Electro-optics. Chapter 7 Physics 208, Electro-optics Peter Beyersdorf. Document info

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1 Electro-optics Chapter 7 Physics 208, Electro-optics Peter Beyersdorf Document info 1

2 Impermeability Tensor It is convenient to consider the inverse of the relative permitivity, which we call the impermeability η = ( ɛ ɛ 0 ) 1 this allows the equation for the index ellipsoid in the principle coordinate system to be written as x 2 n 2 x + y2 n 2 y + z2 n 2 z η ii ( E)x 2 i = 1 = 1 2

3 Electro-Optic Effect The impermeability depends on the distribution of charges in the crystal, which can be modified by an externally applied electric field η ij ( E) = η ij (0) + ( ηij E k ) E=0 E k ( 2 η ij E k E l η ij η ij ( E) η ij (0) = r ijk E k + s ijkl E k E l ( ηij resulting in an index ellipsoid that depends on the applied electric field ) E=0 with r s ijkl = 1 ijk = and 2 E k ) E=0 Pockel s effect 3 η ij ( E)x i x j = 1 E k E l +... ( 2 η ij Kerr effect E k E l ) E=0

4 Electro-optic coefficients The tensors r ijk and s ijkl are difficult to represent in matrix form, but using transpose symmetry of η (η ij =η ji ) and freedom in choosing the order of partial differentiation ( i j = j i ) allows us to say r ijk = ( ηij E k ) E=0 = ( ηji E k ) E=0 = r jik s ijkl = 1 2 ( 2 η ij ) = 1 E k E l E=0 2 ) ( 2 η ji ) E k E l ) E=0 = s jikl s ijkl = 1 2 ( 2 η ij E k E l E=0 = 1 2 ( 2 η ij E l E k E=0 = s ijlk greatly reducing the number of independent tensor elements that we need to represent these tensors 4

5 Electro-optic coefficients Contracting the subscript notation for the elements of a 3x3 tensor as η 11 η 12 η 13 η 21 η 22 η 23 η 1 η 6 η 5 η 6 η 2 η 4 η 31 η 32 η 33 η 5 η 4 η 3 allows us to write a 6x3 matrix for the r- coefficient and a 6x6 matrix for the s-coefficient. These contracted matrices are only shorthand for keeping track of the independent coefficients of the r and s tensors, but do not obey normal mathematical matrix operations. 5 η 1 η 2 η 3 η 4 η 5 η 6

6 Contracted Notation For the linear electro-optic effect the 27 coefficients have 18 independent values that can be expressed in the contracted notation as r 1k = r 11k r 2k = r 22k r 3k = r 33k r 4k = r 23k = r 32k r 5k = r 13k = r 31k r 6k = r 12k = r 21k for k=1,2,3 Similar relations exist for the 36 independent values of the 81 terms for the quadratic electrooptic effect 6

7 Modified Index Ellipsoid [ ( 1 n 2 k x ) ( 1 + r 1k E k x 2 + n 2 y ) ( 1 + r 2k E k y 2 + n 2 z + r 3k E k ) z 2 +2yzr 4k E k + 2xzr 5k E k + 2xyr 6k E k ] = 1 which we write as ( 1 n 2 x ) ( 1 + r 1k E k x 2 + n 2 y ) ( 1 + r 2k E k y yzr 4k E k + 2xzr 5k E k + 2xyr 6k E k = 1 n 2 z + r 3k E k ) z 2 where summation over k is assumed because it is a repeated index (note equation in book has error). The index ellipsoid is distorted by the presence of an electric field. A coordinate rotation is necessary to produce a new set of principle axis for the modified index ellipsoid. 7

8 Symmetry Considerations Symmetry properties in various crystal classes dictate that certain terms in the electrooptic coefficient are zero. In a centro-symmetric crystal, inversion around the point of symmetry should reproduce the original crystal lattice, giving r ijk =r ijk where r ijk is the electrooptic coefficient in the original crystal and r ijk is that in the inverted crystal. Because linear electrooptic coefficient is related to the linear displacement of charges in the crystal, an inversion of the crystal lattice should invert the electrooptic coefficient r ijk =- r ijk These two requirements dictate that r ijk =0 in any centrosymmetric crystal 8

9 Crystal Point Groups All crystals can be categorized into one of 32 point groups defined by seven symmetry properties including translational, rotational, inversion and mirror symmetries. See for example njpg/symmetry/solids.html for visualization of these 32 classes 9

10 Point Group Symmetries C 1 C 2 C 3 C 4 C 6 D 2 D 3 D 4 D 6 C 2v C 3v C 4v C 6v D 2d D 3d C s C 2h C 3h C 4h C 6h D 2h D 3h D 4h D 6h C i S 4 S 6 T T h O T d O h University of Newcastle upon Tyne 10

11 Point Group Symmetries C 4 4 S 4 4 Tetragonal a="b" c = = = /2 C 4h 8 D 2d 8 C 4v 8 D 4 8 D 4h 16 Triclinic Point Conditions Groups Order Orthographic projection a b c C 1 1 C i 2 Trigonal a=b="c" = <2 /3, /2 C 3 3 S 6 6 C 3v 6 D 3 6 D 3d 12 Monoclinic Orthorhombic a b c C s (C 1h ) 2 C = = /2 2 2 C 2h 4 a b c C 2v 4 D = = = /2 2 4 D 2h 8 ell Hexagonal a="b" c = = /2, C 3h 6 C 6 6 C 6h 12 D 3h 12 =2 /3 C 6v 12 D 6 12 D 6h 24 09/10/ :33 Cubic * a=b="c" = = = /2 T 12 T h 24 T d 24 O 24 O h 48 11

12 Form of Electro-Optic Tensor Elements of electrooptic tensor rik which are zero by symmetry considerations are shown in table 7.2 of Yariv and Yeh See For description of Hermann-Mauguin notation 12

13 KDP Example Find the principle axes and E-field dependent indices of refraction for a KDP crystal with an electric field applied along the z-direction 13

14 KDP Example Potassium Dihidrogen Phosphate (KH 2 PO 4 ) _ 42m crystal group (tetragonal) [table 7.3] Form of r ik matrix is found from table 7.2 r ik = r r r 63 Electrooptic and optical properties [table 7.3] r 41 =8x10-12 m/v, r 63 =11x10-12 nm n x =n y =n o =1.5115, n z =n 14 e =1.4698

15 KDP Example equation for index ellipsoid is ( 1 n 2 x ) ( 1 + r 1 ke k x 2 + n 2 y ) ( 1 + r 2 ke k y yzr 4k E k + 2xzr 5k E k + 2xyr 6k E k = 1 n 2 z + r 3 ke k ) z 2 which becomes ( 1 n 2 x ) ( 1 + r 1 ke k x 2 + n 2 y ) ( 1 + r 2 ke k y yzr 41 E x + 2xzr 52 E y + 2xyr 63 E z = 1 n 2 z + r 3 ke k ) z 2 or x 2 n 2 o + y2 n 2 o + z2 n 2 e x 2 n 2 o + y2 n 2 o + 2yzr 41 E x + 2xzr 41 E y + 2xyr 63 E z = 1 for a field applied entirely along z: + z2 n 2 e + 2xyr 63 E z = 1 15

16 KDP Example x 2 n 2 o + y2 n 2 o + z2 n 2 e + 2xyr 63 E z = 1 We can choose a new set of principle coordinates x, y and z that recast this equation into the form x 2 n 2 x + y 2 n 2 y + z 2 n 2 z = 1 From inspection z=z, so let x y and xy differ by a rotation of θ around the z-axis: x=x cosθ-y sinθ y=x sinθ+y cosθ (x cos θ y sin θ) 2 n 2 o + (x sin θ + y cos θ) 2 n 2 o + z2 n 2 e + 2(x cos θ y sin θ)(x sin θ + y cos θ)r 63 E z = 1 16

17 KDP Example (x cos θ y sin θ) 2 x 2 n 2 o n 2 o Evaluating gives or + y 2 n 2 o + (x sin θ + y cos θ) 2 n 2 o + z2 n 2 e + 2(x cos θ y sin θ)(x sin θ + y cos θ)r 63 E z = 1 cross term 2x y cos(2θ) 0 for θ=45 giving or + z 2 n 2 e ( 1 + z 2 n 2 e n 2 o x 2 n 2 o x y sin 2θ n 2 o + y 2 n 2 o + z 2 n 2 e ) ( 1 + r 63 E z x x y sin 2θ n 2 o + (x 2 y 2 )r 63 E z = 1 n 2 o r 63 E z ) 17 + (x 2 sin 2θ + 2x y cos 2θ y 2 sin 2θ)r 63 E y 2 + z 2 n 2 e = 1 y y θ x x

18 KDP Example Equating to gives so ( 1 n 2 o ) ( 1 + r 63 E z x 2 + ( 1 n 2 o n 2 x = n 2 o n x n o ( n2 or 63 E z ) + r 63 E z ) ( n 2 o = 1 n 2 x 1 r 63 E z ) x 2 n 2 x + y 2 n 2 y + z 2 n 2 z = n 2 or 63 E z n 2 x n 2 o(1 n 2 or 63 E z ) ) y 2 + z 2 n 2 e n y n o ( n2 or 63 E z ) 18 = 1 for E z <<1/(n o2 r 63 )=4x10 10 V/m n z = n e

19 Quartz Example Show that quartz remains uniaxial in the presence of an electric field applied along the z- axis SiO 2 32 class trigonal crystal r ij = r r r r r

20 Quartz Example For E x =E y =0 Δη=0 η ij = r r r r r E z Thus the index ellipsod is not changed by the presence of an electric field in the z direction, so quartz, a positive uniaxial crystal, remains uniaxial 20

21 Modulation Since an externally applied electric field can alter the index ellipsoid of a crystal, it alters the birefringence. If the principle indices in the presence of an electric field are n x (E), n y (E) and n z (E) then the birefringence (for a wave propagating in the z- direction) is n y (E)-n x (E). If the material has a thickness d the relative phase retardation will be Γ = ω c ( n y (E) n x(e) ) d 21

22 Modulation For KDP with an electric field applied along the z-axis (see example on page 14 of notes) n x n o ( n2 or 63 E z ) n y n o ( n2 or 63 E z ) n z = n e thus" " " " " " " " " Γ = ω c ( n y (E) n x(e) ) d becomes Γ = ω c n3 or 63 Ed = ω c n3 or 63 V where V is the voltage across d, the thickness of the crystal. This geometry is realized in a Pockel s cell (electrooptic modulator) 22

23 Modulation The voltage for which the relative phase retardation increases by π is V π and is called the half-wave voltage with Γ = ω c n3 or 63 V the half-wave voltage is V π = λ 2n 3 or 63 this is independent of the crystal dimensions. 23

24 Mathematical Representation of Modulation To model the effects of birefringence it is necessary to have a method to keep track of polarization states Jones Calculus To keep track of time dependent modulation it is helpful to describe modulated wave in terms of single frequency components Carrier and sidebands 24

25 Describing Polarization Mathematically Since polarization, like vectors, can be described by the value of two orthogonal components, we use vectors to represent polarization, with each component a phasor Amplitude of the components can be complex to represent a time delay between the components of the waves Jones calculus allows us to keep track of the polarization of waves as they propagate through a system

26 Jones Vectors Defining the phasor representation for the amplitude of a plane wave propagating in the z- direction as Ẽ(z, t) = Ẽ0e i(kz ωt) the polarization can be included by defining the amplitude of the x and y components: ) (Ẽx (z, t) Ẽ(z, t) = Ẽ y (z, t) = (Ẽ0x Ẽ 0y ) e i(kz ωt) the polarization of the wave is described by called the Jones vector (Ẽ0x Ẽ 0y ) 26

27 Jones Vectors Expressing polarization in terms of two orthogonal states with complex amplitude (i.e. amplitude and phase) of each component expressed in vector form vertical horizontal linear at +45 linear at θ right circular * left circular [ ]. 0 1 [ ] [ ] [ ] cos θ sin θ 1 2 [ i [ 1 i ]. ].. * Right (left) circular polarization has a CW (CCW) rotation at a fixed point in space as seen by an observer looking towards the source of the wave

28 Jones Matrices An optical element that transforms one polarization state into another can be treated as a 2x2 matrix acting on a jones vector [ ] Ex,out E y,out = [ ] [ ] M11 M 12 Ex,in M 21 M 22 E y,in The Jones Matrices for a series of optical elements can be multiplied together to find how the optical system transforms the polarization of an input beam 28

29 Polarizers Selectively attenuates one polarization unpolarized input light polarizer polarized output light Malus discovered that an analyzer oriented at θ with respect to a polarizer would have a transmission of I(θ) = I 0 cos 2 θ Wire Grid Polarizer known as Malus Law 29

30 Jones Matrix of a Polarizer What is the Jones matrix for a polarizer that transmits horizontal polarization? ] [ Ex,out 0 = [???? ] [ ] Ex,in E y,in What if it is rotated at an angle θ? [ cos θ sin θ R(θ)M R( θ) R(θ) sin θ cos θ ] Use this to verify Malus Law 30

31 Malus Law [ ] Ex,1 = E y,1 [ ] Ex,2 = E y,2 [ [ cos θ sin θ sin θ cos θ ] [ ] Ex,0 E y,0 = [ Ex,0 ] [ ] 0 ] [ cos θ sin θ sin θ cos θ E 0 E 1 E 2 ] [ ] Ex,1 E y,1 [ ] Ex,2 E y,2 = = = = [ ] [ ] [ cos θ sin θ 1 0 cos θ sin θ sin θ cos θ 0 0 sin θ cos θ [ ] [ ] [ ] cos θ sin θ 1 0 cos θex,0 sin θ cos θ 0 0 sin θe x,0 [ ] [ ] cos θ sin θ cos θex,0 sin θ cos θ 0 [ ] cos 2 θe x,0 sin θ cos θe x,0 ] [ Ex,0 0 ] E 2 = cos θ cos E 2 θ + sin 2 θ = cos θ 0,x 31 I 2 I 1 = cos 2 θ

32 Retarders Devices which delays one polarization component with respect to the other For a birefringent material of thickness d φ = (kn y d kn x d) = 2π λ nd Thus the Jones matrix is [ e i φ ] 32

33 Quarter Wave Plate For a retarder with Δφ=π/2 (i.e. a retardation of λ/4) [ ] i And when a wave with linear polarization at ±45 passes through the retarder it gets converted to left (right) circular polarization 1 2 [ 1 i 1 2i [ i 1 ] ] = 1 2 [ i = 1 2 [ i ] [ ] ] [ 1 1 ]

34 Half Wave Plate For a retarder with Δφ=π (i.e. a retardation of λ/2) [ 1 0 ] 0 1 And when a wave with linear polarization at θ passes through the retarder it gets converted to linear polarization at -θ [ ] [ cos θ sin θ ] = = [ ] cos θ sin θ [ ] cos( θ) sin( θ) 34

35 Compensator Variable waveplate with Jones matrix [ e i φ ] Babinet compensator Berek compensator 35

36 Waveplate Order Recall that the relative delay between the two polzrization states is φ = 2π λ nd For a zero-order quartz waveplate if we wish Δφ π we need d 100μm For a typical multi-order waveplate Δφ 11π so d 1mm which is easier to manufacture and has the same retardation (modulo 2π) as a zero order waveplate 36

37 Waveplate Order dispersion of a waveplate is the sensitivity of the retardation to wavelength φ = 2π λ nd d φ dλ = 2π λ 2 nd which is minimized in a zero order waveplate. Typical zero-order waveplates are actually made of two multiorder waveplates cemented together oriented at 90 so that Δφ=Δφ 1 -Δφ 2 π but the total thickness d 2mm 37

38 Birefringence and Polarization Consider light propagating in the z-direction of a z-cut KDP plate with an electric field applied along the z-direction. From the example on page 19 of these notes we have n x n o ( n2 or 63 E z ) as the principle indices of refraction, thus the net birefringence is and the output polarization is n y n o ( n2 or 63 E z ) Γ = ω c (n y n x)d = 2π λ n3 or 63 V [ e iγ/2 0 ] [ ] E0x 0 e iγ/2 E 0y 38 n z = n e

39 Birefringence and Polarization For linearly polarized light at the input with the output state is E 0 = 1 [ ] E out = 1 [ ] e iγ/2 2 e iγ/2 So external electric field affects the polarization state of the transmitted beam 39

40 Birefringence and Polarization For linearly polarized light at the input with [ 1 0] " " " " " " E " 0 = " " or the output state is [ ] e iγ/2 "" " " " E out " = " " " or " E out " = " " " " respectively 0 E 0 = So external electric field affects only the phase of input light polarized along the principle axes [ 0 1] [ 0 e iγ/2 ] 40

41 Birefringence and Polarization For linearly polarized light at the input with and the crystal followed by a polarizer at -45, the output state is [ E 0 = 1 [ ] ] [ e iγ/2 0 0 e iγ/2] [ 1 1 For a relative transmitted power of ] = i [ 1 sin(γ/2) 2 1] = i sin(γ/2) E 0 P out P o = sin 2 (Γ/2) An additional birefringence of π/2 can bias the output for a linear response 41

42 Birefringence and Polarization P out P o = sin 2 (Γ/2 + π/4) P out Birefringence for Γ<<1 P out P o = Γ 42

43 Phase Modulation modulation waveform unmodulated wave modulated wave 43

44 Phase Modulation Sidebands Consider a phase modulation of Δφ=m cos(ωt) on a field of amplitude E 0 : using the Jacobi-Anger identity: e iz cos φ = with E in the form i(ωt+m cos Ωt) E = E 0 e n= E = E 0 e iωt im cos Ωt e i n J n (z)e inφ gives E = E 0 e iωt n= i n J n (m)e inωt 44

45 Phase Modulation Sidebands For m<<1 we can write and with the relation this becomes E = E 0 e iωt n= J n (z) = ( 1) n J n (z) i n J n (m)e inωt E E 0 ( ij 1 (m)e i(ωt Ωt) + J 0 (m)e iωt + ij 1 (m)e i(ωt+ωt)) E E 0 (J 0 (m)e iωt + ij 1 (m) (e i(ωt Ωt) + e i(ωt+ωt))) 45

46 Phase Modulation Sidebands E E 0 (J 0 (m)e iωt + ij 1 (m) (e i(ωt Ωt) + e i(ωt+ωt))) Phasor representation of sidebands in a frame rotating at ω 46

47 Amplitude Modulation Sidebands E E 0 (J 0 (m)e iωt + J 1 (m) (e i(ωt Ωt) + e i(ωt+ωt))) Phasor representation of sidebands in a frame rotating at ω 47

48 Modulation Depth The phase acquired by a beam passing through an optic of thickness L is phase modulation comes from the index being a function of time n(t), in which case the phase acquired is where φ = k 0 nl n(t) = 1 L/c φ = k 0 n(t)l t t L/c n(t )dt is the index of retraction of the optic averaged over the time the light is in the optic. If the index changes appreciably before the light traverses the optic the modulation depth will be reduced 48

49 Modulation Depth n(t) = 1 L/v p L/vp 0 n(t, z(t))dτ with n(t) = n + δn sin ( Ω(t + τ z ) ) v m and z = v p τ where the modulation is a traveling wave propagating at speed v m n = n δnv mv p Ω(v m v p )L n(t) = n + δn sin Which can be evaluated using ( Ωt + Ωτ ( 1 v p v m )) [ ( ( cos Ω t v p L + L )) ] cos (Ωt) v m v p v p ( ) ( ) A + B A B cos A cos B = 2 sin sin

50 Modulation Depth n = n δnv mv p Ω(v m v p )L δnv m v p n = n + 2 Ω(v m v p )L sin ( ΩL n = n + δn sinc 2 [ ( ( cos Ω t v p L + L )) ] cos (Ωt) v m v p v p ( Ωt ΩL 2 [ 1 1 ]) sin v p v m [ 1 1 v p v m ( Ωt ΩL 2 ]) ( [ ΩL 1 sin 1 2 v p [ 1 1 ]) v p v m v m ]) 50

51 Modulation Depth ( ΩL n = n + δn sinc 2 [ 1 1 ]) ( sin Ωt ΩL v p v m 2 [ 1 1 ]) v p v m δn eff n = n + δn sinc( κl) sin (Ωt κl) κ Ω 2 [ 1 1 ] v p v m Modulation Depth (in units of!n) "#=0 Crystal Length (in units of "#) 0 0.5!! 1.5! 2! 2.5! 3! -0.8 Thus the modulation depth is reduced by sinc(δκl). At low modulation frequency Ω 0 the modulation depth can be increased by increasing the crystal length, but at frequencies Δκ>π/2L there is no improvement in modulation depth with increased crystal length 51

52 LiNbO 3 phase modualtor Consider a Lithium Niobate phase modulator r ik = 0 r 22 r 13 0 r 22 r r 33 0 r 51 0 r r We want pure phase modulation, so need i=1,2 or 3. Since r 33 >r 13 >r 22 use an applied field along z to modulate light polarized in z-direction. 52

53 LiNbO 3 phase modualtor Taking r 33 = m/v, v m = m/s, n e =2.2 and L=40 mm and d=4 mm we have κ Ω 2 [ 1 1 ] v p v m = 10 9 (s/m)ω for V max =15V we can find the modulation depth (in rad) for 100 MHz modulation and 100 GHz modulation Γ = k 0 δn eff L ( ) Γ = 1 k V max 0 2 n3 er 33 sinc( κl) L d for 100 MHz sinc(δκl) 1, and Γ=0.15 for 100 GHz sinc(δκl)=-0.03 and Γ= L d

54 Kerr Effect In centro-symmetric and amorphous materials where the linear electro-optic effect vanishes, a second order effect can be observed η ij = s ijkl E k E l The second order electrooptic tensor s ijkl can be written with contracted notation as s ik and has 6x6=36 unique elements 54

55 Kerr Effect Example Consider an electric field applied to a glass s ik = s 11 s 12 s s 12 s 11 s s 12 s 12 s (s 11 s 12 ) (s 11 s 12 ) (s 11 s 12 ) Taking the direction of the applied field as z gives an index ellipsoid defined by ( ) ( ) ( ) x 2 n 2 + s 12E 2 + y 2 n 2 + s 12E 2 + z 2 n 2 + s 11E 2 = 1 55

56 Kerr Effect Example This is the index ellipsoid of a uniaxial crystals with n o = n 1 2 n3 s 12 E 2 n e = n 1 2 n3 s 11 E 2 giving a birefringence of n e n o = 1 2 n3 (s 12 s 11 )E 2 which is often written as n e n o = Kλ 0 E 2 where K is called the Kerr constant and λ 0 is the vacuum wavelength 56

57 References Yariv & Yeh Optical Waves in Crystals chapter 7 57