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):
∂2 E
= c 2∇2E,
2
∂t
and
∂2 B
= c 2∇2B .
2
∂t
(9.1)
Monochromatic electromagnetic wave:
E(r,t) = Re[E(r)e iωt ] = E0 cos(ωt − k ⋅ r + ϕ0 ),
B(r,t) = Re[B(r)e iωt ] = B0 cos(ωt − k ⋅ r + ϕ0 ).
(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=
c
E×B,
4π
and S =
c 2
E .
4π 0
(9.3)
Note: we use the Gaussian units throughout this class.
On the other hand, S = c
therefore, E 02 = N ⋅
Nω
, where Ω is the volume of the normalization big box;
Ω
4πω
, and the field is quantized. When N is very large, the classical
Ω
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.
148
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,
1 ∂B
∇×E = −
,
c ∂t
∇ ⋅ B = 0,
1 ∂E 4π
∇×B =
+
j.
c ∂t
c
(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 + St .
(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)
⎡
1 ∂A ⎤⎥
∇× ⎢ E +
=0.
⎢
⎥
c
∂t
⎣
⎦
(9.8)
Then
If S = E +
1 ∂A
, one can write
c ∂t
∇×(−∇φ + ∇×m) = 0 = ∇×(∇×m) .
The above equation is satisfied if we set m = 0 , therefore
1 ∂A
E = −∇φ −
,
c ∂t
B = ∇×A .
(9.9)
(9.10)
(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 +
1 ∂(∇ ⋅ A)
= −4πρ ,
c
∂t
149
(9.12)
∇×(∇×A) +
1 ∂ 2 A 1 ⎛⎜ ∂A ⎞⎟ 4π
⎟⎟ =
+ ∇⎜
j.
c
c 2 ∂t 2 c ⎜⎝ ∂t ⎟⎠
(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 (Ax , Ay , Az , φ) are not uniquely
determined by E and B, since E and B remain unchanged under the gauge transformation:
A → A + ∇Λ
E→−
and φ → φ −
1 ∂Λ
,
c ∂t
⎛
1 ∂(A + ∇Λ)
1 ∂Λ ⎞⎟
1 ∂A
⎟⎟ = −
− ∇ ⎜⎜⎜φ −
− ∇φ ,
c
∂t
c ∂t ⎟⎠
c ∂t
⎝
B → ∇×(A + ∇Λ) = ∇×A ,
(9.14)
(9.15)
(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) 3
d r′ ,
r − r′
(9.18)
because Eq. (9.12) becomes
∇2φ = −4πρ .
(9.19)
∇×(∇×A) = −∇2A + ∇(∇ ⋅ A) ,
(9.20)
Using this identity,
one can show that under Coulomb gauge, Eq. (9.13) for vector potential becomes
∇2A −
1 ∂2 A
4π
=−
j (r,t) ,
2
2
c t
c ∂t
(9.21)
where the transverse part of current density jt is defined by
⎡
⎤
1
j(r′,t) 3 ⎥
⎢
jt (r,t) =
∇× ⎢∇× ∫
d r′ ⎥ .
4π
r − r′
⎢
⎥
⎣
⎦
150
(9.22)
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 +
1 ∂φ
= 0,
c ∂t
(9.23)
which causes Eqs. (9.12) and (9.13) change to
1 ∂2 ψ
∇ φ− 2
= −4πρ ,
c ∂t 2
(9.24)
1 ∂2 A
4π
=−
j.
2
2
c
c ∂t
(9.25)
2
∇2A −
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,
2
p2
1 ⎛⎜
e ⎞
H=
→H =
⎜⎜p − A⎟⎟⎟ +eφ ,
2m
2m ⎝
c ⎟⎠
with e the charge (negative for an electron). Semi-classically,
2
1 ⎛⎜
e ⎞⎟
Ĥ =
⎜p̂ − A⎟⎟ +eφ .
2m ⎜⎝
c ⎟⎠
(9.26)
(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
A2 can be omitted for linear optical properties. In addition, we
2
2mc
consider the case of the free electromagnetic field ( ρ = 0 , j = 0 ), in which the gauge invariance
allows we choose the simplest Coulomb gauge:
151
∇ ⋅ A = 0;
φ = 0.
(9.28)
Then
H=
⎞ p2
1 ⎛⎜ 2 e
e
e
−
A⋅p.
⎜⎜p − A ⋅ p − p ⋅ A⎟⎟⎟ =
⎟⎠ 2m mc
2m ⎝
c
c
(9.29)
Here H = H 0 + H cR , and the charge-radiation interaction Hamiltonian.
H cR = −
e
A⋅p .
mc
(9.30)
Within electric dipole approximation, i.e., for electromagnetic wave with small wave number k
(long wavelength),
Ĥ 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(qi , pi ,t) = ∑ (piqi ) − L(qi , qi ,t) .
(9.32)
i
Here L = T −V is the Lagrangian, with T and V kinetic and potential energies, respectively.
qi are generalized coordinates, qi =
dqi
dt
, and the generalized momenta,
pi =
∂L(qi , qi ,t)
,
∂qi
(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 = mxi , and
H=
1 3 2
∑ p +V(xi ) .
2m i=1 i
(9.34)
Ĥ =
1 3 2 ˆ
∑ p̂ +V(x̂i ) .
2m i=1 i
(9.35)
Then in quantum mechanics,
In construction of classical Hamiltonian, we need canonical coordinates qi and pi , which
satisfy the following Poisson bracket relations:
152
{qi ,q j } = 0;
{pi , p j } = 0;
{qi , p j } = δij ,
(9.36)
where the Poisson bracket is defined as
⎛ ∂A ∂B ∂A ∂B ⎞⎟
⎟⎟ .
{A(p,q),B(q, p)}PB ≡ ∑ ⎜⎜⎜
−
⎟⎟
⎜
∂q
∂p
∂p
∂q
s ⎝
s
s
s
s⎠
(9.37)
The quantum mechanical version of the commutators:
⎡q̂ , q̂ ⎤ = 0;
⎣⎢ i j ⎦⎥
⎡ p̂ , p̂ ⎤ = 0;
⎣⎢ i j ⎦⎥
⎡q̂ , p̂ ⎤ = iδ ,
ij
⎣⎢ i j ⎦⎥
(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):
1
e
1
L = mv 2 −[eφ(r)− v ⋅ A(r)]+
(| E |2 − | B |2 )d 3r
2
c
8π ∫
1
e
1
1 !
= mv 2 −[eφ(r)− v ⋅ A(r)]+
(| − A
− ∇φ |2 − | ∇×A |2 )d 3r .
∫
2
c
8π
c
(9.39)
Denote the conjugate momenta to variables r, A and φ as p, ∏ and Pφ , respectively.
∂L
e
= mv − A .
∂ r
c
(9.40)
∂L
1 1 
E
=
(
A
+
∇φ)
=
−
,

∂A
4πc c
4πc
(9.41)
p=
∏=
which is essentially the electric field.
Pφ =
∂L
= 0,
∂φ
(9.42)
i.e., the generalized conjugate momentum Pφ vanishes. Then the corresponding Hamiltonian:
 ⋅ ∏+φP
 )− L
H = (r ⋅ p + A
φ
=
1
e
1
(p − A)2 +eφ +
(16π 2c 2 | ∏ |2 + | ∇×A |2 )d 3r .
∫
2m
c
8π
153
(9.43)
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,
{qi , p j } = δij ;
{Ai (r,t), Π j (r′,t)} = δij δ(r − r′).
(9.44)
In quantum mechanics, the correspondence principle suggests that
[qi , p j ] = iδij ;
[Ai (r,t), Π j (r′,t)] = iδij δ(r − r′) .
(9.45)
First, let’s work in the classical theory of electrodynamics. The Fourier transformations:
A(r) = ∫ [b(k)e ik⋅r +b*(k)e −ik⋅r ]d 3k ,
∏(r) =
1
k[b(k)e ik⋅r −b*(k)e −ik⋅r ]d 3k .
∫
4πic
(9.46)
(9.47)
Define dimensionless variables a:
1/2
⎛ c2 ⎞ 
b(k) = ∑ ⎜⎜⎜ 2 ⎟⎟⎟ ξ(kλ)a(kλ) .
⎜ 4π ω ⎟⎠
λ=1 ⎝
3
(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.
(9.49)
1/2
1 ⎛⎜ ω ⎞⎟ 
E(r)
⎟ ξ(kλ)[a(kλ)e ik⋅r −a *(kλ)e −ik⋅r ]d 3k = −
.
⎜⎜
4 2⎟
i ⎝ 64π c ⎟⎠
4πc
(9.50)
A(r) = ∑ ∫
λ=1
2
∏(r) = ∑ ∫
λ=1
1/2
⎛ c 2 ⎞⎟ 
ik⋅r
*
−ik⋅r
3
⎜⎜
⎟
⎜⎜ 4π 2ω ⎟⎟ ξ(kλ)[a(kλ)e + a (kλ)e ]d k ,
⎝
⎠
2
The Poisson brackets are:
{Ai (r),Aj (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′)} = δij δ(r − r′),
{a(kλ),a *(k′λ′)} = −iδλλ′δ(k − k′)
154
(9.51)
The Hamiltonian for electromagnetic field is
H field =
2
1
2 2
2
2
3
(16π
c
|
∏
|
+
|
∇×A
|
)d
r
=
ω[a(kλ)a *(kλ)]d 3k .
∑
∫
8π ∫
λ=1
(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):
2
⎡1
⎤
1
H field = ∑ ∫ ⎢ p 2(kλ) + ω 2q 2(kλ)⎥d 3k .
⎢2
⎥
2
λ=1
⎣
⎦
(9.55)
Now simply change the classical variables p and q to the quantum mechanical operators p̂
and q̂ , we can quantize the electromagnetic field! First, the fundamental commutator is:
[q̂(kλ), p̂(k′λ′)] = i{q(kλ), p(k′λ′)} = iδλλ′δ(k − k′) .
(9.56)
Defining the corresponding lowering ( â ) and raising ( â † ) operators based on the generalized
coordinate ( q̂ ) and momentum ( p̂ ) operators:
â(kλ) =
ω
1
q̂ + i
p̂ ,
2
2ω
(9.57)
â †(kλ) =
ω
1
q̂ −i
p̂ .
2
2ω
(9.58)
[â(kλ), â †(k′λ′)] = δλλ′δ(k − k′)
(9.59)
The fundamental commutator becomes:
The quantized vector potential A and its conjugate momentum ∏ (essentially the electric field E) are:
2
Â(r) = ∑ ∫
λ=1
1/2
⎛ c 2 ⎞⎟ 
ik⋅r
†
−ik⋅r
3
⎜⎜
⎟
⎜⎜ 4π 2ω ⎟⎟ ξ(kλ)[â(kλ)e + â (kλ)e ]d k ,
⎝
⎠
155
(9.60)
2
ˆ =
∏(r)
∑∫
λ=1
1/2
1 ⎛⎜ ω ⎞⎟ 
⎟⎟ ξ(kλ)[â(kλ)e ik⋅r − â †(kλ)e −ik⋅r ]d 3k .
⎜
i ⎜⎝ 64π 4c 2 ⎟⎠
(9.61)
Finally the Hamiltonian for the quantized electromagnetic field
2
⎡
1⎤
Ĥ field = ∑ ∫ ⎢â †(kλ)â(kλ) + ⎥ ω d 3k
⎢
2 ⎥⎦
λ=1
⎣
(9.62)
The field operators:
2
Ê(r,t) = ∑ ∫
λ=1
2
(9.63)
1/2
B̂(r,t) = ∑ ∫
λ=1
1/2
⎛ ω ⎞⎟ 
⎜
i ⎜ 2 ⎟⎟ ξ(kλ)[â(kλ)e i(k⋅r−ωt ) − â †(kλ)e −i(k⋅r−ωt ) ]d 3k ,
⎜⎝ 4π ⎟⎠
⎛ c 2 ⎞

i ⎜⎜⎜ 2 ⎟⎟⎟ [k × ξ(kλ)][â(kλ)e i(k⋅r−ωt ) − â †(kλ)e −i(k⋅r−ωt ) ]d 3k .
⎝⎜ 4π ω ⎟⎠
(9.64)
Notes:
(1) In Eqs. (9.60) and (9.61), the time-dependency is omitted, and the full expression is
2

Â(r,t) = ∑ ∫ Ak ξ(kλ)[â(kλ)e i(k⋅r−ωt ) + â †(kλ)e −i(k⋅r−ωt ) ]d 3k
(9.65)
λ=1
where Ak =
c 2
=
4π 2ω
c
and ω = ck .
4π 2k
(2) The above expression uses infinite volume and thus Dirac δ(k − k′) to normalize the
planewave e ik⋅k . In literature, the big-box normalization with volume Ω is also widely used:
1
1
↔
∑
Ω k
(2π)3
∫ d k,
3
(9.66)
which is normalized to δk,k′ . But here we shall use
1
Ω
Why? Then
∑
k
↔
1
(2π)3
3

Â(r,t) = ∑ Ak ξ(kλ)[â(kλ)e i(k⋅r−ωt ) + â †(kλ)e −i(k⋅r−ωt ) ] ,
λ,k
where Ak =
∫d k,
2πc 2
=
Ωω
2πc
.
Ωk
156
(9.67)
(9.68)
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 n kλ , where n kλ is an integer that is
the number of photons in that state. The state 0 is the (photon) vacuum.
â †kλ 0 = kλ = 1kλ ,
(9.69)
â †kλ n kλ = n kλ + 1 n kλ + 1 ,
(9.70)
â kλ 0 = 0,
â kλ n kλ = n kλ n kλ −1 ,
(â )
=
n kλ
†
kλ
n kλ
n kλ
0 ,
n kλ !
N̂ kλ ≡ â †kλâ kλ ,
0
(â )
=
kλ
n kλ !
n kλ ,
(9.71)
N̂ kλ n kλ = n kλ n kλ
(9.72)
Here N̂ is the photon number operator. Energy of the vacuum state [derived from Eq. (9.62)]:
2
E0 = ∑ ∫
λ=1
1
ωd 3k = c ∫ kd 3k .
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 n k λ n k λ …n k
1 1
Ĥ n k λ n k λ …n k
1 1
2 2
λ
m m
2 2
λ
m m
, then
1
= ∑ (n kλ + )ω n k λ n k λ …n k λ .
1 1
2 2
m m
2
kλ
(9.74)
Classical momentum
p=
1
(E×B)d 3r ,
∫
4πc
(9.75)
Momentum operator in quantum mechanics using the correspondence principle:
p̂ =
2
2
1
3
†
3
(
Ê×
B̂)d
r
=
â
â
kd
k
=
N̂ kλkd 3k ,
∑
∑
∫
∫
∫
kλ kλ
4πc
λ=1
λ=1
(9.76)
Which is obviously corrected and expected, but mathematically it needs some work to prove it.
p̂ kλ = k kλ ,
p̂ n k λ n k λ …n k
1 1
2 2
λ
m m
(9.77)
= ∑ n kλk n k λ n k λ …n k
kλ
1 1
2 2
λ
m m
.
(9.78)
Summary:
⎧ quantum state of
⎪
⎪
⎨
⎪
⎪
⎪
⎩ electromagnetic field
⎧
⎫ ⎧
⎪
⎪
⎪
⎪ n kλ : number of
quantum state of ⎫
⎪↔⎪
⎪
⎬ ⎪
⎨
⎬↔⎪
⎨
⎪
⎪
⎪
⎪ photons at (k,λ)
oscillators
⎪
⎪
⎪
⎭ ⎪
⎪
⎭ ⎩
⎪
⎩
157
⎫
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎭
Next we derive the wave functions of photons. A general photon state by superposition:
Φ = ∑ c kλ kλ = ∑ c kλâ kλ 0 .
†
kλ
(9.79)
kλ
The normalization condition Φ Φ = 1 ⇒
∑ |c
kλ
2
kλ
| = 1.
(9.80)
Here φkλ is the scalar photon wave function.
Let’s focus on a photon state with specific (k, λ):
⎧
⎪
⎪ δλλ′δk,k′
kλ k′λ′ = ⎪
⎨
⎪
δ δ(k − k′)
⎪
⎪
⎩ λλ′
(Big-box volume Ω)
(Ω → ∞)
(9.81)
(Q: how to derive it?) The Real-space representation:
φkλ (r) ≡ r kλ =
e ik⋅r
Ω
,
(9.82)
because p̂ kλ = k kλ and in the r − representation , p̂ = −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:

 e ik⋅r

,
φkλ (r) ≡ ξkλφkλ (r) = ξkλ
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 :
158
2πω 
ξkλc kλe i(k⋅r−ωt ) ,
Ω
kλ

2πc 2
(k × ξkλ )c kλe i(k⋅r−ωt ) .
ωΩ
E(r,t) ≡ 0 Ê(r,t) Φ = i ∑
B(r,t) ≡ 0 B̂(r,t) Φ = i ∑
kλ
(9.84)
(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) = −
2
Â(r,t) = ∑ ∫
λ=1
(9.86)

Ak ξ(kλ)[â(kλ)e i(k⋅r−ωt ) + â †(kλ)e −i(k⋅r−ωt ) ]d 3k,
The initial state: i = 2lm ⊗ 0 ,
The final state:
e
 ⋅ p̂ ,
m ec
1/2
⎛ c 2 ⎞
with Ak ≡ ⎜⎜⎜ 2 ⎟⎟⎟ .
⎜⎝ 4π ω ⎟⎠
(9.87)
Ei = E 2lm .
f = 100 ⊗ kλ , E f = E100 + ω .
This is a periodic perturbation because H ′(t) ~ H ′ cos(ωt) , so we use Fermi’s Golden Rule to
calculate the transition rate
Ri→f ≡ limt→∞
d
2π
Pi→f (t) =
f H′ i
dt

2
δ(E f − Ei ) .
The (time-independent) perturbation matrix element
e
H fi′ ≡ f H ′ i = −
f Â(r)⋅ p̂ i ,
m ec
(9.88)
(9.89)
where
Because
f Â(r)⋅ p̂ i = 100 kλ Â(r) 0 ⋅ p̂ 2lm .
(9.90)

kλ Â(r) 0 = Ak ξkλe −ik⋅r ,
(9.91)

*
f Â(r)⋅ p̂ i = Ak ∫ ψ100
(r)e −ik⋅rξkλ ⋅(−i∇)ψ2lm (r)d 3r .
(9.92)
we obtain that
159


Using the electric dipole approximation, i.e., e −ik⋅r ≈ 1 , and replacing ξ ⋅ p̂ by (im e ω)ξ ⋅ r :

*
f Â(r)⋅ p̂ i = Ak (im e ω)∫ ψ100
(r)[ξkλ ⋅ r]ψ2lm (r)d 3r .
(9.93)
We need to evaluate:
!
!
*
*
I kλ = ∫ ψ100
(r)[ξkλ ⋅ r]ψ2lm
(r)d 3r = 100 ξkλ ⋅ r 2lm .
!
n ′l ′m ′ ξ ⋅ r nlm = sin θ cosφ n ′l ′m ′ x nlm
+ sin θ sin φ n ′l ′m ′ y nlm + cos θ n ′l ′m ′ z nlm
(9.94)
(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
215
following: 100 z 210 = 5 a 0 , where a 0 is the Bohr radius.
3
Similarly, 100 x 21 ± 1 = ±
27
27
a
100
y
21
±
1
=
i
a .
,
and
35 0
35 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 :
all
Ri→f
=
2
2π
H fi′ (kλ) δ(E f − Ei )d 3k
∑
∫
! λ
2
⎛ e ⎞⎟
2π
⎜⎜−
⎟⎟ A2 | im ω |2 I 2 δ(E + !ck − E )dΩk 2 dk
=
∑
∫∫
e
kλ
1s
2p
⎜
⎜⎝ m ec ⎟⎟⎠ k
! λ
⎡ 1
⎤
e2
2
⎢
⎥ ωδ(E + !ck − E )4πk 2 dk
=
I
dΩ
∑
∫
∫
kλ
1s
2p
⎢
⎥
2π λ
⎣ 4π
⎦
(9.98)
Let’s do the angular part first, and define
I2≡
1
2
I kλ
dΩ .
∫
4π

Here I 2 is taken as the average over (1) m = −1, 0,1 ; (2) all the orientations of ξkλ :
160
(9.99)
I2=
1
[I x2 sin 2(θ)cos2(φ) + I y2 sin 2(θ)sin 2(φ) + I z2 cos2(θ)]sin(θ)dθdφ ,
∫
4π
(9.100)
where
2
2
⎡
⎛ 27 ⎞⎟ ⎤⎥ 2 215 2
1 ⎢⎛⎜ 27 ⎞⎟⎟
2
I = ⎢⎜⎜− 5 ⎟ + 0 + ⎜⎜⎜ 5 ⎟⎟ ⎥ a 0 = 11 a 0 ,
⎜⎝ 3 ⎟⎠ ⎥
3 ⎢⎜⎝ 3 ⎟⎠
3
⎢⎣
⎥⎦
215
I y2 = I z 2 = I x 2 = 11 a 02 .
3
2
x
This shows that I kλ
(9.101)
215 2
is independent of the orientation, thus I = 11 a 0 . The transition rate
3
2
for this spontaneous radiation ( 2p → 1s ) thus is (Note: 2 below is due to summation over λ ):
all
Ri→f
=
e2
215
⋅ 2 ⋅ 11 a 02 ∫ ωδ(E1s + !ck − E 2p )4πk 2 dk .
2π
3
(9.102)
Note that 2 in the above expression is due to summation over λ . We evaluate the integral:
∫
4πc(k ′)3
ωδ(E1s + !ck − E 2p )4πk dk =
,
!c
2
(9.103)
where ω = ck and
k′ =
E 2p − E1s
!c
=
3e 2
.
8a 0!c
(9.104)
Finally we obtain the spontaneous transition rate from 2p to 1s of H atom,
8
all
i→f
R
⎛ 2 ⎞⎟ 5 m c 2
= ⎜⎜⎜ ⎟⎟ α e = 6.27 ×108 s−1 ,
!
⎝ 3 ⎟⎠
where the fine structure constant α =
(9.105)
e2
1
=
. The corresponding lifetime
c 137.036
τ = 1 / R = 1.60 ns , in excellent agreement with experimental data τ = 1.600 ± 0.004 ns.
[Phys. Rev. 148, 1 (1966)]
161
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 N A .
The perturbing Hamiltonian still is H ′(t) =
−e
 ⋅ p̂ .
m ec
The initial state: I = i ⊗ n kλ ,
The final state:
F = f ⊗ n kλ −1 .
Fermi’s Golden Rule:
2
2π ⎛⎜ e ⎞⎟⎟
⎜
Rkλ =
⎟
 ⎜⎜⎝ m ec ⎟⎟⎠
f ; n kλ −1 Â(r)⋅ p̂ i; n kλ
2
δ(E f − Ei − 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
Thus
M fi ≡ f ; n kλ −1 Â(r)⋅ p̂ i; n kλ = f n kλ −1 Â(r) n kλ ⋅ p̂ i ,
(9.107)

n kλ −1 Â(r) n kλ = Ak ξkλe ik⋅r n kλ .
(9.108)

M fi = n kλ Ak (ξkλ ⋅ p fi ),
(9.109)
where p fi ≡ f e ik⋅r p̂ 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) :
162
A(r,t) = A0[e i(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 = | A0 | (ξkλ ⋅ p fi ) ,
(9.110)
therefore, | A0 |2 = n kλAk2 ∝ n kλ .
The total transition rate for a gas is obtained by (1) summation of all possible transitions and
(2) multiplying the number of atoms N A :
R = N A ∑ Rkλ .
(9.111)
kλ
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 =
S kλ
ω
=
cn kλ
Ω
Fig. 9.1: Geometry for
Beer’s law.
,
(9.112)
where Ω is the renormalization volume. Then
n kλ = Fkλ
Ω
,
c
(9.113)
2
2π ⎛⎜ e ⎞⎟⎟ 2 
2
⎜⎜
Rkλ = n kλ
⎟⎟ Ak | ξkλ ⋅ p fi | δ(E f − Ei − ω)
 ⎜⎝ m ec ⎟⎠
2
Ω 2π ⎛⎜ e ⎞⎟⎟ 2 
2
⎜
= Fkλ
⎟ A | ξ ⋅ p | δ(E f − Ei − ω)
c  ⎜⎜⎝ m ec ⎟⎟⎠ k kλ fi
(9.114)
FkλA = number of photons with mode (kλ) per second going through the slab. Note: A is area.
−AdFkλ = N ARkλ ,
⇒
−(Adx)
dFkλ
dx
163
= N ARkλ ,
(9.115)
(9.116)
⇒
2
Ω 2π ⎛⎜ e ⎞⎟⎟ 2 
2
⎜⎜
−Ω
= N ARkλ = N AFkλ
⎟⎟ Ak | ξkλ ⋅ p fi | δ(E f − Ei − ω) , (9.117)
⎟
dx
c  ⎜⎝ m ec ⎠
dFkλ
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
2
2π ⎛⎜ e ⎞⎟⎟ 2 
2
⎜⎜
α(kλ) = N A
⎟⎟ Ak | ξkλ ⋅ p fi | δ(E f − Ei − ω)
c ⎜⎝ m ec ⎟⎠
2
n ⎛ 2πe ⎞⎟ 
⎟⎟ | ξ ⋅ p |2 δ(E − E − ω).
= A ⎜⎜⎜
f
i
cω ⎜⎝ m ec ⎟⎟⎠ kλ fi
1/2
⎛ 2πc 2 ⎞⎟
⎟⎟
We used Ak ≡ ⎜⎜⎜
⎝⎜ ωΩ ⎟⎠
and nA ≡
NA
(9.120)
. Then the total absorption coefficient
Ω
2
α(ω) = ∑ ∫ α(kλ)d 3k .
(9.121)
λ=1
Without losing generality, set the direction of matrix element p fi along z-axis, then
2

∑| ξ
λ=1
kλ
⋅ p fi |2 = sin 2(θ)p fi2
(9.122)
where θ is the angle between k and p fi . The absorption coefficient
nA 4 e 2
p 2 (Ei − E f ) .
2 2 3 fi
c 3m c
(9.123)
α(ω) 4 e 2
=
p 2 (E − E f ) .
nA
3 m 2 2c 4 fi i
(9.124)
α(ω) =
Absorption cross section (per atom):
σ(ω) =
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
164
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:
2
2π ⎛⎜ e ⎞⎟⎟
⎜
Rkλ =
⎟
 ⎜⎜⎝ m ec ⎟⎟⎠
f ; n kλ + 1 Â(r)⋅ p̂ i; n kλ
2
δ(E f + ω − Ei ) ,
M fi ≡ F Â(r)⋅ p̂ I = f n kλ + 1 Â(r) n kλ ⋅ p̂ i

n kλ + 1 Â(r) n kλ = Ak ξkλe −ik⋅r n kλ + 1

M fi = n kλ + 1 Ak (ξkλ ⋅ p fi ),
(9.125)
(9.126)
(9.127)
(9.128)
where p fi ≡ f e −ik⋅r p̂ i . So the FGR is
Rkλ = (n kλ + 1)

2π e 2
2
A
|
ξ
⋅ p fi |2 δ(E f + ω − Ei ) .
k
kλ
2 2
 m ec
(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 (n kλ ) , while Rkλ for spontaneous radiation is independent of n kλ .
Consider two electronic states i and f , with E f = Ei + ω . Optical absorption rate:
abs
Ri→f
= Bi→f N i ρ(ω) ,
(9.130)
where N i the number of atoms in the i state, ρ(ω) the energy density of photons with
frequency ω , and Bi→f the Einstein B coefficient. The radiation rate is
Rfrad
= B f →i N f ρ(ω) + Af →i N f ,
→i
where N f the number of atoms in the f
(9.131)
state, and Af →i the Einstein A coefficient.
From comment (1),
Bi→f = B f →i ≡ B .
165
(9.132)
Let A ≡ Af →i . The equilibrium condition is:
abs
Ri→f
= R rad f →i ,
(9.133)
and one can derive that
ρ(ω) =
A
1
.
B (N i / N f )−1
(9.134)
Statistical mechanics show that
Ni
Nf
=
e
−Ei /kBT
e
−E f /kBT
=e
−(Ei −E f )/kBT
=e
ω/kBT
,
(9.135)
where T the temperature and kB the Boltzmann constant.
Bose factor:
nω =
ρ(ω) =
1
e
ω/kBT
−1
,
(9.136)
Ωω 2
1
,
2
3 ω/kBT
π c e
−1
(9.137)
we find that
A
Ωω 2
= 2 3.
B π c
(9.138)
Based on Einstein relations, one can write down the balance as
dI
= const ⋅I(N f − N i ) ,
dx
where I is light density. Since E f > Ei , N f < N i at equilibrium, thus
(9.139)
dI
< 0 , light is
dx
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.
166