012334563789:9;<<=
4.4
4>18
Rate of Absorption and Stimulated Emission
The rate of absorption induced by the field is
wk  (ω ) =
π
2
2
E0 ( ω )
2
ˆ µ 
k ∈⋅
2
δ ( ωk  − ω )
(4.56)
The rate is clearly dependent on the strength of the field. The variable that you can most easily
measure is the intensity I (energy flux through a unit area), which is the time-averaged value of the
Poynting vector, S
S=
c
( E × B)
4π
I= S =
(4.57)
c
c 2
E2 =
E0
4π
8π
(4.58)
Another representation of the amplitude of the field is the energy density
U=
I 1 2
=
E0
c 8π
(for a monochromatic field)
(4.59)
Using this we can write
4π2
ˆ 
w k = 2 U ( ω) k ∈⋅µ

2
δ ( ωk − ω)
or for an isotropic field where E 0 ⋅ xˆ = E 0 ⋅ yˆ = E 0 ⋅ zˆ =
(4.60)
1
E0
3
2
4π2
2
w k = 2 U ( ω) µ k δ ( ωk − ω)
3
(4.61)
or more commonly
w k = Bk U ( ωk )
4π2
B k = 2 µ k
3
2
(4.62)
Einstein B coefficient
(4.63)
4-19
(this is sometimes written as Bk = ( 2π 3 2 ) µ k
2
when the energy density is in ν).
U can also be written in a quantum form, by writing it in terms of the number of photons N
Nω =
E 02
8π
U=N
 ω3
π 2 c3
(4.64)
B is independent of the properties of the field. It can be related to the absorption cross-section, σA.
σA =
=
σA =
total energy absorbed / unit time
total incident intensity ( energy / unit time / area )
ω ⋅ w k ω ⋅ Bk U ( ωk )
=
I
c U ( ωk )
(4.65)
ω
B k
c
More generally you may have a frequency dependent absorption coefficient
σA ( ω) ∝ Bk ( ω) = Bk g ( ω) where g(ω) is a lineshape function.
The golden rule rate for absorption also gives the same rate for stimulated emission. We find for
two levels m and n :
wnm = wmn
Bnm U (ωnm ) = Bnm U (ωnm )
since U (ωnm ) = U ( ωmn )
(4.66)
Bnm = Bmn
The absorption probability per unit time equals the stimulated emission probability per unit time.
Also, the cross-section for absorption is equal to an equivalent cross-section for stimulated
emission, ( σA )nm = ( σSE ) mn .
4-20
Now let’s calculate the change in the intensity of incident light, due to absorption/stimulated
emission passing through sample (length L) where the levels are thermally populated.
dI
= − N n σA dx + N m σSE dx
I
(4.67)
dI
= − ( N n − N m ) σa dx
I
(4.68)
N n , N m These are population of the upper and lower states, but expressed as a
population densities. If N is the molecule density,
 e−βEn 
Nn = N 

 Z 
(4.69)
∆N=N n − N m is the thermal population difference between states.
Integrating over a pathlength L:
I
= e−∆Nσa L
I0
≈ e− Nσa L
for high freq. ∆N ≈ N
N : cm −3
σn : cm 2
(4.70)
L :cm
or written as Beer’s Law:
A = − log
I
=C∈L
I0
C : mol / liter
∈ = 2303 N σA
∈:liter / mol cm (4.71)
0123345637?9@A9;<<?
4.5
4>21
SPONTANEOUS EMISSION
What doesn’t come naturally out of semi-classical treatments is spontaneous emission—transitions
when the field isn’t present.
To treat it properly requires a quantum mechanical treatment of the field, where energy is
conserved, such that annihilation of a quantum leads to creation of a photon with the same energy.
We need to treat the particles and photons both as quantized objects.
You can deduce the rates for spontaneous emission from statistical arguments (Einstein).
For a sample with a large number of molecules, we will consider transitions between two states
m and n with Em > En .
The Boltzmann distribution gives us the number of molecules in each state.
N m / N n = e − ωmn / kT
(4.72)
For the system to be at equilibrium, the time-averaged transitions up Wmn must equal those down
Wnm . In the presence of a field, we would want to write for an ensemble
?
N m Bnm U (ωmn ) = N n Bmn U (ωmn )
(4.73)
but clearly this can’t hold for finite temperature, where N m < N n , so there must be another type of
emission independent of the field.
So we write
Wnm = Wmn
(4.74)
N m ( A nm + Bnm U ( ωmn ) ) = N n Bmn U ( ωmn )
4-22
If we substitute the Boltzmann equation into this and use Bmn = Bnm , we can solve for Anm :
(
)
Anm = Bnm U ( ωmn ) e ωmn / kT − 1
(4.75)
For the energy density we will use Planck’s blackbody radiation distribution:
U ( ω) =
ω3
1
ωmn /kT
2 3
πc e
−1


Uω
Nω
(4.76)
U ω is the energy density per photon of frequency ω.
N ω is the mean number of photons at a frequency ω.
∴ A nm =
ω3
Bnm
π 2 c3
Einstein A coefficient
(4.77)
ω3
using U ( ωnm ) = N 2 3
π C
(4.78)
The total rate of emission from the excited state is
w nm = Bnm U ( ωnm ) + A nm
=
 ω3
Bnm ( N + 1)
π 2 c3
(4.79)
Notice, even when the field vanishes ( N → 0 ) , we still have emission.
Remember, for the semiclassical treatment, the total rate of stimulated emission was
w nm =
ω3
Bnm ( N )
π 2 c3
(4.80)
If we use the statistical analysis to calculate rates of absorption we have
w mn
ω3
= 2 3 Bmn N
π c
(4.81)
The A coefficient gives the rate of emission in the absence of a field, and thus is the inverse of the
radiative lifetime:
1
τrad =
(4.82)
A