SoSe 19 Laboratory of Nano-Optics

Utrafast and Nonlinear Optics

Exercise 1

1 - Gain from a three-level system

Derive expressions for the saturated and small-signal gain of a three-level atom. Your analysis should parallel the analysis given in Sec. 1.3.2 of Ultrafast Optics for a four-level atom, but with level 4 missing. Assume that stimulated emission occurs between levels 3 and 1 and that the population in level 3 relaxes spontaneously to level 1 with time constant τG. Plot a family of curves for the saturated gain as a function of W (different curves represent different values of S) for both three- and four-level atoms. Comment on the main differences in gain behavior of three- and four-level atoms and on the implications for laser operation.

� Fig. 1 Level scheme and transitions of a three-level system gain medium.

The three-level system is defined by the quantum states � , � and � . These are coupled by transitions with rates kij, where i and j denote two levels. In addition, the laser transition from to � is affected by spontaneous emission represented by the rate � (see Fig. 1). The transition k13 refers to optical pumping and the transition k21 to stimulated emission. The transition k32 is typically nonradiative and much larger than k31 to achieve population inversion between the levels � and � . The gain is obtained by solving the rate equations:

� ,

� ,

� ,

�

N1, N2 and N3 are the populations of the three quantum states, respectively, and their sum must remain constant. Because k32 is much larger than k13, it is safe to assume that the term k13 N3 ~ 0 in the rate equations and that N1 + N2 = N. Imposing the steady-state condition , after a few simple steps one obtains:

� .

|1⟩ |2⟩ |3⟩|2⟩

|1⟩ 1/τG

|2⟩ |1⟩·N1 = − (k13 + k12) N1 + (k12 + 1/τG) N2 + k13N3

·N2 = k12N1 − (k12 + 1/τG) N2 + k32N3

·N3 = k13N1 − (k13 + k32) N3

N1 + N2 + N3 = N .

·Ni = 0

N1 = N ( k12 + 1/τG

k13 + 2k12 + 1/τG ), N2 = N ( k12 + k13

k13 + 2k12 + 1/τG )

The gain is then � and using the book notation

(pumping rate W and lasing rate S) the expression reads:

� , which clearly differs from the case of a

four-level system (see eq. 1.51). The small signal gain is also different from the case of a four-level

system (see eq. 1.52): � . Specifically, the gain exhibits a threshold value

associated with the spontaneous emission � , � , because the ground state is shared by optical pumping and lasing. For the same reason, the gain saturates faster than for a four-level system.

2 - Peak amplitude of a focused ultrafast pulse

A mode-locked laser generates pulses with 105 W of peak power. Spatially the laser output is a Gaussian beam. If the beam is focused in air to a diameter of 10 microns (at e−2 points of the intensity), give the peak intensity and the corresponding peak electric field amplitude.

The problem can be easily solved applying textbook expressions.

3 - Characteristic quantities of an ultrafast pulse

A mode-locked laser generates Gaussian pulses with a 1.06-microns center wavelength, 10-ns cavity round-trip time, 1-W average power, and 20-ps FWHM duration. Give values for the number of cavity modes that are oscillating, the optical bandwidth, the peak power, and the pulse energy.

The number of cavity modes N is given by ratio between the cavity round trip � time and the pulse duration � , � .

The optical bandwidth � is given by the the number of modes times the mode spacing � , � GHz.

The pulse energy � is given by the average power P divided by the repetition rate, � nJ.

The peak power � is approximately given by the pulse energy divided by the pulse duration, � W.

4 - Envelope function under mode-locking

A mode-locked laser usually has a smooth frequency spectrum. Consider a Gaussianspectrum given by � , where � .

g =σlG2

(N2 − N1) =σlGN

2k13 − 1/τG

k13 + 2k12 + 1/τG

g =σlGN

2W − 1/τG

W + 2S + 1/τG=

σlGN2

WτG − 1(W + 2S )τG + 1

g0 =σlGN

2WτG − 1WτG + 1

τG WτG > 1

tctp N = t /tp = 500

Δωωm = 2π /tc Δω = 2πN/tc ≃ 300

EpEp = Ptc = 10

PpPp ≃ Ep /tp = 500

A(ω̃) = e−ω̃2/ΔΩ2

∑m

δ(ω̃ − mΔω)eiϕm ω̃ = ω − ω0

� Fig. 2 Spectral amplitude.

(a) Assuming constant spectral phase, � = 0, work out the expression for the time-domain complex envelope function a(t). (Hint: The Fourier transform of a periodic train of evenly spaced, equal-amplitude delta functions is a periodic train of evenly spaced, equal-amplitude delta functions. This is shown most easily using Fourier series.)

Using the property that � and the possibility to swap sum with

integral, one finds � , where m is summed on the cavity

modes , i.e. -(N-1)/2, ..., (N-1)/2. As shown in the book, the sum leads to a train of equidistant pulses (see Fig. 3 for the parameters given in (b)).

(b) Use a computer and an FFT algorithm to evaluate the time-domain envelope function a(t) for � = 10 GHz, � = 100 MHz, and � = 0. Plot the spectral amplitude function � and the temporal intensity and phase. Give the pulse duration (intensity FWHM). Discuss how the setup of your array representing � determines the number of pulses in the time-domain plot.

An implementation of the FFT is available in most numerical software packages, like Matlab, Mathematica, ...

� Fig. 3 Pulse train resulting the FFT of the spectral amplitude for � (mode locking).

ϕm

∫∞

−∞f (x)δ(x − y)d x = f (y)

a(t) =1

2π ∑m

e−m2Δω2/ΔΩ2e−imΔωt

ΔΩ /2π Δω /2π ϕm A(ω̃)

A(ω̃)

ϕm = 0

(c) Now obtain the phases � from a random number generator. Plot two examples of the temporal intensity and phase and comment on all the key differences compared to the uniform phase case.

� Fig. 4 The FFT of the spectral amplitude for a random phase does not lead to the generation of

ultrashort pulses, because mode locking is not achieved.

5 - Passive mode-locking with fast saturable absorbers

Consider fast saturable absorber mode-locking with the following parameters: = 2.5×10−19 J (transition energy), � = 5% (relative loss), � = 20% (relative loss due to the saturable absorber), T = 10 ns (cavity round trip time), AA =AG =10−4 cm2 (beam cross section), σA =10−16 cm2, σG =10−19 cm2 (absorber and laser transition cross section), τA =100 fs (absorber relaxation time), τG = 1 μs (laser medium relaxation time), ωc =2π×1012 rads−1 (cavity resonance angular frequency).

The starting point is the mode-locking equation for the pulse envelope function,

� , which has the solution

� , where � is the field amplitude and � is the pulse duration, if the following conditions are met:

� (delay time),

� ,

� .

Combining these conditions and using the expression for the gain g as a function of the small-signal gain � , one obtains an equation that can be solved graphically:

ϕm

ℏω0l0 l(i)

( 1ω2

c

d2

dt2− δT

ddt

+ (g − l0 − l(i)) + l(i) |a(t) |2

PA ) a(t) = 0

a(t) = a0sech(t /tp) a0 tp

δT = 0

1ω2

c t2p

+ g − l0 − l(i) = 0

−2

ω2c t2

p+

l(i)a 20

PA= 0

g0

� , where � are the saturation powers for

the absorber and for the lasing medium.

(a) Determine the minimum and maximum values of small-signal gain g0 that yield a stable single-pulse solution. Setting g0 to the average of these values, what are the values of tp and a0 for the (stable) longer-pulse solution? Give numbers for the peak power, pulse energy, average power, and FWHM pulse width.

For the graphical solution (see Fig. 5), one plots the left-hand side and the right-hand side of the previous equation and considers the possible solutions for different values of the small-signal gain. One finds a minimum value � and a maximum value � , which provide a boundary for � and, consequently, one for � . � is obtained by equating the first derivative of the two curves.

� Fig. 5 Graphical solution of the mode-locking parameters for the case of a fast saturable absorber.

Using the parameters in the exercise, one finds � and � such that the average value � to which corresponds � .

The pulse width is obtained from � , which leads to �

fs. The peak power is simply � W. The pulse energy is given by integrating the envelope function over one pulse, � J. The average power is the pulse energy times the repetition rate, � W. Finally, the full width at half maximum (FWHM) for a sech pulse corresponds to � ps.

(b) Using the same g0 as in part (a), write a program to simulate the pulse evolution as it runs repeatedly through the laser cavity. As a suggestion, start your simulation with a noise burst roughly 100 times longer than the steady-state pulse width from above. Do you get a steady-state solution, and does it agree with the solution from theory? Plot pulse parameters (e.g., pulse width, peak power, energy) vs. number of iterations through the laser. Also show a few representative plots of the temporal intensity at different stages during your simulation. [Hint: After each pass

l0 + l(i) −g0

1 +2PA

l(i)

2a0

ωcPGT

=l(i)a 2

0

2PAPA,G =

ℏω0 AA,G

σA,GτA,G

g(min)0 = l0 + l(i) g(max)

0g0 a0 g(max)

0

g(min)0 = 0.25 g(max)

0g(ave)

0 ≃ 0.4 a0 ≃ 103

−2

ω2c t2

p+

l(i)a 20

PA= 0 tp =

1ωca0

2PA

l(i)≃ 800

Pp = a20 = 106

Ep = 2a20 tp = 1.6 × 10−6

P = Ep /T = 1601.763tp ≃ 1.5

through the laser you should recalculate the pulse energy and the gain to be used for the next iteration. The gain may be calculated, for example, from Eq. (2.54) of Ultrafast Optics].

The numerical solution of the mode-locking equation can be implemented using central

differences, � for the time derivatives. After rearranging

the terms at equal times, one finds an expression for � as a function of � and � . The time evolution is simply a do-loop that updates � , with the boundary condition � .

Exercises selected from chapters 1 and 2 of Ultrafast Optics by A.M. Wiener (J. Wiley & Sons, 2009). The FFT in Ex. 4 has been performed by Philipp Rauschel, who is kindly acknowledged also for providing the notes taken during the exercise session.

d2a(t)dt2

≃a(t + Δt) − 2a(t) + a(t − Δt)

Δt2

a(t + Δt) a(t)a(t − Δt) a(t)a(t + T ) = a(t)