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

M L LMH H H H= + + . (4.1)

The Hamiltonian for the matter MH is generally (although not necessarily) time independent,

whereas the electromagnetic field LH and its interaction with the matter LMH 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 LH , and we have a Hamiltonian that can be solved in

the interaction picture representation:

( )

( )0

M LMH H H t

H 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

ˆp ix x→− ∇→

(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 ( ),r tϕ . 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 andϕ .

> 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

( , ,x y zA A A andϕ ), 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 0ϕ = (Coulomb gauge) leads to a plane-wave description of E and B :

( ) ( )22

2 2

,1, 0A r t

A r tc t

∂−∇ + =

∂ (4.4)

0A∇⋅ = (4.5)

1 Jackson, J. D. Classical Electrodynamics (John Wiley and Sons, New York, 1975). 2 Cohen-Tannoudji, C., Diu, B. & Lalöe, F. Quantum Mechanics (Wiley-Interscience, Paris, 1977), Appendix III.

4-3

This wave equation allows the vector potential to be written as a set of plane waves:

( ) ( ) ( )*0 0ˆ ˆ,

i k r t i k r tA r t A Ae eω ω

ε ε⋅ − − ⋅ −

= + . (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 2 kλ π= . The wave has an

amplitude 0A which is directed along the polarization unit vector ε . Since 0A∇⋅ = , we see

that ˆ 0k ε⋅ = or ˆk ε⊥ . From the vector potential we can obtain E and B

( ) ( )

0 ˆ i k r t i k r t

AEt

i A e eω ωω ε

⋅ − − ⋅ −

∂= −

∂⎛ ⎞= −⎜ ⎟⎝ ⎠

(4.7)

( ) ( ) ( )0ˆ i k r t i k r t

B A

i k A e eω ω

ε⋅ − − ⋅ −

= ∇×

⎛ ⎞= × −⎜ ⎟⎝ ⎠

(4.8)

If we define a unit vector along the magnetic field polarization as ( )ˆ ˆˆ ˆb k k kε ε= × = × , we

see that the wavevector, the electric field polarization

and magnetic field polarization are mutually

orthogonal ˆ ˆˆk bε⊥ ⊥ .

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

0 012

E i Aω= (4.9)

0 012

B i k A= (4.10)

then, ( ) ( )0 ˆ, sinE r t E k r tε ω= ⋅ − (4.11)

( ) ( )0ˆ, sinB r t B b k r tω= ⋅ − . (4.12)

Note, 0 0E B k cω= = .

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:

( )x x y z z yF q E v B v B= + − . (4.14)

In Lagrangian mechanics, this force can be expressed in terms of the total potential energy U as

xx

U d UFx dt v

⎛ ⎞∂ ∂= − + ⎜ ⎟∂ ∂⎝ ⎠

(4.15)

Using the relationships that describe E and B in terms of A andϕ , inserting into eq. (4.14), and

working it into the form of eq. (4.15), we can show that:

U q qv Aϕ= − ⋅ (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= − (4.17)

212

L mv qv A qϕ= + ⋅ − (4.18)

The classical Hamiltonian is related to the Lagrangian as

212

H p v Lp v mv q v A qϕ

= ⋅ −

= ⋅ − − ⋅ − (4.19)

Recognizing Lp mv qAv∂

= = +∂

(4.20)

we write ( )1mv p qA= − . (4.21)

Now substituting (4.21) into (4.19), we have: 3 See Schatz and Ratner, p.82-83. 4 Cohen-Tannoudji, et al. app. III, p. 1492.

4-5

( ) ( ) ( )21 12

qm mmH p p qA p qA p qA A qϕ= ⋅ − − − − − ⋅ + (4.22)

( ) ( )21 , ,2

H p qA r t q r tm

ϕ⎡ ⎤= − +⎣ ⎦ (4.23)

This is the classical Hamiltonian for a particle in an electromagnetic field. In the Coulomb

gauge ( )0ϕ = , the last term is dropped.

We can write a Hamiltonian for a collection of particles in the absence of a external field

( )2

0 02i

ii i

pH V rm

⎛ ⎞= +⎜ ⎟

⎝ ⎠∑ . (4.24)

and in the presence of the EM field:

( )( ) ( )2

01

2 i i i ii i

H p q A r V rm

⎛ ⎞= − +⎜ ⎟

⎝ ⎠∑ . (4.25)

Expanding: ( ) 2

0 2 2i i

i iii i

q qH H p A A p Am m

= − ⋅ + ⋅ + ∑i∑ (4.26)

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:

( )0H H V t= + (4.27)

( ) ( )2i

i ii i

qV t p A A pm

= ⋅ + ⋅∑ . (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= − ∇ (4.29)

( ) ( )2 i i ii i

iV t q A Am

= ∇ ⋅ + ⋅∇∑ (4.30)

We can show that A A∇⋅ = ⋅∇ . Notice ( )A A A∇⋅ = ∇⋅ + ⋅∇ (chain rule). For instance, if we are

operating on a wavefunction ( ) ( )A A Aψ ψ ψ∇⋅ = ∇⋅ + ⋅ ∇ . The first term is zero since we are

working in the Coulomb gauge ( )0A∇⋅ = . Now we have:

( ) i

ii i

ii

i i

i qV t Am

q A pm

= ⋅∇

= − ⋅

∑

∑ (4.31)

For a single charge particle our interaction Hamiltonian is

( )

( )0 ˆ c.c.

i k r t

qV t A pmq A pm

e ωε

⋅ −

= − ⋅

⎡ ⎤= − ⋅ +⎢ ⎥⎣ ⎦

(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 ( ) ( )0kλ →∞ → , then 1ik re ⋅ ≈ . 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 0r as the center of mass of a molecule and expand

( )

( )

00

001

ii

i

ik r rik rik r

ik r ik r r

e e ee

⋅ −⋅⋅

⋅

=

⎡ ⎤= + ⋅ − +⎣ ⎦… (4.33)

For interactions, with UV, visible, and infrared (but not X-ray) radiation, 0 1ik r r− << , and setting

0 0r = means that 1ik re ⋅ → . We retain the second term for quadrupole transitions: charge

distribution interacting with gradient of electric field and magnetic dipole.

Now, using 0 0 2A iE ω= , we write (4.32) as

( ) 0 ˆ ˆ2

i t i tiqEV t p e p em

ω ωε εω

− +− ⎡ ⎤= ⋅ − ⋅⎣ ⎦ (4.34)

4-7

( ) ( )

( )

0 ˆ sin

( )

qEV t p tmq E t p

m

ε ωω

ω

−= ⋅

−= ⋅

(4.35)

or for a collection of charged particles (molecules):

( ) ( ) 0ˆ sinii

i i

q EV t p tm

ε ωω

⎛ ⎞= − ⋅⎜ ⎟

⎝ ⎠∑ (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 ( ) 0 sinV t V tω= the rate of

transitions induced by field is

( ) ( )2

2k k k kw V E E E Eπ δ ω δ ω= − − + − +⎡ ⎤⎣ ⎦ (4.37)

Now we evaluate the matrix elements of the EDH in the eigenstates for H0:

00 ˆk

qEV k V k pm

εω

−= = ⋅ (4.38)

We can evaluate the matrix element k p using an expression that holds for any one-particle Hamiltonian:

[ ]0, i pr Hm

= . (4.39)

This expression gives

( )

0 0

.

k

k

mk p k r H H rim k r E E k riim k rω

= −

= −

=

(4.40)

So we have

0 ˆkkV iqE k rω ε

ω= − ⋅ (4.41)

4-8

or for a collection of particles

0

0

0

ˆ

ˆ

kk i i

i

k

kkl

V iE k q r

iE k

iE

ω εω

ω ε μωω μω

⎛ ⎞= − ⋅⎜ ⎟⎝ ⎠

= − ⋅

= −

∑

(4.42)

μ is the dipole operator, and klμ 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:

( )dr r rμ ρ= ∫ . (4.43)

These expressions allow us to write in simplified form the well known interaction potential

for a dipole in a field:

( ) ( )V t E tμ= − ⋅ (4.44)

Then the rate of transitions between quantum states induced by the electric field is

( ) ( )

( ) ( )

22 2

0 2

2 202

2

2

kk kl k k

kl k k

w E E E E E

E

ωπ μ δ ω ωω

π μ δ ω ω δ ω ω

= − − + − +⎡ ⎤⎣ ⎦

= − + +⎡ ⎤⎣ ⎦

(4.45)

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:

( ) ( )2202

,fi fi fi fi

i fw Eπ μ δ ω ω δ ω ω⎡ ⎤= − + +⎣ ⎦∑ (4.46)

The strength of interaction between light and matter is given by the matrix

element ˆfi f iμ μ ε≡ ⋅ . The scalar part f iμ 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?

We can start with the first-order expression:

0

t

k t

ib d k V tτ−= ∫ (4.47)

or equivalently ( )ki tk k

ib e V tt

ω∂= −

∂ (4.48)

We can add irreversible relaxation to the description of kb , following our early approach:

( )2

ki t nkk k k

wib e V t bt

ω∂= − −

∂ (4.49)

Or using ( ) 0 sinkV t iE tμ ω= −

( )

( ) ( ) ( )0

sin2

2 2

k

k k

i t nkk k k

k nkk k

i i t

wib e t V b tt

E w b ti

e e

ω

ω ω ω ω

ω

ω μω

+ −

∂ −= −

∂

⎡ ⎤= − −⎢ ⎥⎣ ⎦

(4.50)

The solution to the differential equation

i ty ay be α+ = (4.51)

is ( )i t

at b ey t Aa i

eα

α−= +

+. (4.52)

k relaxes exponentially ... for instance by coupling to continuum

[ ]expk nkP w t∝ −

4-10

( )( )

( )( )

( )/2 0

2 / 2 / 2

k knk

i t i tw t k

knk k nk k

Eb t Ai w i w i

e eeω ω ω ωμ

ω ω ω ω

+ −− ⎡ ⎤

= + −⎢ ⎥+ + + −⎣ ⎦ (4.53)

Let’s look at absorption only, in the long time limit:

( )( )

0

2 / 2

ki tk

kk nk

E eb tiw

ω ωμω ω

−⎡ ⎤= ⎢ ⎥− −⎣ ⎦

(4.54)

For which the probability of transition to k is

( )

222 0

22 2

14 / 4

kk k

k nk

EP b

wμ

ω ω= =

− + (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) 0B∇⋅ = (2) 0/E ρ∇⋅ = ∈

(3) BEt

∂∇× = −

∂

(4) 0 0 0EB Jt

μ μ ∂∇× = +∈

∂

E : electric field; B : magnetic field; J : current density; ρ : charge density; 0∈ :

electrical permittivity; 0μ : 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) yx zFF FF

x y z∂∂ ∂

∇⋅ = + +∂ ∂ ∂

is a scalar. For a scalar field φ , the gradient

(6) ˆ ˆ ˆx y zx y zφ φ φφ ∂ ∂ ∂

∇ = + +∂ ∂ ∂

is a vector for the rate of change at on point in space. Here 2 2 2 2ˆ ˆ ˆˆx y z r+ + = are unit

vectors.

Also, the curl

Andrei Tokmakoff, MIT Department of Chemistry, 5/19/05

4-12

(7) ˆ ˆ ˆ

x y z

x y z

x y zF

F F F

∂ ∂ ∂∂ ∂ ∂∇× =

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) ( ) 0F∇⋅ ∇× = (9) ( ) 0φ∇× ∇ = (10) ( ) ( ) 2F F F∇× ∇× = ∇ ∇⋅ − ∇ We now introduce a vector potential ( ),A r t and a scalar potential ( ),r tϕ , which we will

relate to E and B

Since 0B∇⋅ = and ( ) 0A∇ ∇× = : (11) B A= ∇× Using (3), we have:

AEt

∂∇× = −∇×

∂

or

(12) 0AEt

⎡ ⎤∂∇× + =⎢ ⎥∂⎣ ⎦

From (9), we see that a scalar product exists with:

(13) ( ),AE r tt

ϕ∂+ = −∇∂

or

convention

4-13

(14) AEt

ϕ∂= −∇∂

So we see that the potentials A and ϕ determine the fields B and E : (15) ( ) ( ), ,B r t A r t= ∇×

(16) ( ) ( ) ( ), , ,E r t r t A r tt

ϕ ∂= −∇ −

∂

We are interested in determining the wave equation for A and ϕ . Using (15) and

differentiating (16) and substituting into (4):

(17) ( )2

0 0 02

AA Jt t

ϕμ μ⎛ ⎞∂ ∂

∇× ∇× +∈ +∇ =⎜ ⎟∂ ∂⎝ ⎠

Using (10):

(18) 2

20 0 0 0 02

AA A Jt t

ϕμ μ μ⎡ ⎤∂ ∂⎛ ⎞−∇ +∈ + ∇ ∇⋅ +∈ =⎜ ⎟⎢ ⎥∂ ∂⎝ ⎠⎣ ⎦

From (14), we have:

2AEt

ϕ∂∇⋅∇⋅ = − −∇

∂

and using (2):

(19) 20/V A

tϕ ρ−∂ ⋅

− ∇ = ∈∂

Notice from (15) and (16) that we only need to specify four field components

( , , ,x y zA A A 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 ( )( )0B∇× ∇⋅ = . It will change E by ( )t χ∂∂− ∇ , but we

can change ϕ to tχϕ ϕ ∂′ = −∂

. Then E and B will both be unchanged. This property of

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 r tχ′ = +∇ ⋅

(21) ( ) ( ) ( ), , ,r t r t r tt

ϕ ϕ χ∂′ = −∂

Up to this point, A′ and Q are undetermined. Let’s choose a χ such that:

(22) 0 0 0Atϕμ ∂

∇⋅ +∈ =∂

Lorentz condition

then from (17):

(23) 2

20 0 02

AA Jt

μ μ∂−∇ +∈ =

∂

The RHS can be set to zero for no currents. From (19), we have:

(24) 2

20 0 2

0tϕ ρμ ϕ∂

∈ − ∇ =∂ ∈

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

20 0 2 0

tχχ μ ∂

∇ −∈ =∂

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, 0J = and 0ρ = .

(26) 2

20 0 2 0AA

tμ ∂

−∇ +∈ =∂

gauge transformation

4-15

(27) 2

20 0 2 0

tϕϕ μ ∂

−∇ +∈ =∂

We now choose 0ϕ = (Coulomb gauge), and from (22) we see: (28) 0A∇⋅ = So, the wave equation for our vector potential is:

(29) 2

20 0 2 0AA

tμ ∂

−∇ +∈ =∂

The solutions to this equation are plane waves. (30) ( )0 sinA A t k rω α= − ⋅ + α : phase (31) ( )0 cosA t k rω α′= − ⋅ + k is the wave vector which points along the direction of propagation and has a magnitude: (32) 2 2 2 2

0 0 /k cω μ ω= ∈ = Since (28) 0A∇ ⋅ = ( )0 cos 0k A t k rω α− ⋅ − ⋅ + = (33) 0 0k A∴ ⋅ = 0k A⊥ 0A is the direction of the potential → polarization. From (15) and (16), we see that for

0ϕ = :

( )

( ) ( )

0

0

cos

cos

AE A t k rt

B A k A t k r

k E B

ω ω α

ω α

∂= − = − − ⋅ +

∂

= ∇× = − × − ⋅ +

∴ ⊥ ⊥

4-16

4.3. EINSTEIN B COEFFICIENT AND ABSORPTION CROSS-SECTION The rate of absorption induced by the field is

( ) ( ) ( )2202

ˆ2k kw E kπω ω ε μ δ ω 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

( )4cS E Bπ

= × (4.57)

2 204 8

c cI S E Eπ π

= = = . (4.58)

(Note, I’ve rather abruptly switched units to cgs). Using this we can write

( ) ( )2

2

2

4 ˆ3k kw I kcπ ω ε μ δ ω ω= ⋅ − , (4.59)

where I have also made use of the uniform distribution of polarizations applicable to an isotropic

field: 20 0 0 0

1ˆ ˆ ˆ3

E x E y E z E⋅ = ⋅ = ⋅ = . An equivalent representation of the amplitude of a

monochromatic field is the energy density

20

18

IU Ec π

= = . (4.60)

which allows the rates of transition to be written as ( )k k kw B U ω= (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

2

22

43k kB π μ= . (4.62)

You may see this written elsewhere as ( ) 222 3k kB π μ= , which holds when the energy density of a wave is expressed in Hz instead of angular frequency.

Andrei Tokmakoff, MIT Department of Chemistry, 5/19/2005

Andrei Tokmakoff, MIT Department of Chemistry, 2/12/2008

4-17

If we associate the energy density with a number of photons N, then U can also be written in a

quantum form

2 30

2 38EN U N

cωω

π π= = . (4.63)

Now let’s relate the rates of absorption to a quantity that is directly measured, an

absorption cross-section α:

( )( )

( )2

2

/int / /

4

k kk

k

k k

total energy absorbed unit timetotal incident ensity energy unit time area

B UwI cU

Bc c

α

ω ωωω

π ωμ

=

⋅⋅= =

= =

(4.64)

More generally, you may have a frequency-dependent absorption coefficient ( ) ( )kBα ω ω∝

( )kB 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 :

( ) ( ) ( ) ( )since

nm mn

nm nm nm nm nm mn

nm mn

w w

B U B U U U

B B

ω ω ω ω

=

= =

=

(4.65)

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

α α= .

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

n A m SEdI N dx N dxI

α α= − + (4.66)

( )n mdI N N dxI

α= − − (4.67)

Here nN and mN 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 β−⎛ ⎞= ⎜ ⎟

⎝ ⎠ (4.68)

Integrating (4.67) over a pathlength L we have

0

N LITI

e α−Δ= = (4.69)

N Le α−≈ 3 2: : :N cm cm L cmα− We see that the transmission of light through the sample decays exponentially as a function of path

length. N = n mN NΔ − 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: N NΔ ≈ . Written as the familiar Beer-Lambert Law:

0

log IA CLI

ε= − = . (4.70)

: / : /C mol liter liter mol cmε 2303 Nε α=

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 m nE E> .

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

m nN N e ω−= (4.71)

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 Uω ω= (4.72)

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 Uω ω

=

+ =

(4.73)

Andrei Tokmakoff, MIT Department of Chemistry, 5/19/2005

4-20

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 ωω= − (4.74)

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

( )3

/2 3

11mn kT

NU

Uc e ω

ωω

ωωπ

=−

(4.75)

Uω is the energy density per photon of frequency ω.

Nω is the mean number of photons at a frequency ω.

3

2 3nm nmA Bcω

π∴ = Einstein A coefficient (4.76)

The total rate of emission from the excited state is

( )nm nm nm nmw B U Aω= + ( )3

2 3using nmU NCωω

π= (4.77)

( )3

2 3 1nmB Ncω

π= + (4.78)

Notice, even when the field vanishes ( )0N → , we still have emission. Remember, for the semiclassical treatment, the total rate of stimulated emission was

( )3

2 3nm nmw B Ncω

π= (4.79)

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

3

2 3mn mnw B Ncω

π= (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:

1rad Aτ = (4.81)

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 / c= ω 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

( ) /kT

1fe 1ωω =

−

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),

( )2

2 3gc

ωω =

π

Using the energy of a photon, the energy density per mode is

( )3

2 3gcω

ω ω =π

and so the probability distribution that describes the quantum frequency dependent energy density is

( ) ( ) ( )3

2 3 /kT

1u g fc e 1ω

ωω = ω ω ω =

π −

Andrei Tokmakoff, MIT Department of Chemistry, 4/2003

4-22

The Quantum Vector Potential

So, for a quantized field, the field will be described by a photon number kjN , which represents the

number of photons in a particular mode ( ,k j ) with frequency ckω = in a cavity of volume v. For

light of a particular frequency, the energy of the light will be kjN ω . So, the state of the

electromagnetic field can be written:

1 1 2 2 3 3EM k , j k , j k , jN , N , N ,ϕ = …

If my matter absorbs a photon from mode 2k , then the state of my system would be

1 2 2 3EM k , j k , j k , jN , N 1, N ,′ϕ = − …

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:

0 EM MH H H= +

If the matter is described by Mϕ , then the total state of the E.M. field and matter can be

expressed as product states:

EM Mϕ = ϕ ϕ

And we have eigenenergies

EM ME E E= +

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:

I 1 2 3 i; N , N , N , N ,ϕ = … … F 1 2 3 ik; N , N , N , , N 1,ϕ = ±… …

: emission

: absorption+⎛ ⎞

⎜ ⎟−⎝ ⎠

Where I have abbreviated ,i ii k jN N≡ , the energies of these two states are: ( )I j j

jE E N ω= +∑ j jckω =

( )F k j j ij

E E N ω ω= + ±∑

So looking at absorption k⎛ ⎞

⎜ ⎟⎝ ⎠

, we can write the Golden Rule Rate for transitions between

states as:

( ) ( )2

k k F I2w E E V tπ

= δ − − ω ϕ ϕ

Now, let’s compare this to the absorption rate in terms of the classical vector potential:

( )2 22

k k jk , j2 2kj

2 q ˆw A k pv mπ

= δ ω −ω ∈ ⋅∑

If these are to be the same, then clearly ( )V t must have part that looks like ( )ˆ p∈⋅ that acts on the 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:

( ) ( )† * †k j k jk, j k, j

k, j

q 1 ˆ ˆˆ ˆV t p A p Am v−

= ⋅∈ + ⋅∈∑

k iN

fieldmatter

i 1N −

4-24

where †k, j k , j

ˆ ˆA and A are lowering/raising operators for photons in mode k . These are operators in

the field states, whereas kp remains only an operator in the matter states. So, we can write out the

matrix elements of V as

( )

( )

F I k i i i

k i

q 1 ˆˆV t k p , N 1, A , N ,m v

1 ˆk Av

−

ϕ ϕ = − ⋅∈ −

= ω ∈⋅μ

… … … …

Comparing with our Golden Rule expression for absorption,

( )2

2202 22

kk k kw Eωπ δ ω ω μ

ω= −

We see that the matrix element

( )2 20 0

i 2

E EA but N4v 8

2 Nv

− = = ωω π

π=

ω

So we can write

,

,

k , j k j

† †k, j k j

2A av

2A av

π=

ω

π=

ω

where †a,a are lowering, raising operators. So

( ) ( )( )i k r t i k r t†j kj kj

k, j

2ˆ ˆA a av

e e⋅ −ω − ⋅ −ωπ= ∈ +

ω∑

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:

( ) ( )

( )

( )

( )

( ) ( ) ( )

( ) ( )

EM M 0

† 1EM 2k kj kj

k, j

2i

M ii i

i k r ti k r t †j k, j k , j

k, j k

H H H V t H V t

H a a

pH V r , t2m

qV t A pm

q 2 ˆ p a am v

V V

e e− ⋅ −ω⋅ −ω

− +

= + + = +

= ω +

= +

−= ⋅

π ⎡ ⎤= ∈ ⋅ +⎣ ⎦ω

= +

∑

∑

∑

Let’s look at the matrix elements for absorption ( )0kω >

( ) ( )i i i i

i

i k

q 2 ˆk, N 1 V , N k, N 1 p a , Nm v

q 2 ˆN k pm v

2 ˆi Nv

− − π− = − ∈⋅

ω

− π= ∈⋅

ω

π ω= − ∈⋅μ

and for stimulated emission ( )0kω <

4-26

( ) ( ) †i i i i

i

i k

q 2 ˆk, N 1 V , N k, N 1 p a , Nm v

q 2 ˆˆN 1 k pm v

2 ˆi N 1v

+ − π+ = + ∈⋅

ω

− π= + ∈⋅

ω

π ω= − + ∈⋅μ

We have spontaneous emission! Even if there are no photons in the mode ( )0kN = , 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.

( )( ) ( )

( )( )( )

( )

( )

2 2

k k i i32 2 j

22

i k32

averageover polarizationnumber densityper mode

32i

k3

3

k i 2 3

energydensityper mode

2w d d k, N 1 V , N2

2 82 N 132 c

4 N 13 c

B N 1c

+π ω= ω δ ω −ω Ω +

π

π ω π= π ω + μ

π

+ ω= μ

ω= +

π

∑∫ ∫

So we have the result we deduced before.

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 kw induced by an isotropic, broadband light source ( ) ( )k k Ew w dω ρ ω ω= ∫ where, for a monochromatic light source

( ) ( ) ( )22022k kˆw E kπω ω μ δ ω ω= ∈⋅ −

For a broadband isotropic light source ( )dρ ω ω represents a number density of electromagnetic modes in a frequency range dω —this is the number of standing electromagnetic waves in a unit volume. For one frequency we wrote:

( )0 ˆ c.c.i k r tA A e ω⋅ −= ∈ +

but more generally:

( )k

i k r t0 j

k, j

ˆA A c.c.e ⋅ −ω= ∈ +∑

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 dω radiated into a solid angle dΩ is

( )

2

3 3

1dN d dc2ω

= ω Ωπ

and we get ρE by integrating over all Ω

( )( )

2 2

E 3 3 2 3

4

1d d d dc 2 c2

π

ω ωρ ω ω = ω Ω = ω

ππ ∫

number density at ω

4-28

We can now write the total transition rate between two discrete levels summed over all frequencies, direction, polarizations

( ) ( )( )

( )

2k

2 22k 0 k j32 3

j

83

2 220 k

k2 3

1 ˆw d E d k2 c2

E6 c

πμ

π ω= ω ω δ ω −ω Ω ∈ ⋅μ

π

ω ω= μ

π

∑∫ ∫

We can write an energy density which is the number density in a range dω × # of polarization components × energy density per mode.

( )

( )

220

k 2 3

k k k

22

k k2

EU 22 c 8

w B U

4B is the Einstein B coefficient for the rate of absorption3

ωω = ⋅ ⋅

π π

= ω

π= μ

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

( )2 30

k 2 3

EN U N8 c

ωω = ω =

π π

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

ω ω ω ω

=

= =

=

The absorption probability per unit time equals the stimulated emission probability per unit time.

rate of energy flow/c

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 .

2 2 2 2

0 0c A Ec ˆS E B k4 8 2

ω ω= × = =

π π π

and the energy density for this single mode wave is the time average of S / c . The vector potential for a single mode is ( )

0i k r tˆA A c.c.e ω⋅ −= ∈ +

with ckω = . More generally any wave can be expressed as a sum over Fourier components of the wave vector:

( )

j

i k r t

jkk, j

eˆA A c.c.V

⋅ −ω

= ∈ +∑

The factor of V normalizes for the energy density of the wave—which depends on k . The interaction Hamiltonian for a single particle is:

( ) qV t Am

ρ−= ⋅

or for a collection of particles

( ) ii

i i

qV t A pm

= − ⋅∑

Now, the momentum depends on the position of particles, and we can express p in terms of an integral over the distribution of particles: ( ) ( ) ( )3

i ii

p d r p r p r p r r= = δ −∑∫

So if we assume that all particles have the same mass and charge—say electrons:

4-30

( ) ( ) ( )3 ,qV t d r A r t p rm−

= ⋅∫

The rate of transitions induced by a single mode is:

( ) ( ) ( )2 22

k k jk, j2 2k, j

2 q ˆw A k p rV mπ

= δ ω −ω ∈ ⋅

And the total transition rate for an isotropic broadband source is: ( )k k k, j

k, jw w=∑

We can replace the sum over modes for a fixed volume with an integral over k :

( ) ( ) ( )

3 2 2

3 3 31

2 2 2k

d k dk k d d dV C

ωωπ π π

Ω Ω⇒ → →∑ ∫ ∫ ∫

So for the rate we have: sind d døθ θΩ=

( )

( ) ( )2 2 2 2

k k3 k, j2 2j

2 q ˆw d d k p r Am2 c

π ω= ω δ ω −ω Ω ∈⋅

π∑∫ ∫

The matrix element can be evaluated in a manner similar to before:

( ) ( )

[ ] ( )

j i ii

1 0 ii

k 1i

k i ii

q qˆ ˆk p r k p r rm m

i ˆq q k r ,H r r

ˆi q k r

ˆi k where q r

=−

∈ ⋅ ∈⋅ δ ⋅

−= ∈⋅ δ −

= − ω ∈⋅

= − ω ∈⋅μ μ =

∑

∑

∑

∑

For the field

2

220

22 4ij

jij ij

kk

k k i

E EAω ω

= =∑ ∑

can be written as k

4-31

( )( )

22 22032 2

22

02 3

28 3

24 2

6

kk k j

j

k

k/for isotropic

ˆW d E d kc

Ec

π μ

ωπ ωω δ ω ω μωπ

ω μπ

= − Ω ∈ ⋅

=

∑∫ ∫

For a broadband source, the energy density of the light

( )

2 20

3 3

22

2

8

43k k k k k

EIUc c

W B U B

ωπ

πω μ

= =

= =

We can also write the incident energy density in terms of the quantum energy per photon. For N photons in a single mode:

3

2 3kN B Ncωω

π=

where kB has molecular quantities and no dependence or field. Note k kB B= —ratio of S.E. = absorption. The ratio of absorption can be related to the absorption cross-section, Aδ

AP total energy absorbed/unit timeI total intensity (energy/unit time/area)

σ = =

( )

( )k k k

k

a k

P W B U

I cU

Bc

= ω⋅ = ω ω

= ω

ωσ =

or more generally, when you have a frequency-dependent absorption coefficient described by a lineshape function ( )g ω

( ) ( ) 2a kB g units of cm

cω

σ ω = ω