(Chapter head:)Nonlinear Wave Mixing Processes

There are the rushing waves... mountains of molecules,each stupidly minding its own business...trillions apart...

yet forming white surf in unison."

-Richard Feynman

The waves that mix are improtant, so are the ways in which they mix.

Learning objectives

Revisit the basics of electromagnetic wave propagation in linear media Establish the conditions for optimal energy transfer in nonlinear media. Analyse three wave mixing process and energy transfer in detail. Recongize the need for phase matching and device schemes to chgiece this

1

CVRectangle

CVText BoxSecond Order Wave Mixing Processes

1 Introduction

Electromagnetic waves travel without interaction amongst themselves in a mate-rial with a linear optical response. There would be no energy exchange betweenthe waves. On the other hand, if the material exhibits a signicant opticalnonlinearity, that would enable energy transfer between the waves. We refer tosuch energy transfer as a wave mixing process. Depending on the order of thenonlinearity, the mixing may involve three waves, four waves or more. Threewave mixing results in optical parametric amplication and second harmonicgeneration. Four wave mixing produces, among other things, third harmonicgeneration and optical phase conjugation. To set the stage for discussing theseeects, we rst review the basics of electromagnetism in linear dielectrics andthen show how nonlinear polarization changes the propagation of the waves andprovides coupling between them, ringing about this wave mixing.

2 Elements of Electromagnetism

Gausslaw of electrostatics relates the closed surface integral of the electric uxwith the net charge qencl enclosed in the volume bounded by the surface:I

S

!E !dS = qencl

0(1)

The left hand side is convertible into a volume integral according to thedivergence theorem: I

S

!E !dS =

ZV

!r !E dV (2)The right hand side of Eq. 1 can also be written as a volume integral:

1

0qencl =

1

0

ZV

dV (3)

where is the volume charge density. This makes it possible to re-writeGaussequation as Z

V

!r !E dV = ZV

0dV (4)

Being true for any volume, the integrands on either side of the above equationmust therefore be equal: !r !E =

0(5)

In a polarized material, the charge density has two parts: (a) a free chargedensity f and (b) a bound charge density b. We can consider how the bound

charge density is related to the polarization!P . To see this, let us write the

potential at a point !r due to a dipole !p located at the point!r0:

V (!r ) = 140

!p brr2

(6)

2

where br = !r !r0r , r = !r !r0 ; dening the the unit vector in the directionof the position vector of the eld point relative to the source dipole. Now ifthere is a volume distribution of dipoles in a material, then the potential dueto such a distribution is given by

V (!r ) = 140

Zv

!P!r0 br

r2dV 0 (7)

because the dipole moment in the volume element dV 0 at!r0 is

!P!r0dV 0.

Since!r0 1r = brr2 ( the dierentiation being with respect to the source position!

r0 ), this can be written out as

V (!r ) = 140

Zv

!P :!r01

r

dV 0 (8)

Now we have

r0: !P

r

!=!P :!r01

r

+1

r

!r0 !P (9)

Using this relation, we re-write Eq.4.8 as

V (!r ) = 140

"ZV

!r0 !P

rdV 0

ZV

1

r

!r0 !P dV 0# (10)Using the divergence theorem, this simplies to

V (!r ) = 140

IS

1

r

!P dS0

ZV

1

r

!r0 !P dV 0 (11)The rst term on the right hand side of the above equation is the potential

due to a surface charge of density

b =!P :bn (12)

The second term is the potential due to a volume charge of density

b = !r !P (13)

with the result we may re-write Eq.11 as

V (!r ) = 140

IS

brdS +

1

40

ZV

brdV 0 (14)

The total charge density is the sum of that due to the free charges f andthat b due to the bound charges: = f + b. Gauss law now becomes

!r !E = f + b0

(15)

3

Using the expression for the bound charge density given by Eq.13, this resultsin !r

0!E +

!P= f (16)

thereby leading to the concept of the electric displacement!D dened by

!D = 0

!E +

!P (17)

modifying the Gauss law for electric elds in the presence of the polarizabledielectrics to !r !D = f (18)For linear dielectrics, of course, we have

!P = 0

!E (19)

!D = 0 (1 + )

!E =

!E (20)

= r0 (21)

r = 1 + (22)

dening the fundamental relationships between the dielectric permittivity, the relative permittivity r and the susceptibility . Note while the relativepermittivity and susceptibility are dimensionless, the permittivity has the samedimensions as the free space permittivity 0.

3 Travelling ElectromagneticWaves in Free Space

In free space devoid of any material, there is no charge. Maxwells equations forthe elds take the form

!r !E = 0 (23)!r !E = @

!B

@t(24)

!r !B = 0 (25)!r !B = 00

@!E

@t(26)

We now wish to develop the dierential equation for the travelling electricand magnetic elds. Toward this, we use the vector identity

!r !r !E = !r !r !E!r2!E (27)

In view of the Eq.23, this becomes

!r !r !E = r2!E (28)

4

Taking the curl of the right hand side of the Eq.4.20b, and making use ofEq.26, we get

!r @!B

@t

!=

@!r !B@t

= 00@2!E

@t2(29)

Thus sing Eq. 28 for the left hand side and Eq.29 for the right hand side,Eq.24 transforms to

r2!E = 00@2!E

@t2(30)

In a similar way, we transform Eq.26 to give

r2!B = 00@2!B

@t2(31)

These are the two wave equations for the electric and magnetic elds, prop-agating at the common wave velocity of magnitude

c =1p00

(32)

As solutions of the wave equations for the electric and magnetic elds asgiven above, let us consider a monochromatic plane wave travelling in the +Xdirection with the elds given by

!E =

!E0e

i(!tkx) (33)

!B =

!B0e

i(!tkx) (34)

where the eld amplitudes!E0 and

!B0 are complex and have no variation in

the Y Z plane. @!E 0@y =

@!E 0@z = 0,

@!B 0@y =

@!B 0@z = 0. Since

!r !E = 0, @Ex@x = 0,@Ey@y = 0,

@Ez@z = 0 it follows that on dierentiating partially with respect to x,

E0x = 0. Similar use of!r !B = 0 leads to B0x = 0, establishing the wave as

purely transverse. Further, from Eq. 24, B0 = E0c . Accordingly we write thetravelling elds as

!E (!r ; t) = beE0ei!t!k !r (35)

!B (!r ; t) =

!k be E0

cei!t!k !r

(36)

Taking the real parts, the elds of the wave are given by

!E (!r ; t) = beE0 cos!t!k !r (37)

!B (!r ; t) =

!k be E0

ccos!t!k !r

(38)

5

3.1 Energy Density in the Travelling Wave

Per unit volume of space, the energy available in the combined electric andmagnetic elds is given by

u =1

20E

2 +1

2

B2

0(39)

Since B = Ec andp00 =

1c , the two energy densities are equal giving for

the total energy densityu = 0E

2 (40)

On using the real part of the electric eld as given by Eq.37, we have

u = 0E20 cos

2 (!t kx) (41)

Poynting vector!S representing the energy transport dened by

!S =

!E !B

0

(42)

becomes now !S = bic0E20 cos2 (!t kx) (43)

or !S = bicu (44)

The linear momentum density of the eld (per unit volume) equals

!p = uc

!i (45)

Averaging over one cycle of the wave (

cos2 (!t kx) = 1=2) these quanti-

ties are given by

hui = 120E

20 (46)D!

SE=1

2c0E

20bi (47)

h!p i = 12c0E

20

!i (48)

The intensity of the wave is given by

I = hSi = 12c0E

20 (49)

6

4 Propagation of Electromagnetic Waves in Lin-ear Materials

If an electromagnetic wave is travelling in a material medium, the propagationis to be described dierently. The Maxwells equation undergo changes dueto the eects of the applied elds in the material. The eects are representedmacroscopically by the electric polarization

!P and the magnetization

!M . We

will consider only materials which are non-conducting, i.e. dielectric materials.With no net charge and no current, Maxwells equations for the electric andmagnetic elds in the material become

!r !D = 0 (50)!r !B = 0 (51)

!r !E = @!B

@t(52)

!r !H = @!D

@t(53)

For a linear material medium, we have

!D =

!E (54)

!H =

!B

(55)

If the material is homogeneous, with no variation of the electric permittivity and magnetic permeability with position, then the Maxwells equationsbecome !r !E = 0 (56)

!r !B = 0 (57)!r !E = @

!B

@t(58)

!r !B = @!E

@t(59)

Now let us take the curl of Eq.58 to give

!r !r !E = @!r !B@t

(60)

In view of Eq.59, this gives

!r !r !E = @@t@!E

@t(61)

7

Using the identity of Eq.27, and on making use of Eq.56, we get

r2!E = @2!E@t2

(62)

This is the wave equation for the electric eld propagating in a linear materialcharacterized by the linear properties and . There is also a similar equationfor the magnetic eld

!B . The velocity of the waves has the magnitude given by

=1p

(63)

If, further the material is non-magnetic (i.e. non-ferromagnetic, non-antiferromagnetic, non-ferrimagnetic!), i.e. diamagnetic or paramagnetic, with close to 0, we have the wave speed given by

v =1p0r0

=cpr

(64)

The refractive index n of the material dened by n = c is thus equal topr.

The energy density, the Poynting vector and the intensity of the wave now aregiven by

u =1

2E2 +

1

2

B2

(65)

!S =

!E !B

(66)

I =1

2E20 (67)

5 Propagation of ElectromagneticWaves in Non-linear Materials

5.1 The Wave Equation

We revert to the Maxwells equations given by Eq.??. Taking the curl of Eq.58,we have

!r !r !E = @

!r !B@t

= @!r !H

@t(68)

Using Eq.59, this gives

!r !r !E = @2!D

@t2(69)

or on using the earlier vector identity27, we have

r2!E = @2!D@t2

(70)

8

Recalling the denition of the displacement Eq.17to get

r2!E = 0 @2!E@t2

+ @2!P

@t2(71)

Since our material now is nonlinear, we write the polarization!P as a sum

of the linear part!P (1) and the higher order nonlinear contributions

!P (N):

!P (N) =

!P (2) +

!P (3) +

!P (4) + ::: (72)

Thus Eq.71 is re-written as

r2!E = 0 @2!E@t2

+ @2!P (1)

@t2+

@2!P (N)

@t2(73)

The rst two terms on the right hand side of the above equation can becombined and brought to the left hand side as

r2!E 0@2!E

@t2= 0

@2!P (N)

@t2(74)

This is the wave equation valid for nonlinear materials. Note that we changed to 0, as the materials we choose are nonmagnetic. This nonlinear waveequation diers from the linear wave equation in the occurrence of the term on

the right hand arising from the nonlinear polarization!P (N). It is this term acting

as source responsible for energy transfer between the coupled propagatingwaves.

5.2 Energy Transfer Rate

The rate at which work is done by an electric eld!E in creating the polarization!

P of a material is given by

dW

dt=

*!E d

!P

dt

+(75)

where the averaging is over one cycle of the alternating electric eld. Thereal parts of the electric eld and polarization may be written as

!E (!r ; t) = 1

2

h!E (!r ) ei!t +!E (!r ) e+i!t

i(76)

!P (!r ; t) = 1

2

h!P (!r ) ei!t +!P (!r ) e+i!t

i(77)

Note the amplitudes of the eld and polarization are written as complex andposition-dependent (which could come from the nonlinear wave coupling). Theequation for the average power now becomes on use of the above two equations,

dW

dt=1

4ha+ b+ c+ di (78)

9

where the bracketed terms a; b; c; d are:

a =!E (!r ) ei!t (i!)!P (!r ) ei!t (79)

b =!E (!r ) ei!t (i!)!P (!r ) e+i!t (80)

c =!E (!r ) e+i!t (i!)!P (!r ) ei!t (81)

d =!E (!r ) e+i!t (+i!)!P (!r ) e+i!t (82)

Of the four terms on the right, two of them (the rst and the last) haveangular frequencies 2! and 2!. On averaging over a cycle these give vanishingcontributions. The second term (b) and the third term (c) are the ones whichneed to be considered. Then

dW

dt=i!

4

D!E !P !E !P

E(83)

In a linear material, the polarization has the same frequency as the electriceld that creates it so that both the above terms give zero contribution. Thatis not however the case with a nonlinear material. We have learnt in an earlierchapter how to write the Fourier components of polarization of the n - th orderdue to a number of electric elds:

!P (n) (!r ; !) = 0(n) (!;!1; !2; ::; !n) be1be2::benE1::Enei!k 1+::!k n!r (84)where each electric eld is given by

!E (!r ; !i) = beiEieibkibr (85)

and the angular frequency ! of the polarization satises the photon energyconservation law:

! = !1 + !2 + :::+ !n (86)

For eective energy transfer to occur over the non-linear material sample,

the power which is proportional to!E !P !E !P

should not be a varying

function of position !r . In other words, the wave vector k should satisfy thelinear momentum conservation condition:

!k

!k 1 +

!k 2 + ::+

!k n

= 0 (87)

This condition is the well-known phase matching condition.

6 Three Wave Mixing

Recall the wave equation Eq.74 for nonlinear materials

r2!E = 0@2!E

@t2+ 0

@2!P (N)

@t2(88)

10

We will now consider how three plane monochromatic waves propagatingalong the X - axis in such a material interact. Let these waves be labelled1; 2; 3 have angular frequencies !1; !2; !3 respectively and respective angularwavenumbers k1; k2; k3. Further we label their Cartesian components usingsu xes i; j; and k. The equations of these waves are

E1i (x; t) =1

2

hE1i (x) e

i(!1tk1x) + c:c:i

(89)

E2j (x; t) =1

2

hE2k (x) e

i(!2tk2x) + c:c:i

(90)

E3k (x; t) =1

2

hE3j (x) e

i(!3tk3x) + c:c:i

(91)

where c.c. means the complex conjugate of the preceding term. Since thewaves are all travelling in the +X direction , the eld Cartesian componentsi; j; k can only refer to either y or z components.At this stage let us emphasize that the waves would be travelling without any

interaction between them if the material were linear. Further, if the material isnon-conducting, there would be no change in the eld amplitudes with positionx, as there is no current. On the other hand, if the material were a conductingmaterial ( 6= 0), there would be energy loss from the dissipation mechanismseven in a linear material. And the eld amplitudes would decay exponentiallyas the wave travels forward. By taking only nonconducting materials, we areavoiding the complication due to such linear energy changes with position.The amplitudes of our waves do change with position because we have a

nonlinear material and the nonlinear polarization couples the waves permittingenergy transfer between them. To deduce the position dependence of the am-plitudes, we proceed as follows. The Cartesian component i of the nonlinearpolarization at the angular frequency !1 = !3 !2 is given by

P(!1)i (x; t) = 0ijkE3jE

2ke

if(!3!2)t(k3k2)xg + c:c: (92)

To come back to the nonlinear wave equation, the term r2E1i (x; t) of thenonlinear wave equation, we may note that since the plane wave is travellingin the x direction, there would be no variation with the y and z directions:@@yE1i (x; t) =

@@zE1i (x; t) = 0, with the result

@2

@x2E1i (x; t) =

@2

@x2

1

2E1i (x) e

i(!1tk1x) + c:c:

(93)

or

r2E1i = 12

@2E1i@x2

k21E1i 2ik1@E1i@x

ei(!1tk1x) + c:c: (94)

11

6.1 An Approximation

As a simplifying approximation, we assert that the amplitude of the waveschange at a rate such that the change in the gradient of the wave envelope withrespect to x over one wavelength is negligible:

@@x

@E@x

@E@x or in terms of

the angular wavenumber k, the approximation implies that

@2E1i@x2

k1 @E1i@x

(95)

Thus in Eq.94, we omit the @2E1i@x2 term in comparison to the rst derivative

term. The nonlinear wave equation Eq.88 has the left hand side then given by

r2!E1i = 12

k21E1i + 2ik1

@E1i@x

ei(!1tk1x) + c:c: (96)

We turn to the terms on the right hand side of Eq.88: The rst term is

01@2!E 1i@t2

= 01@

@t

1

2

hi!1E1ie

i(!1tk1x) + c:c:i

(97)

= 12

h01!

21E1ie

i(!1tk1x) + c:c:i

(98)

where 1 is the permittivity of the material at the angular frequency !1.Since the phase velocity of the wave 1 is given by v1 = 1p01 and v1 =

!1k1, we

have 01!21 = k

21 and the above equation becomes

01@2!E 1i@t2

= 12

hk21E1ie

i(!1tk1x) + c:ci

(99)

This is just equal to the rst term on the right hand side of Eq.96. Sincethese appear on either side of the nonlinear wave equation they cancel eachother. Now what is left to be considered is the second term on right hand sideof the wave equation Eq.88. Using Eq.92, we have

0@2P1i@t2

= 00ijkE3j (x)E2k (x) fi (!3 !2)g2 eif(!3!2)t(k3k2)xg + c:c:

(100)

= 00!21ijkE3j (x)E2k (x) eif!1t(k3k2)xg + c:c: (101)

wherein we have made use of the relation !3 !2 = !1. Now equating thetwo remaining terms on either side of the nonlinear wave equation, (i.e. we areomitting the rst term k21E1i term of Eq.96 and the term given by Eq.97), weget

ik1 @E1i@x

ei(!1tk1x) + c:c: = 00!21ijkE3j (x)E2k (x) eif!1t(k3k2)xg + c:c:(102)

12

Or upon rearrangement and cancelling the common factor,

@E1i@x

=00!

21

ik1ijkE3j (x)E

2k (x) e

i(k3k2k1)x (103)

Introducing the phase mismatch term k = k3 k2 k1, and using therelation 01!

21 = k

21, the above equation becomes

@E1i@x

= i!1r010ijkE3j (x)E

2k (x) e

ikx (104)

In a similar manner we derive the gradients of the elds for the other twowaves, using the same approximations and the same mathematical manipula-tions:

@E2k@x

= i!2

r020kijE1i (x)E

3j (x) e

ikx (105)

@E3j@x

= i!3r030jikE1i (x)E2k (x) e

ikx (106)

In arriving at the above equations, we used the relations !1 = !3 !2 inthe equation for the wave 1, !2 = !1 !3 in the equation for the wave 2 and!3 = !1 + !2 in the equation for the wave 3.

7 Second Harmonic Generation

We are now ready to discuss second harmonic generation as a special case ofthe three wave mixing processes, with !1 = !2 = ! and !3 = 2!. In order todeduce the growth of the second harmonic eld, we need consider only one ofthe two elds given by Eq.?? and the Eq.106. The analysis becomes simplerif can make the assumption that the eld amplitude of the fundamental wavesuers negligible depletion, i.e. @E1i@x = 0. This is an oversimplication in manya practical case but the analysis has instructive features and we will take up thiscase rst and later remove this restriction by allowing for the depletion of thefundamental pump wave. Since this assumption means that the eld amplitudeE1i (x) remains constant not varying with position x, we need to integrate onlythe third equation for E3j . Writing !3 as 2!, this equation becomes

dE3jdx

= i2!r030jikE1iE2ke

ikx (107)

We now wish to integrate the above equation from x0 = 0 to x0 = x. Remem-bering that the two elds E1 and E2 both belong to the fundamental frequencyand so do not change with position, we haveZ

dE3j = i2!r030jikE1iE2k

Z x0

eikx0dx

0(108)

13

Integration yields

E3j = E3j (0) i2!r030jikE1iE2k

eikx0

ikjx0 (109)

If at the input end, x0= 0, there is zero second harmonic eld, E3j (0) = 0.

To obtain the wave intensity, we need the square of the eld amplitude:

E3j (x)E3j (x) = 4!

20320

2jikE

21iE

22k

eikx 1ik

eikx 1ik

(110)

E23j (x) = 4!20320

2jikE

21iE

22kx

2 sin2kx2

kx2

2 (111)We can now write down the intensity of second harmonic wave as a function

of position:

I2! =1

2

r30E22! (112)

Eq.111 then gives

I2! = 2

r03!220

2jikE

21iE

21kx

2 sin2kx2

kx2

2 (113)To nd what fraction of the energy of the incident wave is transferred to

the harmonic wave, we need the intensity of the fundamental frequency wave,which is

I! =1

2

r10E1iE

1k (114)

so that the e ciency of the conversion under the approximation of no deple-tion of the pump wave, is given by

I2!I!

= 8

r00!22jikn3

I!sin2

kx2

kx2

2 x2 (115)We note that for a given length x of the crystal sample, the e ciency of

conversion is proportional to 2, the square of the nonlinear susceptibility andto the intensity I! of the fundamental wave. Further, the e ciency oscillates asa function of the position x for k 6= 0. Maximum e ciency is realized if thephase matching conditionk = 0 is satised. Sincek = k3k2k1 = k32k1,the phase matching condition may be phrased as

k2! = 2k! (116)

which may be alternatively written as

k =2!

c(n2! n!) (117)

14

on using the relation k2! =2!v2!

= 2!cn2! for the harmonic and k! =!v!= !cn!

for the fundamental frequency waves. The factor f =sin2(kx2 )(kx2 )

2 has an oscillation

period called the coherence length lc dened by lc = 2k . Eq.117 then leadsto the expression for the coherence length

lc =

n2! n! (118)

What is this coherence length? We can see that when the phase matchingcondition is not satised, the length of the crystal over which we can realize themaximum cumulative energy transfer, is the spatial period of the oscillationsarising due the factor f .

7.1 Phase Matching Schemes

Phase matching in second harmonic generation facilitates maximum energytransfer from the pump eld to the harmonic eld. Eq.116 tells us that thewavenumber of the harmonic wave must be double the wavenumber of the fun-damental wave. Or the wavelength of the second harmonic wave must be halfthe wavelength of the pump wave. Which is the case when the phase velocitiesare equal for these two waves. Accordingly the refractive indices for the twofrequencies must also be equal:

n2! = n! (119)

There are two schemes, denoted as Scheme I and Scheme II, possible forrealizing the phase matching condition. First we detail Scheme I.

Scheme I

In this scheme, the optical anisotropy of uniaxial crystals is exploited. Ina negative uniaxial crystal, like KH2PO4, the refractive index of the extraor-dinary ray is less than that of the ordinary ray at any wavelength for a givendirection of travel. As may be expected, in normal dispersion, the refractiveindex (for both the rays) increases with increasing frequency. As a consequence,when both the fundamental and harmonic waves are both of the same kind (ei-ther ordinary or extraordinary), phase matching can not occur. However, if welaunch one of the waves, say, the pump wave as an ordinary ray, and the secondharmonic wave is extraordinary ray, then the refractive indices can be matchedfor a certain direction in the crystal, because the speed of the ordinary ray isisotropic whereas the extraordinary ray speed depends on the direction of travelin the crystal. Let be the angle made by the rays with the optical axis of thecrystal. Then the refractive index of the extraordinary ray ne () depends onthe angle according to the relation

1

fne ()g2 =cos2

(n0)2 +

sin2

(ne)2 (120)

15

We can choose an angle c, at which the refractive index ne2! (c) of theextraordinary ray at the second harmonic frequency 2! equals the refractiveindex n0! () of the ordinary ray at the fundamental frequency !. This meansthe angle c should satisfy the equation

cos2 c

(n02!)2 +

sin2 c

(ne2!)2 =

1

fn0!g2(121)

Solving for this phase matching angle, c, we nd that

sin2 c =

1(n0!)

2 1(n02!)

2

1

(ne2!)2 1

(n02!)2

(122)

We illustrate this scheme by considering the fundamental wavelength 1 =1:00 so that the wavelength of the second harmonic is 2 = 0:50 . At thesewavelengths, the refractive indices of KDP crystal are:1 (!) = 1:00 : n0! = 1:496044, n

e! = 1:460993

2 (2!) = 0:50 : n02! = 1:514928, ne2! = 1:472486

Using these data in Eq.122, we get c = 41:20.

Scheme II

In an alternative method, phase matching is accomplished by choosing oneof the input waves at the angular frequency ! as an ordinary ray and the othertwo as extraordinary rays. The frequency-doubled wave will be extraordinary.Now the phase matching condition (k = k3 k2 k1 = 0) becomes

ke3;2! ko2;! ke1;! = 0 (123)Since the refractive index of the extraordinary ray depends on the direction

of propagation, which makes an angle wit the optic axis of the crystal, thiscondition may be written as

ke2! () = ko2;! + k

e1;! () (124)

With the relation k = !nc , this becomes

2!

cne2! () =

!

cno! +

!

cne! () (125)

orne2! () =

1

2[no! + n

e! ()] (126)

Substituting the expressions for the angle-dependence of the refractive in-dices in the above equation yields as the matching condition of Scheme II as"

cos2 c

fno2!g2+sin2 c

fne2!g2#1=2

=1

2

(no! +

"cos2 c

fno!g2+sin2 c

fne!g2#)1=2

(127)

16

7.2 Accurate Treatment of Second Harmonic Generation

We have solved the coupled equations Eq.??, describing three wave mixing underthe approximation that the depletion of the pump wave is negligible. We shallnow remove this unnecessary limitation and solve the equations more generallywith the pump wave transferring appreciable amounts of energy to the harmonicwave.In order to facilitate an easy handling of the equations, we introduce new

eld terms F1; F2; F3 dened by

F1 =

rn1!1E1; F2 =

rn2!2E2; F3 =

rn3!3E3 (128)

We recall that the intensity of a wave is the energy per unit area owingper second. Therefore intensity is also equal to the photon ux multiplied byphoton energy. Photon ux is the number of photons per unit area passing persecond. As such, intensity is proportional to photon ux at a given frequency.

Intensity being proportional to the square of the eld amplitude!E 2, we see

our new eld variables are such that F 2 / E2! and so are proportional to thephoton ux!In terms of these new variables F , we may re-write the coupled wave equa-

tions Eq.??:dF1dx

= ir0!1!2!3n1n2n3

0F1 F3e

ikx (129)

dF2dx

= i

r0!1!2!3n1n2n3

0F1F3eikx (130)

dF3dx

= ir0!1!2!3n1n2n3

0F1F2eikx (131)

Introducing the parameter

=

r0!1!2!3n1n2n3

0 (132)

the above equations may appear as

dF1dx

= iF 1 F3eikx (133)

dF2dx

= iF1F3eikx (134)

dF3dx

= iF1F2eikx (135)For analyzing second harmonic generation, we know that we need to con-

sider only the rst and last of the above three equations. Further, if the phasematching condition is met, i.e. k = 0, then these equations become

dF1dx

= iF 1 F3 (136)

17

dF3dx

= iF 21 (137)Let us choose the pump wave at the input end (x = 0) to be real. Then

F1 (0) and hence F1 (x) become real giving

dF1dx

= iF1F3 (138)

dF3dx

= iF 21 (139)To obtain the solutions of these equations, we set S3 iF3. The two above

equations become thendF1dx

= F1S3 (140)

dS3dx

= F 21 (141)These are consistent with

d

dx

F 21 + S

23

= 0 (142)

orF 21 + S

23 = F

21 (0) (143)

since there is no harmonic eld at the input end (S23 (0) = 0). This resultsin the equation

dS3dx

= F 21 (0) S23 (144)or

dS3F 21 (0) S23

= dx (145)

Integrating from x = 0, S3 (0) = 0, to x = x, S3 (x), we have

1

F1 (0)tanh1

S23

F 21 (0)

= x (146)

orS3 (x) = F1 (0)

ptanh (xF1 (0)) (147)

The conversion e ciency is thus given by

jS3 (x)j2jF1 (0)j2

= tanh2 [F1 (0)x] (148)

The maximum conversion e ciency (unity) is reached when the argumentof the tanh2 function tends to innity.

18