Applied Nonlinear Optics. David C. Hutchings Dept. of Electronics and Electrical Engineering University of Glasgow

Transcription

1 Applied Nonlinear Optics David C. Hutchings Dept. of Electronics and Electrical Engineering University of Glasgow

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3 Preface This course will cover aspects of birefringence, the electro-optic effect and optical frequency conversion. It will be necessary to have a good grounding in electro-magnetic theory and optics prior to this course. Recommended textbooks for this material are Shen [], Zernike and Midwinter [] and Yariv [3]. Like the bulk of the literature in nonlinear optics, these books do not always employ a consistent SI notation. I would recommend Butcher and Cotter [4], although not as applied as the above references, for its consistent notation (SI units) and formalism. iii

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5 Chapter Birefringence. Dielectrics (revision) In a vacuum, Gauss theorem states that the closed surface integral of the normal component of the electric field to the surface is proportional to the total charge enclosed by the surface, E ds = q ε 0 = ε 0 ρdv, (.) where ρ is the charge density. The divergence theorem can be used to change this surface integral to a volume integral, E ds = EdV, (.) which allows us to relate two volume integrals. Now since these are over the same region which we can arbitrarily choose, the integrands must be equal, i.e., E = ρ ε 0. (.3) Hence we have used the divergence theorem to transform the original integral equation [Eq. (.)] for the electric field into a differential one. Let us now address the issue of an electric field in a dielectric medium. The electric field will induce dipoles in the medium, for example by distortion of the electron clouds or aligning polar molecules preferentially along the direction of the field. In an extended medium and a uniform field the average charge density due to these induced dipoles is zero except at the surfaces. This is because the charge on one end of an induced dipole will be neutralised by the opposite charge on the end of an adjacent dipole. At the surface there is no adjacent dipole to cancel the charge. Thus additional polarisation charges have been generated at the surface which must be accounted for. Similar polarisation charges are generated in the case of a nonuniform field. A polarisation field P can be defined equal to the dipole moment per unit volume. If there are N dipoles per unit volume consisting of charges +q and q separated by r then

6 CHAPTER. BIREFRINGENCE P = Nqr. If we assume for a moment that the induced charge separation r E, then the polarisation field is proportional and parallel to the electric field. Conventionally this is written in SI units as P = ε 0 χe, where the dimensionless constant of proportionality χ is called the optical susceptibility. It can be shown that the effective charge density due to the polarisation of a medium is given by, ρ p = P. Inserting this into the differential form of Gauss theorem gives, E = ε 0 (ρ + ρ p ) = ε 0 (ρ P), (.4) where ρ denotes the free (not polarisation) charge density. This can be rearranged to give (ε 0 E + P) = ρ. For convenience a new field is introduced at this point equal to the quantity inside the bracket, D = ε 0 E + P, conventionally known as the electric displacement. Note that unlike E and P the electric displacement is not immediately related to a physical quantity, but allows various electro-magnetic relations to be written in a far simpler form. However, D can be thought of as a vacuum field, consisting of the contribution to the electric field with the effect of the dielectric medium subtracted. The simpler form of Eq. (.4) is D = ρ and is one of the set of differential relations known as Maxwell s equations. Using the susceptibility to substitute for the polarisation field provides, D = ε 0 ( + χ)e = ε 0 ε r E, (.5) where ε r = + χ is known as the dielectric constant or relative permittivity. Note that the vacuum values of these dimensionless quantities are χ = 0 and ε r =. The divergence theorem and Stokes theorem are employed in deriving the remainder of Maxwell s equations. We shall state these without proof here as we will be making frequent use of them throughout this course. In SI units they are given by: D = ρ, B = 0, E = B t H = j + D t The current is related to the electric field by the conductivity, j = σe. In this course the usual case will be of zero current j = 0 and non-magnetic material B = µ 0 H.. The Dielectric Tensor In an anisotropic media the dipoles may be constrained so their direction is different to that of the electric field. One can think of numerous mechanical analogies where the motion of a mass This corresponds to the simplest case of a medium which is linear and isotropic. Most of the course considers the extension of this to cases where the two vectors are (i) not parallel or (ii) not simply proportional.,.

7 .. THE DIELECTRIC TENSOR 3 is constrained by e.g. ramps or string so that acceleration is not parallel to the force. In such systems, the constraint reduces the symmetry. If the dipoles are not parallel to the field, then the polarisation P and the electric displacement D are also not parallel to the field E. Hence Eq. (.5) cannot be written with the relative permittivity as a scaler quantity. To generalise Eq. (.5) to anisotropic media (such that D and E can have different directions), the scaler dielectric constant ε r is replaced with a second rank tensor, ε 0 In a more compact notation, D x D y D z = ε xx ε xy ε xz ε yx ε yy ε yz ε zx ε zy ε zz E x E y E z. (.6) D i = ε 0 ε i j E j. (.7) j=x,y,z We can prove that the tensor ε i j is symmetric using energy conservation. The energy density of an electric field in a dielectric is, W e = E D = i, j E i ε i j E j. (.8) Differentiating gives the power flow into a unit volume, W e t = i j ( ) Ei ε i j t E E j j + E i t. (.9) We can also calculate power flow using the Poynting vector which is the power flow across a unit area, S = E H. Using the divergence theorem gives the power flow per unit volume, S = (E H) = E H H E. (.0) We can substitute for the vector curls using Maxwell s equations and assuming no currents j = 0 gives a power flow per unit volume, E D t + H B t. (.) The first of these terms relates to the rate of change of energy of the electric field, W e t = E D t E j = ε i j E i i j t. (.) Now these two forms of power flow must be the same and hence Eqs. (.9) and (.) must be equal. This can only be the case if ε ji = ε i j and the dielectric tensor is symmetric having a total of 6 different elements. Note that we have not yet specified the axes for our cartesian co-ordinate system {x,y,z}. It is always possible to choose a set such that the (symmetric) dielectric tensor is diagonal. The

8 4 CHAPTER. BIREFRINGENCE basis of this is that a symmetric matrix is specific case of a Hermitian matrix and has pure real eigenvalues and eigenvectors. Thus if we change to a new co-ordinate system where the axes are parallel to these eigenvectors, the dielectric tensor becomes diagonal, ε 0 D x D y D z = ε x ε y ε z E x E y E z, (.3) where ε x = ε xx etc. The set of axes in this case is known as the principal dielectric axes, which may be different from the usual crystal axes. The dielectric tensor may be further simplified by considering the crystal symmetry. There are three distinctive cases summarised in table.. It can be seen that generally the higher the degree of symmetry, the lower the degree of birefringence. In most of the above crystal systems the principal dielectric axes correspond to the usual cartesian crystalline axes. The two exceptions to this are the monoclinic and triclinic crystal systems. No birefringence Uniaxial birefringence ( optic axis) Biaxial birefringence ( optic axes) Isotropic Cubic Hexagonal Tetragonal Trigonal Orthorhombic Monoclinic Triclinic ε ε ε ε x ε x ε z ε x ε y ε z Table.: The form of the dielectric tensor for the various crystal symmetries indicated..3 EM Wave propagation Now consider the propagation of a monochromatic electro-magnetic plane wave through the medium. Such a wave has electric and magnetic fields given by, E = [ E 0 e i(ωt k r) + E 0e i(ωt k r)], (.4) H = [ H 0 e i(ωt k r) + H 0e i(ωt k r)]. (.5) The wavevector k = nωŝ/c where ŝ is a unit vector in the direction of propagation of the wave. The phase velocity is given by v p = cŝ/n. Inserting these into Maxwell s equation E =

9 .3. EM WAVE PROPAGATION 5 B/ t with a nonmagnetic medium (B = µ 0 H) gives, H 0 = ωµ 0 k E 0 = n µ 0 cŝ E 0. (.6) Thus the direction of the magnetic field amplitude is perpendicular to both the direction of propagation ŝ and electric field E 0. Similarly we can use the Maxwell equation H = j + D/ t with j = 0 to obtain, D 0 = ω k H 0 = n c ŝ H 0. (.7) Hence the electric displacement is perpendicular to the direction of propagation ŝ and the magnetic field H 0. However, as we have seen for anisotropic media, E 0 and D 0 are not necessarily parallel. We also note that the power flow is given by the Poynting vector S = E H which is perpendicular to E 0 and H 0. The Poynting vector also provides the direction for the group velocity. Fig.. shows the relative geometry of these vectors in an anisotropic medium. This illustrates that the propagation direction ŝ and the Poynting vector are not necessarily parallel and hence the phase and group velocities may also not be parallel. E D H S=ExH k v v g p Figure.: Relative geometry of the field vectors and the phase and group velocities for an em-wave in an anisotropic medium. The magnetic field H is directed out of the page. Combining Eqs. (.6) and (.7) gives, D 0 = n µ 0 c ŝ ŝ E 0 = n ε 0 [E 0 ŝ(ŝ E 0 )]. (.8) Consider just one component of this expression and introduce the dielectric tensor, [ ] D 0i = n D0i ε 0 s i (ŝ E 0 ), ε i ( D 0i = ε 0 s i (ŝ E 0 ) ) ε i n. (.9)

10 6 CHAPTER. BIREFRINGENCE Since D 0 and ŝ are perpendicular, D 0 ŝ = 0. This provides the relation, ( n s ) i = 0, (.0) i ε i which is known as Fresnel s equation. This is quadratic in the square of the refractive index (n ) and therefore provides two possible solutions given a propagation direction ŝ. Hence the origin of the name of this phenomenon; birefringence which means literally double refraction. It can be shown that these two roots for the refractive index correspond to orthogonal polarisations. Let the two roots be n and n with electric displacement amplitudes D 0 respectively. Using Eq. (.9), the scaler product of these amplitudes can be written, D 0 D 0 = ( ε 0 n n ŝ E 0 ) i = (ε 0n n ŝ E 0 ) n n i s i (n /ε i )(n /ε i ) ( n n /ε i n n /ε i, and D 0 ) s i, (.) where partial fractions have been employed to obtain the final form. Now Fresnel s equation (.0) states that the summation over each of the terms is zero. Hence D 0 D 0 = 0 and therfore are mutually perpendicular. D 0 and D 0.4 The Index Ellipsoid Using Fresnel s equation based on the propagation direction is rather cumbersome when analysing birefringence. A much easier method is based on the direction of the electric displacement. The energy density of an electric field in a birefringent dielectric is, ( ) W e = E D = D x + D y + D y. (.) ε x ε y ε y Substituting ε r = n for the various directions and setting D x / W e = x, etc. gives the following, x n x + y n + z y n z =. (.3) This equation represents an ellipsoid and is conventionally referred to as the index ellipsoid or optical indicatrix. The intercepts of this surface with the cartesian axes are at ±n x for the x-axis, etc. This ellipsoid can be used to find the two allowed directions for the polarisation and their associated refractive indices. The remaining discussion will focus on uniaxial birefringence where n y = n x. This means that the index ellipsoid will have cylindrical symmetry round the z-axis (optic axis). There are two distinct cases of uniaxial birefringence; n z > n x known as positive unixial birefringence and

11 .4. THE INDEX ELLIPSOID 7 n z < n x known as negative unixial birefringence. These are illustrated in Fig... The use of the index ellipsoid is as follows. The direction of propagation k is drawn from the origin and makes an angle of θ to the optic axis. The intersection of the plane normal to the direction of propagation and the index ellipsoid generates an ellipse with semi-axes a and b. For a uniaxial crystal, the semi-axis a (minor semi-axis in the case of a positive uniaxial crystal, major semiaxis in the case of a negative uniaxial crystal) always lies in the xy-plane and therefore has a length a = n x = n o independent of the angle θ; a ray with this polarisation is called the ordinary ray. The length of the other semi-axis is dependent on the angle θ, b = n e (θ); a ray with this polarisation is called the extraordinary ray. When the direction of propagation is parallel or perpendicular to the optic axis, the refractive index of the extraordinary ray can be written down immediately, n e (θ = 0) = n z = n e and n e (θ = π/) = n x = n o. For the general case, we decompose b = n e (θ) into components parallel and perpendicular to the optic axis, x + y = (n e (θ)cosθ) and z = n e (θ)sinθ, and insert into the equation for the index ellipsoid to give for uniaxial crystals, n e(θ) = cos θ n + sin θ o n. (.4) e (a) z k (b) z b θ x x a n >n z x n z <n x Figure.: The index ellipsoid or optical indicatrix for (a) a positive uniaxial crystal and (b) a negative uniaxial crystal. The y-axis is directed into the page. If we re-examine the derivation of Eq. (.9), it can be seen that if n = ε i for i = x, y or z then ŝ E 0 = 0 and the electric field amplitude E 0 is perpendicular to the propagation direction and hence parallel to the electric displacement amplitude D 0. For the ordinary ray in a uniaxial crystal this is satisfied for any propagation direction since n 0 = ε x = ε y. Hence the derivation of the term ordinary since the phase and group velocities are parallel as in a non-birefringent medium. For the extraordinary ray, this condition only occurs for propagation parallel or perpendicular to the optic axis. In general, the phase and group velocities are not parallel. Calcite (crystalline calcium carbonate) is a common uniaxial birefringent crystal and is useful in demonstrating the properties of birefringence. Place a piece of calcite over a piece of

12 8 CHAPTER. BIREFRINGENCE paper with text on it and you will see a double image. If you have a polariser, place it over the calcite and rotate it. You should find at some point that only one image will be visible. Rotate the polariser by 90 and only the other image will be visible. This demonstrates that the two images correspond to orthogonal polarisations. Now remove the polariser and rotate the calcite. You should find that one of the images remains stationary and the other rotates about it. The stationary image corresponds to the ordinary ray where the phase and group velocities are parallel and correspondingly the rotating image corresponds to the extraordinary ray..5 Wave Plates So far we have only considered light to be exclusively of ordinary or extraordinary polarisation. What happens in the general case where the light is some combination of these polarisation states? Suppose we initially have a linear polarisation D in 0 = D 0(ôcosφ + êsinφ) where ô and ê are unit vectors parallel to the ordinary and extraordinary polarisation directions respectively. Now suppose the crystal thickness d is such that the difference in optical path length between the ordinary and extraordinary rays is an odd integer of half-wavelengths (n e (θ) n o )d = (N + )λ/. A crystal of this thickness is termed a half-wave plate. On exit the ordinary and extraordinary rays will be out of phase, D out 0 = D 0 (ôcosφ êsinφ). It can be seen that this is equivalent to the transform φ φ and corresponds to flipping the polarisation angle around the optic axis. In particular for φ = 45, the output (linear) polarisation is perpendicular to the input. Now let us consider a crystal of thickness such that the optical path length difference is an odd number of quarter wave-lengths (n e (θ) n o )d = (N + )λ/4 known as a quarter wave plate. Now the ordinary and extraordinary rays will be π/ out of phase with each other on output from the crystal. Again if we consider the particular case of an input linear polarisation at 45 to the optic axis D in 0 = D 0(ô ê)/ then the output polarisation will be D out 0 = D 0 (ô ± iê)/. The positive and negative cases correspond to the separate cases of Nλ + λ/4 and Nλ + 3λ/4 optical path length differences. In this case we have generated circularly polarised light from a linearly polarised input. A quarter wave plate can also perform the reverse, changing circularly polarised light to linear (since two consecutive quarter wave plates make a half wave plate). For simplicity the common phase factor has been omitted since the polarisation state only depends on the relative phase of the polarisation components

13 Chapter The Electro-optic Effect. The Electro-optic tensor The change in refractive index with a DC electric field is known as the electro-optic effect. In this chapter we will consider the linear electro-optic effect or Pockel s effect where n E. There also exists the quadratic electro-optic effect or Kerr effect which is associated with higher order effects considered later in this course. The principal application of the electro-optic effect is in optical modulators where we use an external influence (applied voltage) to change the optical properties of a material. The general form of the optical indicatrix if the axes do not necessarily correspond to the principal dielectric axes is, x ε xx + y ε yy + z ε zz + yz ε yz + xz ε xz + xy ε xy =, (.) i j or in shorthand notation, i, j=x,y,z ε i j =. Now conventionally the electro-optic coefficient r relates the change in /ε (i.e. /n ) to the electric field E. Generalising this to anisotropic crystals, ( ) ( ) ( ) = ε i j ε i j E ε i j E=0 = r i jk E k. (.) k The electro-optic coefficient r i jk is a third rank tensor as it relates a second rank tensor (/ε i j ) to a vector (E). It has 7 elements but only 8 are independent since ε ji = ε i j and therefore r jik = r i jk. This symmetry is employed to write the electro-optic tensor in a contracted notation where the i j subscripts are replaced by: xx =, yy =, zz = 3, yz = zy = 4, xz = zx = 5 and xy = yx = 6 and the k subscripts by: x =, y = and z = 3. This allows the electro-optic tensor 9

14 0 CHAPTER. THE ELECTRO-OPTIC EFFECT to be written as a 6 3 matrix: (/ε ) (/ε ) (/ε 3 ) (/ε 4 ) = (/ε 5 ) (/ε 6 ) r r r 3 r r r 3 r 3 r 3 r 33 r 4 r 4 r 43 r 5 r 5 r 53 r 6 r 6 r 63 E E E 3. (.3) As in the case of the dielectric tensor, symmetry considerations provide information as to which electro-optic coefficient tensor elements are non-zero and independent. An important property is that materials with inversion symmetry exhibit no electro-optic effect. Consider Eq. (.) under the application of the inversion operator. If the material has inversion symmetry then the material parameters ε i j and r i jk are unchanged. However, under inversion E k E k. Hence k r i jk E k = k r i jk E k for any specified E which requires that all the electro-optical coefficients are zero, r i jk = 0. For other symmetry classes, group theory can be employed to investigate the form of the electro-optic tensor, as shown in table... Examples.. KDP Potassium dihydrogen phosphate KH PO 4 (known as KDP) belongs to the symmetry class 4m and hence exhibits uniaxial birefringence with the principal dielectric axes being the same as the conventional crystal axes. The electro-optic tensor has three non-zero components, two of which are equal. Including these in the modified optical indicatrix gives, x n o + y n + z o n + r 4 E x yz + r 4 E y xz + r 63 E z xy =. (.4) e Now let us consider the case where the electric field is parallel to z such that E x = 0 and E y = 0 so the 4th and 5th terms can be ignored. Now as we stated in the previous chapter on birefringence, by selecting a new co-ordinate set, this equation can be transformed so only the diagonal components remain. In this case we shall employ a cartesian co-ordinate system that is rotated by 45 about the z-axis. Thus we have x (x y )/, y (x + y )/ and z z. On inserting these in Eq. (.4) we obtain for the new optical indicatrix, ( ( ) n + r 63 E z )x + o n r 63 E z y + z o n =. (.5) e This is the same form as the optical indicatrix for a biaxial crystal, x /n x +y /n y +z /n z = where, n ( x = n o + r63 E z n ) / o no r 63n 3 oe z,

15 .. EXAMPLES n ( y = n o r63 E z n ) / o no + r 63n 3 oe z, n z = n e. (.6) Here we have used the binomial expansion to first order only since the refractive index changes induced by the electro-optic effect are generally orders of magnitude smaller than the refractive index itself. If we consider light propagating parallel to the z-axis then the relevant cross-section of the optical indicatrix is shown before and after the application of an electric field in Fig... A ray of general polarisation will be split into components along the x and y directions and these will have different phase velocities. y n o n o x E z y n y y n x x x Figure.: Change in the cross-section of the optical indicatrix for KDP on application of an electric field parallel to the optic axis. The electro-optic effect in this example can be considered as an induced birefringence. Hence this effect can be employed in such applications as half and quarter wave plates. For a half wave plate we require (n y n x)d = λ/, where d is the crystal thickness. Thus we require the application of an electric field E (π) z where, r 63 n 3 oe z (π) d = λ/. Since the DC field is being applied along the z-axis which is the same as the direction of propagation, the applied field is the applied voltage divided by the crystal length, E z = V /d. Hence we require for an induced half wave plate a voltage V (π) = λ/(n 3 or 63 ). For KDP at a visible wavelength of λ = 550 nm, n 3 or pmv and a voltage of V (π) 8 kv is required. We found above that the electro-optic coefficient appears in the factor n 3 r. This is true in all cases and not just the above example. Hence in selecting materials for the electro-optic effect it should be the factor n 3 r which is compared and not just the raw electro-optic coefficient. Fortunately most tabulations of electro-optic coefficients include the refractive index. With the inclusion of polarisers at the input and output, this KDP example can be used to construct an electro-optic amplitude modulator as shown in Fig..(a). This geometry is termed a longitudinal modulator as the electric field is applied along the direction of propagation. With the change in polarisation, the transmission of the output polariser is given by

16 CHAPTER. THE ELECTRO-OPTIC EFFECT T = sin πv /(V (π) ) and is shown in Fig..(b). We note that for small voltages, the transmission will depend quadratically on the applied voltage. In many cases a linear modulator is required. This could be achieved by biasing the voltage to the V π/ point. Since this bias point is equivalent to a quarter wave plate, a more elegant solution is to incorporate an additional conventional quarter wave plate. (a) x (b) T 0.5 Input y z Output 0 0 π/ π V Polariser V Polariser Figure.: Longitudinal geometry for an electro-optic modulator based on KDP... GaAs Gallium Arsenide and other semiconductors of zinc-blende (cubic) symmetry belong to the class 43m and do not normally exhibit birefringence. There are three non-zero electro-optic tensor elements all of which are equal. GaAs is not transparent at visible wavelengths but can be employed as an electro-optic modulator in the infrared (e.g. 0 µm). The KDP modulator is longitudinal which means that changing the crystal length does not change the voltage required since it equally affects the electric field and the optical path length. This means that the large required voltage quoted in the example has no prospect of being reduced. Furthermore the light has to pass through the electrodes so these have the complication of being transparent or having a hole in them. A much more attractive geometry is the transverse one where the electric field is applied perpendicular to the propagation direction. An example transverse amplitude modulator geometry for GaAs and other zinc-blende semiconductors is shown in Fig..3. For this example with a crystal of length L and thickness d, the required voltage to achieve a π phase difference between the two polarisation components is V (π) = λd/(ln 3 r 4 ). Note that we have gained a factor of d/l in comparison to the longitudinal modulator and thus the voltage can be reduced by increasing the length or decreasing the thickness. For GaAs at 0 µm, n=3.3, r 4 =.6 pmv and taking L=5 cm and d=0.5 cm, we obtain V (π) 9 kv. The two modulators discussed so far are examples of an amplitude modulator. These are basically using the induced birefringence to change the polarisation, which is then sent through a polariser. Also of use is a phase modulator. This is conceptually easier to follow as it directly uses the change in refractive index to change the optical path length and hence the optical phase. For the KDP case, the input polarisation would be aligned with either the x or y axes (so only the ordinary or extraordinary ray is input and the polarisation is maintained), and then the change in optical path length is, nd = n 3 or 63 V /. A possible geometry for a GaAs phase

17 .. EXAMPLES 3 [0] Input Polariser [00] V Output Polariser Figure.3: Transverse geometry for an electro-optic modulator based on GaAs. electro-optic modulator is to use a transverse field parallel to [00] to modulate an optical beam with linear polarisation parallel to [0]: the change in optical path length is given by, nl = n 3 r 4 V L/(d). If a time varying voltage is applied to a phase modulator, the time varying phase is equivalent to altering the frequency of the optical beam. In particular, if an optical beam of single frequency ω 0 is modulated with a sinusoidal voltage of frequency ω m, then the frequency spectrum of the transmitted light develops side bands at ω 0 ± ω m, ω 0 ± ω m, etc. What we have accomplished here is mixing of two frequencies, one optical and one electrical. This brings us neatly to the subject of the next chapter: optical frequency mixing.

18 4 CHAPTER. THE ELECTRO-OPTIC EFFECT Triclinic Orthorhombic Tetragonal 4 Cubic 43m 3 Trigonal 3 Hexagonal 6 r r r3 r r r3 r3 r3 r33 r4 r4 r43 r5 r5 r53 r6 r6 r63 mm r r r r3 0 0 r3 0 0 r33 4 r4 r5 0 r5 r r r r4 r r r3 r r r3 0 0 r33 3 r4 r5 0 r5 r4 0 r r r3 0 0 r3 0 0 r33 6mm r4 r5 0 r5 r4 0 Monoclinic 0 0 r3 0 0 r3 0 0 r33 0 r4 0 r r3 0 0 r3 r4 r5 0 r5 r4 0 r 0 0 r 0 0 r r4 0 0 r 0 4 3m 0 0 r3 0 0 r3 0 0 r r5 0 r mm r r4 0 0 r r3 0 r r3 0 0 r33 0 r5 0 r5 0 0 r 0 0 r r r 0 0 r 0 0 r3 0 r4 0 r43 0 r5 0 r6 0 r63 m 0 0 r3 0 0 r3 0 0 r33 4m 0 r5 0 r5 0 0 r r 0 r r 0 6m r r 0 r 0 r3 r 0 r3 r3 0 r33 0 r4 0 r5 0 r53 0 r6 0 r r r63 0 r 0 0 r 0 r 0 0 Table.: The form of the electro-optic tensor for the crystal symmetry classes with no inversion symmetry. A bar over an entry indicates the negative.

19 Chapter 3 Optical Frequency Mixing 3. Lorentz model In the Lorentz model the motion of the electrons in the medium is treated as a harmonic oscillator. This can be pictured as electrons attached to their nuclei by springs with resonant frequency Ω and damping γ. An optical wave provides a forcing term through the dipole interaction with the electron and hence the motion of the electron around its equilibrium position can be described by the linear differential equation, d r(t) dt + γ dr(t) dt + Ω r(t) = e m 0 E(t). (3.) One way to solve this differential equation is to Fourier transform it: r(t) = r(ω)e iωt dω and hence the solution of Eq. 3. is straightforward, r(ω) = e m 0 E(ω) Ω ω iγω. (3.) Now the polarisation is given P = Ner and the optical susceptibility is defined P(ω) = ε 0 χ(ω)e(ω), so the optical susceptibility can be written, χ(ω) = Ne ε 0 m 0 Ω ω iγω. (3.3) Neglecting absorption, the refractive index is given by n(ω) = + Reχ(ω) and is shown in Fig. 3. for frequencies around resonance Ω. This is the characteristic shape of the refractive index around a resonance consisting of a maximum at a frequency just below the transition and a minimum just above; apart from a small frequency regime around the resonance, dn(ω)/dω is positive which is termed normal dispersion. Now suppose the potential for the electron is cannot strictly be described as that of a harmonic oscillator. We aim to describe the situation that the motion of the electron is large enough such that the Taylor series expansion of the restoring form has significant terms of quadratic or 5

20 6 CHAPTER 3. OPTICAL FREQUENCY MIXING Figure 3.: The form of the dispersion of the refractive index around a resonance in the Lorentz model. The frequency (horizontal axis) has been scaled as (ω ω 0 )/γ. higher order. The differential equation describing the electron s motion then has additional anharmonic terms which we will consider to r, d r(t) dt + γ dr(t) dt + Ω r(t) ξr = e m 0 E(t). (3.4) The differential equation is now nonlinear and is complicated to solve. However it is usual that the anharmonic term is small compared with the other terms and a perturbation analysis can be performed to gain some insight. The displacement from equilibrium position is expanded as r = r 0 + r +... The differential equation taken to the lowest order (in r 0 ) is just the harmonic Lorentz equation (3.). The next highest order gives, d r (t) dt + γ dr (t) dt + Ω r (t) = ξr 0(t), (3.5) where Eq. (3.) is used to provide the form for r 0 (t). Consider the case where the optical field is monochromatic and can be described E(t) = [E 0 e iω 0t +E 0 eiω 0t ]/. The Fourier transform of this is E(ω) = [E 0 δ(ω ω 0 )+E 0 δ(ω+ω 0)]/. The linear solution r 0 will be driven at the same frequency and hence we obtain, r 0 (t) = e [ E 0 e iω 0t m 0 Ω ω 0 iγω 0 E ] 0 + eiω 0t Ω ω 0 iγω 0. (3.6) When this expression is squared, the terms have time dependencies of e ±iω 0t or are constant. Thus there exists driving terms for r (t) is at frequencies ±ω 0. Hence the anharmonic term has resulted in the motion of the electron having a frequency component at twice the optical frequency and so will the polarisation. Similarly there will also be DC components produced from the constant term. What has happened here is that the anharmonic term has resulted in

21 3.. LORENTZ MODEL 7 frequency mixing. Since we have only one optical frequency present the only combinations are at ω 0 ± ω 0, i.e. ω 0 and 0. The frequency mixing aspect can be seen more clearly if we take an optical input field that is bichromatic, E(t) = [E e iω t + E e iω t + cc]/ where the cc denotes the complex conjugate. The linear solution r 0 (t) will contain terms oscillating at the same frequencies, r 0 (t) = e [ E e iω t m 0 Ω ω iγω E e iω t + Ω ω + iγω On squaring this will produce a term oscillating at ω + ω, ] + cc. (3.7) r0(t) ω = e +ω m E E e i(ω +ω )t [ Ω ω ] [ iγω Ω ω ] iγω. (3.8) 0 On inserting this into Eq. (3.5) this provides for r (t) (e.g. by the Fourier transform method described previously), r (t) ω +ω = ξe m 0 E E e i(ω +ω )t [ Ω (ω + ω ) iγ(ω + ω ) ] [ Ω ω iγω ] [ Ω ω iγω ]. (3.9) Hence since the polarisation is given by P = Ner, there is also a polarisation component oscillating at the sum frequency, P(ω) ω +ω = Nξe3 m 0 E E δ(ω ω ω ) [ Ω (ω + ω ) iγ(ω + ω ) ] [ Ω ω iγω ] [ Ω ω iγω ]. (3.0) We can extend the definition of the optical susceptibility to include these higher order effects: P(ω) ω +ω = ε 0 χ () (ω,ω )E E δ(ω ω ω ), (3.) where χ () is referred to as the second-order optical susceptibility. The factor of is included since we must allow for both orderings E E and E E. Eq. 3. is specific to the interaction of two monochromatic sources. It can be generalised by summing over all possible frequencies, P () (ω) = ε 0 dω a dω b χ () (ω a,ω b )E(ω a )E(ω b )δ(ω ω a ω b ). (3.) The delta function ensures that the output polarisation oscillates at summations of the frequencies. Although this form looks a bit different from the linear definition of the optical susceptibility, we can write, P () (ω) = ε 0 dω a χ () (ω a )E(ω a )δ(ω ω a ) = ε 0 χ () (ω)e(ω), (3.3)

22 8 CHAPTER 3. OPTICAL FREQUENCY MIXING which shows the same form. The total contribution to the polarisation is then P(ω) = P () (ω)+ P () (ω). By comparing Eqs. (3.0) and (3.) we obtain, χ () (ω,ω ) = Nξe3 4ε 0 m 0 [ Ω (ω + ω ) iγ(ω + ω ) ] [ Ω ω iγω ] [ Ω ω iγω ], = ξε 0 m 0 4N e 3 χ() (ω + ω )χ () (ω )χ () (ω ), (3.4) where we have substituted the linear susceptibility in the final form. This shows that if we have a resonance in the linear susceptibility either at one of the frequency components or their sum, there is likely to be a resonance in the second-order susceptibility. Although this result is specific to the anharmonic Lorentz model, there is an empirical rule called Miller s rule which states χ () (ω,ω ) = χ () (ω + ω )χ () (ω )χ () (ω ), where the variation of with frequency and material is much smaller than in the linear and nonlinear susceptibilities themselves. 3. The Nonlinear Susceptibility Tensor In the previous section, the vectorial nature of the electric field and polarisation was ignored. As in the case of birefringence, this can be allowed for by using tensors for the linear and nonlinear susceptibilities. We can also expand the nonlinear polarisation beyond second-order so that P i (ω) = P () i (ω) + P () i (ω) + P (3) i (ω) +... and the nonlinear polarisation contributions can be written, P () i = ε 0 P () i = ε 0 P (3) i = ε 0 dω a χ () i j (ω a )E j (ω a )δ(ω ω a ), dω a dω a dω b dω b χ () i jk (ω a,ω b )E j (ω a )E k (ω b )δ(ω ω a ω b ), δ(ω ω a ω b ω c ). dω c χ () i jkl (ω a,ω b,ω c )E j (ω a )E k (ω b )E l (ω c ). =. (3.5) Summation over the repeated indices j, k and l is implicit in the above. 3.. st order χ () i j is a second rank tensor (9 elements) and is equal to the dielectric tensor less the identity matrix (the tensor form allows birefringence to be described). The frequency of the polarisation has to be the same as that of the electric field. Depending on the relative phase of the polarisation and the electric field, the interference gives rise to optical absorption or refraction; Reχ () corresponds to refraction and Imχ () to absorption.

23 3.. THE NONLINEAR SUSCEPTIBILITY TENSOR nd order χ () i jk is a third rank tensor (7 elements). If we use a monochromatic input E(t) = (E 0e iω 0t + E 0 eiω 0t )/ then evaluating the integrals gives for the second-order polarisation, P () i (ω) = ε 0 4 {[ χ () i jk (ω 0, ω 0 )E 0 j E 0k + χ() i jk ( ω 0,ω 0 )E 0 je 0k ]δ(ω 0) +χ () i jk (ω 0,ω 0 )E 0 j E 0k δ(ω ω 0 ) +χ () i jk ( ω 0, ω 0 )E 0 je 0k δ(ω + ω 0) }. (3.6) As we indicated in the Lorentz anharmonic model, for a monochromatic input of frequency ω 0, the second-order polarisation has frequency components at ±ω 0 and 0. These give rise to second harmonic generation and optical rectification respectively. Note that there are no components at the original frequency ω 0. If we have a combination of frequencies present (ω and ω say) we can produce the sum (ω = ω + ω ) and difference (ω = ω ω ). This even applies if ω = 0 i.e. DC and gives an alternative description of the electro-optic effect rd order χ (3) i jkl is a fourth rank tensor (8 elements). For a monochromatic input of frequency ω 0, evaluation of the integrals gives frequency components of the third-order polarisation at ω = ±3ω 0 (which describes third harmonic generation) and at ω = ±ω 0. Since we have a component at the same frequency this will act like an absorption or refraction, only in this case the effect will be nonlinear because of the additional electric field terms Properties of the susceptibility tensor Intrinsic permutation symmetry In Eq. (3.5) we are at liberty to exchange pairs of indices j, k, etc. since these are just dummy indices which are summed over the directions x, y and z. Since the electric field component product commutates, then we must have, for example, χ () ik j (ω,ω ) = χ () i jk (ω,ω ). (3.7) This property where the direction indices and frequency arguments can be permutated [e.g. ( j,ω ) (k,ω )] is called intrinsic permutation symmetry. Reality condition The field E(t) and the polarsiation P(t) are physical quantities and hence must be real. This then requires of the susceptibility, for example, χ () i jk ( ω, ω ) = χ () i jk (ω,ω ), (3.8) that is the conjugate of the susceptibility is equivalent to negating the frequencies.

24 0 CHAPTER 3. OPTICAL FREQUENCY MIXING Overall permutation symmetry If all the optical frequencies and their combinations are well removed from any material resonance then in addition to the intrinsic permutation symmetry, overall permutation symmetry also applies where the combination (i, ω ω ) is included in the sets which can be permutated leaving the nonlinear susceptibility invariant. So, for example, χ () jik ( ω ω,ω ) χ () i jk (ω,ω ). (3.9) Note that unlike intrinsic permutation symmetry, overall permutation symmetry is only an approximation, but one which is valid in most cases of interest. Now consider the low frequency limit such that the dispersion in the susceptibility can be ignored. In this situation, all frequencies could be replaced by zero which implies that all frequencies are equivalent. Thus the susceptibility will be invariant if the frequency arguments alone are permuted. Combining this with overall permutation symmetry means that the nonlinear susceptibility is invariant under permutations of the direction indices. This gives, for example, χ () ik j (ω,ω ) χ () i jk (ω,ω ). (3.0) This property is called Kleinmann symmetry. Once again it is important to note that this is just an approximation that applies in the low frequency limit, far from any material resonances. Causality So far we have been using the frequency domain for describing the optical polarisation. The linear susceptibility was defined P(ω) = ε 0 χ(ω)e(ω) which is a product. Now under Fourier transforming, a product becomes a convolution so we have in the time domain, P(t) = ε 0 χ(τ)e(t τ) dτ. (3.) π Now the principle of causality states that any feature in the input (electric field in this case) cannot affect the output (polarisation) at earlier times. That is effect cannot precede cause. Hence in the above convolution E(t τ) cannot influence P(t) if t < t τ i.e. τ < 0. This then requires χ(τ) = 0 for τ < 0. One way of expressing this is to set χ(τ) = χ(τ)θ(τ) where θ(τ) is the Heaviside (step) function. If we Fourier transform this expression, the product becomes a convolution and we obtain the relationship, χ(ω) = iπ P χ(ω) dω Ω ω, (3.) where P is used to denote a principal parts integral. Because of the extra factor i in this relation, by separating this equation into real and imaginary parts, one can relate the real part of χ solely in terms of its imaginary part and vice versa. Hence if only Imχ is supplied (across the entire spectral range), Reχ can be generated. This is one example of a dispersion relation of which the best known is the Kramers-Krönig relation relating refractive index to absorption coefficient. For a wider discussion on dispersion relations and their applicability to the nonlinear case see [5].

25 Chapter 4 Second-Order Optical Nonlinearities 4. Contracted tensor for nd-order nonlinearities In determining the second-order polarisation P () (ω), summation takes place over all permutations, e.g. in the case of second harmonic generation, if the electric field has components along the y and z axes, there will be a polarisation generated parallel to the x axis: χ () xyze y E z + χ () xzye z E y. Since the same combination of fields occurs in both terms then we can contract these to a single term. If we use P () (ω) = [P ω 0δ(ω ω 0 ) + P ω 0δ(ω + ω 0 )]/ and E(ω) = [E ω 0δ(ω ω 0 ) + E ω 0δ(ω + ω 0 )]/ then we can re-write the second-order polarisation as, P ω +ω x P ω +ω y P ω +ω z P ω 0 x P ω 0 y P ω 0 z = ε 0 = ε 0 d d d 3 d 4 d 5 d 6 d d d 3 d 4 d 5 d 6 d 3 d 3 d 33 d 34 d 35 d 36 d d d 3 d 4 d 5 d 6 d d d 3 d 4 d 5 d 6 d 3 d 3 d 33 d 34 d 35 d 36 (E ω 0 x ) (E ω 0 y ) (E ω 0 z ) E ω 0 y E ω 0 z E ω 0 x E ω 0 z E ω 0 x E ω 0 y, (4.) where the second subscript on the coefficient d relates to the conventional axes by, :xx, :yy, 3:zz, 4:yz or zy, 5:zx or xz and 6:xy or yx. It can be seen that the contracted d-tensor for SHG is simply related to the conventional susceptibility tensor by d i jk (ω,ω) = χ () i jk (ω,ω)/. Note that we have used intrinsic permutation symmetry to reduce the 7 elements to 8 independent ones. If Kleinmann symmetry can be applied then the contracted tensor notation can be extended to nondegenerate interactions (ω ω ), E ω x E ω x E ω y E ω y E ω z E ω z E ω y E ω z + E ω y E ω z E ω z E ω x + E ω z E ω x E ω x E ω y + E ω x E ω y, (4.)

26 CHAPTER 4. SECOND-ORDER OPTICAL NONLINEARITIES Triclinic Monoclinic Orthorhombic Tetragonal 4 4 4m Cubic 43 Trigonal 3 Hexagonal 6 3m 6 6m d d d 3 d 4 d 5 d 6 d d d 3 d 4 d 5 d 6 d 3 d 3 d 33 d 34 d 35 d 36 d 4 0 d 6 d d d 3 0 d 5 0 d 34 0 d 36 d d d 36 d 4 d 5 0 d 5 d 4 0 d 3 d 3 d d d 4 0 d d d 36 d d 0 d 4 d 5 d d d 0 d 5 d 4 d d 3 d 3 d 33 0 d 5 d d d 0 d d 3 d 3 d 33 d 4 d 5 0 d 5 d 4 0 d 3 d 3 d d d d d d 0 m 4 mm 4mm 43m mm d d d 3 0 d 5 0 d 4 0 d 6 d 3 d 3 d 33 0 d d 5 0 d d 3 d 3 d 33 d 4 d 5 0 d 5 d 4 0 d 3 d 3 d 36 0 d 5 0 d d 3 d 3 d 33 d d d 4 d d 0 d d 4 d d d d d d d 0 d 5 0 d d 3 d 3 d 33 Table 4.: The form of the d-tensor for the crystal symmetry classes which do not possess inversion symmetry. A bar over an entry indicates the negative. where d i jk (ω,ω ) = χ () i jk (ω,ω )/ = χ () i jk (ω,ω )/. As in the case of the electro-optic coefficient, the number of independent elements can be further reduced with symmetry considerations. First of all if the material exhibits inversion symmetry then all the d tensor elements are zero. If we start from Pi ω = ε 0 d i jk E ω j Eω k and then apply the inversion operator, for materials with inversion symmetry d i jk is unaltered and, = ε 0 d i jk ( E ω j )( Eω k ). This is only consistent if d i jk = d i jk and hence d i jk = 0. Group theory can be applied to investigate other symmetry properties. Table 4. shows the form of the d-tensor for the 8 crystal classes that do not exhibit inversion symmetry. Pi ω If Kleinmann symmetry can be applied (low frequency, well removed from resonances) the

27 4.. EM PROPAGATION WITH A SECOND-ORDER NONLINEARITY 3 8 tensor elements are further reduced to 0. The relevant equalities are summarised in Table 4.. As an example consider KTP (KTiOPO 4 ) which has orthorhombic symmetry of class mm and hence by Table 4. has 5 independent, non-zero components. At a wavelength of 880 nm, these have been measured as [6]: d 5 =.04 pmv, d 3 =.76 pmv, d 4 = 3.9 pmv, d 3 = 4.74 pmv and d 33 = 8.5 pmv. Kleinmann symmetry specifies d 5 d 3 and d 4 d 3 which is only approximately true (within around 30%) in this case. Note that in this example the on-diagonal element d 33 d zzz is several times larger than the off-diagonal elements. This behaviour is quite common among materials which exhibit a second order nonlinearity. d i jk d i j d xyy = d yxy d = d 6 d xzz = d zxz d 3 = d 35 d xyz = d yzx = d zxy d 4 = d 5 = d 36 d xzx = d zxx d 5 = d 3 d xxy = d yxx d 6 = d d yzz = d zyz d 3 = d 34 d yyz = d zyy d 4 = d 3 Table 4.: Equalities among the d-tensor elements under application of Kleinmann symmetry. Rather than completely write out Eqs. (4.) and (4.) every time, we will denote this by the shorthand tensor multiplication, P ω +ω = ε 0 d(ω,ω ) : E ω E ω. (4.3) Furthermore, it is quite common to split the fields into magnitude and direction, i.e. E ω = ê E ω etc. where ê is a unit vector. A scaler quantity d eff is commonly used which hides the details of the tensor multiplication: d eff = ê p [d(ω,ω ) : ê ê ]. This allows the vector equation (4.3) to be written in scaler form. 4. EM Propagation with a Second-Order Nonlinearity 4.. Slowly Varying Envelope Approximation Start from the Maxwell curl equations E(t) = B(t)/ t and H(t) = j(t) + D(t)/ t, assume a non-magnetic material B = µ 0 H and take j = σe. We will split the polarisation into linear and nonlinear components, D = ε 0 E + P = ε 0 ε r E + P NL. Combining the two curl equations then gives the second order PDE, E(t) = µ 0 σ E(t) t ε r E(t) P NL (t) c t µ 0 t. (4.4) Now the properties of the vector differential operators gives that E = ( E) E and if there are no free charges, we also have E = 0. The time derivatives can also be

28 4 CHAPTER 4. SECOND-ORDER OPTICAL NONLINEARITIES simplified by Fourier transforming to the frequency domain to give, ( ) E(ω) = iωµ 0 σ ω ε r c E(ω) µ 0 ω P NL (ω). (4.5) Now let us insert a plane wave propagating in one direction only, which we will take to be forward (taken as parallel to the z-axis here). We will allow the amplitude of this wave to vary on propagation and take E(ω) = Ê(ω,z)e ikz where the value of the wavevector k = ε r ω/c is obtained from solution of the wave equation [Eq. 4.5 without the nonlinear polarisation term]. On taking the space differential, [ E(ω) d E(ω) d Ê(ω,z) dz = dz + ik dê(ω,z) ] k Ê(ω,z) e ikz. (4.6) dz Note that the final term proportional to Ê will exactly cancel a term in Eq Now let us assume that the nonlinear term is small (which is the case except in extreme circumstances which will not be dealt with here) and hence the modification from the linear case will be small. This tells us that the envelope will depart only slightly from its linear (constant) value. It will certainly vary much more slowly than the oscillation of the underlying wave hence this approximation is conventionally termed the slowly varying envelope (or amplitude) approximation. This allows us to use d Ê/dz kdê/dz and neglect the second-order derivative in the envelope. Hence we reduce the second order differential equation to the first order one, dê(ω, z) dz = α Ê(ω,z) + iω µ 0 k PNL (ω)e ikz, (4.7) where we have substituted the absorption coefficient α = ωµ 0 σ/k. Although this is termed the slowly varying envelope approximation, the key approximation is the unidirectional nature of the wave propagation. If propagation both in the forward and backward directions occurs, then coupled terms arise which prevent this simplification. On the right-hand side of Eq. 4.7 the first term describes linear loss and the second term describes a nonlinear polarisation source for the wave. Eq. (4.7) is relevant for any order of nonlinearity. In this chapter we are concentrating on second order nonlinearities for frequency mixing. Let us consider the interaction between two waves of frequency ω and ω, producing a polarisation at the sum frequency ω 3 = ω + ω. Using the contracted d-tensor notation we have, P ω 3 = ε 0 d(ω,ω ) : E ω E ω. Inserting this into Eq. (4.7) gives, dê ω 3 dz dê ω dz dê ω dz = α 3 Êω 3 + iω 3 k 3 c d(ω,ω ) : Ê ω Ê ω e i kz, = α Êω + iω k c d(ω 3, ω ) : Ê ω 3 Ê ω e i kz, = α Êω + iω k c d(ω 3, ω ) : Ê ω 3 Ê ω e i kz, (4.8)

29 4.. EM PROPAGATION WITH A SECOND-ORDER NONLINEARITY 5 where we have used k = k + k k 3 and k 3 = k(ω 3 ) etc. Also of relevance the interactions where the difference between ω 3 and ω produces ω and the difference between ω 3 and ω produces ω which are also listed. Note that Ê ω = (Ê ω ). 4.. Manley-Rowe Relations Now consider the case that we are considering frequencies well removed from any material resonances. This is equivalent to stating we are concerned with a frequency region that is transparent and hence α, α, α 3 = 0. We can also apply overall permutation symmetry which states d(ω 3, ω ) = d(ω 3, ω ) = d(ω,ω ). For simplicity we use the substitution d eff = d(ω,ω ) : ê ê. Hence Eqs. (4.8) can be rewritten as, dê ω 3 dz dê ω dz dê ω dz = iω 3 k 3 c d effêω Ê ω e i kz, = iω k c d effêω 3 (Ê ω ) e i kz, = iω k c d effêω 3 (Ê ω ) e i kz. (4.9) In SI units the irradiance is defined in terms of the electric field amplitude as, I ω = ε 0 cn(ω) E ω /, where n(ω) is the refractive index. Differentiating this gives, di ω dz = ε 0cn 0 (ω) [(E ω ) de ω ( de ω ) ] dz + Eω. (4.0) dz Inserting the electric field derivatives and using k = nω/c then gives, di ω 3 = ε 0ω 3 dz di ω = ε 0ω dz di ω = ε 0ω dz We can see from the above that, [ id eff Ê ω Ê ω ( Ê ω 3 ) e i kz id eff (Êω Ê ω ) Ê ω 3 e i kz], [ id eff (Êω Ê ω ) Ê ω 3 e i kz id eff Ê ω Ê ω ( Ê ω 3 ) e i kz ], [ id eff (Êω Ê ω ) Ê ω 3 e i kz id eff Ê ω Ê ω ( Ê ω 3 ) e i kz ]. (4.) di ω3 ω 3 dz = ω di ω dz = ω di ω dz. (4.) Now irradiance is defined as optical energy flowing through a unit area per unit time. The photon energy is hω and hence the ratio of the irradiance to the optical frequency is proportional to the number of photons passing through a unit area per unit time or in other words the photon flux. Hence Eq. (4.) can be restated as the change in the number of photons at ω 3 is the negative of the change in the number of photons at ω or ω (which are equal): N 3 = N = N. These are known as the Manley-Rowe relations. These seem intuitively correct from simple energy conservation; we are far from material resonances so the only energy flow can be between the waves of different frequency.

30 6 CHAPTER 4. SECOND-ORDER OPTICAL NONLINEARITIES 4..3 Sum frequency generation: ω + ω ω 3 Let us assume that there is initially no light of the sum frequency ω 3 present. Let us also initially examine the case of low conversion efficiency such that any depletion of the frequencies ω and ω can be neglected, i.e. de ω /dz, de ω /dz 0. This simplification allows just one of the differential equations to be studied instead of the complete set. Let us also assume that the linear loss can be neglected, α 3 = 0 and hence we have for the evolution of the light at frequency ω 3, dê ω 3 dz = iω 3 cn 3 d eff Ê ω Ê ω e i kz, (4.3) where we have used as shorthand for the refractive index n 3 = n(ω 3 ). Since Ê ω and Ê ω are treated as constant (low conversion efficiency), this can be easily integrated along the length of the crystal to give, Ê ω 3 (L) = ω 3d eff Ê ω Ê ω (ei kl ) cn 3 k = ω 3d eff L Ê ω Ê ω e i kl/ sinc kl cn 3, (4.4) where sincx = (sinx)/x. Now writing this instead in terms of the irradiance, I ω = ε 0 cn(ω) E ω /, I ω 3 (L) = ω 3 deff ε 0 c 3 L I ω I ω sinc kl. (4.5) n n n 3 There are several points to make about the generation of the sum frequency described by Eq. (4.5): the irradiance of the generated light is () proportional to a material factor d eff /n3 (we will see this is the usual factor in second order processes) () proportional to both the irradiance at ω and at ω (3) grows quadratically with distance (L ) and (4) depends on the factor sinc kl/. This last factor is shown plotted in Fig. 4.. It can be seen that this has a maximum value of at kl/ = 0 but falls off rapidly away from this. Hence it is important for efficient generation to operate at k = 0 i.e. k + k = k 3. This is termed phasematching since we are matching the phase velocity of the existing wave (k 3 ) to that of the nonlinear polarisation (k + k ). Phase-matching can be thought of as photon momentum conservation (since the photon s momentum is given by hk) just as the Manley-Rowe relations describe photon energy conservation Second harmonic generation: ω + ω ω Second harmonic generation is just a special case of sum frequency generation where an optical wave interacts with itself to generate the sum frequency. Instead of three coupled differential equations to consider, in this case we require just two. Let us assume that we are removed from resonances so that linear loss can be neglected and overall permutation symmetry applied. In the low conversion efficiency approximation we get a result similar to the previous case with a