Lorentz Oscillator model of dilute matter Free...

Lorentz Oscillator model of dilute matter

electron bound to nucleus by spring (spring const Ks)

N oscillators per unit volume

x describes the position of an e− in a 1-D potential well V (x)

γ is a frictional damping term, force prop to velocity Fd = mγv

FE is the forcing function from an optical E-field

FE

m=

e

m

E0

2(e−iωt + eiωt)

Force is gradient

of potential

V(x)= ω0x212

Linear restoring force for quadratic potential well V (x) = 12ω20x

2

Fs = ma = mdV

dx= mω2

0x = Ksx ⇒ a =dV

dx= ω2

0x ≡ acceleration

Eqn of motion derived by balancing forces ma + Fd + Fs = FE

x + γx + ω20x =

e

m

E0

2(e−iωt + eiωt)

ma+mγv+mω20x=eE0 cosω0t F = qE

Kelvin Wagner, University of Colorado Physical Optics 2018 1

Free Oscillation

x + γx + ω20x = 0

soln is of the form of damped exponentialx = Aeκt x = κAeκt x = κ2Aeκt

⇒ (κ2 + γκ + ω20)Ae

κt = 0

For nontrivial soln ( ) = 0 ω′ frequency pulling

κ = −b

2± 1

2

√b2 − 4ac = −γ

2± 1

2

√γ2 − 4ω2

0 = −γ

2± i

︷︸︸︷√ω20 − γ2/4 ≈ −γ

2± iω2

0

x(t) = A1e−γ

2 teiω′t + A2e

−γ2 te−iω′t

ω′ =√ω20 − γ2/4 is pulled frequency and initial conditions used to find A1 and A2.

Q = ω0γ is ratio of frequency to damping where Q/4π cycles reduces amplitude to e−1


Harmonic Oscillator Solution

Proper quantum mechanical treatment of spontaneous emission is quite involved. Re-quires frictional force for scattering to be proportional to

γ =e2ω2

0

6πǫomc3≡ 1

tspont

Periodic driving function eiωt gives periodic solution x(t) = X(ω)eiωt

−ω2X + iωγX + ω20X = − e

mE0

Solve for frequency domain response

X(ω) =− e

mE0

ω20 − ω2 + iωγ

For a real driving field two harmonic sideband solutions add giving x(t) = ℜX(ω)eiωt

x(t) =emE0

ω20 − ω2 + iωγ

eiωt

2+ cc


Induced Polarization vector field

microscopic dipole moment

p = −ex =e2

m(ω20 − ω2 + iωγ)

E0eiωt + cc

atomic polarizability relates ~p to ~E

~p = ǫoαrE = α~E = α~E

Linearly DisplacedElectron Cloud

less than 1% radius

Nucleus

Unperturbed Electron cloud

Large displacementcan deform cloud

α =e2

m(ω20 − ω2 + iωγ)

αr =e2

ǫom(ω20 − ω2 + iωγ)

Macroscopic Polarization vector field ~P(~r, t) proportional to ~p and density of dipolesper unit volume

~P = N~p = Nα~E = Nǫoαr~E = ǫoχ~E = ǫoχ~E

⇒ χ =Ne2/m

ǫoD0(ω)

where resonant denominator is abbreviated as D0(ω) = ω20 − ω2 + iωγ

Remember D = ǫE = ǫoεrE = ǫon2E = ǫo(1 + χ)E = ǫoE + P so P = ǫoχE


Index and absorption resonances

n2 = 1 + χ = 1 +Nα

ǫo= 1 +

Ne2

ǫom(ω20 − ω2 − iωγ)

Dilute media: use√1 + δ = 1 + δ

2

n = 1 +Ne2

2ǫom(ω20 − ω2 − iωγ)

= n + iκ

= 1 +Ne2(ω2

0 − ω2)

2ǫom[(ω20 − ω2)2 + ω2γ2)]

+ iNe2γω

2ǫom[(ω20 − ω2)2 + ω2γ2)]

Wave eqn soln for plane wave

~E(~r, t) = E0pe−i(ωt−Kz) + cc = pE0e

−i(ωt−kz)e−2πλ κz + cc

where complex wavevector K = ωcN = 2π

λ(n + iκ) is frequency dependent

Intensity attenuatuation

I(z) =|E(z)|22η

= e−αIzI(0) where αI = 22π

λκ is attenuation coefficient

Complex propagation const K = ~k + i~α |~k| = 2πλ |~α| = αI

2


Two Driving Frequencies

F =e

m

∑

k

Ek

2(eiωkt + e−iωkt)

Harmonic Oscillator is linear so individual solns will add

x + γx + ω20x = F

x(t) = em

∑k

Ek2 eiωkt

ω20−ω2k+iωkγ+ cc =

∑k ck cos(ωkt− φk)

x(t) = em

∑k

iωkEk2 eiωkt


∑k ckωk sin(ωkt− φk)

x(t) = em

∑k

−ω2kEk2 eiωkt


∑k −ω2

kck cos(ωkt− φk)

ak =em

Ek2

ω20 − ω2

k + iωkγck =

em Ek

[(ω2k − ω2

0) + ω2kγ

2]1/2φk = tan−1 γωk

ω2k − ω2

0

plug in ∑

k

(−ω2kak + iωkγak + ω2

0ak)eiωkt + cc =

e

m

∑

k

Ek

2(eiωkt + cc)

LHS = RHS ∀k


Multiple Resonances

nDC = 1+Ne2

mǫo

∑

q

fqω2q

Visible transparencywindow in betwee IRand UV resonances

When Medium consists of N oscillators per unit volume divided into fractions fq withfrequencies ωq and damping γq

χ(ω) =Ne2

ǫom

∑

q

fqw2

q − ω2 − iωγq

Oscillator sum rules∑

q fq = 1∑

qp fqp = Z #electronsQuantum Theory gives

fqp =2m(ωq − ωp)

3~

∣∣∣∣∫

Ψ∗qDΨpdV

∣∣∣∣2

where electron position operator D =∑Z

m=1 rmR. Loudon, The Quantum Theory of Light, Oxford 1973. Ch 4.


Susceptibility χ(ω)

Macroscopic polarization ~P proportional to ~p and density of dipoles N

P (t) = Np = NαE = NǫoαrE = ǫoχE(t)

At a particular frequency P (t) = ℜP (ω)eiωt

P (ω) = −NeX(ω) =Ne2E0

m(ω20 − ω2 + iωγ)

= ǫoχ(ω)E(ω)

Solving for the susceptibility

χ(ω) =Ne2/ǫom

ω20 − ω2 + iωγ

= χ′(ω)− iχ”(ω) =Ne2

ǫom

ω20 − ω2 − iωγ

(ω20 − ω2)2 + ω2γ2

Introduce complex index N = n + iκ so that N 2 = n2 − κ2 + i2nκ

εr = N 2 = 1 + χ = 1 + (χ′ − iχ”)

1 + χ′ = n2 − κ2 = 1 +(ω2

0 − ω2)ω20S

(ω20 − ω2)2 + ω2γ2

χ” = 2nκ =ωγω2

0S

(ω20 − ω2)2 + ω2γ2

where S = Ne2

ǫomω20is dimensionless oscillator strength


Lorentzian approximation

when ω0 ≫ γ can approximate resonance with ω0 = ω in all terms except ω0 − ω(ω2

0 − ω2)2 = [(ω0 − ω)(ω0 + ω)]2 ≈ (ω0 − ω)2(2ω)2

χ” =Sω0γ/4

(ω0 − ω)2 + (γ/2)2=

Ne22πν0ǫom(2πν0)2

∆ν2π/4

(2π)2[(ν − ν0)2 + (∆ν/2)2]

=Ne2

ǫomν016π2

∆ν

(ν − ν0)2 + (∆ν/2)2Lorentzian

χ′ =Sω0(ω0 − ω)/2

(ω0 − ω)2 + (γ/2)2=

Ne2

ǫom8π2ν0

ν0 − ν

(ν0 − ν)2 +(∆ν2

)2 =2(ν0 − ν)

∆νχ”(ν)


Experimental quantities: Index andabsorption

N 2 = 1 + χ = n2 − κ2 + i2nκ = 1 + χ′ − iχ”

χ′ =2(ν0 − ν)

∆νχ”(ν)

2nκ = χ” ⇒ κ = χ”2n

n2 − κ2 = 1 + χ′ = n2 − χ”2

(2n)2⇒ n4 − n2(1 + χ′)− χ”/4 = 0

n2 =1 + χ′

2± 1

2

√(1 + χ′) + 4χ”/4

Simplify for dilute media

N = 1 +Ne2

2ǫom

1

(ω20 − ω2 − iγω)

=

(1 +

Ne2

2ǫom

ω20 − ω2

(ω20 − ω2)2 + γ2ω2

)+ i

(Ne2

2ǫom

γω

(ω20 − ω2)2 + γ2ω2

)= n + iκ


K = Nk0 = k0√1 + χ = k0

√1 + χ′ − iχ” ≈

(1 +

χ′

2− i

χ”

2

)k0

=2π

λ(n + iκ) = β − i

α

2Dilute media

n = 1 +χ′

2α/2 = k0χ”/2

Plane wave solnE = eE0e

i(ωt−Kz) = eE0ei(ωt−βz)e−αz/2

Intensity

I(z) =|E(z)|22η

= I0e−αz

Conservation of energy requires

dI

dz= −power absorbed

volume=

ωǫo2χ”|E|2 = ωǫo

2χ”

2

ǫocnI = −αI

Thus α = −ωχ”/cn = −k0χ”/n


Local Field – Dielectric Constant of DenseMedia

Electrostatics: Charge can not act on itselfField it sees is what would be seen at its position if it were removed without alteringother chargesHowever if it were removed other charges would realign. Difference is local field

Examine by cutting a slot in a dielectric capacitor

C

++++++++++++++++++++++++++++++

--------------------------------------------------

E

Field on two sides of narrow slotcancels since there is no charge

∇× ~E = 0 ⇒∮

C

~E · d~l = 0

~El = ~E

SE ------------------

+++++++++++

++++++++++++++++++++++++++++++

--------------------------------------------------

Gauss’ Lawx

S

~E · d~l = σpol = ~P

ǫo~El = ǫo~E + ~P = ~D~El = ~E + ~P/ǫo


Atom fits in a spherical hole in the dielectricFeynman, Lectures on Physics, vol 2,ch 11.

E++++

- - - -

~El = ~E +~P

3ǫo

E P++++

- - - -

Uniform E field inside sphereDipole field outside

Fields due to uniformly polarized charged sphere

~Es = −~P

3ǫo~E = ~El + ~Es

θ

σ=P n

Charge density per unit area of cavity surface = −P cos θ

charge on annular ring dQ = −P cos θ 2πr2 sin θdθ =~P·~rrdS

Coulombs law ∇ · ~D = ρ gives field z · dE2 = −dQ cos θ/r24πǫototal field

E2 =

∫ π

0

P cos2 θ sin θ

2ǫodθ =

P

2ǫo

[−1

3cos3 θ

]π

0

=P

3ǫo

Two ways a slot can be parallel to ~E and one perpendicular. Av-erage gives local field correction P

3ǫo


Local Field and Lorentz-Lorenz correction

microscopic polarization induced by ~Elocal

~p = αrǫo~El = α~El

Macroscopic polarization given by N number density of microscopic dipoles

~P = N~p = Nαrǫo

(~E +

~P

3ǫo

)= ǫoχ~E

~P

(1− Nαr

3

)= Nαrǫo~E ⇒ χ =

Nαr

1−Nαr/3= ǫr − 1 = n2 − 1

In terms of dielectric constant

ǫ = ǫoεr = ǫo(1 + χ) = ǫo1 + 2

3Nαr

1− 13Nαr

= n2ǫo

solving for microscopic polarizability αr

(1− Nαr

3

)εr = 1 +

2

3Nαr ⇒ εr − 1 =

Nαr

3((2 + εr))

αr =3

N

εr − 1

εr + 2=

3

N

n2 − 1

n2 + 2=

α

ǫoLorentz-Lorenz fomula (Clausius-Mosotti)


Lorentz-Lorenz fomula (Clausius-Mosotti)

n2 − 1

n2 + 2=

αrN

3=

αN

3ǫo=

n2 − 1

n2 − 1 + 3⇒ n2 − 1 =

αrN

3(n2 − 1 + 3)

(n2 − 1)(1− αrN

3) = αrN n2 − 1 =

αrN

1− αrN3

=3αrN

3− αrN

Index of liquids and gases

Gases are dilute

εr − 1 = Nαr

Liquids must be local field corrected

εr − 1 =Nαr

1−Nαr/3

Gas LiquidSubstance εr Nαr Density Density Nαr εr (predict) εr (exp)

CS2 1.0029 0.0029 .00339 1.293 1.11 2.76 2.64O2 1.000523 .000523 .00143 1.19 0.435 1.509 1.507


Index of Mixtures: Refractive Index ofSugar (molec wt 342) in Water (molec wt

18)Feynman, Lectures on Physics, vol 2,ch 32.Consider two materials of index n1 and n2 and polarizabilities α1 and α2.

mixed with molar fractions N1/N0 and N2/N0 (N0 = 6.23× 1023 Avogadros number)

Is the overall index just the weighted average

n?=

N1

N0n1 +

N2

N0n2 no!

Instead it is given by the sum of the polarizabilities

3

(n2 − 1

n2 + 2

)=∑

j

Njαrj = N1α

r1 +N2α

r2

Assume αrH2O

(molec wt 18) is const and solve for αrsugar (molec wt 342) in sugar water.

Data from Handbook AB342

(1−A)B18

from C F (0) EE(0)

F (1) DD(1)

sum

A B C D E F G H G+H

wt Fraction density index N2N0

moles N1N0

moles 3(n2−1n2+2

)N1α

r1 N2α

r2 compare N0α

r2

sucrose n589 sucrose/l water/l to F H/D

0 .9982 1.333 0 998.2/18 0.617 0.617 0 0.617 –

0.30 1.127 1.3811 .970 43.8 0.698 0.487 .211 0.698 .211

0.50 1.2296 1.4200 1.798 34.15 0.759 0.379 0.380 0.759 .211

0.85 1.4454 1.5033 3.59 12.02 0.886 0.1335 0.752 0.8855 .210

1.00 1.588 1.5577 4.64 0 0.960 0 0.96 0.96 .208


Local Field Correction Including Dispersion

α =αr

ǫo=

e2

ǫom(ω20 − ω2 − iγω)

=e2

ǫom

1

D0(ω)

Lorentz-Lorenz formula becomes

n2 − 1

n2 + 2=

N

3

e2

ǫom

1

D0(ω)

In the presence of multiple resonances

n2 − 1

n2 + 2=

Ne2

3ǫom

∑

k

fkDk(ω)

=Ne2

3ǫom

∑

k

fk(ω2

k − ω2 − iγkω)

Rearranging as before and decompose by partial fractions using roots of denominator

n2 − 1 =3N e2

ǫom

∑k

fkDk(ω)

3−N e2

ǫom

∑k

fkDk(ω)

=∑

k

ρkν2k − ν2

=∑

k

Akλ20

λ20 − λ2

k︸︷︷︸Bk

ω0 =√ω20 − ω2

p/3 ν0 = ω0/2π λ0 =cν0

ωp =Ne2

ǫom

Or Taylor expand in transparency regime in terms of frequency ν or wavelength λ = c/ν

n2 − 1 ≈ A +Bν2 + Cν4 + · · · − B′

ν2− C ′

ν4≈ A +B

c2

λ2+ C

c2

λ4+ · · · − B′λ2

c2− C ′λ4

c4Kelvin Wagner, University of Colorado Physical Optics 2018 17

Sellmeier Equations to approximate indexvs wavelength across limited band of

transparencySellmeier = Sellmeir = Selmeir = Selmeier

Sellmeir formulas occur in a wide variety of forms, but most inlude only 1-2 terms or atmost 3. For visible sellmeiers, terms account for UV resonances and Far IR resonance.

n2 = 1 +∑

k

fkν2k − ν2

= 1 +∑

k

bkλ2k − λ2

= 1 +∑

k

Akν2k −

∑

j

Bj

ν2j

where the coefficients for a thin film, glass, or crystal may depend on composition orgrowth condition as well as temperature.

Also note, index is temperature dependent so some formulas include T dependence

∆n ≈ 10−6/C glasses

∆n ≈ 10−5/C crystals

∆n ≈ 10−4/C plastics


Sellmeier Formulas for various glasses takea variety of forms

Cauchy

n = A +B

λ2+

C

λ4

Conrady

n = A +B

λ+

C

λ7/2

Resonance

n2=A+B

λ2 − C2+

D

λ2 − E2+

F

λ2 −G2

Binomial Expansion gives 4 term approximation or more accurate in IR 6 term formula

n2 = aλ2 + b +c

λ2+

d

λ4≡ fλ6 + eλ4 + aλ2 + b +

c

λ2+

d

λ4

Herzberger’s formula used combination with resonance at λ0 = .187

n = A +Bλ2 +C

λ2 − λ20

+D

(λ2 − λ20)

2

Schott Catalog gives expersssions good out to 1µm (very good in UV and blue)

n2 = A0 + A1λ2 +

A2

λ2+

A3

λ4+A4

λ6+

A5

λ8

Bausch and Lhomb gives 7 term formula

n2 = a + bλ2 + cλ4 +d

λ2+

eλ2

(λ2 − f) + gλ2/(λ2 − f)


Sellmeir for Silica

Fit for index, wavelength in µm

n(λ0) = C0 + C1λ2 + C2λ

40 +

C3

λ20 − l

+C4

(λ20 − l)2

+C5

(λ20 − l)3

C0 = 1.4508554 C1 = −.0031268 C2 = −.0000381 C3 = −.0031268C4 = .0030270 C5 = .0000018 l = .035

√l = .187µm

Form of Sellmeier for index squared

n2(λ0)− 1 =A1λ

2

λ2 − λ21

+A2λ

2

λ2 − λ22

+A3λ

2

λ2 − λ23

=∑

k

Akλ2

λ2 − λ2k

material A1 λ1 A2 λ2 A3 λ3

SiO2 .6961663 .0684043 .4079426 .1162414 .8974794 9.8961613.5% GeO2 .711040 .064270 .451885 .129408 .704048 9.425478


Dispersion can be found from derivative ofSellmeier

d(n2 − 1)

dλ= 2n

dn

dλ

d

dλ

λ2

λ2 − λ2j

=f ′g − g′f

g2=

2λ(λ2 − λ2j)− 2λλ2

(λ2 − λ2j)

2=

−2λλ2j

(λ2 − λ2j)

2

dn

dλ=

−2λ

2n

(∑

k

Ajλ2j

(λ2 − λ2j)

2

)

Group index

ng = n− λdn

dλGroup Velocity Dispersion found from derivative of group index

dng

dλ=

dn

dλ− 1 · dn

dλ− λ

d2n

dλ2

= −λ

[(1 · n− dn

dλ· λ

n2

)∑ Ajλ2j

(λ2 − λ2j)

2− λ

n

∑−2(λ2 − λ2j)2λ

(λ2 − λ2j)

4Ajλ

2j

]

= −λng

nSS2

(−

ZZn

λ

dn

dλ

)− 4λ3

n

∑ Ajλ2j

(λ2 − λ2j)

3


Dispersion and GDD for Silica493

Figure D.1. Linear refractive index n and group index ng for fused silica.The zero dispersion wavelength zd = 1:27 m divides the boundary betweenthe normal dispersion regime ( < zd) and the anomalous dispersion regime( > zd). The curves are marked for their values at 1.55 m.is also referred to here as the phase delay coecient, since it is related to theinverse phase velocity. Of more relevance to this thesis, though, are the groupvelocity and the group delay coecient. The group velocity is the velocity atwhich a quasi-monochromatic wavepacket travels, which is in general dierentthan the phase velocity, and is most easily calculated from the group delaycoecient. The group delay coecient is the frequency derivative of the phasedelay: k0 = @k@! = 1c "n+ !@n@!# ngc ; (D.2)

495

Figure D.2. Group-delay dispersion k00 for fused silica. The zero dispersionwavelength zd = 1:27 m divides the boundary between the normal dispersionregime ( < zd) and the anomalous dispersion regime ( > zd). The curveis marked for its value at 1.55 m.Higher-order dispersion becomes important for broad temporal band-widths. The simulations of spatio-temporal solitons in this thesis use thehigher-order dispersion coecients because such broad bandwidths are present.Third-order dispersion is calculated fromk000 = @3k@!3 = 1c "3@2n@!2 + !@3n@!3# (D.7)= 442c3 "3@2n@2 + @3n@3 # :Fourth-order dispersion is calculated fromk0000 = @4k@!4 = 1c "4@3n@!3 + !@4n@!4# (D.8)Kelvin Wagner, University of Colorado Physical Optics 2018 22

Group Velocity and Dispersion

Consider optical pulses (no longer simply harmonic in time)For simplicity, start with a pulsed plane wave in free space

A(~r, t) = e−i(kxx′+kyy

′+kzz′)s

(t−

~k′ ·~rv

)eiω0t

Rotate coordinates to align z with direction of propagation

A(z, t) = e−i(kz−ωt)s(t− z

v

)

Now put into dispersive medium n(ω), thus v = ω(k)k = c

n(ω). Phase velocity is

vp =ω

k=

2πν

2πn(ω)/λ=

c

n(ω)

Utilize a Fourier representation of the frozen spatial representation of pulse s() at t = 0

A(z, 0) = eik0zs(−vz) =

∫ ∞

−∞S(k)eikzdk


Fourier propagation of each spectralcomponent

Thus, pulse at later time is given by propagating each Fourier component with appro-priate ω(k)

A(z, t) =

∫ ∞

−∞S(k)e−ik(vt−z)dk =

∫ ∞

−∞S(k)e−i(ω(k)t−kz)dk

Now ω(k) = k cn(ω) since k(ω) =

n(ω)c ω

Taylor expand about central wavevector k0

ω(k) = ω|k0+dω

dk

∣∣∣∣k0

(k−k0)+d2ω

dk2

∣∣∣∣k0

(k − k0)2

2+· · · = ω0+vg(k−k0)+D

−112(k−k0)

2+. . .

Plug back in Fourier integral

A(z, t) =

∫ ∞

−∞S(k)e−i([ω0+vg(k−k0)+D−11

2(k−k0)2+... ]t−(k−k0)z−k0z)dk

≈ e−i(ω0t−k0z)

∫ ∞

−∞S(k)e−i[ω′t−z](k−k0)dk

︸︷︷︸= e−i(ω0t−k0z)︸︷︷︸E(z − ω′t)︸︷︷︸

function of z − ω′t phase factor Envelope


Phase and Group Velocity

Phase Velocity

vp =ω0

k0=

c

n

Group Velocity

vg = ω′ =dω

dk

∣∣∣∣k0

Since we represent k as a function of ω rather than vice versa, it’s easier to calculate

dk

dω=

d

dω

[ωcn(ω)

]=

n(ω)

c+ω

c

dn

dωvg =

c

n(ω) + ωdn(ω)dω

=dω

dk=

(dk

dω

)−1

In 3-D A(~r, t) = A(~r, t)e−i(ωt−gω(~r))

vp =1∣∣∇ gω

∣∣ vg =1∣∣∣∇∂g∂ω

∣∣∣ω

∣∣∣where ω is the mean frequencyfor a plane wave g = ~k ·~r, thus

∇g

ω= ∇n(ω)ω/c

ωk ·~r = n

ck


Group Velocity, Dispersion, GVD

In terms of free space wavelength λ0 =cν= 2πc

ω

ωdn

dω= ω

dn

dλ0

dλ0

dω= ω

dn

dλ0

−2πc

ω2=

dn

dλ0

−2πc

2πc/λ0= −λ0

dn

dλ0

1

vg=

1

c

(n(λ0)− λ0

dn

dλ0

) PropagationTime τ =

L

vg=

L

c

(n− λ0

dn

dλ0

)

︸︷︷︸group index = ng

Dispersion is due to variation of propagation time with λ0 (or ω)

∆τ =dτ

dλ∆λ=

L

c

(

dn

dλ0− 1 ·

dn

dλ0− λ0

d2n

dλ20

)∆λ=−λ0

L

c

d2n

dλ20

∆λ=−L

c

(λ20

d2n

dλ20

)∆λ

λ

Material dispersion proportional to L and ∆λ. Dispersion parameter

Dm =∆τ

L∆λ= − 1

λ0c

(λ20

d2n

dλ20

)[ s

m m

]=

1012ps

10−3km 109nm︸︷︷︸×109 ps

nm km with λ0 (µm), c(kms )


Group Velocity near Lorentzian

Michelson measured difference between vp and vg.CS2: n = 1.64 using refractometer and ng = 1.76

ω =c

nk

dω = c

(dk

n− k

n2dn

)

dk

dω=

1

c

[n + ω

dn

dω

]

dω

dk=

c

n + ω dndω

For large ω where n < 1 and dndω

> 0 the denominator > 1 and vg < c

Anomalous dispersion where dndω < 0 and where n < 1 ⇒ vg > 1

Near ω0 absorption is highest. Trailing edge of the pulse is preferentially absorbed,causing the peak to be displaced forward, so that energy in the leading edge travels ata velocity less than c. But the peak travels faster than c.


Propagation of Gaussians and complexGaussians

Gaussian pulse and its Fourier decomposition

E(0, t) = e−t2/σ2

︸︷︷︸ eiω0t︸︷︷︸ = eiω0t

∫ ∞

−∞

σ√4π

e−Ω2σ2/4eiΩtdΩ

Each frequency component propagates with its own phase factor

E(z, t) = ei(ω0t−β0z)

∫ ∞

−∞e−iβ(Ω)z σ√

4πe−Ω2σ2/4eiΩtdΩ

And expand the remaining terms in β around the center freq

β(ω0 + Ω) = β(ω0) +dβ

dω

∣∣∣∣ω0

Ω + 12

d2β

dω2

∣∣∣∣ω0

Ω2 = β0 + v−1g Ω + 1

2DΩ2

Leads to an integral with a complex Gaussian, which is a chirped Gaussian pulseUse the very useful definite integral

∫ ∞

−∞e−ax2+bxdx =

√π

aeb

2/4a

and do not worry that the constants a or b may be complex


Complex Gaussian propagation

chirp

Ec(0, t) = e−t2/σ2︷︸︸︷e−ibct

2eiω0t = e

−t2( 1σ2

+ibc)eiω0t

= eiω0t1√

4π(1/σ2 + ibc)

∫ ∞

−∞e− Ω2

4(1/σ2+ibc)eiΩtdΩ

Instantaneous frequency b = 2bc2π

[GHzns

]

fi = I

1

2π

dφ

dt

=

−2bct

2π+

ω0

2π= f0 + bt

Now can proceed as before to propagate the Fourier decomposition by just replacing thewidth σ2 with the complex “width” 1

1/σ2+ibc

E(z, t) = ei(ω0t−β0z)1√

4π(1/σ2 + ibc)

∫ ∞

−∞e−iβ(Ω)ze

− Ω2

4(1/σ2+ibc)eiΩtdΩ

Solve integral as before, or just generalize the previous time domain result.Must isolate resulting real and imaginary parts of quadratic exponential to identify widthof complex Gaussian.


Dielectric Response Theory andKramers-Kronig relations

Frohlich, Theory of Dielectrics, Oxford 1958, Appendix A.1

Electric field driven polarization response of material is1) Linear2) Time Invariant3) Causal

u = t− v v = t−u when u = (−∞, t) → v = (∞, 0)

P (t) = ǫ0

∫ t

−∞E(u)χ(t− u)du = −ǫ0

∫ 0

∞E(t− v)χ(v)dv

D(t) =

∫ t

−∞E(u)ǫ(t− u)du = ǫ0

∫ ∞

0

E(t− v)εr(v)dv

4) Finite response time

χ(t), εr(t) → 0 as t → ∞

instantaneous response component

ǫ(ω) → ǫ∞ + i0 ω → ∞

static component

ǫ(ω) → ǫs + i0 ω → 0

E(t)

P(t) χ Es

8χ E


Harmonically time varying field

E(t) = E0 cosωt = E0e−iωt

2+ cc

x= t−u dt=−dxt→0

−∞→∞ cosω(t− x) = cosωt cosωx− sinωt sinωx

D(t)=E0ǫo

∫ t

−∞cos(ωu)εr(t−u)du+ ǫ∞E(t) = ǫoE0

∫ ∞

0

︷︸︸︷cosω(t− x) εr(x)dx + ǫ∞E(t)

= ǫoE0 cosωt

[∫ ∞

0

εr(x) cosωx dx + εr∞

]+ ǫoE0 sinωt

∫ ∞

0

εr(x) sinωx dx

= ǫ′E0 cosωt︸︷︷︸+ ǫ”E0 sinωt︸︷︷︸in-phase quadrature

Complex dielectric constant at ω

ǫ(ω) = ǫ′(ω) + iǫ”(ω) = ǫo(ε′r(ω) + iε”r(ω))

ǫ′(ω) =

∫ ∞

0

ǫ(t) cosωt dt + ǫ∞ (1)

ǫ”(ω) =

∫ ∞

0

ǫ(t) sinωt dt (2)


Evaluate Fourier transforms

Take Fourier transform (cos transform of 1, sin transform of 2 )

ǫ(t) =2

π

∫ ∞

0

[ǫ′(ω)− ǫ∞] cosωt dω (3)

ǫ(t) =2

π

∫ ∞

0

ǫ”(ω) sinωt dω (4)

Substitute 4 into 1

ǫ′(ω)− ǫ∞ =

∫ ∞

0

cosωt2

π

∫ ∞

0

ǫ”(ω′) sinω′t dω′ dt

=2

πlimR→∞

∫ ∞

0

ǫ”(ω′)

∫ R

0

cosωt sinω′t dt dω′

=2

π

∫ ∞

0

ǫ”(ω′) limR→∞

1

2

(1− cos(ω′ + ω)R

ω′ + ω+1− cos(ω′ − ω)R

ω′ − ω

)dω′

=2

π

∫ ∞

0

ǫ”(ω′)1

2

[1

ω′ + ω+

1

ω′ − ω

]dω′ =

2

π

∫ ∞

0

ǫ”(ω′)1

2

(ω′ − ω) + (ω′ + ω)

ω′2 − ω2dω′

Although taking the limit and just ignoring the oscillating component seems unjustified


Kramers-Kronig relations

When frequency data is available only for measured (positive) frequencies

ǫ′(ω)− 1 =2

π

∫ ∞

0

ǫ”(ω′)ω′

ω′2 − ω2dω′

ǫ”(ω) =2

π

∫ ∞

0

(ǫ′(ω′)− 1)ω

ω′2 − ω2dω′

Suprisingly good agreement when only a band of frequencies is available

Alternatively the Hilbert transform pairs integrating across all frequencies

ǫ′(ω)− 1 =1

πPV

∫ ∞

−∞

ǫ”(ω′)

ω′ − ωdω′

ǫ”(ω) =1

πPV

∫ ∞

−∞

ǫ′(ω′)− 1

ω′ − ωdω′

Similar relationships exist for other pairs of real and imaginary parts of frequency re-sponses for linear systems that are causal

n(ω)Kramers−−−−−−−−−−Kronig

κ(ω) χ′(ω)Kramers−−−−−−−−−−Kronig

χ”(ω) ln ρ(ω)Kramers−−−−−−−−−−Kronig

φ(ω)


Complex Analysis

f(z) is analytic in domain D if it is differentiable at every point in D

w = f(z) = f(x + iy) = u(x, y) + iv(x, y)

f ′(z) = u′x + iv′x = v′y − iu′yCauchy-Rieman eqn

∂u

∂x=

∂v

∂y

∂u

∂y= −∂v

∂x

Complex IntegrationCauchy Thm If f(z) analytic on and inside contour C

∮

C

f(z)dz = 0

Cauchy Integral Formula (f(z) analytic on and inside contour C)

f(a) =1

2πi

∮

C

f(z)

z − adz


Proof of Cauchy-Rieman condition foranalyticity

f ′(z) = lim∆z→0

f(z +∆z)− f(z)

∆zwhere f(z) = u(x, y) + iv(x, y)Now the path of the infinitesimal difference can be along real or imaginary axis incomplex plane (or along any direction)∆z = ∆x

f ′(z) =u(x +∆x, y) + iv(x + ∆x, y)− [u(x, y) + iv(x, y)]

∆x=

∂u

∂x+ i

∂v

∂x

∆z = i∆y

f ′(z)=u(x, y+∆y) + iv(x, y+∆y)− [u(x, y) + iv(x, y)]

i∆y=1

i

[∂u

∂y+i

∂v

∂y

]=−i

∂u

∂y+∂v

∂y

Setting these two results equal gives Cauchy Rieman condition for analyticty

∂u

∂x=

∂v

∂y

∂v

∂x= −∂u

∂y


Alternative proof

z = x + iyz∗ = x− iy

x =z + z∗

2⇒ ∂x

∂z= 1

2y =

z − z∗

2i⇒ ∂y

∂z=

1

2iChain rule

∂f

∂z=

∂f

∂x

∂x

∂z+∂f

∂y

∂y

∂z= 1

2

(∂f

∂x+1

i

∂f

∂y

)

∂f

∂z∗=

∂f

∂x

∂x

∂z∗+∂f

∂y

∂y

∂z∗= 1

2

(∂f

∂x− 1

i

∂f

∂y

)= 0 since f(z) not z∗

⇒ ∂f

∂x= −i

∂f

∂y

ux + ivx = −i(uy + ivy)

which gives by setting real and imaginary parts equal ux = vy vx = −uy

And in polar coordinates ∂u

∂r=

1

r

∂v

∂θ

∂v

∂r= −1

r

∂u

∂θ


Residues

Resz=af(z) = c−1 Laurent expansion f(z) =∑∞

n=−∞ cn(z − a)n

Residue Theorem1

2πi

∮

C

f(z)dz =n∑

k=1

Resz=akf(z)

Calculation of Residues

1. Determine c−1 from Laurent expansion

2. Simple poles Resz=af(z) = limz→a(z − a)f(z)

Use l’Hopital’s rule limx→af(x)g(x)

= limx→af ′(x)g′(x)

eg f(z) = n(z)d(z) n(z) , d(z) analytic. n(a) 6= 0, d(a) = 0, d′(a) 6= 0

Resz=af(z)

g(z)=

f(a)

g′(a)

3. Poles of Order m

Resz=af(z) = limz→a

1

(m− 1)!

(d

dz

)m−1

[(z − a)mf(z)]


Examples of poles and residues

functions zeroes polesz = x + iy 0 m = 1 ∞ (m = 1)z2 = x2 − y2 + i2xy 0 m = 2 ∞ (m = 2)1/z = x/r2 − iy/r2 ∞ m = 1 0 (m = 1)1/z2 ∞ m = 2 0 (m = 2)log z = ln r = iθ + 2nπ 1 m = 1(principal branch) 0,∞ branch points√z 0, branch point ∞, branch point

ez = ex cos y + iex sin y ∞ essential singularity

Example: Definite integral using CCI f(z) = eiz

z2+a2= eiz

(z+ia)(z−ia)

I =

∫ ∞

−∞

cosx

x2 + a2dx = R

∫ ∞

−∞

eix

x2 + a2dx a > 0

Resz=ia

f(z) = limz→ia

(z−ia)eiz

z2+a2= e−a lim

z→ia

z−iaz2+a2

l′Hop= e−a lim

z→ia

12z

= e−a

2ia⇒ I=R

2πie

−a

2ia

=πe−a

a

f(z) analytic in UHP except finite ak I =∫∞−∞ f(x)dx = 2πi

residues∑k

Resz=ak

f(z)


Lorentzian Complex Plane

Write Lorentzian using partial frac-tions

χ(ω) =f0Ne2

ǫom

1

ω20 − ω2 − iΓω

=ω2p

ω1 − ω2

(1

ω − ω1− 1

ω − ω2

)

=ω2p[ω − ω2 − (ω − ω1)]

(ω1 − ω2)(ω − ω1)(ω − ω2)

=ω2p

(ω − ω1)(ω − ω2)

Where the poles are at

ω1 = −12i[Γ− (Γ2 − 4ω2

0)12 ]

ω2 = −12i[Γ + (Γ2 − 4ω2

0)12 ]

ε′ ε” n(ω)


Dielectric Constant and Index in theComplex Plane


Multiple Lorentzians in Dispersion relation


Relative Dielectric Constants and ComplexIndex

N2 = n2 − κ2 + i2nκ = 1 + χ′ + iχ′′ N =√1 + χ =

√ǫr


Dispersion between UV and IR Lorentzianresonances

ng = n + ωdn

dω= n− λ

dn

dλ k =ω

cn(ω) vp =

ω0

k0>

dω

dk

∣∣∣∣k0

= vg


Kramers-Kronig with Complex ContourIntegration

Kramers-Kronig dispersion relation can be derived from the contour integral(Cauchy Principal value Integral)

P∫ ∞

−∞

χ(ω′)

ω′ − ωdω′ = lim

δ→0

(∫ ω−δ

−∞+

∫ ∞

ω+δ

χ(ω′)

ω′ − ωdω′)

=

∫

A

−∫

B

−∫

C

C

BA

ω2ω1

ωω−δ ω+δ 88−

∫A = 0 since it contains no poles∫B = 0 integrand → 1

ω′2 for large radius length ≈ π|ω′|∫C

χ(ω′)ω′−ω

dω′ = −i2π2residue = −iπχ(ω)

iπχ(ω) = P∫ ∞

−∞

χ(ω′)

ω′ − ωdω′

iπ (χ′(ω) + iχ”(ω)) = P∫ ∞

−∞

χ′(ω) + iχ”(ω)

ω′ − ωdω′


Kramers-Kronig

Equate real and imaginary parts

χ”(ω) =−1

πP∫ ∞

−∞

χ′(ω′)

ω′ − ωdω′ χ′(ω) =

1

πP∫ ∞

−∞

χ”(ω′)

ω′ − ωdω′

Since ~P(t) and ~E(t) are real, χ(t) is real

⇒ χ(−ω) = χ∗(ω) χ′(−ω) + iχ”(−ω) = χ′(ω)− iχ”(ω)

χ′(−ω) = χ′(ω) real part is evenχ”(−ω) = −χ”(ω) imaginary part is odd

χ(ω) =1

iπP∫ 0

−∞

χ∗(ω′)

ω′ + ωdω′ +

1

iπP∫ ∞

0

χ(ω′)

ω′ − ωdω′

χ′(ω)+iχ”(ω) =−i

πP∫ ∞

0

1

ω′2 − ω2[− (χ′(ω)− iχ”(ω)) [ω′ − ω] + (χ′(ω) + iχ”(ω)) [ω′

χ′(ω)[ω − ω′ +

ω′ + ω] + iχ”(ω)[ω′ − ω + ω′ + ω]

χ′(ω) =2

πP∫ ∞

0

ω′χ”(ω′)

ω′2 − ω2dω′ χ”(ω) =

−2

πP∫ ∞

0

ωχ′(ω′)

ω′2 − ω2dω′


Propagation of a pulse through a dispersivemedium

Following Jackson, Classical Electrodynamics, sec 7.11

Represent a pulse by its Fourier spectrumui(0, t) incident on surface

A(ω) =1

2π

∫ ∞

−∞ui(0, t)e

iωtdt

note this is using alternative sign defn of FT!Fresnel transmission t01 =

2n0n0+n1

= 21+n(ω)

n2=n0n1(ω)n0=1

z=0

Propagation described by different frequency components propagating at different phasevelocities with different k(ω) = ω

cn(ω) = kR + ikI . For this sign, kI > 0 is absorption

u(z, t) =

∫ ∞

−∞t01(ω)A(ω)e

i(k(ω)z−ωt)dω

output into nondispersive free space t12 =2n1

n1+n2= 2n(ω)

n(ω)+1

u(z, t) =

∫ ∞

−∞

4n(ω)

[1 + n(ω)]2A(ω)ei(k(ω)z−ωt)dω (5)


Approximate the wavevector by its Taylor expansion about the central frequency ω0

k(ω) = k|ω0+∂k

∂ω

∣∣∣∣ω0

(ω−ω0)+∂2k

∂ω2

∣∣∣∣ω0

12(ω−ω0)

2+· · · = k0+v−1g (ω−ω0)+D

12(ω−ω0)

2+. . .

v−1g = 1

c(n + ω ∂n∂ω) D = 1

c

[2∂n∂ω + ω ∂2n

∂ω2

]

Assume ui(0, t) = 0 t < 0⇒ A(ω) analytic in upper ω-plane

UHP ω = α + iβ, β > 0 eiωt → e−βt

t < 0 (divergent unless u = 0)t > 0 (convergent for u finite)

ui smoothly turns on

ui(0, t) = atm

m!H(t) t → 0

A(ω) → a

2π

(i

ω

)m+1

x as |w| → ∞

t3

21

ttt 0

t

u(t)

t=0

integration in time ⇒ 1iω in Fourier domain using standard FT, 1

−iω = iω using + sign


Medium with a single Lorentzian resonance

εr(ω) = n2(ω) = 1 +ω2p

ω20 − ω2 − iγω

ω2p =

Ne2

mǫoplasma freq, as ω → ∞ εr(ω) = 1− ω2p

ω2

when ω±z = ±ω1 − iγ/2 with ω2

1 = ω20 + ω2

p − γ2/4 thus (ω±z )

2 = ω21 + γ2/4∓ 2ω1

iγ2

n(ω±z ) = 0 = 1 +

ω2p

ω20 − (

ω20 + ω2

p − γ2

4 +γ2

4 ∓ HHHHH2ω1

iγ2 )∓

HHHHHiω1γ −γ2

2

= 1− 1 = 0

so ω±z are the zeroes of εr(ω) and n(ω)

when ω±p = ±ω2 − iγ/2 with ω2

2 = ω20 − γ2/4

n(ω±p ) → ∞ give poles

0 =ω20 − (

ω20 −

AAAAA

γ2

4∓ ω2iγ −

AAAAA

γ2

4)∓ iω2γ −

AAAAA

γ2

2

n2(ω) =(ω − ω+

z )(ω − ω−z )

(ω − ω+p )(ω − ω−

p )

iγ/2

Analytic inupper half plane

pole zero

branchcut

ω1ω2−ω1 −ω2


Propagation Velocity Limit

n(ω) =

√(ω − ω+

z )(ω − ω−z )

(ω − ω+p )(ω − ω−

p )

choose branch cuts so that n(ω) → 1 as |ω| → ∞Evaluate 5 by contour integration ω = ωr + iωi n(ω) → 1 as |ω| → ∞

ei(k(ω)z−ωt) = eiω(n(ω)c z−t

)→ ei

ωc (z−ct) = e−ωi(z−ct)eiωr(z−ct)

z > ct use upper half plane contour

z < ct use lower half plane contour

no poles in upper half plane for either n(ω) or A(ω) ⇒∮= 0 z > ct

u(z, t) = 0 z > ct


Fourier representations

sinω0t ⇐⇒ iδ(ω − ω0)− iδ(ω + ω0)

Π

(t

T

)⇐⇒ iT sinc(T (ω − ω0))− iT sinc(T (ω + ω0))

f(t) =

0 t < 0sinω0t t > 0

=−ω

2π

∫e−iωt

ω2 − ω20

dω

t < 0 ∮→ 0

e−iωt → e−ωIt → 0ωΙ

ωRω0−ω0t > 0∮

= 2πi∑

residues = ω0ie−iω0t − ie+iω0t

2ω0= sinωt

e−iωt → e−ωIt → 0ωΙ

ωR

ω0−ω0

cancels

Dispersive propagation

f(x, t) =−ω

2π

∫ei(k(ω)z−ωt) dω

ω2 − ω20

ωΙ

ωR

ω0−ω0

cancels

−ωp −ωz ωpωz


Stationary Phase

Integrals of the form

I =

∫F (ω)eiφ(ωdω

F (ω) slowly varying with ω

φ(ω) phase that results in rapid variation of eiφ(ω)

integrand averages to zero when phase varies rapidly except when φ(ω) is stationary

φ′s =

dφdω

∣∣∣ωs

= 0 ωs is stationary point

⇒ φ(ω) = φs + 0 · (ω − ωs) +12φ′′s(ω − ωs)

2 + . . . nearby to ωs

Since F (ω) is slowly varying it is constant near ωs

Is = F (ωs)eiφs

∫eiφ

′′s(ω−ωs)

2/2dω = F (ωs)eiφs

√2πi

φ′′s

more than one stationary point yields a sum of terms

When φ(ω) is real [φ = ωcn− ωt ] stationary phase works, but in regions of absorption

where N = n + iκ it can produce erroneous results. Use steepest descent instead


Propagation near dispersive resonance

φ(ω) = k(ω)z − ωt where k(ω) = ωcn(ω)

Stationary points (k = ωtz )

∂φ∂ω = k′z − t ⇒ cdkdω = ct

z for ct > z t0 =zc

n2(ω) = 1 +ω2p

ω20 − ω2 − iωγ

|ω|→∞−−−−→ 1− ω2p

ω2

⇒ n(ω) → 1− ω2p

2ω2|ω| → ∞

⇒ k(ω) → ω

c

(1− ω2

p

2ω2

)|ω| → ∞

cdk

dω=

(1− ω2

p

2ω2

)−ω(−2)

ω2p

2ω−3 = 1+

ω2p

2ω2

1

n(0)

ω0+ωp2 2

dkdω

c

n(ω)ωs

ωs

SommerfeldPrecursor

BrilluoinPrecursor

ct/z

ω0

stationary point ωs when cdk(ωs)dω

= ctz= 1 +

ω2p2ω2

ct− z

z=

ω2p

2ω2s

⇒ ωs = ωp

√z/c

2(t− z/c)= ωp

√t0

2(t− t0)


Sommerfeld and Brillouin Precursor

u(z, t) =

∫ ∞

−∞

2

1 + n(ω)A(ω)ei(k(ω)z−ωt)dω =

2

1 + n(ωs)A(ωs)e

i[k(ωs)z−ωst]

√2πi

φ′′s

Complex contour integration and bessel integral gives

u(z, t) = a

(t− t0ω2pt0/2

)m/2

Jm

[2

√ω2pt0

2(t− t0)

]

Sommerfeld precursor: pushes x-raysthrough media at c

Brilluion precursor: low frequenciespropagate through at DC phase velocity

since this is a point where φ′′ = 0 must ex-pand to φ′′′. Leads to Airy integral

t

zt=z/c

tB=zn(0)/c t=zn(ω0)/c t=zng/c

very low freqs

x-rays

These precursors are due to presence of frequency components at all frequencies due tostep in time, with all frequencies present.More realistic laser pulses are bandlimited ⇒ no presursors observed.


Dispersing Prism

A

αβ γ δ

A

Snell’s law at inputsinα = n(λ) sinβ

Prism anglesβ + γ + (180− A) = 180 =⇒ γ = A− β

Snell’s law at outputsin δ = n(λ) sin γ

Total deviation angleD = α + δ − A


Dispersion in a thin prism

B=AL

W

A θ

nAW

W(A+θ)

s

L

A is apex angle. For prism height L base is B = 2L tan A2 ≈ AL

∂D

∂λ=

∂δ

∂λ=

∂δ

∂n

∂n

∂λ= A

∂n

∂λ

nAL = W (A + θ)

Solving for θ and for the thin prism approximating W ≈ L

θ =1

W(nAL−WA) ≈ (n− 1)A

Taking the derivative wrt λ gives

∂θ

∂λ= A

∂n

∂λSo the variation of deviation angle D with wavelength λ is apex angleA× disperison, and ∂δ

∂n= A

While in a thick prism at minimum deviation dδdλ = dδ

dndndλ

For instance at minimum deviation with n = sin δ/ sin γ thus dδdn

= sin γcos δ

For refraction at both faces with γ = A/2 and expressing in terms of lengths

dδ

dn= 2

sin γ

cos δ=

2 sin A2

cos δ=

2s sin A2

s cos δ=

B

WKelvin Wagner, University of Colorado Physical Optics 2018 55

Deviation Angle

sin δ = n(λ) sin γ = n sin(A− β) = n(λ)[sinA cos β − cosA sin β]

n sin β = sinα

n cos β =√n2 − sin2 α

sin δ = sinA√n2 − sin2α− cosA sinα

D = α + δ − A

αβ

∂D∂α = 1 + ∂δ

∂α − 0 = 0 at minimum deviation

∂δ

∂α= −1 =

∂δ

∂γ

∂γ

∂β

∂β

∂α=

n cos γ

cos δ· (−1) · cosα

n cos β=⇒ cosα cos γ = cos β cos δ

=⇒ α = δ β = γ = 12A Symmetric Ray


Dispersion

Index measured at spectroscopic line

Atom Line wavelength Crown FlintK-50 BaFN-10

H C 656.3 1.51992 1.6658Na D 589.3He d 587.6 1.52257 1.67003H F 486.1 1.52861 1.68000

522602 670472

Abbe

number V

V =nd − 1

nF − nC60.1 47.2

UV

IR

Visible

Absorption

Index

486

H

587

He

656

H

nF n d n c

λ

λ

Glass designation3 digits of index (d-line) -1 + 3 digits of V-number[(nd − 1) ∗ 1000] ∗ 1000 + V ∗ 10


Prism Dispersion

Change of refractive index with wavelength

∂δ

∂λ=

∂δ

∂n

∂n

∂λsince

sin δ = sinA√n2 − sin2α− cosA sinα

1.5

2.0

2.5

3.0

400 500 600 700

TiO2

FusedSilica

Flint

∂ sin δ

∂n= cos δ

∂δ

∂n= sinA

1

2(n2 − sin2 α)−1/2 · 2n =

sinA

cos β

=⇒ ∂δ

∂n=

sinA

cosβ cos δ

dispersion =∂δ

∂λ=

sinA

cosβ cos δ

∂n

∂λ

For a 60 prism at minimum deviation sinAcos β = sin 2β

cos β = 2 sin β = 1

dispersion = 1cos δ

∂n∂λ

∆n∆λ ≈ 2.7−2.3

.6−.4 = 2.0µm−1


Simple resolution argument

λλ+δλ

λa a

B

One extra wave OPD gives Rayleigh limit

λ : 2a = nB

λ + δλ : 2a + λ = (n + δn)B

subtracting: λ = δn · B

− λ

δλ= B

dn

dλ


Prism Spectroscope

Source

Slit

B

Dispersing Prism

Collimating Lens

Focusing lens

Detector Plane

α

γ

δ

• Source must be focussed through a narrow slit. Slit width limits resolution.

Image of slit determines resolution. Diffraction limited or geometrical?

• Radiometric throughput of incoherent source dramatically limited by slit width

•Wide slit ⇒ Photon efficient.

• Narrow slit ⇒ High resolution.

• Angles of refraction determined by Snell’s law and glass dispersion. k-space.

• Blue refracted by larger angle than red, nb > nr


Resolving Power of a Prism

λλ+δλ

λa a

BSlit

D2

D1

W

F2

S

SW

δ

Each wavelength gives rise to an image of the slit broadened by diffraction

∆θ =λ

W∆x = F2∆θ

Just resolved spectral lines λ, λ + dλ give separation dδ

λW

= ∂δ∂λ

· dλ =⇒ Resolving Power = λdλ

= W ∂δ∂λ

Dispersion - Change of angle with wavelength B

∂δ

∂λ=

sinA

cos β cos δ

∂n

∂λ=⇒ λ

dλ=

W

cos δ

sinA

cos β

∂n

∂λ=

︷︸︸︷S · 2 sin A

2

∂n

∂λ


Minimum Deviation Resolving Power

At minimum deviation A = β/2

sinAcos β = sin 2β

cos β = 2 sin β

2∆ = 2 · S sin A2 = D2 −D1

2D /2

D /21

∆

SA/2

Resolving Power

RP =λ

dλ= (D2 −D1)

∂n

∂λ

To resolve Na D lines (588.9 and 589.3) required RP > 980

D2 −D1 > 980/.12µm−1 = 8170µm ≈ 9mm


Prisms, dispersion, and pulse tilt

Angular dispersion gener-ates tilted pulses

Real and momentum spaceprofiles of tilted pulses

2δ

1δA

x

Tilted Pulse

Dispersion broadened

θ

n

l(x)

x

z3

3

z2

x2

1z

x

1

Prism

Input Pulse

Kishore Y, Department of Electrical Engineering, University of Colorado, Boulder -p.18/40Kelvin Wagner, University of Colorado Physical Optics 2018 63

Titled pulses out of prisms with 1st orderdispersion and including either 2nd order

dispersion or 3rd order dispersion


Prism curved slit image


Grating Spectroscopy

θλ

d

θλ

d

sin θ =λ

d

d

x2x1

dθiθ

d

x2x1

dθiθ

sin θi =x1d

sin θi =x2d

Resonace condition: OPD = integer number of wavelengths

x1 + x2 = kλ = d(sin θi + sin θd) order k

Dispersion

d(kλ)

dλ=

d

dλ(d(sin θi + sin θd)) ⇒ k = d cos θd

dθddλ

dθddλ

=k

d cos θdKelvin Wagner, University of Colorado Physical Optics 2018 66

Grating Resolution

Diffracted beam at angle θd has width W = Nd cos θd = L cos θdN=no. grating periods illuminated, L = Nd= total illuminated widthDiffraction of uniform illuminated width leads to a beam spreading angle

∆θ =λ

W=

λ

Nd cos θdAngle between two adjacent wavelengths sparated by by ∆λ

∆θd =k

d cos θd∆λ

When spectral components are separated by ∆θthey are barely resolved

∆θ =λ

Nd cos θd≡ k

d cos θd∆λ = ∆θd

Resolving PowerR =

λ

∆λ= kN


Grating Resolution and Higher Orders

• One extra wave OPD gives just resolvable angle

∆θ across beam width Nd

• Lens focuses just resolvable beams to just resolv-

able spots

• Higher order lead to higher resolution

• Just resolvable in m = 1

• Well resolved in m ≫ 1Jenkins and White, Fundamentals of Optics, 1950.2001


Array Theorem

*

comb

window

element

# zeroes =N-1 # subpeaks =N-2To represent diffraction gratings and other finite width periodic structures

Consider a finite width periodic structure

g(x) = w(x) ·[f(x) ∗ 1

Xcomb

( x

X

)] G(u) = W (u) ∗ [F (u) · comb (uX)]

2D

g(x, y) = w(x, y) ·[f(x, y) ∗ ∗ 1

XYcomb

( x

X,y

Y

)]

G(u, v) = W (u, v) ∗ [F (u, v) · comb (uX, vY )]


Square Wave Amplitude Grating

1/2+tm

1/2-tm

1.0

0

1/2

L

x

Square Wave Amplitude Transmission

Amplitude square wave of period L grating can be represented in various ways

t(x) = comb L(x) ∗[(12 − tm)Π

(xL

)+ 2tmΠ

(xL

)]

Thus the FT is given by

T (u) =1

Lcomb 1/L(u)

[(12− tm)Lsinc(Lu) + 2tmL/2sinc

(Lu

2

)]

With Fourier orders

cn = (12− tm)sinc(n) + tmsinc

(n2

)

Fraction of light power diffracted into each first order is given by

|c1|2 =∣∣∣∣(12 − tm)

sin 1π

1π+ tm

sinπ/2

π/2

∣∣∣∣2

=

∣∣∣∣(12 − tm)0

1π+ tm

1

π/2

∣∣∣∣2

=

(2tmπ

)2

< 10.13%


Square Wave Amplitude Grating

T (u) =1

Lcomb 1/L(u)

[(12 − tm)Lsinc(Lu) + tmLsinc

(Lu

2

)]

tm = 0.5 tm = 0.2


Square Wave Phase Grating

1

L

ϕ1

iϕe

real

ima

g

-1

1

When ϕ=π, becomes a bipolar square wave

Phase grating jumps back and forth between 1 (phase 0) and phase φ

Unit amplitude Square Wave Fourier coefficients

cn =1

LFΠ

(x

L/2

)∣∣∣∣fx=n/L

= 12sinc

(n2

)

Square wave phase grating

t(x) = 1− [(1− eiφ)× square wave] = 1−[(1− eiφ)×

∞∑

n=−∞cne

i2πnx/L

]

FT is given by

T (u) = δ(fx)− (1− eiφ)

∞∑

n=−∞

12sinc

(n2

)δ(fx −

n

L

)

To maximize DE into 1st order choose cosφ = −1 ⇒ φ = π

η±1 = −212

sin(1π/2)

1π/2=

2

π⇒ |η±1|2 =

4

π2= 40.5%


Square Wave Phase Grating

T (u) = δ(fx)− (1− eiφ)

∞∑

n=−∞

12sinc

(n2

)δ(fx −

n

L

)

φ = π φ = π/2


Sinusoidal Phase grating

t(x) = eim2 sin(2πf0x)Π

(xw

)

Use Bessel identity

eim2 sin(2πf0x) =

∞∑

n=−∞Jq

(m2

)ei2πqf0x

T (u)=

∞∑

n=−∞Jq

(m2

)w2sinc(w(u− qf0))

PhaseGrating

corrugatedwavefron

resolved intoplane wave k-space

Λ=

2π ΛK =---t θ

Fraunhoffer diffraction: U(x, y) = eikz

iλz ei k2z (x

2o+y2o)T

(xλz ,

yλz

)

I(x) =1

λ2z2

∞∑

n=−∞J2q

(m2

)w4sinc2(w(u− qf0))


Grating Orders and Blazing to IncreaseEfficiency

β δ

Nd

Transmission Grating

• Twin symmetric ± orders

• 25% limit to Efficiency

• red diffracted more than blue

Blazed Transmission Grating

0 d x

OPD

λ0

0th order

+1 order

+2 order

-2 order

-1 order

BlazedTransmission

Grating

• Less common than reflection

• Steers light into desired order

• Efficiency peaks at blaze λbl

•Higher order: echelons


Blazed Reflection Grating

β

βm=0

m=1

δd

Reflection Grating (Blazed)

Nd

k-spacekx

kz


Blazed Gratings in k-spaceand the Littrow condition

Element

Function


Blazed grating Diffraction Efficiency andOrders

In either reflection or transmission geometry, linear phase across element function

t(x) = combL(x) ∗ Π[xL

]eiφ0x/L

φ

x

L/2-L/2

φ0/2

−φ0/2Fourier spectrum gives DE of various orders

T (u) = 1Lcomb1

L(u)Lsinc[L (u + φ0/2πL)] =

∑

n

δ(u− nL)sinc

[Lu + φ0

2π

]

=∑

n

δ(u− nL) sinc

[n + φ0

2π

]

︸︷︷︸an

DE into nth order ηn = |an|2 goes to 100% when φ0

is 1 wave

|T (u)|2 =∑

n

δ(u− nL)sinc

2[n + φ0

2π

]

φ0=2π⇒∑

n

δ(u− nL)sinc

2[n + 1]

n=−1⇒ 100%


Monochrometer: Rotatable grating and aslit

MonochromaterRotatable Grating

Entrance Slit

Exit Slit


High resolution grating: skimming incidence


Zero dispersion line geometry

Zer

o D

ispe

rsio

nge

omet

ry


Mickelson’s Echelon

Michelson’s Transmission Echelon, m=20000,N=20 => R=400,000

t=1cm

Entrance Slit, H or V

SpectrometerEntrance slitcan be H or V

ReflectionEchelon

Gasinlet

Win

dowPressure tuned


Echelle Grating

Tilt can be in or out of plane of dispersion

Echelle Gratingin reflection

100λ

m=100,N=100-1000

Littrow MountedBlazed Grating

λ/2

m=1 or 2N=10 -10

4 5

Convention: Grating tiltangles have integer tan45, 63.43, 71.56, 75.96, 78.69

45

63.4

71.5675.9678.7

• Steps order of 100λ in Echelle, 1λ in blazed grating

Massively overlapping spectral orders

• Litrow mounting in reflection is similar geometry

conventional Echelle at angles with integer tan θ = 2− 8, 63.4,71.5,76,78.7,80.5,81.9

• 31.6,79,158,316Groovesmm ⇒ corresponds to 10,25,50,100 waves at 6328nm per face

• resolution at blaze angle of 63 degree 90% of max achievable

• sorting overlapping orders leads to crossed dispersion prism or grating


Crossed Echelles for High-ResolutionSpectroscopy: Keck HIRES Solar Spectrum


Blackbody Distribution

Spectral Energy per unit wavelength

Bλ(T ) =2hc2

λ5

1

ehc/λkBT − 1W/Sr/cm2/nm

Spectral Energy per unit frequency

Bν(T ) =2hν3

c21

ehν/kBT − 1W/Sr/cm2/Hz

Power in differential bandwidths are equal Bλdλ=Bνdν and∫∞0 Bλdλ=

∫∞0 Bνdν=

σT 4

π

integrating over hemisphere solid angle removes 2π half sided integral gives 12

and σ = 5.6704× 10−8Jm−2s−1K−4 is Stefan Boltzman constant

Since ν = cλ, thus dν

dλ= − c

λ2, giving Bλdλ = −Bν

cλ2dλ or Bν = −Bλ

λ2

c

In terms of number of photons of energy hν = hc/λ per unit bandwidth

Bλ(N, T ) =2c

λ4

1

ehc/λkBT − 1hν/Sr/cm2/nm


Temperature variation of BlackbodyDistributions vs different scales

B.H. Soffer and D.K.Lynch, Some paradoxes, error, and resolutions concerning the spectral optimization of human vision, Am. J. Phys., 67,946, 1999

As the temperature heats up the peak wavelength shifts to higher energy (Wien’s law)But the peak is different when represented as a distribution in different units!In wavelength the solar peak for T=5800K is 500nm (near 560nm eye peak sensitivity)In freqency the solar peak for T = 5800K is at 3.5× 1014 corresponding to 880nmIn photons/sec/nm peak for T = 5800K is at 633nmIn log λ or log ν the peak is at 720nm ( d(log λ) = −d(log ν) )

This discrepancy is due to the Jacobian of the transformationUnitless filter transmission functions dont need Jacobian transformation (ratios cancel)


Transformation of Spectral Energy Distribution

Density Distribution 6= UnitlessTransmission

Spectra measured as a function of wavelength using grating with linear wavelength dispersion can be

transformed to nonequally spaced frequency (or energy or wavenumber scale), but must also include

Jacobian Bλdλ=Bνdν λ = c/ν ⇒ dλ = −c/ν2dν


Blackbody Spectral Energy DistributionExample of Solar blackbody

Wavelength Peak at 500nm is only coincidentally related to yellow color

Integral of Energy dλ gives same value as integral across Jacobian weighted dν


Spectral Energy per Fractional BandwidthDistribution. Solar blackbody at T=5800

Dividing by photon energy hν = hc/λ gives photon number

Fractional bandwidth resolution (eg grating lines) should be plotted on d log λ scale


Pulse propagation Equations

3+1 dimensional vector wave equation(

∂2

∂x2+

∂2

∂y2+

∂2

∂z2− 1

c2∂2

∂t2

)~E(x, y, z, t) = µ0

∂2

∂t2~P(x, y, z, t)

Describes diffraction, dispersion, refraction, linear loss and gain. Breaking up ~P = ~PL+~PNL allows inclusion of nonlinear effects such as nonlinear absorption and refraction,harmonic generation, Raman and Brilluoin processes and more. We will neglect thesenonlinear terms since they only become important for very high fields.

Linearly polarized field propagating along z leads to reduced 1+1D wave eqn(

∂2

∂z2− 1

c2∂2

∂t2

)E(z, t) = µ0

∂2

∂t2P (z, t)

Now the linear polarization can be represented in the frequency domain as a susceptibility

P (ω, z) = ǫoχ(ω)E(ω, z)

Or by the convolution thm (and Kramers-Kronig causality conditions between χ′(ω)and χ”(ω))

P (t, z) = ǫo

∫ t

−∞χ(t′)E(z, t− t′)dt′


1D Wave eqn in temporal frequency domain

Finite response time leads to dispersion in the frequency domain

χ(ω) =

∫ ∞

−∞χ(t)e−iωtdt

temporal Fourier transform of 1+1D wave eqn leads to the wave eqn for each spectralcomponent [

∂2

∂z2+

ω2

c2

]E(z, ω) = −µ0ǫoχ(ω)E(z, ω)

ǫ(ω)[∂2

∂z2+ω2

c2

︷︸︸︷(1 + χ(ω))

]E(z, ω) = 0

Solution of the 2nd order wave eqn leads to ±z propagating solutions

E(z, ω) = E(0, ω)e±ik(ω)z

where propagation constant k(ω) is given by the square root of

k2(ω) =ω2

c2εr(ω) =

ω2

c2n2(ω)


We will Taylor expand k(ω) about the center frequency ωo

k(ω) = k0 +∂k

∂ω

∣∣∣∣ωo

(ω − ωo) +∂2k

∂ω2

∣∣∣∣ωo

(ω − ωo)2

2!+ · · · = k0 + δk

and write equation for +z propagating monochromatic solution with rapidly varyingpart separated out

E(z, ω) = E(0, ω)e−ik(ω)z = E(0, ω)e−iδk(ω)z

︸︷︷︸ e−ik0z

E(z, ω) is slowly varying partwith k20 = ω2o n

2(ωo)/c2

For a small bandwidth∣∣∣∆kk0

∣∣∣≪ 1 the envelope varies slowly with z∣∣∣∣d

dzE(z, ω)

∣∣∣∣≪ k0|E(z, ω)|

So that in one wavelength of propagation the envelope varies sufficiently slowly. FT intotime domain

E(t, z) =1

2π

[∫ ∞

−∞E(0, ω)e−iδk(ω)zei(ω−ωo)tdω

]ei(ωot−k0z) = 1

2E(t, z)ei(ωot−k0z)


Taylor expanding χ(ω) (or ǫ(ω)) about ωo

χ(ω) = χ(ωo) +∞∑

n=1

1

n!

∂nχ

∂ωn

∣∣∣∣ωo

(ω − ωo)n ǫ(ω) = ǫ(ωo) +

∞∑

n=1

1

n!

∂nǫ

∂ωn

∣∣∣∣ωo

(ω − ωo)n

Gives the polarization in the frequency domain

P (ω, z) = ǫo

(ǫ(ωo)− 1 +

∞∑

n=1

1

n!

∂nǫ

∂ωn

∣∣∣∣ωo

(ω − ωo)n

)E(ω, z)

Which can be transformed back to the time domain using ∂n

∂tn (−iω)n

P (t, z) = 12

[ǫo[ǫ(ωo)− 1]E(t, z) + ǫo

∞∑

n=1

(−i)nǫ(n)(ωo)

n!

∂n

∂tnE(t, z)

]ei(ωot−k0z)

where ǫ(n)(ωo) =∂nǫ∂ωn

∣∣ωo. Now substitute this representation into the polarization driven

wave eqn.


(∂2

∂z2− 1

c2∂2

∂t2

)1

2E(t, z)ei(ωot−k0z) = µo

∂2

∂t2P (t, z)

1

2

(∂2E∂z2

− 2ik0

∂E∂z

−k20E − 1

c2

(SSSSS

∂2E∂t2

+ 2iωo

∂E∂t

−ω2

o E))

ei(ωot−k0z)

=1

2

1

c2

[(ǫ(ω)− SS1)

∂2E(t, z)∂t2

+∂2

∂t2

( ∞∑

n=1

(−i)nǫ(n)(ωo)

n!

∂nE(t, z)∂tn

ei(ωot−k0z)

)]

=1

2

1

c2

[ǫ(ω)

∂2E(t, z)∂t2

+ (−i)ǫ(1)(ωo)

1!

(∂3E∂t3

+ 2iωo

∂2E∂t2

− ω2o

∂E∂t

)

+(−i)2ǫ(2)(ωo)

2!

(∂4E∂t4

+2iωo

∂3E∂t3

−ω2o

∂2E∂t2

)+

∂2

∂t2

(∞∑

n=3

(−i)nǫ(n)(ωo)

n!

∂nE(t, z)∂tn

)]ei(ωot−k0z)

Simplifying to second order

−ik0

(∂E∂z

+(c−1+

ω2o ǫ

(1)(ωo)

k02c2

)

︸︷︷︸

∂E∂t

)+1

2

∂2E∂z2

=1

2

1

c2

(ǫ(ωo)+2ωoǫ

(1)(ωo)+ǫ(2)(ωo)

2ω2o

)

︸︷︷︸

∂2E∂t2

k′ k0k”


Group Velocity Frame

Transform to a coordinate system moving with group velocity vg =(dkdω

)−1

Z = z T = t− z

vg

Partial derivatives become1 0

∂

∂t=

︷︸︸︷∂T

∂t

∂

∂T+

︷︸︸︷∂Z

∂t

∂

∂Z=

∂

∂T

− 1vg

1

∂

∂z=

︷︸︸︷∂T

∂z

∂

∂T+

︷︸︸︷∂Z

∂z

∂

∂Z=

∂

∂Z− 1

vg

∂

∂T

Leading term becomes simple evolution in stationary coordinate system(∂

∂z+

1

vg

∂

∂t

)E =

(∂

∂Z−

1

vg

∂

∂T

)E +

1

vg

∂

∂TE =

∂E∂Z


Slowly Varying Envelope Approximation

Equation for forward envelope implicitly neglecting reflections becomes

∂

∂ZE − i

2k0”

∂2

∂T 2E +

i

3k0c2

∞∑

n=3

(−i)n

n!

[ω2o ǫ

(n)(ωo) + 2nωoǫ(n−1)(ωo)

+n(n− 1)ωoǫ(n−2)(ωo)

]∂nE∂T n

= − i

2k0

∂

∂Z

(∂

∂Z− 2

vg

∂

∂T

)E

Group Velocity Dispersion (GVD) parameter is

k0” =∂2k

∂ω2

∣∣∣∣ωo

= − 1

v2g

∂vg∂ω

∣∣∣∣ωo

=1

2k0

[2

v2g− 2

c2ǫ(ωo)−

4ωo

c2ǫ(1)(ωo)−

ω2o

c2ǫ(2)(ωo)

]

Slowly varying conditions∣∣∣dE(t)dt

∣∣∣≪ ωo |E(t)| and∣∣∣dE(ω,z)dz

∣∣∣≪ k0

∣∣E(ω, z)∣∣ become

∣∣∣∣1

k0

(∂

∂Z− 2

vg

∂

∂T

)E∣∣∣∣ =

∣∣∣∣1

k0

(∂

∂z− 1

vg

∂

∂t

)E∣∣∣∣≪ |E|

Thus this SVEA allows us to neglect right hand side and gives first order evolution eqn.Neglecting 3rd and higher order dispersions gives GVD dispersion eqn

∂

∂ZE − i

2k0”

∂2

∂T 2E = 0


Gaussian Pulse Propagation Snapshots


Rectangular Pulse Propagation Snapshots


Gaussian and Rectangular PulsePropagation in space z


Gaussian and Rectangular PulsePropagation in reduced group velocity

coordinate frame (Z, T )


Gaussian and Rectangular PulsePropagation

in space and time Z, T )


Dispersive Pulse Propagation Equations

Dispersive vector wave eqn

∇2~E−∇(∇ · ~E)− 1

c2∂2~E

∂t2= µ0

∂2~P

∂t2

First need to expand linear dispersive Polarization response of medium

Pj(~r, t) = ǫo

∫ ∞

0

Rjk(~r, τ )Ek(~r, t− τ )dτ

Linear causal response function Rjk(~r, τ ) anisotropic and spatially varyingAssume quasi-monochromatic SSB Electric Field input centered at ω0 (and −ω0)

~E(~r, t) = 12

[~A(~r, t)e−iω0t + ~A∗(~r, t)eiω0t

]

Fourier transform in time

~E(~r, ω) = 12

[~A(~r, ω − ω0) + ~A∗(~r, ω + ω0)

]

This allows causal temporal response to be written as a product in the frequency domainand inverse FT

Pj(~r, t) = ǫo1

2π

∫χjk(~r, ω)Ek(~r, ω)e+iωtdω


Polarization in frequency and time domain

Taylor expand χjk(~r, ω) about center freq of pulse ω0

Pj(~r, ω) = ǫoχjk(~r, ω)Ek(~r, ω) =ǫo2χjk(~r, ω)

[Ak(~r, ω − ω0) + A∗

k(~r, ω + ω0)]

=ǫo2

[χjk(~r, ω0) +

∞∑

s=1

[ω − ω0]s

s!

∂sχjk(~r, ω)

∂ωs

∣∣∣∣∣ω=ω0

]Ak(~r, ω − ω0)

+ǫo2

[χjk(~r,−ω0) +

∞∑

s=1

[ω + ω0]s

s!

∂sχjk(~r, ω)

∂ωs

∣∣∣∣∣ω=−ω0

]A∗

k(~r, ω + ω0)

Consider freqs near ω0 as Ω′=ω−ω0 and Ω=ω+ω0. Inverse FT back to time domain ( ∂

n

∂tn ⇐⇒ (−iω)n)

Pj(~r, t) =ǫo2

1

2π

∫ 0

−∞

χjk(~r, ω0) +

∞∑

s=1

[Ω]s

s!

∂sχjk(~r, ω)

∂ωs

∣∣∣∣∣ω=−ω0

Ak(~r,Ω)e

+i(

ω︷︸︸︷Ω− ωo)tdΩ + cc

=ǫo2

χjk(~r, ω0) +

∞∑

s=1

∂sχjk(~r, ω)

∂ωs

∣∣∣∣∣ω=−ω0

12π

∫ 0

−∞

[ iiΩ]s

s!Ak(~r,Ω)e

+iΩtdΩ

e−iωot + cc

=ǫo2

χjk(~r, ω0)Ak(~r, t) +

∞∑

s=1

is

s!

∂sχjk(~r, ω)

∂ωs

∣∣∣∣∣ω=ω0

∂sAk(~r, t)

∂ts

e−iω0t + cc

+ǫo2

[χjk

(~r, ω0 + i ∂

∂t

)Ak(~r, t)

]e−iω0t + cc


Vector Helmholtz Eq

Spatial Derivatives

∇×∇× ~E = ∇2~E−∇(∇ · ~E) =[

∂2

∂x2∂2

∂y2∂2

∂z2

]Ex

Ey

Ez

−

∂∂x∂∂y∂∂z

[

∂∂x

∂∂y

∂∂z

]Ex

Ey

Ez

Components

∇2Ej −∂

∂xj

∑

k

∂Ek

∂xk

vector Helmholtz eqn for Single Side Band (SSB) quasi monochromatic envelope

∇2Aj(~r, t)e−iω0t − ∂

∂xj

∑

k

∂Ak(~r, t)

∂xke−iω0t

=1

c2∂2

∂t2[Aj(~r, t)e

−iω0t + χjk(~r, ω0 − Ω)Ak(~r, t)e−iω0t

]

=1

c2∂2

∂t2[Aj(~r, t)e

−iω0t + χjk

(~r, ω0 + i ∂∂t

)Ak(~r, t)e

−iω0t]

Only for fractional BW η = ∆ωFWHMω0

< 1


SpatioTemporal Vector Helmholtz OperatorEquation

Using the linear dielectric tensor ǫjk(~r, ω) = 1 + χjk(~r, ω)

∇2Aj(~r, t)e−iω0t − ∂

∂xj

∑

k

∂Ak(~r, t)

∂xke−iω0t =

1

c2∂2

∂t2(ǫjk(~r, ω0 + i ∂∂t

)Ak(~r, t)e

−iω0t)

=1

c2

[∂2

∂t2− 2iω0

∂

∂t− ω2

0

] ǫjk(~r, ω0 + i ∂

∂t

)Ak(~r, t)

e−iω0t

= − 1

c2

[ω0 + i

∂

∂t

]2 ǫjk(~r, ω0 + i ∂∂t

)Ak(~r, t)

e−iω0t

In a weakly absorptive isotropic medium with possibly inhomogeneous variation[k(~r, ω) + i

α(~r, ω)

2

]2≡ ω2ǫ(~r, ω)

c2⇒ k2(ω) + ik(ω)α(ω) ≈ ω2

c2[ǫR(ω) + iǫI(ω)]

∇2Aj(~r, t)e−iω0t − ∂

∂xj

∑

k

∂Ak(~r, t)

∂xke−iω0t =

−[k(~r, ωo + i ∂

∂t

)[k(~r, ωo + i ∂

∂t

)+ iα

(~r, ωo + i ∂

∂t

)]Aj(~r, t)

]e−iω0t


Operator expansion

Instead of plane wave use wave packet centered around ~k0 = (0, 0, k0)

E(x, y, z, t) = B(x, y, z, t)ei(~k0·~r−ω0t)

Evaluation of partial derivatives yields 2 terms that can be regrouped as operator

∂

∂tB(x, y, z, t)ei(k0z−ω0t) = −iωoe

i(k0z−ω0t)B(x, y, z, t) + ei(k0z−ω0t)∂

∂tB(x, y, z, t)

= ei(k0z−ω0t)(−i)[ωo + i ∂∂t

]B(x, y, z, t)

∂

∂xB(x, y, z, t)ei(k0z−ω0t) = ei(k0z−ω0t) (i)

[0 − i ∂

∂x

]B(x, y, z, t)

∂

∂yB(x, y, z, t)ei(k0z−ω0t) = ei(k0z−ω0t) (i)

[0 − i ∂

∂y

]B(x, y, z, t)

∂

∂zB(x, y, z, t)ei(k0z−ω0t) = ei(k0z−ω0t) (i)

[k0 − i ∂

∂z

]B(x, y, z, t)


Operator equations

Helmholtz eqn∇2E + k2E = 0

Using these operator representation frequency domain dispersion operator

D(kx, ky, kz, ω)B(kx, ky, kz, ω) =

(k2x + k2y + k2z −

n2(ω)ω2

c2

)B(kx, ky, kz, ω) = 0

will become back in the time domain

D(−i ∂

∂x,−i ∂∂y , k0 − i ∂

∂z , ωo + i ∂∂t

)B(x, y, z, t)

=

(−i ∂

∂x

)2+(−i ∂

∂y

)2+(k0 − i ∂

∂z

)2−([

ωo + i ∂∂t]n(ωo + i ∂∂t

)

c

)2B(x, y, z, t) = 0

Algebraic expansion of

1

c2[ωo + i ∂∂t

]2=

1

c2

[ω2o + 2iωo

∂∂t − ∂2

∂t2

]=[k20 +

2ik0c

∂∂t − 1

c2∂2

∂t2

]


Interpret the Operator

interpret the dispersive wave number operator

k2(ωo + i ∂

∂t

)=

([ωo + i ∂∂t

]n(ωo + i ∂∂t

)

c

)2=

[ωo + i ∂∂t

]2n2(ωo + i ∂∂t

)

c2

with the Fourier domain Taylor expansion about the center frequency Ω = ω − n0

k(ω) = k0 +11!k

′0Ω + 1

2!k′′0Ω

2 + 13!k

′′′0 Ω

3 + 14!k

′′′′0 Ω4 + . . .

Square it

k2(ω)=k20+2k0k′0Ω+(k

′20 +

22!k0k

′′0 )Ω

2+( 22!k′0k

′′0+

23!k0k

′′′0 )Ω

3+( 12!2k′′20 + 2

3!k′0k

′′′0 +

14!k′′′′0 )Ω4+. . .

Fourier relation ∂∂t −iω gives ∆ω +i ∂∂t transforms back into time domain as

k2(ωo + i ∂

∂t

)= k20 + 2ik0k

′0∂∂t

k′0 = 1/vg−(k′20 + k0k

′′0 )

∂2

∂t2GVD

−i2

3(3k′0k

′′0 + k0k

′′′0 )

∂3

∂t3TOD

+1

12(3k′′20 + 4k′0k

′′′0 + k0k

′′′′0 ) ∂

4

∂t4+ . . . FOD


Expansion of kz(kx, ky, ω)

Fourier expanding the z-component of the wavevector can be extended to both the tem-poral frequency and transverse spatial frequnecies and will include all mixed derivativesas well

kz(kx, ky, ω) = kz(0, 0, ωo) +∂kz∂kx

∣∣∣0kx +

∂kz∂ky

∣∣∣0ky +

∂kz∂ω

∣∣∣ωo(ω − ωo)

+12

∂2kz∂k2x

∣∣∣0k2x +

12

∂2kz∂k2y

∣∣∣0k2y +

∂2kz∂ω2

∣∣∣ωo(ω − ωo)

2

+∂2kz

∂kx∂ky

∣∣∣0,ωo

kxky +∂2kz∂kx∂ω

∣∣∣0,ωo

kx(ω − ωo) +∂2kz∂ky∂ω

∣∣∣0,ωo

ky(ω − ωo)

= kz(0, 0, ωo) +∇kkz · ~kt +12∇k∇k : ~k~k

Terms can be identified as spatial walkoff, group velocity, diffraction, dispersion,anisotropic mixed diffraction (tilted ellipse) and space-time coupling.


Anisotropic Nonlinear Wave Eqn

∇2~E− 1

c2~E−∇(∇ · ~E) = 1

ǫoc2

(~PL + ~PNL

)

[∂2

∂x2+

∂2

∂y2+

∂2

∂x2− 1

c2∂2

∂t2

]~E−

∂∂x∂∂y∂∂z

[

∂∂x

∂∂y

∂∂z

]Ex

Ey

Ez

= 1

ǫoc2

(χ~E + χ

(2)~E~E + χ

(3)~E~E~E

)

Linear term leads to determinental eqn for plane wave SSB

~E(x, y, z, t) = A0ei(~k·~r−ωt)

∂∂x → ikx

∂∂y → iky

∂∂z → ikz

∂∂t → −iω

−k2x−k2y−k2z+

ω2

c2−

kxkx kxky kxkzkykx kyky kykzkzkx kzky kzkz

− (−ω)2

ǫoc2

χxx

χyy

χzz

Ex

Ey

Ez

= 1

ǫoc2∂2

∂t2

(χ

(2)~E~E)

Determinental soln leads to quadratic for k as fnc of m ~k = km = ωcn(ω)m.

In an arbitrary coordinate system this gives

n4~k · ǫ · ~k + n2(~k · [adjǫ− tr[adjǫ]I ] · ~k

)+ |ǫ| = 0


Uniaxial medium

ǫ = ǫoI + (ǫe − ǫo)cc

This factors into two terms

(~k2 − k20n2o)(

~k · ǫ · ~k− k20n2on

2e) = 0

Two rootsk+ = k0no

k− =k0none√m · ǫ · m

=k0none√

n2o(m× c)2 + n2

e(m · c)2Booker Quartic: Planar boundary value solutions

~k = ~kT + qn

Ordinary

(~kT + qn) · (~kT + qn)− k20n2o = 0 |~kT |2 + q2 − k20n

2o = 0 q2 = k20n

2o − k2T

q = ±√k20n

2o − k2T This is a circle. Walkoff given by derivative

∂q

∂kx= 1

2

−2kx√k20n

2o − k2T

=−kxkz


Extraordinary wave

~k · ǫ · ~k = k20n2on

2e

(~kT + qn) · ǫ · (~kT + qn)− k20n2on

2e = 0

⇒ q2 n · ǫn︸︷︷︸+q(n · ǫ · ~kT + ~kT · ǫ · n︸︷︷︸) +~kT · ǫ · ~kT − k20n

2on

2e︸︷︷︸ = 0

A B C

q =−B ±

√B2 − 4AC

2Atilted//ellipse

For a near normal incidence this can be Taylor expanded

kez = kz(0) +∂kz∂kx

∣∣∣0kx +

∂kz∂ky

∣∣∣0ky +

∂2kz∂k2x

∣∣∣0

k2x2+∂2kz∂k2y

∣∣∣0

k2y2+

∂2kz∂kx∂ky

∣∣∣0

kxky2

For an ellipse tilted by θ n = (0, sin θ, cos θ) ~kT = (kx, kt cos θ,−kt sin θ)

(n2o sin

2 θ+n2e cos

2 θ)q2+(n2o−n2

e)kt sin 2θq+k2t (n2o cos

2 θ+n2e sin

2 θ)+n2okt−k20n

2on

2e = 0


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