Pockels Effect

Guoqing (Noah) Chang, October 15, 2015 Nonlinear optics: a back-to-basics primer Lecture 4: three-wave mixing

2nd order nonlinearity: optical rectification ω ω ω Example: three wave mixing at three different frequencies, 1 , 2 , 3 and ω3 = ω1 +ω2

(2) (2) Pi (ω3 ,ω1,ω2 ) = ε 0 χijk E j (ω1)Ek (ω2 ) ω1 ∗ ω optical rectification 0 = ω + (−ω) Ek (−ω) = Ek (ω) 2

∗ (2) (2) ω3 Pi (0,ω,−ω) = ε 0 χijk E j (ω)Ek (ω)

Positive and negative charges in the Response of a medium with 2nd- Optical rectification signal (top nonlinear medium are displaced, order nonlinearity to a single trace) and laser power (lower causing a potential difference frequency light wave. It trace). It can be used for THz between the plates. generates a DC term. wave generation.

Geoffrey New, Introduction to nonlinear optics, chapter 1 2 2nd order nonlinearity: Pockels effect

(2) (2) DC Pockels effect ω = ω + 0 Pi (ω) = 2ε 0 χijk E j (ω)Ek

(1) (1) Note that Pi (ω) = ε 0 χij E j (ω)

(1) (2) DC (2) DC Di = ε 0[δij + χij + 2χijk Ek ]E j (ω) = ε 0[ε ij + 2χijk Ek ]E j (ω) Pockels effect

In linear optics as we have reviewed in lecture 1, an anisotropic medium is associated with an index ellipsoid.

2 = = nnnxyo nnze= ε xx 0 0nx 00 2 2 22 0ε 00= n 0 xyz yy y ++= Index ellipsoid for 2221 ε 2 uniaxial crystal 0 0zz  00nz nnnooe

Pockels effect leads to a change for the dielectric constant, and therefore modifies the index ellipsoid.

(2) DC ∆ε ij = 2χijk Ek

3 2nd order nonlinearity: Pockels effect

−1 (ε ) ij is more convenient to calculate the index ellipsoid.

−2  −2  nx 0 0  nx 0 0 x −1  −2   −2   (ε ) = 0 n 0 = ij  y  [x y z] 0 ny 0 y 1  0 0 n−2   −2    z   0 0 nz z

−1 DC −1 ∆(ε )ij ≡ rijk Ek ∑ (ε ) ij is symmetric and thus the order of the k indices ij is irrelevant, a similar contracted notation can be introduced:

−1 1 ∆ ε = ∆ = DC = − = − ( ) p ( 2 ) p ∑ rpn En (n 1 3, p 1 6) n n

(2) DC From ∆ ε ij = 2 χ ijk E k we can find

2χijk (ω,ω,0) rpn = rijk = rjik = − ε iiε jj 4 Pockels effect: linear electro-optic effect Take KDP crystal as an example:  0  y DC    0 0 0  E =  0    x 0 0 0 V / L   z  0 0 0  −1 1 r V rpn =   Assume an ordinary wave polarized in the x- 63 ∆(ε )6 = ∆( )6 = r41 0 0  y plane propagating in the +z-direction. n2 L   0 r41 0 Without applying a DC field, the light wave   1 1 r V maintain its polarization. Now we add a DC ∆ = ∆ = 63  0 0 r63  ( 2 )xy ( 2 ) yx voltage V between the exit and entrance n n L surfaces. The crystal’s length is L. y ' x2 + y 2 z 2 2xyr V y Now the index + + 63 =1 ellipsoid becomes 2 2 no ne L

The plane through the origin and perpendicular to the propagation direction (+z) cut the new ellipsoid into an ellipse. The lengths of x its semi-axes correspond to the 1 refractive indices of the two 2 n± ≅ n0 (1± n0 r63V / L) ' orthogonally polarized waves: 2 x Pockels effect cause a refractive index change linearly proportional to the electric field. So it has another name—linear electro-optic effect. r 63 is called the linear electro-optic coefficient. 5 Pockels effect: application examples

y ω ' Pockels cell: Electrically control the direction of an optical beam. E( ) y

As the voltage varies with time, the optical power on the Voltage “off” path x changes accordingly. In this scenario, the device functions as an EO x' intensity modulator. That is, we can use a low 1 2 frequency RF or n± ≅ n0 (1± n0 r63V / L) microwave signal to 2 http://www.photonics.com/Article.aspx y modulate an optical signal. ?AID=25650 E(ω) y' EO phase modulation: phase modulation of an optical beam can be achieved if we align the input light field with its polarization along the x’ axis. x The Pockels effect provides a means to interact a low frequency electromagnetic wave (RF, microwave, or THz wave) with an ' optical wave. EO modulator enabled by this effect is one of the x most important devices in modern optics. 6 Who is Pockels? Friedrich Carl Alwin Pockels (1865–1913) was a German physicist. From 1900 to 1913 he was professor of theoretical physics at the University of Heidelberg. In 1893 he discovered that a steady electric field applied to certain birefringent materials causes the refractive index to vary, approximately in proportion to the strength of the field. This phenomenon is now called the Pockels effect. -- From Wiki

Friedrich Carl Alwin Pockels (1865–1913) 7 Difference frequency generation (DFG) ω dA1 2 1deff ∗ j∆kz ω1 = − j A3 A2e dz n(ω )c (1) ω2 = ω3 − ω1 1 0 ω ω 3 dA2 2 2deff ∗ j∆kz = − j A3 A1 e (2) dz n(ω2 )c0 ω dA3 2 3deff − j∆kz = − j A1 A2e (3) dz n(ω3 )c0 We assume that A 3 is nondepleted (i.e., constant) and ∆ k = 0 . Differentiating both sides of Eq. (2) and using Eq. (1)(3), we can get 2 2 d A 4ω1ω2deff 2 2 = ≡ κ 2 2 2 A3 A2 A2 dz n(ω1)n(ω2 )c0 4ω ω d 2 A1(z) κ 2 = 1 2 eff 2 2 A3 n(ω1)n(ω2 )c0 Amplitude

1 κz −κz A1(z) = A1(0)(e + e ) A (z) 2 2 j n(ω )ω A A (z) = − [ 1 2 ]1/ 2 3 A∗ (0)(eκz − e−κz ) 2 1 z 2 n(ω2 )ω1 A3 DFG: optical parametric generation, amplification, oscillation 1 κz For κ z >> 1 , A 1 ( z ) = A 1 ( 0 ) e corresponding to an exponential growth. 2 So such a DFG process can be used to amplify a weak field. But it is completely different from a laser amplifier based on stimulated emission.

ω1 ω1 "signal" ω1 ω3 ω3 ω 2 "idler" Laser amplifier involves electron transition and the amplifier stores energy by absorbing pump photons. Signal is amplified by Optical Parametric By convention: stimulated emission. Only work at certain wavelengths limited by Generation (OPG) ωsignal > ωidler gain materials.

ω1 ω1 ω1 ω ω3 3 ω2 ω2

During optical parametric amplification (OPA) , the crystal does mirror mirror not absorb photons and no energy is stored in the crystal. Signal is amplified by wave mixing. Working wavelength is limited by Optical Parametric Oscillation (OPO) crystal transmission. How a laser works

Continuous wave output Back Output mirror Ip mirror power gain medium Time

Continuous wave (CW) laser power jω0t ∗ − jω0t jω0t E(t) = 2Ecosω0t = Ee + E e = Ee + c.c. ω0 Frequency . Gain medium - Enable stimulated emission to produce identical copies of photons - Determine the light wavelength . Pump - Inject power into the gain medium - Achieve population inversion . Resonator cavity - Make light wave oscillating to efficiently extract energy stored in the gain medium - Improve directionality and color purity of the light How an ultrafast laser works

Ultrashort Back Output pulse Output mirror Ip mirror Mode gain medium locker Ultrafast laser t t jω0t ∗ − jω0t jω0t E(t) = 2A(t)cosω0t = A(t)e + A (t)e = A(t)e + c.c.

ω → ω j 0t A(t) A( ) A(t)e → A(ω −ω0 )

Longer pulse corresponds to narrower spectrum. time frequency

Shorter pulse corresponds to broader spectrum. time frequency 11 How an ultrafast laser works

Ultrashort Back Output pulse Output mirror Ip mirror Mode gain medium locker Ultrafast laser t t

10ps Nd:glass Nd:YLF Nd:YAG τ Dye S-P Dye Diode

1ps t CW Dye Color Center World shortest pulse: 67 attoseconds. The 100fs Cr:LiS(C)AF center wavelength is 20 nm. It is generated CPM Dye Er:fiber Nd:fiber Cr:YAG by high harmonic generation. SHORTEST PULSEDURATION Cr:forsterite 10fs w/Compression K. Zhao et al., “Tailoring a 67 attosecond pulse Ti:sapphire through advantageous phase-mismatch,” Opt. Lett. 37, 3891 (2012) 1965 1970 1975 1980 1985 1990 1995 2000 2005 YEAR 12 Ultrafast laser: the 4th element—mode locker

Ultrashort Back Output pulse Output mirror Ip mirror Mode gain medium locker Ultrafast laser t t

t Ultrafast laser: the 4th element—mode locker

Ultrashort Back Output pulse Output mirror Ip mirror Mode gain medium locker Ultrafast laser t t

t The metric system

We meet with both small and large numbers when we talk about femtosecond pulses—they are incredibly short with the powers and intensities incredibly high.

Prefixes: Small Big Milli (m) 10-3 Kilo (k) 10+3 Micro (µ) 10-6 Mega (M) 10+6 Nano (n) 10-9 Giga (G) 10+9 Pico (p) 10-12 Tera (T) 10+12 Femto (f) 10-15 Peta (P) 10+15 Atto (a) 10-18 Exa (E) 10+18

Adapted from Rick Trebino’s course slides 15 Some physical quantities

Average power:

Repetition rate:

Pulse energy: (Average power = Rep-rate X Pulse energy)

Pulse width (duration):

Peak power: (Peak power = Pulse energy / duration)

(peak) Intensity:

If an optical beam with 1 PW peak power is focused to 1 um2 area, the peak intensity is 1023 W/cm2.

16 The Gaussian pulse

For almost all calculations, a good first 1 approximation for any ultrashort pulse is that 0.8 the pulse has a pulse envelope of Gaussian shape. t 2 = − 0.6 A(t) A0 exp( 2 ) 2T0 100 fs 0.4 2 Intensity 2 t ∝ − The intensity is: I(t) A0 exp( 2 ) 0.2 T0 Intensity full-width-half-maximum 0 -200 -100 0 100 200 1/ 2 Time [fs] TFWHM = 2(ln 2) T0 1 The optical spectrum 0.8 2 2 2 1/ 2 T0 ω A(ω) = (2πT ) A exp(− ) 0.6 0 0 2 0.4 Spectral bandwidth Spectralintensity 0.2 2(ln 2)1/ 2 ∆ω = FWHM 0 T0 1 1.01 1.02 1.03 1.04 1.05 1.06 Wavelength [µm] Phase matching for oe SHG in BBO

θ = ° = ° ne (0.515um, 23.3 ) no (1.03um) ne (0.4um,θ = 29.2 ) = no (0.8um)

1.7

1.68

1.66 n o n (29.2o) n (23.3o) 1.64 e e

1.62

1.6 Refractive index

1.58 n e 1.56

1.54 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 wavelength [µm] 18 Phase matching bandwidth for a pulse

1 For type I SHG from a femtosecond pulse centered at 1 . 03 um , the center wavelength 0.8 experiences a perfect phase matching if the 0.6 BBO crystal is cut at θ = 23 . 3 °. That is, the coherence length at the center wavelength is 0.4 infinite.

Spectralintensity 0.2 How about the coherence length at other wavelength at this angle? 0 1 1.01 1.02 1.03 1.04 1.05 1.06 µ π πc0 Wavelength [ m] Lc = = ° ∆k 2ω[no (ω) − ne (2ω,23.3 )] 5

4 Roughly speaking, 2 mm BBO corresponds to 60 nm phase-matching bandwidth, and 1 mm 3 BBO 220 nm. 2 In general, using thinner crystal leads to broader 1 phase matching bandwidth at the expense of coherencelength [mm] conversion efficiency. 0 0.9 0.95 1 1.05 1.1 1.15 wavelength [µm] Linear propagation of a pulse

22 j(ω0t−k0 z) 1 ∂∂P E(z,t) = A(z,t)e Fourier ω0n(ω0 ) ()∇−2 E =µ k = 22 0 2 transform 0 c ct∂∂ t -jk0 z 0 0 E(z,ω) = A(z,ω −ω0 )e

Take the Fourier transform of both sides of the wave equation:

P(ω) = ε χ (1) (ω)E(ω) 2 0 2 2 2 ω 2 n (ω)ω (∇ + )E = −ω µ P(ω) ∇2 + ω = 2 0 ( 2 )E( ) 0 c0 c0 Plane wave propagating in +z direction: -jk0 z E(z,ω) = A(z,ω −ω0 )e 2 2 d 2 d A dA 2 2 + ω ω = − 2 jk +[k (ω) − k ]A = 0 [ 2 k ( )]E(z, ) 0 dz 2 0 dz 0 dz n(ω)ω k(ω) = c d 2 A dA 0 − + ω + ω − = 2 2 jk0 [k( ) k0 ][k( ) k0 ]A 0 dz dz 2 ≈ Slowly varying amplitude d A dA 2k0 << 2k approximation dz 2 0 dz dA = − j[k(ω) − k ]A dz 0 Group velocity Vs phase velocity

2 dA(z,ω −ω ) dk 1 d k 2 0 k(ω) = k + (ω −ω ) + (ω −ω ) +... = − j[k(ω) − k0 ]A(z,ω −ω0 ) 0 0 2 0 dω ω=ω 2 dω dz 0 ω=ω0

ω = + (1) ω −ω (1) dk Let’s first take a look at the effect of the first two terms k( ) k0 k ( 0 ) k = dω ω=ω0 (1) dA(z,ω −ω0 ) (1) -jk (ω−ω0 )z = − jk (ω −ω )A(z,ω −ω ) A(z,ω −ω0 ) = A(0,ω −ω0 )e dz 0 0 (1) -jk0 z -jk (ω)z -j[k0 +k (ω−ω0 )]z E(z,ω) = A(z,ω −ω0 )e = A(0,ω −ω0 )e = A(0,ω −ω0 )e z jω0 (t− ) ω − z V E(z,t) = A(0,t − k (1) z)e j( 0t k0 z) = A(0,t − )e p Vg

Group velocity: travelling speed of the pulse envelope. 1 dk c = = = 0 Vg (1) 1/ k dω ω=ω dn(ω) 0 n(ω ) +ω 0 0 dω ω=ω0

Phase velocity: travelling speed of the carrier wave under the pulse envelope. ω c = 0 = 0 Electric field and pulse envelope Vp in time domain k0 n(ω0 ) 21 Group velocity Vs phase velocity

In vacuum Unrealistic

Most common Possible case

Unrealistic Unrealistic

22 Adapted from Rick Trebino’s course slides Time domain description dA(z,ω −ω ) 0 = − (1) ω −ω ω −ω ∂A(z,t) (1) ∂A(z,t) jk ( 0 )A(z, 0 ) + k = 0 dz Fourier transform ∂z ∂t Now take into account nonlinearity:

Coupled wave equations for Coupled wave equations for three-wave mixing of three-wave mixing of CW ultrashort pulses waves

∂A (z,t) 1 ∂A (z,t) 2ω1deff ∗ j∆kz dA 2ω1deff ∗ ∆ 1 1 1 j kz + = − j A3 A2e = − j A3 A2e ∂z Vg,1 ∂t n(ω1)c0 dz n(ω1)c0 ω dA 2ω2deff ∂A (z,t) 1 ∂A (z,t) 2 2deff ∗ j∆kz 2 = − j A A∗e j∆kz 2 + 2 = − j A A e 3 1 ∂ ∂ ω 3 1 dz n(ω2 )c0 z Vg,2 t n( 2 )c0

dA 2ω d ∂A (z,t) 1 ∂A (z,t) 2ω3deff − j∆kz 3 = − j 3 eff A A e− j∆kz 3 + 3 = − j A A e 1 2 ∂ ∂ ω 1 2 dz n(ω3 )c0 z Vg,3 t n( 3 )c0

1 = = V ≠ V ≠ V Vg,i (i) ,i 1,2,3 In general, g ,1 g , 2 g ,3 . It leads temporal walk-off of k these three pulses. 23 Numerical example: SFG between 1.5 um and 0.8 um pulses Saturation of the SFG pulse power due to temporal walk-off

120

100

60 P (mW)P 40

0 0 1 2 3 4 5 6 7 8 Crystal Length(mm) 25 Group-velocity mismatch: take SHG as an example

Fundamental pulse and SH pulse propagate at different group velocity, which introduce a walk-off between them and decrease the conversion efficiency.

1 1 GVM ≡ − Vg (2ω) Vg (ω) SH pulse

Fundamental pulse at input

SH pulse left behind

Fundamental pulse at exit Group-velocity mismatch results in a longer SH pulse. 26 20 W, 40 fs green light via SHG in 0.5 mm BBO crystal

27 Take-home message

. Linear EO effect is a second-order nonlinear effect.

. OPA—a special case of DFG—can be used to amplify weak signal.

. Ultrafast lasers generate femtosecond pulses.

. Thinner crystals are required to achieve broadband phase matching for nonlinear phenomena involving shorter pulses.

. Group velocity mismatch reduces the conversion efficiency and shapes the generated pulse.

28 Suggested reading

Three wave mixing

-- Geoffrey New, Introduction to nonlinear optics, chapter 1,2 -- George Stegemann and Robert Stegemann, Nonlinear optics, chapter 6 -- Robert Boyd, Nonlinear optics, chapter 2

Dispersion and ultrashort pulse

-- Geoffrey New, Introduction to nonlinear optics, chapter 6