Overview - Previous lecture 1/2

Transcription

1 Overview - Previous lecture 1/2 Derived the wave equation with solutions of the form We found that the polarization of the material affects wave propagation, and found the dispersion relation ω(k) with the corresponding refractive index given by (approximately because: ignores magnetization, assumes low frequency) slide 1

2 Overview - Previous lecture 2/2 Found that polarization of materials (atoms) has finite response time Harmonic fields will give harmonic response, but with finite phase lag electric susceptibility must contain imaginary contribution: This in turn implied that the refractive index contains an imaginary component which was found to lead to a complex wavevector that describes attenuation α In short: finite polarization response time (related to n) refractive index n(ω) must be linked to attenuation α(ω) slide 2 attenuation

3 Overview - Today s lecture Derive Kramers-Kronig relations linking - real refractive index n to an absorption spectrum α - real susceptibility X to a spectrum of the imaginary susceptibility X - a phase shift upon reflection ϕ to a reflectivity spectrum R slide 3

4 A simple truth and an essential trick! X(t) doesn t change if you multiply it by a step function : This has implications about the symmetry of the frequency dependence (Fourier Tr.) slide 4

5 A simple consequence and some simple math If then also with the Fourier transform of X(t), given by If you delve deep into your memory for Fourier theory, you ll remember (yes?) where the symbol denotes convolution, which is defined as: slide 5

6 The starting point We now have found the relation The left side of the equation is simply given by In the following slides, we will work out the right side of the equation. First: look at, then convolute with slide 6

7 Fourier transform of the step function around t=0 described by finite cosine amplitude at zero frequency described by opposite Fourier amplitude for ω and -ω infinite collection of sines with weighing factor. Note that sine waves have the correct symmetry around t=0 slide 7

8 Getting ready to convolute We now have and as well as the relation we ll now write this convolution in terms of the known Fourier components above slide 8

9 Convolute! Using the known Fourier components, we find is equal to which we can split into two components one relating to the real part of F[Θ(t)] and one related to the imaginary part of F[Θ(t)] : which we can in turn write as slide 9

10 Mixing the ingredients Using the result derived on the previous slide, we found that the relation corresponds to the relation which after some serious math magically turns into Time to scratch your head! Taking an infinite integral of X with a frequency dependent imaginary weighing factor gives you X again! we can link imaginary parts of X to real parts of X and vice versa slide 10

11 Separating imaginary and real parts Final steps: with and we find that Grouping real and imaginary terms yields These Kramers-Kronig relations for X(ω) relate X (ω) to X (ω) and vice versa slide 11

12 Rewrite to positive frequencies only Reality condition: X (ω) is an even function, X (ω) is an odd function This allows Kramers-Kronig relations to be written as : (see homework) Note that these integrals run over positive angular frequencies only slide 12

13 Simplifications for dilute media Previously we derived the frequency dependent complex refractive index: For dilute media or weak susceptibility : X and X are small. Using the approximation for small x this leads to Comparing real and imaginary terms respectively, this gives and We also derived so slide 13

14 Kramers-Kronig relations for n and α With and the relation can be written as One can get the frequency dependent refractive index from an absorption spectrum We simplified the math by assuming dilute media, but the result is true in general! slide 14

15 Example for a single absorption line 1/3 To get n(ω) at a specific frequency, integrate the entire absorption spectrum with a frequency dependent weighing factor 1/(ω 2 -ω 2 ) absorption band Here: example for ω = 0.9 y=1/(ω )) α(ω') If ω < ω res then: Below ω α small and positive weighing factor negative small negative contribution Above ω α large and positive weighing factor positive large positive contribution ω'/ω res For frequencies below resonance, KK relations yield large n(ω) slide 15

16 Example for a single absorption line 2/3 To get n(ω) at a specific frequency, integrate the entire absorption spectrum with a frequency dependent weighing factor 1/(ω 2 -ω 2 ) absorption band y=1/(ω ) α(ω') If ω > ω res then: Below ω α large and positive weighing factor negative large negative contribution Above ω α small and positive weighing factor positive small positive contribution ω'/ω res For frequencies above resonance, KK relations yield small n(ω) slide 16

17 Example for a single absorption line 3/3 Repeating this integration for many ω yields n(ω): [n(ω) - 1] α(ω') ω'/ω res Between resonances: n slowly increases = normal dispersion Close to resonances: n rapidly decreases = anomalous dispersion At high frequencies (~ x-ray wavelengths) n can become slightly smaller than 1 slide 17

18 A quicker but less intuitive route - Cauchy We can also use Cauchy s integral theorem to derive the KK relations for X(ω) This is equivalent to proving (with apologies for the parameter changes!) We will show this by solving the integral using Cauchy s integral theorem slide 18

19 Integrating in the complex plane Cauchy s integral theorem ( CIT ): the closed path integral of an analytic function yields zero. (with some caveats) More info, see e.g. MathWorld = 0 1 c = 0 If we make ω complex, the CIT still holds. We now have a closed path of which - the value of the integrand goes to zero for large ω - the integration is tricky around Ω - the integration is over real frequencies and very familiar looking.. slide 19

20 Integration at large values of ω = 0 c Half-circle section: the value of the integrand goes to zero for large ω = 0 This can be seen by looking at X(ω) at fixed complex and large ω goes to zero For large real part of ω, this term oscillates fast over small Δt slide 20

21 Integration around a pole That leaves us with + = 0 c = 0 We can calculate the small path integral around Ω using the residue theorem: (the essential step) where is the vanishingly small length of the half circle around Ω That leaves us with = slide 21

22 The final result 2 3 We have found - = 3 2 We found: integration over real ω (except small region) = We set out to prove: ~ DONE! It turns out that avoiding the infinitely small section around omega is reasonable (recall the example integrations earlier this lecture) slide 22

23 Kramers-Kronig relations relating reflectivity to phase shift 1/3 A light wave incident on a planar surface of a dispersive material generates a reflected wave with a frequency dependent amplitude and phase shift The fraction of reflected irradiance scales with the reflectance R: with r(ω) the amplitude reflection coefficient: Note that for a low absorption situation (e.g. glass at visible frequencies) this gives: slide 23

24 Kramers-Kronig relations relating reflectivity to phase shift 2/3 It is difficult to measure ϕ vs. ω, but we can derive a KK relation for ϕ and R First step: express r(ω) as a complex exponent: We want to arrive at a relation of the form with X of the form X + i X (this linked the X integral to X and vice versa) To enable this, we will derive a KK relation for ln (r(ω) ) slide 24

25 Kramers-Kronig relations relating reflectivity to phase shift 3/3 Using the Cauchy Integral Theorem and the residue theorem we find The latter integral can be (see handouts) written in the form we can find the frequency dependent phase shift upon reflection by measuring a reflection spectrum! slide 25

26 Summary - Today s lecture Using reality condition OR Cauchy, found Kramers-Kronig relations linking - real refractive index n to an absorption spectrum α - real susceptibility X to a spectrum of the imaginary susceptibility X - a phase shift upon reflection ϕ to a reflectivity spectrum R slide 26

27 Next week Show that oscillating electrons radiate, and show that this can slow down light microscopic view of refractive index Low frequency Amplitude in phase with E in Lecture 5: model dipole radiation Lecture 6: derive simple model for oscillating dipoles (Lorentz model) slide 27