4: birefringence and phase matching

Transcription

1 /3/7 4: birefringence and phase matching Polarization states in EM Linear anisotropic response χ () tensor and its symmetry properties Working with the index ellipsoid: angle tuning Phase matching in crystals In vacuum, D = ε E E = x y z Polarization in EM Plane wave state, arbitrary direction: ( ) i E ( ) i E x E = x y z ( ) ei k r ωt ei k r ωt E = x E x e i k r ωt ( ) + y E y e i k r ωt ( ) + z E z e i k r ωt ( ) x E x e i( k r ωt ) = E x x e i ( k x x+k y y+k z z ωt ) = ikx E x e i( k r ωt ) E( r,t) = E e i( k r ωt ) ( ) E y e i( k r ωt ) E z e i( k r ωt ) E = i( k x E x + k y E y + k z E z )e i( k r ωt ) = ik E e i( k r ωt ) = From this we can say that k E = and k E Therefore, the electric field lies in a plane perpendicular to k The polarization direction can take on any linear combination of horizontal and vertical states (this includes circular polarization).

2 /3/7 Other vector relations Similarly, E = B ik E = +iω B t so B k,e Energy flow is given by the Poynting vector: S = E H = µ E B so S E,B and S k These relations hold in any isotropic medium. But if the medium is anisotropic, the vector relations must be modified. E = B = E = B t D B = µ t General elliptical polarization state For light propagating in the z-direction: Ex(z, t) = E E (z, t) = E y x y cos(kz -ωt) x cos(kz -ωt + ε) y Pick a plane (z = ), plot E-field as f(t) An ellipse can be represented by 4 quantities:. size of minor axis. size of major axis 3. orientation (angle) 4. sense (CW, CCW)

3 /3/7 Linear polarization state For light propagating in the z-direction, ε=: E x (z,t)= E x cos(kz-ωt) x E y (z,t)= E y cos(kz-ωt) y E x E y Vertical E 45 o E Circular polarization state For light propagating in the z-direction, ε=±π/: E x (z,t)= E x cos(kz-ωt) x E y (z,t)= E y cos(kz-ωt ±π /) y E x (z,t)= E x e i( kz-ω t) x i kz-ω t±π / E y (z,t)= E y e ( ) y=ey ( ±i)e i( kz-ω t) y E ±i 3

4 /3/7 Jones vector representation Use when light is fully polarized, coherent E a b Horizontal: Vertical: With a, b complex, normalized: a + b = E E 45 o : R/L circular: E E ±i Other representation Fields: Stokes vectors (4 components) Optical components: Mueller matrices (4x4) Good for partially polarized light Does not track phase info for coherent light Manipulating polarization: polarizers S: e-field is in plane of surface (skim) P: e-field has out of plane component (plunge) Effects to work with Brewster angle: S reflects some, P does not reflect Total internal reflection (TIR): full reflection if AOI is greater than critical angle Linear dichroism: material absorbs one polarization much more than other Wire-grid polarizer Polaroid 4

5 /3/7 Crystal polarizers Crystal polarizers used as: Beam displacers, Beam splitters, Polarizers, Analyzers,... Examples: Nicol prism, Glan- Thomson polarizer, Glan or Glan- Foucault prism, Wollaston prism Non-crystal: cube, thin-film polarizer Jones matrix: Horizontal polarizer Vertical polarizer Retarder: half wave plate Retard the phase of one polarization component to change the polarization state half-wave plate: retard by π = 8 o = λ/ enter with 45 o polarization relative to crystal axes E = E ( ˆx + ŷ)e i ( kz ωt ) = E propagate through thickness L E = E ˆxe ikn o L + ŷe ikn e L ( )e i ( kz ωt ) = E e ikn o L = E e ikn e L eikn o L e ik ( n e n o)l set L so that Δφ = kl( n e n o ) = π + m π m = order Net effect is that y component changes sign Jones matrix for half wave plate 5

6 /3/7 Retarder: quarter wave plate Convert linear to circular quarter-wave plate: retard by π/ = 9 o = λ/4 enter with 45 o polarization relative to crystal axes E = E ( ˆx + ŷ)e i ( kz ωt ) = E now set thickness L so that ( ) = π Δφ = kl n e n o Net effect is that y component changes by i + m π m = order Jones matrix for quarter wave plate i Rotated polarization vector Arbitrary orientation of linear polarization E = E ( ˆxcosϕ + ŷsinϕ )e i ( kz ωt ) cosϕ = E = E v ϕ sinϕ Using rotation matrix cosϕ sinϕ sinϕ cosϕ Half wave plate: = cosϕ = sinϕ polarization rotates by φ cosϕ M ϕ v x = v ϕ sinϕ cosϕ sinϕ y θ x 6

7 /3/7 Rotated Jones matrix In practice, the waveplate is rotated in its mount Earlier, we had M λ/ M ϕ v x Multiply both sides by cosϕ sinϕ cosϕ = sinϕ inverse rotation matrix sinϕ cosϕ sinϕ cosϕ cosϕ sinϕ ( ) = M λ/ v ϕ = v ϕ sinϕ cosϕ cos ϕ sin ϕ cosϕ = sinϕ cosϕ sinϕ ( M ϕ M λ/ M ϕ ) v x = M ϕ v ϕ = v ϕ = cosϕ sinϕ cosϕ sinϕ cosϕ sinϕ This has the effect of rotating the matrix cosϕ sinϕ sinϕ cosϕ cosϕ sinϕ sinϕ = cosϕ cosϕ sinϕ sinϕ cosϕ Maxwell's Equations: linear anisotropic medium The induced polarization, P, contains the effect of the medium: D = B = In an anisotropic medium: P E E = B t D B = µ t ( ) = ε χ E, D = ε E + P = ε + χ So now D and E are not necessarily parallel. D = ε ε E D x D y D z ε xx ε xy ε xz = ε ε yx ε yy ε yz ε zx ε zy ε zz Define the displacement vector ( ) E = ε ε E E x E y E z D = ε E + P 7

8 /3/7 Linear anisotropic response For anisotropic linear response: D i = ε In a basis aligned with the crystal axes: D X D Y D Z ε XX = ε YY ε ZZ E X E Y E Z j ε ij E j = ε ε ij E j ε o = ε o ε e D is not parallel to E if E projects onto different axes D X D Y D Z ε o = ε o ε e E Y E Z = ε o E Y ε e E Z E X E Y E Z Uniaxial case: o = ordinary Contracted notation Repeated indices are summed e = extraordinary Linear tensor χ () biaxial uniaxial isotropic 8

9 /3/7 The dielectric tensor ε ij is symmetric in a nonabsorbing medium. U = S Continuity equation: t Rate of change of energy density = - div of power flow S = (E H) = H ( E) E ( H) E = B t, H = D S = E D H B t U = U E + U H = S = E D + H B Look at the E component of the energy density: U E = E D = E i ε ij E j But we also know that: Take the time derivative: U E = E D = ε ij E i E j U E = ε ij ( E i E j + E i E j ) = (ε ji + ε ij )E i E j Therefore, the dielectric tensor is symmetric: ε ji = ε ij This is an example of an intrinsic symmetry. This does not require the symmetry of the crystal, or the linearity of the response. Energy density inside the medium: see Davis 8.3 for derivation The index ellipsoid U E = D i E = ε The index ellipsoid is a surface of constant energy density (in the crystal basis): U E = D X + D Y + D Z ε ε XX ε YY ε ZZ D X + D Y + D Z ε U ε XX ε YY ε ZZ = ij ε ij E i E j Write this with new variables to make the ellipse equation more clear: X = ε U E / D X etc. = X ε XX + Y ε YY + Z ε ZZ = X In an arbitrary basis, the ellipse equation looks like: x + n n y + n z + n 3 4 yz + n 5 n o + Y + Z n o n e xz + n 6 xy = The indices -6 are like the contracted notation we will use for second-order NLO 9

10 /3/7 Wave propagation in birefringent crystals Inside the medium, D = So D = ik D = and k D The wave is described by the D-field inside the medium. If a wave is linearly polarized, and the D-field is oriented along one of the crystal axes, the wave sees only the refractive index corresponding to the direction of D. ( ) = D e i k r ωt D r,t ( ) ẑ D z e i ( k xx ωt ), kx = ω c n e If k is parallel to one of the axes, but D is not, the input polarization can be resolved along o- and e- axes: ( ) = D e i k r ωt D r,t ( ) ẑ D z e i ω c n ex ωt + ŷ Dy e i ω In Jones vector notation, a D( r,t) = D b D c n ox ωt ae iω c n ox be iω c n ex The vector components get a relative phase shift Δφ = ω ( c n n e )x Plane wave propagation: general direction In an anisotropic medium, the phase velocity of light depends on its polarization state and its propagation direction. For a given propagation direction, there exist in general two waves, each having its own refractive index (or equivalently phase velocity) and polarization. All light traveling in that direction can be decomposed onto the two eigenwaves. define wave unit vector s = k k. B n E = H = s E t µ c D n H = D = s H t c Then i ω n c s, t iω H (B) Relations between the directions of the vectors: ) D, H, and s are mutually perpendicular. ) D, E, s, and E H (energy flow) lie in the same plane. 3) The Poynting vector S=E H is generally not along s. α E D α k (s) S=E H (t)

11 /3/7 Spatial walk-off Angle α between k (wavefront normal) and S (power flow) Beam polarized along o-axis does not deviate from k Beam at angle θ to e-axis will walk off double refraction α E D k (s) H (B) α S=E H (t) θ α Walk-off angle: tanα = ± n o n tanθ e Use + for negative unixial for positive uniaxial Using the index ellipsoid: tuning the refractive index Optic axis The role of the index ellipsoid: D For a given arbitrary wave normal direction s, the index ellipsoid can be used to ) Find the indices of refraction of the two eigenwaves. ) Find the corresponding directions of the D vectors of the two eigen waves. The prescription is as follows: ) Draw a plane that is through the origin and is perpendicular to s. This plane intersects the index ellipsoid surface with a particular intersection ellipse. ) The lengths of the two semiaxes of the intersection ellipse, n and n, are the two indices of refraction of the eigenwaves. 3) The two axes of the intersection ellipse are each parallel to the allowed D vectors of the eigenwaves. n n D s

12 /3/7 Computation of the angle-dependent refractive index z The index ellipsoid: s The equation of the index ellipsoid of a uniaxial crystal is x + y n n n o e e e o = = n > n z + n e = ε / ε = ε / ε, ordinary refractive index ε / ε, extraordinary refractive index o n < n o x z y : positive uniaxial crystal prolate spheroid : negative uniaxial crystal oblate spheroid D e n e (θ) θ n o D o The uniaxial index ellipsoid is rotationally symmetric around the z-axis. Let s be in the y-z plane with a polar angle q. The two polarization directions of the D vectors are: D o is parallel to the x-axis, D e is in the y-z plane and is perpendicular to s. The corresponding refractive indices are: n o = n o, n e (θ)cosθ + n (θ)sinθ / e = n e (θ) = cos θ + sin θ n o n e n o n e. When s is on the z direction, n e ( ) = n o. Therefore the z-axis is the optic axis. y A derivation of the angle-dependent refractive index Let k = ˆx k sinθ + ẑ k cosθ ( ) k = k n e θ Find the components of D ˆx ŷ ẑ that correspond to k H = D k H = k t iω D = i k H x k z H D x = k H z ω = H ω k n e ( θ )cosθ = H c n e ( θ )cosθ D z = k H x ω = H ω k n e ( θ )sinθ = H c n e ( θ )sinθ D x θ k = ˆxk z H + ẑk x H z

13 /3/7 A derivation of the angle-dependent refractive index Put these into equation for ellipsoid: D X + D Y + D Z ε U E ε XX ε YY ε ZZ = = ε U E D X n o + D Y n o + D Z n e H ε U E c ( ) n e θ cos θ n o + sin θ n e = Magnetic energy density U H = µ H Is equal to electric energy density U H = U E H U = H ε U E c ε U E µ c = U H U E µ ε c = Finally: n e ( θ ) = cos θ + sin θ n o n e The refractive index can be angle-tuned anywhere between n e and n o. Phase matching: an important application for the the angle-dependent refractive index Recall that for SHG: A ( z ei k z ω t) ω = i d k c A e i ( k z ω t) Energy must be conserved: ω + ω = ω ω = ω Energy w w w Momentum may or may not be conserved: A where Δk = k k z = i ω d k c A e i Δk z Conversion will be most efficient if Δk = ω c n(ω ) = ω c n(ω ) k k n( ω ) = n( ω ) This is the phase-matching condition for SHG k 3

14 /3/7 Phase-matching Second-Harmonic Generation The phase-matching condition for SHG: n( ω) = n( ω) Unfortunately, dispersion prevents this from ever happening Refractive index ω Frequency ω First Demonstration of Second-Harmonic Generation P.A. Franken, et al, Physical Review Letters 7, p. 8 (96) The second-harmonic beam was very weak because the process wasn t phase-matched. 4

15 /3/7 First demonstration of SHG: The Data The actual published result The second harmonic Input beam Note that the very weak spot due to the second harmonic is missing. It was removed by an overzealous Physical Review Letters editor, who thought it was a speck of dirt. SHG without phase-matching Non-depleted pump approximation: treat A as constant A z = i ω d k c A A e i Δk z ( ) ( L) = i ω d k c A L ei Δk L i Δk L Convert to intensity ε n c I z I ( L) = ε n c ( ) = Integrate: I ω d ε n n c I 3 L sinc ( Δk L / ) A I = ε n c A d ( L) = i ω k c A ( ) ω d sin Δk L / n c L Δk L / As a function of L and fixed Δk >: I ( L) = Yield oscillates: Period = coherence length Amplitude proportional to L coh = π / Δk max I L ( ) / Δk e i Δk z dz ω d ε n n c I 4 3 ( Δk sin Δk L / ) 5

16 /3/7 Light created in real crystals Far from phase-matching: SHG crystal Input beam Output beam Closer to phase-matching: SHG crystal Input beam Output beam Note that SH beam is brighter as phase-matching is achieved. Phase-matching Second-Harmonic Generation using birefringence Birefringent materials have different refractive indices for different polarizations. Ordinary and Extraordinary refractive indices can be different by up to. for SHG crystals. We can now satisfy the phase-matching condition. Put the highest frequency on the lowest index: for negative uniaxial use the extraordinary polarization for ω and the ordinary for ω: n e (ω,θ) = n o (ω ) Refractive index ω n e n o Frequency ω n e depends on propagation angle, so we can tune for a given ω. Some crystals have n e < n o, so the opposite polarizations work. 6

17 /3/7 Real crystal dispersion data Best resource: refractiveindex.info Others: crystal manufacturers, Handbook of Optics Example: β-bbo = barium borate, BaB O 4 n.84 o = λ n e = λ λ is in micrometers.7.65 n o n e < n o everywhere, so we need to angle tune.6.55 n e..5. λ (µm) Types of phase matching Type : ω on low index (n e ) = ω on high (n o ) ω c n o ( ω ) n e ( ω,θ ) Opposite polarizations (χ () tensor allows this) Type : ω on low index (n e ) Δk = ω c n o ( ω ) ω c n e ( ω,θ ) ( ) Δk = ω c n o ( ω ) + ω c n e ( ω,θ ) ω c n e ( ω,θ ) Project E equally on both axes (n o and n e ) Type 3: non-critical or 9 o phase matching Temperature-tuned Δk = ω c n o ( ω,t ) n e ( ω,9,t ) Only for particular crystals and wavelengths ( ) 7

18 /3/7 Practical issues Phase matching bandwidth Type has more BW, choose L of crystal Group velocity walk-off (for short pulses) Angular acceptance Birefringent beam walk-off Strength of nonlinearity Crystal damage threshold Thermal stability: typically angle-tuned, temperature stabilized Available size of crystals, $$ 8