Macroscopic dielectric theory

Transcription

1 Macroscopic dielectric theory

2 Maxwellʼs equations E = 1 c E =4πρ B t B = 4π c J + 1 c B = E t

3 In a medium it is convenient to explicitly introduce induced charges and currents E = 1 B c t D =4πρ H = 4π J + Jext c B = + 1 c D t D = E +4π P H = B +4π M With no external charges and currents E = 1 c D = B t H = 4π c J + 1 c B = D t

4 In a medium it is convenient to explicitly introduce induced charges and currents Electric polarization E = 1 B c t D =4πρ H = 4π J + Jext c B = + 1 c D t D = E +4π P H = B +4π M With no external charges and currents E = 1 c D = B t H = 4π c J + 1 c B = D t

5 In a medium it is convenient to explicitly introduce induced charges and currents Electric polarization E = 1 B c t D =4πρ H = 4π J + Jext c B = + 1 c D t D = E +4π P H = B +4π M With no external charges and currents Magnetic polarization E = 1 c D = B t H = 4π c J + 1 c B = D t

6 In a medium it is convenient to explicitly introduce induced charges and currents Electric polarization E = 1 B c t D =4πρ H = 4π J + Jext c B = + 1 c D t D = E +4π P H = B +4π M With no external charges and currents Magnetic polarization Induced current density E = 1 c D = B t H = 4π c J + 1 c B = D t

7 P, M and J represent the response of the medium to the electromagnetic field If we assume the response is linear, we can write: P = αe J = σe M = χb where α,σ e χ are the polarizability, conductivity and susceptibility tensors, respectively. For optical frequencies we can neglect susceptibility (χ=) so that B = H

8 By introducing into the fourth of Maxwellʼs equations we get: B = 4π c σ E + α E t + 1 c E t Conductivity (σ) and polarizability (α) both contain the material response giving rise to two induced currents, a polarization current: and a ohmic current J pol = α E t J ohm = σ E

9 The absorbed power per unit volume is given by: W t abs = J E = σe 2 If we consider solutions of Maxwellʼs equations in the form sin(ωt) we see immediately that Johm is in phase with E while Jpol is out phase by π/2. Polarization current does not give rise to absorption and is responsible of dispersion (refraction index). Ohmic current, in turn produce an absorpion: W t abs = J ohm E = σe 2

10 We introduce the dielectric tensor ε D = E =1+4πα As with simple a.c. circuits, absorption and dispersion can be treated with a single complex quantity. If we consider a field in the form E = E e iωt we get B = 4π c J + 1 c D t = 1 c E t 4π E = c E t in which we have introduced the complex dielectric function: = 1 + i 2 = + i 4πσ ω

11 B = 4π c σ E + α E t + 1 c E t E t E = E e iωt = iω E e iωt = iω E B = 4π c σ E + 1 c (4πα +1) E t = = 4π c σ E + E c t 4π = c σ +( iω) c = c + i4πσ = c c i ω E t E t =

12 By introducing the complex dielectric constant into Maxwellʼs equations we get 2 E c 2 2 E t 2 = As well known a class of solutions of the wave equations are plane waves: E = E e i(q r ωt) + c.c.

13 whose dispersion law ω 2 = c2 q2 suggest a complex refraction ñ = n + iκ so that ñ 2 = and therefore 1 = n 2 κ 2 2 =2nκ

14 A plane wave propagating along the x axis can be written in terms of the complex dielectric index: E = E e i(q r ω t) + c.c. = E e iω (t ñx c ) + c.c. = E e ωκx c e iω (t nx c ) + c.c. E x c ωκ

15 Phase and group velocity v p = ω k = λ T v g = ω k

16 Phase and group velocity v p = ω k = λ T v g = ω k

17 The absorption coefficient η is defined by the equation: W = W e ηx which describes the attenuation of the average energy flux, which is proportional to the square of the field, so: η = 2ωκ c = ω 2 nc = 4πσ nc

18 The complex reflection coefficient is given by the generalization of Frenel law: So the reflectivity is: r = ñ 1 ñ +1 = (n 1) + iκ (n +1)+iκ R = r r = (n 1)2 + κ 2 (n + 1) 2 + κ 2 If κ>>n, R 1 and thw wave is rapidly attenuated in the medium. This happens, for example in metals at low frequency (high σ)

19 The real and imaginary part of the dielectric function and in general of any function describing the linear response to a physical stimulus are not independent of each other. This is a consequence of the causality principle, which states that the response cannot occur in time preceding the stimulus. Let us focus on polarization (response) and the electric field (stimulus). The following dependence will hold: P (t) = t G (t t ) E (t ) dt i.e. the polarization at a given time t weighted by the Green function G. depends on the history of the electric field

20 For a sinusoidal field (and a substitution of a variable) we ca write: P (t) = t G (t t ) E e iωt (t ) dt = E e iωt G (τ) e iωτ dτ so we the complex response function (polarizability in this case) can always be expressed as: α (ω) = G (τ) e iωτ dτ

21 If α(ω) decreases to zero away from the real axis, the function α(ω)/(ω ω ) has a single pole in ω. And its integral along the path in the figure is zero, i.e. by applying Cauchyʼs theorem: P + α (ω) ω ω dω = iπα (ω ) Im ω where P indicates the principal part of the integral. For the dielectric function we obtain the dispersion relation: (ω) =1+ 1 iπ P + (ω ) 1 ω ω dω ω Re ω

22 From: α (ω) = G (τ) e iωτ dτ we immediately obtain that ( ω) = (ω) we can therefore rewrite the integral and write the Kramers-Kronig dispersion relations between the real and imaginary part of the dielectric function: 1 (ω) =1+ 2 π P 2 (ω) = 2ω π P ω 2 (ω ) ω 2 ω 2 dω 1 (ω ) 1 ω 2 ω 2 dω

23 Dispersion relation for reflectivity r (ω) = (n 1) + iκ (n + 1) + iκ = r (ω) eiθ R(ω) = r(ω) 2 = (n 1) + iκ (n + 1) + iκ 2 ln (r (ω)) = ln ( r (ω) )+iθ (ω) = 1 2 ln (R (ω)) + iθ (ω) θ (ω) = 2ω π P ln ( r (ω ) ) ω 2 ω 2 dω = ω π P ln (R (ω )) ω 2 ω 2 dω

24 Dispersion relation for reflectivity r (ω) = (n 1) + iκ (n + 1) + iκ = r (ω) eiθ We measure: R(ω) = r(ω) 2 = (n 1) + iκ (n + 1) + iκ 2 ln (r (ω)) = ln ( r (ω) )+iθ (ω) = 1 2 ln (R (ω)) + iθ (ω) θ (ω) = 2ω π P ln ( r (ω ) ) ω 2 ω 2 dω = ω π P ln (R (ω )) ω 2 ω 2 dω

25 Dispersion relation for reflectivity r (ω) = (n 1) + iκ (n + 1) + iκ = r (ω) eiθ We measure: R(ω) = r(ω) 2 = (n 1) + iκ (n + 1) + iκ 2 If we consider the natural logarithm of r(ω): ln (r (ω)) = ln ( r (ω) )+iθ (ω) = 1 2 ln (R (ω)) + iθ (ω) θ (ω) = 2ω π P ln ( r (ω ) ) ω 2 ω 2 dω = ω π P ln (R (ω )) ω 2 ω 2 dω

26 Static sum rule. For ω= we have: 1 () = π P 2 (ω ) dω ω so, if the static dielectric constant is 1, the imaginary part wil be in some part of the spectrum, i.e the medium must absorb radiation. Because of the 1/ω factor in the integrand, the static dielectric function will be greater if the absorption occurs at low frequency.

27 Absorption localized at ω= ω 2 (ω) =Aδ (ω ω ) 1 () = π A ω therefore: 1 (ω) =1+ ω2 [ 1 () 1] ω 2 ω 2 2 (ω) = π 2 ω [ 1 () 1] δ (ω ω )

28 Drude-Lorentz model. We describe the medium as an ensemble of harmonic oscillators whose resonance frequency and damping are ω and γ, respectively:

29 Drude-Lorentz model. We describe the medium as an ensemble of harmonic oscillators whose resonance frequency and damping are ω and γ, respectively: m d2 y dy + mγ dt2 dt + ω2 my = ee e iωt

30 Drude-Lorentz model. We describe the medium as an ensemble of harmonic oscillators whose resonance frequency and damping are ω and γ, respectively: whose solution is m d2 y dy + mγ dt2 dt + ω2 my = ee e iωt

31 Drude-Lorentz model. We describe the medium as an ensemble of harmonic oscillators whose resonance frequency and damping are ω and γ, respectively: m d2 y dy + mγ dt2 dt + ω2 my = ee e iωt whose solution is y = ee e iωt m [(ω 2 ω 2 ) iγω]

32 Drude-Lorentz model. We describe the medium as an ensemble of harmonic oscillators whose resonance frequency and damping are ω and γ, respectively: m d2 y dy + mγ dt2 dt + ω2 my = ee e iωt whose solution is y = ee e iωt m [(ω 2 ω 2 ) iγω] We obtain the dipole moment per unit volume by multiplying y by the charge e and by the oscillator density N. The dielectric function is therefore:

33 Drude-Lorentz model. We describe the medium as an ensemble of harmonic oscillators whose resonance frequency and damping are ω and γ, respectively: m d2 y dy + mγ dt2 dt + ω2 my = ee e iωt whose solution is y = ee e iωt m [(ω 2 ω 2 ) iγω] We obtain the dipole moment per unit volume by multiplying y by the charge e and by the oscillator density N. The dielectric function is therefore: (ω) =1+4π α =1+4π P E =1+4πe2 N m 1 (ω 2 ω 2 ) iγω

34 So: 1 =1+ 4πe2 N m 2 = 4πe2 N m ω 2 ω 2 (ω 2 ω 2 ) 2 +(γω) 2 γω (ω 2 ω 2 ) 2 +(γω) ! 1! 2 ω= 4eV γ= 1eV Photon Energy (ev) 12 14

35 5 4 eps1 eps Photon energy (ev) (ω) = 2ω π P 1 (ω ) 1 ω 2 ω 2 dω Integration by parts 2 (ω) = 1 π P d1 (ω ) dω ln ω + ω dω ω ω

36 5 4 eps1 eps2 3 2 General behaviour! Photon energy (ev) (ω) = 2ω π P 1 (ω ) 1 ω 2 ω 2 dω Integration by parts 2 (ω) = 1 π P d1 (ω ) dω ln ω + ω dω ω ω

37 By considering the complex refraction index, one can identify regions in which transmissivity, absorption or reflectivity are the main effects: 2. n! 1.5 T A R n,κ Photon Energy (ev)

38 By considering the complex refraction index, one can identify regions in which transmissivity, absorption or reflectivity are the main effects: 2. n! T A R R n,κ 1..2 R Photon Energy (ev)

39 Optical properties of metals can be obtained from the Lorentz-Drude model by setting ω = and by defining the plasma frequency: ω 2 p = 4πNe2 m The dielectric functions becomes: = 1 ω2 p ω 2 + γ 2 γω 2 p + i ω 3 + γ 2 ω

40 Free electron metal with ω p =8eV and γ=.5 ev

41 The conductivity, σ is given by: σ = ω 2 4π which in the statc limit becomes: ω 2 σ = lim ω 4π = Ne2 mγ which can be compared with the expression derived in the transport theory to find that γ=1/τ

42 At very low frequencies (ω<<1/τ) the reflectivity is: R ω = 1 2 2πσ =1 2 2ω ωpτ 2

43 Free electron metal with ω p =8eV and γ=.5 ev

44 At frequencies much higher than the plasma frequency we get: 1 = 1 ω 2 p ω 2 2 = ω 2 pγ ω 3 which represent the behaviour of any material at high energy. If we now consider the KK relation: 1 (ω) =1+ 2 π P 1 (ω) ω = 1 2 ω 2 π ω 2 (ω ) ω 2 ω 2 dω At high energy we can neglect ω 2 in the integrand denominator because ϵ 2 decreases very rapidly (as 1/ω 3, superconvergence theorem) and therefore: ω 2 (ω )dω

45 At frequencies much higher than the plasma frequency we get: 1 = 1 ω 2 p ω 2 2 = ω 2 pγ ω 3 which represent the behaviour of any material at high energy. If we now consider the KK relation: 1 (ω) =1+ 2 π P 1 (ω) ω = 1 2 ω 2 π ω 2 (ω ) ω 2 ω 2 dω At high energy we can neglect ω 2 in the integrand denominator because ϵ 2 decreases very rapidly (as 1/ω 3, superconvergence theorem) and therefore: ω 2 (ω )dω

46 so we get the sum rule: ω 2 (ω )dω = π 2 ω2 p = 2π2 e 2 N m which relates he dielectric function to the total number of electrons contributing to it. We can define the effective number of electrons contributing to the dielectric function up to a frequency ω max : ω max ω 2 (ω )dω = π 2 4πe 2 m N eff

47 The other KK relation 2 (ω) = 2ω π P 1 (ω ) 1 ω 2 ω 2 dω at high frequency becomes 2 (ω) ω = 2 πω ( 1 (ω ) 1) dω and by comparison with the expression of the dielectric function at high energy we get: ( 1 (ω) 1) dω =

48 Free electron metal with ω p =8eV and γ=.5 ev 3 2.8! 1,! Reflectivity Photon energy (ev)

49 Normal incidence Ag experimental Reflectivity 1..8 Reflectivity Photon energy (ev) 2

50 Experimental dielectric function of Ag: 6 4 2! 1,! ! 1! Photon energy (ev)

51 Experimental Reflectivity of Al Reflectivity (%)

52 Reflection and refraction at an arbitrary angle E = E e i (ωt k r) E = E e i (ωt k r) E = E e i (ωt k r) All waves have the same frequency, ω, and The refracted wave has phase velocity k = k = ω c v φ = ω k = c n k = k ω = (1 δ + iβ) c

53 Kinematic boundary conditions: at the boundary (z=): nothing occurs along x so along z k r = k r = k r k x = k x = k x k z = k z = k z k sin φ = k sin φ = k sin φ therefore: φ = φ ; sin φ sin φ = n

54 If we write the complex index of refraction as ñ = (1 δ + iβ) and assume β then n 1-δ and we can have total external reflection for angles above the critical angle φ c = arcsin (1 δ) or, in terms of the glancing incidence θ c = 2δ

55 E = E e i (ωt k r) Dynamic boundary conditions for an s polarized wave: E = E e i (ωt k r) E = E e i (ωt k r)

56 E = E e i (ωt k r) Dynamic boundary conditions for an s polarized wave: E = E e i (ωt k r) E = E e i (ωt k r) tangential electric and magnetic fields are continuos: E = E = E H cos φ H cos φ = H cos φ

57 E = E e i (ωt k r) Dynamic boundary conditions for an s polarized wave: E = E e i (ωt k r) E = E e i (ωt k r) tangential electric and magnetic fields are continuos: E = E = E H cos φ H cos φ = H cos φ since H (r, t) =ñ k k E (r, t) we obtain (E E ) cos φ =ñe cos φ

58 For an s polarized wave: E E = 2 cos φ cos φ + n 2 sin 2 φ E = cos φ n2 sin 2 φ E cos φ + n 2 sin 2 φ so the reflectivity is R s = cos φ n 2 sin 2 φ cos φ + n 2 sin 2 φ 2 2

59 For a p polarized wave: E E = 2n cos φ n 2 cos φ + n 2 sin 2 φ E = n2 cos φ n2 sin 2 φ E n 2 cos φ + n 2 sin 2 φ so the reflectivity is R p = n 2 cos φ n 2 sin 2 φ n 2 cos φ + n 2 sin 2 φ 2 2

60 Reflectivity as a function of ϕ and n R s R p Broken: n=1.5 Continuos:n=1.1 Reflectivity Brewster angle Angle of incidence(degrees)

61 Brewsterʼs angle When the reflected and refracted waves form an angle of 9 i.e. when: n2 θ b = arctan n 1 the reflected wave is 1% s polarized (if no absorption occurs!)

62 Ag s- and p- polarized reflectivity at ω=2ev Ag R s R p Reflectivity Angle of Incidence (Degrees)

63 Experimental reflectivity of Ag for p-polarized light..2.4 Reflectivity scale Photon Energy (ev) Incidence angle (degrees) 8

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