1 0 0]; >> length = [1 0 0 0]; >> [astar,apow]=atomic_unit(length); >> astar 5.2917e-11 >> apow 2 -2 -1 1 Natural units

Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • recall length=[1 0 0 0] • recall [astar,apow]=atomic unit(length) • recall astar = 5.2971e-11 • recall apow = [2 -2 -1 1] • we have just calculated the atomic • i. e. Bohr’s radius 4π 2 a = 0~ = 5.2917 × 10−11 m ∗ me2

Natural units e. g. atomic length unit Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • recall [astar,apow]=atomic unit(length) • recall astar = 5.2971e-11 • recall apow = [2 -2 -1 1] • we have just calculated the atomic unit of length • i. e. Bohr’s radius 4π 2 a = 0~ = 5.2917 × 10−11 m ∗ me2

Natural units e. g. atomic length unit Peter Hertel

Overview

Normal matter

Atomic units Length and • recall energy length=[1 0 0 0]

Electric and magnetic fields

More examples • recall astar = 5.2971e-11 • recall apow = [2 -2 -1 1] • we have just calculated the atomic unit of length • i. e. Bohr’s radius 4π 2 a = 0~ = 5.2917 × 10−11 m ∗ me2

Natural units e. g. atomic length unit Peter Hertel

Overview

Normal matter

Atomic units Length and • recall energy length=[1 0 0 0] Electric and • recall [astar,apow]=atomic unit(length) magnetic fields

More examples • recall apow = [2 -2 -1 1] • we have just calculated the atomic unit of length • i. e. Bohr’s radius 4π 2 a = 0~ = 5.2917 × 10−11 m ∗ me2

Natural units e. g. atomic length unit Peter Hertel

Overview

Normal matter

Atomic units Length and • recall energy length=[1 0 0 0] Electric and • recall [astar,apow]=atomic unit(length) magnetic fields • recall astar = 5.2971e-11 More examples • we have just calculated the atomic unit of length • i. e. Bohr’s radius 4π 2 a = 0~ = 5.2917 × 10−11 m ∗ me2

Natural units e. g. atomic length unit Peter Hertel

Overview

Normal matter

Atomic units Length and • recall energy length=[1 0 0 0] Electric and • recall [astar,apow]=atomic unit(length) magnetic fields • recall astar = 5.2971e-11 More examples • recall apow = [2 -2 -1 1] • i. e. Bohr’s radius 4π 2 a = 0~ = 5.2917 × 10−11 m ∗ me2

Natural units e. g. atomic length unit Peter Hertel

Overview

Normal matter

Atomic units Length and • recall energy length=[1 0 0 0] Electric and • recall [astar,apow]=atomic unit(length) magnetic fields • recall astar = 5.2971e-11 More examples • recall apow = [2 -2 -1 1] • we have just calculated the atomic unit of length Natural units e. g. atomic length unit Peter Hertel

Overview

Normal matter

Atomic units Length and • recall energy length=[1 0 0 0] Electric and • recall [astar,apow]=atomic unit(length) magnetic fields • recall astar = 5.2971e-11 More examples • recall apow = [2 -2 -1 1] • we have just calculated the atomic unit of length • i. e. Bohr’s radius 4π 2 a = 0~ = 5.2917 × 10−11 m ∗ me2 • Schr¨odingerequation for 2 1 −e2 − ~ ∆φ + φ = Eφ 2m 4π0 r • in atomic units 1 1 − ∆φ − φ = Eφ 2 r • −r 1 φ ∝ e with E = − 2 • radius −r −r R ∞ dr r2 e r e hRi = 0 = 1 R ∞ 2 −r −r 0 dr r e 1 e • Bohr’s result 4π 2 hRi = a = 0~ ∗ me2

Natural units Bohr’s radius Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • in atomic units 1 1 − ∆φ − φ = Eφ 2 r • ground state −r 1 φ ∝ e with E = − 2 • radius −r −r R ∞ dr r2 e r e hRi = 0 = 1 R ∞ 2 −r −r 0 dr r e 1 e • Bohr’s result 4π 2 hRi = a = 0~ ∗ me2

Natural units Bohr’s radius Peter Hertel

Overview • Schr¨odingerequation for Normal matter 2 2 ~ 1 −e Atomic units − ∆φ + φ = Eφ 2m 4π r Length and 0 energy

Electric and magnetic fields

More examples • ground state −r 1 φ ∝ e with E = − 2 • radius −r −r R ∞ dr r2 e r e hRi = 0 = 1 R ∞ 2 −r −r 0 dr r e 1 e • Bohr’s result 4π 2 hRi = a = 0~ ∗ me2

Natural units Bohr’s radius Peter Hertel

Overview • Schr¨odingerequation for hydrogen atom Normal matter 2 2 ~ 1 −e Atomic units − ∆φ + φ = Eφ 2m 4π r Length and 0 energy • in atomic units Electric and 1 1 magnetic − ∆φ − φ = Eφ fields 2 r More examples • radius −r −r R ∞ dr r2 e r e hRi = 0 = 1 R ∞ 2 −r −r 0 dr r e 1 e • Bohr’s result 4π 2 hRi = a = 0~ ∗ me2

Natural units Bohr’s radius Peter Hertel

Overview • Schr¨odingerequation for hydrogen atom Normal matter 2 2 ~ 1 −e Atomic units − ∆φ + φ = Eφ 2m 4π r Length and 0 energy • in atomic units Electric and 1 1 magnetic − ∆φ − φ = Eφ fields 2 r More • ground state examples −r 1 φ ∝ e with E = − 2 • Bohr’s result 4π 2 hRi = a = 0~ ∗ me2

Natural units Bohr’s radius Peter Hertel

Overview • Schr¨odingerequation for hydrogen atom Normal matter 2 2 ~ 1 −e Atomic units − ∆φ + φ = Eφ 2m 4π r Length and 0 energy • in atomic units Electric and 1 1 magnetic − ∆φ − φ = Eφ fields 2 r More • ground state examples −r 1 φ ∝ e with E = − 2 • radius −r −r R ∞ dr r2 e r e hRi = 0 = 1 R ∞ 2 −r −r 0 dr r e 1 e Natural units Bohr’s radius Peter Hertel

Overview • Schr¨odingerequation for hydrogen atom Normal matter 2 2 ~ 1 −e Atomic units − ∆φ + φ = Eφ 2m 4π r Length and 0 energy • in atomic units Electric and 1 1 magnetic − ∆φ − φ = Eφ fields 2 r More • ground state examples −r 1 φ ∝ e with E = − 2 • radius −r −r R ∞ dr r2 e r e hRi = 0 = 1 R ∞ 2 −r −r 0 dr r e 1 e • Bohr’s result 4π 2 hRi = a = 0~ ∗ me2 • MKSA for energy is [2 1 -2 0] • [Estar,Epow]=atomic unit(energy) −18 • E∗ = 4.3598 × 10 J = 27.21 eV • atomic energy unit • H atom ground state energy is E = −13.6 eV • photon are in the eV range • He-Ne laser: λ = 633 nm

• ~ω = 0.072 E∗ = 1.96 eV

Natural units Hartree Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • [Estar,Epow]=atomic unit(energy) −18 • E∗ = 4.3598 × 10 J = 27.21 eV • atomic energy unit Hartree • H atom ground state energy is E = −13.6 eV • photon energies are in the eV range • He-Ne laser: λ = 633 nm

• ~ω = 0.072 E∗ = 1.96 eV

Natural units Hartree Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • MKSA for energy is [2 1 -2 0]

Electric and magnetic fields

More examples −18 • E∗ = 4.3598 × 10 J = 27.21 eV • atomic energy unit Hartree • H atom ground state energy is E = −13.6 eV • photon energies are in the eV range • He-Ne laser: λ = 633 nm

• ~ω = 0.072 E∗ = 1.96 eV

Natural units Hartree Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • MKSA for energy is [2 1 -2 0] Electric and • [Estar,Epow]=atomic unit(energy) magnetic fields

More examples • atomic energy unit Hartree • H atom ground state energy is E = −13.6 eV • photon energies are in the eV range • He-Ne laser: λ = 633 nm

• ~ω = 0.072 E∗ = 1.96 eV

Natural units Hartree Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • MKSA for energy is [2 1 -2 0] Electric and • [Estar,Epow]=atomic unit(energy) magnetic fields −18 • E∗ = 4.3598 × 10 J = 27.21 eV More examples • H atom ground state energy is E = −13.6 eV • photon energies are in the eV range • He-Ne laser: λ = 633 nm

• ~ω = 0.072 E∗ = 1.96 eV

Natural units Hartree Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • MKSA for energy is [2 1 -2 0] Electric and • [Estar,Epow]=atomic unit(energy) magnetic fields −18 • E∗ = 4.3598 × 10 J = 27.21 eV More examples • atomic energy unit Hartree • photon energies are in the eV range • He-Ne laser: λ = 633 nm

• ~ω = 0.072 E∗ = 1.96 eV

Natural units Hartree Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • MKSA for energy is [2 1 -2 0] Electric and • [Estar,Epow]=atomic unit(energy) magnetic fields −18 • E∗ = 4.3598 × 10 J = 27.21 eV More examples • atomic energy unit Hartree • H atom ground state energy is E = −13.6 eV • He-Ne laser: λ = 633 nm

• ~ω = 0.072 E∗ = 1.96 eV

Natural units Hartree Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • MKSA for energy is [2 1 -2 0] Electric and • [Estar,Epow]=atomic unit(energy) magnetic fields −18 • E∗ = 4.3598 × 10 J = 27.21 eV More examples • atomic energy unit Hartree • H atom ground state energy is E = −13.6 eV • photon energies are in the eV range • ~ω = 0.072 E∗ = 1.96 eV

Natural units Hartree Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • MKSA for energy is [2 1 -2 0] Electric and • [Estar,Epow]=atomic unit(energy) magnetic fields −18 • E∗ = 4.3598 × 10 J = 27.21 eV More examples • atomic energy unit Hartree • H atom ground state energy is E = −13.6 eV • photon energies are in the eV range • He-Ne laser: λ = 633 nm Natural units Hartree Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • MKSA for energy is [2 1 -2 0] Electric and • [Estar,Epow]=atomic unit(energy) magnetic fields −18 • E∗ = 4.3598 × 10 J = 27.21 eV More examples • atomic energy unit Hartree • H atom ground state energy is E = −13.6 eV • photon energies are in the eV range • He-Ne laser: λ = 633 nm

• ~ω = 0.072 E∗ = 1.96 eV • =[2 1 -3 -1] • el str=[1 1 -3 -1] • the atomic (natural) unit of electric field strength is 2 5 m e 11 −1 E∗ = 3 4 = 5.1423 × 10 V m (4π0) ~ • law, Pockels effect, Starck effect . . . • external field strength is always very small

Natural units Electric field strength Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • el field str=[1 1 -3 -1] • the atomic (natural) unit of electric field strength is 2 5 m e 11 −1 E∗ = 3 4 = 5.1423 × 10 V m (4π0) ~ • Ohms law, Pockels effect, Starck effect . . . • external field strength is always very small

Natural units Electric field strength Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • voltage=[2 1 -3 -1] Electric and magnetic fields

More examples • the atomic (natural) unit of electric field strength is 2 5 m e 11 −1 E∗ = 3 4 = 5.1423 × 10 V m (4π0) ~ • Ohms law, Pockels effect, Starck effect . . . • external field strength is always very small

Natural units Electric field strength Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • voltage=[2 1 -3 -1] Electric and magnetic • el field str=[1 1 -3 -1] fields

More examples • Ohms law, Pockels effect, Starck effect . . . • external field strength is always very small

Natural units Electric field strength Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • voltage=[2 1 -3 -1] Electric and magnetic • el field str=[1 1 -3 -1] fields

More • the atomic (natural) unit of electric field strength is examples 2 5 m e 11 −1 E∗ = 3 4 = 5.1423 × 10 V m (4π0) ~ • external field strength is always very small

Natural units Electric field strength Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • voltage=[2 1 -3 -1] Electric and magnetic • el field str=[1 1 -3 -1] fields

More • the atomic (natural) unit of electric field strength is examples 2 5 m e 11 −1 E∗ = 3 4 = 5.1423 × 10 V m (4π0) ~ • Ohms law, Pockels effect, Starck effect . . . Natural units Electric field strength Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • voltage=[2 1 -3 -1] Electric and magnetic • el field str=[1 1 -3 -1] fields

More • the atomic (natural) unit of electric field strength is examples 2 5 m e 11 −1 E∗ = 3 4 = 5.1423 × 10 V m (4π0) ~ • Ohms law, Pockels effect, Starck effect . . . • external field strength is always very small • induction=[0 1 -2 -1] • the atomic (natural) unit of magnetic induction is 2 3 m e 5 B∗ = 2 3 = 2.3505 × 10 T (4π0) ~ • Faraday effect, Hall effect . . . • external field induction is always very small

Natural units Magnetic induction Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • the atomic (natural) unit of magnetic induction is 2 3 m e 5 B∗ = 2 3 = 2.3505 × 10 T (4π0) ~ • Faraday effect, Hall effect . . . • external field induction is always very small

Natural units Magnetic induction Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and • induction=[0 1 -2 -1] magnetic fields

More examples • Faraday effect, Hall effect . . . • external field induction is always very small

Natural units Magnetic induction Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and • induction=[0 1 -2 -1] magnetic fields • the atomic (natural) unit of magnetic induction is More 2 3 examples m e 5 B∗ = 2 3 = 2.3505 × 10 T (4π0) ~ • external field induction is always very small

Natural units Magnetic induction Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and • induction=[0 1 -2 -1] magnetic fields • the atomic (natural) unit of magnetic induction is More 2 3 examples m e 5 B∗ = 2 3 = 2.3505 × 10 T (4π0) ~ • Faraday effect, Hall effect . . . Natural units Magnetic induction Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and • induction=[0 1 -2 -1] magnetic fields • the atomic (natural) unit of magnetic induction is More 2 3 examples m e 5 B∗ = 2 3 = 2.3505 × 10 T (4π0) ~ • Faraday effect, Hall effect . . . • external field induction is always very small • rho=[-3 1 0 0] • [rhoval,rhopow]=atomic unit(rho) • rhoval = 6.174 • rhopow=[-6 6 4 -3] • 4 6 m e m −3 %∗ = 3 6 = 3 = 6.147 kg m (4π0) ~ a∗ • mass is mass of , therefore must be multiplied by approximately 4000

Natural units Mass Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • [rhoval,rhopow]=atomic unit(rho) • rhoval = 6.174 • rhopow=[-6 6 4 -3] • 4 6 m e m −3 %∗ = 3 6 = 3 = 6.147 kg m (4π0) ~ a∗ • mass is mass of nucleons, therefore must be multiplied by approximately 4000

Natural units Mass density Peter Hertel

Overview

Normal matter

Atomic units

Length and • rho=[-3 1 0 0] energy

Electric and magnetic fields

More examples • rhoval = 6.174 • rhopow=[-6 6 4 -3] • 4 6 m e m −3 %∗ = 3 6 = 3 = 6.147 kg m (4π0) ~ a∗ • mass is mass of nucleons, therefore must be multiplied by approximately 4000

Natural units Mass density Peter Hertel

Overview

Normal matter

Atomic units

Length and • rho=[-3 1 0 0] energy • [rhoval,rhopow]=atomic unit(rho) Electric and magnetic fields

More examples • rhopow=[-6 6 4 -3] • 4 6 m e m −3 %∗ = 3 6 = 3 = 6.147 kg m (4π0) ~ a∗ • mass is mass of nucleons, therefore must be multiplied by approximately 4000

Natural units Mass density Peter Hertel

Overview

Normal matter

Atomic units

Length and • rho=[-3 1 0 0] energy • [rhoval,rhopow]=atomic unit(rho) Electric and magnetic fields • rhoval = 6.174

More examples • 4 6 m e m −3 %∗ = 3 6 = 3 = 6.147 kg m (4π0) ~ a∗ • mass is mass of nucleons, therefore must be multiplied by approximately 4000

Natural units Mass density Peter Hertel

Overview

Normal matter

Atomic units

Length and • rho=[-3 1 0 0] energy • [rhoval,rhopow]=atomic unit(rho) Electric and magnetic fields • rhoval = 6.174 More • rhopow=[-6 6 4 -3] examples • mass is mass of nucleons, therefore must be multiplied by approximately 4000

Natural units Mass density Peter Hertel

Overview

Normal matter

Atomic units

Length and • rho=[-3 1 0 0] energy • [rhoval,rhopow]=atomic unit(rho) Electric and magnetic fields • rhoval = 6.174 More • rhopow=[-6 6 4 -3] examples • 4 6 m e m −3 %∗ = 3 6 = 3 = 6.147 kg m (4π0) ~ a∗ Natural units Mass density Peter Hertel

Overview

Normal matter

Atomic units

Length and • rho=[-3 1 0 0] energy • [rhoval,rhopow]=atomic unit(rho) Electric and magnetic fields • rhoval = 6.174 More • rhopow=[-6 6 4 -3] examples • 4 6 m e m −3 %∗ = 3 6 = 3 = 6.147 kg m (4π0) ~ a∗ • mass is mass of nucleons, therefore must be multiplied by approximately 4000 • =[1 0 -1 0] • atomic unit of velocity is 2 e 6 −1 v∗ = = 2.1877 × 10 m s 4π0~ −1 • velocity of light is c = α v∗ • fine structure constant α = 1/137.035999074 • can be measured directly (quantum Hall effect) • relativistic corrections

Natural units Light velocity Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • atomic unit of velocity is 2 e 6 −1 v∗ = = 2.1877 × 10 m s 4π0~ −1 • velocity of light is c = α v∗ • fine structure constant α = 1/137.035999074 • can be measured directly (quantum Hall effect) • relativistic corrections

Natural units Light velocity Peter Hertel

Overview

Normal matter

Atomic units Length and • velocity=[1 0 -1 0] energy

Electric and magnetic fields

More examples −1 • velocity of light is c = α v∗ • fine structure constant α = 1/137.035999074 • can be measured directly (quantum Hall effect) • relativistic corrections

Natural units Light velocity Peter Hertel

Overview

Normal matter

Atomic units Length and • velocity=[1 0 -1 0] energy • Electric and atomic unit of velocity is magnetic 2 fields e 6 −1 v∗ = = 2.1877 × 10 m s More 4π0~ examples • fine structure constant α = 1/137.035999074 • can be measured directly (quantum Hall effect) • relativistic corrections

Natural units Light velocity Peter Hertel

Overview

Normal matter

Atomic units Length and • velocity=[1 0 -1 0] energy • Electric and atomic unit of velocity is magnetic 2 fields e 6 −1 v∗ = = 2.1877 × 10 m s More 4π0~ examples −1 • velocity of light is c = α v∗ • can be measured directly (quantum Hall effect) • relativistic corrections

Natural units Light velocity Peter Hertel

Overview

Normal matter

Atomic units Length and • velocity=[1 0 -1 0] energy • Electric and atomic unit of velocity is magnetic 2 fields e 6 −1 v∗ = = 2.1877 × 10 m s More 4π0~ examples −1 • velocity of light is c = α v∗ • fine structure constant α = 1/137.035999074 • relativistic corrections

Natural units Light velocity Peter Hertel

Overview

Normal matter

Atomic units Length and • velocity=[1 0 -1 0] energy • Electric and atomic unit of velocity is magnetic 2 fields e 6 −1 v∗ = = 2.1877 × 10 m s More 4π0~ examples −1 • velocity of light is c = α v∗ • fine structure constant α = 1/137.035999074 • can be measured directly (quantum Hall effect) Natural units Light velocity Peter Hertel

Overview

Normal matter

Atomic units Length and • velocity=[1 0 -1 0] energy • Electric and atomic unit of velocity is magnetic 2 fields e 6 −1 v∗ = = 2.1877 × 10 m s More 4π0~ examples −1 • velocity of light is c = α v∗ • fine structure constant α = 1/137.035999074 • can be measured directly (quantum Hall effect) • relativistic corrections • external electric field E • change −1 −1 ((ω; E) )ij = ((ω; 0) )ij + rijk(ω)Ek • Pockels coefficients vanish if crystal has an inversion center • niobate has no inversion center −1 • largest coefficient r333 = 30 pm V • small or large?

• r333 E∗ ≈ 16 • lithium niobate is rather resilient to electrical fields

Natural units Pockels coefficients Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • permittivity change −1 −1 ((ω; E) )ij = ((ω; 0) )ij + rijk(ω)Ek • Pockels coefficients vanish if crystal has an inversion center • lithium niobate has no inversion center −1 • largest coefficient r333 = 30 pm V • small or large?

• r333 E∗ ≈ 16 • lithium niobate is rather resilient to electrical fields

Natural units Pockels coefficients Peter Hertel

Overview

Normal matter Atomic units • external electric field E Length and energy

Electric and magnetic fields

More examples • Pockels coefficients vanish if crystal has an inversion center • lithium niobate has no inversion center −1 • largest coefficient r333 = 30 pm V • small or large?

• r333 E∗ ≈ 16 • lithium niobate is rather resilient to electrical fields

Natural units Pockels coefficients Peter Hertel

Overview

Normal matter Atomic units • external electric field E Length and energy • permittivity change Electric and −1 −1 magnetic ((ω; E) )ij = ((ω; 0) )ij + rijk(ω)Ek fields

More examples • lithium niobate has no inversion center −1 • largest coefficient r333 = 30 pm V • small or large?

• r333 E∗ ≈ 16 • lithium niobate is rather resilient to electrical fields

Natural units Pockels coefficients Peter Hertel

Overview

Normal matter Atomic units • external electric field E Length and energy • permittivity change Electric and −1 −1 magnetic ((ω; E) )ij = ((ω; 0) )ij + rijk(ω)Ek fields • More Pockels coefficients vanish if crystal has an examples inversion center −1 • largest coefficient r333 = 30 pm V • small or large?

• r333 E∗ ≈ 16 • lithium niobate is rather resilient to electrical fields

Natural units Pockels coefficients Peter Hertel

Overview

Normal matter Atomic units • external electric field E Length and energy • permittivity change Electric and −1 −1 magnetic ((ω; E) )ij = ((ω; 0) )ij + rijk(ω)Ek fields • More Pockels coefficients vanish if crystal has an examples inversion center • lithium niobate has no inversion center • small or large?

• r333 E∗ ≈ 16 • lithium niobate is rather resilient to electrical fields

Natural units Pockels coefficients Peter Hertel

Overview

Normal matter Atomic units • external electric field E Length and energy • permittivity change Electric and −1 −1 magnetic ((ω; E) )ij = ((ω; 0) )ij + rijk(ω)Ek fields • More Pockels coefficients vanish if crystal has an examples inversion center • lithium niobate has no inversion center −1 • largest coefficient r333 = 30 pm V • r333 E∗ ≈ 16 • lithium niobate is rather resilient to electrical fields

Natural units Pockels coefficients Peter Hertel

Overview

Normal matter Atomic units • external electric field E Length and energy • permittivity change Electric and −1 −1 magnetic ((ω; E) )ij = ((ω; 0) )ij + rijk(ω)Ek fields • More Pockels coefficients vanish if crystal has an examples inversion center • lithium niobate has no inversion center −1 • largest coefficient r333 = 30 pm V • small or large? • lithium niobate is rather resilient to electrical fields

Natural units Pockels coefficients Peter Hertel

Overview

Normal matter Atomic units • external electric field E Length and energy • permittivity change Electric and −1 −1 magnetic ((ω; E) )ij = ((ω; 0) )ij + rijk(ω)Ek fields • More Pockels coefficients vanish if crystal has an examples inversion center • lithium niobate has no inversion center −1 • largest coefficient r333 = 30 pm V • small or large?

• r333 E∗ ≈ 16 Natural units Pockels coefficients Peter Hertel

Overview

Normal matter Atomic units • external electric field E Length and energy • permittivity change Electric and −1 −1 magnetic ((ω; E) )ij = ((ω; 0) )ij + rijk(ω)Ek fields • More Pockels coefficients vanish if crystal has an examples inversion center • lithium niobate has no inversion center −1 • largest coefficient r333 = 30 pm V • small or large?

• r333 E∗ ≈ 16 • lithium niobate is rather resilient to electrical fields • An elastic medium is described by and strain

• strain tensor Sij describes deformation

• stress tensor Tij describes on element

• dFj = dAiTij • Hooke’s law E  ν  T = S + δ S ij 1 + ν ij 1 − 2ν ij kk • Poisson’s ratio between 0 and 1/2 • elasticity module E −3 • expect E = eV A˚ = 160 GPa • . . . as order of magnitude • steel : E = 200 GPa

Natural units Elasticity module Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • strain tensor Sij describes deformation

• stress tensor Tij describes force on area element

• dFj = dAiTij • Hooke’s law E  ν  T = S + δ S ij 1 + ν ij 1 − 2ν ij kk • Poisson’s ratio between 0 and 1/2 • elasticity module E −3 • expect E = eV A˚ = 160 GPa • . . . as order of magnitude • steel : E = 200 GPa

Natural units Elasticity module Peter Hertel

Overview Normal matter • An elastic medium is described by stress and strain Atomic units

Length and energy

Electric and magnetic fields

More examples • stress tensor Tij describes force on area element

• dFj = dAiTij • Hooke’s law E  ν  T = S + δ S ij 1 + ν ij 1 − 2ν ij kk • Poisson’s ratio between 0 and 1/2 • elasticity module E −3 • expect E = eV A˚ = 160 GPa • . . . as order of magnitude • steel : E = 200 GPa

Natural units Elasticity module Peter Hertel

Overview Normal matter • An elastic medium is described by stress and strain Atomic units • strain tensor S describes deformation Length and ij energy

Electric and magnetic fields

More examples • dFj = dAiTij • Hooke’s law E  ν  T = S + δ S ij 1 + ν ij 1 − 2ν ij kk • Poisson’s ratio between 0 and 1/2 • elasticity module E −3 • expect E = eV A˚ = 160 GPa • . . . as order of magnitude • steel : E = 200 GPa

Natural units Elasticity module Peter Hertel

Overview Normal matter • An elastic medium is described by stress and strain Atomic units • strain tensor S describes deformation Length and ij energy • stress tensor Tij describes force on area element Electric and magnetic fields

More examples • Hooke’s law E  ν  T = S + δ S ij 1 + ν ij 1 − 2ν ij kk • Poisson’s ratio between 0 and 1/2 • elasticity module E −3 • expect E = eV A˚ = 160 GPa • . . . as order of magnitude • steel : E = 200 GPa

Natural units Elasticity module Peter Hertel

Overview Normal matter • An elastic medium is described by stress and strain Atomic units • strain tensor S describes deformation Length and ij energy • stress tensor Tij describes force on area element Electric and magnetic • dF = dA T fields j i ij

More examples • Poisson’s ratio between 0 and 1/2 • elasticity module E −3 • expect E = eV A˚ = 160 GPa • . . . as order of magnitude • steel : E = 200 GPa

Natural units Elasticity module Peter Hertel

Overview Normal matter • An elastic medium is described by stress and strain Atomic units • strain tensor S describes deformation Length and ij energy • stress tensor Tij describes force on area element Electric and magnetic • dF = dA T fields j i ij

More • Hooke’s law examples E  ν  T = S + δ S ij 1 + ν ij 1 − 2ν ij kk • elasticity module E −3 • expect E = eV A˚ = 160 GPa • . . . as order of magnitude • steel : E = 200 GPa

Natural units Elasticity module Peter Hertel

Overview Normal matter • An elastic medium is described by stress and strain Atomic units • strain tensor S describes deformation Length and ij energy • stress tensor Tij describes force on area element Electric and magnetic • dF = dA T fields j i ij

More • Hooke’s law examples E  ν  T = S + δ S ij 1 + ν ij 1 − 2ν ij kk • Poisson’s ratio between 0 and 1/2 −3 • expect E = eV A˚ = 160 GPa • . . . as order of magnitude • steel : E = 200 GPa

Natural units Elasticity module Peter Hertel

Overview Normal matter • An elastic medium is described by stress and strain Atomic units • strain tensor S describes deformation Length and ij energy • stress tensor Tij describes force on area element Electric and magnetic • dF = dA T fields j i ij

More • Hooke’s law examples E  ν  T = S + δ S ij 1 + ν ij 1 − 2ν ij kk • Poisson’s ratio between 0 and 1/2 • elasticity module E • . . . as order of magnitude • steel : E = 200 GPa

Natural units Elasticity module Peter Hertel

Overview Normal matter • An elastic medium is described by stress and strain Atomic units • strain tensor S describes deformation Length and ij energy • stress tensor Tij describes force on area element Electric and magnetic • dF = dA T fields j i ij

More • Hooke’s law examples E  ν  T = S + δ S ij 1 + ν ij 1 − 2ν ij kk • Poisson’s ratio between 0 and 1/2 • elasticity module E −3 • expect E = eV A˚ = 160 GPa • steel : E = 200 GPa

Natural units Elasticity module Peter Hertel

Overview Normal matter • An elastic medium is described by stress and strain Atomic units • strain tensor S describes deformation Length and ij energy • stress tensor Tij describes force on area element Electric and magnetic • dF = dA T fields j i ij

More • Hooke’s law examples E  ν  T = S + δ S ij 1 + ν ij 1 − 2ν ij kk • Poisson’s ratio between 0 and 1/2 • elasticity module E −3 • expect E = eV A˚ = 160 GPa • . . . as order of magnitude Natural units Elasticity module Peter Hertel

Overview Normal matter • An elastic medium is described by stress and strain Atomic units • strain tensor S describes deformation Length and ij energy • stress tensor Tij describes force on area element Electric and magnetic • dF = dA T fields j i ij

More • Hooke’s law examples E  ν  T = S + δ S ij 1 + ν ij 1 − 2ν ij kk • Poisson’s ratio between 0 and 1/2 • elasticity module E −3 • expect E = eV A˚ = 160 GPa • . . . as order of magnitude • steel : E = 200 GPa • Recall the Drude model result Ω2 χ(ω) = χ(0) Ω2 − ω2 − iΓω • where Nq2 χ(0) = 2 0mΩ • recall m(x¨ + Γx˙ + Ω2x) = qE • N ≈ 1 and Ω ≈ 1 in natural units!

Natural units Susceptibility Peter Hertel

Overview

Normal matter

Atomic units

Length and energy

Electric and magnetic fields

More examples • where Nq2 χ(0) = 2 0mΩ • recall m(x¨ + Γx˙ + Ω2x) = qE • N ≈ 1 and Ω ≈ 1 in natural units!

Natural units Susceptibility Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • Recall the Drude model result 2 Electric and Ω magnetic χ(ω) = χ(0) 2 2 fields Ω − ω − iΓω

More examples • recall m(x¨ + Γx˙ + Ω2x) = qE • N ≈ 1 and Ω ≈ 1 in natural units!

Natural units Susceptibility Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • Recall the Drude model result 2 Electric and Ω magnetic χ(ω) = χ(0) 2 2 fields Ω − ω − iΓω

More • where examples Nq2 χ(0) = 2 0mΩ • N ≈ 1 and Ω ≈ 1 in natural units!

Natural units Susceptibility Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • Recall the Drude model result 2 Electric and Ω magnetic χ(ω) = χ(0) 2 2 fields Ω − ω − iΓω

More • where examples Nq2 χ(0) = 2 0mΩ • recall m(x¨ + Γx˙ + Ω2x) = qE Natural units Susceptibility Peter Hertel

Overview

Normal matter

Atomic units

Length and energy • Recall the Drude model result 2 Electric and Ω magnetic χ(ω) = χ(0) 2 2 fields Ω − ω − iΓω

More • where examples Nq2 χ(0) = 2 0mΩ • recall m(x¨ + Γx˙ + Ω2x) = qE • N ≈ 1 and Ω ≈ 1 in natural units!