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6/14/20151
412 PHYS
Lasers and their
Applications
Department of Physics
Faculty of Science
Jazan University
KSA
Lecture-1
6/14/20152
Recommended texts
The lectures and notes should give you a good base from which to start your
study of the subject. However, you will need to do some further reading. The
following books are at about the right level, and contain sections on almost
everything that we will cover:
1. “Principles of Lasers,” Orazio Svelto, fourth edition,
Plenum Press.
2. “Lasers and Electro-Optics: Fundamentals and
Engineering,”Christopher Davies Cambridge University
Press.
3. “Laser Fundamentals,” William Silfvast, Cambridge
University Press.
4. “Lasers,” Anthony Siegman, University Science Books.
14/06/2015
2
LASER SPECTRUM
10-13 10-12 10-11 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 1 10 102
LASERS
200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 10600
Ultraviolet Visible Near Infrared Far Infrared
Gamma Rays X-Rays Ultra- Visible Infrared Micro- Radar TV Radio
violet waves waves waves waves
Wavelength (m)
Wavelength (nm)
Nd:YAG
1064
GaAs
905
HeNe
633
Ar
488/515
CO2
10600
XeCl
308
KrF
248
2w
Nd:YAG
532
Retinal Hazard Region
ArF
193
Communication
Diode
1550
Ruby
694
Alexandrite
755
6/14/2015
6/14/20154
Introduction (Brief history of laser)
The laser is essentially an optical amplifier. The word laser is an acronym that stands
for “light amplification by the stimulated emission of radiation”.
The theoretical background of laser action as the basis for an optical amplifier was
made possible by Albert Einstein, as early as 1917, when he first predicted the
existence of a new irradiative process called “stimulated emission”.
His theoretical work, however, remained largely unexploited until 1954, when C.H.
Townes and Co-workers developed a microwave amplifier based on stimulated
emission radiation. It was called a MASER
14/06/2015
3
6/14/20155
Others devices followed in rapid succession, each with a different laser
medium and a different wavelength emission.
- In 1960, T.H.Maiman built the first laser device (ruby laser) which
emitted deep red light at a wavelength of 694.3 nm.
- Ali Javan and associates developed the first gas laser (He-Ne
laser), which emitted light in both the infrared (at 1.15mm) and
visible (at 632.8 nm) spectral regions.
6
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7
8
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5
9
10
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A laser consists of three parts:
1. A gain medium that can amplify light by means of the basic process of stimulated
emission;
2. A pump source, which creates a population inversion in the gain medium;
3. Two mirrors that form a resonator or optical cavity in which light is trapped,
traveling back and forth between the mirrors.
11 6/14/2015
Brewster Angle Gain region
Examples of Electrical and Optical pumping
12
6/14/2015
Dr. Mohamed Fadhali
Dr. Mohamed Fadhali
14/06/2015
7
Lasers are quantum devices, requiring understanding of the gain medium.
Laser light usually generated from discrete atomic transitions
13
Laser: Light amplification by stimulated emission of radiation
A laser converts electricity or incoherent light to coherent light.
Laser-matter interaction
Is a radiation emitted from a hot body. It's anything but black!
The name comes from the assumption that the body absorbs at every frequency
and hence would look black at low temperature . It results from a combination of
spontaneous emission, stimulated emission, and absorption occurring in a
medium at a given temperature.
It assumes that the box is filled with molecules that, together, have
transitions at every wavelength.
14
Blackbody radiation
14/06/2015
8
(perfect blackbody:  reflectivity = trasmissivity = 0
 emissivity = absorptivity = 1 )
Model: Assume a hole in large box with reflective interior walls: incident light from ~all
angles will make multiple passes inside box, resulting in thermal equilibrium inside box.
15
Blackbody radiation
Approach:
1. Calculate all possible ways EM radiation ‘fits in the box’
depending on the wavelength (density of states calculation)
2. First (wrongly) assume that each radiation mode has E=KBT/2
energy (this was the approach before the photon was known)
results in paradox
3. Fix this by assuming energy in field can only exist in energy
quanta & apply Maxwell-Boltzmann statistics  problem solved
For three dimensional case, and taking cavity with dimensions a×a×a →(V = a3),
we find allowed modes with equally spaced k values
We can now calculate the density of states as a function of several parameters,
e.g. number of states within a k-vector interval dk.
Each allowed k-vector occupies volume k in k-space (reciprocal space):
3
3x y zk k k k
a a a a
   
        
16
ALLOWED MODES AND DENSITY OF STATES
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9
Number of k-vectors in k range of magnitude dk depends on k:
In two dimensions: number of allowed
k-vectors goes up linearly with k
In three dimensions: number of allowed
k-vectors goes up quadratically with k
width dk
17
Mode density
volume 1/8 sphere = Vs
33
1 4 1 4 2
8 3 8 3
s
k n
V
c
    
   
 
Number of modes in this volume = 2 × Vs / k and k = (/a)3
3
3
33
3
3
3
33
3
8
3
4
2
2
3
4
8
1
2
a
c
n
a
c
n
a
c
n
N

















The mode density (modes per unit volume) in frequency range d becomes
With the group index, which we set as ngn
Mode density
14/06/2015
10
Classically (e.g. in gases), it was known that each degree of freedom had E = KBT/2
(e.g. atom moving freely: three degrees of freedom: E=3/2 KBT.
Applying this to the calculated mode density gives (incorrectly!) the energy density:
This gives rise to the Ultraviolet catastrophe
6/14/2015
0 2
0
2x10
7
4x10
7
6x10
7
8x10
7
1x10
8
T = 5000 K
T = 6000 K
T = 3000 K
SpectralRadianceExitance
(W/m
2
-mm)
Wavelength (mm)
M = T
Cosmic black body background
radiation, T = 3K.
Rayleigh-Jeans
law
6/14/2015
14/06/2015
11
one photon energy × probability of
having 1 photon present in mode
two photons × probability of
having 2 photons present in mode
normalization factor
The effect of energy quantization
* Analytical solution for blackbody radiation
The equation for energy per mode can be solved analytically:
Giving the following energy density inside the cavity at a given frequency  :
This is the Planck blackbody radiation formula
6/14/2015
HW: Prove that?
14/06/2015
12
Short wavelength behavior:
Result of quantum nature of light
mode density thermal population
6/14/2015
23
Blackbody Radiation
• (Stimulated) Absorption
• Spontaneous Emission
• Stimulated Emission
All light-matter interactions can be described by one of three
quantum mechanical processes:
…We will now look at each.
6/14/201524
Fundamentals of Light-Matter Interactions
14/06/2015
13
Interaction of Radiation with Atoms and Molecules:
The Two-Level System
The concept of stimulated emission was first developed by Albert Einstein
from thermodynamic considerations. Consider a system comprised of a two-
level atom and a blackbody radiation field, both at temperature T.
6/14/201525
This is, of
course,
absorption.
Energy
Ground level
Excited level
Absorption lines in an
otherwise continuous
light spectrum due to a
cold atomic gas in front
of a hot source.
Atoms and molecules can also absorb photons, making a
transition from a lower level to a more excited one
14/06/2015
14
When an atom in an excited state falls to a lower energy level, it emits a
photon of light.
Molecules typically remain excited for no longer than a few nanoseconds.
This is often also called fluorescence or, when it takes longer,
phosphorescence.
Energy
Ground level
Excited level
Excited atoms emit photons spontaneously
Ni is the number density of
molecules in state i (i.e.,
the number of molecules
per cm3).
T is the temperature, and
kB Boltzmann’s constant
= 1.38x10-16 erg / degree
= 1.38x10-23 j/K
 exp /i i BN E k T 
Energy
Population density
N1
N3
N2
E3
E1
E2
28
In what energy levels do molecules reside?
Boltzmann population factors
14/06/2015
15
*
In the absence of collisions, molecules
tend to remain in the lowest energy state
available.
Collisions can knock a molecule
into a higher-energy state.
The higher the temperature,
the more this happens.
 
 
22
1 1
exp /
exp /
B
B
E k TN
N E k T



Low T High T
Energy
Energy
Molecules
3
2
1
2
1
3
Boltzmann Population Factors
2
1
Calculating the gain: Einstein A and B coefficients
Recall the various processes that occur in the laser medium:
Absorption rate = B N1 r()
Spontaneous emission rate = A N2
Stimulated emission rate = B N2 r()
14/06/2015
16
Interaction of Radiation with Atoms and Molecules:
The Two-Level System
The processes of spontaneous emission and (stimulated) absorption were well
known. Einstein had to postulate a new process, stimulated emission in
order for thermodynamic equilibrium to be established.
2
1
Spontaneous
Emission
Stimulated
Absorption
Stimulated
Emission
2 21N A
2 21 2 21 ( )N W N B r 1 12 1 12 ( )N W N B r 
  3
( ) . /J s mr  
From thermodynamic equilibrium
  3
2 21 2 21 1 12( ) ( ) ( ) . /N A N B N B J s mr  r  r   
Units of B must be consistent with units of r() units of A are sec-1.
Absorption calculations are best done using A to avoid confusion on units.
3 3
3
8 1
( )
1
h
k TB
hn
c e

 
r  

2 1( )
2 2
1 1
B
E E
K TN g
e
N g


and
3 3
21 21 2 21 1 123
8
,
hn
A B g B g B
c
 
 
2 21 2 21 1 12N A N W N W 
HW: How to
reach to this
expression ???
14/06/2015
17
Absorption, emission, amplification depend on number of atoms in various states
Define concentration of atoms in state 2 as N2 (units often cm-3)
To find N1(t) and N2(t), you need to model time dependence of all processes
Process 1: spontaneous emission
Chance of spontaneous emission per unit time is A (Einstein coefficient)
If there are N2 atoms excited per volume, then at later time t we will have less, or
2 2
2
spsp
dN N
AN
dt 
 
    
 
2 2N A t N   
In differential form this becomes a rate equation of the form
where A is the rate constant for spontaneous emission and sp is the
life time for spontaneous emission given by sp=1/A
33
Rate equations: spontaneous emission
Suppose you can bring atoms in the excited state by some energy input
look at time dependence of N2 after the energy input is turned off at t=0:
( ) ( ) /2 2
2 2 0 spt
spsp
dN N
N t N e
dt


 
    
 
Note that the N2 drops to 1/e of its original
value when t=sp.
We have solved our first rate equation to
calculate the time dependent
concentration of excited atoms
6/14/201534
Rate equations: spontaneous emission
14/06/2015
18
2
2 21 ( )
st
dN
N B
dt
r 
 
  
 
B21 is the Einstein coefficient for stimulated emission
Under monochromatic illumination at frequency  we can write this as
2
2 21( )
st
IdN
N
dt h

 

 
  
 
Rate equations – Stimulated emission
More complex situations, add more processes to the rate equations
Scales with the electromagnetic spectral energy density r()
where r()d  is the energy per unit volume in the frequency range {,  +d }
Process 2: Stimulated emission
with I /(h ) is the photon flux given and 21() the cross section for stimulated emission
Note that ()I /(h ) is the rate constant for stimulated emission (units again s-1)
Process 3: Absorption
for absorption (‘stimulated absorption’) we obtain a similar rate
equation:
2
1 12 1 12( ) ( )
abs
IdN
N B N
dt h

r   

 
  
 
with B12 the Einstein coefficient for absorption
and 12 the absorption cross section
6/14/201536
Rate equations – Absorption
14/06/2015
19
We now have the rate equations describing the population of levels 1 and 2
Population is the ‘amount of occupation’ of the different energy levels
2
1 12 2 21 2( ) ( )
dN
N B N B AN
dt
r  r   
1
1 12 2 21 2( ) ( )
dN
N B N B AN
dt
r  r    
Since we have only two states, we find
(atoms that leave state 2 must end up in state 1)
N1 and N2 should add up to the total amount of atoms: N1+N2=N
We can now solve the time dependent population N2 under illumination
Before doing that, let’s look at the relations between A, B12, and B21
dt
dN
dt
dN 21

Rate equations for two-level system
Hypothetical situation: closed system at temperature T with collection of
two-level atoms and no external illumination:
All processes together will result in a thermal equilibrium with a
population distribution described by a Boltzmann factor:
/2
1
Bh k TN
e
N


In equilibrium, on average 021

dt
dN
dt
dN
This implies that in equilibrium
2 12
1 21
( )
( )
B
h
k TN B
e
N A B

r 
r 

 

1 12 2 21 2( ) ( ) 0N B N B ANr  r   
Resulting in an equation relating the Einstein coefficients to the thermal distribution:
6/14/201538
Einstein coefficients in thermal equilibrium
14/06/2015
20
2 12 12 12
12 21
( )
1 21 21 21
( ) ( )
1
( ) ( )
B
h
k T
T
N B B B
e B B B
N A B A B B

r  r 
r  r 


      
 
1
2
2 2
12 1 21 2 1 2
1
( )
1N
N
AN NA A
B N B N B N N B
r    
  
1
( )
1
h
k TB
A
B e
r  

Conversely, we can derive the ‘emission spectrum’ from our two-level atom
Substituting the thermal distribution over the available energy levels we obtain
which looks very similar to the Planck blackbody radiation formula :
3 3
3
8 1
( )
1
h
k TB
hn
c e

 
r  

3 3 3
3 3
8 8A hn hn
B c
  

 implying
At high temperatures e-h /k
B
T →1, and the radiation density becomes large:
Ratio between rates of spontaneous and stimulated emission
21 21
21 21
1
( )
h
k TB
A A
e
W B

r 
    
14/06/2015
21
HW. 1
Compare the rates of spontaneous and stimulated emission at room temperature (T
= 300K) for an atomic transition where the frequency associated with the transition
is about 3 × 1010 Hz, which is in the microwave region (KB=1.38 × 1023 J/K)
(b) What will be the wavelength of the line spectrum resulting from the transition of
an electron from an energy level of 40 × 10−20 J to a level of 15 × 10−20 J?
HW. 2
Consider the energy levels E1 and E2 of a two-level system. Determine the
population ratio of the two levels if they are in thermal equilibrium at room
temperature, 27◦C, and the transition frequency associated with this system is
at 1015 Hz
HW. 3
The oscillating wavelengths of the He–Ne, Nd:YAG, andCO2 lasers are 0.6328,
1.06, and 10.6µ m, respectively. Determine the corresponding oscillating
frequencies. What energy is associated with each transition?

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Laser lecture 01

  • 1. 14/06/2015 1 6/14/20151 412 PHYS Lasers and their Applications Department of Physics Faculty of Science Jazan University KSA Lecture-1 6/14/20152 Recommended texts The lectures and notes should give you a good base from which to start your study of the subject. However, you will need to do some further reading. The following books are at about the right level, and contain sections on almost everything that we will cover: 1. “Principles of Lasers,” Orazio Svelto, fourth edition, Plenum Press. 2. “Lasers and Electro-Optics: Fundamentals and Engineering,”Christopher Davies Cambridge University Press. 3. “Laser Fundamentals,” William Silfvast, Cambridge University Press. 4. “Lasers,” Anthony Siegman, University Science Books.
  • 2. 14/06/2015 2 LASER SPECTRUM 10-13 10-12 10-11 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 1 10 102 LASERS 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 10600 Ultraviolet Visible Near Infrared Far Infrared Gamma Rays X-Rays Ultra- Visible Infrared Micro- Radar TV Radio violet waves waves waves waves Wavelength (m) Wavelength (nm) Nd:YAG 1064 GaAs 905 HeNe 633 Ar 488/515 CO2 10600 XeCl 308 KrF 248 2w Nd:YAG 532 Retinal Hazard Region ArF 193 Communication Diode 1550 Ruby 694 Alexandrite 755 6/14/2015 6/14/20154 Introduction (Brief history of laser) The laser is essentially an optical amplifier. The word laser is an acronym that stands for “light amplification by the stimulated emission of radiation”. The theoretical background of laser action as the basis for an optical amplifier was made possible by Albert Einstein, as early as 1917, when he first predicted the existence of a new irradiative process called “stimulated emission”. His theoretical work, however, remained largely unexploited until 1954, when C.H. Townes and Co-workers developed a microwave amplifier based on stimulated emission radiation. It was called a MASER
  • 3. 14/06/2015 3 6/14/20155 Others devices followed in rapid succession, each with a different laser medium and a different wavelength emission. - In 1960, T.H.Maiman built the first laser device (ruby laser) which emitted deep red light at a wavelength of 694.3 nm. - Ali Javan and associates developed the first gas laser (He-Ne laser), which emitted light in both the infrared (at 1.15mm) and visible (at 632.8 nm) spectral regions. 6
  • 6. 14/06/2015 6 A laser consists of three parts: 1. A gain medium that can amplify light by means of the basic process of stimulated emission; 2. A pump source, which creates a population inversion in the gain medium; 3. Two mirrors that form a resonator or optical cavity in which light is trapped, traveling back and forth between the mirrors. 11 6/14/2015 Brewster Angle Gain region Examples of Electrical and Optical pumping 12 6/14/2015 Dr. Mohamed Fadhali Dr. Mohamed Fadhali
  • 7. 14/06/2015 7 Lasers are quantum devices, requiring understanding of the gain medium. Laser light usually generated from discrete atomic transitions 13 Laser: Light amplification by stimulated emission of radiation A laser converts electricity or incoherent light to coherent light. Laser-matter interaction Is a radiation emitted from a hot body. It's anything but black! The name comes from the assumption that the body absorbs at every frequency and hence would look black at low temperature . It results from a combination of spontaneous emission, stimulated emission, and absorption occurring in a medium at a given temperature. It assumes that the box is filled with molecules that, together, have transitions at every wavelength. 14 Blackbody radiation
  • 8. 14/06/2015 8 (perfect blackbody:  reflectivity = trasmissivity = 0  emissivity = absorptivity = 1 ) Model: Assume a hole in large box with reflective interior walls: incident light from ~all angles will make multiple passes inside box, resulting in thermal equilibrium inside box. 15 Blackbody radiation Approach: 1. Calculate all possible ways EM radiation ‘fits in the box’ depending on the wavelength (density of states calculation) 2. First (wrongly) assume that each radiation mode has E=KBT/2 energy (this was the approach before the photon was known) results in paradox 3. Fix this by assuming energy in field can only exist in energy quanta & apply Maxwell-Boltzmann statistics  problem solved For three dimensional case, and taking cavity with dimensions a×a×a →(V = a3), we find allowed modes with equally spaced k values We can now calculate the density of states as a function of several parameters, e.g. number of states within a k-vector interval dk. Each allowed k-vector occupies volume k in k-space (reciprocal space): 3 3x y zk k k k a a a a              16 ALLOWED MODES AND DENSITY OF STATES
  • 9. 14/06/2015 9 Number of k-vectors in k range of magnitude dk depends on k: In two dimensions: number of allowed k-vectors goes up linearly with k In three dimensions: number of allowed k-vectors goes up quadratically with k width dk 17 Mode density volume 1/8 sphere = Vs 33 1 4 1 4 2 8 3 8 3 s k n V c            Number of modes in this volume = 2 × Vs / k and k = (/a)3 3 3 33 3 3 3 33 3 8 3 4 2 2 3 4 8 1 2 a c n a c n a c n N                  The mode density (modes per unit volume) in frequency range d becomes With the group index, which we set as ngn Mode density
  • 10. 14/06/2015 10 Classically (e.g. in gases), it was known that each degree of freedom had E = KBT/2 (e.g. atom moving freely: three degrees of freedom: E=3/2 KBT. Applying this to the calculated mode density gives (incorrectly!) the energy density: This gives rise to the Ultraviolet catastrophe 6/14/2015 0 2 0 2x10 7 4x10 7 6x10 7 8x10 7 1x10 8 T = 5000 K T = 6000 K T = 3000 K SpectralRadianceExitance (W/m 2 -mm) Wavelength (mm) M = T Cosmic black body background radiation, T = 3K. Rayleigh-Jeans law 6/14/2015
  • 11. 14/06/2015 11 one photon energy × probability of having 1 photon present in mode two photons × probability of having 2 photons present in mode normalization factor The effect of energy quantization * Analytical solution for blackbody radiation The equation for energy per mode can be solved analytically: Giving the following energy density inside the cavity at a given frequency  : This is the Planck blackbody radiation formula 6/14/2015 HW: Prove that?
  • 12. 14/06/2015 12 Short wavelength behavior: Result of quantum nature of light mode density thermal population 6/14/2015 23 Blackbody Radiation • (Stimulated) Absorption • Spontaneous Emission • Stimulated Emission All light-matter interactions can be described by one of three quantum mechanical processes: …We will now look at each. 6/14/201524 Fundamentals of Light-Matter Interactions
  • 13. 14/06/2015 13 Interaction of Radiation with Atoms and Molecules: The Two-Level System The concept of stimulated emission was first developed by Albert Einstein from thermodynamic considerations. Consider a system comprised of a two- level atom and a blackbody radiation field, both at temperature T. 6/14/201525 This is, of course, absorption. Energy Ground level Excited level Absorption lines in an otherwise continuous light spectrum due to a cold atomic gas in front of a hot source. Atoms and molecules can also absorb photons, making a transition from a lower level to a more excited one
  • 14. 14/06/2015 14 When an atom in an excited state falls to a lower energy level, it emits a photon of light. Molecules typically remain excited for no longer than a few nanoseconds. This is often also called fluorescence or, when it takes longer, phosphorescence. Energy Ground level Excited level Excited atoms emit photons spontaneously Ni is the number density of molecules in state i (i.e., the number of molecules per cm3). T is the temperature, and kB Boltzmann’s constant = 1.38x10-16 erg / degree = 1.38x10-23 j/K  exp /i i BN E k T  Energy Population density N1 N3 N2 E3 E1 E2 28 In what energy levels do molecules reside? Boltzmann population factors
  • 15. 14/06/2015 15 * In the absence of collisions, molecules tend to remain in the lowest energy state available. Collisions can knock a molecule into a higher-energy state. The higher the temperature, the more this happens.     22 1 1 exp / exp / B B E k TN N E k T    Low T High T Energy Energy Molecules 3 2 1 2 1 3 Boltzmann Population Factors 2 1 Calculating the gain: Einstein A and B coefficients Recall the various processes that occur in the laser medium: Absorption rate = B N1 r() Spontaneous emission rate = A N2 Stimulated emission rate = B N2 r()
  • 16. 14/06/2015 16 Interaction of Radiation with Atoms and Molecules: The Two-Level System The processes of spontaneous emission and (stimulated) absorption were well known. Einstein had to postulate a new process, stimulated emission in order for thermodynamic equilibrium to be established. 2 1 Spontaneous Emission Stimulated Absorption Stimulated Emission 2 21N A 2 21 2 21 ( )N W N B r 1 12 1 12 ( )N W N B r    3 ( ) . /J s mr   From thermodynamic equilibrium   3 2 21 2 21 1 12( ) ( ) ( ) . /N A N B N B J s mr  r  r    Units of B must be consistent with units of r() units of A are sec-1. Absorption calculations are best done using A to avoid confusion on units. 3 3 3 8 1 ( ) 1 h k TB hn c e    r    2 1( ) 2 2 1 1 B E E K TN g e N g   and 3 3 21 21 2 21 1 123 8 , hn A B g B g B c     2 21 2 21 1 12N A N W N W  HW: How to reach to this expression ???
  • 17. 14/06/2015 17 Absorption, emission, amplification depend on number of atoms in various states Define concentration of atoms in state 2 as N2 (units often cm-3) To find N1(t) and N2(t), you need to model time dependence of all processes Process 1: spontaneous emission Chance of spontaneous emission per unit time is A (Einstein coefficient) If there are N2 atoms excited per volume, then at later time t we will have less, or 2 2 2 spsp dN N AN dt           2 2N A t N    In differential form this becomes a rate equation of the form where A is the rate constant for spontaneous emission and sp is the life time for spontaneous emission given by sp=1/A 33 Rate equations: spontaneous emission Suppose you can bring atoms in the excited state by some energy input look at time dependence of N2 after the energy input is turned off at t=0: ( ) ( ) /2 2 2 2 0 spt spsp dN N N t N e dt            Note that the N2 drops to 1/e of its original value when t=sp. We have solved our first rate equation to calculate the time dependent concentration of excited atoms 6/14/201534 Rate equations: spontaneous emission
  • 18. 14/06/2015 18 2 2 21 ( ) st dN N B dt r         B21 is the Einstein coefficient for stimulated emission Under monochromatic illumination at frequency  we can write this as 2 2 21( ) st IdN N dt h            Rate equations – Stimulated emission More complex situations, add more processes to the rate equations Scales with the electromagnetic spectral energy density r() where r()d  is the energy per unit volume in the frequency range {,  +d } Process 2: Stimulated emission with I /(h ) is the photon flux given and 21() the cross section for stimulated emission Note that ()I /(h ) is the rate constant for stimulated emission (units again s-1) Process 3: Absorption for absorption (‘stimulated absorption’) we obtain a similar rate equation: 2 1 12 1 12( ) ( ) abs IdN N B N dt h  r            with B12 the Einstein coefficient for absorption and 12 the absorption cross section 6/14/201536 Rate equations – Absorption
  • 19. 14/06/2015 19 We now have the rate equations describing the population of levels 1 and 2 Population is the ‘amount of occupation’ of the different energy levels 2 1 12 2 21 2( ) ( ) dN N B N B AN dt r  r    1 1 12 2 21 2( ) ( ) dN N B N B AN dt r  r     Since we have only two states, we find (atoms that leave state 2 must end up in state 1) N1 and N2 should add up to the total amount of atoms: N1+N2=N We can now solve the time dependent population N2 under illumination Before doing that, let’s look at the relations between A, B12, and B21 dt dN dt dN 21  Rate equations for two-level system Hypothetical situation: closed system at temperature T with collection of two-level atoms and no external illumination: All processes together will result in a thermal equilibrium with a population distribution described by a Boltzmann factor: /2 1 Bh k TN e N   In equilibrium, on average 021  dt dN dt dN This implies that in equilibrium 2 12 1 21 ( ) ( ) B h k TN B e N A B  r  r      1 12 2 21 2( ) ( ) 0N B N B ANr  r    Resulting in an equation relating the Einstein coefficients to the thermal distribution: 6/14/201538 Einstein coefficients in thermal equilibrium
  • 20. 14/06/2015 20 2 12 12 12 12 21 ( ) 1 21 21 21 ( ) ( ) 1 ( ) ( ) B h k T T N B B B e B B B N A B A B B  r  r  r  r             1 2 2 2 12 1 21 2 1 2 1 ( ) 1N N AN NA A B N B N B N N B r        1 ( ) 1 h k TB A B e r    Conversely, we can derive the ‘emission spectrum’ from our two-level atom Substituting the thermal distribution over the available energy levels we obtain which looks very similar to the Planck blackbody radiation formula : 3 3 3 8 1 ( ) 1 h k TB hn c e    r    3 3 3 3 3 8 8A hn hn B c      implying At high temperatures e-h /k B T →1, and the radiation density becomes large: Ratio between rates of spontaneous and stimulated emission 21 21 21 21 1 ( ) h k TB A A e W B  r      
  • 21. 14/06/2015 21 HW. 1 Compare the rates of spontaneous and stimulated emission at room temperature (T = 300K) for an atomic transition where the frequency associated with the transition is about 3 × 1010 Hz, which is in the microwave region (KB=1.38 × 1023 J/K) (b) What will be the wavelength of the line spectrum resulting from the transition of an electron from an energy level of 40 × 10−20 J to a level of 15 × 10−20 J? HW. 2 Consider the energy levels E1 and E2 of a two-level system. Determine the population ratio of the two levels if they are in thermal equilibrium at room temperature, 27◦C, and the transition frequency associated with this system is at 1015 Hz HW. 3 The oscillating wavelengths of the He–Ne, Nd:YAG, andCO2 lasers are 0.6328, 1.06, and 10.6µ m, respectively. Determine the corresponding oscillating frequencies. What energy is associated with each transition?