012334563789:9;<<= 4>18

4.4 Rate of Absorption and Stimulated Emission The rate of absorption induced by the field is

( ) ( ) ( )22

022k k

ˆw E k!

" " µ # " "= $% &

(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

( )cS E B

4= '

! (4.57)

2 2

0

c cI S E E

4 8= = =

! ! (4.58)

Another representation of the amplitude of the field is the energy density

2

0

I 1U E

c 8= =

! (for a monochromatic field) (4.59)

Using this we can write

( ) ( )2

2

k k2

4ˆw U k

!= " $%µ # " & "

(4.60)

or for an isotropic field where 2

0 0 0 0

1ˆ ˆ ˆE x E y E z E

3% = % = % =

( ) ( )2

2

k k k2

4w U

3

!= " µ # " & "

(4.61)

or more commonly

( )k k kw B U= " (4.62)

2

2

k k2

4B

3

!= µ

Einstein B coefficient (4.63)

4-19

(this is sometimes written as ( ) 22

k kB 2 3= ! µ when the energy density is in ().

U can also be written in a quantum form, by writing it in terms of the number of photons N

2 30

2 3

EN U N

8 c

"" = =

! !

(4.64)

B is independent of the properties of the field. It can be related to the absorption cross-section, )A.

( )( )

( )

A

k kk

k

A k

total energy absorbed / unit time

total incident intensity energy / unit time / area

B Uw

I c U

Bc

) =

" % "" %= =

"

") =

(4.65)

More generally you may have a frequency dependent absorption coefficient

( ) ( ) ( )A k kB B g) " * " = " where g(") 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 :

( ) ( ) ( ) ( )since

nm mn

nm nm nm nm nm mn

nm mn

w w

B U B U U U

B B

" " " "

=

= =

=

(4.66)

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, ( ) ( )A SEnm 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.

n A m SE

dIN dx N dx

I= & ) + ) (4.67)

( )n m a

dIN N dx

I= & & ) (4.68)

n mN , N These are population of the upper and lower states, but expressed as a

population densities. If N is the molecule density,

n

n

E

N NZ

e&+ =

(4.69)

n mN=N N, & is the thermal population difference between states.

Integrating over a pathlength L:

0

aN LI

Ie&, )= for high freq. N N, - (4.70)

aN Le& )- 3 2

nN : cm : cm L :cm& )

or written as Beer’s Law:

0

IA log C L

I= & = $ C : mol / liter :liter / mol cm$ (4.71)

A2303 N$ = )

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4.5 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 m nE E> .

The Boltzmann distribution gives us the number of molecules in each state.

// mn kT

m nN N e "&= (4.72)

For the system to be at equilibrium, the time-averaged transitions up mnW must equal those down

nmW . In the presence of a field, we would want to write for an ensemble

( ) ( )m nm mn n mn mnN B U N B U?

" "= (4.73)

but clearly this can’t hold for finite temperature, where m nN N< , so there must be another type of

emission independent of the field. So we write

( )( ) ( )

nm mn

m nm nm mn n mn mn

W W

N A B U N B U

=

+ " = "

(4.74)

4-22

If we substitute the Boltzmann equation into this and use mn nmB B= , we can solve for nmA :

( )( )/ 1mn kT

nm nm mnA B U e ""= & (4.75)

For the energy density we will use Planck’s blackbody radiation distribution:

( ) mn

3

/kT2 3

NU

1U

c 1e "

""

"" =

! &

(4.76)

U" is the energy density per photon of frequency ".

N" is the mean number of photons at a frequency ".

3

nm nm2 3A B

c

". =

!

Einstein A coefficient (4.77)

The total rate of emission from the excited state is

( )nm nm nm nmw B U A= " + ( )3

nm 2 3using U N

C

"" =

!

(4.78)

( )3

nm2 3B N 1

c

"= +

!

(4.79)

Notice, even when the field vanishes ( )0N / , we still have emission.

Remember, for the semiclassical treatment, the total rate of stimulated emission was

( )3

nm nm2 3w B N

c

"=

!

(4.80)

If we use the statistical analysis to calculate rates of absorption we have

3

mn mn2 3w B N

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:

rad

1

A0 = (4.82)