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Intensities and Distributions
We want to calculate what is expected as far as the excitation of vibrational states and
spontaneous vs. stimulated emission.
Before we leave the Einstein coefficients, let’s see what we can do with them.
Go back to Planck’s Law of radiation:
Remember that the density of modes (light) with their frequency
between ω and ω + d ω is:
( ) ( )
2 2 3
ρ ω d ω =ω d ω π/ c
At thermal equilibrium the probability that the mode oscillator is thermally excited to its
n
th state is:
[ ]
[ ]
n
n
exp -E / kT P( ) exp -E / kT n
n = ∑
n
E n 2
= (zero point energy cancels)
Define u = exp -[ = ω/ kT]
( )
n n
n
n 0
u P( ) u 1 u
u
n ∞
=
= ⎢^ ⎥= −
∑
The mean number of photons excited in the field mode at temperature T is:
( )
n n nP( ) 1 u nu n n
< >= (^) ∑ n = − ∑
( )
n u n 1 u u u u n 1 u
∑
[ ]
u 1 n
1 u exp ω/ kT 1
This is the “Planck thermal excitation function.”
The mean energy density of the radiation in the modes of frequency ω and temperature T
is:
< WT ( ω ) > d ω=< n> =ωρ ω ( ) d ω
[ ]
3
2 3
c exp / kT 1
T
d W d
< WT ( ω )> d ω= energy / unit volume / unit angular frequency range
Planck’s Law of Blackbody Radiation
We can rewrite the expression (^) N
mean # of photons
u n 1 u
to get
n u 1 n
Remember (^) ( )
n P( ) n = 1 −u u
So we have:
( )
n
1 n
n P( ) 1 n
n (^) +
n = 0 has the largest probability.
Characterize a probability distribution by its factorial moments. The r
th factorial moment
is:
n n ( 1 )( n (^2) )( n r (^1) ) n n( 1 )( n (^2) )... (^) ( n r 1 P( )) n
< − − − + >= (^) ∑ − − − + n
r = r! < n>
Use this to calculate the fluctuations:
( ) ( ) ( )
(^2 22 2 ) n n n P( ) n n ; but n n 1 2 n n
Δ = (^) ∑ − < > n =< > − < > < − >= < >
So (^) ( ) ( )
(^2 ) Δn = < n > + < n>
Fluctuation: ( )
1/ 2 2 Δn = < n > + < n>
>> 1 Δn ∼
We can only measure this with perfect detectors and instantaneous measurements.
Let’s look at the characteristics of the Einstein coefficients.
- Note, because
n
exp 1 kT
; g B 1 12 =g B 2 21
- And
21 T 1 12 21 2
A
<W ( )
N
B B
N
; and
1
2
N
exp N kT
We have: <W (T ω ) >= A 21 < n > / B 21 Thermal energy density
So, the sum of the emission rates is:
B 21 < W (T ω ) > +A 21 = A (^21) ( < n > + (^1) ) < W (T ω) > B 21 / A 21 =< n>
Note we have quantized the field using a back door!
Now: using 21 T
21
B W ( )
n A
When the two rates are equal, (^) ( ( B 21 < W (T ω ) >) = A 21 ), then < n >≈ 1. However,
because we know that
n
exp 1 kT
, when < n >≈ 1 , then 0. kT
This happens at room temperature (in a cavity) at
13 -
ω = 3 × 10 s (this is far infrared)
When = ω <<kT; A 21 << (^) ( B 21 < W (T ω)>)
When = ω >>kT; A 21 >> (^) ( B 21 < W (T ω)>)
At optical frequencies
15 - ω ∼ 3 × 10 s
= ω ∼ 3 × 10 J
For calculations:
23 5
34 19
34 15
k (^) B x J K x eV K
x J s eV x J
h x J s x eV s
− −
− −
− −
Light Beams with External Sources
We have in general (not just at thermal equilibrium)
( )
3 2 3 21 21
= ω / π c B =A ; but at steady state.
Define a “ saturation radiative energy density”.
( )
3 2 3
Ws = = ω / π c
then W Bs 21 =A 21
So, at
15 -
ω ≈ 3 × 10 s
-14 - s
W d ω≅ 10 d ωJm
The intensity of the saturating light source:
3 21 s (^2 ) 21
A
I cW c B c
s
-6 -
I s d ω≅ 3 × 10 d ωWm
If the conventional light source has
10 -
Δ ω ∼ 6 × 10 s
5 - I (^) s 1.8 10 Wm
strongest
d ω ≈ ×
∫ Å Total Beam Intensity!
All the strongest conventional sources are < 1/10 of this intensity. So all molecules
decay by 21 A for conventional sources!!
Directions of the 3 Einstein processes:
So, the spontaneous emission, A 21 , is what leads to the absorption of the incoming
intensity.
Mean rates of the Einstein transitions for steady state:
( )
( )
N B 1 NA
s
s s
W W W
W
W W W
absorption
( )
( )
N B 2 NA
s
s s
W W W
W
W W W
stimulated emission
( )
N A 2 NA
s^2
W
W W
spontaneous emission
So, (^2) 2
N
N A A
at full saturation. Spontaneous emission cannot happen faster than a
particular limit. Stimulated emission can increase without limit.
( )
1 2 2 2 1
N N
- N A N N B W
t t
The general time dependent solution of the equation above:
N = N 1 +N 2
N ( ) 1 N 1 ( 0 ) N exp - A( 2B (^) ) N 2 2
s s
s s
W W W W
t W t W W W W
N ( ) from N 2 t (^) 2 = N −N 1
If all N molecules are in the ground state at t = 0; light turned on at t = 0, and stays on:
N ( ) 2 N { 1 exp - A ( 2B ) }
s^2
W
t W t W W
If the light turned off after steady state,
ss -(A)t N ( ) 2 t = N 2 e. This is only for the two state
condition.
So, note that the time constant at the beginning is:
Light on
Light off
A 2B
A
exc
em
With 2-state you always have N / N 2 < 0.5 No inversion.
The three Einstein radiative transitions
