Physics of Condensed Matter I - PDF Free Download

Physics of Condensed Matter I 1100-4INZ`PC Faculty of Physics UW Jacek.Szczytko@fuw.edu.pl

Dictionary D = εe ε 0 vacuum permittivity, permittivity of free space (przenikalność elektryczna próżni) ε r relative permittivity (względna przenikalność elektryczna) ε = ε 0 ε r permittivity (przenikalność elektryczna) B = μh μ 0 vacuum permeability, permeability of free space (przenikalność magnetyczna) μ 0 = 4π 10 7 H/m μ r relative permeability (względna przenikalność magnetyczna) μ = μ 0 μ r permeability (przenikalność magnetyczna) magnetic susceptibility χ m = μ r 1 electric field E and the magnetic field B displacement field D and the magnetizing field H 2015-11-06 2

Summary Fermi golden rule The probability of transition per unit time: W t = W 0 t τ P mn = w mn τ = 2π ħ m W n 2 δ E m E n Transitions are possible only for states, for which E m = E n W t = w ± e ±iωt 0 t τ P nm = w nm τ = 2π ħ n w± m 2 δ E n E m ± ħω Transitions are possible only for states, for which The perturbation in a form of an electromagnetic wave: E m = E n ± ħω A nm = ω nm 3 e 2 3πε 0 ħc 3 m Ԧr n 2 = 4α 3 ω nm 3 c 2 m Ԧr n 2 P nm = A nm δ E n E m ± ħω 2015-11-06 3

Summary Fermi golden rule The transition rate the probability of transition per unit time from the initial state ȁi to final ȁf is given by: Szybkość zmian czyli prawdopodobieństwo przejścia na jednostkę czasu ze stanu początkowego ȁi do końcowego ȁf dane jest wzorem: P mn = 2π ħ f W i 2 ρ E f W- interaction with the field ρ E f - the density of final states Perturbation W does not have to be in the form of an electromagnetic wave. 2015-11-06 4

Summary Fermi golden rule Hydrogen A nm = 4α 3 ω nm 3 c 2 m Ԧr n 2 P nm = A nm δ E n E m ± ħω 2015-11-06 5

The history In the XIX century: the matter is granular, the energy (mostly e-m) is a wave Problems NOT solved Black body radiation Photoelectric effect Origin of spectral lines of atoms 2015-11-06 6

The history In the XIX century: the matter is granular, the energy (mostly e-m) is a wave Problems NOT solved Black body radiation spectrum Photoelectric effect Origin of spectral lines of atoms 2015-11-06 7

Ultraviolet catastrophe Rayleigh Jeans law The spectral distribution of blackbody radiation: Classically The theorem of equipartition of energy: average energy of the standing wave is independent of frequency E = kt Radiation energy density (r) is the number of waves of a particular frequency range ν dν times the average energy E, divided by the volume of the cavity: ρ ν, T dν = 8πν2 ktdν The total radiation energy density at a given temperature is given by the sum of all frequencies: c 3 ρ T = න ρ ν, T dν = 8π 0 c 3 kt න ν 2 dν = 0 2015-11-06 8

Ultraviolet catastrophe Rayleigh Jeans law 2015-11-06 9

Ultraviolet catastrophe Rayleigh Jeans law https://en.wikipedia.org/wiki/ultraviolet_catastrophe#/media/file:black_body.svg 2015-11-06 10

The history In the XX century: the matter is (also) a wave and the energy is (also) granular (corpuscular) Solved problems: Black body radiation spectrum (Planck 1900, Nobel 1918) Photoelectric effect (Einstein 1905, Nobel 1922) Origin of spectral lines of atoms (Bohr 1913, Nobel 1922) Photons energy: E = h ν (h = 6.626 10-32 J s = 4.136 10-15 ev s) momentum: p = E / c = h / λ p = h / l Count Dooku's Geonosian solar sailer light mill - Crookes radiometer 2015-11-06 11

S. Harris Derivation of Planck's law. Lasers. Everything should be made as simple as possible, but not simpler Albert Einstein 2015-11-06 12

The spontaneous and stimulated emission Let s consider the transition between two states E 2 2 What are the parameters describing the number of transitions from the state 1 to 2 and vice versa? hν ħω = hν = E 2 E 1 E 1 1 1. Absorption dn dt the number of transitions abs = N 1 B 12 ρ The number of availabale states Radiation energy density The proportionality factor 2015-11-06 15

The spontaneous and stimulated emission Let s consider the transition between two states E 2 2 What are the parameters describing the number of transitions from the state 1 to 2 and vice versa? hν ħω = hν = E 2 E 1 1. Absorption E 1 1 dn dt abs = N 1 B 12 ρ 2. Spontaneous emission 2015-11-06 16

The spontaneous and stimulated emission Let s consider the transition between two states E 2 hν 2 What are the parameters describing the number of transitions from the state 1 to 2 and vice versa? ħω = hν = E 2 E 1 1. Absorption E 1 1 dn dt 2. Spontaneous emission abs = N 1 B 12 ρ dn dt spon = AN 2 The proportionality factor the number of transitions The number of availabale states 2015-11-06 17

The spontaneous and stimulated emission Let s consider the transition between two states E 2 2 What are the parameters describing the number of transitions from the state 1 to 2 and vice versa? hν hν ħω = hν = E 2 E 1 hν 1. Absorption E 1 1 dn dt abs = N 1 B 12 ρ 2. Spontaneous emission dn dt spon = AN 2 3. Stimulated emission 2015-11-06 18

The spontaneous and stimulated emission Let s consider the transition between two states E 2 hν hν hν 2 What are the parameters describing the number of transitions from the state 1 to 2 and vice versa? ħω = hν = E 2 E 1 1. Absorption E 1 1 dn dt 2. Spontaneous emission abs = N 1 B 12 ρ 3. Stimulated emission dn dt spon = AN 2 the number dn of transitions = N dt 2 B 21 ρ wym The number of availabale states The proportionality factor Radiation energy density 2015-11-06 19

The spontaneous and stimulated emission Let s consider the transition between two states E 2 2 What are the parameters describing the number of transitions from the state 1 to 2 and vice versa? N 1 B 12 ρ AN 2 N 2 B 21 ρ E 1 1 ħω = hν = E 2 E 1 1. Absorption dn dt 2. Spontaneous emission abs = N 1 B 12 ρ 3. Stimulated emission dn dt spon = AN 2 dn dt wym = N 2 B 21 ρ 2015-11-06 20

The spontaneous and stimulated emission Let s consider the transition between two states E 2 2 dn dt abs = N 1 B 12 ρ AN 2 N 2 B 21 ρ dn dt spon = AN 2 E 1 1 dn dt wym = N 2 B 21 ρ In thermal equilibrium conditions (a necessary condition, but it is also true in states far from equilibrium, eg. in lasers!) dn = dn + dn dt abs dt spon dt wym N 1 B 12 ρ = AN 2 + N 2 B 21 ρ 2015-11-06 21

The spontaneous and stimulated emission Let s consider the transition between two states E 2 2 N 1 B 12 ρ = AN 2 + N 2 B 21 ρ E 1 AN 2 N 2 B 21 ρ 1 ρ = A B 21 1 N 1 B 12 N 2 B 21 1 The occupations of N 1 and N 2 in thermal equilibrium conditions are given by Boltzman distribution N 1 = const e E 1 kt N 2 = const e E 2 N kt 1 = e (E 1 E 2 ) kt = e hν kt N 2 What happens with ρ when T? 2015-11-06 23

The spontaneous and stimulated emission Let s consider the transition between two states E 2 E 1 AN 2 2 N 2 B 21 ρ 1 ρ ν, T = A 1 B 21 N 1 B 12 1 N 2 B 21 = A B 1 exp hν kt 1 In turn, for hν kt we have Reileigh-Jeans law ρ ν, T dν = 8πν2 c 3 ktdν Expanding the exponential function 2015-11-06 25

The spontaneous and stimulated emission Let s consider the transition between two states E 2 AN 2 2 N 2 B 21 ρ ρ ν, T = 1 8πν 2 exp hν kt 1 c 3 hν Planck equation E 1 1 A and B are Einstein coefficients. Units of A: dn dt spon = AN 2 A = 1 τ and: B = τd ν hν 1 2015-11-06 27

Electromagnetic wave The perturbation in a form of an electromagnetic wave. H e m Ԧ A Ԧp ԦA = A 0 e i(ωt k Ԧr) + ei(ωt k Ԧr) P nm = w nm τ = 2π ħ n w± m 2 δ E n E m ± ħω expanding a series Ԧp e i(k Ԧr) Ԧp 1 + ik Ԧr + ik Ԧr 2 2! + after laborious calculations we get the probability of emission of electromagnetic radiation dipole (described by the operator e Ԧr) A nm = w nm τ = ω nm 3 e 2 3πε 0 ħc 3 n Ԧr m 2 = 4α 3 ω nm 3 c 2 n Ԧr m 2 α = e2 4πε 0 ħc 1 137 It is one of the Einstein coefficients (lasers, etc. - next week!) for nondegenerated states. 2015-11-06 28

Fala elektromagnetyczna The perturbation in a form of an electromagnetic wave. A nm = ω nm 3 e 2 3πε 0 ħc 3 m Ԧr n 2 = 4α 3 ω nm 3 c 2 m Ԧr n 2 In the case o degenerated states we introduce oscillator strength A nm = 4α 3 ω nm 3 c 2 S mn g m S nm = the degeneracy of the initial state i j n i Ԧr m j 2 In the case of the hydrogen atom states it is convenient to represent operator Ԧr in the circular form: 2 2 1 n i Ԧr m j = ni z m j + 2 n 2 1 i x + iy m j + 2 n 2 i x iy m j it is easy to then integrate spherical harmonics, because: z = r cos θ x ± iy = re ±iφ sin θ 2015-11-06 29

Czesław Radzewicz Fala elektromagnetyczna Laser needs minimum 3 states Rzadko stosowany laser 3-poziomowy 2015-11-06 30

The revision of optics Maxwell equations E = B t H = Ԧj sw + D t D = ρ sw B = 0 D = εε 0 E B = μμ 0 H Ԧj sw = σe 2015-11-06 31

The revision of optics Electromagnetic wave in vacuum Maxwell equations: E = rot E = B t B = rot B = ε 0 μ 0 E t Wave equation: ΔE = ε 0 μ 0 2 E t 2 ΔB = ε 0 μ 0 2 B t 2 The speed of the electromagnetic wave: c = 1, c 3 10 8 m ε 0 μ 0 s Refractive index: n = 1 k = ω c Electromagnetic wave in dielectric Maxwell equations: E = rot E = B t B = rot B = ε 0 μ 0 εμ E t Wave equation: ΔE = ε 0 μ 0 εμ 2 E t 2 ΔB = ε 0 μ 0 εμ 2 B t 2 The speed of the electromagnetic wave: 1 c = ε 0 μ 0 εμ = c n Refractive index: n = c v = εμ k = nω c 2015-11-06 32

The revision of optics Electromagnetic wave in vacuum Electromagnetic wave in dielectric Maxwell equations: E = rot E = B t E B = rot B = ε 0 μ 0 t Wave equation: 2 E ΔE = ε 0 μ 0 t 2 2 B ΔB = ε 0 μ 0 t 2 The speed of the electromagnetic wave: c = 1, c 3 10 8 m ε 0 μ 0 s Refractive index: n = 1 Maxwell equations: E = rot E = B t B = rot B = ε 0 μ 0 εμ E t Wave equation: ΔE = ε 0 μ 0 εμ 2 E t 2 ΔB = ε 0 μ 0 εμ 2 B t 2 But how the medium interacts with an electromagnetic wave? Is e (so n) constant? k = ω c The speed of the electromagnetic wave: 1 c = ε 0 μ 0 εμ = c n Refractive index: n = c v = εμ k = nω c 2015-11-06 33

Classical theory for the index of refraction 2015-11-06 34

Classical theory for the index of refraction The Lorentz Oscillator model A polarizable medium - dielectric: E 2015-11-06 35

Classical theory for the index of refraction The Lorentz Oscillator model A polarizable medium - dielectric: + + + + + + + + + + + + + + + + + + + + + + + + E P polarization D = ε 0 E + P Ԧx -q +q Ԧp = q Ԧx The dipole moment of an atom (molecule) 2015-11-06 36

Classical theory for the index of refraction The Lorentz Oscillator model We consider A space filled with oscillators of the resonant frequency 0 and damping factor ; Oscillators have a mass m, the charge q They are moved by the oscillating electric field E. Ԧx -q +q Ԧp = q Ԧx the dipole moment of an atom (molecule) polarization P = N Ԧp = N ε 0 αe = ε 0 χe polarisability polaryzowalność dielectric susceptibility 2015-11-06 37

Classical theory for the index of refraction The Lorentz Oscillator model We consider A space filled with oscillators of the resonant frequency 0 and damping factor ; Oscillators have a mass m, the charge q They are moved by the oscillating electric field E. Ԧx -q +q Ԧp = q Ԧx Thus D = ε 0 E + P = ε 0 1 + χ E = ε 0 εe We are determining n 2 = ε = 1 + χ polarization P t = N Ԧp t = Nq Ԧx t = ε 0 χe t We have to determine Ԧx(t) 11/6/2015 38

Classical theory for the index of refraction The Lorentz Oscillator model We consider A space filled with oscillators of the resonant frequency 0 and damping factor ; Oscillators have a mass m, the charge q They are moved by the oscillating electric field E. We are determining n 2 = ε = 1 + χ Ԧx -q +q Ԧp = q Ԧx d 2 Ԧx d Ԧx + γ dt2 dt + ω 0 2 Ԧx = q m Eeiωt damping the elastic force driving force the steady state solution: Ԧx t = Ԧx 0 e iωt 11/6/2015 39

Classical theory for the index of refraction The Lorentz Oscillator model We consider A space filled with oscillators of the resonant frequency 0 and damping factor ; Oscillators have a mass m, the charge q They are moved by the oscillating electric field E. We are determining n 2 = ε = 1 + χ Ԧx -q +q Ԧp = q Ԧx d 2 Ԧx d Ԧx + γ dt2 dt + ω 0 2 Ԧx = q m Eeiωt damping the elastic force driving force the steady state solution: Ԧx t = Ԧx 0 e iωt 11/6/2015 40

Classical theory for the index of refraction The wave in the media (different): d 2 Ԧx d Ԧx dt2 + γ dt + ω 0 2 Ԧx = q m Eeiωt Lorentz model d 2 Ԧx d Ԧx dt2 + γ dt + ω 0 2 Ԧx = 0 Emission spectrum d 2 Ԧx dt 2 + 0 + 0 = q m Eeiωt Plasma waves the steady state solution: Ԧx t = Ԧx 0 e iωt 11/6/2015 41

Classical theory for the index of refraction The Lorentz Oscillator model We obtain: Kramers Kronig dispersion relations a) anomalous dispersion κ = Nq2 2ε 0 m γω ω 0 2 ω 2 2 + γ 2 ω 2 b) n = 1 + Nq2 2ε 0 m ω 0 2 ω 2 ω 0 2 ω 2 2 + γ 2 ω 2 2015-11-06 42

Classical theory for the index of refraction The Lorentz Oscillator model (dispersion medium) a) b) n is the refractive index and indicates the phase velocity real refractive index. Without absorption n = n. the imaginary part κ is called the "extinction coefficient and indicates absorption of the wave propagating through the material The quantity dn dω is called the dispersion of the medium dn dω Far from the resonance this is positive function normal dispersion At the frequencies close to the resonance it is negative anomalous dispersion 11/6/2015 43

Classical theory for the index of refraction The Lorentz Oscillator model Several resonances in the medium H 2 O a). b) 2015-11-06 44

Classical theory for the index of refraction The Lorentz Oscillator model Several resonances in the medium H 2 O a). b) For one oscillator frequency ω 0 ε L = 1, but for many frequencies it is approximately constant sum of the contributions from all the other. 2015-11-06 45

Classical theory for the index of refraction The Lorentz Oscillator model Water example:. 2015-11-06 46

Classical theory for the index of refraction Lambert-Beer law : The electromagnetic field of the wave passing through the medium: E z, t = E 0 exp i ω 0 t k 0 n z + ik 0 κz = E 0 exp 2π λ κz exp i ω 0t k 0 n z ω 0, k 0 of the wave in vacuum The intensity I z E z 2 = E 0 2 exp 4π λ κz I z = I 0 e αz 2 k 0 Absorption coefficuent α = 2κk 0 2015-11-06 47

S. Harris Derivation of Planck's law. Lasers. 2015-11-06 48