Derivations

Question 1

Derive einstein A and B coefficients

Answer 1

Rate of spontaneous emission N2A21

Rate of absorption N1B12p(w21)

Rate of stimulated emission N2B21p(w21)

Consider thermal equilibrium

=> Absorption rate = stimulated rate + spontaneous rate

divide by N2B21

rearrange for p(w21)

Boltzmann’s law Pi ∝ gi exp[-Ei/kbT]

N2/N1 = g2/g1 exp[-E2/kbT] / exp[-E1/kbT]

ΔE = ħω

Assuming radiation is blackbody radiation given by Planck’s law

A21/B21 = ħω^3/π^2c^3

and g1B12/g2B21 = 1

Question 2

Condition for Optical Gain

Answer 2

We require that stimulated emission > rate of absorption

N2B21p(w21) > N1B12p(w21)

Einsteins relation g1B12 = g2B21

so

N2p(w21) > N1p(w21) B12/B21

B12/B21 = g2/g1

N2/g2 > N1/g1

Question 3

Achieving steady state population inversion

Answer 3

The rate of transitions between states

dN2/dt = R2 - N2/τ2

dN1/dt = R1 + N2A21 - N1/τ1

Steady state implies dN1/dt = dN2/dt = 0

N2 = R2τ2

N1 = τ1(R1+R2τ2A21)

N2/g2 > N1/g1

rearrange for R2τ2g1/R1τ1g2

Question 4

Gain coefficient

Answer 4

dI(ω,z)/dz = α[ω-ω0] I(ω,z)

∫ dI(ω,z)/I(ω,z) = ∫ α[ω-ω0]z

ln(I(ω,0)) = α[ω-ω0]z + c

to determine c, consider z = 0

=> c = ln(I(ω,0)

=> ln(I(ω,ω0)) = α(ω,ω0)z + ln(I(ω,0)

ln(I(ω,z)) = α(ω,ω0)z + ln(I(ω,0)

ln(I(ω,z)/I(ω,0)) = α(ω-ω0)z

(I(ω,z)/I(ω,0)) = exp[α(ω-ω0)z]

Question 5

Light field in a cold cavity

Answer 5

uin(r,ω,t) = ain(z,ω)ulmp+(r,ω,t)

u+(r,ω,t) = a+(z,ω)u+lmp(r,ω,t)

u-(r,ω,t) = a-(z,ω)u-lmp(r,ω,t)

u+ = t1 uin + r1exp(iφ1) u-

u+ = u-

left wave - phase shift of δrt - φ1
change of amplitude due to r2 the reflectivity of M2

=> a-(0,ω) = t1ain(0,ω) + r1exp(i[δrt-φ1])

substitute it into a+(0,ω)

rearrange for a+(0,ω)

intensity is proportional amplitude^2

R = r1r2, T1 = |t1|^2 and F = 4R/(1-R)^2

I+(0,ω) = Iin(0,ω) = T1/(1-R)^2 1/(1+Fsin^2(δrt/2))

Question 6

Ray transfer matrices - cavity stability

Answer 6

γ = m ± i √ (1-m^2)

= cosθ ± isinθ
= e^iθ

after N round trips, a ray r0 = c(a)r(a) + c(b)r(b) becomes,

rN = c(a)e^iNθr(a) + c(b)e^-iNθr(b)

rewriting

rN = c(a)[cos(Nθ) + isin(Nθ)]r(a) + c(b)[cos(Nθ) - isin(Nθ)]r(b)

rN = (c(a)r(a) + c(b)r(b)) cos(Nθ) + i(c(a)r(a)-c(b)r(b)) sin(Nθ)

defining s0 = i(c(a)r(a)-c(b)r(b)) we have

rN = r0cos(Nθ) + s0sin(Nθ)

Question 7

Number of photons left in the cavity after time t

Answer 7

If Np(t) = Np,0 exp(-t/τ)

The probability of a photon being in a cavity after one round trip = R2 - R1

after one round trip Np = Np,0 R1R2

After x round trips Np = Np,0 (R1R2)^x

The round trip duration is given by

t(R.T) = 2L/(c/n)

In time t there are x roundtrips. Therefore the probability of being in a cavity after x round trips is (R1R2)^(tc/2Ln)

Therefore the number of photons left in cavity after x roundtrips is

Np(t). = Np,0(R1R2)^(tc/2Ln)

Equating

Np(t) = Np,0 (R1R2)^(tc/2Ln)
Np(t) = Np,0 exp(-t/τ)

rearrange for τ then taylor expand logx ~ x-1 0 < x < 2

τ = 2Ln/c / (1-R1R2)

Question 8

3rd Order Harmonic Distortion

Answer 8

consider a sine wave y = sinω0t

in terms of exponents y = (e^iω0t - e^-ω0t)/2i

consider a third-order non-linear operation

y’ = y - 0.3y^3

substitute in y

expand the ^3

separate into orders

3.1/4 sinω0t + 0.3/4 sin3ω0t

Question 9

2nd order harmonic distortion

Answer 9

consider a sine wave y = sinω0t

in terms of exponents y = (e^iω0t - e^-ω0t)/2i

consider a second-order non-linear operation

y’ = y - 0.1y^2

same steps as 3rd order

y’ = sin(ω0t) + 0.05cos(2ω0t) - 0.05

Question 10

Phase mismatch

Answer 10

draw diagram

first consider a more general input of two waves @ ω1 and ω2

Consider a non-linear field generated at position z

-> at z, field is in phase with input wave
-> field oscillates at ω3 = ω2 + ω1

@ z we have : E1exp(-ik1z)
@ z we have : E2exp(-ik2z)

want to find the polarisation field @ z

P^(ω3)(z) = const x E1exp(-ik1z)E2exp(-ik2z)

P^(ω3)(z) = const x E1E2 exp(-i(k1+k2)z)

The electric field at z propagates and exits the crystal at position L

E^(ω3)(Lz) ∝ P^(ω3)(z)exp(-ik3(L-z))

Substitute P^(ω3)(z)

E^(ω3)(Lz) ∝ E1E2exp(iΔkz)

This is only E^(ω3) generated at z only-integrate over crystal

E^(ω3)(Lz) ∝ (L ∫ z=0) E1E2 exp(iΔkz) dz

Assuming negligble depletion of pump (ω) light in crystal i.e. E1 and E2 are not functions of z

E^(ω3)(L) ∝ E1E2 (L ∫ z=0) exp(iΔkz) dz

For SHG E1 = E2 = E^ω

and E^(ω3) = E^(2ω)

Δk = 2^(2ω) - 2k^(ω)

Question 11

Sum and difference generation

Answer 11

two input waves E^(ω1) = E1cosω1t, E^(ω2) = E2cos(ω2t)

interact with λ^(2) susceptibility

2E^(ω1) E^(ω2) = 2 . 1/2 E1 E2(cos(ω1 - ω2) + cos(ω1+ ω2))

Question 12

complex amplitude?

Answer 12

u(x,y,z) = u0 exp(ik(ysinθ+zcosθ))

where k = 2π/λ

transverse plane when z = z0

u(x,y) = u0 exp(ik(ysinθ+zcosθ))

1 round trip

u’(x,y) = u0 exp(ik(ysinθ+(z0+2L)cosθ))

u’(x,y) = u(x,y)exp(i2klcosθ)