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012334563789:;9<8
4-23
4.6
Quantized Radiation Field
Background
Our treatment of the vector potential has drawn on the monochromatic plane-wave solution to the
wave-equation for A. The quantum treatment of light as a particle describes the energy of the light
source as proportional to the frequency ω , and the photon of this frequency is associated with a
cavity mode with wavevector k = ω / c that describes the number of oscillations that the wave can
make in a cube with length L. For a very large cavity you have a continuous range of allowed k.
The cavity is important for considering the energy density of a light field, since the
electromagnetic field energy per unit volume will clearly depend on the wavelength λ = 2π/|k| of
the light.
Boltzmann used a description of the light radiated from a blackbody source of finite volume at
constant temperature in terms of a superposition of cavity modes to come up with the statistics for
photons. The classical treatment of this problem says that the energy density (modes per unit
volume) increases rapidly with increasing wavelength. For an equilibrium body, the energy
absorbed has to equal the energy radiated, but clearly as frequency increases, the energy of the
radiated light should diverge. Boltzmann used the detailed balance condition to show that the
particles that made up light must obey Bose-Einstein statistics. That is the equilibrium probability
of finding a photon in a particular cavity mode is given by
f ( ω) =
1
e
ω/ kT
−1
(4.83)
From our perspective (in retrospect), this should be expected, because the quantum treatment of
any particle has to follow either Bose-Einstein statistics or Fermi-Dirac statistics, and clearly light
energy is something that we want to be able to increase arbitrarily. That is, we want to be able to
add mode and more photons into a given cavity mode. By summing over the number of cavity
modes in a cubical box (using periodic boundary conditions) we can determine that the density of
cavity modes (a photon density of states),
ω2
g ( ω) = 2 3
(4.84)
πc
Using the energy of a photon, the energy density per mode is
ω3
ω g ( ω ) = 2 3
πc
(4.85)
and so the probability distribution that describes the quantum frequency dependent energy density
is
u ( ω ) = ω g ( ω ) f ( ω ) =
ω3
1
2 3 ω /kT
πc e
−1
(4.86)
4-24
The Quantum Vector Potential
So, for a quantized field, the field will be described by a photon number N kj , which represents the
number of photons in a particular mode ( k , j ) with frequency ω = ck in a cavity of volume v. For
light of a particular frequency, the energy of the light will be N kjω . So, the state of the
electromagnetic field can be written:
ϕEM = N k , j , N k
1 1
2 , j2
, Nk
3 , j3
,
(4.87)
If my matter absorbs a photon from mode k2 , then the state of my system would be
ϕ′EM = N k , j , N k
1
2 , j2
− 1, N k , j , 
3
(4.88)
What I want to do is to write a quantum mechanical Hamiltonian that includes both the matter and
the field, and then use first order perturbation theory to tell me about the rates of absorption and
stimulated emission. So, I am going to partition my Hamiltonian as a sum of a contribution from
the matter and the field:
H 0 = H EM + H M
(4.89)
If the matter is described by ϕM , then the total state of the E.M. field and matter can be
expressed as product states:
ϕ = ϕEM ϕM
(4.90)
E = E EM + E M
(4.91)
And we have eigenenergies
4-25
Now, if I am watching transitions from an initial state  to a final state k , then I can express
the initial and final states as:
matter
field
ϕI = ; N1 , N 2 , N 3  , N i , 
ϕF = k; N1 , N 2 , N3 ,  , Ni ± 1, 
k
Ni

N i −1
 + : emission 


 − : absorption 
(4.92), (4.93)
Where I have abbreviated N i ≡ N ki , ji , the energies of these two states are:
EI = E + ' N j ( ω j )
ω j = ck j
(4.94)
j
EF = Ek + ' N j ( ω j ) ± ωi
j
k
So looking at absorption 
 
states as:

 , we can write the Golden Rule Rate for transitions between

w k =
2π
δ ( E k − E  − ω ) ϕ F V ( t ) ϕ I

2
(4.95)
Now, let’s compare this to the absorption rate in terms of the classical vector potential:
2π
w k =
v 2
'
kj
2
q2
ˆ j ⋅p 
δ ( ωk − ω) 2 A k, j k ∈
m
2
(4.96)
ˆ p ) that acts on the
If these are to be the same, then clearly V ( t ) must have part that looks like (∈⋅
matter, but it will also need another part that acts to lower and raise the photons in the field. Based
on analogy with our electric dipole Hamiltonian, we write:
4-26
V(t) =
−q 1
ˆ + p † ⋅∈
ˆ†
ˆjA
ˆ *j A
pk ⋅∈
'
k
k,j
k,j
m v k,j
(
)
(4.97)
ˆ and A
ˆ † are lowering/raising operators for photons in mode k . These are operators in
where A
k, j
k,j
the field states, whereas pk remains only an operator in the matter states. So, we can write out the
matrix elements of V as
ϕF V ( t ) ϕI = −
q 1
ˆ
k pk ⋅∈
m v
ˆ , N , 
 , N i − 1, A
i
i
(4.98)
=
1
ˆ 
ωk k ∈⋅µ
v
A (i
−)
Comparing with our Golden Rule expression for absorption,
wk  =
π
2
δ ( ωk  − ω )
2
ωk2 2
2
E0 µk 
2
ω
(4.99)
We see that the matrix element
A (i
−)
=
E 02
4vω2
but
E 02
= N ω
8π
(4.100)
=
2π
vω
N
So we can write
 k, j =
2π
a
v ω k, j
(4.101)
†k, j =
2π †
a
v ω k, j
where a, a † are lowering, raising operators. So
4-27
2π
i k ⋅ r −ωt )
− i( k ⋅ r −ωt )
ˆ j a kj e (
∈
+ a †kj e
vω
(
ˆ =
A
'
k, j
)
So what we have here is a system where the light field looks like an infinite number of harmonic
oscillators, one per mode, and the field raises and lowers the number of quanta in the field while
the momentum operator lowers and raises the matter:
H = H EM + H M + V ( t ) = H 0 + V ( t )
H EM = ' ωk a †kj a kj +
(
1
2
)
k,j
HM = '
i
V (t) =
pi2
+ Vi ( r , t )
2mi
−q
A⋅p
m
='
k,j
q 2π
− i k ⋅ r −ωt )

ˆ j ⋅p ) a k, j ei( k ⋅r −ωt ) + a †k, j e (
∈
(

+
m v ωk
= V( −) + V( + )
Let’s look at the matrix elements for absorption (ωk  > 0 )
k, Ni − 1 V(
−)
, N i =
=
−q 2π
ˆ p ) a  , Ni
k, N i − 1 (∈⋅
m vω
−q 2π
m vω
= −i
ˆ p
Ni k ∈⋅
2πω
ˆ k
N i ∈⋅µ
v
4-28
and for stimulated emission (ωk  < 0 )
k, Ni + 1 V (
+)
, N i =
=
−q 2π
ˆ p ) a † , N i
k, Ni + 1 (∈⋅
m vω
−q 2π
m vω
= −i
ˆ pˆ 
N i + 1 k ∈⋅
2πω
ˆ k
Ni + 1 ∈⋅µ
v
We have spontaneous emission! Even if there are no photons in the mode ( N k = 0 ) , you can still
have transitions downward in the matter which creates a photon.
Let’s play this back into the summation-over-modes expression for the rates of
absorption/emission by isotropic field.
w k = , dω
=
2π ω2
+
δ ( ωk − ω) , dΩ ' k, Ni + 1 V ( ) , N i
3
2
2
 ( 2π )
j
2π ω2
 2 ( 2πc )3

( 2πω)( Ni + 1)
averageover polarization
number density per mode
=
4 ( N i + 1) ω3
3c
3
= Bk ( Ni + 1)
µ k
8π
2
µ k
3


2
ω3
π 2 c3

energy density
per mode
So we have the result we deduced before.
2
4-29
Appendix: Rates of Absorption and Stimulated Emission
Here are a couple of more detailed derivations:
Version 1:
Let’s look a little more carefully at the rate of absorption wk  induced by an isotropic, broadband
light source
wk  = , wk  (ω ) ρ E ( ω ) d ω
where, for a monochromatic light source
wk  (ω ) =
π
2
2
E0 (ω )
2
ˆ µ 
k ∈⋅
2
δ ( ωk  − ω )
For a broadband isotropic light source ρ (ω ) dω represents a number density of electromagnetic
modes in a frequency range d ω —this is the number of standing electromagnetic waves in a unit
volume.
For one frequency we wrote:
ˆe
A = A0 ∈
i ( k ⋅r −ωt )
+ c.c.
but more generally:
ˆje
A = ' A 0k ∈
i ( k ⋅ r −ωt )
+ c.c.
k, j
where the sum is over the k modes and j is the polarization component.
By summing over wave vectors for a box of fixed volume, the number density of modes in a
frequency range d ω radiated into a solid angle d Ω is
dN =
1
( 2π )
3
ω2
dω dΩ
c3
and we get ρE by integrating over all Ω
ρ E ( ω ) dω =
1
( 2π )
3
ω2
ω2
d
ω
d
Ω
=
, 2π2 c3 dω
c3
4π
number density at ω
4-30
We can now write the total transition rate between two discrete levels summed over all
frequencies, direction, polarizations
w k = , dω
2
π
1 ω2
E
ω
δ
ω
−
ω
(
)
(
)
0
k
3
2 2
( 2 π ) c3
' , dΩ
2
ˆ j ⋅µ 
k ∈
j

8π
2
µ k
3
2
=
E 0 ( ωk ) ω2
6π  c
2
3
µ k
2
We can write an energy density which is the number density in a range d ω × # of polarization
components × energy density per mode.
rate of energy flow/c
E2
ω2
U ( ωk ) = 2 3 ⋅ 2 ⋅ 0
2π c
8π
w k = Bk U ( ωk )
B k =
4π2
µ k
3 2
2
is the Einstein B coefficient for the rate of absorption
U is the energy density and 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 ( ωk ) = N
 ω3
π 2 c3
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 )
Bnm = Bmn
The absorption probability per unit time equals the stimulated emission probability per unit time.
4-31
Version 2:
Let’s calculate the rate of transitions induced by an isotropic broadband source—we’ll do it a bit
differently this time. The units are cgs.
The power transported through a surface is given by the Poynting vector and depends on k .
S=
c ω2 A 20 ˆ ω2 E 02
c
E×B =
k=
4π
8π
2π
and the energy density for this single mode wave is the time average of S / c .
The vector potential for a single mode is
ˆe
A = A0 ∈
i ( k ⋅r −ω t )
+ c.c.
with ω = ck . More generally any wave can be expressed as a sum over Fourier components of the
wave vector:
ˆj
A = ' Ak j ∈
e
i ( k ⋅ r −ωt )
k, j
V
+ c.c.
The factor of V normalizes for the energy density of the wave—which depends on k .
The interaction Hamiltonian for a single particle is:
V (t ) =
−q
A⋅ ρ
m
or for a collection of particles
V (t ) = − '
i
qi
A ⋅ pi
mi
Now, the momentum depends on the position of particles, and we can express p in terms of an
integral over the distribution of particles:
p = , d3 r p ( r )
p ( r ) = ' pi δ ( r − ri )
i
So if we assume that all particles have the same mass and charge—say electrons:
4-32
V (t ) =
−q
d 3r A ( r , t ) ⋅ p ( r )
,
m
The rate of transitions induced by a single mode is:
( w k )k, j =
2
2π
q2
ˆ j ⋅ p (r) 
δ
ω
−
ω
A
k∈
(
)
k
V 2
m2 k , j
2
And the total transition rate for an isotropic broadband source is:
w k = ' ( w k ) k , j
k,j
We can replace the sum over modes for a fixed volume with an integral over k :
1
V
'
k
-,
d 3k
( 2π )
→,
3
dk k 2 d Ω
( 2π )
3
→,
dω ω 2 d Ω
( 2π C )
3
d Ω = sin θ dθ dø
So for the rate we have:
2π ω2
q2
ˆ p (r) 
w k = , dω 2
δ ( ωk − ω) 2 , dΩ ' k ∈⋅
 ( 2 π c )3
m
j
2
Ak, j
2
can be written as k
The matrix element can be evaluated in a manner similar to before:
q
−q
ˆ j ⋅p ( r )  =
ˆ pi δ ( r ⋅ ri ) 
k∈
' k ∈⋅
m
m i
=
−i
ˆ k [ r1 , H 0 ] δ ( r − ri ) 
q ' q ∈⋅
 i
ˆ k r1 
= −iωk ' q ∈⋅
i
ˆ 
= −iωk k ∈⋅µ
where µ = ' q i ri
i
For the field
'A
kij
kj
2
=
Ekij
' 2ω
kij
i
2
=
E02
4ω 2
4-33
Wk  = , d ω
2
ωk2 2
2π ω 2
ˆ j ⋅µ 
δ
ω
−
ω
E0 , d Ω' k ∈
(
)
k
3
2
2
4 ( 2π c )
ω
j


2
8π / 3 µk
for isotropic
ω2
2
E0 µ k 
=
2 3
6π  c
For a broadband source, the energy density of the light
U=
I ω 2 E02
=
c 8π 3c 3
Wk  = Bk  U (ωk  )
Bk  =
4π 2
µk 
3 2
2
We can also write the incident energy density in terms of the quantum energy per photon. For N
photons in a single mode:
N ω = Bk  N
ω 3
π 2 c3
where Bk  has molecular quantities and no dependence or field. Note Bk  = Bk —ratio of S.E. =
absorption.
The ratio of absorption can be related to the absorption cross-section, δ A
σA =
P
total energy absorbed/unit time
=
I total intensity (energy/unit time/area)
P = ω⋅ Wk = ω Bk U ( ωk )
I = cU ( ωk )
σa =
ω
B k
c
or more generally, when you have a frequency-dependent absorption coefficient described by a
lineshape function g (ω )
4-34
σa ( ω ) =
ω
B k g ( ω )
c
units of cm 2
012334563789:;9=<<8
4>35
The Boltzmann distribution gives us the number of molecules in each state.
N m / N n = e − ωmn / kT
(4.102)
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 )
(4.103)
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
(4.104)
N m ( A nm + Bnm U ( ωmn ) ) = N n Bmn U ( ωmn )
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
(4.105)
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
(4.106)
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
(4.107)
The total rate of emission from the excited state is
w nm = Bnm U ( ωnm ) + A nm
using U ( ωnm ) = N
ω3
π 2 C3
(4.108)
4-36
=
 ω3
Bnm ( N + 1)
π 2 c3
(4.109)
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
(4.110)
If we use the statistical analysis to calculate rates of absorption we have
w mn =
ω3
Bmn N
π 2 c3
(4.111)
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 =
(4.112)
A
Now, if I am watching transitions from an initial state  to a final state k , then I can express
the initial and final states as:
matter
field
ϕI = ; N1 , N 2 , N 3  , N i , 
ϕF = k; N1 , N 2 , N3 ,  , Ni ± 1, 
k
Ni

N i −1
 + : emission 


 − : absorption 
(4.113), (4.114)
Where I have abbreviated N i ≡ N ki , ji , the energies of these two states are:
EI = E + ' N j ( ω j )
j
ω j = ck j
(4.115)
4-37
EF = Ek + ' N j ( ω j ) ± ωi
j
k
So looking at absorption 
 
states as:

 , we can write the Golden Rule Rate for transitions between

w k =
2π
δ ( E k − E  − ω ) ϕ F V ( t ) ϕ I

2
(4.116)
Now, let’s compare this to the absorption rate in terms of the classical vector potential:
w k =
2π
v 2
'
δ ( ωk − ω)
kj
2
q2
ˆ j ⋅p 
A k, j k ∈
2
m
2
(4.117)
ˆ p ) that acts on the
If these are to be the same, then clearly V ( t ) must have part that looks like (∈⋅
matter, but it will also need another part that acts to lower and raise the photons in the field. Based
on analogy with our electric dipole Hamiltonian, we write:
V(t) =
−q 1
ˆ + p † ⋅∈
ˆ†
ˆjA
ˆ *j A
pk ⋅∈
'
k
k,j
k,j
m v k,j
(
)
(4.118)
ˆ and A
ˆ † are lowering/raising operators for photons in mode k . These are operators in
where A
k, j
k,j
the field states, whereas pk remains only an operator in the matter states. So, we can write out the
matrix elements of V as
ϕF V ( t ) ϕI = −
q 1
ˆ
k pk ⋅∈
m v
ˆ , N , 
 , N i − 1, A
i
i
(4.119)
=
1
ˆ 
ωk k ∈⋅µ
v
A (i
−)
Comparing with our Golden Rule expression for absorption,
wk  =
π
2
δ ( ωk  − ω )
2
ωk2 2
2
E0 µk 
2
ω
(4.120)
4-38
We see that the matrix element
A (i
−)
=
E 02
4vω2
but
E 02
= Nω
8π
(4.121)
=
2π
vω
N
So we can write
 k, j =
2π
a
v ω k, j
(4.122)
†k, j =
2π †
a
v ω k, j
where a, a † are lowering, raising operators. So
ˆ =
A
'
k, j
2π
i k ⋅ r −ωt )
− i( k ⋅ r −ωt )
ˆ j a kj e (
∈
+ a †kj e
vω
(
)
So what we have here is a system where the light field looks like an infinite number of harmonic
oscillators, one per mode, and the field raises and lowers the number of quanta in the field while
the momentum operator lowers and raises the matter:
4-39
H = H EM + H M + V ( t ) = H 0 + V ( t )
H EM = ' ωk a †kj a kj +
(
1
2
)
k,j
HM = '
i
V (t) =
pi2
+ Vi ( r , t )
2mi
−q
A⋅p
m
='
k,j
q 2π
− i k ⋅ r −ωt )

ˆ j ⋅p ) a k, j ei( k ⋅r −ωt ) + a †k, j e (
∈
(

+
m v ωk
= V( −) + V( + )
Let’s look at the matrix elements for absorption (ωk  > 0 )
k, Ni − 1 V(
−)
, N i =
=
−q 2π
ˆ p ) a  , Ni
k, N i − 1 (∈⋅
m vω
−q 2π
m vω
= −i
ˆ p
Ni k ∈⋅
2πω
ˆ k
N i ∈⋅µ
v
and for stimulated emission (ωk  < 0 )
4-40
k, Ni + 1 V (
+)
, N i =
=
−q 2π
ˆ p ) a † , N i
k, Ni + 1 (∈⋅
m vω
−q 2π
m vω
= −i
ˆ pˆ 
N i + 1 k ∈⋅
2πω
ˆ k
Ni + 1 ∈⋅µ
v
We have spontaneous emission! Even if there are no photons in the mode ( N k = 0 ) , you can still
have transitions downward in the matter which creates a photon.
Let’s play this back into the summation-over-modes expression for the rates of
absorption/emission by isotropic field.
w k = , dω
=
2π ω2
+
δ ( ωk − ω) , dΩ ' k, Ni + 1 V ( ) , N i
3
2
 ( 2π2 )
j
2π ω2
 2 ( 2πc )3

( 2πω)( Ni + 1)
averageover polarization
number density per mode
=
4 ( N i + 1) ω3
3c
3
= Bk ( Ni + 1)
µ k
8π
2
µ k
3


2
ω3
π 2 c3

energy density
per mode
So we have the result we deduced before.
2
4-41
Appendix: Rates of Absorption and Stimulated Emission
Here are a couple of more detailed derivations:
Version 1:
Let’s look a little more carefully at the rate of absorption wk  induced by an isotropic, broadband
light source
wk  = , wk  (ω ) ρ E ( ω ) d ω
where, for a monochromatic light source
wk  (ω ) =
π
2
2
E0 (ω )
2
ˆ µ 
k ∈⋅
2
δ ( ωk  − ω )
For a broadband isotropic light source ρ (ω ) dω represents a number density of electromagnetic
modes in a frequency range d ω —this is the number of standing electromagnetic waves in a unit
volume.
For one frequency we wrote:
ˆe
A = A0 ∈
i ( k ⋅r −ωt )
+ c.c.
but more generally:
ˆje
A = ' A 0k ∈
i ( k ⋅ r −ωt )
+ c.c.
k, j
where the sum is over the k modes and j is the polarization component.
By summing over wave vectors for a box of fixed volume, the number density of modes in a
frequency range d ω radiated into a solid angle d Ω is
dN =
1
( 2π )
3
ω2
dω dΩ
c3
and we get ρE by integrating over all Ω
ρ E ( ω ) dω =
1
( 2π )
3
ω2
ω2
d
ω
d
Ω
=
, 2π2 c3 dω
c3
4π
number density at ω
4-42
We can now write the total transition rate between two discrete levels summed over all
frequencies, direction, polarizations
w k = , dω
2
π
1 ω2
E
ω
δ
ω
−
ω
(
)
(
)
0
k
3
2 2
( 2 π ) c3
' , dΩ
2
ˆ j ⋅µ 
k ∈
j

8π
2
µ k
3
2
=
E 0 ( ωk ) ω2
6π  c
2
3
µ k
2
We can write an energy density which is the number density in a range d ω × # of polarization
components × energy density per mode.
rate of energy flow/c
E2
ω2
U ( ωk ) = 2 3 ⋅ 2 ⋅ 0
2π c
8π
w k = Bk U ( ωk )
B k =
4π2
µ k
3 2
2
is the Einstein B coefficient for the rate of absorption
U is the energy density and 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 ( ωk ) = N
 ω3
π 2 c3
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 )
Bnm = Bmn
The absorption probability per unit time equals the stimulated emission probability per unit time.
4-43
Version 2:
Let’s calculate the rate of transitions induced by an isotropic broadband source—we’ll do it a bit
differently this time. The units are cgs.
The power transported through a surface is given by the Poynting vector and depends on k .
S=
c ω2 A 20 ˆ ω2 E 02
c
E×B =
k=
4π
8π
2π
and the energy density for this single mode wave is the time average of S / c .
The vector potential for a single mode is
ˆe
A = A0 ∈
i ( k ⋅r −ω t )
+ c.c.
with ω = ck . More generally any wave can be expressed as a sum over Fourier components of the
wave vector:
ˆj
A = ' Ak j ∈
e
i ( k ⋅ r −ωt )
k, j
V
+ c.c.
The factor of V normalizes for the energy density of the wave—which depends on k .
The interaction Hamiltonian for a single particle is:
V (t ) =
−q
A⋅ ρ
m
or for a collection of particles
V (t ) = − '
i
qi
A ⋅ pi
mi
Now, the momentum depends on the position of particles, and we can express p in terms of an
integral over the distribution of particles:
p = , d3 r p ( r )
p ( r ) = ' pi δ ( r − ri )
i
So if we assume that all particles have the same mass and charge—say electrons:
4-44
V (t ) =
−q
d 3r A ( r , t ) ⋅ p ( r )
,
m
The rate of transitions induced by a single mode is:
( w k )k, j =
2
2π
q2
ˆ j ⋅ p (r) 
δ
ω
−
ω
A
k∈
(
)
k
V 2
m2 k , j
2
And the total transition rate for an isotropic broadband source is:
w k = ' ( w k ) k , j
k,j
We can replace the sum over modes for a fixed volume with an integral over k :
1
V
'
k
-,
d 3k
( 2π )
→,
3
dk k 2 d Ω
( 2π )
3
→,
dω ω 2 d Ω
( 2π C )
3
d Ω = sin θ dθ dø
So for the rate we have:
2π ω2
q2
ˆ p (r) 
w k = , dω 2
δ ( ωk − ω) 2 , dΩ ' k ∈⋅
 ( 2 π c )3
m
j
2
Ak, j
2
can be written as k
The matrix element can be evaluated in a manner similar to before:
q
−q
ˆ j ⋅p ( r )  =
ˆ pi δ ( r ⋅ ri ) 
k∈
' k ∈⋅
m
m i
=
−i
ˆ k [ r1 , H 0 ] δ ( r − ri ) 
q ' q ∈⋅
 i
ˆ k r1 
= −iωk ' q ∈⋅
i
ˆ 
= −iωk k ∈⋅µ
where µ = ' q i ri
i
For the field
'A
kij
kj
2
=
Ekij
' 2ω
kij
i
2
=
E02
4ω 2
4-45
Wk  = , d ω
2
ωk2 2
2π ω 2
ˆ j ⋅µ 
δ
ω
−
ω
E0 , d Ω' k ∈
(
)
k
3
2
2
4 ( 2π c )
ω
j


2
8π / 3 µk
for isotropic
ω2
2
E0 µ k 
=
2 3
6π  c
For a broadband source, the energy density of the light
U=
I ω 2 E02
=
c 8π 3c 3
Wk  = Bk  U (ωk  )
Bk  =
4π 2
µk 
3 2
2
We can also write the incident energy density in terms of the quantum energy per photon. For N
photons in a single mode:
N ω = Bk  N
ω 3
π 2 c3
where Bk  has molecular quantities and no dependence or field. Note Bk  = Bk —ratio of S.E. =
absorption.
The ratio of absorption can be related to the absorption cross-section, δ A
σA =
P
total energy absorbed/unit time
=
I total intensity (energy/unit time/area)
P = ω⋅ Wk = ω Bk U ( ωk )
I = cU ( ωk )
σa =
ω
B k
c
or more generally, when you have a frequency-dependent absorption coefficient described by a
lineshape function g (ω )
4-46
σa ( ω ) =
ω
B k g ( ω )
c
units of cm 2
Readings
1
Jackson, J. D. Classical Electrodynamics (John Wiley and Sons, New York, 1975).
Cohen-Tannoudji, C., Diu, B. & Lalöe, F. Quantum Mechanics (Wiley-Interscience, Paris, 1977),
Appendix III.
3
Cohen-Tannoudji, et al. app. III, p. 1492.
2