4-1
4.1. INTERACTION OF LIGHT WITH MATTER
One of the most important topics in time-dependent quantum mechanics for chemists is the
description of spectroscopy, which refers to the study of matter through its interaction with light
fields (electromagnetic radiation). Classically, light-matter interactions are a result of an oscillating
electromagnetic field resonantly interacting with charged particles. Quantum mechanically, light
fields will act to couple quantum states of the matter, as we have discussed earlier.
Like every other problem, our starting point is to derive a Hamiltonian for the light-matter
interaction, which in the most general sense would be of the form
H = H M + H L + H LM .
(4.1)
The Hamiltonian for the matter H M is generally (although not necessarily) time independent,
whereas the electromagnetic field H L and its interaction with the matter H LM are time-dependent.
A quantum mechanical treatment of the light would describe the light in terms of photons for
different modes of electromagnetic radiation, which we will describe later.
We will start with a common semiclassical treatment of the problem. For this approach we
treat the matter quantum mechanically, and treat the field classically. For the field we assume that
the light only presents a time-dependent interaction potential that acts on the matter, but the matter
doesn’t influence the light. (Quantum mechanical energy conservation says that we expect that the
change in the matter to raise the quantum state of the system and annihilate a photon from the
field. We won’t deal with this right now). We are just interested in the effect that the light has on
the matter. In that case, we can really ignore H L , and we have a Hamiltonian that can be solved in
the interaction picture representation:
H ≈ H M + H LM ( t )
= H0 + V (t )
(4.2)
Here, we’ll derive the Hamiltonian for the light-matter interaction, the Electric Dipole
Hamiltonian. It is obtained by starting with the force experienced by a charged particle in an
electromagnetic field, developing a classical Hamiltonian for this system, and then substituting
quantum operators for the matter:
Andrei Tokmakoff, MIT Department of Chemistry, 2/7/2008
4-2
ˆ
p → −i ∇
x → xˆ
(4.3)
In order to get the classical Hamiltonian, we need to work through two steps: (1) We need
to describe electromagnetic fields, specifically in terms of a vector potential, and (2) we need to
describe how the electromagnetic field interacts with charged particles.
Brief summary of electrodynamics
Let’s summarize the description of electromagnetic fields that we will use. A derivation of the
plane wave solutions to the electric and magnetic fields and vector potential is described in the
appendix. Also, it is helpful to review this material in Jackson1 or Cohen-Tannoudji, et al.2
> Maxwell’s Equations describe electric and magnetic fields ( E , B ) .
> To construct a Hamiltonian, we must describe the time-dependent interaction potential (rather
than a field).
> To construct the potential representation of E and B , you need a vector potential A ( r , t ) and
a scalar potential ϕ ( r , t ) . For electrostatics we normally think of the field being related to the
electrostatic potential through E = −∇ϕ , but for a field that varies in time and in space, the
electrodynamic potential must be expressed in terms of both A and ϕ .
> In general an electromagnetic wave written in terms of the electric and magnetic fields requires
6 variables (the x,y, and z components of E and B). This is an overdetermined problem;
Maxwell’s equations constrain these. The potential representation has four variables
( Ax , Ay , Az and ϕ ), but these are still not uniquely determined. We choose a constraint – a
representation or guage – that allows us to uniquely describe the wave. Choosing a gauge such
that ϕ = 0 (Coulomb gauge) leads to a plane-wave description of E and B :
−∇ 2 A ( r , t ) +
2
1 ∂ A(r ,t )
=0
c2
∂t 2
∇⋅ A = 0
1
2
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.
(4.4)
(4.5)
4-3
This wave equation allows the vector potential to be written as a set of plane waves:
A ( r , t ) = A0εˆ e
(
i k ⋅r −ωt
) + A*εˆ e−i( k ⋅r −ωt ) .
0
(4.6)
This describes the wave oscillating in time at an angular frequency ω and propagating in space
in the direction along the wavevector k , with a spatial period λ = 2π k . The wave has an
amplitude A0 which is directed along the polarization unit vector εˆ . Since ∇ ⋅ A = 0 , we see
that k ⋅ εˆ = 0 or k ⊥ εˆ . From the vector potential we can obtain E and B
E=−
∂A
∂t
i ( k ⋅r −ωt )
− i ( k ⋅r −ωt ) ⎞
= iω A0 εˆ ⎛⎜ e
−e
⎟
⎝
⎠
(4.7)
B = ∇× A
i ( k ⋅r −ωt )
− i ( k ⋅r −ωt ) ⎞
= i ( k × εˆ ) A0 ⎛⎜ e
−e
⎟
⎝
⎠
(4.8)
If we define a unit vector along the magnetic field polarization as bˆ = ( k × εˆ ) k = kˆ × εˆ , we
see that the wavevector, the electric field polarization
and
magnetic
field
polarization
are
mutually
orthogonal kˆ ⊥ εˆ ⊥ bˆ .
Also, by comparing eq. (4.6) and (4.7) we see that
the vector potential oscillates as cos(ωt), whereas the
field oscillates as sin(ωt). If we define
then,
Note, E0 B0 = ω k = c .
1
E0 = iω A0
2
(4.9)
1
B0 = i k A0
2
(4.10)
E ( r , t ) = E0 εˆ sin ( k ⋅ r − ωt )
(4.11)
B ( r , t ) = B0 bˆ sin ( k ⋅ r − ωt ) .
(4.12)
4-4
Classical Hamiltonian for radiation field interacting with charged particle
Now, let’s find a classical Hamiltonian that describes charged particles in a field in terms of the
vector potential. Start with Lorentz force3 on a particle with charge q:
F = q(E +v ×B) .
(4.13)
Here v is the velocity of the particle. Writing this for one direction (x) in terms of the Cartesian
components of E , v and B , we have:
Fx = q ( Ex + v y Bz − vz By ) .
(4.14)
In Lagrangian mechanics, this force can be expressed in terms of the total potential energy U as
∂U d ⎛ ∂U ⎞
+ ⎜
⎟
∂x dt ⎝ ∂vx ⎠
Fx = −
(4.15)
Using the relationships that describe E and B in terms of A and ϕ , inserting into eq. (4.14), and
working it into the form of eq. (4.15), we can show that:
U = qϕ − qv ⋅ A
(4.16)
This is derived in CTDL,4 and you can confirm by replacing it into eq. (4.15).
Now we can write a Lagrangian in terms of the kinetic and potential energy of the particle
L = T −U
L=
1
mv 2 + qv ⋅ A − qϕ
2
(4.17)
(4.18)
The classical Hamiltonian is related to the Lagrangian as
H = p ⋅v − L
= p ⋅ v − 12 m v 2 − q v ⋅ A − qϕ
Recognizing
we write
∂L
= mv + qA
∂v
(4.20)
1 p − qA .
v=m
(
)
(4.21)
p=
Now substituting (4.21) into (4.19), we have:
3
4
See Schatz and Ratner, p.82-83.
Cohen-Tannoudji, et al. app. III, p. 1492.
(4.19)
4-5
H=
1
m
p ⋅ ( p − qA ) −
H=
1
2m
( p − qA )
2
−
q
m
( p − qA ) ⋅ A + qϕ
2
1
⎡⎣ p − qA ( r , t ) ⎤⎦ + qϕ ( r , t )
2m
(4.22)
(4.23)
This is the classical Hamiltonian for a particle in an electromagnetic field. In the Coulomb
gauge (ϕ = 0 ) , the last term is dropped.
We can write a Hamiltonian for a collection of particles in the absence of a external field
⎛ p2
⎞
H 0 = ∑ ⎜ i + V0 ( ri ) ⎟ .
i ⎝ 2mi
⎠
(4.24)
2
⎛ 1
⎞
H = ∑⎜
pi − qi A ( ri ) ) + V0 ( ri ) ⎟ .
(
i ⎝ 2mi
⎠
(4.25)
and in the presence of the EM field:
Expanding:
H = H0 − ∑
i
qi
( pi ⋅ A + A ⋅ pi ) +
2mi
qi
∑ 2m
i
A
2
(4.26)
i
Generally the last term is considered small compared to the cross term. This term should be
considered for extremely high field strength, which is nonperturbative and significantly distorts the
potential binding molecules together. One can estimate that this would start to play a role at
intensity levels >1015 W/cm2, which may be observed for very high energy and tightly focused
pulsed femtosecond lasers. So, for weak fields we have an expression that maps directly onto
solutions we can formulate in the interaction picture:
H = H0 + V (t )
V (t ) = ∑
i
qi
( pi ⋅ A + A ⋅ pi ) .
2mi
(4.27)
(4.28)
Quantum mechanical Electric Dipole Hamiltonian
Now we are in a position to substitute the quantum mechanical momentum for the classical. Here
the vector potential remains classical, and only modulates the interaction strength.
4-6
p = −i ∇
V (t ) = ∑
i
(4.29)
i
qi ( ∇i ⋅ A + A ⋅∇i )
2mi
(4.30)
We can show that ∇⋅ A = A ⋅∇ . Notice ∇ ⋅ A = ( ∇⋅ A ) + A ⋅∇ (chain rule). For instance, if we are
operating on a wavefunction ∇⋅ A ψ = ( ∇ ⋅ A ) ψ + A ⋅ ( ∇ ψ
working in the Coulomb gauge ( ∇⋅ A = 0 ) . Now we have:
V (t ) = ∑
i
) . The first term is zero since we are
i qi
A ⋅∇i
mi
(4.31)
q
= − ∑ i A ⋅ pi
i mi
For a single charge particle our interaction Hamiltonian is
q
A⋅ p
m
i ( k ⋅r −ωt )
q
= − ⎡⎢ A0εˆ ⋅ p e
+ c.c.⎤⎥
m⎣
⎦
V (t ) = −
(4.32)
Under most circumstances, we can neglect the wavevector dependence of the interaction
potential.
If
the
wavelength
of
the
field
is
much
larger
than
the
molecular
dimension ( λ → ∞ ) ( k → 0 ) , then eik ⋅r ≈1 . This is known as the electric dipole approximation.
We do retain the spatial dependence for certain types of light-matter interactions. In that
case we define r0 as the center of mass of a molecule and expand
eik ⋅ri = eik ⋅r0 eik ⋅( ri −r0 )
= eik ⋅r0 ⎡⎣1 + ik ⋅ ( ri − r0 ) + … ⎤⎦
(4.33)
For interactions, with UV, visible, and infrared (but not X-ray) radiation, k ri − r0 << 1 , and setting
r0 = 0 means that eik ⋅r → 1 . We retain the second term for quadrupole transitions: charge
distribution interacting with gradient of electric field and magnetic dipole.
Now, using A0 = iE0 2ω , we write (4.32) as
V (t ) =
−iqE0
⎡⎣εˆ ⋅ p e−iωt − εˆ ⋅ p e+ iωt ⎤⎦
2mω
(4.34)
4-7
− qE0
(εˆ ⋅ p ) sin ωt
mω
−q
=
( E (t ) ⋅ p )
mω
V (t ) =
(4.35)
or for a collection of charged particles (molecules):
⎛
⎞E
q
V ( t ) = − ⎜ ∑ i ( εˆ ⋅ pi ) ⎟ 0 sin ωt
⎝ i mi
⎠ω
(4.36)
This is known as the electric dipole Hamiltonian (EDH).
Transition dipole matrix elements
We are seeking to use this Hamiltonian to evaluate the transition rates induced by V(t) from our
first-order perturbation theory expression. For a perturbation V ( t ) = V0 sin ω t the rate of
transitions induced by field is
wk =
π
2
Vk
2
⎡⎣δ ( Ek − E − ω ) + δ ( Ek − E + ω ) ⎤⎦
(4.37)
Now we evaluate the matrix elements of the EDH in the eigenstates for H0:
Vk = k V0
We can evaluate the matrix element k p
=
− qE0
k εˆ ⋅ p
mω
(4.38)
using an expression that holds for any one-particle
Hamiltonian:
[r , H0 ] =
i p
.
m
(4.39)
This expression gives
k p
m
k r H0 − H0 r
i
m
=
k r E − Ek k r
i
= imω k k r .
=
(
)
(4.40)
So we have
Vk = − iqE0
ωk
k εˆ ⋅ r
ω
(4.41)
4-8
or for a collection of particles
Vk = − iE0
ωk
⎛
⎞
k εˆ ⋅ ⎜ ∑ qi ri ⎟
ω
⎝ i
⎠
ωk
k εˆ ⋅ μ
ω
ω
= −iE0 k μkl
ω
= − iE0
(4.42)
μ is the dipole operator, and μkl is the transition dipole matrix element. We can see that it is the
quantum analog of the classical dipole moment, which describes the distribution of charge density
ρ in the molecule:
μ = ∫ dr r ρ ( r ) .
(4.43)
These expressions allow us to write in simplified form the well known interaction potential
for a dipole in a field:
V ( t ) = − μ ⋅ E (t )
(4.44)
Then the rate of transitions between quantum states induced by the electric field is
wk =
=
π
E0
2
π
2
2
2
E0
ωk2
2
μkl ⎡⎣δ ( Ek − E − ω ) + ( Ek − E + ω ) ⎤⎦
2
ω
(4.45)
μkl ⎡⎣δ (ωk − ω ) + δ (ωk + ω ) ⎤⎦
2
2
Equation (4.45) is an expression for the absorption spectrum since the rate of transitions
can be related to the power absorbed from the field. More generally we would express the
absorption spectrum in terms of a sum over all initial and final states, the eigenstates of H0:
w fi = ∑
i, f
The
strength
of
interaction
element μ fi ≡ f μ ⋅ εˆ i .
π
2
E0
2
μ fi ⎡⎣δ (ω fi − ω ) + δ (ω fi + ω ) ⎤⎦
between
The scalar part
2
light
and
f μi
matter
is
(4.46)
given
by
the
matrix
says that you need a change of charge
distribution between f and i to get effective absorption. This matrix element is the basis of
selection rules based on the symmetry of the states. The vector part says that the light field must
project onto the dipole moment. This allows information to be obtained on the orientation of
molecules, and forms the basis of rotational transitions.
4-9
Relaxation Leads to Line-broadening
Let’s combine the results from the last two lectures, and describe absorption to a state that is
coupled to a continuum. What happens to the probability of absorption if the excited state decays
exponentially?
k relaxes exponentially
... for instance by coupling to continuum
Pk ∝ exp [ − wnk t ]
We can start with the first-order expression:
bk =
−i
∫
t
t0
dτ k V t
(4.47)
∂
i
bk = − eiωk t Vk ( t )
∂t
or equivalently
(4.48)
We can add irreversible relaxation to the description of bk , following our early approach:
w
∂
i
bk = − eiωk t Vk ( t ) − nk bk
2
∂t
(4.49)
Or using V ( t ) = −iE0 μ k sin ωt
w
∂
−i
bk = eiωk t sin ωt Vk − nk bk ( t )
2
∂t
(4.50)
=
E0 ωk ⎡ i(ωk
e
2i ω ⎣⎢
+ω )
−e
i (ωk −ω )t ⎤
⎥ μk −
⎦
wnk
bk ( t )
2
The solution to the differential equation
y + ay = b eiα t
is
y ( t ) = A e− at +
b eiα t
.
a + iα
(4.51)
(4.52)
4-10
bk ( t ) = A e− wnk t /2 +
⎤
E0 μk ⎡
e i ( ωk + ω ) t
ei(ωk −ω )t
−
⎢
⎥
2i ⎣ wnk / 2 + i (ωk + ω ) wnk / 2 + i (ωk − ω ) ⎦
(4.53)
Let’s look at absorption only, in the long time limit:
bk ( t ) =
i ω −ω t
⎤
E0 μk ⎡
e( k )
⎢
⎥
2 ⎣ ωk − ω − iwnk / 2 ⎦
(4.54)
For which the probability of transition to k is
Pk = bk
2
E02 μk
=
4 2
2
1
( ωk − ω )
2
+ wnk 2 / 4
(4.55)
The frequency dependence of the transition probability has a Lorentzian form:
The linewidth is related to the relaxation rate from k into the continuum n. Also the linewidth is
related to the system rather than the manner in which we introduced the perturbation.
4-11
4.2. SUPPLEMENT: REVIEW OF FREE ELECTROMAGNETIC FIELD
Maxwell’s Equations (SI):
(1)
∇⋅ B = 0
(2)
∇⋅ E = ρ / ∈0
(3)
∇× E = −
(4)
∇× B = μ0 J +∈0 μ0
∂B
∂t
∂E
∂t
E : electric field; B : magnetic field; J : current density; ρ : charge density; ∈0 :
electrical permittivity; μ0 : magnetic permittivity
We are interested in describing E and B in terms of a scalar and vector potential. This is
required for our interaction Hamiltonian.
Generally: A vector field F assigns a vector to each point in space. The divergence of the
field
(5)
∇⋅ F =
∂Fx ∂Fy ∂Fz
+
+
∂x
∂y
∂z
is a scalar. For a scalar field φ , the gradient
(6)
∇φ =
∂φ
∂φ
∂φ
xˆ +
yˆ +
zˆ
∂x
∂y
∂z
is a vector for the rate of change at on point in space. Here xˆ 2 + yˆ 2 + zˆ 2 = rˆ 2 are unit
vectors.
Also, the curl
Andrei Tokmakoff, MIT Department of Chemistry, 5/19/05
4-12
(7)
∇× F =
xˆ
yˆ
zˆ
∂
∂x
∂
∂y
∂
∂z
Fx
Fy
Fz
is a vector whose x, y, and z components are the circulation of the field about that
component.
Some useful identities from vector calculus are:
(8)
∇ ⋅ ( ∇× F ) = 0
(9)
∇× ( ∇φ ) = 0
(10)
∇× ( ∇× F ) = ∇ ( ∇ ⋅ F ) − ∇ 2 F
We now introduce a vector potential A ( r , t ) and a scalar potential ϕ ( r , t ) , which we will
relate to E and B
Since ∇⋅ B = 0 and ∇ ( ∇× A ) = 0 :
(11)
B = ∇× A
Using (3), we have:
∇× E = − ∇×
∂A
∂t
or
(12)
⎡
∂A ⎤
∇× ⎢ E + ⎥ = 0
∂t ⎦
⎣
From (9), we see that a scalar product exists with:
(13)
E+
or
∂A
= − ∇ϕ ( r , t )
∂t
convention
4-13
∂A
− ∇ϕ
∂t
(14)
E=
(15)
So we see that the potentials A and ϕ determine the fields B and E :
B ( r , t ) = ∇× A ( r , t )
(16)
E ( r , t ) = − ∇ϕ ( r , t ) −
∂
A(r ,t )
∂t
We are interested in determining the wave equation for A and ϕ .
Using (15) and
differentiating (16) and substituting into (4):
(17)
(18)
⎛ ∂2 A
∂ϕ ⎞
∇× ( ∇× A ) + ∈0 μ0 ⎜ 2 + ∇
⎟ = μ0 J
∂t ⎠
⎝ ∂t
Using (10):
⎡ 2
∂2 A ⎤
∂ϕ ⎞
⎛
⎢ −∇ A + ∈0 μ0 ∂t 2 ⎥ + ∇ ⎜ ∇ ⋅ A + ∈0 μ0 ∂t ⎟ = μ0 J
⎝
⎠
⎣
⎦
From (14), we have:
∇⋅ E = −
∂∇ ⋅ A
− ∇ 2ϕ
∂t
and using (2):
(19)
−∂V ⋅ A
− ∇ 2ϕ = ρ / ∈0
∂t
Notice from (15) and (16) that we only need to specify four field components
( Ax , Ay , Az , ϕ ) to determine all six E and B components. But E and B do not uniquely
determine A and ϕ . So, we can construct A and ϕ in any number of ways without
changing E and B . Notice that if we change A by adding ∇χ where χ is any function
of r and t , this won’t change B ( ∇× ( ∇ ⋅ B ) = 0 ) . It will change E by ( − ∂∂t ∇χ ) , but we
can change ϕ to ϕ ′ = ϕ −
∂χ
. Then E and B will both be unchanged. This property of
∂t
4-14
changing representation (gauge) without changing E and B is gauge invariance. We can
transform between gauges with:
(20)
A′ ( r , t ) = A ( r , t ) + ∇ ⋅ χ ( r , t )
(21)
ϕ′(r , t ) = ϕ ( r , t ) −
∂
χ (r ,t )
∂t
gauge
transformation
Up to this point, A′ and Q are undetermined. Let’s choose a χ such that:
(22)
∇⋅ A + ∈0 μ0
∂ϕ
=0
∂t
Lorentz condition
then from (17):
(23)
−∇ 2 A + ∈0 μ0
∂2 A
= μ0 J
∂t 2
The RHS can be set to zero for no currents.
From (19), we have:
(24)
∈0 μ0
∂ 2ϕ
ρ
− ∇ 2ϕ =
2
∂t
∈0
Eqns. (23) and (24) are wave equations for A and ϕ . Within the Lorentz gauge, we can
still arbitrarily add another χ (it must only satisfy 22). If we substitute (20) and (21) into
(24), we see:
(25)
∇ 2 χ −∈0 μ0
∂2 χ
=0
∂t 2
So we can make further choices/constraints on A and ϕ as long as it obeys (25).
For a field far from charges and currents, J = 0 and ρ = 0 .
(26)
−∇ 2 A +∈0 μ0
∂2 A
=0
∂t 2
4-15
(27)
−∇ 2ϕ +∈0 μ0
∂ 2ϕ
=0
∂t 2
We now choose ϕ = 0 (Coulomb gauge), and from (22) we see:
(28)
∇⋅ A = 0
So, the wave equation for our vector potential is:
(29)
−∇ 2 A +∈0 μ0
∂2 A
=0
∂t 2
The solutions to this equation are plane waves.
(30)
(31)
A = A0 sin (ωt − k ⋅ r + α )
= A0 cos (ωt − k ⋅ r + α ′ )
α : phase
k is the wave vector which points along the direction of propagation and has a magnitude:
(32)
k 2 = ω 2 μ0 ∈0 = ω 2 / c 2
Since (28) ∇ ⋅ A = 0
−k ⋅ A0 cos (ωt − k ⋅ r + α ) = 0
(33)
∴ k ⋅ A0 = 0
k ⊥ A0
A0 is the direction of the potential → polarization. From (15) and (16), we see that for
ϕ =0:
E=−
∂A
= − ω A0 cos (ωt − k ⋅ r + α )
∂t
B = ∇× A = − ( k × A0 ) cos (ωt − k ⋅ r + α )
∴ k ⊥E⊥B
Andrei Tokmakoff, MIT Department of Chemistry, 5/19/2005
4-16
4.3. EINSTEIN B COEFFICIENT AND ABSORPTION CROSS-SECTION
The rate of absorption induced by the field is
wk (ω ) =
π
2
2
E0 (ω )
2
k εˆ ⋅ μ
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, the energy flux through a unit area, which is the time-averaged value of
the Poynting vector, S
c
(E × B)
4π
c
c 2
I= S =
E2 =
E0 .
4π
8π
S=
(4.57)
(4.58)
(Note, I’ve rather abruptly switched units to cgs). Using this we can write
wk =
4π 2
I (ω ) k εˆ ⋅ μ
3c 2
2
δ ( ωk − ω ) ,
(4.59)
where I have also made use of the uniform distribution of polarizations applicable to an isotropic
1
2
field: E0 ⋅ xˆ = E0 ⋅ yˆ = E0 ⋅ zˆ = E0 . An equivalent representation of the amplitude of a
3
monochromatic field is the energy density
U=
I
1 2
=
E0 .
c 8π
(4.60)
which allows the rates of transition to be written as
wk = Bk U (ωk
)
(4.61)
The first factor contains the terms in the matter that dictate the absorption rate. B is independent of
the properties of the field and is called the Einstein B coefficient
4π 2
2
μk .
2
3
You may see this written elsewhere as Bk = ( 2π 3
Bk =
(4.62)
2
)μ
2
k
of a wave is expressed in Hz instead of angular frequency.
Andrei Tokmakoff, MIT Department of Chemistry, 2/12/2008
, which holds when the energy density
4-17
If we associate the energy density with a number of photons N, then U can also be written in a
quantum form
N ω=
E02
8π
U=N
ω3
π 2c3
.
(4.63)
Now let’s relate the rates of absorption to a quantity that is directly measured, an
absorption cross-section α:
α=
=
=
total energy absorbed / unit time
total incident intensity ( energy / unit time / area )
ω ⋅ wk
I
4π 2
μk
c
ω ⋅ Bk U (ωk
cU (ωk )
=
2
=
ω
c
)
(4.64)
Bk
More generally, you may have a frequency-dependent absorption coefficient α (ω ) ∝ Bk (ω )
= Bk g (ω ) where g(ω) is a unit normalized lineshape function. The golden rule rate for absorption
also gives the same rate for stimulated emission. Given two levels m and n :
wnm = wmn
Bnm U (ωnm ) = Bnm U (ωnm )
since U (ωnm ) = U (ωmn )
(4.65)
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, (α A )nm = (α SE )mn .
4-18
We can now use a phenomenological approach to calculate the change in the intensity of
incident light, due to absorption and stimulated emission passing through a sample of length L
where the levels are thermally populated. Given that we have a thermal distribution of identical
non-interacting particles, with quantum states such that the level m is higher in energy than n :
k
dI
= − N n α A dx + N m α SE dx
I
(4.66)
dI
= − ( N n − N m ) α dx
I
(4.67)
Here N n and N m are population of the upper and lower states, but expressed as a population
densities. If N is the molecule density,
⎛ e − β En ⎞
Nn = N ⎜
⎟
⎝ Z ⎠
(4.68)
Integrating (4.67) over a pathlength L we have
T=
I
= e−ΔNα L
I0
≈ e − Nα L
(4.69)
N : cm −3
α : cm 2
L : cm
We see that the transmission of light through the sample decays exponentially as a function of path
length. ΔN =N n − N m is the thermal population difference between states. The second expression
comes from the high frequency approximation applicable to optical spectroscopy, but certainly not
for magnetic resonance: ΔN ≈ N . Written as the familiar Beer-Lambert Law:
A = − log
I
= ε CL .
I0
(4.70)
C : mol / liter
ε = 2303 N α
ε : liter / mol cm
4-19
4.4. 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
(4.71)
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 )
(4.72)
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
(4.73)
N m ( Anm + Bnm U (ωmn ) ) = N n Bmn U (ωmn )
Andrei Tokmakoff, MIT Department of Chemistry, 5/19/2005
4-20
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
(4.74)
For the energy density we will use Planck’s blackbody radiation distribution:
U (ω ) =
ω3
1
(4.75)
ωmn / kT
−1
π c e
Uω
Nω
2 3
U ω is the energy density per photon of frequency ω.
Nω is the mean number of photons at a frequency ω.
∴ Anm =
ω3
π 2 c3
Bnm
Einstein A coefficient
(4.76)
The total rate of emission from the excited state is
wnm = Bnm U (ωnm ) + Anm
=
ω3
π 2 c3
using U (ωnm ) = N
ω3
π 2 C3
Bnm ( N + 1)
(4.77)
(4.78)
Notice, even when the field vanishes ( N → 0 ) , we still have emission.
Remember, for the semiclassical treatment, the total rate of stimulated emission was
wnm =
ω3
π 2 c3
Bnm ( N )
(4.79)
If we use the statistical analysis to calculate rates of absorption we have
wmn =
ω3
π 2 c3
Bmn N
(4.80)
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 rad =
(4.81)
A
4-21
4.5. 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
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
πc
Using the energy of a photon, the energy density per mode is
ω3
ω g ( ω) = 2 3
πc
and so the probability distribution that describes the quantum frequency dependent energy density
is
u ( ω) = ω g ( ω) f ( ω) =
Andrei Tokmakoff, MIT Department of Chemistry, 4/2003
ω3
1
ω /kT
2 3
πc e
−1
4-22
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
,…
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
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
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
And we have eigenenergies
E = E EM + E M
4-23
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
ϕI = ; N1 , N 2 , N 3 … , N i , …
field
Ni
k
ϕF = k; N1 , N 2 , N 3 , … , N i ± 1, …
N i −1
⎛ + : emission ⎞
⎜
⎟
⎝ − : absorption ⎠
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
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
⎠
wk =
2π
δ ( E k − E − ω ) ϕ F V ( t ) ϕI
2
Now, let’s compare this to the absorption rate in terms of the classical vector potential:
wk =
2π
v 2
∑ δ (ω
k
− ω)
kj
2
q2
ˆ j ⋅p
A
k∈
m2 k , j
2
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
(
)
4-24
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
ϕ F V ( t ) ϕI = −
=
q 1
ˆ
k pk ⋅∈
m v
1
ˆ
ωk k ∈⋅μ
v
ˆ …, N , …
… , N i − 1,… A
i
i
A (i − )
Comparing with our Golden Rule expression for absorption,
wk =
π
2
δ ( ωk − ω )
2
We see that the matrix element
(−)
Ai
E 02
=
4vω2
=
2π
vω
E 02
=N ω
but
8π
N
So we can write
 k , j =
2π
a
v ω k, j
 †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
)
ωk2 2
E μ
ω2 0 k
2
4-25
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, N i − 1 V ( − )
, Ni =
=
−q 2 π
ˆ p ) a , Ni
k, N i − 1 (∈⋅
m vω
−q 2 π
m vω
= −i
ˆ p
N i k ∈⋅
2π ω
ˆ k
N i ∈⋅μ
v
and for stimulated emission (ωk < 0 )
4-26
k, N i + 1 V ( + )
, Ni =
=
−q 2π
ˆ p ) a † , Ni
k, N i + 1 (∈⋅
m vω
−q 2π
m vω
= −i
ˆ pˆ
N i + 1 k ∈⋅
2π ω
ˆ k
N i + 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
2
2π
2
ω2
( 2π )
2 3
ω2
( 2πc )
δ ( ωk − ω) ∫ dΩ ∑ k, N i + 1 V ( + ) , N i
j
( 2π ω)( Ni + 1)
3
= Bk ( N i + 1)
2
average over polarization
number density per mode
4 ( N i + 1) ω3
=
μk
3 c3
8π
μk
3
2
ω3
π 2 c3
energy density
per mode
So we have the result we deduced before.
2
4-27
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
2
ˆ μ
k ∈⋅
δ ( ωk − ω )
For a broadband isotropic light source ρ (ω ) d ω 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
d ω dΩ
c3
and we get ρE by integrating over all Ω
ρ E ( ω ) dω =
1
( 2π )
3
ω2
ω2
d
ω
d
Ω
=
dω
∫
c3
2π 2 c3
4π
number density at ω
4-28
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
3
2 2
( 2π ) c
∑ ∫ dΩ
ˆ j ⋅μ
k ∈
j
8π
μk
3
=
E 0 ( ωk
6π
)
2
2
c
ω2
3
μk
2
2
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
Bk =
4π 2
μk
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.
4-29
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 02 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:
4-30
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, j k ∈
(
)
k
2
2
V
m
2
And the total transition rate for an isotropic broadband source is:
wk = ∑(wk
)k, j
k, j
We can replace the sum over modes for a fixed volume with an integral over k :
1
d 3k
dk k 2 d Ω
dω ω 2 d Ω
⇒
→
→
∑ ∫ 2π 3 ∫ 2π 3 ∫ 2π C 3
V k
( )
( )
(
)
d Ω = sin θ dθ dø
So for the rate we have:
w k = ∫ dω
2π
2
ω2
( 2π c )
3
q2
ˆ p (r)
δ ( ωk − ω) 2 ∫ dΩ ∑ k ∈⋅
m
j
2
Ak, j
2
can be written as k
The matrix element can be evaluated in a manner similar to before:
q
ˆ j ⋅p ( r )
k∈
m
=
=
−q
ˆ pi δ ( r ⋅ ri )
∑ k ∈⋅
m i
−i
ˆ k [ r1 , H 0 ] δ ( r − ri )
q ∑ q ∈⋅
i
= −iωk
ˆ
∑ q ∈⋅
k r1
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
4-31
Wk = ∫ d ω
ωk2 2
2π ω 2
ˆ j ⋅μ
E d Ω∑ k ∈
−
δ
ω
ω
(
)
k
ω2 0 ∫
4 2 ( 2π c )3
j
2
2
8π / 3 μk
for isotropic
=
ω2
2
E0 μk
2 3
6π c
For a broadband source, the energy density of the light
I ω 2 E02
U= = 3 3
c 8π c
Wk = Bk U (ωk
)
4π 2
Bk = 2 μk
3
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
ω3
π 2c3
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
Bk
c
or more generally, when you have a frequency-dependent absorption coefficient described by a
lineshape function g (ω )
σa ( ω) =
ω
Bk g ( ω )
c
units of cm 2