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More Accurate Solution for the Two-Level System

Einstein Coefficients and Light Amplification

4.8 More Accurate Solution for the Two-Level System

For strongly allowed transitions, f is of the order of unity, for example, for the 2P→1S transition in the hydrogen atom, f = 0.416. On the other hand, for the transitions from the upper laser level, f∼10–3–10–6.

4.7.2 Interaction of a Near-Monochromatic Wave with an Atom

Similarly

idC2

dt = 1

2E0D12C1(t)

ei(ω+ω)t+ei(ωω)t

(4.104) where use has been made of the fact that

ψ1rψ1dτ =

ψ2rψ2dτ =0 and

ω=(E2E1)

=ω21 = −ω12 (4.105)

In the rotating wave approximation, considering absorption we neglect the terms ei(ω+ω)tand ei(ω+ω)tin Eqs. (4.103) and (4.104) and obtain

dC1

dt = − i

2E0D12C2(t)ei(ωω)t (4.106) dC2

dt = − i

2E0D12C1(t)ei(ωω)t (4.107) If we assume a solution of the form

C1(t)=ei t (4.108)

then from Eq. (4.107),

C2(t)= − 2 E0D12

ei( ω+ω)t (4.109) Substituting in Eq. (4.106), we get

i 2 E0D12

ω+ω

= − i 2E0D21

or

ω+ω

2 0

4 =0 (4.110)

where

2

0= E20D12D21

2 =E20D2

2 (4.111)

4.8 More Accurate Solution for the Two-Level System 93 and

D=D12=D21 (4.112)

Equation (4.110) gives

1,2=1 2

&

ωω

±

ωω2

+ 201/2'

(4.113) Thus the general solution will be

C1(t)=A1ei 1t+A2ei 2t (4.114) C2(t)= − 2

0

ei(ωω)t

A1 1ei 1t+A2 2ei 2t

(4.115) If we now assume that the atom is initially in the ground state, i.e.,

C1(0)=1, C2(0)=0 (4.116)

then

A1= − 2

1

A2 (4.117)

and

1=A1+A2=A2

12

1 =

ωω2

+ 201/2 A2 1

or

A2= 1 (4.118)

where

=

ωω2

+ 20

1/2

(4.119) On substitution we finally obtain

C2(t)= −i 0ei(ωω)t/2 sin

t

2 (4.120)

Thus the transition probability for absorption is given by

|C2(t)|2=

sin t

2

2

2

0

2

2

(4.121)

Fig. 4.12 Variation of the transition probability with time for a two-level system for different frequencies of the electromagnetic field. The curves correspond to the function DE0/=0.1ω. The solid line corresponds to Eq. (4.121) and the dotted curve corresponds to an accurate numerical

computation (Reprinted with permission from Salzman (1917). © 1971 American Institute of Physics)

which has been plotted in Fig.4.12. Also shown in the figure are the results of the exact numerical calculations without resorting to the rotating wave approximation.

At resonanceω=ωand one obtains

|C2(t)|2=sin2

0t

2 (4.122)

which shows that the system flip flops between states 1 and 2. A comparison of Eqs.

(4.122) and (4.77) shows that the perturbation theory result is valid if D21E0

t

2

<<1 or

D21E0

2 1

ω)2 <<1 (4.123) It may be of interest to note that the solutions obtained in this section are exact whenω=0 (i.e., a constant electric field) and if D21 is replaced by 2D21 in the solution given by Eq. (4.121). This follows from the fact that forω=0, the exact equations [Eqs. (4.103) and (4.104)] are the same as Eqs. (4.106) and (4.107) with D21replaced by 2D21.

Problems 95

Problems

Problem 4.3 Consider the two-level system shown in Fig.4.1with E1= −13.6 eV and E2= −3.4 eV.

Assume A21 6×108s1. (a) What is the frequency of light emitted due to transitions from E2and E1? Assuming the emission to have only natural broadening, what is the FWHM of the emission? What is the population ratio N2/N1at T=300 K?

[Answer : (a)v2.5×1015Hz,v=A21/2π=108Hz, N2/N1e394

Problem 4.4 Given that the gain coefficient in a Doppler-broadened line is α(v)=α(v0) exp

4 ln 2(vv0)2/(v0)2

where v0is the centre frequency andv0is the FWHM and that the gain coefficient at the line centre is twice the loss averaged per unit length, calculate the bandwidth over which oscillation can take place.

[Answer:v0].

Problem 4.5 Consider an atomic system as shown below:

3——— E3=3 eV 2——— E2=1 eV 1——— E1=0 eV The A coefficient of the various transitions are given by

A32=7×107s−1, A31=107s−1, A21=108s−1 (a) What is the spontaneous lifetime of level 3?

(b) If the steady-state population of level 3 is 1015atoms/cm3, what is the power emitted spontaneously in the 32 transition? [Answer: (a) tsp=1.2×10−8s (b) 2.2×1010W/m3]

Problem 4.6 Consider the transition in neon that emits 632.8 nm in the He–Ne laser and assume a temperature of 300 K. For a collision time of 500 ns, and a lifetime of 30 ns, obtain the broadening due to collisions, lifetime, and Doppler and show that the Doppler broadening is the dominant mechanism.

Problem 4.7 Consider an atomic system under thermal equilibrium at T=1000 K. The number of absorptions per unit time corresponding to a wavelength of 1μm is found to be 1022s–1. What would be the number of stimulated emissions per unit time between the two energy levels? [Ans: 10225−1]

Problem 4.8 Consider a laser with plane mirrors having reflectivities of 0.9 each and of length 50 cm filled with the gain medium. Neglecting scattering and other cavity losses, estimate the threshold gain coefficient (in m1) required to start laser oscillation. [Ans: 0.21 m1]

Problem 4.9 An atomic transition has a linewidth ofν=108Hz. Estimate the approximate value of g(ω) at the center of the line. [Ans: 1.6×10951]

Problem 4.10 There is a 10% loss per round trip in a ruby laser resonator having a 10 cm long ruby crystal as the active medium. Calculate the cavity lifetime, assuming that the mirrors are coated on the ends of the ruby crystal. Given: Refractive index of ruby at the laser wavelength is 1.78 [Ans: 11.3 ns]

Problem 4.11 In a ruby crystal, a population inversion density of (N2– N1)=5×1017cm–3is generated by pumping. Assuming g(ν0)=5×10–12s, tsp=3×10–3s, wavelength of 694.3 nm and a refractive index of 1.78, obtain the gain coefficientγ(ν0). By what factor will a beam get amplified if it passes through 5 cm of such a crystal? [Ans: 5×10–2cm–1, 1.28]

Problem 4.12 An optical amplifier of length 10 cm amplifies an input power of 1 to 1.1 W. Calculate the gain coefficient in m–1. [Ans: 0.95 m−1]

Problem 4.13 Doppler broadening leads to a linewidth given by νD=2ν0

2kBT Mc2 ln 2

Estimate the broadening for the 632.8 nm transition of Ne (used in the He–Ne laser) assuming T=300 K and atomic mass of Ne to be 20. What would be the corresponding linewidth of the 10.6μm transition of the CO2molecule? [Ans: 1.6×109Hz, 6×107Hz]

Problem 4.14 In a typical He–Ne laser the threshold population inversion density is 109cm–3. What is the value of the population inversion density when the laser is oscillating in steady state with an output power of 2 mW?

Problem 4.15 Given that the gain coefficient in a Doppler-broadened line is γ(ν)=γ0exp

4 ln 2(νν0)2 0)2

and that the gain coefficient at the center of the line is four times the loss averaged per unit length, obtain the bandwidth over which oscillation will take place. [Ans:

2υ0]

Problem 4.16 A laser resonator 1 m long is filled with a medium having a gain coefficient of 0.02 m-1. If one of the mirrors is 100 % reflecting, what should be the minimum reflectivity of the other mirror so that the laser may oscillate? [Ans: 96%]

Chapter 5