MEFT / Quantum Optics and Lasers
Suggested problems – Set 3
Gonçalo Figueira
Note: some problems are taken or adapted from “Fundamentals of Photonics”, in which case
the corresponding number is indicated. At the end of the list you may find the solutions to
selected problems.
Interaction between light and matter
Einstein’s coefficients
Consider the (nondegenerate) energy levels 𝐸! and 𝐸! of an atom placed in a resonator that
can sustain a number of modes, such that 𝐸! − 𝐸! = ℎ𝜈. The respective population densities
-3
at a given temperature 𝑇 and time 𝑡 are 𝑁! (𝑡) and 𝑁! (𝑡) (m ).
a) Considering only spontaneous emission, obtain an expression for 𝑑𝑁! /𝑑𝑡 and show that
𝑁! (𝑡) decays exponentially with time.
b) Introducing now absorption and stimulated emission, and assuming 𝑁! 0 = 0, obtain
again an expression for 𝑑𝑁! /𝑑𝑡 and show that
𝑁! 𝑡 =
𝐵𝜌𝑁
1 − exp(−𝐴𝑡 − 2𝐵𝜌𝑡)
𝐴 + 2𝐵𝜌
where 𝑁 = 𝑁! + 𝑁! is the total (constant) population density, 𝐴 = 𝐴!" and 𝐵 = 𝐵!" = 𝐵!" as
defined in the classroom.
c) In the above conditions, what is the maximum value for the population density of level 2?
d) Using Boltzmann’s distribution for the ratio 𝑁! /𝑁! at a given temperature 𝑇, show that the
steady state spectral energy density at the frequency 𝜈 inside the resonator is
𝜌=
𝐴/𝐵
exp
!!
!"
−1
e) Compare the above expression with Planck’s law for the blackbody radiation in order to
obtain the expressions for Einstein’s 𝐴 and 𝐵 coefficients .
Doppler-broadened lineshape function (based on 13.3-2)
A collection of atoms in a gas has a component of velocity v along a particular direction that
obeys the Gaussian probability density function
p(v ) =
" v2 %
1
exp $ − 2 '
2πσ v
# 2σ v &
2
where σv = kT / M and M is the atomic mass.
1
a) If each atom has a Lorentzian natural lineshape function of width Δν and central
frequency ν0, derive an expression for the average lineshape function g(ν ) .
b) Show that in the limit Δν → 0 a general Lorentzian lineshape function g(ν) becomes
a Dirac delta, i.e. it obeys the following two conditions:
+∞
(i ) ∫ −∞ g (ν ) d ν = 1, for all Δν
g (ν ) = 0
(ii ) Δlim
ν →0
c) Use the previous result to show that if Δν << ν 0σ v / c, g(ν ) may be approximated by
the Gaussian lineshape function
g (ν) =
⎡ (ν − ν0 )2 ⎤
1
exp⎢−
⎥
2σD2 ⎦
2πσD
⎣
where the new, Doppler-broadeded linewidth is given by
σD = ν0
σ v 1 kT
=
c
λ M
d) Show that the FWHM Doppler-broadened linewidth is ΔνD =
8ln2σ D ≈ 2.35 σ D .
e) Compute the Doppler linewidth for the λ0 = 632.8 nm transition in Ne and for the λ0 =
10.6 µm transition in CO2 at room temperature, assuming that Δν << ν 0σ v / c . These
transitions are used in He-Ne and CO2 gas lasers, respectively.
-27
[M(Ne) = 20.18 amu; M(CO2) = 44 amu; 1 amu = 1.66 × 10 kg]
f)
Show that the maximum value of the transition cross section for the Gaussian
lineshape function in (c) is
σ0 =
λ2
8π
4ln2 1
λ2 1
≈ 0.94
π t sp Δν D
8π t sp Δν D
Compare with expression (13.3-35) (see pg. 512) for the Lorentzian lineshape
function.
Laser amplifiers
(14.1-2) Amplifier Gain and Rod Length
A commercially available ruby laser amplifier using a 15-cm-Iong rod has a small-signal gain
of 12. What is the small-signal gain of a 20-cm-Iong rod? Neglect gain saturation effects.
(14.1-3) Laser Amplifier Gain and Population Difference
2
3+
A 15-cm-Iong rod of Nd :glass used as a laser amplifier has a total small-signal gain of 10 at
3+
λ0 = 1.06 µm. Use the data in the table below to determine the population difference N (Nd
3
ions per cm ) required to achieve this gain.
Amplification of a Broadband Signal (based on 14.1-4)
The transition between two energy levels exhibits a Lorentzian lineshape of central frequency
14
ν0 = 5 × 10 Hz with a linewidth Δν = 1 THz. The population is inverted so that the maximum
-1
-1
gain coefficient γ(ν0) = 0.1 cm . The medium has an additional loss coefficient αs = 0.05 cm ,
which is independent of ν.
Consider a light wave having a uniform power spectral density centered about ν0 with a
bandwidth 2Δν, propagating through this medium.
a) Write the expression for the frequency-dependent intensity of the light wave at the
input, I(ν,0), assuming that the total intensity at z = 0 is I0.
b) Write the expression for the frequency-dependent intensity of the light wave at a
position z inside the medium, I(ν,z).
c) Sketch a graph of the gain coefficient γ(ν) and the absorption coefficient αs as a
function of ν inside the bandwidth region of the light wave.
d) Estimate the loss or gain encountered by the light wave in 1 cm.
x
Hint: Consider whether for the values involved you may use the approximation e ≈ 1+ x .
The following relation may also be useful:
∫
a
1
dy = 2 tan −1 a
2
y
−a 1+
3
(14.2-2) Saturation Time Constant
Show that if
t sp << τ nr ,
t sp << τ 20 , and
t sp >> τ 1
then τ s ≈ t sp .
Spectral Broadening of a Saturated Amplifier (based on 14.4-2)
Consider a homogeneously broadened amplifying medium with a Lorentzian lineshape of
width Δν,
g ( ν) =
Δν / 2π
(ν − ν0 )2 + (Δν / 2)2
.
a) Write the lineshape g(ν), the gain coefficient γ(ν) and the transition cross section σ(ν)
in the form
g (ν ) = g 0
1
,
1+ y 2
#
ν − ν0
,
%where y =
$
Δν / 2
γ (ν ) = γ 0
1
1+ y 2
g 0 = g (ν 0 ) ,
and
γ 0 = γ (ν 0 ) ,
1
1+ y 2
&
σ 0 = σ (ν 0 )(
'
σ (ν ) = σ 0
Determine the respective coefficients g0, γ0 and σ0.
b) Determine the expression for the saturated gain coefficient γs(ν).
c) Show that for a photon-flux density φ(ν), the saturated gain coefficient γs(ν) assumes
a Lorentzian lineshape with a broadened width
Δν s = Δν 1+
φ
φ (ν 0 )
Hint: compare the expression obtained in (b) with a general Lorentzian lineshape of the type
γ (ν ) = γ 0!
1
,
1+ y s2
γ 0! =
γ0
1+ a
and match the corresponding terms.
(14.4-6) Resonant Absorption of a Medium in Thermal Equilibrium
3
23
A unity refractive-index medium of volume 1 cm contains Na = 10 atoms in thermal
equilibrium. The ground state is energy level 1; level 2 has energy 2.48 eV above the ground
state (λ0 = 0.5 µm). The transition between these two levels is characterized by a
spontaneous lifetime tsp = 1 ms, and a Lorentzian lineshape of width Δν = 1 GHz.
Consider the situation for two temperatures, T1 and T2 , such that kT1 = 0.026 eV and kT2 =
0.26 eV.
4
a) Determine the populations N1 and N2 at the two temperatures.
b) Determine the number of photons emitted spontaneously every second.
c) Determine the attenuation coefficient of this medium at λ0 = 0.5 µm assuming that the
incident photon flux is small.
d) Find the value of photon-flux density at which the attenuation coefficient decreases by
a factor of 2 (i.e., the saturation photon-flux density).
(14.4-7) Gain in a Saturated Amplifying Medium
Consider a homogeneously broadened laser amplifying medium of length d = 10 cm and a
18
2
saturation photon flux density φs = 4 × 10 photons/cm .s. It is known that a photon-flux
15
2
density at the input φ(0) = 4 × 10 photons/cm .s produces a photon-flux density at the output
16
2
φ(d) = 4 × 10 photons/cm .s.
a) Determine the small-signal gain of the system G0.
b) Determine the small-signal gain coefficient γ0.
c) What is the photon-flux density at which the gain coefficient decreases by a factor of
5?
19
d) Determine the gain coefficient when the input photon-flux density is φ(0) = 4 × 10
2
photons/cm .s. Under these conditions, is the gain of the system greater than, less
than, or the same as the small-signal gain determined in (a)?
5
t
Selected Solutions
Einstein’s coefficients
a)
dN 2
= −A21N 2
dt
N 2 (t ) = N 2 (0)exp(−A21t ) = N 2 (0)exp(−t / τsp )
b)
dN 2
= −A21N 2 − B21ρ(ν)N 2 + B12ρ(ν)N1
dt
= −AN 2 − BρN 2 + BρN1
= −AN 2 − BρN 2 + Bρ(N − N 2 )
= −(A + 2Bρ)N 2 + BρN
The differential equation may be solved by defining an auxiliary function 𝑁! ’ = 𝑁! −
!"#
!!!!"
such that 𝑑𝑁!! /𝑑𝑡 = 𝑑𝑁! /𝑑𝑡. Replacing 𝑁! in the differential equation above we arrive at the
solution
⎛
dN 2′
BρN ⎞
= −(A + 2Bρ) ⎜ N 2′ +
+ BρN = −(A + 2Bρ)N 2′
dt
A + 2Bρ ⎟⎠
⎝
N 2′ (t ) = c exp [ −(A + 2Bρ)t ]
So for 𝑁! (𝑡), and considering the initial condition 𝑁! (0) = 0, we have
BρN
A + 2Bρ
BρN
BρN
N 2 (0) = c +
→c = −
A + 2Bρ
A + 2Bρ
BρN
N 2 (t ) =
1− e −(A+2Bρ)t
A + 2Bρ
N 2 (t ) = c exp [ −(A + 2Bρ)t ] +
(
)
c) The long term solution (𝑡 ≫ 1/(𝐴 + 2𝐵𝜌)) tends towards the maximum value:
N 2 (t → ∞) =
Bρ
N
A + 2Bρ
d) In a steady state situation the population densities remain constant: 𝑑𝑁! /𝑑𝑡 = 𝑑𝑁! /𝑑𝑡 = 0.
Using the expression from (b):
0 = −AN 2 − BρN 2 + BρN1
ρB(N1 − N 2 ) = AN 2
ρ=
AB
AB
=
N1 N 2 − 1 exp(hν kT ) − 1
6
e) Comparing the latest result with Planck’s law,
ρ(ν) =
8πhν3
1
3
c
exp(hν kT ) − 1
and using the earlier result of 𝐴 ≡ 𝐴!" = 1/𝜏!" we obtain
A
1
8πhν3
c3
=
=
⇔
B
=
B Bτsp
c3
8πhν3t sp
Doppler-broadened lineshape function
a)
g (ν ) =
∫
∞
−∞
"
v%
g $ν − ν 0 ' p (v ) dv
#
c&
−1
13 + ν − ν 1+v c .2 53
) 0 6 ⋅ exp "$ − v 2 %' dv
0(
=
1+ 2
∫
2
Δν 2
# 2σ v &
/ 37
2π 3 σ v ( Δν 2) −∞ 43 ,
1
∞
c) Make the following replacement of variable in the integral:
y = ν0
v
c
v=
c
y
ν0
dv =
c
dy
ν0
Using the result of (b) the integral becomes
(
+c
c2
#
%
δ
y
−
ν
−
ν
⋅
exp
−
y 2 - dy
*
(
)
∫
$
&
0
2
2
) 2σ v ν 0 , ν 0
2π 3 σ v −∞
# ν −ν 2 %
1
( 0) 1
=
exp 0−
3
2σ D2 1&
0$
2π σ D
g (ν ) ≈
1
∞
e)
Ne: Δν D = 1.31×109 Hz = 1.31 GHz
CO2: Δν D = 5.29 ×107 Hz = 52.9 MHz
14.1-2
G(ν) = 27.7
14.1-3
N = γ (ν 0 ) σ (ν 0 ) ≈ 3.84 ×1018 cm−3
7
Amplification of a Broadband Signal (based on 14.1-4)
$ I 2Δν
& 0
a) I (ν ,0) = %
&' 0
ν − ν 0 ≤ Δν
ν − ν 0 > Δν
b) I (ν ,z ) = I (ν ,0) exp "#(γ (ν ) − αs ) z $%
c) Let us write the expressions for the total intensity (for all frequencies) at each position:
+∞
I ( 0) =
∫ I (ν ,0)d ν = I
0
−∞
+∞
I (z ) =
∫ I (ν ,z )d ν
−∞
+∞
= I (ν ,0) ∫ exp $%(γ (ν ) − αs ) z &'d ν
−∞
ν 0 +Δν
=
I0
∫ exp$%(γ (ν ) − αs ) z &'d ν
2Δν ν 0 −Δν
-1
-1
Note that the parameters inside the exponential are small (γ = 0.1 cm , αs = 0.05 cm , z = 1
cm), so we can use the approximation and write
ν +Δν
I (z ) ≈
=
I0 0
∫ (1+ γ (ν )z − αs z ) d ν
2Δν ν 0 −Δν
ν 0 +Δν
(
I0 %
'2Δν (1− αs z ) + z ∫ γ (ν )d ν *
2Δν '&
*)
ν 0 −Δν
You now need to calculate the integral, which corresponds to integrating a Lorentzian
lineshape between two finite limits. Make the following replacement of variable:
y=
∫
ν − ν0
Δν 2
ν =ν 0 +Δν
ν =ν 0 −Δν
→
ν = ν0 +
∫
Δν
y
2
dν =
Δν
dy
2
2
−2
The integral now becomes
ν 0 +Δν
∫
ν 0 −Δν
2
1 Δν
dy
1+
y2 2
−2
γ (ν )d ν = γ (ν 0 ) ∫
= tan−1 2 γ (ν 0 )Δν ≈ 1.1 γ (ν 0 )Δν
Finally, we can compute the gain at a position z as the quotient
8
G (z ) =
I (z )
I0
= 1+
(
1
2
≈
1 $
2Δν (1− αs z ) + tan−1 2 γ (ν 0 )Δν z &'
%
2Δν
)
tan−1 2 γ (ν 0 ) − αs z
Obvioulsy for z = 0 we have G(0) = 1. Using the values provided:
G ( z = 1 cm) ≈ 1+ ( 21 ×1.1× 0.1− 0.05) ×1
≈ 1.005
9
Spectral Broadening of a Saturated Amplifier (based on 14.4-2)
b) Starting with
γ s (ν ) =
γ (ν )
1+ φ φs (ν )
1
= τ sσ (ν )
φs (ν )
we may write
! γ $
γ s (ν ) = # 0 2 &
" 1+ y %
=
1
!
1 $
1+ φτ s #σ 0
2&
" 1+ y %
γ0
1+ y + φτ sσ 0
2
c) Compare the previous expression with a general Lorentzian curve of the form
γ (ν ) = γ 0!
1
,
1+ y s2
γ 0! =
γ0
1+ a
! γ $ 1
γ0
γ0
=# 0 &
=
2
"
%
1+ y + φτ sσ 0
1+ a 1+ y s 1+ a + (1+ a ) y s2
2
By direct comparison we may write
a = φτ sσ 0
y 2 = (1+ φτ sσ 0 ) y s2
By noting the following identities
ν − ν0
ν − ν0
, ys = 2
Δν
Δν s
1
τ sσ 0 =
φs (ν 0 )
y =2
→
Δν s y
=
Δν y s
we finally arrive at
Δν s = Δν 1+
φ
.
φs (ν 0 )
(14.4-7) Gain in a Saturated Amplifying Medium
a)
b)
c)
d)
G = 10
-1
γ = 0.23 cm
19
2
φ = 4φs = 1.6 ×10 photons/cm .s
-1
γ = 0.021 cm
10