Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Lecture 8: Light propagation in anisotropic media Petr Kužel • Tensors — classification of anisotropic media • Wave equation • Eigenmodes — polarization eigenstates • Normal surface (surface of refractive indices) • Indicatrix (ellipsoid of refractive indices) Anisotropic response Isotropic medium, linear response: P = ε0χ E D = ε E = ε 0 (1 + χ ) E Anisotropic medium: the polarization direction does not coincide with the field direction: ( ) Py = ε 0 (χ 21E x + χ 22 E y + χ 23 E z ), Pz = ε 0 (χ 31E x + χ 32 E y + χ 33 E z ). Px = ε 0 χ11E x + χ12 E y + χ13 E z , Pi = ε 0 χ ij E j ( ) Di = ε ij E j = ε 0 1 + χij E j ! Tensors k⊥D k is not parallel to S S ⊥E Tensors Transformation of a basis: f j = Aij ei Corresponding transformation of a tensor: xk = Aki xi′ xi′ = Aik−1 xk 1st rank (vector) tkl′ = Aik−1 A−jl1tij tij = Aki Alj t kl′ 2nd rank pi1 ,i2!in = Ak1i1 Ak2i2 ! Aknin pk′1 ,k2!kn nth rank Tensors: symmetry considerations Intrinsic symmetry: reflects the character of the physical phenomenon represented by the tensor (usually related to the energetic considerations) ε ik = ε∗ki Extrinsic symmetry: reflects the symmetry of the medium pi1 ,i2!in = Ak1i1 Ak2i2 ! Aknin pk1 ,k2!kn ε ij = Aki Alj ε kl Example: two-fold axis along z: −1 0 0 A = 0 −1 0 0 0 1 ε13 = A11 A33 ε13 = – ε13 ⇒ ε13 = ε31 = 0 ε23 = A22 A33 ε23 = – ε23 ⇒ ε23 = ε32 = 0 Tensors: symmetry considerations Second example: higher order axis along z: cos α sin α 0 A = − sin α cos α 0 0 0 1 ε11 = c 2ε11 + s 2 ε 22 − 2 scε12 ε 22 = s 2 ε11 + c 2ε 22 + 2scε12 The above system of equations leads to: (ε11 − ε 22 )s 2 = −2scε12 (ε11 − ε 22 )sc = 2s 2ε12 Solution for a higher (than 2) order axis: sinα ≠ 0 ε11 = ε22, ε12 = 0 ( ) ε12 = scε11 − scε 22 + c 2 − s 2 ε12 εij Optical symmetry Crystallographic system Dielectric tensor optical system of axes Axes systems: ç isotropic cubic e = ç e 0 ç ç • crystallographic • optical (dielectric) n 0 0 2 æ è uniaxial • laboratory biaxial 0 n 0 2 0 0 n 2 ö ÷ ÷ ÷ ÷ ø crystallographic system of axes n2 0 0 2 0 ε = ε 0 0 n 2 0 0 n hexagonal tetragonal trigonal no2 0 0 ε = ε 0 0 no2 0 0 ne2 0 n o2 0 0 ε = ε 0 0 n o2 0 0 ne2 0 orthorhombic n x2 0 0 ε = ε 0 0 n 2y 0 0 n z2 0 n x2 0 0 ε = ε 0 0 n 2y 0 0 n z2 0 monoclinic n x2 0 0 2 ε = ε 0 0 n y 0 0 n z2 0 ε11 ε12 0 ε = ε12 ε 22 0 0 0 ε 33 triclinic n x2 0 0 2 ε = ε 0 0 n y 0 2 0 0 n z ε11 ε 12 ε 13 ε = ε12 ε 22 ε 23 ε 13 ε 23 ε 33 Wave equation Maxwell equations for harmonic plane waves: k ∧ E = ωµ0 H k ∧ H = −ω ε ⋅ E in the system of principal optical axes nx2 ε = ε0 0 0 0 n 2y 0 0 0 nz2 Wave equation: k ∧ (k ∧ E ) + ω2µ 0 ε ⋅ E = k (k ⋅ E ) − k 2 E + ω2µ 0 ε ⋅ E = 0 or ω2 s(s ⋅ E ) − E + 2 2 ε r ⋅ E = 0 k c Wave equation: continued We define effective refractive index: n= kc ω Wave equation in the matrix form: ( nx2 − n 2 s y2 + s z2 n2 sx s y 2 n sx sz ) n2 sx s y ( n 2y − n 2 s x2 + s z2 n2 s y sz ) Ex 2 Ey = 0 n s y sz nz2 − n 2 s x2 + s 2y E z n 2 sx sz ( ) Solutions of this homogeneous system: det = 0. • Effective refractive index n for a given propagation direction • Frequency ω for a given wave vector k • Surface of accepted wave vector moduli for a given ω (normal surface or surface of refractive indices) Wave equation: continued After having developed the determinant one obtains: 2 2 2 2 2 2 2 2 n ny ω ω ω2 nx2 n ω y 2 2 2 2 2 ω nz 2 z 2 −k + k k k k k − − + − − c2 c2 x c2 c2 c 2 2 2 ω nx k y 2 c 2 ω 2 2 nz 2 − k c −k 2 2 2 ω nx + kz 2 2 c 2 n 2y 2 − k −k = 0. c2 2 ω or in terms of the effective refractive index: (n 2 x )( )( ) − n 2 n 2y − n 2 nz2 − n 2 + [ ( )( ) ( )( ) ( )( )] n 2 s x2 n 2y − n 2 nz2 − n 2 + s 2y nx2 − n 2 nz2 − n 2 + s z2 nx2 − n 2 n 2y − n 2 = 0 . This is a quadratic equation in n2, thus it provides two eigenmodes for a given direction of propagation s. Wave equation: continued In the case when n ≠ ni, the wave equation can be further simplified: s 2y s x2 s z2 1 + + = n 2 − nx2 n 2 − n 2y n 2 − nz2 n 2 Polarization of the electric field for the eigenmodes: ( nx2 − n 2 s y2 + s z2 n2 sx s y 2 n sx sz ) n2 sx s y ( n 2y − n 2 s x2 + s z2 n2 s y sz ) Ex 2 Ey = 0 n s y sz nz2 − n 2 s x2 + s 2y E z n 2 sx sz ( ) For n ≠ ni: sx 2 2 N − nx sy E = 2 2 N n − y sz 2 2 N n − z Isotropic case nx = ny = nz ≡ n0 wave equation: (n 2 0 ) 2 − n2 n2 = 0 Normal surface: degenerated sphere Eigen-polarization: arbitrary (Isotropic materials, cubic crystals) Uniaxial case nx = ny ≡ no and nz ≡ ne ≠ no wave equation: (n 2 o − n2 ) 1 s x2 + s 2y s 2 − − z2 = 0 2 2 n ne no Normal surface: sphere + ellipsoid with one common point in the z-direction Sections of the normal surface: positive uniaxial crystal sy no sz no ne ne sx sx Uniaxial case: continued sz no S s or k E D H ne sx Uniaxial case: continued Propagation (z ≡ optical axis) ordinary ray extraordinary ray degenerated case (equivalent to isotropic medium); index n , E ⊥ k o index n , e E // z index n , o E in the (xy) plane, index n(θ), E in the (kz) plane, E⊥k E⋅k≠0 k // z k⊥z k: angle θ with z with sin 2 θ cos 2 θ 1 + = ne2 no2 n 2 (θ ) Biaxial case nx < ny < nz Normal surface has a complicated form Biaxial case: continued sy Sections of the normal surface by the planes xy, xz, and yz: sx = 0: sy = 0: nz nx 1 s z2 + s y2 − 2 n2 n x 1 s z2 s 2y − − =0 n 2 n 2y n z2 ny sx sz ny 1 s x2 + s z2 − 2 2 n n y 1 s x2 s z2 2 − 2 − 2=0 n nz n x nz sx sz n y sz = 0: 1 s x2 + s 2y − 2 2 n nz 1 s x2 s 2y − − =0 n 2 n 2y nx2 nx nz sy Group velocity in anisotropic media Maxwell equations in the Fourier space: k ∧ E = ω µ0 H /⋅ H k ∧ H = −ω ε ⋅ E /⋅ E One can finally obtain: ω= 2k ⋅ (E ∧ H ) S =k⋅ U E⋅D+ H ⋅B This means: v g = ∇k ω = S = ve U Birefringence Isotropic medium: incident plane wave, reflected plane wave Anisotropic medium: refracted wave is decomposed into the eigenmodes • Tangential components should be conserved on the interface: ki sin α i = k1 sin α1 = k 2 sin α 2 • k1 and k2 are not constant but depend on the propagation direction • Thus we get generalized Snell’s law for an ordinary and an extraordinary beam: ni sin α i = no sin α1 = n(α 2 )sin α 2 • Analytic solution only for special cases • Numerical solution • Graphic solution Birefringence: graphic solution Birefringence: uniaxial crystals positive (ne > no) negative (ne < no) O optical axis E optical axis O E O optical axis optical axis E E O optical axis E O E optical axis O Indicatrix: ellipsoid of indices z s DII, NII DI, NI x2 y2 z 2 + + =1 nx2 n 2y nz2