Bloch waves and Bandgaps Chapter 6 Physics 208, Electro-optics Peter Beyersdorf
Document info Ch 6, 1 Bloch Waves
There are various classes of boundary conditions for which solutions to the wave equation are not plane waves
Planar conductor results in standing waves
E(z) = 2E0 sin (kzz) cos (ωt) Waveguide and cavities results in modal structure ikz E(x, y, z) = Enm(x, y)e−
Periodic materials result in Bloch waves 2π/Λ iK" "r E! (!r) = E(K! , !r)e− · dK! !0 Ch 6, 2 Class Outline
Types of periodic media dispersion relation in layered materials
Bragg reflection
Coupled mode theory
Surface waves
Ch 6, 3 Periodic Media
Many useful materials and devices may have an inhmogenous index of refraction profile that is periodic
Dielectric stack optical coatings
Diffraction gratings
Holograms
Acousto-optic devices
Photonic bandgap crystals
Ch 6, 4 Periodic Media
Many useful materials and devices may have an inhmogenous index of refraction profile that is periodic
Dielectric stack optical coatings
Diffraction gratings
Holograms
Acousto-optic devices
Photonic bandgap crystals
Ch 6, 5 Periodic Media
Many useful materials and devices may have an inhmogenous index of refraction profile that is periodic
Dielectric stack optical coatings
Diffraction gratings
Holograms
Acousto-optic devices
Photonic bandgap crystals
Ch 6, 6 Periodic Media
Many useful materials and devices may have an inhmogenous index of refraction profile that is periodic
Dielectric stack optical coatings
Diffraction gratings
Holograms
Acousto-optic devices
Photonic bandgap crystals
Ch 6, 7 Periodic Media
Many useful materials and devices may have an inhmogenous index of refraction profile that is periodic
Dielectric stack optical coatings
Diffraction gratings
Holograms
Acousto-optic devices
Photonic bandgap crystals
Ch 6, 8 Bloch’s Theorem
2π/Λ Λ iK" "r E! (!r) = E(K! , !r)e− · dK! !0 Wave solutions in a periodic medium (Bloch waves) are different than in a homogenous medium (plane waves)
A correction factor E(K,r) accounts for the difference between plane wave solutions and Bloch wave solutions
Wave amplitude has a periodicity defined by the underlying medium, Ek(K,r)=Ek(K,r+Λ)
E(K,r) are normal modes of propagation
Ch 6, 9 Waves in Layered Media
For a wave normally incident on an Λ isotropic layered material, we’ll find the “dispersion relationship” (ω vs K curve) for Bloch waves. This will tell us about the behavior of waves in the material.
We’ll see the periodic structure reflects certain wavelengths. This is referred to as Bragg reflection.
Ch 6,10 Wave Equation in Layered Media
Starting with the wave equation
∂2E! ! ! E! + µ" = 0 ∇ × ∇ × ∂t2 ! " we will plug in the dielectric tensor written as a Fourier series with periodicity Λ
i 2πl z !(z) = !le− Λ !l and an arbitrary wave
i(kz+ωt) E! = E!0(k)e− dk to get !
2 ikz 2 i 2πl z ikz k E!0(k)e− dk + ω µ #le− Λ E!0(k)e− dk = 0 Ch 6,11 ! "l ! Wave Equation in Layered Media
2πl 2 ! ikz 2 i Λ z ! ikz with k E0(k)e− dk + ω µ #le− E0(k)e− dk = 0 ! "l ! 2πl k! = k + and defining Λ
2 ikz 2 2πl ik!z gives k E! (k)e− dk ω µ # E! (k" )e− dk" = 0 0 − l 0 − Λ ! ! "l
or 2 2 2πl ikz k E!0(k) ω µ #lE!0(k ) e− dk = 0 " − − Λ $ ! #l
so 2πl k2E! (k) ω2µ # E! (k ) = 0 for all k 0 − l 0 − Λ !l
Ch 6,12 Wave Equation in Layered Media
2πl k2E! (k) ω2µ # E! (k ) = 0 0 − l 0 − Λ for all k !l Is an infinite set of equations. Consider equations for: kΛ/2π=0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3 …
K=0 coupled K=0.2π/Λ K=0.4π/Λ 2 Let K be the value value of k±2πl/Λ closest to ω με0 in a series of coupled equations, where ε0 is the zeroth order Fourier coefficient of ε(z). The whole series of equations for -∞ i(kz+ωt) 2π i(K l 2π )z iωt E! (z) = E!0(k)e dk E! (K, z) = E!0(K l )e − Λ − → − Λ Ch 6,13 ! !l Bloch Waves in Layered Media The Bloch waves are normal modes of propagation so 2π/Λ iKz E! (z) = E(K, z)e− dK !0 and each mode is composed of plane wave components of amplitude E(K±2πl/Λ). To find these amplitudes we consider the dispersion relation 2πl k2E! (k) ω2µ # E! (k ) = 0 0 − l 0 − Λ !l Since this represents an infinite set of coupled equations, we will examine this expression and isolate the equations that couple most strongly to E0(K), ignore the rest and solve for E0(k) Ch 6,14 Field Components in Layered Media 2πl k2E! (k) ω2µ # E! (k ) = 0 0 − l 0 − Λ !l 2 2 2 2π 2 2π k E!0(k) ω µ#øE!0(k) ω µ#1E!0(k ) ω µ# 1E!0(k + ) . . . = 0 or − − − Λ − − Λ − Allowing us to express the E0(K-2πl/Λ) amplitudes as 1 2 2π 2 2π E!0(K) = ω µ#1E!0(K ) + ω µ# 1E!0(K + ) . . . K2 ω2µ# − Λ − Λ − − ø ! " ! 2π 1 2 ! 2π 2 ! E0(K ) = 2 ω µ$1E0(K 2 ) + ω µ$ 1E0(K) . . . − Λ K 2π ω2µ$ − Λ − − − Λ − ø # $ ! " ! 2π 1 2 ! 2 ! 2π E0(K + ) = 2 ω µ$1E0(K) + ω µ$ 1E0(K + 2 ) . . . Λ K + 2π ω2µ$ − Λ − Λ − ø # $ ! " … Ch 6,15 Resonant Coupling of Waves Momentum of forwards wave k with wavenumber k is ℏk, for -k backwards wave it is -ℏk. kg Grating can be thought of as superposition of forwards and backwards going waves, with momenta ±ℏkg, where kg=2π/Λ. For light to couple between forwards and backwards waves, momentum must be conserved ℏk+mℏkg=-ℏk This is like a collision of a forward photon with m phonons producing a backwards photon Ch 6,16 Field Components in Layered Media 1 2 2π 2 2π E!0(K) = ω µ#1E!0(K ) + ω µ# 1E!0(K + ) . . . K2 ω2µ# − Λ − Λ − − ø ! " ! 2π 1 2 ! 2π 2 ! E0(K ) = 2 ω µ$1E0(K 2 ) + ω µ$ 1E0(K) . . . − Λ K 2π ω2µ$ − Λ − − − Λ − ø # $ ! " ! 2π 1 2 ! 2 ! 2π E0(K + ) = 2 ω µ$1E0(K) + ω µ$ 1E0(K + 2 ) . . . Λ K + 2π ω2µ$ − Λ − Λ − ø # $ 2 ! " 2π 2 In the case where K l ! ω µ# ø i.e. the forward wave − Λ momentum cannot be! conve"rted to the backward wave momentum by the addition of a kick from the grating in the layered material, only the l=0 term is significant and the 2πl dispersion relation K2E! (K) ω2µ # E! (K ) = 0 0 − l 0 − Λ for any value of K is uncoupled t!ol that for other values of K, 2 2 and gives K ω µ "ø = 0 meaning the phase velocity is that due − to the average index of refraction for the medium Ch 6,17 Coupling of Field Components ! 2πl 1 2 ! 2π(l 1) 2 ! 2πl E0(K + ) = 2 ω µ$1E0(K + − ) + ω µ$ 1E0(K + 2 ) . . . Λ K + 2πl ω2µ$ Λ − Λ − Λ − ø # $ l=0 ! " 1 2 2π 2 2π E!0(K) = ω µ#1E!0(K ) + ω µ# 1E!0(K + ) . . . K2 ω2µ# − Λ − Λ − − ø ! " ! 2π 1 2 ! 2π 2 ! l=-1 E0(K ) = 2 ω µ$1E0(K 2 ) + ω µ$ 1E0(K) . . . − Λ K 2π ω2µ$ − Λ − − − Λ − ø # $ 2π 1 2π l=+1 ! ! " 2 ! 2 ! E0(K + ) = 2 ω µ$1E0(K) + ω µ$ 1E0(K + 2 ) . . . Λ K + 2π ω2µ$ − Λ − Λ − ø # $ ! " 2π 2 K l ω2µ# In the case where − Λ ≈ ø for some non-zero value of l=m this l=m term is! also sig" nificant and for the dispersion 2πl relation K2E! (K) ω2µ # E! (K ) = 0 0 − l 0 − Λ !l Λ we need only consider two values of K, i.e. K and K-2πm/ . Ch 6,18 Dispersion Relation in Layered Media 2 2 2πl The dispersion relation k E!0(k) ω µ #lE!0(k ) = 0 − − Λ !l Considering only terms with E(K) and E(K-2πm/Λ) gives 2πm K2 ω2µ" E# (K) ω2µ" E# (K ) = 0 − ø 0 − m 0 − Λ 2! " 2πm 2 2πm 2 K ω µ"ø E#0 K ω µ" mE#0(K) = 0 − Λ − − Λ − − #$ % & $ % A nontrivial solution to these coupled equations only exists if 2 2 2 K ω µ"ø ω µ"m −2 2−πm 2 2 = 0 ω µ" m K ω µ"ø ! − − − Λ − ! ! ! Λ ! " # ! 1 i2πmz/Λ and for ε-m=εm*! in a lossless medium, since!!m = !(z)e− dz 2 Λ 0 2πm 2! K2 ω2µ" K ω2µ" ω2µ " = 0 − ø − Λ − ø − | m| Ch 6,19 #$ % & ! " ! " Dispersion Relation in Layered Media 2 2πm 2 K2 ω2µ" K ω2µ" ω2µ " = 0 − ø − Λ − ø − | m| #$ % & ! " ! " which can be solved for K, the Bloch wave vector for a wave of frequency ω 2 2 mπ mπ 2 2 2 2πm 2 K = ! + "øµω ( "m µω ) + "øµω Λ ± " Λ ± & | | Λ "$ % ' ( # graph of dispersion relationship (figure 6.2) Ch 6,20 Bandgaps in Layered Media 2 2 mπ mπ 2 2 2 2πm 2 K = ! + "øµω ( "m µω ) + "øµω Λ ± " Λ ± & | | Λ "$ % ' ( # 2 When the Bragg condition is met (K-2πm/Λ=ω ε0 μ) K2 K2 ω2 < ω2 > real solutions exist for µ ( " + " ) and µ (" " ) ø | m| ø − | m| K2 K2 < ω2 < Solutions for µ (! + ! ) µ (! ! ) ø | m| ø − | m| are complex, this region is called the forbidden band. At the center of the forbidden band 2 2 2π 2 2 2 2 (mπ) K m ω µ# K ω µ"ø = 0 ω = where − Λ ≈ ø and − , i.e. Λ2µ# ! " ø mπ " the dispersion relation gives K = 1 i | 1| Λ ± 2" ! ø " Ch 6,21 Bandgap Properties The forbidden band has a width in ω, called the bandgap that is "m ∆ωgap = ω| | "ø And at its center has an attenuation coefficient mπ ∆ωgap Im[k] = 2Λ ω Thus, the greater the Fourier coefficient |εm| the larger the bandgap and the stronger the attenuation in the gap. Ch 6,22 Bloch Waveform With our calculated dispersion relationship 2 2 mπ mπ 2 2 2πm 2 K = ! + "øµω ( "m µω) + "øµω Λ ± " Λ ± & | | Λ "$ % ' ( we can choo#se a frequency ω, calculate K(ω) and solve 2πm K2 ω2µ" E# (K) ω2µ" E# (K ) = 0 − ø 0 − m 0 − Λ 2! " 2πm 2 2πm 2 K ω µ"ø E#0 K ω µ" mE#0(K) = 0 − Λ − − Λ − − #$ % & $ % for E(K) and E(K-2πm/Λ) giving the waveform of the Bloch mode at frequency ω (or wavenumber K(ω)) considering only these two components 2π i(K l 2π )z iωt E! (K, z) E!0(K l )e− − Λ − ≈ − Λ Ch 6,23 l=0!,m Bloch Waveform Example Consider a periodic structure nh nl consisting of alternating layers of high index and low index material (nh=1.8, nl=1.5). Find the waveform in the material for a wave of wavelength λ=2Λ Λ=λ/2 Ch 6,24 Bloch Waveform Example Example solved using “Mathematica” Ch 6,25 Bloch Waveform Example Ch 6,26 Bloch Waveform Example DispersionRelation 0.60362 0.6036 0.60358 0.60356 0.60354 0.60352 0.9999 0.99995 1 1.00005 1.0001 kêk0 Dispersion relation without Ch 6,27 exaggeration Bloch Waveform Example Ch 6,28 Bloch Waveform Example Bloch wave 1 Bloch wave 2 20 0.75 0.5 10 0.25 0 0 - 0.25 - 10 - 0.5 - 20 - 0.75 -6 - 6 -6 - 6 -6 0 1¥10-6 2 ¥10- 6 3¥10-6 4 ¥10- 6 5¥10-6 0 1¥10 2 ¥10 3¥10 4 ¥10 5¥10 z HmL z HmL Ch 6,29 High-Reflector Stack A series of alternating high-index, low-index layers, each λ/4 in thickness has 2π 2 π2 k = ω2µ# ω2 = − Λ ø and ε1≠0, and Λ2µ# ! " ø therefor light with wavenumber k=2π/λ cannot propagate through the medium. Instead it is resonantly coupled to a wave with wavenumber k=-2π/λ, i.e. a backwards traveling wave: The medium acts as a reflector for specific wavelengths, this is the principle behind high- reflectivity dielectric coatings. Ch 6,30 Alternative Methods In our previous method we solved the 1D wave equation for all frequencies of plane waves to get the dispersion relation. Alternatively we can consider wave propagation in the material, impose boundary conditions at the interfaces and require self-consistent solutions to get the dispersion relation - We will apply this method for a solution in 2D to the dielectric stack problem Ch 6,31 Boundary Conditions When an EM wave propagates across an interface, Maxwell’s equations must be satisfied at the interface as well as in the bulk materials. The constraints necessary for this to occur are called the “boundary conditions” ^ !1, µ1 !2, µ2 y !E dA = q z^ · ! "d E ds = B dA · −dt · ! " B dA = 0 · ! B d ds = J dA + !E dA µ · · dt · ! " " 6.32 Boundary Conditions Gauss’ law can be used to find the boundary conditions on the component of the electric field that is perpendicular to the interface. If the materials are dielectrics there will be no free charge on the surface (q=0) !1, µ1 !2, µ2 !E dA = q · ! "d E ds = B dA · −dt · ! " B dA = 0 · ! B d ds = J dA + !E dA µ · · dt · ! " " 0 !1E1zA !2E2zA = q !1E1z = !2E2z − ∴ 6.33 !→ Boundary Conditions Faraday’s law can be applied at the interface. If the loop around which the electric field is computed is made to have an infintesimal area the right side will go to zero giving a relationship between the parallel components of the electric field !1, µ1 !2, µ2 !E dA = q · ! "d E ds = B dA · −dt · ! " B dA = 0 · ! B d ds = J dA + !E dA µ · · dt · ! " d " 0 E2x,yL E1x,yL = B dA E1x,y = E2x,y − −dt · ∴ 6.34 ! → Boundary Conditions Gauss’ law for magnetism gives a relationship between the perpendicular components of the magnetic field at the interface !1, µ1 !2, µ2 !E dA = q · ! "d E ds = B dA · −dt · ! " B dA = 0 · ! B d ds = J dA + !E dA µ · · dt · ! " " B1z = B2z B1zA B2zA = 0 ∴ 6. − 35 Boundary Conditions Ampere’s law applied to a loop at the interface that has an infintesimal area gives a relationship between the parallel components of the magnetic field. (Note that in most common materials μ=μo) !1, µ1 !2, µ2 !E dA = q · ! "d E ds = B dA · −dt · ! " B dA = 0 · ! B d ds = J dA + !E dA µ · · dt · ! " " B1x,y B2x,y B1x,y B2x,y 0 d 0 L L = J dA + !E dA = µ − µ · dt · ∴ µ1 µ2 6.36 1 2 ! → ! Reflection at a Boundary The reflection and transmission coefficients at an interface can be found using the boundary conditions, but they depend on the polarization of the incident light D1z = D2z E E E = E i r ni 1x,y 2x,y “s” polarization (senkrecht, aka TE or θi θr vertical) has an E field that is InPlaneterf aofc ethe interface (here the perpendicular to the plane of incidence B1z = B2z yz plane) (perpendicular to page) y “p” polarization (parallel aka TM or θt horizontal) has an E field that is H1x,y = H2x,y nt z x parallel to the plane of incidence Et 6.37 Unit Cell Construct Label the forwards and backwards going waves in the nth b a “unit cell” cn an an forward going wave at right side of nth unit cell inside material 1 of form dn bn bn backward going wave at right side of nth unit cell inside material 1 n2 n1 cn forward going wave at right side of nth unit cell inside material 2 y^ Λ=a+b dn backward going wave at right side of nth unit cell z^ inside material 2 a thickness of layer for material 1 b thickness of layer for material 2 Λ total thickness of unit cell n number of unit cells to the right of some arbitrary origin forward waves phase factor are expressed as e-ikz+iωt Λ Ch 6,38 TE Continuity Equations b a For TE waves (E is out of the page) Ex and Hy∝∂Ex/∂z must be continuous at cn an each interface dn bn ik2z b ik2z b an 1 + bn 1 = cne + dne− n2 n1 − − ik2z b ik2z b ik1zan 1 + ik1zbn 1 = ik2zcne + ik2zdne− − − − − ik1z a ik1z a Λ=a+b cn + dn = ane + bne− ik a ik a ik c ik d = ik a e 1z ik b e− 2z 2z n − 2z n 1z n − 1z n There are 4 equations and 6 unknowns, so these can be manipulated to eliminate cn and dn in an expression of the form an 1 A B an − = bn 1 C D bn Ch 6,39 ! − " ! " ! " Unit Cell Equation (TE) For TE waves b a an 1 A B an cn an − = bn 1 C D bn ! − " ! " ! " with dn bn n2 n1 i k2z k1z A = eik1z a cos(k b) + + sin(k b) 2z 2 k k 2z ! " 1z 2z # $ Λ=a+b ik a i k2z k1z B = e− 1z sin(k b) 2 k − k 2z ! " 1z 2z # $ i k2z k1z C = eik1z a sin(k b) −2 k − k 2z ! " 1z 2z # $ ik a i k2z k1z D = e− 1z cos(k b) + sin(k b) 2z − 2 k k 2z ! " 1z 2z # $ Ch 6,40 Unit Cell Equation (TM) For TM waves b a an 1 A B an cn an − = bn 1 C D bn ! − " ! " ! " with dn bn n2 n1 2 2 i n k2z n k1z A = eik1z a cos(k b) + 1 + 2 sin(k b) 2z 2 n2k n2k 2z Λ=a+b ! " 2 1z 1 2z # $ 2 2 ik a i n2k1z n1k2z B = e− 1z sin(k b) 2 n2k − n2k 2z ! " 1 2z 2 1z # $ 2 2 i n k1z n k2z C = eik1z a 2 1 sin(k b) −2 n2k − n2k 2z ! " 1 2z 2 1z # $ 2 2 ik a i n1k2z n2k1z D = e− 1z cos(k b) + sin(k b) 2z − 2 n2k n2k 2z ! " 2 1z 1 2z # $ Ch 6,41 Multiple Cell Propagation 2 From conservation of energy |an| + |bn| b a 2 2 2 = |an-1| + |bn-1| which means the ABCD cn an matrix is “unimodular”. dn bn For 2x2 matrices n2 n1 1 Λ=a+b A B − 1 D B = − C D AD BC C A ! " − !− " For unimodular matrices the determinant is one, AD B C = 1, so − n a D B a n = − 0 b C A b Ch 6,42 ! n " !− " ! 0 " Bloch Wave Solutions The propagating waves in the medium are bloch waves with an amplitude that is periodic in Λ and a phase given by Kz, so a bloch wave should obey A B a a n = eiKΛ n C D b b ! " ! n " ! n " requiring A + D A + D 2 eiKΛ = 1 2 ± ! 2 − " # A + D A + D 2 or eiKΛ = i 1 2 ± − 2 ! " # iKΛ iψ A + D with the form e = cos ψ i sin ψ = e± where cos ψ = ± 2 1 1 A + D K = cos− Ch 6,43 giving Λ 2 ! " Material Bandgaps Contour plot Sign[Im[K(θ,ω]] (stop bands are in black) Brewster’s angle Ch 6,44 Bragg Reflection If a wave is incident on a layered b a material and cannot propagate cn an because it is within the bandgap, the energy of the wave is reflected. dn bn n2 n1 In the notation where the fields in the nth unit cell of layer 1 are an, bn,y^ Λ=a+b z^ the reflection of a wave a0 incident aN on the structure in material 1 will b0 have a reflection coefficient, b0 a0 rN = ao Λ Ch 6,45 Bragg Reflection By requiring bN=0, i.e. no input at b a the far end of the dielectric stack cn an of N layers, we can solve for bo and ao in terms of an. dn bn N a A B a n2 n1 0 = N b0 C D bN ! " ! " ! " y^ Λ=a+b where z^ an 1 A B an − = aN bn 1 C D bn b0 ! − " ! " ! " with the values of A, B, C and D a0 previously found. Λ Ch 6,46 Bragg Reflection For unimodular matrices, Chebyshev’ identity N A B 1 A sin NKΛ sin (N 1)KΛ B sin NKΛ = − − C D sin KΛ C sin NKΛ D sin NKΛ sin (N 1)KΛ ! " ! − − " 1 A + D with KΛ = cos− 2 ! " so a 1 A sin NKΛ sin (N 1)KΛ B sin NKΛ a 0 = − − n b0 sin KΛ C sin NKΛ D sin NKΛ sin (N 1)KΛ 0 ! " ! − − " ! " b C sin NKΛ r = 0 = N a A sin NKΛ sin(N 1)KΛ o − − Ch 6,47 Structure Reflectivity in Air Consider an infintesimal thickness of Ein r,t Ec r,t Et material 1 on top of the layered stucture. It has reflectivity ra1 on the air side (from air to material 1), Er and reflectivity rN on the structure side. l=0 E = tE r r E E = r E + r t E c in − a1 N c and r a1 in N a1 c tEin t2 giving E = and a1 c Er = ra1 + Ein 1 + ra1rN 1 + r r a1 N r + r r = a1 N so the reflectivity of the structure in air is 1 + ra1rN Ch 6,48 Spectral Reflectivity Spectral reflectivity RN at normal incidence of an N layer stack (quarter wave at ω0, nh=2.5, nl=1.5) C sin NKΛ 2 R = r 2 = n | N | A sin NKΛ sin(N 1)KΛ ! ! ! − − ! ! ! N=1 ! ! N=2 N=4 N=8 Ch 6,49 Summary Ch 6,50 References Yariv & Yeh “Optical Waves in Crystals” chapter 6 http://www.tf.uni-kiel.de/matwis/amat/semi_en/ kap_2/backbone/r2_1_4.html Ch 6,51