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1
Lasers – Basic Concepts
• HMY 645• Lecture 05• Fall Semester 2015
Stavros IezekielDepartment of Electrical and
Computer EngineeringUniversity of Cyprus
LASER BASICS
2
3
There are many different types of lasers
4
Gas lasers (e.g. HeNe) Solid‐state lasers (e.g. ruby rod)
Diode lasers Fiber lasers
5
Pump
gPlanar mirror
A laser is an oscillator; all oscillators have gain and feedback.
Three basic “ingredients” are needed to make a laser:
(1) An optical medium that has gain (provides stimulated emission)(2) A pump to excite the optical medium(3) An optical resonator to provide feedback
• LASER is an acronym for:
light amplification by stimulated emission of radiation
6
• Simple model of a laser:
1 Take a medium (e.g. one with three energy levels), and create population inversion by pumping (either optical or electrical).
Pump
Optical gain = g
Pin Pout = gPin
7
2 Put the optical gain medium inside an optical cavity, e.g. one which is formed from a parallel pair of flat mirrors:
Pump
g
This is similar to a positive feedback system:
G
Feedback network
Planar mirror
Feedback is provided by the optical cavity
Need:
0180)(
1)(
jGH
jGH
+
+
8
Optical Gain
(Stimulated emission)
PhotonAmplified once
First reflectionAmplified for
second time
Second reflection
Amplified a third time
A simplified representation of laser action
Output
This mirror is partially reflecting, to allow some photons to escape.
Third reflection
9
Spatial coherence
Temporal coherence
• The radiation that is emitted by a laser is said to be spatially and temporally coherent, i.e. the light is highly monochromatic (but not perfectly) and very directional (again, not perfectly).
10
z
E0
E
Laser
)sin(),( 0 tkzEtzEPerfectly monochromatic E‐field (travelling wave) emitted by “perfect” laser:
Temporal coherence: and are constant (i.e. time‐independent)
EMISSION PROCESSES IN LASERS
11 12
How to generate identical photons? Clone them…..
Laser relies on concept of stimulated emission
13
Bohr’s model of the hydrogen atom • Electrons can only reside in specific “orbits”; angular momentum in orbital n is nh/2. (Only three orbitals are shown here.)
• Each orbital is a function of a quantised amount of energy. Less energy is required to remain in the ground state (orbital n=1) compared to higher levels.
• An electron can jump to a higher level by absorbing a photon with the correct quantum of energy.
• If an electron jumps from a higher level to a lower one, it will emit a photon with an energy equal to the corresponding energy difference between levels.
Proton
n = 1
n = 2
n = 3
Electron
14
Proton1
2
PhotonEnergy = E2 - E1
Absorption
(Only two energy levels shown)
15
Proton1
2
PhotonEnergy = E2 - E1
Emission
Emission can either happen spontaneously, or it can be stimulated by another photon of the same energy.
16
Absorption and emission in a two‐level system
In equilibrium, most atoms are in their ground state:
Energy
1E
2E
hchfEE
12
Ground state
Excited state2N
1N
Ni = number of atoms per unit volume with energy Ei
17
Absorption in a two‐level system
The atoms can be excited to the higher level, for example by absorption of light
Energy
1E
2E
hchfEE
12
Photon Energy = E2 - E1
2N
1N
18
Spontaneous emission in a two‐level system
The atoms can then return spontaneously to the ground state, and emit photons as they do so. The photons are emitted in random directions.
Energy
1E
2E
hchfEE
12
Photon Energy = E2 - E1
2N
1N
19
Stimulated emission in a two‐level system
The atoms can also be stimulated to return to the ground state by incoming photons. This process of stimulated emission can be used for optical amplification.
Energy
1E
2E
hchfEE
12Photons: • in phase• same direction• same energy
2N
1N
We will show that for amplification by stimulated emission, we need N2 > N1 (population inversion) otherwise the incoming photons will be absorbed
20
To aid in explanation, consider following symbols:
Atom in excited state, i.e. initial state is:
1E
2E
Atom in ground state, i.e. initial state is:
1E
2E
Now, the chance of stimulated emission from an excited atom is exactly the same as the chance of absorption by an atom in the ground state.
21
Example
Consider a material that contains nine atoms as shown, with a mix of excited atoms and atoms in ground state. Three photons enter this material from the left:
What happens next?
22
Fewer photons come out.
Why? Because from the initial state (see previous slide):
• 3 atoms can emit light spontaneously• 6 atoms can absorb light
• Hence absorption will dominate
23
To increase the number of photons when going through the atoms, more atoms need to be in the excited state than the ground state.
We need population inversion. (Atoms however will decay down to the ground state relatively quickly, so we need to come up with a strategy for population inversion –we will see that best approach is a four‐level system.)
N2 > N1 – more “cloned” rather than “eaten”
N2 < N1 – more “eaten” rather than “cloned”
24
Remember ‐‐
1E
2EIn
1E
2EOutIn
Absorption
Atom absorbs photon (and becomes excited)
Stimulated Emission
“Clone” the photon
Chance of stimulated emission from excited atom is exactly the same as chance of absorption by atom in lower state.
Hence we need population inversion
25
Population inversion is NOT possible in a two‐level system:
(i) Imagine that we begin with more atoms in the ground state.
(ii) Now assume that we “pump” some of the ground‐state atoms into an excited state such that we have equal numbers of excited and ground‐state atoms. For example:
26
(iii) We now have a situation in which an incoming photon now has as much chance of stimulating an excited atom to fall to the ground state as it does of being absorbed.
i.e. the equal probability of stimulated emission and absorption is a problem because in the steady‐state we will end up with the rate of spontaneous emission being equal to the rate of absorption.
In fact, if one factors in the existence of spontaneous emission as well, the population N1 will always be bigger than N2.
http://phet.colorado.edu/simulations/lasers/lasers.jnlp
27
Population inversion and pumping
• Population inversion (and therefore lasing) cannot be achieved in a two‐level system. In steady‐state, the incoming photons will cause equal numbers of absorptions and stimulated emissions.
• A three‐level system allows population inversion through pumping:
2E
1E
3E
PUMP
hcEE
13
METASTABLELEVEL
RAPID DECAY FROM 3 TO 2
28
Stimulated emission can lead to an avalanche of photons
If an optical medium contains many atoms at an excited level, one photon can end up stimulating the emission of many others through an “avalanche”:
In practice there will also be some losses, but overall we will have a material with optical gain (i.e. one input photon leads to many photons at the output)
29
Optical medium with gain
If the optical gain medium is placed inside a resonator, we can have lasing
MirrorR = 100%
Partially reflecting mirror
R < 100%
Output
In addition to gain, there will also be losses (e.g. due to absorption). For there to be lasing, we must satisfy the threshold condition:
Gain Loss
1I 2I
3I4I
Over one round‐trip, we require:
14 II
30
Blackbody Radiation
• When matter is heated, it emits radiation.
• A blackbody is a cavity with a material that only emits thermal radiation. The radiation properties are independent of the material.
• Incoming radiation is absorbed in the cavity.
• Planck showed that he could model the experimental spectral distribution by using the quantum hypothesis.
2 52 /
exp / 1B
hcI
hc k T
kB = Boltzmann constant = 1.38×10‐23 JK‐1
31
Blackbody Radiation
For a cavity with a refractive index n, the energy density inside the cavity as a function of frequency (energy per unit volume per unit frequency) is:
1exp
8)(3
33
Tkhfc
hfnhf
B
This quantity is directly proportional to the number of photons per unit volume that have an energy of hf.
EINSTEIN COEFFICIENTS
32
33
The Einstein A and B Coefficients
The Einstein coefficients are used to examine the transition rates (for absorption, spontaneous and stimulated emission) between energy levels. Consider a two‐level system:
1E
2E 2N
1N
AbsorptionStimulatedemission Spontaneous
emission)(112 hfNB
221NA)(221 hfNB
AB: Absorption, SP: Spontaneous emission, ST: Stimulated emission
34
A21: Spontaneous Emission
1E
2E 2N
1N
Spontaneousemission
221NA
• Atoms decay spontaneously from level 2 to level 1. If the population density for level 2 is N2, then it will decrease as follows:
2212 NA
dtdN
SP
• If there were no other transition processes,then N2 would decay exponentially with a time constant of A21.
• Since the decay is from 2 to 1, it follows that N1will increase at the same rate as N2 decreases:
SPSP dtdN
dtdN 12
Rate of emission
221NA
35
B12: Absorption
1E
2E 2N
1N
Absorption
• Incoming photons raise atoms in level 1 up to level 2, so N2 will increase and the rate of increase will depend on the number of absorbing atoms and also the number of incoming photons:
)(1122 hfNB
dtdN
AB
)(112 hfNB
Proportional to the number of photons with energy E2-E1
Proportional to number of atoms available for absorption
Again, we have:
ABAB dtdN
dtdN 12
36
B21: Stimulated emission
1E
2E 2N
1N
Stimulated emission
• Incoming photons stimulate atoms in level 2 down to level 1, so N2 will decrease and the rate of decrease will depend on the number of excited atoms and also the number of incoming photons:
)(2212 hfNB
dtdN
ST
)(221 hfNB
Proportional to the number of photons with energy E2-E1
Proportional to number of atoms available for stimulation
Again, we have:
STST dtdN
dtdN 12
Rate of emission
)(221 hfNB
37
Relationship between the Einstein coefficients
If we look at all three radiative processes, adding them together we have:
)()( 2211122212 hfNBhfNBNA
dtdN
spontaneous emission absorption
stimulatedemission
)()( 2211122211 hfNBhfNBNA
dtdN
38
In thermal equilibrium (no external excitation), decreases in N1 will be balanced by increases in N1 (and the same is true for N2). So the rate of change with time must be zero:
00 21 dt
dNdt
dN
0)()( 221112221 hfNBhfNBNA
)()( 21212112 hfBANhfNB
212
112
21
221112
212)(B
NNB
ANBNB
ANhf
39
Boltzmann statistics
Ni is the number density of atoms in state i (i.e., the number of atoms per cm3).
T is the temperature, and kB is Boltzmann’s constant.
exp /i i BN E k T
Ene
rgy
Population density
N1
N3
N2
E3
E1
E2
40
The Maxwell‐Boltzmann distribution
• As a result, higher‐energy states are always less populated than the ground state, and absorption is stronger than stimulated emission unless we have population inversion.
22
1 1
exp /exp /
B
B
E k TNN E k T
1
23
Ene
rgy
Low T
1
23
Ene
rgy
High T
• In equilibrium, the ratio of the populations of the two states is:
Tkhf
TkEE
NN
BB
expexp 12
1
2
41
1exp
8)(3
33
Tkhfc
hfnhf
B
Tkhf
NN
B
exp1
2
So we have:
Energy level populations:Boltzmann Blackbody radiation ‐ Planck
212
112
21)(B
NNB
Ahf
Rate equations ‐ Einstein
2112
21
3
33
exp1exp
8
BTk
hfB
A
Tkhfc
hfn
BB
This can only be satisfied if: 2112 BB
42
1exp1exp
8
21
21
3
33
TkhfB
A
Tkhfc
hfn
BB
3
33
21
21 8c
hfnBA
We find that stimulated emission exceeds spontaneous emission provided:
hnfchf 8
3
The implication is that lasing is more difficult to achieve at shorter wavelengths (e.g. X‐ray and UV)
43
In summary:
1
2
112
221
2
1
2112 )()(
NN
hfNBhfNB
dtdNdt
dN
BB
AB
ST
If we want stimulated emission to exceed absorption, then we must have population inversion:
12 NN
According to Boltzmann statistics:
Tkhf
NN
B
exp1
2
Population inversion requires negative absolute temperature (which we cannot have), so this implies that the laser is based on non‐thermal equilibrium.
PUMPING
44
45
Achieving inversion: Pumping the laser medium
Now let I be the intensity of (flash lamp) light used to pump energy into the laser medium:
R = 100% R < 100%
I0 I1
I2I3 Laser medium
I
Will this intensity be sufficient to achieve inversion, N2 > N1?
46
Rate equations for a two‐level system
Rate equations for the densities of the two states:
21 2 2( )dN BI N N AN
dt
12 1 2( )dN BI N N AN
dt
22 2d N BI N ANdt
Absorption Stimulated emission Spontaneous emission
1 2N N N 1 2N N N
If the total number of atoms is N:
2 1 2 1 22 ( ) ( )N N N N NN N
2d N BI N AN A Ndt
2
1
N2
N1
LaserPump
Pump intensity
BBB 2112 AA 21
47
Inversion is impossible in a two‐level system
0 2BI N AN A N
1 / sat
NNI I
/ 2satI A B
In steady‐state:
( 2 )A BI N AN
where:
N is always positive, no matter how high I is.
It is impossible to achieve an inversion in a two‐level system!
2d N BI N AN A Ndt
/( 2 )N AN A BI
/(1 2 / )N N BI A
Isat is the saturation intensity.
2
1
N2
N1
Laser
48
Rate equations for a three‐level system
Assume we pump to a state 3 that rapidly decays to level 2.
21 2
dN BIN ANdt
11 2
dN BIN ANdt
1 22 2d N BIN ANdt
Absorption
Spontaneous emission
1 2N N N 1 2N N N
The total number of atoms is N:
22N N N
d N BIN BI N AN A Ndt
Fast decay
Laser Transition
Pump Transition
1
23
Level 3 decays fast and so is zero.
12N N N
49
Why inversion is possible in a three‐level system
1 /1 /
sat
sat
I IN NI I
In steady‐state:
( ) ( )A BI N A BI N
Now if I > Isat, N is negative!
( ) /( )N N A BI A BI
d N BIN BI N AN A Ndt
0 BIN BI N AN A N
Fast decay
Laser Transition
Pump Transition
1
23
50
Rate equations for a four‐level system
Now assume the lower laser level 1 also rapidly decays to a ground level 0.
20 2
dN BIN ANdt
1 0,N
22 2( )dN BI N N AN
dt
2N N
Laser Transition
Pump Transition
Fast decay
Fast decay
1
2
3
0As before:
Because0 2N N N
The total number of atoms is N :
0 2N N N d N BIN BI N A N
dt
At steady state: 0 BIN BI N A N
51
Why inversion is easy in a four‐level system
0 BIN BI N A N
/1 /
sat
sat
I IN NI I
Now, N is negative—always!
Laser Transition
Pump Transition
Fast decay
Fast decay
1
2
3
0
( / ) /(1 / )N BIN A BI A
/( )N BIN A BI
( )A BI N BIN
52
Summary of population inversion:
OPTICAL RESONATORS
53
)(8
)(emission sspontaneou of rate
emission stimulated of rate3
3
21
21 hfhfc
AhfB
For a particular frequency “f” we need a high density (hf) of photons with energy hffor this ratio to be >> 1.
So we must confine photons in the region where population inversion is also present.
Earlier we showed that:
1
2
absorption of rateemission stimulated of rate
NN
Also:
Hence we need population inversion
Photon Density (hf)
Photons reflect backwards and forwards between the two parallel mirrors
(the photons ideally will be travelling normal to the mirrors)
This setup is known as a Fabry‐Perot resonator
FABRY‐PEROT RESONATOR
56
57
• We now consider a passive cavity (there is no optical gain element):
Planar mirrorr2
Terminology
Planar mirrorr1
• This structure is called a Fabry‐Perot cavity (or Fabry‐Perot interferometer). The analysis has similarities with the structure below:
11
2
1
rr
Ei Et
Er
Fabry-Perotetalon
Fabry-Perotcavity
Ei
Et1 Et2 Et3
Er1 Er2 Er3
12 nn
1n
1n
2n
58
Analysis of Fabry‐Perot etalon: Calculation of refelected and transmitted waves
We will perform an analysis using normal incidence:
1n 1n2n
1t
't1r
'r21
12nn
nt
21
21
nnnnr
21
22'nn
nt
21
12'nnnnr
rr ' 1'2 ttrStokes relationships
L
A B
jAeB
LnkL 22
Phase change along cavity length (half round trip).
1',' rrr
59
a jeta jetat '
ra jetar 2'
jetart 2''
Values are referred to vector tips inside the etalon and for intial vector, to vector bases outside.
jetar 32' jetart 32''
jetar 43' jetart 43''
jetar 54' jetart 54''
jetar 65' jetart 65''
60
Adding the transmitted waves we get:
0
22
42
5432
''
...''1'
....'''''
i
ijj
jjj
jjjT
eretat
ereretat
taetrtaetretata
Infinite geometric series: 11
10
xx
xi
i
j
j
T eretata
22'1'
61
j
j
T eReTaa 21
ttT ' 222 '' rrrR
1'2 ttrFrom the Stokes relations:
1 TR
Adding the reflected waves we get:
j
j
jjj
jjjR
ertaetrra
erertaetrra
taerttaerttaertraa
22
2
44222
65432
'1''
...''1''
....''''''
62
j
j
j
j
R ertaetrar
ertaetrraa
22
2
22
2
'1'''
'1''
RrrrrR ',' 22
j
j
j
j
j
j
j
j
R
eReRa
eReRRa
eRTeRRa
ertaetrara
2
2
2
2
2
2
22
2
1)1(
1)1(1
1
'1'''
63
j
j
j
jT
eReR
eReT
aa
22 1)1(
1
j
jR
eReR
aa
2
2
1)1(
The ratio between the transmitted and incident waves is:
The ratio between the reflected and incident waves is:
64
The irradiance for a complex field a is:
221*
21 acaacI
The irradiance transmission coefficient is therefore:
22
2
22
2
2
22
1
)1(
1
)1(
jj
jTT
eR
R
eR
eR
aa
aa
and the irradiance reflection coefficient is:
22
222
1
1
j
jR
eR
eRaa
65
22
22
2
222
22
*22
*2222
sin4)1()sin21(21
2cos21)(1
1111
111
RRRRRR
eeRReReReReR
eReReR
jj
jj
jj
jjj
222 sin41 je
66
22
22
sin4)1()1(RR
RaaT
22
22
sin4)1(sin4
RRR
aaR
Transmission coefficient
Reflection coefficient
67
1
1
0
0
0.5
0.5
Tran
smis
sion
5.0RExample:
Ref
lect
ion
0
0
2 3 4
2 3 4
68
1
1
0
0
0.5
0.5
Tran
smis
sion
95.0RExample:
Ref
lect
ion
0
0
2 3 4
2 3 4
69
1
0.1
0.5
0.2
0.3
0.4
0.6
0.7
0.8
0.9
Lnmc
22 Lncm
22)1(
Lncm
22)2( Frequency
Tran
smis
sion
Fabry‐Perot Terminology
LncfFSR22
70
Fabry‐Perot Terminology
FSRf
FWHMf
FWHM bandwidth is proportional to the finesse (Quality factor) of the cavity:
RRF
1
LFncfFWHM22
Fff FSR
FWHM
Bandwidth = FSRFinesse
CAVITY WITH GAIN
71
g
72
Inclusion of gain in the cavity
1n 1n2n
1t
't1r
'r
L
A B
jgAeB
Lnkd 22
The picture will be similar to the one we used before, but this time, for each pass through the cavity, we have a gain of g.
73
a jegta jetagt '
ra jetarg 22 ' jetartg 22 ''
jetarg 323 ' jetartg 323 ''
jetarg 434 ' jetartg 434 ''
jetarg 545 ' jetartg 545 ''
jetarg 656 ' jetartg 656 ''
74
j
j
j
jT
eReR
eReT
aa
22 1)1(
1
In other words, without gain, for each pass through the cavity the field is multiplied by exp(-j), and when we include gain this changes to g exp(-j):
Without gain:
With gain:
j
jT
egRgeR
aa
221)1(
If we repeat the previous analysis, this time we have:
2222
222
sin4)1()1(RgRg
RgaaT
75
22
22
sin4)1()1(
GRGRRG
aaT
2gG Let: (Gain in irradiance per pass through cavity)
The maximum transmission coefficient will occur when sin = 0:
2
2
max
2
)1()1(
GRRG
aaT
If: max
21aa
RG T
76
Photon Lifetime
The Fabry‐Perot cavity can store an electric field if power is continuously provided by an external source. If the source is then switched off, it takes some time for the energy in the cavity to decay to zero. We can analyse this by considering photons in the cavity:
NP
NP = initial photon number
2r1r
L
PNr 22
PNrr 22
21
2n
77
In one round trip (distance of 2L), the number of photons that have been lost is:
PP NRNrr 222
21 11
where we assume that r1 = r2. If the light travels at c/n2 in the cavity, then the time taken for one round trip is:
cLn22 time tripround
Hence the rate of change of photons is:
P
PPP NLn
NRcdt
dN
2
2
21
22
12
RcLn
P
78
The solution is:
PPP
tNtN
exp)( 0
The photon lifetime is:
trip-roundper lost photons offraction timetrip-round
12
22
RcLn
P
We can relate the photon lifetime to the cavity Q.
lostpower averageresonanceat cavity in the storedenergy
period onein lost energy resonanceat cavity in the storedenergy 2
0
Q
79
Let W = stored energy. We have:
QW
dtdW
dtdWWQ 0
0 /
t
QWtW 0
0 exp)(
hfNW P00 energy stored of valueinitial
W and NP must decay at the same rate, hence:
0 Q
P
LASING MODES
80
Amplifying mediumopticaloutput
mirrorreflectivity = R2
L
• This structure acts as a Fabry‐Perot resonator; optical cavityprovides positive feedback, i.e. gives oscillation:
mirrorreflectivity = R1
gain coefficient = g(hf)absorption coeff. = (hf)
Threshold Conditions
• Optical amplification of selected modes is provided by the feedback mechanism of the optical cavity. In repeated passes between the two partially reflecting mirrors, a portion of the radiation associated with those modes having the highest optical gain coefficient is retained and further amplified.
• Lasing occurs when the gain of one or more guided modes is sufficient to exceed the total optical loss. Hence for a cavity of length L, the light travels z = 2L per round‐trip. During the round‐trip, fractions R1 and R2 are reflected from the mirrors.
• Optical intensity is given by:
I(z) = I(0) exp { [g(hf) - (hf)] z }
• The intensity at the end of the round‐trip (z = 2L) must match that at the start (z = 0), hence:
I(2L) = I(0) R1R2 exp { [g(hf) - (hf)] 2L } = I(0)
• This gives the optical gain threshold for lasing:
gth = (2L)-1 ln [(R1R2)-1] +
• Also, the round‐trip phase satisfies: exp(-j2βL) = 1
Mode Selection
• Since the laser has a resonant cavity, standing waves can exist between the mirrors if there is sufficient gain. These standing waves can only exist if L is an integral number (m) of half wavelengths:
where n is the refractive index of the cavity. Condition is obtained from:2L = 2m.
L = m2n
• Hence the mth longitudinal mode is related to its frequency of oscillation fm by:
m = 2nL = 2nL fm
cm
• For the next mode, m + 1, we have:
m + 1 = 2nL fm+1
c
• If we denote the spacing between adjacent modes as f, we have f = fm+1 - fm
• Hence taking the difference of the above two equations:
2nL f = 1c
i.e. f = c2nL
• In terms of wavelength spacing between modes, we have:
= m + 1 - m = c - c = c ffmfm+1 fm+1 fm
• If the peak frequencyof the laser is f, then for a semiconductor laser we have fm+1 fm f
• Hence:
= c f = 2 ff 2 c
• Therefore: = 2
2nL
• This tells us about mode spacing, but how many modesactually lase? Will the laser be single or multimode?
Gain - Wavelength Profile
• Answer depends on relationship between optical gain gand wavelength . This takes a gaussian form:
g() = g(0) exp [ - ( - 0)2 / 22 ]
• 0 is centre wavelength, is spectral width of the gain,and g(0) is the peak gain, which occurs at = 0
89
Optical gain betweenFWHM points
m
(a) 5 modes
(b)4 modes
.
Cavity modes
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Number of laser modes ‐ depends on how the cavity modes intersect the optical gain curve
90
( c )
Relative intensity
Gain Curve
Allowed Modes