4 18
4.4 Rate of Absorption and Stimulated Emission
The rate of absorption induced by the field is
π 2 2 w (ω) = E (ω) k ∈ˆ ⋅µ δ (ω −ω) (4.56) k 22 0 k
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
c S = (E× B) (4.57) 4π
c c I = S = E2 = E2 (4.58) 4π 8π 0
Another representation of the amplitude of the field is the energy density
I 1 U = = E2 (for a monochromatic field) (4.59) c 8π 0
Using this we can write
2 4π 2 w = U(ω) k ∈ˆ ⋅µ δ(ω − ω) (4.60) k 2 k
1 2 or for an isotropic field where E ⋅ xˆ = E ⋅ yˆ = E ⋅ zˆ = E 0 0 0 3 0
2 4π 2 w = U(ω) µ δ(ω − ω) (4.61) k 32 k k
or more commonly
w k = Bk U(ωk ) (4.62)
2 4π 2 B = µ Einstein B coefficient (4.63) k 32 k
4-19
2 2 (this is sometimes written as Bk = (2π 3 ) µk 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
E2 ω3 Nω = 0 U = N (4.64) 8π π2c3
B is independent of the properties of the field. It can be related to the absorption cross-section, σA.
total energy absorbed / unit time σ = A total incident intensity (energy / unit time / area) ω ⋅ w ω ⋅ B U(ω ) = k = k k I c U (ωk ) (4.65)
ω σ = B A c k
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 σ dx + N σ dx (4.67) I n A m SE
dI = − (N − N )σ dx (4.68) I n m a
Nn , Nm 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 (4.69) Z
∆N=Nn − Nm is the thermal population difference between states.
Integrating over a pathlength L:
I = e−∆NσaL for high freq. ∆N ≈ N (4.70) I0
−NσaL −3 2 ≈ e N : cm σn : cm L :cm
or written as Beer’s Law:
I A = −log = C ∈ L C : mol / liter ∈:liter / mol cm (4.71) I0
∈ = 2303 N σA 4 21
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.
−ωmn /kT Nm / Nn = e (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
? Nm Bnm U (ωmn ) = Nn Bmn U (ωmn ) (4.73)
but clearly this can’t hold for finite temperature, where Nm < Nn , so there must be another type of emission independent of the field.
So we write
Wnm = Wmn (4.74)
Nm (Anm + Bnm U(ωmn )) = Nn Bmn U(ωmn ) 4-22
If we substitute the Boltzmann equation into this and use Bmn = Bnm , we can solve for Anm :
ωmn /kT Anm = Bnm U (ωmn )(e − 1) (4.75)
For the energy density we will use Planck’s blackbody radiation distribution:
ω3 1 U(ω) = (4.76) π2c3 ωmn /kT −1 e U ω Nω
Uω is the energy density per photon of frequency ω.
Nω is the mean number of photons at a frequency ω.
ω3 ∴ A = B Einstein A coefficient (4.77) nm π2 c3 nm
The total rate of emission from the excited state is
ω3 w = B U(ω ) + A using U (ω ) = N (4.78) nm nm nm nm nm π2 C3
ω3 = B (N +1) (4.79) π2 c3 nm
Notice, even when the field vanishes ( N → 0) , we still have emission. Remember, for the semiclassical treatment, the total rate of stimulated emission was
ω3 w = B (N) (4.80) nm π2 c3 nm
If we use the statistical analysis to calculate rates of absorption we have
ω3 w = B N (4.81) mn π2 c3 mn
The A coefficient gives the rate of emission in the absence of a field, and thus is the inverse of the radiative lifetime: 1 τ = (4.82) rad A