148

Chapter 9. Electromagnetic Radiation

9.1 Photons and Electromagnetic Wave

Electromagnetic radiation is composed of elementary particles called photons. The

correspondence between the classical electric field and the quantum picture of photons is that the

intensity of light is proportional to the number of photons.

Properties of photon: (rest) mass m = 0 , energy ε = ω = cp , momentum p = k , angular

frequency ω = ck , spin s = 1 , so that photons are bosons. Light has two choices of polarization,

which are given the label λ ( λ = 1,2 ). The frequency is the same one that is associated with an

oscillating electric field, which satisfying the electromagnetic wave equation (in vacuum):

∂2E∂t 2

= c2∇2E, and ∂2B∂t 2

= c2∇2B . (9.1)

Monochromatic electromagnetic wave:

E(r,t) = Re[E(r)eiωt ] = E0cos(ωt− k ⋅ r +ϕ

0),

B(r,t) = Re[B(r)eiωt ] = B0cos(ωt− k ⋅ r +ϕ

0).

(9.2)

A field may have classical meaning only after averaging over some spatial volume Ω , for

example, the Poynting vector (directional energy flux density)

S =

c4π

E×B, and S =c4π

E02 . (9.3)

Note: we use the Gaussian units throughout this class.

On the other hand, S = c

NωΩ

, where Ω is the volume of the normalization big box;

therefore, E

02 = N ⋅

4πωΩ

, and the field is quantized. When N is very large, the classical

electromagnetism works extremely well, i.e., you can treat macroscopic electromagnetic fields as

being continuous. However, at the microscopic level, the electromagnetic fields must be

quantized, leading to phenomena which are not consistent with the classical theory.

Examples: (a) Spontaneous radiation; (b) Photoelectric effects. The classical theory of

electromagnetism fails completely. We have treated the photoelectric effect semi-classically, but

the semi-classical theory in which the field is continuous (NOT an operator) cannot explain the

spontaneous radiation.

149

9.2 Gauge Invariance

Let’s begin with the classical theory. Then we use the correspondence principle to construct the

quantum mechanical version of the electromagnetic theory, i.e., field quantization.

Maxwell’s equations

∇⋅E = 4πρ, ∇⋅B = 0,

∇×E =−1c∂B∂t

, ∇×B =1c∂E∂t

+4πc

j . (9.4)

Here ρ the charge density and j the current density. The problem is to convert the above

classical equations into a quantum mechanical description, i.e., to derive the Hamiltonian

(operator) at microscopic level compatible to Maxwell’s equations at macroscopic level.

There is an important theorem: any vector function of position can be written as the sum of

two terms; one is the gradient of a potential and the other is the curl of a vector:

S(r) =∇g +∇×m = Sl+ S

t. (9.5)

Here Sl and St

are longitudinal and transverse parts of S . Assuming B(r) has this form, then

Eq. (9.4) becomes

∇⋅(∇g +∇×A) =∇2g = 0 . (9.6)

We can simply set g(r) = 0 , so that B =∇×A . (9.7)

Then

∇× E +

1c∂A∂t

⎡

⎣⎢⎢

⎤

⎦⎥⎥ = 0 . (9.8)

If S = E +

1c∂A∂t

, one can write

∇×(−∇φ+∇×m) = 0 =∇×(∇×m) . (9.9)

The above equation is satisfied if we set m = 0 , therefore

E =−∇φ−

1c∂A∂t

, (9.10)

B =∇×A . (9.11)

When these two Eqs. [(9.10) and (9.11)] for E(r) and B(r) are put into Maxwell’s equations

(9.4), the equations for the scalar and vector potentials are

∇2φ+

1c∂(∇⋅A)∂t

=−4πρ , (9.12)

150

∇×(∇×A)+

1c2

∂2A∂t 2

+1c∇∂A∂t

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

=4πc

j . (9.13)

E and B are observables, while A and φ are mathematical tools for computing E, B and

other observables. However, the four unknown functions (A

x, A

y, A

z, φ) are not uniquely

determined by E and B, since E and B remain unchanged under the gauge transformation:

A→ A +∇Λ and φ→ φ−

1c∂Λ∂t

, (9.14)

E→−1

c∂(A +∇Λ)∂t

−∇ φ−1c∂Λ∂t

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

=−1c∂A∂t−∇φ , (9.15)

B→∇×(A +∇Λ) =∇×A , (9.16)

where Λ(r,t) is an arbitrary scalar function. Thus it is necessary to impose one additional

condition (or constraint), which is called gauge.

Coulomb gauge is widely used in condensed matter physics and quantum chemistry:

∇⋅A = 0 . (9.17)

Coulomb gauge is also called transverse gauge because Eq. (9.17) implies that the longitudinal

part of A (Al) is zero. Coulomb gauge also implies that the scalar potential acts instantaneously:

φ(r,t) =

ρ( ′r ,t)r− ′r∫ d 3 ′r , (9.18)

because Eq. (9.12) becomes ∇

2φ =−4πρ . (9.19)

Using this identity,

∇×(∇×A) =−∇2A +∇(∇⋅A) , (9.20)

one can show that under Coulomb gauge, Eq. (9.13) for vector potential becomes

∇2A− 1

c2

∂2A∂t 2

=−4πc

jt(r,t) , (9.21)

where the transverse part of current density jt is defined by

jt(r,t) =

14π∇× ∇×

j( ′r ,t)r− ′r

d 3 ′r∫⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥. (9.22)

151

In the Coulomb gauge, since the vector potential A is purely transverse, it should respond only to

the transverse part of the current, and the longitudinal component of A does not occur.

Another widely used one (especially in quantum field theory) is the Lorentz gauge:

∇⋅A +

1c∂φ∂t

= 0 , (9.23)

which causes Eqs. (9.12) and (9.13) change to

∇2φ−

1c2

∂2ψ∂t 2

=−4πρ , (9.24)

∇2A− 1

c2

∂2A∂t 2

=−4πc

j . (9.25)

In the Lorentz gauge, both the vector and scalar potentials obey the retarded wave equation.

They combine to produce a four-vector that is invariant under a Lorentz transformation.

Another gauge that is often used is the condition that the scalar potential is set equal to zero,

φ(r) = 0 . In this case it is found that the longitudinal vector potential is not zero, which leads to

an interaction between charges, the instantaneous Coulomb interaction.

We use Coulomb gauge in this course, so that the Coulomb interaction is un-retarded.

9.3 Semi-Classical Approximation

The choice of gauge won’t affect E and B, but it affects the Hamiltonian.

Classically,

H =

p2

2m→ H =

12m

p−e

cA

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

2

+eφ , (9.26)

with e the charge (negative for an electron). Semi-classically,

H =

12m

p−e

cA

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

2

+eφ . (9.27)

Here A is a classical (continuous) field. As usual in the following we omit except for the case

in which confusion might rise.

The quadratic term

e2

2mc2A2 can be omitted for linear optical properties. In addition, we

consider the case of the free electromagnetic field ( ρ = 0 , j = 0 ), in which the gauge invariance

allows we choose the simplest Coulomb gauge:

152

∇⋅A = 0; φ = 0 . (9.28)

Then

H =

12m

p2−e

cA ⋅p−e

cp ⋅A

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

=p2

2m−

e

mcA ⋅p . (9.29)

Here H = H0

+ HcR

, and the charge-radiation interaction Hamiltonian.

H

cR=−

e

mcA ⋅p . (9.30)

Within electric dipole approximation, i.e., for electromagnetic wave with small wave number k

(long wavelength),

H cR= eE ⋅ r . (9.31)

9.4 Hamiltonian in Classical Mechanics

Field (second) quantization follows the same approach used to quantize energies of bound states

of particles employing Bohr’s correspondence principle. In classical mechanics,

H(q

i,p

i,t) = (p

iqi)

i

∑ −L(qi, q

i,t). (9.32)

Here L =T −V is the Lagrangian, with T and V kinetic and potential energies, respectively.

qi are generalized coordinates,

qi

=dq

i

dt, and the generalized momenta,

p

i=∂L(q

i, q

i,t)

∂ qi

, (9.33)

which are also called the conjugate momenta.

Consider a particle in a conservative force field and use the Cartesian coordinates as

generalized coordinates, i.e., qi= x

i ( i = 1,2,3), pi

= m xi, and

H =

12m

pi

2 +V(xi)

i=1

3

∑ . (9.34)

Then in quantum mechanics,

H =

12m

pi

2 +V(xi)

i=1

3

∑ . (9.35)

In construction of classical Hamiltonian, we need canonical coordinates qi and pi

, which

satisfy the following Poisson bracket relations:

153

{q

i,q

j} = 0; {p

i,p

j} = 0; {q

i,p

j} = δ

ij, (9.36)

where the Poisson bracket is defined as

{A(p,q),B(q,p)}

PB≡

∂A∂q

s

∂B∂p

s

−∂A∂p

s

∂B∂q

s

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟s

∑ . (9.37)

The quantum mechanical version of the commutators:

q

i,q

j⎡⎣⎢

⎤⎦⎥ = 0; p

i, p

j⎡⎣⎢

⎤⎦⎥ = 0; q

i, p

j⎡⎣⎢

⎤⎦⎥ = iδ

ij, (9.38)

which look almost exactly the same as Eq. (9.36). The fundamental commutators in Eq. (9.38)

leads to first quantization; therefore, we seek classical conjugate variables in electromagnetic

field, converting them into quantum mechanical operators to accomplish second quantization.

9.5 Lagrangian and Hamiltonian of Classical Electromagnetic Field

Lagrangian for a particle (mass: m, velocity: v, and charge e) in an electromagnetic field (electric

field E and magnetic field B):

L =12mv 2−[eφ(r)−e

cv ⋅A(r)]+ 1

8π(| E |2 − | B |2)d 3r∫

=12mv 2−[eφ(r)−e

cv ⋅A(r)]+ 1

8π(|−

1c!A−∇φ |2 − |∇×A |2)d 3r∫ .

(9.39)

Denote the conjugate momenta to variables r, A and φ as p, ∏ and Pφ , respectively.

p =∂L∂ r

= mv−ec

A . (9.40)

∏=

∂L∂ A

=1

4πc(1cA +∇φ) =−

E4πc

, (9.41)

which is essentially the electric field.

Pφ

=∂L∂ φ

= 0 , (9.42)

i.e., the generalized conjugate momentum Pφ vanishes. Then the corresponding Hamiltonian:

H = (r ⋅p + A ⋅∏+ φPφ)−L

=1

2m(p−e

cA)2 +eφ+

18π

(16π2c

2 |∏ |2 + |∇×A |2)d 3r∫ . (9.43)

154

Since Pφ

= 0 , we can use the gauge φ = 0 so that only variables ∏ and A are in H . This

gauge imposes the transversality constraints of ∇⋅A = 0 and ∇⋅∏= 0 .

9.6 Field Quantization

In order to quantize the field, we require ∏ and A to be canonical, i.e.,

Classically,

{q

i,p

j} = δ

ij; {A

i(r,t),Π

j( ′r ,t)} = δ

ijδ(r− ′r ). (9.44)

In quantum mechanics, the correspondence principle suggests that

[q

i,p

j] = iδ

ij; [A

i(r,t),Π

j( ′r ,t)] = iδ

ijδ(r− ′r ) . (9.45)

First, let’s work in the classical theory of electrodynamics. The Fourier transformations:

A(r) = [b(k)eik⋅r +∫ b*(k)e−ik⋅r ]d 3k , (9.46)

∏(r) =

14πic

k[b(k)eik⋅r −∫ b*(k)e−ik⋅r ]d 3k . (9.47)

Define dimensionless variables a:

b(k) =

c2

4π2ω

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

1/2 ξ(kλ)

λ=1

3

∑ a(kλ) . (9.48)

Imposing the transversality constraints of ∇⋅A = 0 and ∇⋅∏= 0 , one can derive

k ⋅[b(k)+ b*(k)] = 0 , and k ⋅[b(k)−b*(k)] = 0 , respectively; thus k ⋅b(k) = 0⇒ a(k3) = 0.

A(r) =

c2

4π2ω

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

1/2

∫ξ(kλ)

λ=1

2

∑ [a(kλ)eik⋅r +a*(kλ)e−ik⋅r ]d 3k , (9.49)

∏(r) =

1i

ω64π4c2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟∫1/2 ξ(kλ)

λ=1

2

∑ [a(kλ)eik⋅r −a*(kλ)e−ik⋅r ]d 3k =−E(r)4πc

. (9.50)

The Poisson brackets are:

{Ai(r),A

j( ′r )} = 0, {a(kλ),a( ′k ′λ )} = 0

{Πi(r),Π

j( ′r )} = 0, {a*(kλ),a*( ′k ′λ )} = 0

{Ai(r),Π

j( ′r )} = δ

ijδ(r− ′r ), {a(kλ),a*( ′k ′λ )} =−iδ

λ ′λδ(k− ′k )

(9.51)

155

The Hamiltonian for electromagnetic field is

H

field=

18π

(16π2c

2 |∏ |2 + |∇×A |2)d 3r∫ = ω[a(kλ)a*∫λ=1

2

∑ (kλ)]d 3k . (9.52)

Define

q(kλ)≡ 1

2ω[a(kλ)+a*(kλ)] , (9.53)

and

p(kλ)≡−i ω / 2 [a(kλ)−a*(kλ)] , (9.54)

the familiar form consisting oscillators is recovered from Eq. (9.52):

H

field=

12

p2(kλ)+

12ω2

q2(kλ)

⎡

⎣⎢⎢

⎤

⎦⎥⎥∫

λ=1

2

∑ d3k . (9.55)

Now simply change the classical variables p and q to the quantum mechanical operators p

and q , we can quantize the electromagnetic field! First, the fundamental commutator is:

[q(kλ), p(′k ′λ )] = i{q(kλ),p( ′k ′λ )} = iδ

λ ′λδ(k− ′k ) . (9.56)

Defining the corresponding lowering ( a ) and raising ( a† ) operators based on the generalized

coordinate ( q ) and momentum ( p ) operators:

a(kλ) =

ω2

q + i1

2ωp , (9.57)

a†(kλ) =

ω2

q− i1

2ωp . (9.58)

The fundamental commutator becomes:

[a(kλ),a†( ′k ′λ )] = δ

λ ′λδ(k− ′k ) (9.59)

The quantized vector potential A and its conjugate momentum ∏ (essentially the electric field E) are:

A(r) =

c2

4π2ω

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

1/2

∫ξ(kλ)

λ=1

2

∑ [a(kλ)eik⋅r + a†(kλ)e−ik⋅r ]d 3k , (9.60)

156

∏(r) =

1iω

64π4c2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟∫1/2 ξ(kλ)

λ=1

2

∑ [a(kλ)eik⋅r − a†(kλ)e−ik⋅r ]d 3k . (9.61)

Finally the Hamiltonian for the quantized electromagnetic field

H

field= a

†(kλ)a(kλ)+12

⎡

⎣⎢⎢

⎤

⎦⎥⎥ω d 3k∫

λ=1

2

∑ (9.62)

The field operators:

E(r,t) = i

ω4π2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

1/2

∫ξ(kλ)

λ=1

2

∑ [a(kλ)ei(k⋅r−ωt )− a†(kλ)e−i(k⋅r−ωt ) ]d 3k , (9.63)

B(r,t) = i

c2

4π2ω

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟∫1/2

[k×ξ(kλ)]

λ=1

2

∑ [a(kλ)ei(k⋅r−ωt )− a†(kλ)e−i(k⋅r−ωt ) ]d 3k . (9.64)

Notes:

(1) In Eqs. (9.60) and (9.61), the time-dependency is omitted, and the full expression is

A(r,t) = A

k

ξ(kλ)[a(kλ)ei(k⋅r−ωt ) + a†(kλ)e−i(k⋅r−ωt ) ]d 3k∫

λ=1

2

∑ (9.65)

where A

k=

c2

4π2ω=

c4π2k

and ω = ck .

(2) The above expression uses infinite volume and thus Dirac δ(k− ′k ) to normalize the

planewave eik⋅k . In literature, the big-box normalization with volume Ω is also widely used:

1Ω k∑ ↔

1(2π)3

d 3k∫ , (9.66)

which is normalized to δk, ′k . But here we shall use

1

Ω k∑ ↔

1

(2π)3d 3k∫ , (9.67)

Why? Then

A(r,t) = A

k

ξ(kλ)[a(kλ)ei(k⋅r−ωt ) + a†(kλ)e−i(k⋅r−ωt ) ]

λ,k∑ , (9.68)

where A

k=

2πc2

Ωω=

2πcΩk

.

157

9.7 Photon States and Wave Functions

Photons are bosons, behaving as independent simple harmonic oscillator. The ket state for

photons of wave vector k and polarization λ is denoted as nkλ

, where nkλ is an integer that is

the number of photons in that state. The state 0 is the (photon) vacuum.

akλ 0 = 0, a†

kλ 0 = kλ = 1kλ, (9.69)

akλ nkλ = nkλ nkλ −1 , a†

kλ nkλ = nkλ +1 nkλ +1 , (9.70)

nkλ =a†

kλ( )nkλ

nkλ !0 , 0 =

akλ( )nkλ

nkλ !nkλ , (9.71)

N kλ ≡ a†

kλakλ, N kλ nkλ = nkλ nkλ (9.72)

Here N is the photon number operator. Energy of the vacuum state [derived from Eq. (9.62)]:

E

0=

12ωd 3k = c kd 3k∫∫

λ=1

2

∑ . (9.73)

Thus vacuum has infinite energy! Q: Does this even make sense? How to experimentally verify? A ket state with photons of different momenta is designated as

nk1λ1

nk2λ2…nkmλm

, then

H nk1λ1

nk2λ2…nk

mλ

m

= (nkλ +12kλ

∑ )ω nk1λ1nk2λ2

…nkmλ

m

. (9.74)

Classical momentum

p =

14πc

(E×B)d 3r∫ , (9.75)

Momentum operator in quantum mechanics using the correspondence principle:

p =

14πc

(E×B)d 3r∫ = akλ† akλkd 3k∫

λ=1

2

∑ = N kλkd 3k∫λ=1

2

∑ , (9.76)

Which is obviously corrected and expected, but mathematically it needs some work to prove it.

p kλ = k kλ , (9.77)

p nk1λ1

nk2λ2…nkmλm

= nkλk nk1λ1nk2λ2

…nkmλmkλ∑ . (9.78)

Summary:

quantum state ofelectromagnetic field

⎧⎨⎪⎪⎪

⎩⎪⎪⎪

⎫⎬⎪⎪⎪

⎭⎪⎪⎪↔ quantum state of

oscillators

⎧⎨⎪⎪

⎩⎪⎪

⎫⎬⎪⎪

⎭⎪⎪↔

nkλ : number of

photons at (k,λ)

⎧⎨⎪⎪⎪

⎩⎪⎪⎪

⎫⎬⎪⎪⎪

⎭⎪⎪⎪

158

Next we derive the wave functions of photons. A general photon state by superposition:

Φ = ckλ kλ

kλ∑ = ckλakλ

†

0kλ∑ . (9.79)

The normalization condition Φ Φ = 1 ⇒

|ckλ |kλ∑

2= 1. (9.80)

Here φkλ is the scalar photon wave function.

Let’s focus on a photon state with specific (k, λ):

kλ ′k ′λ =δλ ′λδk, ′k (Big-box volume Ω)

δλ ′λδ(k− ′k ) (Ω→∞)

⎧⎨⎪⎪⎪

⎩⎪⎪⎪

(9.81)

(Q: how to derive it?) The Real-space representation:

φkλ(r)≡ r kλ =

eik⋅r

Ω, (9.82)

because p kλ = k kλ and in the r− representation , p =−i∇ . (Q: does λ play a role in

photon wave function?) We can’t use a scalar wave function to fully describe a particle with

spin; e.g., we use spinors for electron wave functions. Here for spin-one photons, we use vector

wave functions:

φkλ(r)≡

ξkλφkλ(r) =

ξkλ

eik⋅r

Ω, (9.83)

where ξkλ indicates that photon has none-zero spin.

A spin-one particle usually has three values along an arbitrary direction; however, it is not

true for photons. Helicity of photons: the components of spin are parallel to momentum.

Mathematically, the tranversality condition ( k ⋅ξ = 0 ) restricts the photon spin to be ± along

k . (Q: Why? Hint: angular momentum including spin is the generator operator for rotation.)

Physically, it is irrelevant to the chosen gauge; instead, it is the consequence of ultra-relativity,

and a full understanding requires quantum field theory.

The electromagnetic fields associated with a photon state Φ :

159

E(r,t)≡ 0 E(r,t) Φ = i

2πωΩ

ξkλckλe

i(k⋅r−ωt )

kλ∑ , (9.84)

B(r,t)≡ 0 B(r,t) Φ = i

2πc2

ωΩ(k×

ξkλ)ckλe

i(k⋅r−ωt )

kλ∑ . (9.85)

9.8 Spontaneous Radiation

We consider the spontaneous decay of H atom from 2lm to

100 . Semi-classically, the

transition rate is zero; however, experimentally the lifetime τ = 1/R = 1.60 ns. Now we use

the full quantum-mechanical treatment, i.e., field quantization, to compute τ .

The perturbing Hamiltonian is

′H (t) =−

e

mec

A ⋅ p , (9.86)

A(r,t) = A

k

ξ(kλ)∫

λ=1

2

∑ [a(kλ)ei(k⋅r−ωt ) + a†(kλ)e−i(k⋅r−ωt ) ]d 3k, with Ak≡c2

4π2ω

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

1/2

. (9.87)

The initial state: i = 2lm ⊗ 0 , Ei

= E2lm

.

The final state: f = 100 ⊗ kλ ,

E

f= E

100+ ω .

This is a periodic perturbation because ′H (t) ~ ′H cos(ωt) , so we use Fermi’s Golden Rule to

calculate the transition rate

R

i→f≡ lim

t→∞

d

dtP

i→f(t) =

2π

f ′H i2δ(E

f−E

i) . (9.88)

The (time-independent) perturbation matrix element

′Hfi≡ f ′H i =−

e

mec

f A(r) ⋅ p i , (9.89)

where

f A(r) ⋅ p i = 100 kλ A(r) 0 ⋅ p 2lm . (9.90)

Because

kλ A(r) 0 = A

k

ξkλe

−ik⋅r , (9.91)

we obtain that

f A(r) ⋅ p i = A

kψ

100* (r)e−ik⋅r

ξkλ ⋅(−i∇)∫ ψ

2lm(r)d 3r . (9.92)

160

Using the electric dipole approximation, i.e., e−ik⋅r ≈ 1 , and replacing ξ ⋅ p by (ime

ω)ξ ⋅ r :

f A(r) ⋅ p i = A

k(im

eω) ψ

100* (r)[

ξkλ ⋅ r]∫ ψ

2lm(r)d 3r . (9.93)

We need to evaluate:

I kλ = ψ

100* (r)[

!ξkλ ⋅ r]∫ ψ

2lm* (r)d 3r = 100

!ξkλ ⋅ r 2lm . (9.94)

′n ′l ′m!ξ ⋅ r nlm = sin θ cosφ ′n ′l ′m x nlm

+ sin θ sinφ ′n ′l ′m y nlm + cosθ ′n ′l ′m z nlm (9.95)

One can use the selection rule (under the electric dipole approximation) to exclude the zero

matrix elements of ′n ′l ′m

ξ ⋅ r nlm :

(1) Δl = ±1 (9.96)

(2) ′n ′l ′m x nlm : Δm = ±1; ′n ′l ′m y nlm : Δm = ±1; ′n ′l ′m z nlm : Δm = 0. (9.97)

For the 2lm → 100 (or specifically 2p→ 1s ) transition, we need to evaluate the

following: 100 z 210 =

215

35a

0, where a0

is the Bohr radius.

Similarly, 100 x 21±1 = ±

27

35a

0, and

100 y 21±1 = i

27

35a

0.

Now we can evaluate the associated transition rate using FGR by replacing the delta-function

by (i) integrating over k and (ii) summing over λ = 1,2 :

Ri→f

all =2π!

′Hfi(kλ)

2δ(E

f−E

i)d 3k∫

λ∑

=2π!

−e

mec

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

Ak

2 | imeω |2 I kλ

2 δ(E1s

+ !ck−E2p

)dΩk2dk∫∫

λ∑

=e

2

2π14π

I kλ2

dΩ∫⎡

⎣⎢⎢

⎤

⎦⎥⎥ωδ(E1s

+ !ck−E2p

)4πk 2dk∫

λ∑

(9.98)

Let’s do the angular part first, and define

I 2 ≡

14π

I kλ2 dΩ∫ . (9.99)

Here I 2 is taken as the average over (1) m =−1,0,1 ; (2) all the orientations of ξkλ :

161

I 2 =

14π

[Ix2 sin2(θ)cos2(φ)+ I

y2 sin2(θ)sin2(φ)+ I

z2 cos2(θ)]sin(θ)dθdφ∫ , (9.100)

where

I 2x

=13−

27

35

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

+ 02 +27

35

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥a

02 =

215

311a

02 ,

I 2y

= Iz2 = I

x2 =

215

311a

02 .

(9.101)

This shows that I kλ is independent of the orientation, thus I 2 =

215

311a

02 . The transition rate

for this spontaneous radiation ( 2p→ 1s ) thus is (Note: 2 below is due to summation over λ ):

R

i→fall =

e2

2π⋅2 ⋅

215

311a

02 ωδ(E

1s+ !ck−E

2p)4πk 2 dk∫ . (9.102)

Note that 2 in the above expression is due to summation over λ . We evaluate the integral:

ωδ(E

1s+ !ck−E

2p)4πk 2 dk∫ =

4πc( ′k )3

!c, (9.103)

where ω = ck and

′k =

E2p−E

1s

!c=

3e2

8a0!c

. (9.104)

Finally we obtain the spontaneous transition rate from 2p to 1s of H atom,

R

i→fall =

23

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

8

α5 mec2

!= 6.27×108 s−1 , (9.105)

where the fine structure constant α =

e2

c=

1137.036

. The corresponding lifetime

τ = 1/R = 1.60 ns , in excellent agreement with experimental data τ = 1.600 ± 0.004 ns.

[Phys. Rev. 148, 1 (1966)]

162

9.9 Optical Absorption and Stimulated Radiation

Now let’s consider the reverse process: optical absorption. The optical absorption by solids is a

crucial topic in condensed matter physics. Here we only study the absorption by gas. There are a

number of processes whereby gases can absorb electromagnetic radiation:

1) Collisions of atoms.

2) Rotation and vibration.

3) Inelastic processes such as Raman scattering.

4) Two-photon absorptions.

But we omit all the above processes, focusing only on the simplest and most crucial case: optical

absorption by a single static atom. The results for gas in a box are multiplied by the number of

atoms NA.

The perturbing Hamiltonian still is ′H (t) =

−em

ec

A ⋅ p .

The initial state: I = i ⊗ nkλ

,

The final state: F = f ⊗ nkλ −1 .

Fermi’s Golden Rule:

Rkλ =

2π

em

ec

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

f ; nkλ −1 A(r) ⋅ p i; nkλ

2δ(E

f−E

i−ω) , (9.106)

with Ei and

E

f the initial and final energies for electronic states

i and

f , respectively.

Here Rkλ is the rate for the specific transition associated with a photon mode (kλ) , and

M

fi≡ f ; nkλ −1 A(r) ⋅ p i; nkλ = f nkλ −1 A(r) nkλ ⋅ p i , (9.107)

nkλ −1 A(r) nkλ = A

k

ξkλe

ik⋅r nkλ. (9.108)

Thus

M

fi= nkλ Ak

(ξkλ ⋅p fi

), (9.109)

where p

fi≡ f eik⋅rp i , the optical transition matrix element.

Note that semi-classically, A is NOT an operator, and the electromagnetic field is classical –

no quantum states, instead it is a continuous field describe by the vector potential A(r,t) :

163

A(r,t) = A0[ei(k⋅r−ωt ) +e−i(k⋅r−ωt ) ] , with A0

a constant vector potential. The matrix elements M

if

have exactly the same form as the quantum version:

M

fi= | A

0| (ξkλ ⋅p fi

), (9.110)

therefore, | A0|2= nkλAk

2 ∝ nkλ .

The total transition rate for a gas is obtained by (1) summation of all possible transitions and

(2) multiplying the number of atoms NA:

R = N

ARkλ

kλ∑ . (9.111)

9.9.1 Absorption Coefficient and Cross Section

Now we need to consider physical quantities, which are

measurable. An important parameter is the flux Fkλ of photons

in the gas, which is proportional to the Poynting vector S. Fkλ

has the dimensional units of number of photons/(cm2·s):

Fkλ =

Skλ

ω=

cnkλ

Ω, (9.112)

where Ω is the renormalization volume. Then

nkλ = Fkλ

Ωc

, (9.113)

Rkλ = nkλ

2π

em

ec

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

Ak2 |ξkλ ⋅p fi

|2 δ(Ef−E

i−ω)

= Fkλ

Ωc

2π

em

ec

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

Ak2 |ξkλ ⋅p fi

|2 δ(Ef−E

i−ω)

(9.114)

FkλA = number of photons with mode (kλ) per second going through the slab. Note: A is area.

−AdFkλ

= NAR

kλ , (9.115)

⇒ −(Adx)

dFkλ

dx= N

AR

kλ, (9.116)

Fig. 9.1: Geometry for Beer’s law.

164

⇒ −Ω

dFkλ

dx= N

AR

kλ= N

AFkλ

Ωc

2π

em

ec

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

Ak2 |ξkλ ⋅p fi

|2 δ(Ef−E

i−ω), (9.117)

⇒

dFkλ

dx=−α(kλ)Fkλ . (9.118)

Eq. (9.118) is Beer’s law for optical absorption:

F(x) = F0e−αx . (9.119)

Here the absorption coefficient

α(kλ) = NA

2πc

em

ec

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

Ak2 |ξkλ ⋅p fi

|2 δ(Ef−E

i−ω)

=n

A

cω2πem

ec

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

|ξkλ ⋅p fi

|2 δ(Ef−E

i−ω).

(9.120)

We used A

k≡

2πc2

ωΩ

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

1/2

and n

A≡

NA

Ω. Then the total absorption coefficient

α(ω) = α(kλ)d 3k∫

λ=1

2

∑ . (9.121)

Without losing generality, set the direction of matrix element p

fi along z-axis, then

|ξkλ ⋅p fi

|2= sin2(θ)pfi2

λ=1

2

∑ (9.122)

where θ is the angle between k and p

fi. The absorption coefficient

α(ω) =

nA

c43

e2

m22c3p

fi2 (E

i−E

f) . (9.123)

Absorption cross section (per atom):

σ(ω) =

α(ω)n

A

=43

e2

m22c 4p

fi2 (E

i−E

f) . (9.124)

9.9.2 Stimulated Radiation

Radiation (emission) is a complimentary experiment to absorption. The atom must first be

excited so that some of the electrons are in excited states. This step is usually accomplished by

165

optical absorption or by bombarding the atoms with a beam of energetic electrons. After an

electron is in an excited state, it can return to the ground state by the emission of a photon.

Fermi’s Golden rule:

Rkλ =

2π

em

ec

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

f ; nkλ +1 A(r) ⋅ p i; nkλ

2δ(E

f+ ω−E

i) , (9.125)

M

fi≡ F A(r) ⋅ p I = f nkλ +1 A(r) nkλ ⋅ p i (9.126)

nkλ +1 A(r) nkλ = A

k

ξkλe

−ik⋅r nkλ +1 (9.127)

M

fi= nkλ +1 A

k(ξkλ ⋅p fi

), (9.128)

where p

fi≡ f e−ik⋅rp i . So the FGR is

Rkλ = (nkλ +1)

2π

e2

m2ec2

Ak2 |ξkλ ⋅p fi

|2 δ(Ef

+ ω−Ei) . (9.129)

Comments:

1) The optical transition matrix elements (p

fi) for stimulated radiation are identical to those for

the corresponding absorption.

2) Transition rate (Rkλ) for stimulated radiation is proportional to the number of available

photons (nkλ) , while Rkλ for spontaneous radiation is independent of nkλ .

Consider two electronic states i and

f , with

E

f= E

i+ ω . Optical absorption rate:

R

i→fabs = B

i→fN

iρ(ω) , (9.130)

where Ni the number of atoms in the

i state, ρ(ω) the energy density of photons with

frequency ω , and B

i→f the Einstein B coefficient. The radiation rate is

R

f→irad = B

f→iN

fρ(ω)+ A

f→iN

f, (9.131)

where N

f the number of atoms in the

f state, and

A

f→i the Einstein A coefficient.

From comment (1),

B

i→f= B

f→i≡ B . (9.132)

166

Let A≡ A

f→i. The equilibrium condition is:

R

i→fabs = Rrad

f→i, (9.133)

and one can derive that

ρ(ω) =

AB

1(N

i/N

f)−1

. (9.134)

Statistical mechanics show that

Ni

Nf

=e−Ei/kBT

e−Ef /kBT

= e−(Ei−Ef )/kBT = eω/kBT , (9.135)

where T the temperature and kB the Boltzmann constant.

Bose factor:

nω

=1

eω/kBT −1, (9.136)

ρ(ω) =

Ωω2

π2c3

1

eω/kBT −1, (9.137)

we find that

AB

=Ωω2

π2c3. (9.138)

Based on Einstein relations, one can write down the balance as

dIdx

= const ⋅I(Nf−N

i) , (9.139)

where I is light density. Since E

f> E

i, N

f< N

i at equilibrium, thus

dIdx

< 0 , light is

diminishing. In order to make a laser (light amplification by stimulated emission of radiation),

inverse population ( N

f> N

i) is required, which means negative temperature. A simplest

implementation is a three-level system with a “forbidden” transition between two states.