quantum and nonlinear optics

8/11/2019 Quantum and Nonlinear Optics

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Contents

1 Introduction 1

2 Classical Theory (Lorentz Atom) 22.1 frequency response of the atom . . . . . 22.2 properties of a dipole ensemble . . . . . 3

2.2.1 n() and () . . . . . . . . . . 3

2.2.2 Susceptibility() . . . . . . . . 4

3 Classical Anharmonic Oscillator 43.1 noncentrosymmetric medium . . . . . . 4

3.1.1 perturbative approach . . . . . . 43.1.2 and P . . . . . . . . . . . . . . 5

3.2 centrosymmetric medium . . . . . . . . 6

4 Driven Wave Equation 74.1 The Paraxial Limit (Boyd 2.10) . . . . . 7

4.1.1 gaussian beam . . . . . . . . . . 8

4.1.2 guided waves (SMF) . . . . . . . 84.2 Nonlinear Processes . . . . . . . . . . . 84.2.1 Sum Frequency Generation (SFG) 94.2.2 Manley-Rowe relations (Boyd 2.5) 114.2.3 Second Harmonic Generation

(SHG) . . . . . . . . . . . . . . . 114.3 Phase Matching (Boyd 2.3) . . . . . . . 12

4.3.1 Birefringent phase matching(angle tuning) . . . . . . . . . . 12

4.3.2 Quasi-Phase Matching . . . . . . 134.4 Intracavity (2) processes (Boyd 2.8-2.9) 14

5 Intensity Dependent Refractive Index 155.1 3rd order nonlinearity . . . . . . . . . . 15

5.1.1 electro-optic Kerr effect . . . . . 155.1.2 orientational Kerr effect . . . . . 155.1.3 electrostriction . . . . . . . . . . 165.1.4 saturable absorption . . . . . . . 16

5.2 nonlinear phase shifts . . . . . . . . . . 165.2.1 self-phase modulation (SPM) . . 165.2.2 cross-phase modulation (XPM) . 175.2.3 self focusing . . . . . . . . . . . . 17

5.3 Phase Conjugation (Boyd 7.2) . . . . . . 195.3.1 Degenerate Four-Wave-Mixing . 195.3.2 nearly degenerate FWM . . . . . 21

5.4 Optical Bistability . . . . . . . . . . . . 215.4.1 Optical Switching (Boyd 7.3.3) . 21

5.5 tensor properties** . . . . . . . . . . . . 225.6 Two Photon Absorption . . . . . . . . . 22

5.6.1 two-photon microscopy . . . . . 235.7 NL phase shifts from(2) processes ** . 23

6 Nonlinear Pulse Propagation 236.1 Nonlinear Schodinger Equation . . . . . 236.2 = 0, k2= 0 . . . . . . . . . . . . . . . 256.3 GVD (= 0) . . . . . . . . . . . . . . . 256.4 nonlinear (k2 = 0) . . . . . . . . . . . . 266.5 (= 0, k2= 0) . . . . . . . . . . . . . . 27

6.5.1 solitons . . . . . . . . . . . . . . 28

6.5.2 Space-Time Duality . . . . . . . 29

6.5.3 Telecom Systems . . . . . . . . . 29

6.6 Generation of Short Pulses . . . . . . . 30

6.6.1 Mode Locking . . . . . . . . . . . 30

6.6.2 pulse generation in fiber . . . . . 31

6.6.3 Additive-Pulse Mode Locking(APM)** . . . . . . . . . . . . . 32

6.7 solitary lasers . . . . . . . . . . . . . . . 326.7.1 Ultrafast Measurment . . . . . . 32

7 Two-Level Systems 33

7.1 Optical Bloch Equations . . . . . . . . . 33

7.1.1 derivation . . . . . . . . . . . . . 33

7.1.2 comparison with classical dipole 36

7.1.3 steady-state solutions and TLAensembles . . . . . . . . . . . . . 36

7.2 density operator for ensembles . . . . . 38

7.3 Rabi Oscillations . . . . . . . . . . . . . 39

7.3.1 Bloch vector/sphere . . . . . . . 397.3.2 self-induced transparency (SIT) . 41

7.4 laser cooling . . . . . . . . . . . . . . . . 41

7.4.1 Bose-Einstein Condensate . . . . 43

7.4.2 semiconductors . . . . . . . . . . 44

7.4.3 photon echo . . . . . . . . . . . . 44

7.4.4 Electromagnetically InducedTransparency (EIT) . . . . . . . 44

1 Introduction

(Boyd 1.1, 1.2, 1.4)

For linear systems in= out, but is non-linear ata high enough E-field intensity.

The equation of motion for e:

m(x + 2x) Frestoring = eE(t) = eE0cos(t)

Frestoring = kx =m20x for a linear system, butif the electron is driven far enough, then Ux2 andFrestoring

x, and we must expand the restoring force

to model an anharmonic oscillator.

Frestoring = m(20x + ax2 + bx3 + . . . )

and the equation of motion is

m(x + 2x + 20x + ax2 + bx3) = eE(t)

Now solutions of cos(2t), cos(3t), etc. can be foundfor x. If the electron is driven by multiple externalfields then we see wave mixing (1+ 2, 1

2,

etc.)

Any material can be modeled as a collection ofdipoles:

P(r, t) =N (r, t) = N ex(r, t)

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where N is the density of atoms. This radiation canbe seen mathematically by putting Pin as the drivingterm for the wave equation.

2E 1c2

2E

t2 =

1

0c22P

t2

For high E, the dipoles created produce new frequen-cies that are radiated and produce yet more new fre-quencies.

Most common source of = 500 nm light comesfrom upconverted = 1m light from Nd:Yg laser.

Third order terms (from b term) require a larger Efield and so are less likely and less efficient. These leadto Third Harmonic Generation, and even a singlebeam can lead to Four Wave Mixing in a mediumwith third order non-linearity.

4= 1+ 2 3

With Four Wave Mixing, we see an intensity-dependent index of refraction n(I), which leads to avariety of effects such as self-focusing. This causes amaterial to act like a lens so that the beam convergesto a small area. We also see self-phase modulationwhere the phase of a wave is shifted due to the changesthat it induces in the material.

In most cases we consider a two-level system far fromresonance. Some examples of resonant nonlinearitiesare two-photon absorption, where two photons whoseenergies add to the resonant energy are absorbed byan electron at once, and Raman scattering, where aphoton, 1, is absorbed and two photons are emitted,one at the electronic vibrational frequency0 and oneat 2= 1

0.

2 Classical Theory (Lorentz Atom)

This works to accurately describe the frequency re-sponse of atoms under propagation of weak Efields.

mr= m20r 2mr e E(r, t)

where the first term on the right of the equals is therestoring force, the second is the damping caused bycollisions with other atoms, for instance, and the third

term is the non-relativistic Lorentz force on the elec-tron.

2.1 frequency response of the atom

To derive the response of the Lorentz atom to an ex-ternal E field, we start with the above equation. Weassume that E is a plane wave polarized in the x di-rection and propagating in the z direction:

E(r, t) = xE(z, t) = x(E0ei(kzt) + E0 e

i(kzt))

The second exponential term is written as c.c..If we consider the Lorentz force only in the x direc-

tion, then

xe+ 2xe+ 20xe =

e

mE(z, t)

We will assume also a monochromatic and a steady-state solution for xe(t):

xe(t) =x0eei(kzt) + cc

This is substituted into the equation of motion, andcanceling the exponential terms on either side:

(2 2i + 20)x0e = e

mE0

x0e = eE0/m

20 2 2iWe then find the dipole moment of the atom = qd:

(t) = exe(t) = ex0eei(kzt)

and 0 =ex0e, which gives as the amplitude of thedipole moment

0= e2E/m

20 2 2i

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Now we find an atomic polarizability a, which isdefined as the induced dipole moment of the atom di-vided by the external Efield:

a= e2/m

20 2 2iThis is an important classical result and a precursor

for non-linear effects to come.

2.2 properties of a dipole ensemble

Now we want to look at a collection of atomic dipoleswhich is how we model most materials. When we con-sider light propagation through such a material, wewant to understand the macroscopic interaction. Forthis we define a polarization Pwhich is the total dipolemoment of the material divided by the volume.

P(r, t) = 1

Vi

i(t)(r Ri)

In solids, the density is typically on the order of1020 atoms/cm3. We can treat the material as contin-uous and the atoms as uniformly distributed so thatall atoms in a small volume d3r experience the sameEfield. The polarization is found to be

P(r, t) = n

d3r(r, t) = N ex(r, t)

This is important as now we dont have to account for

1020 atoms and their local Efields.

2.2.1 n() and ()

We write the equations for light-matter interaction asfollows:

2 E 1c2

2 E

t2 =

1

0c22 P

t2 (Field)

x + 2x + 20x=

e

m

E (Matter)

Assuming propagation along z,2 /z, and as-suming general solutions for E and x, /z ik and/t i. Substituting E and P into the waveequation:

k2E0+ 2

c2E0=

N e2

0c2 x0

and substituting x into the equation of motion to getx0, we find

k2E0+ 2c2

E0=N e22

m0c2E0

20 2 2i

k2 =

c

2 1 +

N e2/m020 2 2i

We want to find a dielectric function for susceptibility,so we start with the permittivity:

() =0(1 + ())

At this point, we should indicate which functionsand values are complex with a tilde. The susceptibilityis defined as

() = N e2/m0

20 2 2i

() =Na

so that

k2 =

c

2[1 + ()]

k=

c

[1 + ()]1/2

If is small enough, i.e. the density is low enough,

then we can make the approximation

k

c

1 +

1

2()

=

c

1 +

1

2Re[] +

1

2Im[]

If we also definek to be

k= k+ i

then we find the real and imaginary parts

k= c 1 +1

2Re[()]= c n()

=

c

12

Im[()] =1

2()

where n() is the index of refraction and () isthe frequency-dependent absorption of the material.Rewriting these, we find

n() = 1 + N e2

20m

20 2(20 2)2 + (2)2

() = N e2

0mc2

2

(20 2)2 + (2)2If we now look at very close to 0, we see that

|0| |0+| and we can make the approximation

20 2 = (0 )(0+ ) 2(0 )

and this simplifies our expressions for n() and ()

n() = 1 + N e2

40m

0 (0 )2 + 2

() = N e2

20mc

(0 )2 + 2() is a Lorentzian and is defined as the absorptionline of the medium.

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If we put the complexkvector back into the Efield,we find that

E(z, t) =E0ez/2ei(kzt) =E(z)ei(kzt)

And using this we can find an intensity I(z)

I(z) =nc0|E(z)|2 =nc0|E0|2ez

I(z) =I(0)ez

2.2.2 Susceptibility ()

We already found the polarization to be

P(r, t) = N ex(r, t) =(N e2/m) E(r, t)

20 2 2i

andP(r, t) =0() E(r, t)

so that = P0/0E0

() = N e2/m0

20 2 2i

as was predicted in the previous section.

It should be noted that we have only been consider-

ing the case of continuous wave light, but in reality wewill have a spectrum of frequencies, which would notapply. To deal with this non-linear case, we decom-pose the signal into its frequency components via theFourier Transform.

P(z, ) =0()E(z, )

Converting back into the time domain

P(z, t) =(t) E(z, t) = d (t )E(z, t )The solutions arent always analytical, but in certainlimits we can find simple solutions, such as the casewhen the frequency is far from resonance ( 0).

3 Classical Anharmonic Oscillator

3.1 noncentrosymmetric medium

Now we add a higher-order correction to the restoring

force (considering only 1-d motion)

Frestoring = m(20x + ax2 + bx3 + . . . )

U(x) =

dxFrestoring =m(1

220x

2 +1

3ax3 +

1

4bx4)

Ifa = 0 then the medium doesnt have inversion sym-metry and U(x) =U(x). We call this type of medianon-centrosymmetric, which is only found in solids.Liquids and gases are isotropic.

We now have an equation of motion that looks like

x + 2x + 0x + ax2 = eE(t)/m

No general solution exists, so we must solve usingperturbation theory, which means that the E field isstrong but not too strong. We will see that even asmall nonlinearity can produce significant effects.

3.1.1 perturbative approach

First we must replace Eby E, where is some smallperturbation to the field. This will also help us keep

track of orders of perturbation. We now can representx(t) by

x(t) =x(1) + 2x(2) . . .

t which we can substitute into the above equation ofmotion and group like terms ofn.

(x(1) + 2x(2) . . . ) + 2(x(1) + 2x(2) . . . )

+ 20(x(1)+2x(2) . . . ) + a(2(x(1))2+23x(1)x(2) . . . )

= eE/m

and collecting terms1 : x(1) + 2x(1) + 20x

(1) = eE(t)/m2 : x(2) + 2x(2) + 20x

(2) + a[x(1)]2 = 0

3 : x(3) + 2x(3) + 20x(3) + 2ax(1)x(2) = 0

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Note thatx(1) becomes the driving factor in the secondorder equation.

Consider an input E field with two frequencies

E(t) =E1(1)ei1t + E2(2)e

i2t

For the first order of perturbation, the atoms will re-spond by oscillating at 1 and 2

x(1)(t) =x(1)(1)ei1t + x(1)(2)ei2t

and

x(1)(j) = eEj/m

20 2j i2jThis denominator will become cumbersome to write asit will appear frequently, so we write the above equa-tion as

x(1)(j) =eEj/m

D(j)

Well now put this result into the equation for thesecond order of perturbation, 2.

x(2) + 2x(2) + 20x(2) =

a{ [x(1)(1)]2ei21t + [x(1)(2)]2ei22t+ 2x(1)(1)[ x

(1)(2)ei(1+2)t + x(1)(2)e

i(12)t]

+ 2|x(1)(1)|2 + 2|x(1)(2)|2 }Thus we have transformed a nonlinear problem into adriven linear problem. The terms can all be superim-posed, so we can pick out a driving term and find its

solution.First well look at the driving term ei22t, which

will give us the equation

x(2) + 2x(2) + 20x(2) = a

eE1/m

D(1)

2ei21t

If we assume the solution for x(2)(t) to be

x(2)(t) =x(2)(21)ei21t

well find that

(421 4 + 20)x(2)(21) = a(eE1/m)

2

D2(1)

and

x(2)(21) = a(e/m)2E21

D(21)D2(1)

We can similarly find the solutions for the other driv-ing terms

x(2)(22) = a(e/m)2E22D(22)D2(2)

x(2)(1+ 2) = 2a(e/m)2E1E2

D(1+ 2)D(1)D(2)

x(2)(1 2) = 2a(e/m)2E1E

2

D(1 2)D(1)D(2)

There is also a DC termx(2)(0) which we call opticalrectification, but we will ignore this as it just shiftsthe AC field. Note the 2 in the x(2)(1 + 2) andx(2)(1 2) terms. This must be included as thereare two ways to to add 1 and 2. These can in turnbe plugged in to find the x(3) solutions.

3.1.2 and P

Polarization will similarly depend upon the input fre-quencies

P = N ex(t) = N e(x(1)(t) + x(2)(t) + . . . )=P(1)(1)e

i1t+P(1)(2)ei2t+P(2)(21)e

i21t+. . .

and

P =0((1) E+ (2) E E+ (3) E E E+ . . . )

sincej is a tensor of rankj

1. Combining these two

relations, we can derive expressions for the nonlinearsusceptibilities

(2)(i+ j, i, j) = N e3a/0m

D(i+ j)D(i)D(j)

where i and j are the input frequencies and i +j is the output. The SFG and SHG second-ordersusceptibilities are as follows

(2)(21, 1, 1) = N e3a/0m

D(21)D2(1)

(2)(1+ 2, 1, 2) = N e3a/0m

D(1+ 2)D(1)D(2)

(2)(1 2, 1, 2) = N e3a/0m

D(1 2)D(1)D(2)Note that D() =D().

Typically we want to the much less than 0 sothat there is little or no absorption. In this limit

1 0: D(1) 2021

0: D(21)

20

so we can come up with an estimate for (2)

(2) N e3a

0m260

Now we need an estimate for the parametera. We canrewrite the potential as

U(x) =1

2m20x

2(1 + 2a

320x + . . . )

We expect the anharmonicity of the electrons to be-

come important when the second term in the expan-sion becomes appreciable to the limit, i.e., |x| becomescomparable to the radius of the atom, a0. That is, if

2a

320a0 1 a 3

20

2a0

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so now we have

(2) 3N e3

20m240a0

which comes out to be around 6.9 1012 m/V formost materials.

3.2 centrosymmetric medium

In a centrosymmetric medium, the atoms are locatedat sites of inversion symmetry. The restoring forcecannot contain any even power terms

Frestoring = m20x + mbx3

U(x) = 1

2m20x

2 14

mbx4

We will see the third order nonlinearities of isotropicmedia (band 0 are independent of direction as in air,glass and liquids). Well also keep the vector nature ofthe electron motion to highlight the vector nature of(3).

Frestoring = m20r+ mb(r r)rSimilarly, the equation of motion is

r+ 2r+ 20r b(r r)r= eE(t)/mAgain well look at the perturbative solutions.

r(t) =r(1) + 2r(2) + 3r(3) + . . .

for E(t) E(t). And matching equal powers ofr(1) + 2r(1) + 20r

(1) = eE(t)/mr(2) + 2r(2) + 20r

(2) = 0

r(3) + 2r(3) + 20r(3) =b[r(1) r(1)]r(1)

The first power of is just the driven linear equation,and we already know the solution to this.

r(1)(t) = n

r(1)(n)eint

where

r(1)(n) =e E(n)/m

D(n)

P(1)(t) = N er(1)(t) =n

0(1)(n) E(n)e

int

In the second power of, the equation of motion isdamped and not driven, so the steady-state solutionwill be zero. The third order solution is more interest-ing. Since the term r(1) appears three times we needto sum over three indices

[r(1)

r(1)

]r(1)

=mnp

e3m3

[ E(m) E(n)] E(p)D(m)D(n)D(p)

ei(m+n+p)t

Now well assume a sum of oscillating solutions forthe position vector

r(3)(t) =q

r(3)(q)eiqt

When we substitute this into the third-order equationof motion and cancel the oscillating part we find

r(3)(q) = q

be3m3

[ E(m) E(n)] E(p)D(q)D(m)D(n)D(p)

and when we sum over this, we must apply the con-straintq =m+n+p. We can find the third-orderpolarization as well

P(3)(q) =mnp

N be4

m3[ E(m) E(n)] E(p)

D(q)D(m)D(n)D(p)

As for the third-order susceptibility, the tensor prop-erties make solutions much more complicated. There

are many ways to form a triple product of fields, andin general they dont take the form we need for anisotropic medium. Using the anharmonic atom model,we can derive much simpler solutions.

(3) can be expressed as a product of(1)s

(3)(q; m, n, p) =

bm30N3e4

(1)(q)

(1)(m)(1)(m)

(1)(p)

For 0 we can make similar approximations aswe did for the non-centrosymmetric case: D() 20

|(3)| N be40m380

The potential energy is

U(x) =1

2m20x

2(1 bx2

220)

so that when x becomes comparable to a0 nonlineareffects become important

ba20220

1 b 20

a20

Takinga0= 0.3 nm and 0= 7 1015 rad/s, we findthat

|(3)| 344 1024 m2/V2which is close to that of glass. (skip from pg 29 to pg36)

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4 Driven Wave Equation

Now that we have P(1),P(2),P(3), etc. we can insertthese into Maxwells equations to get a self-consistenttheory of light propagation when nonlinear effects be-come important. The wave equation becomes

2 E

1

c2

2 E

t2 =

1

0c2

2 P

t2

We can assume the plane wave limit as this provesto be accurate in a few cases, including single-modefibers and in thin crystals.

We now plug into the wave equation

E(r, t) =n

En(r)eint

andP(r, t) =

n

Pn(r)eint

to find

2 En+

2n

c2

En=

2n

0c2

Pn

Splitting Pinto its linear and nonlinear components

Pn= PLn +

PNLn

=0(1)(n) En+ PNLn

2 E+ 2n

c2(1 + (1)) En=

2n

0c2PNLn

We must assume that the Efield consists of a rapidlyvarying phase and a slowly varying amplitude

En(r) =An(z)eiknz PNLn (r) =

PNLn (z)eiknz

Note that in assuming this, we must also assume thatthe nonlinear medium is simple enough that E andPNL can propagate without the polarization changing.The index of refraction n() will also be well-defined

in this case, so kn= n()n/c.

2 En= d2Endz2

= d2An

dz2 + i2kn

dAndz

k2nAn

For the next term we assume also that

1 + (1) =(n) =n2(n)

so that

d2An

dz2 + i2kn

dAn

dz

k

2

nAn+

k

2

nAn= 2n

0c2P

NL

n

This is exact, but if we want to make any process wehave to make an assumption. We say that the envelopefunction is varying much slower than the modulatingfunction, and so we can neglect the first term

| ddz

dAndz

| |2kndAndz

|

This is called the Slowly Varying Envelope Approxi-mation (SVEA). With this, we find that

dAndz

= in

0n(n)cPNLn (S V E A)

Now we have a first order differential equation insteadof a second order, which simplifies things considerably.

4.1 The Paraxial Limit (Boyd 2.10)

In the paraxial limit we can assume that a beam is aplane wave, and this turns out to be a good approxi-

mation close to the confocal parameter b. We go backto the wave equation

2 En+ 2n

c2(1 + (1)) En=

2n

0c2PNLn

and substitute in

En(r) =An(x,y ,z)eiknz

PNLn (r) =Pn(x,y ,z)eiknz

Since E or P are varying slowly in the z direction,2/z2 0, and

2tAn+ i2knAnz

= 2n

0c2PNLn

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4.1.1 gaussian beam

Now we want to look at the free space case, PNLn = 0.

2tAn+ i2knAnz

= 0

This equation can be solved for, and the family of so-lutions founds is called the Hermite-Gaussian modes

An(r, z) = An w0w(z)

er2/w2(z)eiknr

2/2R(z)ei(z)

where

w(z) =w0

1 +

nz

w20

21/2

is the radius of the beam (where A 1/e)

R(z) =z 1 + w20

nz 2

is the radius of curvature of the optical wavefront (zeroat the middle), and

(z) = tan1

nz

w20

the phase of the wavefront.

Atz = w20/n the amplitude A is reduced by 2and the intensityI reduced by a factor of 2. The con-focal parameter b is defined as twice this value. Also,at z = 0 the phase front phase shift (Guoy phaseshift) where the phase fronts switch inflection.

4.1.2 guided waves (SMF)

To get high nonlinear effects, we must have high in-tensity and a long interaction length, but in a beamthere is a tradeoff between I and L. In a waveguide,however, the interaction length is as long as the fiberso its possible to see high nonlinear effects with verylow input powers. ( 1 W).

Usually we are interested in single mode propaga-tion where

#modes 2a

n21 n22

There is a cutoff wavelength c where nothing willpropagate. Only < c can propagate.

The field for the fundamental mode is

E(r, t) =A(z, )F(x, y)ei(zt)

where the traverse intensity distribution is defined in-side and outside of the fiber in terms of Bessel func-tions:

F(x, y) =

J0() a

a/J0(a)e(a) > a

where is the propagation constant. Here we define

2 =

2n1

2 2

2 =2

n22k

20

and similar to the plane wave case, we can derive

dA

dz =

i

20ncPNL

This is done in detail in Agrawal. In this class though,the main importance of fibers is their ability to containthe field for long interaction lengths, which enhancesthe nonlinear processes greatly.

4.2 Nonlinear Processes

In general, PNL PL, but propagation can be dom-inated by nonlinear effects. (2) can lead to sum fre-quency generation and second harmonic generation,which well see next, and (3) leads to an intensity-dependent refractive index n2.

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4.2.1 Sum Frequency Generation (SFG)

We assume that 0 so absorption can be ne-glected, and that (2) is entirely real so the materialhas permeation symmetry. We start with

Ei(z, t) =Ei(z)eiit =Ai(z)e

i(kizit)

Pi(z, t) =Pi(i)eiit

Generally we see that

(n+m) =0jk

mn

(2)ijk(m+n; m, n)Ej(m)Ek(n)

and so

Pi(3) = 20jk

(2)ijk(3; 1, 2)Ej(1)Ek(2)

the factor of two comes from the fact that the twoinput frequencies can be added in two ways.

For simplicity well assume that both input fields arepolarized in the same direction, x

Pi(3) = 20(2)ixxEx(1)Ex(2)

and we define 2d= (2). This will come in handy forderiving the coupled-amplitude equations.

Well start with P1(z, t), which is given generally afew lines above. Plugging in P1(2)

P1(z, t) = 20(2)(1; 3, 2)E3E2 ei1t

= 40dA3A2ei(k3k2)zei1t

Here we define the wave vector mismatch as k =k1+ k2 k3 and using the definition of a wave

P1(z, t) =P(2)1 (z)e

i(k1z1t)

we find that

P(2)1 (z)e

i(k1z1t) = 40dA3A2ei(k3k2)zei1t

P(2)1 (z) = 40dA3A

2ei(k3k2k1)z

P(2)1 (z) = 40dA3A

2eikz

Using the same technique, we find that

P(2)2 (z) = 40dA3A

1eikz

P(2)3 (z) = 40dA1A2e

ikz

We now plug these into the wave equation

dAndz

= in20nc

PNLn

and define eta n = 2dn/nc. Plugging in our nonlin-ear polarizations we found above we have the coupledamplitude equations

dA1dz

=i1A3A2eikz

dA2dz

=i2A3A1eikz

dA3dz

=i3A1A2eikz

4.2.1.1 weak conversion limit Lets first look atthe limit of low conversion from1and 2to3. If any3 photons are produced then a 1 and a 2 photonmust be annihilated, thus depleting the input fields.

In the case of low conversion, we take A1 and A2 tobe constant and we can integrate to find an expressionfor A3 from the coupled amplitude equation above

A3(L) =0

A3(0) + i3A1A2

L/2L/2

eikzdz

=i3A1A21

ik(eikL/2 eikL/2)

=i3A1A2Lsin(kL/2)

kL/2

=i3A1A2Lsinc(kL/2)

But we really measure P or I and

Ii = 2ni0c|Ai|2

so

I3(L) = 2n30c[i3A1A2L sinc(kL/2)]2

= 2n30c23|A1|2|A2|2L2 sinc2(kL/2)

= n3

23

2n1n20cI1I2L

2 sinc2(kL/2)

=642

d2

2

3n23c

2 n32n1n20cI1I2L2 sinc2(kL/2)

=

32223

0c3n1n2n3

(d2I1I2L

2) sinc2(kL/2)

For phase matching, we need all waves to have thesame velocity.

3 = 1+ 2

9


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k3= k1+ k2

As kL goes farther away from zero, the wave vec-tor mismatch gets larger and the interference becomesdestructive. In this case, E3 will not build up. Thephase matching condition for SFG is

n33= n11+ n22

Lets look at a phase matched example to see whatkind of input powers we need to get an appreciableoutput power.

We assume that I1 corresponds to frequency of and I2 has 2 and I3 has 3, all propagating througha crystal of length L = 1 cm. The fundamental fre-quency = 2c/(1050nm)

3 =

322(5 1015)2

(8.85 1012)(3 108)3(1.5)3

(1012)2I1I2(1cm)2

= 1012 m2/W I1I2

and taking I1

= I2

= 1010 W/m2

I3(L) = 108 W/m2

which is only about 1% of the input fields. The onlyway to achieve this kind of power is by using a pulsedlaser.

4.2.1.2 SFG with one strong input Lets con-sider another example where one input field is weakand one is strong.

Well assume that I2 is strong and undepleted, andthat I1 is weak at L = 0, so A2 is constant. We canrewrite the coupled equations as

dA1dz

=i1A3A2eikz =i1A3e

ikz

dA3dz

=i3A1A2eikz =i3A1e

ikz

and we can derive solutions that oscillate out-of-phase

d2A3dz2

=i3

dA1dz

+ ikA1

eikz

=i3

i1A3e

ikz +ik

i3

dA3dz

eikz

eikz

d2A3

dz2

=ikdA3

dz 13A3

Assuming a solution A3 ez, we find that

= i[k/2

(k/2)2 + 12]

=i[k/2 ]

so now we have general solutions

A3(z) =C+e+z + Ce

z

A1(z) = dA3dz eikz

i3

= 1

i3[+C+e

(+ik)z + Ce(ik)z]

and we use the crystal boundary conditions to evaluateC. A1(0) =A

01 and A3(0) = 0 at the input so that

A3(L) = i3

A01e

ikL/2 sin(L)

A1(L) =A01eikL/2 cos(L) + ik2sin(L)

and finally finding the intensities

I3(L) = 31

I01 sin2(

13L) =

31

I01 sin2(

13L)

I1(L) =I01 cos

2(

13L)

From this graph, its easy to see that the energysloshes back and fourth between the 1 and 3 fields.The process first pumps energy into the sum frequencyfield, but when the photons in the input field 1 aredepleted, the process reverses the flow. This is calledback-conversion and must be avoided to produce ap-preciable sum frequency effects.

This is the case of perfect phase matching, but if thefields are not phase matched, then we get a lower con-version efficiency and faster cycles of conversion/back-conversion.

10


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4.2.2 Manley-Rowe relations (Boyd 2.5)

SFG can be described on the photon level by

h1+ h2= h3

This is just conservation of energy, which agrees withQuantum Mechanics, comes from the coupled ampli-tude equations which are derived from classical E&M.

Starting with

Ii = 2ni0cAiAi

the variation in intensity over z is

dIi

dz = 2ni0cAidAi

dz + Ai

dAi

dz and plugging in the coupled equations

dI1dz

= 2n10c2d21k1c2

(iA1A2A3e

ikz)

= 40d1(iA1A

2A3e

ikz)

and similarly for the other two

dI2dz

= 40d2(iA1A

2A3e

ikz)

dI3

dz = 40d3(iA

1A

2A3eikz

)Ensuring that total intensity is not depleted

I=I1+ I2+ I3

dI

dz =

dI1dz

+dI2

dz +

dI3dz

= 0

from this we can see that 1 +2= 3as expected, butnow if we divide each intensity by its correspondingfrequency

d

dz I1

1= d

dz I2

2= d

dz I3

3where Ii/iis proportional to the number of photons inthe field. As1 photons are created, the same numberof2photons must also be created, and for each1and2photon pair created, a 3photon must be destroyed.

4.2.3 Second Harmonic Generation (SHG)

Using SHG we can efficiently produce lasers with wave-lengths much shorter than currently available. For anoutput of 2there is only one distinguishable possibil-ity for the polarization

P2(2) =0(2)(2; , )E1E1

and for an output of there are two distinct configu-rations

P1() = 20(2)(; 2, )E2E1

and from these we can find the coupled amplitudeequations for SHG.

P2(z, t) =P2(2)ei2t =0

(2)E21 ei2t

= 20dA21ei2k1zei2t

=P(2)2 (z)e

i(k2z2t)

where

P(2)2 (z) = 20dA

21ei(2k1k2)z = 20dA

21eikz

In this case the wave vector mismatch is k= 2k1k2.We find P1 in the same way

P1(z, t) =P1(1)ei1t = 20

(2)E2E1 e

i1t

= 40dA2A1ei(k2k1)zei1t

=P(2)1 (z)e

i(k1z1t)

where

P(2)1 (z) = 40dA2A1eikz

These are the nonlinear polarizations that we plug intothe paraxial wave equation to get the coupled equa-tions

dA1dz

=i1A2A1eikz

dA2dz

=i2A21eikz

The general solutions of these can be found and ex-pressed in terms of the Jacobi elliptic functions. (skip

pg 58-61) We skip right to the intensities derived fromthese coupled amplitude equations

I1(L) =I1(0)

1 tanh2

2dL

322

0n1n22c3

I1(0)

11


12/45

I2(L) =I1(0) tanh2

1dL

322

0n21n2c3

I1(0)

In practice, only about 80% efficiency is achieved.This is due to the fact that real laser beams have arange ofk vectors which are not all perfectly phase-matched. The more the wave vectors are mismatched,

the faster the energy will slosh back and fourth fromI1 and I2.

4.3 Phase Matching (Boyd 2.3)

As mentioned before, phase matching allows us to keepthe waves we want in phase so that we can get appre-ciable output intensities. For SFG the phase matchingcondition is

k= k1+ k2 k3= 1c

(n11+ n22 n33)In normal dispersion this cannot be satisfied. Simplyputting waves into the material, we can find the coher-ence length Lc or the length over which the waves areapproximately in phase.

Lc= 2/k

Which means that after propagating Lc the waves areout of phase by . Typically this length will be on theorder of 100 m. In a crystal that is 1 cm in length,we see that I3 varies as follows

The sum frequency never gets the chance to increaseto any appreciable level. Fortunately there are twomain approaches to getting the waves in phase. Thefirst is to use a birefringent material, or one whose in-dex of refraction depends on the polarization, and senddifferent waves down the two axes. The second is touse Quasi-Phase Matching, which requires a materialwith a periodic (2) so that the output frequency may

continue to grow instead of deplete.

4.3.1 Birefringent phase matching (angle tun-ing)

Phase matching is impossible for the normal dispersioncase.

For SHG, n(2)= n() in any case. If we use amaterial with birefringence, the and 2 fields willfeel different indexes of refraction and will thus travelat different speeds. We can arrange the wave totravel down the slow axis and the 2 wave to traveldown the fast axis so that the two are perfectly inphase for the length of the crystal. n() =n(2) will

let k= 0

A uniaxial crystal has one axis that is different fromthe other two, called the extraordinary or c-axis.

Light coming in along k is polarized in the twoperpendicular directions, the ordinary which doesntchange upon rotation of the crystal through , andthe extraordinary which changes depending on . Wedefine the ordinary refractive index as no and the ex-traordinary refractive index asne. ne(= 0) =no andne(= 90) = ne.

12


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For 0 < < 90 we can derive the following fromclassical E&M propagation

1

n2e()=

sin2()

n2e+

cos2()

n2o

Positive Uniaxial Negative Uniaxial(ne > no) (ne < no)

Type I no33= ne11+ ne22 ne33= no11+ no22Type II no33= n

o11+ n

e22 n

e33= n

e11+ n

o22

Well look at the positive uniaxial case in a Type Icrystal so that e + e o. As the crystal is turned, nevaries along an ellipse.

For SHG we have two such ellipses, one for the wave and another for the 2 wave. We find the phasematching angle at the points of intersection of the no2

and ne, which implies that the phase matching condi-

tion in this case is no2 =ne, so we can find the phase

matching angle

1

n2e()=

sin2(m)

n2e() +

cos2(m)

n2o()

1

n2o(2)=

sin2(m)

n2e() +

cos2(m)

n2o()

1

n2o(2)

=sin2(m)

n2e()

+(1 sin2(m))

n2o()

sin(m) =

1n2o(2) 1n2o()1

n2e() 1n2o()

The negative uniaxial case is the same except the sare replaced by 2s.

There are very few materials that exhibit birefrin-gence and a high (2) nonlinearity, however, whichleads to the next type of phase-matching.

4.3.2 Quasi-Phase MatchingMaterials with a high (2) but no birefringence canbe used in this application (e.g. semiconductors andpolymers). In practice we can get a much strongernonlinear effect in this way. The idea is to change the

sign of the nonlinear susceptibility just when the sumfrequency starts to deplete. This happens when thewaves are about out of phase, so the (2) shouldchange sign with period 2Lc, the interaction length.

Note that we call this process Quasi phase match-ing because the waves are globally phase matched butlocally mismatched.

Well start from the coupled amplitude equation

dA2dz

=i(2)2

n2c A22e

ikz

and assume that (2) is periodic in z

(2) =(2)0 sin(2z/a) =

(2)

02i

ei2z/a ei2z/aso we can rewrite

dA2dz

= i2A

22

n2c

(2)0

2i

ei(k+2/a)z ei(k2/a)z

Integrating to get A2, we can make some simplifica-tions.

A2=2A

22

(2)0

2n2c dz

ei(k+2/a)z ei(k2/a)z

Since the material is transparent,n1< n2and so k 1

An applied electric field induces a torque on themolecule, and since E is changing we find energychange per d E

dU=p d E= p3dE3 p1dE1= 3E3dE3 1E1dE1

U= 12

(3E23+ 1E

21 )

but since E1= Esin and E3= Ecos

U= 12

(3E2cos2+ 1E

2sin2)

= 12

E2[1+ (3 1)cos2]

Considering an intensity-dependent refractive index

n2 = 1 + = 1 + Nand to findwe note that

U = 12|E|2

so its clear from our expression for U that

=1+ (3 1)

cos2

where cos2is the expectation value of cos2in ther-mal equilibrium, which we find in terms of the Boltz-mann distribution

cos2

=

dcos2eU()/kT

deU()/kT

15


16/45

which we can find the n experienced in the weak opticalaxis (U= 0)

cos2

0

=

0 dsincos

20 dsin

=1

3

This gives us

= 23

1+1

33

and finally

n20= 1 + N

2

31+

1

33

And through some more rigamarole we can find that

n2= N

45n0

(3 1)2kT

This value is typically on the order of 3 1022m2/W.The response time is on the order of 1 ps, which is 100

times slower than the electronic response.

5.1.3 electrostriction

This is the tendency for materials to become com-pressed under an applied electric field. This changes and thus n. We can find a reduced expression for thenonlinear index

n2= 3

4n200c(3)

The nonlinearity due to electrostriction is not partic-ularly big, accounting for about 20% of(3).

5.1.4 saturable absorption

Considering an input beam at 0, the probabil-ity that atoms will be found in the upper state |2increases while decreasing for the lower state|1. Thetransition probability follows Fermis golden rule

P =2

h|2| He |1|2states

With a high enough intensity, the number of photonsis so high that the excitation rate is higher than thedecay rate, so that no more photons can be absorbed.Thus absorption saturates and decreases.

= 0

1 + I/Is

where Is is the intensity at which absorption is satu-rated.

(1)

1 + I/Is (1)

[1 I

Is + . . . ]

= (1) (1)

IsI

so that (3) (1)/Is.

5.2 nonlinear phase shifts

5.2.1 self-phase modulation (SPM)

Because the third order refractive index is dependenton I, the wave will effective itself, and accumulatephase as it propagates. Upon propagating z thephase will be

= L + NL(I) =2

n0z+

2

n2Iz

Looking at a pulse, the phase shift will vary acrossthe pulse as intensity increases then decreases. Theinstantaneous frequency is inst =/t, whichleads to a change in the frequency spectrum of thepulse.

Lets start with a linearly polarized electric field ina (3) medium

P(3) = 30(3)|E|2E

where E(r, t) =A(z)eik0z

P(3)(z) = 30(3)|A|2A

and plugging this into the canonical wave equation

dA

dz =

i

20ncPNL =

i3(3)

2nc |A|2A

Since there is only one field and no absorption, theenergy cannot dissipate so that I |A|2 is constantthroughout the medium (I(z) =I(0) =I).

dA

dz =i

3(3)

40n2c2I

A

A(z) =A(0)eiNL (z)

where

NL(z) = 3(3)

40n20c2

Iz=

cn2Iz

where n2 is the nonlinear refractive index we found inthe section above. The wave also picks up the usuallinear phase shift, so combined we find that

k=

c(n0+ n2I) E(z) =A(0)eik0zeiNL (z)

16


17/45

5.2.2 cross-phase modulation (XPM)

As with (2) we want to consider multiple incidentbeams. To keep things simple, well start with twobeams at coming in two different directions.

E1= yA1(z)ei(kzt)

E2= yA2(z)ei(kxx+kzzt)

wherekx= k sin and kz = k cosandk = k2x+ k2z .The total field is then

E(r) =A1(z)eikz + A2(z)e

i(kxx+kzz)

which we plug into the expression for polarization tofind

P(3)(r, ) = 30(3){(|A1|2 + 2|A2|2)A1eikz

+ (2|A1|2 + |A2|2)A2ei(kxx+kzz)+ A21A

2ei[kxx+(2kkz)z]

+ A1A22ei[2kxx+(2kzk)z]}

The first line demonstrates that the nonlinear phaseshift on A1 due to A2 is twice as strong as the effectfrom A1. That is, XPM is twice as strong as SPM.This makes physical sense because a beam needs tochange the medium, then feel the change, which mustbe half as strong as feeling the change directly fromanother beam. The same physical intuition applies tothe second line.

The third and fourth lines demonstrate nonlinearpolarizations in new directions

(skipped 94,95,96)

5.2.3 self focusing

Now we are dealing with a wave that is not a planewave, so the intensity in the middle of the beam ishigher than that of the outside. Due to the intensity

dependent index of refraction, a beam will tend to self-focus as it propagates.

In the figure, a beam diffracts at low intensity, but asyou increase the power, the beam will self-focus until itis self-trapped, then collapse as the power increasesmore. The collapse will result in either damage or

generation of other nonlinear processes that keep thebeam from focusing further.

Its also worth noting that a self-trapped beam in 2-D is unstable, but in 1-D its possible to create a beamthat is stable.

5.2.3.1 critical power For the 2-D system, wellstart with the paraxial wave equation to find the powerat which self-focusing exactly balances diffraction in amedium. Well assume a gaussian beam of the form

An(r, z) =A0w0

w(z)

transversefield

distribution er

2/w2(z)

diffraction eikr

2/2R(z) ei(z)

which solves the paraxial wave equation for PNL = 0.Assuming that this beam is incident on a(3) materialright at z = 0

I(r) = 2n00c|A(r, z = 0)|2 =I0e2r2/w20so that

n(r) =n0+ n2I0e2r2/w20

By expanding the exponential, we get a first order ap-proximation ofn(r)

n(r) n0+ n2I0(1 2r2/w20)and we use this to find the nonlinear phase shift

NL(r, z) c

n2I0(1 2r2/w20)z

= NL(0, z) + NL(r, z)

This nonlinear phase shift is added directly to the am-plitude equation as follows

An(r, z) A0 w0w(z)

er2/w2(z)eikr

2/2R(z)

ei(z) eiNL(0,z)eiNL (r,z) nonlinear phase

17


18/45

From the gaussian beam section we have the equa-tions forw(z), R(z) and (z), and at z we find that

w(z) w01

R(z)=

z(1 +

b

2z)

1 4z

b2

and rewriting the amplitude equation

An(r, z) A0 er2/w20 ei2(r2/b2)kz

ei(r2/w20)(2/c)n2I0z ei(z) eiNL (0,z)

Note that the last two factors do not have any r-dependence. Its clear now that the radial term (3rdexponent) can cancel the diffraction term (2nd expo-nent).

2r2kz

b2 =

2r2n2I0z

w20c

n0c

1k2w40

= n2I0

w20c

Where k = 2n0/. Now we want to relate the inten-sity and power of a gaussian beam

I(r) =I0e2r2/w20

P =

2rdrI(r) =

1

2w20I0

and replacingI0 with Pcr

n0

c

2

42n20w40

= n2

w20c

2Pcr

w20

Pcr= 2

8n0n2

Notice that the beam waist cancels so that this condi-tion is entirely based on the power, and not on inten-sity. At = 1 m, in glass Pcr 106W and in CS2Pcr 3 104W.

5.2.3.2 z-scan Using the z-scan, we can determineboth the sign and the magnitude ofn2. A laser beamis focused using a lens, which brings the laser to a fo-cus that then diffracts. A power meter on the rightcaptures some of the light that comes through the pin-hole.

By putting a material at z 0 thelight is either focused or diffracted, making the powerthrough the pinhole in front of the detector more orless.

As the material is translated from left to right, thefocus is moved and the amount of power that reachesthe detector varies as in the figure above (looks alot like n2). http://www.optics.unm.edu/sbahae/z-scan.htm

Now that we have the critical power Pcr, we wantto find the characteristic length of the medium. Wellassume that P Pcr so that the beam comes to afocus in the medium.

Using Pythagorean theorem we have that the hy-potenuse is

h=

w20+ z2sf

and using Fermats principle which states that the op-tical path length

n(r)dl of all rays traveling from a

wavefront at the input face to the self-focus must beequal

w20+ z

2sf (n0+ n/2) zsf(n0+ n)

zsf

1 +12

w20z2sf

+ . . . zsf(n0+ n)

zsf w0n/n0

and in terms of power

zsf w20

1P/Pcr

=LNL

The characteristic length for diffraction goes as

LDF =

b

2 =

n0w20

Diffraction and self-focusing balance when LNL =LDF, which brings us to the same conclusion

Pcr= 2

2n0n2

18


19/45

A more rigorous approach involves using the parax-ial wave equation, which will give rise to the nonlinearschrodinger equation (NLSE)

i

t = 2+ V SE

iA

z =

k22

2A

t2 i|A|2A NLSE

where k2= 2k/2.

5.3 Phase Conjugation (Boyd 7.2)

Consider a wave that travels through a material thatmesses up the perfect wave front (a berating medium).Reflecting this wave off of a mirror and back throughthe same medium, the aberration gets twice as large.

However, using a phase-conjugated mirror, thewavefront is reversed upon reflection as well as its

phase so that propagation through the same mediumwill completely cancel the aberration.

The signal electric field

Es(r, t) =Es(r)eit

where

Es(r) =As(r)eikr = |As|ei(kr+s(r))

and the phase-conjugated field is

Ec(r, t) =Ec(r)eit

where

Ec(r) =rEs (r) =rA

seikr =r|As|ei(kr+s(r))

so that the phase-conjugating mirror (PCM)

1) reverses the wavevector eikr

2) reverses the wavefront eis(r)

3) imparts reflectivity coefficient (r)

r >1 is possible (gain) and r 2 = 105 is observed.

Even if the incident wave is not a plane wave, thewavevector and wavefront will be reversed. Comparinga PCM to a regular mirror, the PCM acts as a time-reversal operator.

As an example of an application of phase conjuga-tion, a pulse may be amplified by sending it througha Nd:YAG medium that is pumped with high-powerwhite light. This light also heats up the medium whichcauses large temperature gradients (T) and hencelarge refractive index gradients (n= n(r)).

Using a phase conjugating mirror to send the pulsethrough the same medium, the aberrations can be re-versed while amplifying further. The PCM is imple-

mented using stimulated Brillouin scattering or Degen-erate Four-Wave-Mixing (DFWM).

5.3.1 Degenerate Four-Wave-Mixing

In DFWM the three input waves are the same so thatthe induced polarization is

P(3) = 30(3)(; , , )|A|2A

19


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To get this to work, we arrange the pump fields tocounter-propagate in a transverse direction to the sig-nal field.

Where the pump fields are at a large enough angleto the signal field so that we can ignore coupling toother field components. Assuming that the fields arephase-matched the nonlinear polarization goes as

P(3)1 = 30

(3)[|E1|2E1+ 2E1|E2|2 + 2E1|E3|2+ 2E1|E4|2 + 2E3E4E2 ]

P(3)2 = 30(3)[|E2|2E2+ 2E2|E1|2 + 2E2|E3|2+ 2E2|E4|2 + 2E3E4E1 ]

P(3)3 = 30

(3)[|E3|2E3+ 2E3|E1|2 + 2E3|E2|2+ 2E3|E4|2 + 2E1E2E4 ]

P(3)4 = 30

(3)[|E4|2E4+ 2E4|E1|2 + 2E4|E2|2+ 2E4|E3|2 + 2E1E2E3 ]

and if we assume that the pump fields E1 and E2 aremuch stronger than the signal fields E3 and E4 then

we can neglect terms inE3and E4of higher order thanone.

P(3)1 = 30

(3)[|E1|2E1+ 2E1|E2|2]P

(3)1 = 30

(3)[|E2|2E2+ 2E2|E1|2]P

(3)1 = 60

(3)[E3|E1|2 + E3|E2|2 + E1E2E4 ]P

(3)1 = 60

(3)[E4|E1|2 + E4|E2|2 + E1E2E3 ]and putting these back into the coupled equations

dA1dz =

i3

2nc cos (3)

[|A1|2

+ 2|A2|2

]A1

dA2dz

= i32nc cos

(3)[|A2|2 + 2|A1|2]A2dA3dz

= i3

nc(3)[(|A1|2 + |A2|2)A3+ A1A2A4]

dA4dz

= i3nc

(3)[(|A1|2 + |A2|2)A4+ A1A2A3]

Weve already solved for A1 and A2. Their solutionsare

A1(z) =A1(0)ei1(z)

A2(z) =A2(0)ei2(z)

and the combined phase is

1,2(z) =

cn2(I1,2+ 2I2,1)

z

cos

Now looking at A3 and A4, if we assume that I1 =I2 = I, then A1 and A2 will have equal and oppositeexponentials and so thatA1A2is a constant. We definethe following

3 =3

nc(3)(|A1|2 + |A2|2)

=3

nc

(3)A1A2

so

dA3dz

=i3A3+ iA4

dA4dz

= i3A4 iA3The first term in each of these equations leads to anonlinear phase shift, so we make the transformations

A3= A3e

i3z A4= A4ei3z

and substituting these back into the coupled equations

dA3dz

= iA4dA4dz

= iA3

d2A4dz2

+ ||2A4 = 0A4(z) =C1cos(||z) + C2sin(||z)

and

A3(z) = i

dA4

dz

=i||

[

C1sin(

|

|z) + C2cos(

|

|z)]

Using boundary conditions, we can solve for the coef-ficients.

A4(L) = 0 A3(0) =i

||

C2

Were assuming thatA4 is not really injected from theother side (z = L). Finally we find that

A4(z) =i||

sin[||(L z)]cos(||L) A

3(0)

A3(z) =cos[||(L z)]

cos(||L) A3(0)

If we put inA1,A2and A3(0), we get outA3(L) (trans-mitted) and A4(0) (reflected).

A4(0) =A4(0) =i

||tan(||L)A3(0)

A3(L) =A3(L)e

i3L =A3(0)e

i3L

cos(||L)We can now define a phase-conjugate reflectivity

Rpc = I4(0)/I3(0) = tan2(||L) which tells us howstrong the phase-conjugate beam is. This can intro-duce either a gain or loss in reflection (0 Rpc ).

Notice that DFWM does in fact act like a PCM,A4(0) A3(0), and the signal is amplified as it travels

20


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through the medium, I3(L) = sec2(||L)I3(0), where

the gain can go to infinity. Also, the phase mismatchis automatically zero since all frequencies are the sameand were looking only at sets of counter-propagatingbeams, k1 +k2 k3 k4 = 0. We can understandthis fundamentally as a photon from each pump beamannihilating with energy 2h and going into a photonin the A4 field and a photon in the A3 field, counter-

propagating so that the initial and final momentum iszero, h(k1+ k2) = h(k3+ k4) = 0.

I3(L) =G I3(0)

I4(0) =Rpc I3(0)

where G sec2(||L) and Rpc tan2(||L).This type of setup acts like a cavity with a gain

material inside. DFWM can be used to make laserssince the medium creates a beam at at very highreflectivity. This is called self-oscillation because the

output field is generated without the need of an inputfield.

5.3.2 nearly degenerate FWM

Researchers have done Nearly-DFWM (NDFWM) bysetting the pumps to a frequency and the input sig-nal to where . Here phase-matching be-comes an issue.

5.4 Optical Bistability

Bistability occurs when one input intensity may giverise to multiple output intensities. Whenn= n0 +n2I,Iinchanges the refractive index, which in turn changesthe input intensity. This creates a feedback loop

As Iin is increased from zero, it follows the bottompath, but once it hits the peak, it jumps upwards as

indicated since there are no other stable states. As Iindecreases from above, it gets to the left-most peak and

jumps down.

If the input starts somewhere on the dotted line,small perturbations will increase rapidly, causing Ioutto jump to one of the two stable solutions.

5.4.1 Optical Switching (Boyd 7.3.3)

Using a weak optical field, a signal field incident on thebeamsplitter can be routed entirely to one output orthe other. The figure below is a representation of anoptical loop mirror which acts as an all-optical switch.

We assume that a signal field amplitude A comesin as the input. The beamsplitter has reflection andtransmission coefficients

t=

T r= i

R

where

R+ T = 1

We can neglect the reflection of the two mirrors sincethey will introduce a total of 2 radians.

After the first reflection or transmission the ampli-tudes are

Aref,i = i

RA0

Atrans,i =

T A0

so that after propagation through the nonlin-ear medium and another transmission or reflection(NL = (/c)n2IL)

Aref,f = RA0ei(/c)n2(20n0c|Aref,i|2)

Atrans,f =T A0ei(/c)n2(20n0c|Atrans,i|

2)

So just looking at the transmitted output, the in-tensity in vs intensity out

IoutIin

=

1 2T R

1 + cos

(1 2T) Iinn2

c

21


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Here output 1 is the transmission and output2 is the reflection, which oscillate out of phase as afunction of the input intensity. The reflection and thetransmission coefficients for the beamsplitter are thesame in this case. (T =R)

5.5 tensor properties**

(123-133)

5.6 Two Photon Absorption

We can find higher orders of absorption too.

= 0+ I

where we can just tack on the extra term

dI

dz = 0I I2

where Im[(3)]. In the NLS

dA

dz = 1

2(0+ I)A

Looking at a hydrogen atoms energy states, theDirac notation for each state is|n l ml , where =(1)l is the parity of the state (1).

The Hamiltonian of the interaction of light withatoms is

H= E= er E

and the matrix element of the probability of a transi-tion from state a to state b is expressed as

Wab = b| H|a = eE

drddr2sinb ra

looking only at the r dependence

drf(r) =

0

drf(r) +

0

drf(r)

=

0

d(r)f(r) +

0drf(r)

=

0drf(r) +

0drf(r)

=

0 iff(r) = f(r)2

0 drf(r) iff(r) =f(r)If Wab = 0 then the transition is forbidden, and ifWab= 0 then a transition is allowed. The first cor-responds to same parity

b (r)(r)3a(r) = b (r)r3a(r)and the second corresponds to opposite parity

b (r)(r)3a(r) =b (r)r3a(r)and since parity is = (1)l, so no transition is al-lowed between states where the quantum numbers lare the same. (l= 1).

So in principle, from the ground state, only half ofthe upper-level states are accessible with one-photonabsorption. With two-photon absorption the onlytransitions that are allowed are ones between sameparity states. One photon takes the atom to an inter-mediate state of opposite parity and the second photontakes the atom from this intermediate state to a higherstate of opposite parity. In total then, TPA only can

access states of same parity.

Similarly, when the atom gets stuck in the 2s state,it would take an infinite amount of time for it to decayto the 1s state, but by emitting two photons, it canreach the ground state.

22


23/45

5.6.1 two-photon microscopy

In traditional fluorescent microscopy, a focused laserwill excite a large volume of atoms so that 3D imageswere blurry. Using TPA a very small point-like volumecan be excited because of the intensity dependence, sothat a better quality 3D image can be scanned.

A high intensity is still required though, and sincemuch of the TPM is performed in living samples, ahigh average power can destroy a sample. Very highintensities can be achieved by producing pulses with

high peak power. A peak power around a kilowatt(Ppeak 1kW) can be focused down to 109 W/cm2.

5.7 NL phase shifts from (2) processes **

(145-148)

6 Nonlinear Pulse Propagation

Up until now, weve only considered continuous wavesat a single frequency . Well assume that a pulse ismany cycles in duration (d/ 1) and this analysisapplies for pulses as short as 10 cycles.

For example: a 100 fs pulse has

0.4/T0 4 1012 Hz 0.010

and c = 1/= /c so

= (2/c)

and plugging in the value for , 10 nm.

Since a pulse is composed of many fourier compo-nents, all of which travel at different group velocities,which gives rise to Group Velocity Dispersion (GVD).This will cause the pulse to broaden as it propagatesand since NLO helps to widen the frequency spectrum,the pulse will be broadened further.

To analyze a pulses propagation, well assume that

the medium is isotropic ((2) = 0 and (3) = 0) andwell neglect the transverse dimensions so that only zwill appear in the equation.

6.1 Nonlinear Schodinger Equation

We start with the wave equation

2E

z 2 1

0c22D

t2 =

2E

z 2 1

0c2

0

2E

t2

2P

t2

= 0

2E

z 2 1

c2

2E

t2 =

1

0c2

2

t2 [P

(1)

+ P

(3)

]

where, neglecting the THG terms,

P(3) = 30(3)|E|2E

We can get rid of the /tterms by Fourier transform-ing in time.

2

z 2E(z, )+

2

c2E(z, ) =

2

0c2[P(1)(z, )+P(3)(z, )]

and since P(1)(z, ) =0(1)()E(z, ) and n20 = 1 +

(1)()

2

z 2E(z, ) +

2

c2n20()E(z, ) =

2

0c2P(3)(z, )

Now we write E(z, ) and P(z, ) as a product of anenvelope and a continuous wave

E(z, t) =A(z, t)ei(k0z0t)

P(3)(z, t) =P(3)(z, t)ei(k0z0t)

k0 and 0 are the central values of the pulse. Now wedo the Fourier transforms

E(z, ) =

dtA(z, t)ei(k0z0t)eit

=A(z, 0)eik0z

23


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and similarly

P(3)(z, ) =P(3)(z, 0)eik0z

Substituting these into the wave equation derivedabove and canceling the eik0z terms

2k0

z

+2

z2

k20+k

2()]A(z,

0) =

2

0c2

P(3)(z,

0)

where k 2() =2n2/c2.If we assume that k() is not varying too rapidly

over the pulse spectrum, we can Taylor expand about0

k() =k(0) + dk

d

0

+n=2

knn!

( 0)n

=k0+ k1( 0) + D( 0)

Where kn= n

k/n

.And using a bit of algebra we can simplify the waveequation by putting things back into the time domain

2 = ( 0)2 + 20( 0) + 20=

i

t

2+ 2

i

t

0+

20

=

0+ i

t

2so that

i2k0

z+

2

z 2 k20

A(z, t)

+

k0+ ik1

t+

n=2

knn!

(i

t)n

2A(z, t)

= 10c2

0+ i

t

2P(3)(z, t)

Now to simplify things further, assume a referenceframe moving along side the pulse = tz/vg =t k1z. Making the coordinate transformations

z =

z

z

z+

z

=

z k1

t=

z

t

z+

t

=

Substituting these back into the wave equationi2k0

z k1

+

z k1

2 k20

A(z, t)

+

k0+ ik1

+

n=2

knn!

(i

)n2

A(z, t)

= 10c2

0+ i

2P(3)(z, t)

Where D =n=2

knn! (i

)n

i2k0

z k1

+

2

z 2+ k21

2

2 2k1

2

z k20

A

+

k20+ i2k0k1

+ 2k0D k21

2

2+ i2k1

D+ D2

A

=

20

0c2 1 + i

0

2

P(3)

and combining terms

i2k0 2k1

z+

2

z 2+ 2

k0+ ik1

D+ D2

A

= 20

0c2

1 +

i

0

2P(3)

Assuming that

P(3)

= 30(3)

|A|2

A= 4n2

0n22

0c|A|2

A

i2k0

1 i k1

k0

A

z+

2A

z 2+

2k0

1 + i

k1k0

D+ D2

A

= 4020n

20n2

c

1 +

i

0

2|A|2A

Now to simplify we need to make three approxima-tions.

k1k0 =

c

n00vg =

1

0

c/n0vg =

1

0

vvg

1

0

since vg v. This is true because n() varies slowlyin a transparent medium.

v =

k vg =

d

dk

and = ck/n() where we can treat n() like a con-stant so that

vg

c

n()

k

=v

Looking at all of these terms, and noticing that theoptical period is much slower the envelope periodTopt env

k1k0

A

1

0

A

env Topt A

env 0

24


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Rewriting the wave equation

i2k0A

z +

2A

z 2 + [2k0D+ D

2]A

= 4020n

20n2

c |A|2A

The second approximation is the Slowly-Varying En-

velope Approximation (SVEA) so we can neglect the2/z2 term.

i2k0A

z = [2k0D+ D2]A 40

20n

20n2

c |A|2A

And for the third approximation, we neglect the D 2Aterm since this corresponds to 4A/4 terms andhigher.

A

z =iDA + i

2020n

20n2

(n00/c)c |A|2

A

Taken to the third order

D= k22

2

2 i k3

6

3

3

= 20n0n20= 003(3)|A0|2

so that we end up with the Nonlinear SchrodingerEquation

dA

dz = i k2

2

2A

2 +

k36

3A

3 GVD

+ i|A|2A NL

where the first two terms have to do with the GVD andthe last term is the intensity-dependent part (nonlinearphase shift and TPA).

For non-transparent materials, we may also add anabsorption term

dAdz

= i k22

2A2

+ k36

3A3

+ i|A|2A 2

A

6.2 = 0, k2 = 0

Just to verify that the base case works, assume that= k2= 0.

dA

dz = 0

and the solution is simple

A(z, ) =A(0, )

which makes sense since a pulse wont have dispersiveor nonlinear effects to change its shape as it propa-gates.

6.3 GVD ( = 0)

First we have to consider only the dispersion (= 0),and well also neglect the third order dispersion.

dA

dz = i k2

2

2A

2

This causes a broadening effect as a pulse propagates.

We define a new variable so that under a Fourier trans-formation, and / i. The Fouriertransform of an arbitrary waveform is

A(z, ) =

A(z, )eid

and so the Schrodinger equation becomes

dA(z, )

dz =i

k22

2 A(z, )

This is easily solved

A(z, ) =A(0, )eik22z/2

and transforming back into the time domain

A(z, ) = 1

2

A(0, )eik22z/2eid

To solve any further, we have to choose a shape forthe input pulse (usually a gaussian).

A(0, ) =A0e2/2T20

whereT0 is the pulse width. In the frequency domain,the initial pulse is solved by completing the square inthe exponent

A(0, ) =

A0e2/2T20 eid

=

A0e2/2T20 +id

=

A0e 1

2T20

(2+i2T20 +i2T40

2i2T40 2)

d

=

A0e 1

2T20

(2+i2T2

0

+i2T4

0

2)

eT20

2

/2d

=A0eT20

2/2

e(+iT20 )

2/2T20 d

=A0eT20

2/2

2T20 =

2A0T0eT20

2/2

And putting this back into the equation above

A(z, ) = 1

2

2A0T0

eT20

2/2eik22z/2eid

= A0T0

2

e(T20ik2z)

2/2eid

and by completing the square again we find that

A(z, ) = A0T0

T20 ik2ze

2

2(T20 ik2z)

25


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We can rewrite this in terms of the characteristiclength LD = T

20 /|k2|

A(z, ) = A01 iz/LD

e

2

2T20

[ 11iz/LD

]

simplifying:

1 + i= Aei =A(cos + isin)

Acos= 1 Asin=

A=

1 + 2 = tan1()

11 + i

= 1Aei/2

= [(1 + 2)1/4]e[i2 tan

1()]

e[ 11+i] =e[1i

1+2]

=2/2T20 and =z/LD. Rewriting the pulsepropagation in terms of an amplitude and phase

A(z, ) = |A(z, )|ei(z,)

|A(z, )| =A0

(1 + (z/LD)2)1/4e

2/2T201+(z/LD)

2

(z, ) = 2

2T20

z/LD1 + (z/LD)2

12

tan1

z

LD

After propagating a distance ofLD the GVD effects

become significant.

The pulse stays gaussian but broadens by a factorof

2 as it propagates a distance ofLD.

In the normal dispersion regime, this is the case,and low frequencies will propagate faster than high fre-quencies (red leads blue). In a medium such as fusedsilica (optical fibers) there exists a zero-dispersionwavelength at which k2 = 0 for the right . For fusedsilica, k2= 0 at = 1.27 m.

Though there is zero dispersion, the pulse cannotpropagate forever since the pulse consists of a range ofwavelengths, so that the pulse propagates longer butstill decays since k3= 0 and k4= 0.

Looking at a typical glass, = 800 nm and k2 =50 ps2/km.

T0= 10 ps LD= 2 km

T0= 0.1 ps LD= 20 cmIn anomalous dispersion (k2 < 0) the red frequencies

will propagate slower than the blue ones.An unchirped initial pulse becomes chirped after

some propagation, and this is seen easily by represent-ing the pulse as follows

A(z, ) = |A(z, )|ei(z,)

where the phase can be found by rearranging the imag-inary number in the exponent and amplitude so that

(z, ) = 2

2T20sgn(k2)(z/LD)

1 + (z/LD)2 +1

2tan1 z

LD

This is the quadratic phase in frequency across thepulse and we define a quantity called the chirp, orthe change in frequency over the pulse

=

=

T20

sgn(k2)(z/LD)

1 + (z/LD)2

6.4 nonlinear (k2 = 0)In this case there is no dispersion so all of the effectsare due to nonlinearity.

dA

dz =i|A|2A

26


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and this is easy to solve giving

A(z, ) =A(0, )ei|A(0,)|2z =A(0, )ei

NL (z,)

Here, the temporal shape stays unchanged, I(z, ) =I(0, ), but the phase varies in time due to SPM. Defin-ing LNL= 1/|A0|2

NL =z/LNL

so that (LNL) = 1. The total phase is

total = k0z 0t + NL

Assuming we know the input gaussian pulse

A(0, ) =A0e2/2T20 NL e2/T20

and so we find the instantaneous frequency (chirp)

= dd

total

=0+ 2|A0|2

z T20 e2/T20

=0+ 2z

T0LNL

T0e

2/T20

The front edge of the pulse is red-shifted and theback is blue-shifted. This looks a lot like the GVDchirp, except that while the GVD chirp was linearin time, the NL chirp is not, and is also intensity-dependent. Note also that the chirp is linear over thecenter of the pulse.

The pulse shape remains unchanged, but in theFourier plane there are changes

A(z, ) =

A(0, )eiNL(z,)eid

This can become difficult to solve, so a rough estimateof the spectral width can be obtained from the gener-ated frequencies

2 |A0|2z

T0

NL

T0

Without nonlinearity, 0.4/T0, so that whenNL/T0gets appreciable to 0.4/T0we start to see spec-tral broadening.

For large NL the solution can only be obtainednumerically. Since this is a (3) process FWM will oc-cur so that two photons at0 can annihilate to createone photon below and one above.

This takes energy from the center of the spectrumand shifts it to the edges. Sidebands start to evolvewhich give rise to their own sidebands.

6.5 (= 0, k2= 0)In reality,k2 is never zero so there will be a compositeof dispersive and nonlinear effects. Since both normalGVD and NL processes put the red frequencies aheadof the blue frequencies, the effects will strengthen one

27


28/45

another. The relative strengths of these effects can beexpressed as

LDS= T20|k2| LNL =

n0c

30(3)|A|2 =

1

(0/c)n2I

Where the frequency and time width look like

The shortest pulse that can be made depends onparameter k, where k t.

Very little can be done analytically, so most ofthe description will be qualitative. Assuming thatLDS LNL and looking at propagation in the nor-mal dispersion regime

but more frequencies are created so that the GVDbroadening effect increases, increasing in turn the ten-dency to spread spectrally.

Notice that with nonlinear propagation, has in-creased, making shorter pulses possible, and also that(t) is linear so it can be undone by anomalous GVDeffects. This will compress the pulse to an even smallersize than the original due to the increase in bandwidth.

In this way, 6 ps pulses have been created and com-pressed to 200 f s.

In the anomalous dispersion regime k2 < 0, blueleads red and so a pulse usually decays but if the con-ditions are just right, the nonlinear effects can exactlycancel the anomalous GVD effects as the pulse propa-gates so that a soliton is formed.

6.5.1 solitons

dAdz

= i k22

2A2

+ i|A|2A

In just one spatial dimension and time, this can besolved analytically, but requires an advanced techniquecalled the inverse scattering method. (Agrawal5.2.1) The solution is found to be

A(z, ) =A0 sech

T0

ei

NL (z)

where

NL(z) =12 |A0|2z = 2c n2I0zand a soliton may only form when the characteristiclengths are balanced

LDS=LNL |A0|2 = k2T20

where = (/c)20n0n2c, which must have the oppo-site sign ofk2.

Typically, k2 and n2 are fixed given , so we can

choose the input pulse duration and power,T0andP0.

|A0|2 =

c

n2I0= k2

T20 I0 = k2c

n2T20

and

P0 = I0(Area) =k2A2n2T20

Assuming some typical numbers in a fiber: =1.5 m, k2 =10 ps2/km, n2 = 2 1016cm2/W,d= 8 m so that

P0 5

T20

where the pulse width is given in picoseconds. So fora 10 ps pulse, we need approximately 0.5 pJ to makea soliton. The characteristic lengths work out to bearound 10 km, so pretty long. Looking at a 40 kmfiber, a low power 10 ps pulse will undergo temporaland spectral broadening, but if the power is turned upthen solitons will form and propagate undisturbed.

28


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I(z, ) = |A0|2 sech2

T0

Even if the pulse starts out at P > P0 it will radiatethe energy so that it decays and stays in stable solitonform. If the input power is high enough, solitons ofhigher order can be made, with periodic stable shapes.

This is seen when the soliton order, N, is greater thanone. N=

LDS/LNL.

6.5.2 Space-Time Duality

In time we have the NLSE

dAdz

= i k22

2

A2

+ i|A|2A

which produces a soliton when the SPM and GVD ef-fects cancel each other. In space we have the paraxialwave equation which governs the propagation of beams

2A + i2kdA

dz =

2

0c2PNL

Well take PNL = 3(3)|A|2A

dAdz

=i 12k

2A + i|A|2A

where = 3(3)2/2k0c2.

Well consider only one dimension for now since soli-tons are unstable in two dimensions. This is imple-mented by making one dimension of the beam muchlonger than the other so that the beam is effectivelyone-dimensional.

dA

dz

=i1

2k

2A

x2

+ i

|A

|2A

The coefficient of diffraction has to be negative, andthis means that the SPM and diffraction effects canbalance so that a solitons can form. This gives a beamthat propagates forever.

In the case where we include both 2/x2 and2/y2 will result in either collapse or diffraction, butthe equilibrium is not stable.

Now if we imagine pulsing a beam, we can add allfour terms together

dA

dz =i

1

2k2A i

k22

2A

2 + i|A|2A

In general, solutions to this equation are complicated.If a soliton exists for this setup, it would be calledan optical bullet since it would be and EM packetlocalized in all dimensions, propagating without decay.

6.5.3 Telecom Systems

For propagation in long fibers used in telecommunica-tion systems, we want to minimize the effects of GVDand SPM so that information keeps its integrity. Orig-inally, telecom designers used light at = 1300 nmbecause there is a local minima in the loss spectrumfor silica here, and there are minimal dispersion effectssince the zero-dispersion wavelength is very close to1300 nm as well.

The local maxima is due to water atoms that aretrapped within the silica. This can be eliminated withmodern techniques so that the loss is low here as well.

29


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The zero-dispersion wavelength can be shifted sothat it goes through 1550 nm, at todays telecom sys-tem wavelength.

Propagation through a long fiber with both disper-sive and nonlinear effects can be tricky. An input pulsewill be spread in time as well as spectrally so that it isimpossible to recover the original pulse. Using a PCMhowever, we can add a second length of fiber to exactly

undo the unwanted effects.

The initial pulse broadens and new frequencies arecreated

A(0, ) A(0, )ei(L+NL )

and by reflecting this pulse off of a PCM

A(0, )ei(L+NL )

and forcing the image to propagate down another fiberthe same length as the first, the pulse that emerges

looks exactly like the input pulse. This can be im-plemented in fiber using a pump beam coupled to theinput beam and sent through a PPLN section to pro-duce the phase-conjugation mirror.

6.6 Generation of Short Pulses

6.6.1 Mode Locking

A laser is a system that uses amplification and feed-back to generate light at a specific frequency.

The amplification can be achieved by pumpingpower into the system and the feedback comes from afully reflecting mirror on one side and a semi-reflecting

(R


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Ec(t) =E0ei0t

N/2n=N/2

ei(nct/L)

where the random fluctuations are ignored. Whennct/L = 2 constructive interference occurs so thatthe sum equals N. Substituting this value to t into theequation

E(t= 2Ln/c) =N E0ei0t

and

I(t= 2Ln/c) =N2absE02

and every multiple of time around the cavity, a pulsecomes out.

This happens naturally since the shortest and mostintense pulses experience the least loss in the saturableabsorber. The saturable absorber causes the modes inthe cavity to be phase locked in this way.

To understand how these pulses form, consider ran-dom noise that passes through the saturable absorber.

Peaks are sharpened as the saturable absorber trans-mits more at higher intensities, and after repeatedlyseeing the same peak through, the energy from all thesurrounding noise will be funneled into a single sharppeak.

To get a saturable absorber, we need a material thatallows light to promote electrons or molecules to anupper state. Once in this upper state, less of theseupper states are available, so that more light is simplytransmitted instead of interacting with the material.

Until the 1990s these were built using dyes, butideally we want a saturable absorber that doesnt ac-tually absorb any of the light. The simplest way toachieve this is to use a Kerr lens, which focuses high-intensity light and diffracts low-intensity light (self-focusing medium). An aperture is then put in front ofthe beam so that only high-intensity light is allowedto propagate.

6.6.2 pulse generation in fiber

We can also generate pulses in fiber using the samecomponents as traditional lasers, i.e. gain, feedbackand a saturable absorber.

The fiber coupler works by overlapping the wave-functions of the photons passing through, so that thereis a probability that they will tunnel to the adjacentfiber. This coupling can be made weaker or strongerbased on the distance between fibers.

Using an optical loop mirror, a field Iin enters inone side and passes through a nonlinear medium (n2)

in the loop. The higher the intensity, the more phaseshift a field picks up and so the intensity out Iout dueto interference at the coupler. This was covered in theoptical switching section.

This is the saturable absorber that will act to passhigh intensities better than low intensities. There isno actual absorption here, part of the field is beingdiscarded out of the input, but we can approximateIoutas the same output that a saturable absorber gives.

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This is only the case when the coupler is not a 50:50splitter. With each pass through the saturable ab-sorber, peaks are sharpened and noise is reduced untila train of pulses emerges.

In this way, a pulsed laser can be made using allpassive components. The distance between pulses de-pends on the round trip of the cavity.

6.6.3 Additive-Pulse Mode Locking (APM)**

(181-185)

6.7 solitary lasers

This system has both anomalous GVD (prisms) andnonlinearity which effectively act at the same time, as

in a fiber. To add the saturable absorption, a Kerr lens(aperture) is needed to cut off the wings introducedby SPM broadening. This kind of system can reachsoliton pulses as small as a few femto-seconds.

The problem here is that the envelope is 4 fs and acycle is 2.7 fs for 800 nm light, so the SVEA breaksdown.

6.7.1 Ultrafast Measurment

We can measure these pulses accurately to a few frac-tions of a femtosecond using an autocorrelator. Using a

Mach-Zhender interferometer with one adjustable arm,a single pulse can be split into two with some delay inbetween.

By putting a nonlinear element after the interferom-eter, such as a crystal that produces second harmonicgeneration or two-photon absorption, the overlap ofthe two pulses can be calculated since the nonlinearoutput depends on the intensity.

A A=

I(t)I(t + )dt

Measuring the second harmonic intensity ISH vs. thedelay

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From a plot like this, the pulse width can be inferredsince ISH 4I0. For instance, a FWHM of 30 mgives an initial pulse of around 100 fs.

7 Two-Level Systems

Until now weve assumed that any frequencies were farfrom the atoms resonant frequencies. In this case, theprobability that the atom will stay in the ground stateis close to one, which justifies the use of the perturba-tive approach to solving the wave equation.

Now were assuming that ab, or the transitionfrequency from one state|ato another|b, where theenergy difference between states is E= h(b a) =hab.

With resonant excitation, effects are much largerthan those induced by non-resonant excitations, so theperturbative approach will quickly become invalid.

The easiest example of a two-level system is anatomic transition such as that in sodium vapor (and

other alkalis), which has one electron on the outermostshell with all other shells filled, like a hydrogen atom.For sodium, the outer electron experiences a transition3s 3p.

The sodium D lines occur between 3s and 3p, whichwill function as our|a and|b states.

It should be noted that bands of allowed energystates exist, so things get much more complicated. Thetwo-level approximation works reasonably well for ourcalculations.

7.1 Optical Bloch Equations

7.1.1 derivation

7.1.1.1 caand cb The optical Bloch equations gov-ern the dynamics of a two-level atom interacting witha light field. Well start with the Schrodinger equation

ih

t| =H|

and well start with the case of no incident light sothat H=H0, the atomic Hamiltonian.

n(r, t) = r|n =un(r)eint

where n= En/h. As usualH0|un =En|un

and|un are stationary eigenstates, all of which areorthogonal.

un|um =nmNow well apply a field with ab. The wave-

functions for|aand|barea= ua(r)e

iat

b= ub(r)eibt

The Hamiltonian is redefined as H = H0+HI whereHI is the interaction Hamiltonian between atom andfield. Applying the dipole approximation ( a0),the interaction is approximated by

HI= E(ratom, t)where = er is the dipole moment of the atom.

|aand|bwill no longer be eigenstates in the pres-ence of applied field. Instead, linear combinations ofthe two will make up the eigenstates ofH.

(r, t) =ca(t)a(r, t) + cb(t)b(r, t)

where ca and cb are probability amplitudes such that

| = |ca| + |cb| = 1The eigenstates are represented by

| =ca|a + cb|b =ca|uaeiat + cb|ubeibt

To find an equation of motion for the probabilityamplitudescaand cb, the above equation is substituted

into the Schrodinger equation

ih

t[ca|uaeiat + cb|ubeibt]

= (H0+ HI)[ca|uaeiat + cb|ubeibt]

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whereH0|ua,b =Ea,b|ua,b

so that the RHS of the above equation is rewritten as

RHS= caEa|uaeiat + cbEb|ubeibt+ caHI|uaeiat + cbHI|ubeibt

and to get an equation for ca, we operate the RHS and

LHS byua| whereua|ub = 0

ih

t[cae

iat] =caEaeiat

+ caeiatua|HI|ua + cbeibtua|HI|ub

Using the approximation forHI= e Ermade before,the second term in the above equation goes to zero

ua|HI|ua = eEua|r|ua = 0Assuming that |ua and |ub have opposite parity(ua|r|ub = 0), the transition dipole moment is de-fined as

ab = ua||ub =

d3rua ub

and we rewrite the Schrodinger equation as

ih(caeiat

iacae

iat) =cahaeiat+cbe

ibtab ERearranging and defining ba = b a, we have anequation of motion for ca and through the analogousprocedure, cb:

dcadt

=i abE(t)

h eibat

dcbdt

=ibaE(t)

h eibat

where ab= ba and ab = ba.

7.1.1.2 the density matrix The density matrixdescribes a system in a mixed state, and particular anensemble of such atoms. Instead of excluding a singleatoms and ignoring its interacts with other atoms, we

can consider all of that in the density matrix.

[]ij =

aa abba bb

where

aa= |ca|2 bb= |cb|2 ab = ba= cacbeibat

The diagonal terms are easy to understand

T r[] = |ca|2 + |cb|2 = 1To understand the off-diagonal terms, well start by

considering the dipole moment.

=(t) = ||=cac

beibatua||ub + cacbeibatub||ua

=baab+ abba

These terms are only non-zero if | is a superpositionof states.

ba= (cacb+ cbc

a ibacacb)eibat

=ibaE(t)

h |ca|2 i

abE(t)

h |cb|2 ibacacbeibat

= i baE(t)h

(bb aa) ibaba= ab

Doing the same thing with the diagonal elements

aa= iE(t)

h (abba baab)

= bb

This is a closed system since the only two accessiblestates are|a and|b.

7.1.1.3 relaxation times T1 and T2 Now welllook at the interaction between atoms, i.e. they col-lide which affects the phases of the wavefunctions aswell as the populations in each state. This results inrelaxation and damping which is modeled by randomHamiltonians.

bb decays at a rate ba= 1/T1, where T1 is the re-laxation time (lifetime) of an atom in the upper state.Decay is due largely to spontaneous emission in aclosed two-level system. ba decays at rate ba= 1/T2where T2 is the dephasing time of the dipoles.

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Putting the relaxation time into the equations

ba= i 2

(bb aa) (iba+ 1T2

)ba

bb= i

2(ab ba) bb

T1

aa= i2

(ab ba) + bbT1

where = 2baE(t)/h is the Rabi flopping frequency.Notice that the first of these equations depends onlyon the population difference W =bb aa.

7.1.1.4 population difference W (W = 1)corresponds to an atom in state|a, and (W= 1) toan atom in state|b. When (1 < W < 1) the atomis in a superposition of the two states.

Looking at the time dependence ofW

W =

d

dt (bb aa) = 2

T1 bb+ i(ab

ba)

and we can see that

2bb= (bb aa) + (bb+ aa) =W+ 1

so that

W = 1T1

(W+ 1) + i(ab ba)

Even in the absence of an applied field, there is anequilibrium population distribution among atom lev-els. This is defined as Weq.

W = 1T1

(W Weq) + i(ab ba)

Weq is the offset from zero of the population and allowsus to consider pumping of population into the excitedstate.

Well now assume that E is composed of a slowlyvarying envelope function and a fast oscillating term,E E(t)eit. The Rabi frequency becomes

(t) = 2ba

h [E(t)eit + E(t)eit]

To get at the meaning ofT1 and T2 well let E= 0so that solving forW(t) gives

W(t) =Weq+ [W(0) Weq]et/T1

andba(t) =ba(0)e

ibatet/T2

which gives a dipole moment of

(t) =abba= [abba(0)eibat]et/T2

So an undriven atom, the dipole oscillates at ba anddecays to zero with characteristic time T2. Thus,T2 issimply the dipole dephasing time.

For a single atom, the only means of relaxation ifspontaneous emission so that there is simultaneouspopulation and dipole relaxation. In an ensemble, forthe two to relax at the same rate, its necessary thatT2= 2T1. Pure dephasing processes, such as collisions,dont cause loss of energy, only decoherence in dipolephases.

The dephasing rate is proportional to the populationrelaxation and also this pure dephasing.

1

T2=

1

2T1+

1

Tcoll

For an atomic vapor, Tcoll depends on the density and

mass of the atoms as well as the temperature of the sys-tem. For Sodium vapor, collisional broadening domi-nates at around T 300 Celsius.

7.1.1.5 envelope(t) Assuming the SVEA,

ba(t) =(t)eit E E(t)eit

and substituting these into the equation for ba(t)

(t) =

1

T2 i( ba)

i ba

h

EW+ EW ei2t

where we define the detuning parameter = ba.The last step is to make the rotating wave approxi-

mation (RWA). When the frequency is tuned to aroundresonance ba, so that all frequencies are much lessthan the optical frequency: , 1T2 , . Thus, theterm ei2t does not couple to since it is oscillatingso fast and can be neglected. In essence, this says thatwe can neglect the part of the interaction Hamiltonianthat oscillates far from resonance

HI= E(t) = ba(Eeit +Eeit)

Now we have two simple expressions for the dipoleand population difference (the Optical Bloch equa-tions)

d

dt =

1

T2 i

i

2W

dW

dt =

W Weq

T1

+ i( )

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with = 2baE(t)/h. These are the basis for much ofquantum optics.

7.1.2 comparison with classical dipole

Well compare the quantum mechanical dipole momentwith the classical dipole moment. Starting with theclassical equation of motion

x + 2x + 20x= eE(t)/m

and assume that E(t) =A(t)eit and x= x(t)eit,combinations of a slowly varying envelope and fastmodulation, we have

x i2x 2x i2x + 2x + 20x= eA/m

Using the SVEA we can neglect x since x x:

2(

i)x= (2

20+ i2)x

eA/m

As with the RWA, we make the

quantum and nonlinear optics

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