A state-of-the-art Titanium Sapphire
laser system
Assignment and Laboratory exercise in
Laser physics VT-2005
Introduction
This assignment and demonstration will focus on the so called "kilohertz" laser
system. This system is part of the Lund Laser Center and is used for ultrafast physics
research. The kHz laser produces femtosecond laser pulses with a duration of about 35 fs
(1fs =10-15 s) and an energy of 1.6 mJ at a central wavelength of 800 nm (infra red).
The kHz system comprises four parts. An oscillator that produces 15 fs pulses in
the IR, a stretcher that streches the pulses to 180 ps, an amplifier stage that increases the
pulses energy up 1.6 mJ and a compressor that compresses the pulses to 35 fs.
Introduction to femtosecond lasers
Pulsed lasers are powerful tools in many research areas and in the industry.Since
the very first days of the laser, there has been a quest to achieve shorter and shorter laser
pulses. Light pulses shorter than 100 fs were observed for the first time in 1981; pulses of
6 fs were obtained by a group at Bell Laboratories in 1987.
These ultra short pulses are of great interest for many applications.
- In the temporal domain: fs light pulses are equivalent to a super-stroboscope and enable
us to look at ultrafast phenomena. For instance, one can observe the first steps of
chemical reactions.
- In nonlinear optics: the energy is released during the very short duration of the pulses.
Very high peak powers (up to Petawatts) can be reached using short pulses.
The generation of ultrashort pulses implies the amplification of a broad frequency
spectrum. The time-frequency uncertainty relation gives t a with a almost equal to
1 (depending on the temporal profile chosen). From that relation we can see that to a
small t corresponds to a large frequency spectra . This important relation will have
many consequences on the choice of the different parts used to build the laser. (The
amplification media should amplify a broad frequency spectrum, the mirrors must be
broadband, etc.)
Titanium sapphire has an energy level diagram, which is typical for solid laser
materials (four level lasers). The unique property of this material is a broad absorption
band, which can be pumped by fixed-frequency lasers and a wide lower laser level. The
bandwidth over which laser can be obtained is wide enough to allow for 3 fs long pulses.
The absorption band has its peak at around 500 nm and the emission spectrum is peaked
around 800 nm. Since the absorption and emission bands are well separated, losses due to
reabsorption of the laser radiation are minimized.
2
Relaxation
3
Pump
Lasing
500 nm
800 nm
4
Relaxation
1
Two conditions determine the frequency components available to build up a wide
frequency spectrum. The first is the (approximately) resonance condition for a standing
wave: n=L or expressed in frequency =nc/2L where L is the length of the cavity. The
second one is the spectral bandwidth of the laser. The laser bandwidth is mainly limited
by the gain profile of the amplifying medium.


Passive Kerr lens mode locking (Svelto 344- 345)
In a cavity consisting only of a gain medium, end mirror and output coupling
mirror, all modes are oscillating in random phase. By introducing additional components
in the cavity, it is possible to phase-lock these modes to each other. Depending on the
method by which the phases are locked one refers to either active or passive modelocking. When the phase difference between the modes is constant, the modes are
oscillating in phase and interfere constructively. If we add the E-fields from the standing
modes, the sum as function of time will be a short pulse propagating back and forth in the
cavity.
At low intensities, the refractive index (for a given wavelength) is constant. At
high intensities, the refractive index vary by n=n1+n2I. Due to the fact that the laser has a
spatial intensity variation, the refractive index will be larger in the center of the beam,
where the intensity is the highest. This makes the crystal behave like a lens (referred to as
a Kerr lens). A Kerr lens can be used to create losses for CW radiation, whereas the
losses for high intensity pulses obtained by mode-locking are minimal. Thus, a Kerr lens
can be used for passive mode-locking. It is common that Kerr lensing in the amplifying
medium is used. The amplifying medium is used both to obtain gain and to obtain modelocking.
Titanium sapphire
crystal
Aperture
Low intensities
large losses
Laser beam
x
n=n1+n2I
I
High intensity
small losses
The beams spatial profile creates the "Kerr lens"
Dispersion compensation (Svelto 347-355)
Due to dispersion, different frequency components propagate with different
velocities in all materials. In solids, glass, crystals etc., the effect can be significant. In
nearly all materials, shorter wavelengths propagate slower than longer wavelengths. This
phenomenon is known as positive group velocity dispersion (GVD). The consequence is
that a short pulse consisting of many wavelengths is stretched in time. In order to
compensate this effect, negative group velocity dispersion is needed. Negative GVD
means that short wavelengths propagate more rapidly than longer wavelengths. This can
be obtained with the prism arrangement shown below.
R
B
Amplification of the laser light
As mentioned previously, the oscillator of the kHz laser system delivers laser light
pulses of 15 fs with an energy of 2 nJ. This power is not enough for the applications we
are aiming at. The laser pulses are then amplified to 2 mJ, using the Chirped Pulse
Amplification technique (1989).
The figure below resumes the main steps of the amplification.
15 fs ; 89 MHz from
the oscillator
Pockel’s cell
Pulsed YLF
laser
Stretching
of pulses
180 ps ; 89 MHz
Grating stretcher
Selection of an
First
impulsion
amplification
180 ps; 1kHz
Second
amplification
180 ps ; 1kHz
Q-switched
Compression
Laser YLF
Grating
compressor
35 fs ; 1kHz
Pulse stretching
If we directly amplify the short pulses from the oscillator, we will reach very high
peak intensities. The high intensities would damage the amplifier medium. In order to
decrease the peak intensities we induce a varying phase (a chirp) on the different spectral
components of the pulse, leading to a stretching of the pulse duration. In the kHz system
a grating stretcher is used. The different frequencies will not have the same optical path
length, and the pulses get as long as 180 ps.
Amplification stage
The stretched pulses (180 ps) are sent into the first amplification stage. This is
performed by using Pockel’s cells and polarizers.
The first amplification step is done by a regenerative amplifier cavity, with two
concave mirors. The selected pulses do a few round trips before they are coupled out. The
Ti:Sapphire crystal in the regenerative cavity is pumped by a frequency doubled Nd-YLF
laser. The beam is then sent to two other amplification stages. In these amplifiers the
pulses pass the gain medium twice (two pass amplifier).
The amplification of a laser pulse, through an amplifier medium – assumed to
have a 4-level energy scheme – is given by:


   
out  satT ln 1  G0  exp  in   1 ,

 sat  

where in , out are the input, output fluence (energy/beam area) of the amplified pulse,
and sat the saturation fluence. G 0 is the unsaturated gain, and is connected to the upper
level population (N0) through
G0  exp lN 0  .
If the energy per unit area absorbed by the amplifying medium is given by p, the pump
wavelength is λp, and the amplified pulse has a wavelength of λa, then the fluence
available for amplification is
gain  p
p
 lN 0sat so that
a
 
G0  exp  gain 
 sat 
To describe a regenerative or multipass amplifier, the above equation has to be solved for
the consecutive passes through the crystal. In that case, T - the transmission of the crystal
- should also represent other losses (e.g. mirror reflection) in the cavity through a
roundtrip. The unsaturated gain coefficient g0 decreases after each pass, with the decrease
of the upper level population. We neglect any loss of the upper population through
spontaneous decay mechanisms, so in the nth round we have

n
gain
n 1
gain

 out n 1
n 1 
 
 in  .
 T

Compression
After the amplification we still have long pulses. In order to achieve high power
we have to recompress the pulses. To achieve that, we use a grating compressor. As in
the stretcher, the different optical frequency components will not have the same path in
the cavity. The compressor compensates for the chirp of the pulses and recompresses the
pulses in time. We then get high intensity peaks.
Measurement techniques
To be able to control the pulses we need to measure and characterize our fs pulses.
The only methods available on that timescale (fs) are optical. The fastest electronic
detector can only detect on a ps timescale. We have to use the pulse itself to measure it.
Preparatory exercises
1) Calculate the average and the peak power at the output of the kHz oscillator (2nJ, 15 fs
at a repetition rate of 89MHz).
2) After the grating stretcher, each pulse have a duration of 180 ps. Calculate the average
and peak power of these pulses (take into account that there is a 50% loss in the grating
stretcher).
3) Calculate the peak intensity in W/cm2 at the output of the amplifier (1.7 mJ, 35 fs) if
the beam is focused on a target with a spot size of 1mm diameter.
4) It is often difficult to have an idea of a femtosecond. To be aware of how short it is,
calculate the distance covered by the light in 1 fs. The cavity of the regenerative amplifier
is approximately 3.6 m long. Calculate the time needed for the seed pulse for a round trip.
To achieve optimum amplification, we have approximately 20 passes. What limit that
presents for the repetition rate of the amplified pulse train? (The real, 1 kHz repetition
rate is imposed by the repetition rate of the pump laser.)
5) Assume a Gaussian shaped pulse in the time domain of 15 fs with a central wavelength
λ0 of 800nm. Determine the corresponding spectrum:
- in the angular frequency domain ω0, Δω.
-
in the wavelength domain λ0, Δλ.
If by spectral narrowing the bandwidth is reduced to half, how will the pulse duration
change?
Assignment
1) Mode locking
The optical length of the oscillator cavity is 1,7 m. Calculate the mode spacing in the
frequency domain.
The bandwidth of a Ti:Sapphire crystal is 100 THz centered on ν0=375 THz , which
enables generation of 3 fs pulses. The spatial extension of a fs pulse is ~10 μm. In this
exercise, it is desired to have a longer pulse to be able to visualize the pulse in the cavity.
We reduce the bandwidth to 10 GHz. The broadening mechanism is assumed to be
mainly inhomogeneous, which implies a Gaussian line shape. The intensities of the
modes in the cavity are proportional to the gain. Plot the electric field amplitude of the
modes as a function of mode frequency.
The cavity modes are standing waves. Plot the superposition of the cavity modes for
random phases of the modes at t=0 in function of the cavity length. The Matlab
command: phi=2π*rand(1,N) is useful. Make a time evolution of the superposed standing
waves (choose an appropriate timescale).
Under mode locked operation the phase difference between consecutive modes is
constant. Plot the electric field in the cavity at time t=0 and make a time evolution.
Calculate the temporal length of the pulse out of spatial extension. The time frequency
dependence of a Gaussian pulse is:
t   
2 ln 2

Does the calculated temporal length fulfill the time frequency dependence?
2) Dispersion
Titanium Sapphire is a broadband amplifier medium. We can assume that in the
frequency domain we get is a Gaussian spectrum:
E  ,0   E 0 e
i (  )
  
0
exp   
  





2




The oscillator of the kHz lab delivers pulses whose spectra is centered on λ 0=800 nm and
a width of Δλ=70nm. With Matlab or any other program, retrieve the shape of the pulse
in the time domain for a phase φ(ω)=0 (no phase difference between the different spectral
components).
Warning: when computing a Fourier transform with Matlab the Fourier
transformed vector should be equally spaced with omega.
When a pulse with a broadband frequency goes through a material of a length L whose
refractive index n(ω) changes with ω the phase can be written as   L n   . That can
c
be written under a Taylor expansion form:

      0      0 

1  
Each coefficient  
n!  n

0
  0 2
 2
2!
 2
 ...
0
n
n
is a constant that can be obtained for a given material.
0
Modify the former program to add a phase to the spectrum. Retrieve the pulse in the time
domain for several α1(ω), and α2(ω). Can you identify the changes induced by these chirp
rates? Plot these results on a graph.
Hint: typical order of magnitude for α1(ω) is 10-15 s, α2(ω) is 10-28s2
A typical dispersive media is a SF57 glass plate whose chirp rate is α2=2,25.10-25 s2/m.
Compute the passage of a pulse (λ0=800, Δλ=70nm) through 1mm of this material. Plot
the incoming and outgoing pulses on a graph and estimate their duration.
3) Amplification
Using the following numerical values to calculate the output energy of the amplified
pulse after each pass, until the amplification reaches saturation, and the pulse energy
starts to decrease. Plot the pulse energy as a function of the number of passes. Vary the
pump energy and observe how the build-up changes.
sat = 1 J/cm2
Ep = 7 mJ or 14 mJ, 95% of what is being absorbed by the medium
d = 0.08 cm diameter of both beams
Ein = 0.2 mJ for the first pass
T = 0.9
λp = 532 nm and λa = 800 nm