Transcript Bound – Bound Transitions

Bound – Bound Transitions
Einstein Relation for BoundBound Transitions
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Lower State i : gi = statistical weight
Upper State j : gj = statistical weight
For a bound-bound transition there are three
direct processes to consider:
#1: Direct absorption: upward i → j
#2: Return Possibility 1: Spontaneous transition
with emission of a photon
#3: Return Possibility 2: Emission induced by
the radiation field
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The Absorption Process
Possibility 1: Upward i → j
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Ni(ν) Rij dω/4π = Ni(ν) Bij Iν dω/4π
Rij is the rate at which the transition occurs
Ni(ν) is the number of absorbers cm-3 in state i
which can absorb in the range (ν, ν + dν)
The equation defines the coefficient Bij
Transitions are not infinitely sharp so there is a
spread of frequencies φν (the absorption profile)


0
 d  1
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Let Us Absorb Some More
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If the total number of atoms is Ni then the number of
atoms capable of absorbing at frequency ν is Ni(ν) = Niφν
In going from i to j the atom absorbs photon of energy hνij
= Ej - Ei
The rate that energy is removed from the beam is:
 a I  N i Bij

h ij
4
 I
aν is a macroscopic absorption coefficient such as Κν
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The Return Process: j → i
Going Down Once
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Spontaneous transition with the emission
of a photon
Probability of a spontaneous emission per
unit time ≡ Aij
Emission energy rate:
ρjν (spontaneous) = Nj Aji (hνij/4π) ψν
The emission profile is normalized:
∫ψνdν = 1
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The Return Process: j → i
Going Down Twice
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Emission induced by the radiation field
ρjν(induced) = Nj Bji ψν Iν (hνij/4π)
Note that the induced emission profile has
the same form as the spontaneous
emission profile.
Spontaneous emission takes place
isotropically.
Induced emission has the same angular
distribution as Iν
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Assume TE
This gives us Iν = Bν and the ratio of the state populations!
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Iν = Bν and the Boltzman Equation
N *j
N

*
i

gj
gi
 h ij
e
kT
Ni* Bij Bν = Nj* Aji + Nj* Bji Bν
()
– Upward = Downward
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What did we do to get ()?
– Integrate over ν and assume that Bν ≠ f(ν) over a line width. The
two pieces that depend on ν are Bν (assumed a constant) and
(φν,ψν) but the latter two integrate to 1!
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Now we do Some Math
Solution of () for Bν
Boltzman Equation
Substitute for Nj* and
note that Ni* falls out!
B 
N *j Aji
N i* Bij  N *j B ji
N *j 
gj
gi
 h ij
N i*e
kT
h ij
g j * kT
N i e Aji
gi
B 
 h ij
g
N i* Bij  j N i*e kT B ji
gi
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Onward!
Use this to multiply the
top and bottom of the
previous equation
But this is just the
Planck function so
h ij
gi 1 kT
e
g j B ji
B 
Bound Bound Transitions
Aji
B ji
 gi Bij hkT ij

e  1


g
B
 j ji

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A Revelation!
The Intersection of Thermodynamics and QM
B 
Aji
B ji
 gi Bij hkT ij
e

 g j B ji
2h 3 1
 2 h
c

e kT  1
 1


Aji 2h 3
 2
B ji
c
 g j Bij 

  1
 gi B ji 
2h 3
Aji  2 B ji
c
g j Bij  gi B ji
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What Does this Mean?
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We have used thermodynamic arguments but
these equations must be related only to the
properties of the atoms and be independent of
the radiation field
==> These are perfectly general equations
==> These arguments predicted the existence of
stimulated emission.
NB: Stimulated (induced) emission is not
intuitively obvious. However, it is now an
observed fact.
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The Equation of Transfer
Note the Following:
 dI
 j   I
 dz
h ij
 a I  N i Bij
 I
4
h ij
 j  N j Aji

4
h ij
 j  N j B ji
  I
4
h ij
h ij
h ij
 dI
 N j Aji
   N j B ji
  I  N i Bij
 I
dz
4
4
4
 dI h ij
 N j Aji   ( N i Bij  N j B ji  ) I 

dz
4
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The Coefficients Are:
Line Absorption Coefficient usually denoted lν
Departure Coefficient
N i (T , N e , J )
bi 
N i* (T , N e )
 N j B ji 
l 
N i Bij 1 
4
N i Bij

h ij
N j b j N *j b j g j kT


e
*
Ni bi N i
bi gi
h ij



h ij
 g j B ji  b j kT

l 
N i Bij 1 
e

4
gi Bij bi


Gather the terms on I
Boltzmann Equation
h ij
h ij
   b j kT

h ij
l 
N i Bij 1 
e

4

b
BoundBound Transitions

 i

giBij = gjBji
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The Line Source Function
Source Function = j/
S 
N j Aji 
Divide by NjBjiψν
N i Bij  N j B ji 
Aji
B ji

N i Bij
1
N j B ji 
2h 3
1
 2
c bi hkT ij
e 1
b j 
 dI

dz
 I  S
Use relations between A & B’s
The equation of transfer for a line!
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Let Us Simplify Some
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Assume φν = ψν (Not unreasonable)
If we assume LTE then bi = bj = 1
Then lν = NiBijφν(1-e-hν/kT)
– The term (1-e-hν/kT) is the correction for
stimulated emission
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Sν = Bν
Bound Bound Transitions
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An Alternate Definition
With Respect to the Energy Density
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Define Bi,j as
uνBij = Bij 8π/c3 (hν3/(ehν/kT-1))
– Bi,j is the probability that an atom in lower level i
will be excited to upper level j by absorption of a
photon of frequency ν = (Ej - Ei) / h
– This B is defined with respect to the energy density U
of the radiation field, not Iν (=Bν) as previously.
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For this definition
– Aji = (8πhν3/c3) Bji
– giBij = gjBji
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Now Let Us Try To Determine Aji,
Bij, and Bji
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Classical Oscillator and EM Field
– Dimensionally correct but can be off by orders of
magnitude
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QM Atom and Classical EM Field
– Correct Expression for Bij
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QM Atom and a Quantized EM Field
– Gives Bij, Bji, and Aji
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However, note that Bij, Bji, and Aji are
interrelated and if you know one you know them
all!
Bound Bound Transitions
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A Classical Approach
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The probability of an event generally depends on
the number of ways that it can happen. Suppose
photons in the range (ν, ν + dν) are involved.
The statistical weight of a free particle: Position
(x, x+dx) and momentum (p, p+dp) is dN/h3 =
dxdydzdpxdpydpz/h3
Then the statistical weight per unit volume of a
particle with total angular momentum p, in
direction dΩ, is p2dpdΩ/h3.
Bound Bound Transitions
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Free Particles
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Momentum of a photon is p = hν/c so the
statistical weight per unit volume of
photons in (ν, ν+dν) is ν2dνdΩ/hc2.
This means high frequency transitions
have higher transition probabilities than
low frequency transitions.
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Semiclassical Treatment of an
Excited Atom
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Oscillating Dipole
(electric) in which the
electron oscillates
about the nucleus.
For the case of no
energy loss the
equation of motion is:
d 2r
m 2  4 2 02 r
dt
– ν0 is the frequency of
the oscillation
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The electric dipole
radiates (classically)
and energy is lost due
to the radiation
2
2
2e d r
I
3 c 3 dt 2
Bound Bound Transitions
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A Damped Oscillator
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The energy loss leads to a further term in the
equation of motion which slows the electron
2
3
2e d r
F
3
3
3 c dt
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The damping force is
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If F is small then the motion can be considered to be
near a simple harmonic r = r0 cos (2πν0t)
r  4 02 r 2 r
2
2
e
mr  4 02 r 2 r  4 2 02
r
3
3c
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The Solution is
Damped Harmonic Oscillator
r  r0e
 t
2
cos(2 0t )
8 e 2
22
2
 
  2.47(10 ) 0
2 0
3me c
2 2

γ is called the classical damping constant
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Electric Field
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The field set up by the electron is proportional
to the electron displacement
E (t )  E0e

 t
2
cos(2 0t )
At time t = (γ/2)-1 we get E(t) = E0/e which
characterizes the time scale of the decay
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A Result!
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The decay time must be proportional to Aji -the
probability of a spontaneous transition
downwards
Aclassical = γ
– For Hα, = 6563Å; ν0 = c/λ so γ  5(107) s-1
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The field decays exponentially so the frequency
is not monochromatic. To get the frequency
dependence do a Fourier analysis
The square of the spectrum is the intensity of the
field!
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The Broadening Profile
Lorentz Profile

I ( )  I 0
2
 
2
2
4 (   0 )   
2   2
2
 I0

4 2
  
(   0 )  

 4 
2
2
I
I0
 I0
2
  
(   0 )  

 4 
2
2
2
  
(   0 ) 2  


4 2
 4 
2
2   
(   0 ) 


2
4
 4 
2
2

4 2
  
(   0 )  

 4 

2
2
Bound Bound Transitions

(   0 ) 
4

FWHM 
2
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More On Broadening
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In QM the natural broadening arises due to
the uncertainty principle
The uncertainty in the time in which an
atom is in a state is its lifetime in that
state.
Average lifetime A-1
Thus the uncertainty E = ht ≈ Ah
This means that ν  A  γ
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Oscillator Strengths
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Since γ is of the same order as A we
usually express exact values of A in terms
of the classical value of γ.
Oscillator Strength nupper → mlower
Anm ≡ 3 (gm/gn) fnm γ
= (gm/gn) (8π2e2ν2/mec3) fnm
= (gm/gn) 7.42(10-22) ν2 fnm
Bmn = (πe2/mehν) fnm
Bound Bound Transitions
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Oscillator Strengths
Balmer Series of H
f nm 
3
2
1  1
1  GI

 2
2 
3 3
3 3 gm  m n  n m
6
g m  2m 2
f nm
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
3 3
3
1  GI
 1
 2  2 3 5
m n  n m
This is known as Kramer’s
Formula. GI is the Gaunt
Factor which is order 0.1 - 10
Bound Bound Transitions
Line U-L
f
Hα
3-2
0.641
Hβ
4-2
0.119
Hγ
5-2
0.044
Hδ
6-2
0.021
Hε
7-2
0.012
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The Absorption Coefficient
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We are now ready to determine the absorption
coefficient aν per atom of an absorption line
centered at ν0.
It is defined so that the probability of absorption
per unit path length of a photon is Naν where N
is the number density of atoms capable of
absorption.
Consider a beam of intensity Iν traveling across a
unit cross-sectional area. In time dt the photons
have traveled cdt.
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Energy Considerations
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The energy removed from the beam in (ν, ν +
dν) is Iν naν cdt. Alternately, the number of
transitions is Iν naν cdt / hν. The volume
occupied by the photons is cdt (unit cross
sectional area).
The number of transitions per unit volume per
unit time per unit frequency due to the beam is Iν
naν / hν.
Integrating over solid angle and assuming
thermal equilibrium (4πIν = cUν) the number of
absorptions is cUν naν / hν.
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Onward

We now integrate over frequency over the line
profile only to obtain the number of transitions
from lower state to upper state per unit volume
per unit time
Transition Rate  cN 

0

U
a d
h
But this is equal to Uν Bnm N so
U Bnm N  cN 

0
U
a d
h
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Line Absorption
The nature of line absorption
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Uν does not
vary across
the line
aν is small
except near
ν0: ν0 = (EnEm) / h
Uν/hν varies
slowly across
the line
cN 

0
U 0
U
a d  cN
h
h 0
U Bnm N  cN
For ν = ν0
U 0
h 0




0
0
a d
a d
 e2
c 
f nm 
a d

me h
h 0 0


0
a d 
Bound Bound Transitions
 e2
me c
f nm
cm2 Hz
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We are about done


0

a d 
mec
f nm
The integral of aν over ν can be thought of as the
total absorption cross section per atom initially
in the lower state. We rewrite aν as
a 

 e2
e
2
mec
f nm
φν is the absorption profile and we expect it to
follow the Lorentz profile
Bound Bound Transitions
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Combine the Profile
a  a0
So
a 


4 2
  
(   0 )2  

 4 
e
2
me c
f nm
2

4 2
  
(   0 ) 2  

 4 
2
In principal fnm ~ γ but there are other broadening
mechanisms.
– Natural width (γ) is small compared to other mechanisms
(Stark, Doppler, van der Waals, etc)
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For ν = ν0: a0 = (πe2/mec) (fnm/γ)
Bound Bound Transitions
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Stimulated Emission
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We have yet to account for stimulated emission
We usually assume emission is isotropic but if it goes as
BmnIν then it correlates with Iν
We treat this as a negative absorption and reduce the
value of aν by the appropriate factor.
In our discussion of the Einstein coefficients we found
the factor to be (1 - e-(hν/kT)) so the corrected aν is
a 
e
2
mec
f nm (1  e
Bound Bound Transitions
 h
kT
)
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