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Fundamentals of Quantum Optics

Marco OrnigottiLukas MaczewskyAlexander Szameit

Summer Semester 2016Friedrich-Schiller Universität - Jena

June 19, 2016

Contents

1 HAMILTONIAN FORMULATION OF THE ELECTROMAGNETICFIELD 11.1 Coulomb Gauge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Field Energy Density . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 QUANTIZATION OF THE FREE ELECTROMAGNETIC FIELD 7

3 QUANTUM STATES OF THE ELECTROMAGNETIC FIELD 103.1 Number States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3.1.1 Properties of Fock States . . . . . . . . . . . . . . . . . . . . . 103.2 Coherent states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.2.1 Expectation Value of the electric Field over Coherent States . 133.2.2 Action of the Number Operator on Coherent States . . . . . . 143.2.3 Photon Probability . . . . . . . . . . . . . . . . . . . . . . . . 153.2.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.3 Phase Space and Quadrature Operators for the Electromagnetic Field 163.3.1 Number states in phase space . . . . . . . . . . . . . . . . . . 173.3.2 Coherent states in phase space . . . . . . . . . . . . . . . . . . 17

3.4 Displacement Operator . . . . . . . . . . . . . . . . . . . . . . . . . . 193.5 Squeezed States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 Interaction of Matter with Quantized Light -I 244.1 Electromagnetic Potentials in Presence of Matter . . . . . . . . . . . 244.2 Introduction of Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.2.1 Charge Density . . . . . . . . . . . . . . . . . . . . . . . . . . 264.2.2 Current Density . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.3 Atom-Field Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . 274.3.1 Electric Part . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.3.2 Magnetic Part . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.3.3 Orders of Magnitude . . . . . . . . . . . . . . . . . . . . . . . 29

4.4 The Interaction Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . 304.5 Second Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.5.1 Second Quantization of the Atomic Hamiltonian . . . . . . . . 324.5.2 Dipole Operator . . . . . . . . . . . . . . . . . . . . . . . . . . 344.5.3 Jaynes-Cummings Model . . . . . . . . . . . . . . . . . . . . . 354.5.4 States Of The Joint System . . . . . . . . . . . . . . . . . . . 37

4.6 Schrödinger, Heisenberg and Interaction Picture . . . . . . . . . . . . 374.6.1 Schrödinger Picture . . . . . . . . . . . . . . . . . . . . . . . . 374.6.2 Heisenberg Picture . . . . . . . . . . . . . . . . . . . . . . . . 384.6.3 Interaction Picture . . . . . . . . . . . . . . . . . . . . . . . . 38

5 Interaction of Matter with Quantized Light - II 395.1 Photon Emission and Absorption Rates . . . . . . . . . . . . . . . . . 39

5.1.1 Absorption . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395.1.2 Emission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405.1.3 Rate Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5.2 Quantum Rabi Oscillations . . . . . . . . . . . . . . . . . . . . . . . . 425.2.1 The Semiclassical Rabi Model in a Nutshell . . . . . . . . . . 425.2.2 The quantum Rabi Model . . . . . . . . . . . . . . . . . . . . 445.2.3 Interaction of an Atom with a Coherent State . . . . . . . . . 46

5.3 Wigner-Weißkopf-Theory [5] . . . . . . . . . . . . . . . . . . . . . . . 51

6 Quantum Mechanics of Beam Splitters 586.1 Classical theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586.2 Quantum Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616.3 Classical VS Quantum Beam Splitter . . . . . . . . . . . . . . . . . . 65

6.3.1 Classical BS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 656.3.2 Quantum BS . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

6.4 Effects of Vacuum in BS: Coherent State Input . . . . . . . . . . . . 676.5 Effects of Vacuum in BS: Single Photon Input . . . . . . . . . . . . . 686.6 Hong-Ou-Mandel Effect [9] . . . . . . . . . . . . . . . . . . . . . . . . 68

7 Quantum Theory of Photodetection 707.1 Direct Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707.2 Homodyne Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

8 Degree of Coherence and Quantum Optics 768.1 First Order Degree of Coherence . . . . . . . . . . . . . . . . . . . . . 768.2 Quantum Theory of g(1) (τ) . . . . . . . . . . . . . . . . . . . . . . . 798.3 Second Order Degree of Coherence . . . . . . . . . . . . . . . . . . . 808.4 Quantum Theory of g(2) (τ) . . . . . . . . . . . . . . . . . . . . . . . 82

8.4.1 Number States . . . . . . . . . . . . . . . . . . . . . . . . . . 828.4.2 Coherent States . . . . . . . . . . . . . . . . . . . . . . . . . . 838.4.3 Chaotic Light . . . . . . . . . . . . . . . . . . . . . . . . . . . 838.4.4 Squeezed Vacuum . . . . . . . . . . . . . . . . . . . . . . . . . 83

8.5 Comparison and Nonclassical Light . . . . . . . . . . . . . . . . . . . 83

9 Elements of Quantum Nonlinear Optics 859.1 Nonlinear Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . 859.2 Quantization of the Electromagnetic Field in Dielectric Media . . . . 879.3 Second Harmonic Generation . . . . . . . . . . . . . . . . . . . . . . 919.4 Parametric Down Conversion . . . . . . . . . . . . . . . . . . . . . . 96

10 Appendix A: The Harmonic Oscillator 98

11 Appendix B: The Quantization of a Paraxial Electromagnetic Field 9911.1 From the Vector Potential to the Paraxial Operator . . . . . . . . . . 9911.2 The Paraxial Dispersion Relation . . . . . . . . . . . . . . . . . . . . 10111.3 Back to Continuous Mode Operators . . . . . . . . . . . . . . . . . . 10311.4 The Maxwell-Paraxial Modes . . . . . . . . . . . . . . . . . . . . . . 10411.5 The Paraxial States of the Electromagnetic Field . . . . . . . . . . . 10511.6 An Example of Paraxial Mode Operators . . . . . . . . . . . . . . . . 105

12 Bibliography 106

HAMILTONIAN FORMULATION OF THE ELECTROMAGNETIC FIELD

1 HAMILTONIAN FORMULATION OF THE ELEC-

TROMAGNETIC FIELD

First of all let us write down the Maxwell equations in vacuum:

∇ · E (r, t) = 0, (1.1a)

∇× E (r, t) = −∂B (r, t)

∂t, (1.1b)

∇ ·B (r, t) = 0, (1.1c)

∇×B (r, t) =1

c2

∂E (r, t)

∂t. (1.1d)

Eq.(1.1c) is also satisfied if B (r, t) is derived from a potential A (r, t), i.e.,

B (r, t) = ∇×A (r, t) . (1.2)

Substituting Eq.(1.2) into Eq.(1.1c) leads to

∇ · (∇×A) = 0. (1.3)

And substituting Eq.(1.2) into Eq.(1.1b) we get

∇× E +∂

∂t(∇×A) = 0. (1.4)

In Eq.(1.4) we can factor out the curl operator to obtain

∇×[E +

∂

∂tA

]= 0. (1.5)

Furthermore, if we define

E +∂A∂t

= −∇φ, (1.6)

we have that∇×

[E +

∂

∂tA

]= −∇×∇φ = 0. (1.7)

So now we have the electric and magnetic field, given as a function of the vectorpotential A and the scalar potential φ. The evolution equations for the potentialswe can get by using eq.(1.1a) and eq.(1.1d).Equation (1.1a) leads to

∇ ·(− ∂

∂tA−∇φ

)= 0

⇒ ∂

∂t(∇ ·A) +∇2φ = 0 (1.8)

1


and from eq.(1.1d), using the vector identity ∇ × ∇ ×A = ∇ (∇ ·A) − ∇2A, weget:

∇ (∇ ·A)−∇2A =1

c2

∂

∂t

(−∂A∂t−∇φ

)⇒ ∇ (∇ ·A)−∇2A = − 1

c2

∂2A

∂t2− 1

c2

∂∇φ∂t

(1.9)

Note:Eq.(1.8) and eq.(1.9) Maxwell’s equations for the field potentials. However A andφ have no physical meaning. In order to show what is meant by this statement letus have a look at the following transformations

A = A′ −∇Λ, (1.10a)

φ = φ′ +∂Λ

∂t, (1.10b)

where Λ is an arbitrary (but well behaved) scalar field, which is called gauge field.We now calculate the fields E′ and B′ generated by A′ and φ′. For the magneticfield B′ we obtain following result:

B′ = ∇×A′

= ∇×A +∇× (∇Λ)

= ∇×A = B, (1.11)

while for the electric field E′ we have:

E′ = −∂A′

∂t−∇φ′

= −∂A∂t

−∂∇Λ

∂t−∇φ

+∂∇Λ

∂t

= −∂A∂t−∇φ = E. (1.12)

So we conclude:

• Different field potentials generate the same E and B (gauge transformations)

• While E and B are manifestly gauge invariant, the field potentials are not. Itis therefore important when dealing with field potentials to fix the gauge byfixing the choice of Λ.

1.1 Coulomb Gauge

Assume to consider the condition

∇ ·A = 0 (1.13)

2


to be valid. Therefore using this gauge we obtain for the scalar potential fromEq.(1.8) that

∇2φ = 0, (1.14)

from which it follows that φ = 0. This result is in accordance with the fact thatthe scalar potential is associated to the charge distribution (and, ultimately, to theelectrostatic field) and therefore since in vacuum there are no charges, the scalarpotential is zero. Substituting this result into Eq.(1.9) then gives

∇ (∇ ·A)−∇2A = − 1

c2

∂2A

∂t2

− 1

c2

∂∇φ∂t

→ ∇2A− 1

c2

∂2A

∂t2= 0. (1.15)

From both, Eq.(1.14) and Eq.(1.15) we can see that in the Coulomb (radiation)gauge in vacuum, E and B are completely determined by the vector potential solely,i.e.,

E = −∂A∂t

(1.16a)

andB = ∇×A. (1.16b)

What about Λ?Once the gauge condition is fixed, it must be respected by all (A, φ). Therefore:

∇ ·A =∇ ·A′ −∇ · (∇Λ) = 0 ⇒ ∇2Λ = 0 (1.17)

andφ = φ

′ +∂Λ

∂t= 0 ⇒ ∂Λ

∂t= 0 (1.18)

This means, that Λ is a harmonic function, i.e., it is a solution of the wave equation

∇2Λ− 1

c2

∂2

∂t2Λ = 0. (1.19)

1.2 Field Energy Density

Let us consider the electromagnetic field Hamiltonian:

H =1

2

∫dV

[ε0|E (r, t) |2 +

1

µ0

|B (r, t) |2]. (1.20)

Our goal is now to rewrite Eq. (1.20) in terms of the vector potential. To do that,however, we make sure to follow the following prescriptions, which will turn out tobe useful for quantization.a) Finite Volume V = L3 =⇒

∫dV →

∑k

This is done because by introducing a finite volume we can imagine to write the

3


field as a superposition of cavity modes, that constitute a finite set of discretebasis vectors;

b) Allowed Modes

k =2π

L(nxx + nyy + nzz) (1.21)

These are the k-vectors of the modes sustained by the fictitious cavity intruducedin a);

c) PolarizationThe electromagnetic field can be characterized by 2 orthogonal states of polar-ization in the transverse plane (e.g., |H〉 , |V 〉 , |L〉 , |R〉 , |TE〉 , |TM〉...).For each mode we then have two possible polarization states, namely ekλ, withλ = 1, 2 and ekλ · ek′λ′ = δλλ′ . The vector potential then can be written in thefollowing way:

A (r, t) =∑k

2∑λ=1

ekλAkλ (r, t) (1.22)

Moreover, A must obey Coulomb gauge condition, i.e., ∇ ·A = 0. This leads to

k · ekλ = 0, (1.23)

which states that the vector potential (and therefore E and B) does not possesscomponents along the propagation direction k. In other words, A is a purely trans-verse field. Another condition that must be fulfilled, is that A must obey the waveequation

∇2A− 1

c2

∂2A

∂t2= 0. (1.24)

Inserting Eq.(1.22) leads to

∑k

2∑λ=1

ekλ

[∇2Akλ (r, t)− 1

c2

∂2Akλ (r, t)

∂t2

]= 0. (1.25)

Let us now assume that

Akλ (r, t) = Akλ (t) eik·r + A∗kλ (t) e−ik·r, (1.26)

thus we have:

∇2Akλ (r, t) = −k · k[Akλ (t) eik·r + A∗kλ (t) e−ik·r

]. (1.27)

Defining ωk = ck we have from Eq.(1.25)

∂2Akλ (t)

∂t2+ ω2

kAkλ (t) = 0 (1.28)

4


and the same expression for the complex conjugate. The solution to Eq.(1.28) isthen

Akλ (t) = Akλe−iωkt, (1.29)

and for the complex conjugate

A∗kλ (t) = A∗kλeiωkt. (1.30)

Using this result and Eqs. (1.16), the electric and magnetic fields can be then writtenas follows:

E (r, t) =∑k

2∑λ=1

iωkekλ[Akλe

−iωkt+ik·r − A∗kλeiωkt−ik·r

], (1.31)

and

B (r, t) =∑k

2∑λ=1

i (k× ekλ)[Akλe


]. (1.32)

Now we can calculate the electromagnetic field Hamiltonian H which is given by Eq.(1.20) and wich we will write in following form:

H =1

2

∫dV

[ε0|E (r, t) |2 +

1

µ0

|B (r, t) |2]

= HE +HB, (1.33)

where HE and HB are the Hamiltonians of the electric and magnetic field, respec-tively. In the following we will calculate explicitly the electric field Hamiltonian,using the vector Potential A. We obtain:

HE =ε0

2

∫dV |E (r, t) |2

=ε0

2

∫dV E∗ (r, t) · E (r, t)

=ε0

2

∑k′

2∑λ′=1

∑k

2∑λ=1

∫dV (−iωk) (iωk′)

[Ak′λ′e

−iωk′ t+ik′·r − A∗k′λ′eiωk′ t−ik′·r

]×

[A∗kλe

iωkt−ik·r − Akλe−iωkt+ik·r

]ek′λ′ · ekλ (1.34)

NOTE: The only quantities that remain inside of the integral are the exponentials.By recalling that ∫

dV e±i(k−k′)·r = V δk,k′ , (1.35a)

and ∫dV e±i(k+k′)·r = V δk,−k′ , (1.35b)

5


we obtain, substituting the above result into Eq. (1.20),

HE = −∑kk′

∑λλ′

V ε0ωkωk′

2A∗k′λ′Akλe

i(ωk′−ωk)tδk,k′ ek′λ′ · ekλ︸︷︷︸δλλ′

+

+ A∗k′λ′A∗kλe

i(ωk′+ωk)tδk,−k′ ek′λ′ · ekλ + Ak′λ′Akλe−i(ωk′+ωk)tδk,−k′ ek′λ′ · ekλ

+ Ak′λ′A∗kλe−i(ωk′−ωk)tδk,k′ ek′λ′ · ekλ︸︷︷︸

δλλ′

= −∑k

∑λ

ε0V ω2k|Akλ|2 −

∑k

∑λ

V ε0ω2k

2

[A∗kλA−k,λ′ e−k,λ′ · ekλei2ωkt+

+ AkλA−k,λ′ e−k,λ′ · ekλe−i2ωkt]. (1.36)

The second term in Eq. (1.36) compensates a similar term in the magnetic fieldHamiltonian, which is given by:

HB = − 1

2µ0

∑kk′

∑λλ′

∫dV

[Ak′λ′e

−iωk′ t−ik′·r + A∗k′λ′eiωk′ t−ik′·r

]×

[A∗kλe

iωkt−ik·r − Akλe−iωkt+ik·r

](k′ × ek′λ′) (k× ekλ) (1.37)

Analogously to the electric field Hamiltonian, the integration gives factors δk,k′ andδk,−k′ . With this we obtain:

δk,k′ → (k× ekλ′) (k× ekλ) = k2δλλ′ (1.38a)

δk,−k′ → (k× ekλ) (k× e−k,λ′) = −k2ekλ · e−k,λ′ (1.38b)

Combining HE and HB we obtain the full electromagnetic field Hamiltonian

H =∑k

2∑λ=1

ε0V ω2k (AkλA

∗kλ + A∗kλAkλ) . (1.39)

For the calculation we have used that

ε0ω2ke−kλ′ · ekλ −

k2

µ0

ekλ · e−kλ′ =

(ε0ω

2k −

k2

µ0

)ekλ · e−kλ′ =

=

(ε0c

2k2 − k2

µ0

)=ε0µ0c

2k2 − k2

µ0

=1c2c2k2 − k2

µ0

= 0, (1.40)

where ωk = ck.NOTE: Although for the classic EM field AA∗ = A∗A we still write this as AA∗ +

A∗A instead of 2|A|2 in order to obtain a direct connection with the quantized form,which we will see later, where A and A∗ will be promoted to operators.

6

QUANTIZATION OF THE FREE ELECTROMAGNETIC FIELD

2 QUANTIZATION OF THE FREE ELECTRO-

MAGNETIC FIELD

We consider the EM field Hamiltonian as a collection of cavity modes, as given byEq. (1.39), i.e.,

H =∑k

2∑λ=1

ε0V ω2k (AkλA

∗kλ + A∗kλAkλ) .

We now introduce the creation and annihilation operators and express the Hamil-tonian in terms of these. If we compare the above expression of the field Hamiltoinanof a harmonic oscillator (see Eq. (10.3)), we see that we can obtain the quantizedversion of the field Hamiltonian if we promote the mode coefficients A and A∗ tooperators in such a way that A ' a† and A∗ ' a, namely

Akλ →√

~2ε0V ωk

a†kλ, (2.1a)

A∗kλ →√

~2ε0V ωk

akλ. (2.1b)

Substituting this into Eq. (1.39) brings us the following form for the quantized fieldHamiltonian:

H =∑k,λ

~ω2

(a†kλakλ + akλa

†kλ

). (2.2)

The following commutation rules are understood[akλ, a

†k′λ′

]= δkk′δλλ′ , (2.3a)

[akλ, ak′λ′ ] = 0 =[a†kλ, a

†k′λ′

]. (2.3b)

If we make use of the above commutation relation, Eq. (2.2) can be written in thefollowing, more inspiring, form:

H =∑k,λ

~2ωk,λ

(a†kλakλ + a†

kλakλ + 1)

=∑k

2∑λ=1

~ωk,λ

(a†kλakλ +

1

2

). (2.4)

This result can be interpreted in a very intuitive way: thanks to the above equation,the total energy of the electromagnetic field can be interpreted as given by the sum ofthe contribution of several uncoupled harmonic oscillators, each one characterized byits own frequency ωk,λ. Since the field is composed by several uncoupled oscillators,it is reasonable to assume that for each oscillator (i.e., each field mode), thereshould be a state |nkλ〉 associated. This allows us to write the global state of the

7


electromagnetic field as a direct product of all such states, i.e.,

|nk1,1, nk1,2, nk2,1, nk2,2, . . .〉 = |nk1,1〉 |nk1,2〉 |nk2,1〉 . . . ≡ |nkλ〉 (2.5)

These states are known in literature as Fock states, and constitute the energy eigen-states of the electromagnetic field. To understand that, let us analyze how theHamilton operator (2.4) acts on such states, namely

H |nkλ〉 =∑k

2∑λ=1

~ωk(a†kλakλ +

1

2

)|nkλ〉

=∑k

2∑λ=1

~ωk[a†kλ

√nkλ |nkλ − 1〉+

1

2|nkλ〉

]

=∑k

2∑λ=1

~ωk(nkλ +

1

2

)|nkλ〉 , (2.6)

where the following relations have been used:

a |n〉 =√n |n− 1〉 (2.7a)

a† |n〉 =√n |n− 1〉 (2.7b)

NOTE: Combining the above relations we can introduce the photon number oper-ator Nkλ as follows

Nkλ = a†kλakλ. (2.8)

The matrix elements of the Hamilton operator of the free field are then obtained bymultiplying eq. (2.6) from the left side with 〈nkλ| we obtain:

〈nkλ| H |nkλ〉 =∑k

2∑λ=1

~ω(nkλ +

1

2

)(2.9)

This is the total energy of the field expressed as the sum of quanta distributed amongall the possible field modes. We used 〈nkλ| nk′λ′〉 = δkk′δλλ′ .IMPORTANT NOTE:

• nkλ is the number of quanta (field excitations - photons) in the mode (k, λ);

• Classical Electromagnetism: NO FIELD = NO ENERGY;

• Quantum Optics: For the vacuum state nkλ = 0 we obtain:

H |0〉 =∑k

2∑λ=1

~ωk(nkλ +

1

2

)|0〉 =

∑k

2∑λ=1

~ωk26= 0. (2.10)

8


So the energy of the vacuum state is∑

k,λ ~ωk/2. It must be nonzero becauseof the Heisenberg principle (compare to particle in a box)

To conclude the quantization procedure, let us now write the vector potential, theelectric and the magnetic field operators in terms of a and a†. The expressions ofthe vector potential, electric and magnetic fields in terms of the vector potentialcoefficients A and A∗ are given as follows:

A (r, t) =∑k

2∑λ=1

ekλ[Akλe

−iωkt+ik·r + A∗kλeiωkt−ik·r

], (2.11a)

E (r, t) =∑k

2∑λ=1

ekλ · iωk[Akλe


], (2.11b)

B (r, t) =∑k

2∑λ=1

i (k× ekλ)[Akλe


]. (2.11c)

By recalling that Akλ →√

~2ε0V ωk

a†kλ and A∗kλ →

√~

2ε0V ωkakλ, we can therefore

write the field operators in the following form:

A (r, t) = A+ (r, t) + A− (r, t) , (2.12a)

E (r, t) = E+ (r, t) + E− (r, t) , (2.12b)

B (r, t) = B+ (r, t) + B− (r, t) . (2.12c)

Where + indicates the positive frequency part and − the negative one. Furthermore:

A+ (r, t) =∑k

2∑λ=1

ekλ

√~

2ε0V ωkakλe

−iωkt+ik·r, (2.13a)

A− (r, t) =∑k

2∑λ=1

ekλ

√~

2ε0V ωka†kλe

iωkt−ik·r, (2.13b)

E+ (r, t) = i∑k

2∑λ=1

ekλ

√~ωk

2ε0Vakλe


E− (r, t) = −i∑k

2∑λ=1

ekλ

√~ωk

2ε0Va†kλe

iωkt−ik·r, (2.14b)

B+ (r, t) = i∑k

2∑λ=1

√~

2ε0V ωk(k× ekλ) akλe


B− (r, t) = −i∑k

2∑λ=1

√~

2ε0V ωk(k× ekλ) a

†kλe

iωkt−ik·r. (2.15b)

9

QUANTUM STATES OF THE ELECTROMAGNETIC FIELD

3 QUANTUM STATES OF THE ELECTROMAG-

NETIC FIELD

We now want to study the possible states of the electromagnetic field and theirproperties. Without loss of generality, we consider only single mode fields, i.e., theelectromagnetic field is represented by a single harmonic oscillator corresponding toa single degree of freedom of the field (polarization, frequency, spatial mode, etc.).To this aim, let us fix the polarization of the field to be x and let us assume thatthe field is characterised by a single k-vector k and that it propagates along thez-direction. . Using these assumptions, the electric field operator defined in Eq.(2.11b) assumes then the following form:

E(z, t) = iE0

[aei(kz−ωt) − a†e−i(kz−ωt)

], (3.1)

where E0 =√

~ω/2ε0V . Notice, moreover, that for monochromatic plane wave fields|B| = |E|/c and the magnetic field can be therefore be neglected.

3.1 Number States

According to the results of the previous section, the Hamiltonian of a single modefieldis given by

H = ~ω(n+

1

2

), (3.2)

where n = a†a is the number operator, which counts the number of excitations in amode, and therefore its eigenvalues n give the number of excitations (i.e., photons)in the mode. It makes then sense to introduce the Fock (number) states as follows:Def:Fock (number) states are eigenstates of the number operator, such that

n |n〉 = n |n〉 . (3.3)

3.1.1 Properties of Fock States

Fock states form a complete set of orthonormal vectors, i.e.,

• Orthogonality:〈n| m〉 = δnm, (3.4)

• Completeness:∞∑n=0

|n〉〈n| = 1. (3.5)

The operators a† and a create and annihilate, respectively, one photon in the

10


field i.e.,a† |n〉 =

√n+ 1 |n+ 1〉 , (3.6a)

a |n〉 =√n |n− 1〉 . (3.6b)

In order to create a state from the vacuum we need to apply this relations recursively:

|n〉 =

(a†)n

√n!|0〉 , (3.7)

while the vacuum state is defined as

a |0〉 = 0. (3.8)

The expectation value of the Hamilton operator of the field over the Fock statesis given by

H = 〈n| H |n〉 = ~ω(〈n| n |n〉+

1

2〈n| n〉

)= ~ω

(n 〈n| n〉+

1

2〈n| n〉

)= ~ω

(n+

1

2

)〈n| n〉 , (3.9)

and therefore:H = ~ω

(n+

1

2

). (3.10)

The expectation value of the electric field operator (3.1) is instead given by

〈n| E (z, t) |n〉 = iE0

[〈n| a |n〉 ei(kz−ωt) − 〈n| a† |n〉 e−i(kz−ωt)

]= iE0

[√nei(kz−ωt) 〈n| n− 1〉 −

√n+ 1 〈n| n+ 1〉 e−i(kz−ωt)

]= 0. (3.11)

The electric field is not well defined for number states!!However

⟨E⟩2

is nonzero since it is connected with the field energy

〈n| E2 |n〉 = 2E20

(n+

1

2

). (3.12)

We can therefore calculate the field variance:⟨∆E (z, t)

⟩2

=⟨

E2 (z, t)⟩−⟨

E (z, t)⟩2

= 2E20

(n+

1

2

). (3.13)

The above equation has an interesting consequence: letting n = 0, i.e. no photons

11


in the field, the variance ⟨∆E (z, t)

⟩2

6= 0.

This means that the vacuum state |0〉 shows field fluctuations (quantum noise).Moreover the number operator and the electric field operator do not commute, i.e.,[

n, E (z, 0)]

= iE0

−aei(kz−ωt) + a†e−i(kz−ωt)

. (3.14)

Summarizing:

• Number states are quantum states of well definite photon number. As a con-sequence, the electric field is not well defined for these states;

• vacuum field fluctuations.

3.2 Coherent states

As we have seen number states do not have a well defined E and therefore are notfitted to describe a classical field. So how can we find then a classical expression fora quantum state of the e.m. field?

Considerations:

1. Large numbers of photons in the field |N〉 , N 1?

FALSE:we just saw, that no matter how big N is, a number state has an ill-definedE and therefore the classical limit cannot come from |N〉

2. Since E ∼ a− a† a possible trial solution could be |ψ〉 ∼ |n〉+ |n+ 1〉?

Good: we generalize this concept and therefore obtain the definition:

Def. Coherent States: Coherent states are eigenstates of the annihilation opera-tor and therefore obtain following eigenvalue equation:

a |α〉 = α |α〉 , α ∈ C (3.15)

(〈α| a† = 〈α|α∗).

Let us write the coherent states in number state basis:

|α〉 =∞∑n=0

cn |n〉 , (3.16)

from this two equations we can calculate the expansion coefficients:

a |α〉 =∞∑n=0

cna |n〉 =∞∑n=1

√ncn |n− 1〉 = α

∞∑n=0

cn |n〉 . (3.17)

12


Now, in order to let both sums on the r.h.s. of eq.(3.17) start from the same number,i.e. n = 1, we need to change the expression

∑∞n=0 cn |n〉 to

∑∞n=1 cn−1 |n− 1〉. This

leads to√ncn = αcn−1. (3.18)

Applying successively the annihilation operator on equation (3.17) it follows that

cn = c0αn√n!, (3.19)

where c0 is defined by normalization, i.e., by imposing 〈α| α〉 = 1. This leads to

〈α| α〉 = |c0|2∞∑n=0

∞∑m=0

(α∗)n αm√n!m!

〈n| m〉︸︷︷︸δnm

= |c0|2∞∑n=0

(|α|2)n

n!= |c0|2 e|α|

2

. (3.20)

Therefore:

|α〉 = exp

[−|α|

2

2

]∞∑n=0

αn√n!|n〉 = exp

[−|α|

2

2

]∞∑n=0

αn(a†)n

√n!

|0〉 . (3.21)

3.2.1 Expectation Value of the electric Field over Coherent States

Let us now calculate the expectation value of the electric field in coherent statebasis, using

〈α|E(z, t)|α〉 = iE0〈α|aei(kz−ωt) − a†e−i(kz−ωt)|α〉 (3.22)

= iE0

[αei(kz−ωt) − α∗e−i(kz−ωt)

](3.23)

= iE0|α|[ei(kz−ωt+φ) − e−i(kz−ωt+φ)

](3.24)

= −2E0|α| sin(kz − ωt+ φ), (3.25)

where in passing from the second to the third line we have used the fact that α =

|α| exp (iφ). As can be seen, moreover, the expectation value of the electric fieldoperator over the coherent state |α〉 gives as result the classical electric field.

Now let us calculate the variance of the electric field operator, i.e.,

〈α|∆E2(z, t)|α〉 = 〈α|E2(z, t)|α〉 −(〈α|E(z, t)|α〉

)2

= E20 =

~ω2ε0V

. (3.26)

A coherent state is a nearly classical state beause it reproduces correctly the classicalfield although it still contains field fluctuations (that are the same of the vacuumfield).Note, moreover, coherent states are also defined for the harmonic oscillator as

13


states of minimal uncertainty.

3.2.2 Action of the Number Operator on Coherent States

Let us now investigate how the number operator acts on coherent states. Thereforewe will calculate the expectation value of the number operator in coherent statebasis, as follows

〈α| n |α〉 = 〈α| a†a |α〉 = |α|2 〈α| α〉 = |α|2 ≡ n. (3.27)

This means that coherent states do not possess a well defined number of photons,as we will see now.In fact, compared to the number states where the variance 〈∆n〉2 = 0, the varianceof the photon number doesn’t vanish for coherent states. In order to convinceourselves, that this is the case, let us calculate

〈∆n〉2 =⟨n2⟩− 〈n〉2

= 〈α| a†aa†a |α〉 − 〈α| a†a |α〉 . (3.28)

Here we are going to use the commutation rule

[a, a†] = aa† − a†a = 1, (3.29)

which leads to

〈α| a†aa†a |α〉 = 〈α| a† (1 + a†a)a |α〉

= 〈α| a†a |α〉+ 〈α| a†a†aa |α〉

= |α|2 + |α|4 . (3.30)

Combining our results and plugging them back into eq.(3.28), we obtain the varianceof the photon number operator for the case of coherent states as follows:

〈∆n〉2 = |α|2 + |α|4 − |α|4 = |α|2 , (3.31)

hence〈n〉 = 〈∆n〉2 = |α|2 (3.32)

14


3.2.3 Photon Probability

The probability of finding n photons in a coherent state |α〉 can be calculated asfollows:

Pn = |〈n| α〉|2 =

∣∣∣∣∣exp

[−|α|

2

2

]∞∑m=0

αm√m!〈n| m〉

∣∣∣∣∣2

= exp[− |α|2

] (|α|2)nn!

= exp[−n2

] nnn!. (3.33)

This satisfies the Poissonian statistic (mean value = variance):

Pn =nn

n!exp[−n2] (3.34)

and moreover:∆n

n=

1√n

(3.35)

3.2.4 Summary

Number States:Number states are described by |n〉. They are eigenstates of the numberoperator n = a†a and fulfil the equation

n |n〉 = n |n〉 .

The mean value of the electric field in number state basis is:

〈n| E |n〉 = 0

but the variance is: ⟨∆E⟩2

∼(n+

1

2

)(quantum noise)

Coherent States:Coherent states are the most classical among the quantum states. Tey aredescribed by |α〉. They are eigenstates of the creation operator a and fulfil theequation

a |α〉 = α |α〉

We can transform them to number state basis by:

|α〉 = exp

[−|α|

2

2

]∞∑n=0

αn√n!|n〉

15


The mean value of the electric field in coherent state basis is:⟨E⟩

= −2E0 |α| sin (kz − ωt+ φ)

and the variance: ⟨∆E⟩

= E0 ≡√

~ω2ε0V

3.3 Phase Space and Quadrature Operators for the Electro-

magnetic Field

Classical mechanics: H (q, p)→ each point in phase space represents a configurationof the system (see fig.3.1)

Figure 3.1: Representation of the system configuration in phase space

We can introduce this also in quantum optics. In order to do this, we firstremember the harmonic oscillator and adapt the same rules to our operator:

x ∼ a+ a† → X1 =1

2

(a+ a†) (3.36a)

p ∼ a− a† → X2 =1

2i

(a− a†) (3.36b)

Now we can build a 2D phase space associated to the single mode quantum fieldusing X1 and X2, with [

X1, X2

]=i

2. (3.37)

16


3.3.1 Number states in phase space

Expectation value:〈n| X1 |n〉 = 0 = 〈n| X2 |n〉 (3.38)

but variance:〈n|∆X2

1 |n〉 =1

4(2n+ 1) = 〈n|∆X2

2 |n〉 (3.39)

Proof: ⟨∆X2

1

⟩=

⟨X2

1

⟩−ZZZZ

⟨X1

⟩2

= 〈n| 14

(a+ a†) (a+ a†) |n〉

=1

4〈n|aa+ aa† + a†a+

a†a† |n〉

=1

4

[〈n| a√n+ 1 |n+ 1〉+ 〈n| a†√n |n− 1〉

]=

1

4[(n+ 1) 〈n| n〉+ n 〈n| n〉] =

1

4(2n+ 1) (3.40)

Since the variances are non-zero and[X1, X2

]6= 0 we can apply an indetermination

relation∆X2

1∆X22 ≥

1

16(3.41)

→ ∆X1 6= 0, even if n = 0 (v hacuum fluctuations). Moreover:⟨∆X2

1

⟩+⟨

∆X22

⟩=

1

2(2n+ 1) = n+

1

2= 〈n| n +

1

2|n〉 , (3.42)

the vacuum fluctuations are automatically taken into account.

Figure 3.2: Representation ofnumber states in phase space

Phase Space Representation(fig.3.2)

• circle of radius√n+ 1/2 (radius is well

defined)

• phase completely undefined (circle)

• vacuum state = circle with radius√1/2

3.3.2 Coherent states in phase space

For coherent states we have a peculiar thing:

〈α| X1 |α〉 =1

2(α + α∗) = Re α (3.43a)

17


〈α| X2 |α〉 =1

2i(α− α∗) = Im α (3.43b)

This means that the complex α-plane and the phase space for a coherent state areisomorph with each other. Therefore it is useful to use the complex polar represen-tation for α, i.e.:

α = |α| eiϑ. (3.44)

The phase space representation of a coherent state is illustrated in Fig. 3.3

Figure 3.3: Representation of co-herent states in phase space

Remember:

〈α| n |α〉 = |α|2

there is no well defined number of pho-tons for a coherent state. Furthermorethere is also a phase uncertainty (be-cause of Heisenberg uncertainty princi-ple)

Moreover: ⟨∆X2

1

⟩= 〈α| X2

1 |α〉 −(〈α| X1 |α〉

)2

=1

4=⟨

∆X22

⟩(3.45)

the variance of the quadrature operator is the same as the quantum vacuum. Espe-cially:

∆X1∆X2 =1

4(3.46)

and therefore coherent states are states of minimum uncertainty. Note: Looking atthe figure, one can imagine (also motivated by the fact, that coherent states havethe same ∆xk as the vacuum state) that coherent states can be viewed as suitablydisplaced vacuum states.

18


3.4 Displacement Operator

Figure 3.4: Schematic illustration of the creation of a coherent state by displacingthe vacuum state

How is this possible?Let’s assume that there is an operator D (α) such that:

D (α) |0〉 = |α〉 . (3.47)

To obtain an expression for D (α) let us note that

|α〉 = exp

[−|α|

2

2

]∞∑n=0

αn√n!|n〉

= exp

[−|α|

2

2

]∞∑n=0

(αa†)n

n!|0〉

= exp

[−|α|

2

2

]exp

[αa†] |0〉 (3.48)

With:

|n〉 =∞∑n=0

(a†)n

√n!|0〉 (3.49)

We

are tempted to say that eq. (3.48) describes the displacement operator. In order tounderstand why this is the case let us consider the Baker-Campbell-Hausdorff-Formula:

eA+B = e12 [A,B]eBeA (3.50)

19


let us choose:A = αa† and B = α∗a,

then we obtain: [A, B

]=[αa†, α∗a

]= |α|2

[a†, a

]= |α|2

However, we are still missing a factor eα∗a. But note that:

e−α∗a |n〉 =

∞∑m=0

(−αa)m

m!|0〉

=∞∑m=0

(−α)m

m!(a)m |0〉︸︷︷︸

=0

. (3.51)

If we therefore defineD (α) = eαa

†−α∗a (3.52)

as the displacement operator, the factor eα∗a has no effect but contributes to makethe displacement operator D (α) more symmetric.Note: The displacement operator is unitary:

• D† (α) = D (−α)

• D (α) D† (α) = D† (α)D (α) = 1

Note:You can prove, that without eα∗a the displacement operator is not unitary!More on D (α): It is useful and constructive to calculate the following quantity

〈α| a |α〉 = 〈0| D† (α) aD (α) |0〉

What is then D† (α) aD (α)?

It is hard to calculate this by hand, but we will use the:Decomposition Theorem for Exponential Operators

eABe−A = B +[A, B

]+

1

2!

[A,[A, B

]]+

1

3!

[A,[A,[A, B

]]]+ . . . (3.53)

For our purpose we chose: A = −αa† + α∗a and B = a. So we obtain:

D† (α) aD (α) = a+[−αa† + α∗a, a

]= a+

−α [a†, a]︸︷︷︸

=−1

+α∗ [a, a]︸︷︷︸=0

= a+ α (3.54)

20


Summarizing:

D† (α) aD (α) = a+ α (3.55)

D† (α) a†D (α) = a† + α∗ (3.56)

In this way we can evaluate each expectation value as expectation value done on thevacuum state. And so we obtain, that:

〈α| f(a, a†) |α〉 = 〈0| D† (α) f

(a, a†) D (α) |0〉

= 〈0| f(a+ α, a† + α∗

)|0〉 (3.57)

Coherent States: Classical field (α) plus quantum fluctuations.Squeezed States: States of the electromagnetic field, for which it is possible fora quadrature to beat the Heisenberg limit. Of course the price to pay is that theuncertainty on the other quadrature gets bigger.

3.5 Squeezed States

So far we have considered number states (well defined n but ill-defined E and B)and coherent states (almost classical, minimal uncertainty) and their phase spacerepresentations (see Fig. 3.3). They fulfil the uncertainty relation:

∆X1∆X2 ≥1

4. (3.58)

Is this the limit for manipulating states or is there another possibility to play aroundbeating the fundamental Heisenberg limit?

Figure 3.5: Comparison of a squeezed state with the vacuum state in phase space

As before ∆X1∆X2 = const but in the certain case the uncertainty on x2 is muchless than in the cases before.How can we define squeezed states and how are they generated?

21


We seek for a state |ξ〉 with ∆X21∆X2

2 ≥ 116, but:

∆X21 =

1

4γ (3.59)

∆X22 =

1

4γ, (3.60)

where γ ≤ 1 so that it is a reduction of the variance (uncertainty) for x1 but at thesame time it is an amplification for the variance of x2 (so that the product is stillconstant).We employ:

1. Since we learned, that every (coherent) state can be displaced from the vac-uum, let us consider the vacuum itself and squeeze it (Squeezed Vacuum,actually this is, what we generate in the lab).

2. Let us assume to be able to define an operator S (ξ) similar to the displacementoperator D (α), namely

S (ξ) = e12

[ξ∗a2−ξ(a†)

2]. (3.61)

Eq.(3.61) is the squeezing operator, where ξ ∈ C and ξ = reiϑ.

Figure 3.6: Schematic illustration of squeezed vacuum

Note: For the squeezed state in fig.3.6 we have ∆X1 <12, ∆X2 >

12but ∆X1∆X2 =

14(For more information on dynamics in phase space see Ref. [4])

Note: S (ξ) contains a2 and(a†)2, i.e. the squeezing operator is a nonlinear op-

erator. If we want to generate squeezed states, therefore, we must do it through a

22


nonlinear process (typically SPDC). Furthermore we require S (ξ) to be unitary:

S† (ξ) = S (−ξ) , (3.62)

S† (ξ) S (ξ) = S (ξ) S† (ξ) = 1. (3.63)

Therefore we can generate a squeezed vacuum state by applying the squeezing op-erator on the vacuum state:

|ξ〉 = S (ξ) |0〉 . (3.64)

The variance of X1 is obtained by⟨∆X2

1

⟩=⟨

X21

⟩−⟨

X1

⟩2

. (3.65)

For this we calculate the expectation value of X1:

〈ξ| x1 |ξ〉 =1

2〈ξ| a+ a† |ξ〉 =

1

2〈0| S† (ξ)

(a+ a†) S (ξ) |0〉 . (3.66)

As we did for the displacement operator for the coherent states, we look here for:

S† (ξ) aS (ξ) = a cosh r − a†eiϑ sinh r, (3.67)

S† (ξ) a†S (ξ) = a†cosh (r)− aeiϑ sinh r. (3.68)

Therefore we obtain:

〈ξ| X1 |ξ〉 = cosh r 〈0| a |0〉 − eiϑ sinh r 〈0| a† |0〉+

+ cosh r 〈0| a† |0〉 − eiϑ sinh r 〈0| a |0〉 = 0. (3.69)

For ϑ = 0 we obtain at the end:

〈ξ|∆X21 |ξ〉 =

1

4e−2r → 〈ξ|∆X2

2 |ξ〉 =1

4e2r, (3.70)

r is then called the "squeezing parameter".Final Notes:

a |0〉 = 0,

but then also:S (ξ) a |0〉 = 0.

Since S (ξ) is unitary and therefore S† (ξ) S (ξ) = 1 and therefore:

S (ξ) a |0〉 = S (ξ) aS† (ξ) S (ξ) |0〉︸︷︷︸=|ξ〉

= S (ξ) aS† (ξ) |ξ〉 = 0. (3.71)

This allows us to define the squeezed states as eigenstates of the operator S (ξ) aS† (ξ)

23

Interaction of Matter with Quantized Light -I

with the eigenvalue zero. To Summarize:

Number States: n |n〉 = n |n〉

Coherent States: a |α〉 = α |α〉

Squeezed States: S (ξ) aS† (ξ) |ξ〉 = 0

4 Interaction of Matter with Quantized Light -I

After studying the free electromagnetic field and introducing its quantum states,the next step is to consider the interaction of the E.M. field with matter. Up tonow we considered the E.M. field in free space, therefore the first step is to rewritethe electromagnetic potentials in presence of matter. Afterwards we will introduceatoms to our calculations and then analyse their interaction with the quantizedfield. As a result of our considerations we will be able to introduce the interactionHamiltonian of the E.M. field with matter at the end of this section.

4.1 Electromagnetic Potentials in Presence of Matter

Since we are not considering the case of vacuum anymore we need to take intoaccount charge and current densities. Therefore the E.M. potentials in presence ofmatter are now given as follows

∇ (∇ ·A)−∇2A +1

c2

∂∇φ∂t

+1

c2

∂2A

∂t2= µ0J (4.1a)

−∇2φ =σ

ε0

(4.1b)

∇ ·A = 0 (Coulomb Gauge) (4.1c)

In general eqs. (4.1a)-(4.1c) are difficult to solve, since we cannot say anymore thatφ = 0. In presence of matter the scalar potential φ (r) is given as

φ (r) =1

ε0

∫σ (r′)

|r− r′|dr′. (4.2)

Moreover we know that since∇A = 0 the vector potentialA needs to be transversal,and so the correspondent E field must also be transversal. In order to separate thetransversal and longitudinal part of a vector field we use the Helmholtz theorem.The Helmholtz theorem states that it is possible to split any vector field in threedimensions into the sum of an irrotational (curl-free) vector field and a solenoidal(divergence-free) vector field. Applying the H.T. on eq. (4.1a) we obtain for thecurrent J that

J = JL + JT , (4.3)

24


with∇JT = 0 (solenoidal) , (4.4a)

∇× JL = 0 (irrotational) . (4.4b)

The Coulomb gauge fixes the vector potential to be transverse. So the µ0JL termdoes not contribute to A. Therefore we obtain after separating the longitudinal andtransverse parts:

−∇2A +1

c2

∂2A

∂t2= µ0JT , (4.5a)

1

c2

∂∇φ∂t

= µ0JL, (4.5b)

−∇2φ =σ

ε0

. (4.5c)

From the last two equations we obtain the continuity equation

∇ · JL = −∂σ∂t. (4.6)

Splitting J in its transversal and longitudinal parts allows us to split the fields inthe same way, namely

E = ET + EL, (4.7a)

B = BT + BL, (4.7b)

with

ET = −∂σ∂t, BT = ∇×A, (4.8)

EL = −∇φ, BL = 0. (4.9)

While the longitudinal field is basically the electrostatic field generated by the chargedistribution σ, the transverse field is the one that propagates. BL = 0 because of∇ ·B = 0.

4.2 Introduction of Atoms

Let us now specify our analysis to the case of an atom (with Ne−, see Fig. 4.1) asa field source and try to solve equations (4.5).

25


Figure 4.1: Illustration of the choice of a coordinate system in presence of an atom,where rα represent the position vector of the α-th electron with respect to the nucleus

4.2.1 Charge Density

All the electrons and protons are treated like point charges, this is why we will writethe charge density σ (r) as

σ (r) = −e∑α

δ (r− rα) + eZδ (r) . (4.10)

This charge density generates a longitudinal fieldEL, which is essentially the Coulombpotential of the atom).

4.2.2 Current Density

Each electron orbits the nucleus, this is why we write the total current density J (r)

as a sum of the currents produced by each electron

J (r) = −e∑α

rαδ (r− rα) , (4.11)

where the dot stands for the time derivative.

26


4.3 Atom-Field Interaction

It is convenient to treat the interaction of the atom with the electromagnetic field interms of the polarization P (r) and magnetization M (r), since they are associatedwith the macroscopic atomic charges and currents. Using the standard results ofelectromagnetism (see, e.g., [3]), the polarization and magnetization are linked tothe charge density and current (respectively) via the following relations:

σ (r) = −∇ ·P (r) , (4.12)

J (r) = P (r) +∇×M (r) . (4.13)

The polarization P can be thought to be the sum of all the contributions comingfrom the dipoles composing the medium. In our case, such elementar dipoles arerepresented by each single electron and the atom’s nucleus. The polarization of suchan elementar dipole is then given by

P = qαrαδ (r− rα) (4.14)

In order to simplify our calculations we will use a trick in the choice of expressing thepolarization, namely we imagine to cover the distance between the nucleus (situatedat r = 0) and the electron at rα with a series of non-point dipoles oriented along0rα, in such a way that −qα of one dipole is compensated by +qα of the precedingone (see Fig. 4.2).

Figure 4.2: Split polarization vector between nucleus and electron

Therefore, since for a single dipole we have that

Pn (r) = (qαrα/n) δ

(r−

p+ 12

nrα

), (4.15)

summing all the contributions as in Fig. 4.2 and taking the limit n → ∞ gives

27


(4.16)

P (r) = −e∑a

rα

1∫0

dξ δ (r− ξrα) . (4.16)

Reasoning in an analogous way for M we obtain:

M (r) = −e∑a

rα × rα

1∫0

dξ ξδ (r− ξrα) . (4.17)

Recalling that potential energy of a permanent electric and a permanent magneticdipole are given by −d · E and −µ · B, respectively, we will define the electric VE

and magnetic VM contribution to the atom’s potential energy as

VE = −∫d3rP (r) · ET (r) = e

∑α

1∫0

dξ rα · ET (ξrα) , (4.18)

and

VM = −∫d3rM (r) ·B (r) = −e

∑α

1∫0

dξ (rα × rα) ·B (ξrα) . (4.19)

Usually the typical scale of atoms (a few Ångström) is much smaller than the wave-length of light (400 to 700 nm), therefore let us now expand the fields around ξrα = 0

in order to evaluate the various interaction contributions:

4.3.1 Electric Part

VE = e∑a

1∫0

dξ

1 + ξ (rα · ∇) +

1

2!ξ2 (rα · ∇)2 + . . .

rα · ET (0) . (4.20)

Here we can perform the ξ-Integration and obtain:

VE = e∑a

1︸︷︷︸dipole

+1

2!(rα · ∇)︸︷︷︸

quadrupole

+1

3!(rα · ∇)2︸︷︷︸hexapole

+ . . .

rα · ET. (0) . (4.21)

Equation (4.21) is the multipole expansion of the electric field potential. In partic-ular we notice that

Dipole: −eD = −∑

α erα ⇒ eD · ET (0)

Quadrupole: Q = −12

∑α erαrα ⇒ −∇ ·Q · ET (0)

28


4.3.2 Magnetic Part

Now we will do the same expansion for the magnetic field.

VM =e

m

∑a

1

2!︸︷︷︸dipole

+2

3!(rα · ∇)︸︷︷︸

quadrupole

+3

4!(rα · ∇)2︸︷︷︸hexapole

+ . . .

Iα ·B (0) (4.22)

where Iα = mrα × drαdt

is the angular momentum of the αth electron. Eq. (4.22) isthen a multipole expansion of the magnetic field potential. The explicit expressionof the magnetic dipole, e.g., is then given by

− eDM = −1

2

∑α

e

mIα ⇒ eDM ·B (0) (4.23)

4.3.3 Orders of Magnitude

In order to get an idea of which contributions of the multipole expansion we haveto take in account in our calculations, let us consider the following argument. Todo that, let us take, approximately, rα ≈ a0 = 10−10m (Bohr radius), Iα ≈ ~ =

10−34 (quantum of angular momentum) and ∇ET (0) ≈ ωcET (0) , ω ≈ 3 · 1015Hz

(visible light). If we substitute these numbers in the definitions of electric dipoleand quadrupole, as well as magnetic dipole we obtain:

• de = eD · ET (0) ≈ 4πε0~2me

ET (0)

• qe = −∇ ·Q · ET (0) ≈ 3e~16mc

ET (0)

• dm = eDM ·B (0) ≈ e~2m

B (0) ≈ e~2mc

ET (0) (plane wave: B ∼ Ec)

If we now compare the different coefficients with each other:

1.qedm

=3e~ET (0)

16mc· 2mc

e~ET (0)=

3

8∼ 1,

i.e., the electric quadrupole and magnetic dipole terms are approximately ofthe same order of magnitude

2.qede

=3e~ET (0)

16mc· me

4πε0~2ET (0)≈ e2

4πε0~c∼ 1

137≡ α,

i.e., the electric dipole is roughly 2 orders of magnitude bigger than thequadrupole (and αn orders of magnitude bigger than the n-th pole). It istherefore also α-times bigger than the magnetic dipole!

Conclusion: It makes sense to consider interactions of an atom with an electro-magnetic field in the dipole approximation, since this is the term that gives the

29


biggest contribution.Note: Some cases forbid the transition to the electric dipole (selection rules). Inall those cases we need to consider contributions from higher order of the electricmultipole expansion, or, if the atom is magnetically active, then take in account themagnetic multipole expansion.

4.4 The Interaction Hamiltonian

Now we are going to discuss, how to construct the dipole-Hamiltonian out of ourtheory. We are interested in quantized fields, so we need to take the approach ofquantizing E through quantizing A. We therefore need to write D ·E starting fromA.

Hamiltonian of the Atom-Light System

Since we still want to have a Hamiltonian representation, we need to take intoaccount the canonical momentum of the system. Furthermore we still want to beable to use the canonical quantization scheme to quantize the field.In our first attempt let us assume that the Hamiltonian we are looking for has thisform:

H(1) =∑α

p2α

2m︸︷︷︸kinetic energy

+1

2

∫d3r σ(r)φ(r)︸︷︷︸

AtomicCoulomb potential

+1

2

∫d3r

[ε0|E (r) |2 +

1

µ0

|B (r) |2]

︸︷︷︸field

(4.24)Considering this Hamiltonian first thing we notice is that there is no interaction part.We could add an interaction term by hand but which kind of? Moreover, it can bedemonstrated, that pα is not a canonical momentum anymore when the field is there.Hence it is better to use the minimal coupling principle where pα →

[pα + eA(r)

].

This definition takes automatically into account the atom-field interaction throughthe vector potential A (r). As a consequence of this consideration our new Hamil-tonian should have following form:

H′ = 1

2m

∑α

[pα + eA(r)

]2

+1

2

∫d3r σ(r)φ(r) +

1

2

∫d3r

[ε0E

2T (r) +

1

µ0

B2 (r)

].

(4.25)Where the first term is the Hamiltonian of a free particle in the presence of anelectromagnetic field. The second term is the Coulomb potential of the atom (com-pletely classical, no operators). Finally the third term is the Hamiltonian of the freeelectromagnetic field.

30


Let us develop the first term of eq. (4.25) to extract the interaction Hamiltonian:

1

2m

∑α

[pα + eA(r)

]2

=∑α

p2α

2m︸︷︷︸KineticEnergy

+e

m

∑α

A(rα)pα+

(e2

2m

)∑α

A2(rα). (4.26)

Since the first term on the right hand side of this equation describes the kineticenergy of the isolated atom, we obtain the interaction Hamiltonian as:

H′I =e

m

∑α

A(rα)pα +

(e2

2m

)∑α

A2(rα). (4.27)

Note:

• The interaction Hamiltonian is written as a function of A and is therefore notgauge invariant

• We want gauge invariance and will write it therefore as a function of E ·D

For this we need to apply the so called Power-Zienau-Wooley-transformation

U = exp

[i

~

∫d3r p (r) A(r)

]. (4.28)

The application of U to the system Hamiltonian is such that

H = U†H′U. (4.29)

In doing that, however, the state |ψ′〉 of the system must also be transformed ac-cording to

|ψ〉 = U† |ψ′〉 . (4.30)

After some calculation we obtain the full Hamiltonian describing the atom-fieldinteraction:

H =∑α

p2α

2m+

1

2

∫d3r σ(r)φ(r)︸︷︷︸

HA

+1

2

∫d3r

[ε0E

2T (r) +

1

µ0

B2 (r)

]︸︷︷︸

HF

+ e∑α

rα · ET (0)︸︷︷︸HED

,

(4.31)where now the interaction Hamiltonian is in the right dipole form:

HED = e∑α

rα · ET (0) = eD · ET (0) .

31


4.5 Second Quantization

4.5.1 Second Quantization of the Atomic Hamiltonian

We have seen (see eq. (4.31)) that the atomic Hamiltonian is given by:

HA =∑α

p2α

2m+

1

2

∫d3r σ(r)φ(r).

This form, however, is not useful to us, since we would prefer a second quantizedapproach, i.e., we would like to express the atomic Hamiltonian in terms of "cre-ation" and "annihilation" operators, so that we can unify the formalism with theone used for describing the electromagnetic field. To do so, we consider the atomicenergy eigenstates |i〉, with eigenvalues ~ωi, so that:

HA |i〉 = ~ωi |i〉 . (4.32)

Here we will use the completeness of |i〉, i.e., 1 =∑

1 |i〉〈i| in order to write

HA =

(∑i

|i〉〈i|

)HA

(∑j

|j〉〈j|

)=

∑ij

|i〉〈i| HA |j〉〈j|

=∑ij

|i〉 HA 〈i| j〉〈j|

=∑ij

~ωiδij |i〉〈j|

= ~ωi |i〉〈i| . (4.33)

The second quantized form of the atomic Hamiltonian is then given by:

HA =∑i

~ωi |i〉〈i| (4.34)

In this form, HA is written as a sum of projections onto the energy eigenstates |i〉,where each of them has the eigenvalue ~ωi. In order to catch the essential physicsbehind the atom-field interaction in a simple and intuitive way, we will considerthe case of a two-level system, i.e., we assume the atomic system only to possess aground state, |1〉, and an excited state, |2〉. The transition frequency is defined asω0 = ω2 − ω1. A pictorial representation of such a system is given in Fig. 4.3.

32


Figure 4.3: Schematic sketch of a two level system

The Hamiltonian of a two level system is then given by:

HA = ~ω1 |1〉〈1|+ ~ω2 |2〉〈2| . (4.35)

For a two level system it is useful to introduce the following operators:

σ− = |1〉〈2| (4.36a)

σ+ = (σ−)† = |2〉〈1| . (4.36b)

To understand their meaning, let us consider how they act on the atomic states,namely:

σ− |1〉 = |1〉〈2| 1〉 = 0, (4.37a)

σ− |2〉 = |1〉〈2| 2〉 = |1〉 , (4.37b)

σ+ |1〉 = |2〉〈1| 1〉 = |2〉 , (4.37c)

σ+ |2〉 = |2〉〈1| 2〉 = 0. (4.37d)

Then, σ+ brings the atom from |1〉 to |2〉, while σ− brings the atom from |2〉 to |1〉.Moreover

σ−σ+ = |1〉〈2| 2〉〈1| = |1〉〈1| , (4.38a)

σ+σ− = |2〉〈1| 1〉〈2| = |2〉〈2| , (4.38b)

i.e., this two different combinations of the operators project directly into the eigen-states of the system. It is also useful to introduce the population inversion operator,σ3, as:

σ3 = |2〉〈2| − |1〉〈1| . (4.39)

Then, the set σ+, σ−, σ3 obeys the usual Pauli algebra:

[σ+, σ−] = σ3, (4.40a)

33


[σ3, σ±] = ±2σ±. (4.40b)

The atomic Hamiltonian can be then rewritten in terms of σ± as follows:

HA = ~ω1 |1〉〈1|+ ~ω2 |2〉〈2| = ~ω1σ−σ+ + ~ω2σ+σ−. (4.41)

Convention: Since the energy is usually defined up to a constant, we have thefreedom to decide where to put the zero of the energy. We have two choices:

1. E = 0 for the ground state (Fig. 4.3). Therefore we obtain:

HA = ~ω0σ+σ−,

with ω0 = ω2.

2. E = 0 in the middle of the two levels (see fig. 4.4). Here we obtain ω1 = −ω0

2

and ω2 = ω0

2.

Figure 4.4: Sketch of two level system with E = 0 between the two states

If not specified explicitly, in the rest of this notes we will use the second convention.The Hamiltonian is therefore given by:

HA = −~ω0

2|1〉〈1|+ ~ω0

2|2〉〈2| = ~ω0

2σ3.

4.5.2 Dipole Operator

It is now easy to second quantize the dipole operator. If we assume that the con-sidered two level system has a transition allowed to the electric dipole then therepresentation of this operator in second quantization must be such that it real-ized the transition |1〉 → |2〉 (and vice versa for the complex conjugate). If we callD12 = D21 ∈ R the dipole moment of the atom (assuming it real), we obtain thedipole operator:

D = D12 (σ+ + σ−) (4.42)

34


4.5.3 Jaynes-Cummings Model

We have now all the elements to build a consistent model of the interaction of a twolevel atom with a quantized field. If we recall that the (single-mode) electric fieldoperator is given by

E (z, 0) = iE0

(aeikz − a†e−ikz

)x, (4.43)

(t = 0→ Schrödinger picture, see below) we can then rewrite HED, in the followingway:

HED = eD · E = ieE0 (D12 · x)︸︷︷︸interaction parameter

(aeikz − a†e−ikz

)(σ+ + σ−) .

If we introduce~g = eE0 (D12 · x) , (4.44)

where g describes the strength of the interaction, we obtain, assuming that the atomis locates at position z = 0,

HED = ~g(aσ+ + aσ− + a†σ+ + a†σ−

). (4.45)

What is the meaning of the 4 terms appearing in (4.45)? To understand that, letus consider the diagrammatic representation of the atom-field interaction given byFig. 4.5, where

Figure 4.5: Interaction of electric field with two level system

the photon is illustrated as a solid line (the arrow shows in which direction thephoton is propagating), the dashed line symbolizes the time evolution of the twolevel system and the atom is represented by a black dot. The initial state of theatom is on the right side of the diagram (at t = −∞), while the final state is on theleft side (at t =∞). We assume for simplicity that the interaction occurs at t = 0.Now we will make sketches, analogous to Fig. 4.5, for all four terms of Eq. (4.45),in order to figure out which terms are contributing to the interaction:

35


Figure 4.6: Atom in excited state,photon gets absorbed

Figure 4.7: Atom in ground state,photon gets absorbed

Figure 4.8: Atom in ground state,photon is emitted

Figure 4.9: Atom in excited state,photon is emitted

Fig. 4.6 :aσ−The atom is initially in |2〉.It absorbs a photon and de-cays to |1〉. This does notconserve energy and istherefore excluded!

Fig. 4.7 :aσ+

The atom is initially in |1〉.It absorbs a photon andgets excited to |2〉.This process is allowed.

Fig. 4.8 :a†σ−

The atom is initially in |1〉.It emits a photon and getsexcited to |2〉.This does not conserveenergy and is therefore ex-cluded!

Fig. 4.9 :a†σ+

The atom is initially in |2〉.It emits a photon and de-cays to |1〉.This process is allowed.

Then, only aσ+ and a†σ+ give contributions to our calculations and the interac-tion Hamiltonian can be written as follows:

HED = ~g(σ+a+ σ−a

†) . (4.46)

36


The full Hamiltonian in the Jaynes-Cummings model is then given by:

H =1

2~ω0σ3 + ~ωa†a+ ~g

(σ+a+ σ−a

†) , (4.47)

together with the commutation relations

[a, a†] = 1, (4.48a)

[σ+, σ−] = σ3, (4.48b)

[σ3, σ±] = ±2σ±, (4.48c)

and the other commutators are zero.

4.5.4 States Of The Joint System

The form of the Hamiltonian (4.47) suggests that the considered system can bethought as the direct product of two disjoint Hilbert spaces, a 2D Hilbert spaceHA belonging to the atom and spanned by |1〉 , |2〉, and an infinite- dimensionalHilbert space describing the e.m. field HE and spanned, e.g. by the Fock states|n〉.Since these two Hilbert spaces are disjointed, one can define the state of the jointsystem atom + field simply as the direct product between these two spaces, i.e.,HJ = HA ⊗ HE. The joint Hilbert space HJ is then spanned by |n, i〉 = |n〉 |i〉.

4.6 Schrödinger, Heisenberg and Interaction Picture

So far we considered all the operators to be time-independent. The question is howto introduce the time dependence in our model. We have three possibilities:

4.6.1 Schrödinger Picture

The Schrödinger picture is typically used in quantum mechanics. Here operatorsare time-independent

dOS

dt= 0, (4.49)

whereas states evolve in time according to the Schrödinger equation

i~∂ΨS (t)

∂t= HΨS (t) . (4.50)

The formal solution is given by

ΨS (t) = e−iHt~ ΨS (0) . (4.51)

and dH/dt = 0.

37


4.6.2 Heisenberg Picture

Operators vary in time, while the states are time independent (typical in quantumfield theory and quantum optics), namely

dΨH

dt= 0,

and the operators evolve according to the Heisenberg equation:

i~dOH

dt=[OH , H

]. (4.52)

The total Hamiltonian, however, is time independent in the Heisenberg picture! Theconnection to the Schrödinger picture is given as:

ΨH = eiHt~ ΨS (t) = ΨS (0) , (4.53)

andOH (t) = e

iHt~ OSe

− iHt~ . (4.54)

Proof:

〈Ψ (t)| OS |Ψ (t)〉 = 〈Ψ (0)| eiHt~ OSe

− iHt~︸︷︷︸=OH(t)

|Ψ (0)〉 = 〈Ψ (0)| OH (t) |Ψ (0)〉

4.6.3 Interaction Picture

Intermediate picture where the time evolution is shared between the state and theoperator. In the interaction picture the wave function is fefined as:

ΨI (t) = eiH0t~ ΨS (t) = e

iH0t~ e−

iHt~ ΨS (0) , (4.55)

where H0 is the free Hamiltonian of the total system. For the case of a two- levelsystem interacting wiht the e.m. field, e.g., H0 = HA + HF . The equation of motionis then given by:

i~dΨI (t)

dt= i~

iH0t

~eiH0t~ ΨS (t) + e

iH0t~dΨS (t)

dt

= −H0ΨI (t) + e

iH0t~

HΨS (t)

= −H0ΨI (t) + HΨS (t)

=(−H0 + H0 + HED

)ΨS (t) .

And therefore:i~dΨI (t)

dt= HEDΨS (t) (4.56)

38

Interaction of Matter with Quantized Light - II

In the interaction picture the time evolution associated to operators is realized bymaking the Schrödinger operator evolving with the free Hamiltonian:

OI = eiH0t~ OSe

− iH0t~ , (4.57)

and accordingly:

i~dOI (t)

dt=[OI (t) , H0

]. (4.58)

The interaction picture is useful because all the dynamics are held by the interactionHamiltonian. The free Hamiltonian only accounts for a global phase factor. We cantherefore employ the Heisenberg representation to solve the atom-field dynamics. Inthis case, the time-evolution of the atomic and field operators is given by

σ± →

σ− (t) = σ−e−iω0t,

σ+ (t) = σ+eiω0t,

(4.59)

and

a, a† →

a (t) = ae−iωt,

a† (t) = a†eiωt.(4.60)

The time dependent interaction Hamiltonian thus becomes:

HED (t) = ~g(σ+ae

i(ω0−ω)t + σ−a†e−i(ω0−ω)t

). (4.61)

Note: We assume ω ≈ ω0 (nearly resonant field). Having eliminated σ−a and σ+a†

corresponding to eliminate the terms that oscillate with ω+ω0 ≈ 2ω0 (rotating waveapproximation).

5 Interaction of Matter with Quantized Light - II

5.1 Photon Emission and Absorption Rates

Let us now evaluate the matrix elements for absorption and emission processes drivenby HED. Here, |i〉 and |f〉 describe the initial and final state of the joint system,respectively.

5.1.1 Absorption

First of all, let us have a look at the absorption process. Let us assume, that atthe beginning of this process, the atom is in its ground state and the e.m. field hasn photons. The absorption one photon of the e.m. field leads to the excitation ofthe atom. Therefore let us write the initial and final joint state for the absorption

39


process|i〉 = |n, 1〉 = |n〉 |1〉 , (5.1a)

|f〉 = |n− 1, 2〉 = |n− 1〉 |2〉 . (5.1b)

Then

〈f | HED (t) |i〉 = ~gei(ω0−ω)t 〈f | σ+a |i〉+ e−i(ω0−ω)t 〈f | σ−a† |i〉

= ~gei(ω0−ω)t 〈n− 1| a |n〉〈2| σ+ |1〉

+ e−i(ω0−ω)t 〈n− 1| a† |n〉︸︷︷︸0

〈2| σ− |1〉︸︷︷︸0

. (5.2)

Using〈2| σ+ |1〉 = 〈2| 2〉〈1| 1〉 = 1, (5.3a)

〈n− 1| a |n〉 =√n 〈n− 1| n− 1〉 =

√n, (5.3b)

we finally obtain〈f | HED (t) |i〉ABS = ~g

√nei(ω0−ω)t. (5.4)

Comment: If there are no photons in the field (n = 0), then 〈f | HED (t) |i〉ABS = 0

(the atom cannot absorb photons). This is in accordance with the result from thesemi-classical theory of light-matter interaction.

5.1.2 Emission

For the emission process we will consider an atom in its excited state and an e.m.field with n photons, i.e., the initial and final joint states of the emission process aregiven as

|i〉 = |n, 2〉 = |n〉 |2〉 , (5.5a)

|f〉 = |n+ 1, 1〉 = |n+ 1〉 |1〉 . (5.5b)

Then

〈f | HED (t) |i〉 = ~gei(ω0−ω)t 〈f | σ+a |i〉+ e−i(ω0−ω)t 〈f | σ−a† |i〉

= ~gei(ω0−ω)t 〈n− 1| a |n〉︸︷︷︸

0

〈2| σ+ |1〉︸︷︷︸0

+ e−i(ω0−ω)t 〈n− 1| a† |n〉〈2| σ− |1〉. (5.6)

Using〈1| σ− |2〉 = 〈1| 1〉〈2| 2〉 = 1, (5.7a)

and〈n+ 1| a |n〉 =

√n+ 1 〈n+ 1| n+ 1〉 =

√n+ 1, (5.7b)

40


we obtain〈f | HED (t) |i〉EMISS = ~g

√n+ 1e−i(ω0−ω)t. (5.8)

Comment: Since the matrix element depends on√n+ 1 it is different from zero

also in the case in which there are no photons in the field! This means that anatom in the excited state can interact with the vacuum state of the field, resultingin decaying by emitting a photon (spontaneous emission)

5.1.3 Rate Equation

In order to derive the rate equations we will have a look at a two level system ina single-mode e.m. field. The time evolution of the population changes due tocontinuous absorption and emission of photons. This change is given by:

dN1

dt= −WABSN1 +WEMISSN2, (5.9a)

dN1

dt= −WABSN1 +WEMISSN2, (5.9b)

where N1 is the population of level 1, N2 describes the population of level 2, WABS

stands for the absorption probability and WEMIS gives the emission probability. Inthermal equilibrium we have:

dN1

dt= 0 =

dN2

dt(5.10)

and therefore:WABSN1 = WEMISSN2 (5.11)

And so:N1

N2

=WEMISS

WABS

=〈f | HED (t) |i〉EMISS

〈f | HED (t) |i〉ABS=n+ 1

n. (5.12)

At thermal equilibrium Boltzmann statistics imposes that

N1

N2

= exp

[~ω0

kT

]. (5.13)

Comparing this two equations leads to

n+ 1

n= exp

[~ω0

kT

], (5.14)

which fulfils the Bose-Einstein-Statistics

n =1

exp[~ω0

kT

]− 1

.5 (5.15)

41


5.2 Quantum Rabi Oscillations

5.2.1 The Semiclassical Rabi Model in a Nutshell

In order to have a comparison case, let us first briefly revise the results and the maindynamics of the semiclassical Rabi model. Assuming a monochromatic classicalelectric field E = E0 cosωt, the semiclassical dipole Hamiltonian can be writtenas Hsc(t) = V0 cosωt, where V0 = −d · E0. The time dependent state vector canbe written as a superposition (with time dependent coefficients) of the ground andexcited states as follows:

|ψ(t)〉 = C1(t)e−iω1t|1〉+ C2(t)e−iω2t|2〉. (5.16)

Using the Schrödinger equation we can obtain a set of two differential equationsdescribing the evolution of the time dependent coefficients (that represent the am-plitude probability of being in the correspondent level). The result is the following:

C1 = −iV~ei(ω−ω0)tC2, (5.17a)

C2 =iV

~e−i(ω−ω0)tC1, (5.17b)

where V = 〈2| − d · E0|1〉 ∈ R and ω0 = ω2 − ω1 is the transition frequency. Inorder to obtain the previous equations, moreover, we assumed that the electromag-netic field is nearly resonant with the atomic transition. With this approximationthe terms oscillating at a frequency (ω + ω0) have been neglected (Rotating WaveApproximation [2]).

Solutions of this system of differential equations can be sought for example bytaking the derivative of the second equation and substituting the first one into it,thus obtaining the equation of motion for the probability amplitude of the excitedstate

C2 + i(ω − ω0)C2 +V 2

4~2C2 = 0. (5.18)

Solution of this second order equation with the initial condition that the groundstate is the only populated state (C1(0) = 1 and C2(0) = 0) brings to the following-expressions for the probability amplitudes C1(t) and C2(t):

C1(t) = ei∆t/2

cos

(ΩRt

2

)− i∆

ΩR

sin

(ΩRt

2

), (5.19a)

C2(t) =iV

~ΩR

ei∆t/2 sin

(ΩRt

2

), (5.19b)

where ∆ = ω−ω0 is the detuning between the electric field frequency and the atomictransition, and the quantity ΩR is known as Rabi frequency and its expression is the

42


0 5 10 15 20 25

0.0

0.2

0.4

0.6

0.8

1.0

WRt

P2HtL D=0

D=0.5

D=1

Figure 5.1: Plot of P2(t) versus t for various values of the detuning ∆. Figureadapted from Ref. [2].

following:

ΩR =

√∆2 +

V 2

~2. (5.20)

The Rabi frequency represents the frequency at which the oscillation of populationbetween the ground and the excited state of the atom takes place, triggered by theinteraction with the electromagnetic field. Theprobability that the atom is in theexcited state is in fat given by

P2(t) = |C2(t)|2 =V 2

~2Ω2R

sin2

(ΩRt

2

), (5.21)

which for the case of perfect detuning (∆ = 0) assumes the easier form

P2(t) = sin2

(V t

2~

). (5.22)

The time evolution of the population probability of the excited state is reportedin Fig. 5.1 for various values of ∆. In the case of perfect resonance, a completetransfer of population between the ground and the excited state takes place at thetime t = π~/V . An electromagnetic field characterized by the frequency ω thatrealized this condition is said to be a π-pulse.

It is also useful to introduce the population inversion probability

W (t) = P2(t)− P1(t) = sin2

(V t

2~

)− cos2

(V t

2~

)= − cos

(V t

~

). (5.23)

43


At the time t = π~/2V , one can note that the population inversion is W (π~/2V ) =

0, and the population is coherently shared between the ground and the excited state,i.e., the state of the system is given by

|ψ(t)〉 =1√2

(|1〉+ i|2〉) . (5.24)

5.2.2 The quantum Rabi Model

We then want to investigate if the process of Rabi oscillations is influenced by thequantization of the field or not. Before starting solving the problem of calculatingthe dynamics of the Jaynes-Cummings model, let us make some preliminary obser-vations. First of all we assume that no electrons (or atomic species in general) arelost by the system, and that therefore the population of the atomic system mustremain constant over time. If we define the electron number operator as

PE = σ−σ+σ+σ− = |1〉〈1|+ |2〉〈2| = 1, (5.25)

then it is immediate to see that [PE, H

]= 0. (5.26)

Since we are considering a closed system, not only the energy is conserved, but alsothe total number of excitations is conserved. This means that whenever a photongets annihilated the atom (in order to conserve energy) must absorb it and makethe transition to the excited astate, and every time a photon is created, the atommust decay to the ground state. We can formalize this by defining the followingexcitation number operator

NE = a†a+ σ−σ+, (5.27)

so that [Ne, H

]= 0. (5.28)

This allows us to rewrite the total Jaynes-Cummings Hamiltonian as the sum oftwo parts, one depending on PE and NE and the other one containing the dipolarinteraction. Without loss of generality, we can consider first the simple case ofperfect resonance, thus putting ∆ = ω − ω0 = 0. As we did for the semiclassicalcase, the state of the system can be written as a superposition of the initial state|i〉 ≡ |n, 2〉 and the final state |f〉 ≡ |n + 1, 1〉 (i.e. we are considering the case inwhich the atom is excited and decays to the ground state by emitting a photon) asfollows:

|ψ(t)〉 = Ci(t)|i〉+ Cf (t)|f〉, (5.29)

44


with the initial conditions imposing that Ci(0) = 1 and Cf (0) = 0. If we now usethe above expression for the state and the interaction Hamiltonian ito solve thethe Schrödinger equation for the considered system, we obtain the following set ofcoupled differential equations for the coefficients Ci(t) and Cf (t):

Ci = −ig√n+ 1Cf , (5.30a)

Cf = −ig√n+ 1Ci. (5.30b)

Taking the derivative of the first equation and eliminating Cf (t) by sung the secondone we obtain the following equation of motion for the amplitude probability Ci(t):

Ci + g2(n+ 1)Ci = 0, (5.31)

whose solution compatible with the initial condition is

Ci(t) = cos

[Ω(n)t

2

], (5.32a)

Cf (t) = −i sin

[Ω(n)t

2

], (5.32b)

where Ω(n) = 2g√n+ 1 is the quantum Rabi frequency, whose expression should be

compared with the semiclassical result (at prefect resonance) ΩR = V/~. In orderto draw some conclusion, we can define also in this case the probabilities that thesystem is (at a certain time t) in the initial or final state as

Pi(t) = |Ci(t)|2 = cos2

[Ω(n)t

2

], (5.33a)

Pf (t) = |Cf (t)|2 = sin2

[Ω(n)t

2

], (5.33b)

and the population inversion probability as

W (t) = 〈ψ(t)|σ3|ψ(t)〉 = Pi(t)− Pf (t) = cos Ω(n)t, (5.34)

that, apart from amines sign due to the different choice of initial condition, coincideswith the result of Eq. (5.23), where the semiclassical Rabi frenquecy ΩR has to besubstituted by its quantum counterpart Ω(n).

In the semiclassical case, the Rabi frequency depends on the amplitude of theelectric field through V = 〈2| − d · E0|1〉, and if the amplitude of the field is zero(meaning vettE0 = 0), there is no field, the Rabi frequency is consequently zeroand there is no population oscillation. In the fully quantized model, however, theRabi frequency depends on the number of photons in the field through the quantity√n+ 1. As can be noted, Ω(0) 6= 0. This has a straightforward physical meaning.

If the electromagnetic field is in its vacuum state, the atom has then a nonzero

45


probability of spontaneously emitting a photon and subsequently re-absorbing itand then re-emitting it etc, originating the so-called vacuum Rabi fluctuations. Thisprocess, of course, is a manifestation of the quantum nature of the electromagneticfield and has no classical counterpart.

However, apart from this difference, the dynamics of an atom interacting withan electromagnetic field with a definite number of photons (i.e., in a Fock state) isvery much similar to the semiclassical case, as it gives rise to a regular and periodicdynamics, namely the Rabi oscillations. This could be a bit confusing, since wesaid in the previous chapters that number states are among the most nonclassicalstates of the electromagnetic field (they have no definite electric field!). However, theinduced dynamics are practically the semiclassical one. One would have expectedthis result by using a coherent state, that is the most classical state of the field.However, in this case, the dynamics is much richer, and very different from thesemiclassical case.

5.2.3 Interaction of an Atom with a Coherent State

To analyze the interaction of a two level system with an electromagnetic field de-scribed by a coherent state, we must first of all generalize the solution given aboveto the case of an atom interacting with a field in a superposition of Fock states.Once we have this result, in fact, writing down the solution for the coherent statecase will be very easy.

To do that we assume the atomic state at the time t = 0 to be in a superpositionof ground and excited state

|ψ(0)〉atom = C1|1〉+ C2|2〉, (5.35)

and the electromagnetic field to be in a superposition of number states

|ψ(0)〉field =∞∑n=0

Cn|n〉. (5.36)

As we said before, the initial state of the joint system atom+field can be written asthe direct product of the corresponding Hilbert spaces, resulting in

|ψ(0)〉 = |ψ(0)〉atom ⊗ |ψ(0)〉field =

∞∑n=0

Cn|n〉

C1|1〉+ C2|2〉 . (5.37)

At the generic time t, the state of the electromagnetic field can only be changed ofone photon, either created (emission) or annihilated (absorption) and therefore

|ψ(t)〉field =∞∑n=0

Cn|n〉+ Cn±1|n± 1〉 . (5.38)

46


For the atom, instead, the coefficients C1 and C2 are time dependent and therefore

|ψ(t)〉atom = C1(t)|1〉+ C2(t)|2〉. (5.39)

The joint state of the system atom+field at the generic time t is therefore given by

|ψ(t)〉 = |ψ(t)〉atom ⊗ |ψ(t)〉field. (5.40)

This state must be a solution of the Schrödinger equation. Therefore:

i~∂

∂t|ψ(t)〉 = i~

∑n

(Cn|n〉+ Cn+1|n+ 1〉+ Cn−1|n− 1〉)[C1(t)|1〉+ C2(t)|2〉

],

(5.41)and

HED|ψ(t)〉 = ~g∑n

(σ+a+ σ−a

†) (Cn|n〉+ Cn+1|n+ 1〉+ Cn−1|n− 1〉)

× [C1(t)|1〉+ C2(t)|2〉]

= ~g∑n

√nCnC1(t)|n− 1, 2〉+

√n+ 1CnC2(t)|n+ 1, 1〉

+√n+ 1Cn+1C1(t)|n, 2〉+

√nCn−1C2(t)|n, 1〉

, (5.42)

where in the last line we neglected the terms containing |n± 2〉 since they representstates of the electromagnetic field that cannot be reached (since at maximum thenumber of photons in the field can change by one unity by assumption). If wewant to obtain the equation of motion for C1(t) and C2(t) we have to project theSchrödinger equation

i~∂

∂t|ψ(t)〉 = HED|ψ(t)〉 (5.43)

onto the four different states |n, 1〉 (corresponding to the atom in the ground stateand the field with n photons), |n, 2〉 (atom in the excited state and n photons in thefield), |n + 1, 1〉 (the atom decayed to the ground state by emitting a photon) and|n− 1, 2〉 (the atom absorbed a photon to go to the excited state). We then obtainthe following four equations of motion:

CnC1(t) + g2nCnC1(t) = 0, (5.44a)

Cn+1C1(t) + g2(n+ 1)Cn+1C1(t) = 0, (5.44b)

CnC2(t) + g2(n+ 1)CnC2(t) = 0, (5.44c)

Cn−1C1(t) + g2nCn−1C1(t) = 0. (5.44d)

These equations, together with the initial condition (5.37) (and after having properlyshifted the indices so that it is possible to write the solution as a single summation

47


in n) bring the following result:

|ψ(t)〉 =∞∑n=0

[C2Cn cos

(gt√n+ 1

)− iC1Cn+1 sin

(gt√n+ 1

)]|2〉

+[−iC2Cn−1 sin

(gt√n)

+ C1Cn cos(gt√n)]|1〉|n〉. (5.45)

In general, this state is an entangled state, meaning that it is not possible anymoreto write this state as a simple product between a state of the field and a stateof the atomic system, as it is done in Eq. (5.37). Therefore, in the general caseof interaction of the field with an atomic system, it does not make sense anymoreto talk about the field and the atom, but rather we must speak of the atom-fieldsystem, as the general wave function is non-separable anymore (entangled).

If we assume that the atom is initially in the excited state (i.e., we set C1(0) = 0

and C2(0) = 1), the previous solution can be still written as a product between anevolved field state and the atomic states as follows:

|ψ(t)〉 = |ψ1(t)〉|1〉+ |ψ2(t)〉|2〉, (5.46)

where

|ψ1(t)〉 = −i∞∑n=0

Cn sin(gt√n+ 1

)|n+ 1〉, (5.47a)

|ψ2(t)〉 =∞∑n=0

Cn cos(gt√n+ 1

)|n〉. (5.47b)

In this case we can calculate analytically the atomic inversion probability as

W (t) = 〈ψ(t)|σ3|ψ(t)〉

= 〈ψ2(t)|ψ2(t)〉 − 〈ψ1(t)|ψ1(t)〉

=∞∑n=0

|Cn|2 cos(

2gt√n+ 1

). (5.48)

We now have a very general expression for the atomic inversion as a function of theexpansion coefficients Cn. If we now want to study the dynamics of the system underthe action of a coherent state, we just have t substitute in the previous expressionthe expression for the expansion coefficients of a coherent state in terms of numberstates, namely

Cn = e−|α|2/2 α

n

√n!. (5.49)

If we do that, eq. (5.48) becomes (we call n = |α|2 as the mean number of photons

48


0 5 10 15 20 25

-0.5

0.0

0.5

1.0

T

WHtL

HaL

0 10 20 30 40 50

-0.5

0.0

0.5

1.0

T

WHtL

HbL

Figure 5.2: (a) Atomic inversion with the field initially in a coherent state withn = 5. (b) Same as (a) but showing the evolution for a longer time, beyond thefirst revival time. Both graphs are plotted against the normalized time T = gt. Thefigures are adapted from Ref. [2].

in the coherent state)

W (t) = e−n∞∑n=0

nn

n!cos(

2gt√n+ 1

). (5.50)

In Fig. 5.2 a plot of the atomic inversion is given. As it appears clear by simply

49


looking at the graph, the dynamics of the system in this case is profoundly differentfrom the semiclassical case. Let us try to understand what are the characteristics ofthis dynamics. First of all, from Fig. 5.2(a) we note that the Rabi oscillation appearsto damp out after a certain time. If we however wait long enough [Fig. 5.2(b)], theoscillations start to revive, although not in phase with the initial oscillations. Thisdynamics repeats almost periodically in time, each time making more difficult todistinguish between a collapse and a revival. We now try to interpret this strangebehavior.

Let us start with the following observation: Eq. (5.50) is basically a series ofcosines, each one characterized by a frequency Ω(n) that depends on the number ofphotons in the field. If the average number of photons is n, then it makes sense tosay that the dominant Rabi frequency will be Ω(n), namely

Ω(n) = 2g√n+1 ' 2g

√n, (5.51)

where the last approximation is valid in the limit n 1, as it is usually the casefor a coherent state, where n ' 106. Since the photon number is not perfectlydetermined, however, there will be a range of frequencies around Ω(n) that willcontribute sensibly to the dynamics. Calling ∆n the uncertainty in the photonnumber, the relevant frequency range should be between Ω(n−∆n) and Ω(n+∆n).If we want to say something about the dynamics of the system, we therefore haveto investigate this frequency range

Collapse The collapse time tC is defined as the instant of time in which the relativedephasing between the various frequency components that are present in the initialexpression of W (t) is such that they all interfere destructively, leaving no atomicinversion, therefore collapsing the dynamics. Using the uncertainty relation existingbetween time and frequency, we can estimate the collapse time tC as follows:

tC [Ω(n+ ∆n)− Ω(n−∆n)] ' 1. (5.52)

Since for a coherent state ∆n =√n, we have that

ω(n±∆n) ' 2g

√n±√n ' 2g

√n

[1± 1√

n

]1/2

' 2g√n

(1± 1

2√n

)= 2g

√n± g, (5.53)

and therefore the collapse time is given by

tC '1

2g. (5.54)

50


Interestingly, the collapse time is independent of the average number of photons inthe coherent state. This derivation is not fully rigorous, nevertheless it catches thesalient features. For a more rigorous derivation, the reader is addressed to Ref. [2].

Revivals We now turn to our attention to determine the revival time tR, corre-sponding to the time in which all the relevant frequencies are oscillating in phase(or most of them are in the case of a fractional revival), thus reproducing the ini-tial oscillatory pattern. By means of simple considerations based on interference ofadjacent frequencies, the revival time can be estimated by means of the followingrelation:

[Ω(n+ 1)− Ω(n)] tR = 2πk, (5.55)

where k ∈ N. By using the approximated expressions of Ω(n + 1) and Ω(n) givenabove, we can estimate the revival time to be given by

tR '(

2π

g

)k√n. (5.56)

In this case, the revival time depends on the average number of photons in the fieldas a monotonically increasing function. The more photons are in the field, therefore,the more is tR close to infinity. As for the case of the collapse time, a more rigorousderivation of tR is given in ref. [2].

5.3 Wigner-Weißkopf-Theory [5]

In this section we will study the interaction of a 2 level atom with a multimode field,where the model multimode field is seen as a continuum of states.In order to study the Wigner-Weißkopf model, we need to introduce the continuouslimit to our discrete model of quantization. This can be done by considering twothings, which are first taking the limit V → ∞ for the cavity volume and then,accordingly we will need to generalize the definition of a, a† to the continuous case.

Figure 5.3: Geometry of optical cavity (taken from [1])

51


Since we consider the case V → ∞, the number of allowed k-vectors inside thecavity rises, so that we can consider them as continuous. Therefore we will substitutethe sum over the k-vectors by an integration over the density of states in k-spaceD (k): ∑

k

2∑λ=1

−→2∑

λ=1

∫d3kD (k) . (5.57)

In order to obtain D (k), we consider a quadratic cavity as shown in Fig. 5.3,therefore:

k =2π

L(nx, ny, nz) , (5.58)

with ni ∈ N, but only one of the ni can be zero, since the field in the cavity vanishesif two or three n’s are zero. For a given combination (nx, ny, nz), there is only onestate in the volume (2π)3

V. Therefore

D (k) =V

(2π)3 . (5.59)

How do we generalize operators?

With this we get:akλ →

√∆ωaλ (ω) (5.60a)

a†kλ →

√∆ωa†

λ (ω) (5.60b)[akλ, a

†kµ

]= δ (ω − ω′) δλµ, (5.60c)

where ∆ω = c∆k = 2πcL

is a normalization factor introduced to preserve the bosonenature of the commutator (5.60c).The electric field operator (for a linearly polarizedfield propagating along z-axis) is given by:

ET (z, t) = i

∫dω

(~ω

4πε0cA

) 12 [a (ω) e−iω(t− zc ) + a† (ω) eiω(t− zc )

]. (5.61)

The magnetic field operator B is defined accordingly.

The Model

Our model considers the interaction of a 2-level atom with a single-mode field. Forsimplicity we only take single photon processes into account. In the beginning of ourconsiderations we assume the atom to be in the excited state and assume no photonsin the field, i.e., |Φ2〉 |0〉. After the interaction we want our atom to be in the groundstate (decay by emission of a photon) |Φ1〉 |Φ (k)〉, and therefore to have one photonin the field. The emitted photon can have any momentum k, meaning that it canbe emitted in any direction! This is due to the fact that the atom is inserted inan isotropic vacuum, and there is no preferred direction of emission. Therefore, to

52


correctly account for all the possibilities, we need to integrate over all the possiblevalues of k.Since both |Φ1〉 , |Φ2〉 and |Φ (k)〉 are orthogonal basis in their respective Hilbertspaces, |Ψ (0)〉 = |Φ2〉 |0〉, and we can write the state at time t as superposition ofthe atomic and field states as follows:

|Ψ (t)〉 = a (t) exp

[−iEt~

]|Φ2〉+

∫d3k b (k, t) exp

[−iε (k) t

~

]|Φ (k)〉 (5.62)

where ε (k) is the energy of the photon with momentum k. In this case, for the sakeof simplicity, we assume to use the ground state as a reference for the energy scale,so that E = E2 − E1 = E2. Eq. (5.62) must solve the Schrödinger equation:

i~∂

∂t|Ψ (t)〉 =

(H0 + V

)|Ψ (t)〉 , (5.63)

where V is the atom-field interaction, e.g. the dipole Hamiltonian. Calculating theleft-hand-side (LHS) and the right-hand-side (RHS) separately, we have:LHS

i~∂

∂t|Ψ (t)〉 = i~

da (t)

dtexp

[−iEt~

]|Φ2〉+ Ea (t) exp

[−iEt~

]|Φ1〉

+ i~∫d3k

db (k, t)

dtexp

[−iE (k) t

~

]|Φ (k)〉

+

∫d3k E (k) b (k, t) exp

[−iE (k) t

~

]|Φ (k)〉

RHS(H0 + V

)|Ψ (t)〉 = Ea (t) exp

[−iEt~

]|Φ2〉+

∫d3k E (k) b (k, t) exp

[−iE (k) t

~

]|Φ (k)〉

+ a (t) exp

[−iEt~

]V |Φ2〉+

∫d3k b (k, t) exp

[−iE (k) t

~

]V |Φ (k)〉

Remember V is a dipole-line separator given by

〈Φk| V |Φk〉 = 0, (5.64)

and the photon states are orthogonal, i.e.,

〈Φ (q)| Φ (k)〉 = δ (q− k) . (5.65)

If we project the Schrödinger eq. (5.63) onto 〈Φ2| and 〈Φ (k)|:

i~da (t)

dt=

∫d3k b (k, t)M (k) e−iω(k)t, (5.66a)

53


i~db (k, t)

dt= a (t)M∗ (k) eiω(k)t, (5.66b)

where ~ω (k) = ε (k) − E is the transmission energy, M (k) = 〈Φ2| V |Φ (k)〉 givesthe matrix element and with the initial conditions a (0) = 1, b (k, 0) = 0.Eq. (5.66b) can be formally solved by multiplying by dt and integrating, namely

t∫0

db (k, t′) =M∗ (k)

i~

t∫0

dt′ a (t′) eiω(k)t′ ,

thus obtaining

b (k, t) =M∗ (k)

i~

t∫0

dt′ a (t′) eiω(k)t′ . (5.67)

Substituting the expression of b (k, t) by Eq. (5.67) we obtain for Eq. (5.66a)

da (t)

dt=

1

i~

∫d3k|M (k)|2

i~eiω(k)t

t∫0

dt′ a (t′) eiω(k)t′ ,

i.e.,

da (t)

dt= − 1

~2

∫d3k |M (k)|2 eiω(k)t

t∫0

dt′ a (t′) eiω(k)t′ . (5.68)

The last term of Eq. (5.68) makes the problem non-Markovian, i.e. the system hasmemory and therefore depends on its past history. We need to make it Markovian,i.e. a (t2 > t1) does not have to depend from all a (t) with t ∈ [0, t1]. Here we usea little trick: we will multiply the equation by e−λt and integrate in time. We thenhave from Eq. (5.68)LHS

∞∫0

dtda (t)

dte−λt =

∞∫0

dt

d

dt

[a (t) e−λt

]+ λa (t) e−λt

= a (t) e−λt∣∣∞0

+ λ

∞∫0

dt a (t) e−λt

= λ

∞∫0

dt a (t) e−λt − 1,

54


RHS

− 1

~2

∫d3k |M (k)|2

∞∫0

dt e−λte−iω(k)t

t∫0

dt′ a (t′) eiω(k)t′

︸︷︷︸= − 1

~2

∫d3k |M (k)|2

t∫0


∞∫t′

dt e−[λ−iω(k)]t

= − 1

~2

∫d3k |M (k)|2

t∫0


−1

λ+ iω (k)e−[λ−iω(k)]t

∣∣∞t′

= − 1

~2

∫d3k

|M (k)|2

λ+ iω (k)

t∫0

dt′ a (t′) eλt′.

Here we used the fact that the underlined integral can be solved in two ways (see Fig.5.4), i.e., whether considering t as fixed and integrating along a vertical strip,

∫∞0dt,

which corresponds to∫∞

0dt∫ t

0dt′, or integration along a horizontal strip (i.e.

∫∞t′dt)

where we first integrate over∫ t

0dt′, which corresponds to swithcing the integrations

in Eq. (5.68), i.e.,∫ t

0dt′∫∞t′dt.

Figure 5.4: Interpretation of the integral dt′dt

In this case, we use the second possibility, to integrate the RHS of Eq. (5.68)andthen obtaining

∞∫0

dt e−λta (t) =

[λ+

1

~2

∫d3k

|M (k)|2

λ+ iω (k)

]−1

. (5.69)

Note: If we go back to Eq. (5.68) and assume |M (k)|2 to be independent of k we

55


obtain:

da (t)

dt= − 1

~2

∫d3k |M (k)|2 e−iω(k)t

t∫0

dt′ a (t′) eiω(k)t′ (5.70)

= −|M |2

~2

∫d3k e−iω(k)t

t∫0

dt′ a (t′) eiω(k)t′ (5.71)

= −|M |2

~2

t∫0

dt′ a (t′)

∫d3k e−iω(k)(t−t′)︸︷︷︸

∼δ(t−t′)

= −|M |2

~2a (t) , (5.72)

and therefore:a (t) = a (0) e−γt, (5.73)

with γ = |M |2~2 . This solution is consistent with the assumption of having an isotropic

vacuum. In this case, in fact, there is no reason why the emitted photon shouldprefer one k vector to the other one, as they are all equally probable. Therefore,the matrix element M (k) cannot (by symmetry considerations) depend on k. Nowlet us assume that this would still be a good solution even for the case that M (k)

depends on k and solve the problem for λ→ 0. Keeping a (t) = e−zt as Ansatz, wethen have:

∞∫0

dt e−λte−zt = − 1

z + λe−(z+λ)t

∣∣∞0

=1

z + λ. (5.74)

and therefore Eq. 5.69 becomes

1

z + λ=

[λ− i

~2

∫d3k

|M (k)|2

ω (k)− iλ

]−1

, (5.75)

where we have used the fact that

1

~2

1

λ+ iω=

1

~2

1

i (ω − iλ)= − i

~2

1

ω − iλ.

As λ is an artificial parameter inserted for facilitating the solution of Eq. (5.68),the real solution must not depend on it. To do so, we then take the limit of (5.75)for λ→ 0, thus obtaining

z = limλ→0+

− i

~2

∫d3k

|M (k)|2

ω (k)− iλ

= limλ→0+

− i

~2

∫d3k|M (k)|2 [ω (k) + iλ]

ω2 (k) + λ2

= limλ→0+

− i

~2

∫d3k

ω (k) |M (k)|2

ω2 (k) + λ2+

1

~2

∫d3k

λ |M (k)|2

ω2 (k) + λ2

, (5.76)

56


By using

limλ→0+

λ

ω2 (k) + λ2= πδ (ω (k)) , (5.77a)

andδ (ax) =

1

|a|δ (x) . (5.77b)

Eq. (5.76) becomes

z = − i

~2

∫d3k|M (k)|2

ω (k)+π

~

∫d3k |M (k)|2 δ (~ω (k)) . (5.78)

In order to better understand this result, we define

E ′ = E +

∫d3k

∣∣∣〈Φ1| V |Φ (k)〉∣∣∣2

E − ε (k), (5.79a)

γ =2π

~

∫d3k

∣∣∣〈Φ1| V |Φ (k)〉∣∣∣2 δ (ε (k)− E) , (5.79b)

so that z = iE ′/~ + πγ/(2~). Then, the solution for a (t) reads as follows:

a (t) = e−iE′t~ e−

γ2t, (5.80)

and the occupation probability can be simply calculated as |a (t)|2, i.e.,

Pl (t) = |a (t)|2 = e−γt. (5.81)

The excited state of the atom decays exponentially in time due to the interactionwith the electromagnetic field in the vacuum state.

Let us then now calculate, what is the probability that at time t → ∞ the sys-tem is in the state |Φ (k)〉 (atom decayed and emitted a photon).To do that, we calculate b (k, t) using the above result a (t) = e−zt, thus obtaining

b (k,∞) =M∗ (k)

i~

∞∫0

dt′ e−[z−iω(k)]t′

=

[M∗ (k)

i~

] [π

~

∫d3k′ |M (k)|2 δ (~ω (k))− i∆

]−1

, (5.82)

where

∆ =ε (k)− E

~− 1

~

∫d3k′

|M (k)|2

E − ε (k′). (5.83)

57

Quantum Mechanics of Beam Splitters

Rearranging the terms gives the following final result:

b (k,∞) =M∗ (k)

ε (k′)− E − 1~

∫d3k′ |M(k)|2

E−ε(k′)

+i~γ2, (5.84)

and therefore the probability that a photon with momentum k has been emitted isgiven by

Pk = |b (k,∞)|2 =|M (k)|2

[ε (k′)− E]2 +(~γ

2

)2 . (5.85)

Usually, for simplicity, in our models we assume that the transition of of electronsbetween energetic level has a delta like resonance function, i.e., the emitted photonis monochromatic. But as seen in Eq. (5.85) this is not true, because the transitionprobability has a Lorentzian shape with non zero bandwidth. Therefore the emittedphotons show a spectral broadening, which we call natural linewidth. The naturallinewidth gives the lower limit to the spectral bandwidth of our photon.

6 Quantum Mechanics of Beam Splitters

6.1 Classical theory

Figure 6.1: Schematic sketch of a classical beam splitter with incoming and outgoingfields

The beam splitter in Fig. 6.1 is assumed to be ideal, i.e., it has two different surfacesand its thickness can be assumed to be small enough to be neglected. The reflectionand transmission coefficients are (r, t) and (r′, t′) for the first (solid line in Fig.6.1) and second (dashed line in Fig. 6.1) surface, respectively. The incoming and

58


outgoing fields can be then written as

EIN (r, t) =1

2e−iωt

[a1e

ik1·r + a2eik2·r

], (6.1a)

EOUT (r, t) =1

2e−iωt

[a3e

ik3·r + a4eik4·r

], (6.1b)

Where ai are the amplitudes of the ingoing (a1, a2) and outgoing (a3, a4) fields,respectively. We can write this in a matrix form, obtaining(

a4

a3

)=

(t r′

r t′

)(a1

a2

)= B

(a1

a2

), (6.2)

where due to energy conservation |a1|2 + |a2|2 = |a3|2 + |a4|2.If we define:

v =

(a1

a2

), (6.3a)

andu = Bv, (6.3b)

then the total energy reads

Ein = |Ein|2 = (v,v) , (6.4)

andEout = (u,u) = (Bv, Bv) =

(v, B†Bv

)= Ein = (v,v) , (6.5)

where we used (AB,C) =(A,B†C

). We get that Eout = Ein if B†B = 1.

What does this mean? IfB†B = 1→ B† = B−1,

then

B =

(t r′

t t′

)→ B† =

(t∗ r∗

r′∗ t′∗

),

and

B−1 =1

tt′ − rr′︸︷︷︸(t′ −r′

−r t

),

where tt′ − rr′ is just a phase factor. Therefore

B†B = 1→

t′ = t∗,

r′ = −r∗,

59


and so the beam splitter can be represented by the following matrix

B =

(t −r∗

r∗ t∗

). (6.6)

Moreover it follows that:t′

r′= −

(t

r

)∗. (6.7)

This result allows us to obtain the following relation for the phases of the transmis-sion and reflexion coefficients:

(Φt′ − Φr′) + (Φt − Φr) = ∓π + 2nπ.

If we chooseΦt − Φr = Φt′ − Φr′ =

π

2,

this leads to

B =

(t ir

ir t

). (6.8)

Note:

1. Energy conservation is fulfilled:

|t|2 + |r|2 = 1

2. because of the first point we can write t = cosϑ, r = sinϑ, so that |t|2 + |r|2 =

cos2 ϑ+ sin2 ϑ = 1 and so

B =

(cosϑ i sinϑ

i sinϑ cosϑ

).

60


6.2 Quantum Theory

Figure 6.2: Schematic sketch of a QM beam splitter with incoming and outgoingfield

Let us consider a beam splitter with very high transmission (ε 1). We wantan operator representation for a beam splitter from a quantum mechanical point ofview, therefore the operator must be unitary and linear in a†, a. Hence let us lookfor an exponential operator. We obtain the simple result:(

a4

a3

)= B

(a1

a2

). (6.9)

Proof:

We want a unitary operator U such that1:

U a1U = a4, (6.10a)

U a2U = a3. (6.10b)1Note that the transformation between the input mode operators a1 and a2 and the correspon-

dent output mode operators a3 and a4 implemented by the unitary operator U is a particular case ofa more general class of transformations, namely the so-called Bogoliubov transformations. Bogoli-ubov transformations are unitary transformations from a unitary representation of a commutation(or anti-commutation) algebra into another, equivalent and unitary representation. They are oftenemployed in theoretical physics and condensed matter theory to diagonalize Hamiltonians, leadingto the steady-state solutions of the correspondent Schrödinger equations. Moreover, Bogoliubovtransformation are an essential ingredient to understand the Unruh effect and the Hawking radi-ation. The interested reader can consult Refs. [6, 7, 8] for more detailed informations about theuse of Bogolyubov transformations in quantum field theory and condensed matter theory.

61


Since U is unitary, we can write it in the form of an exponential operator, namely

U = e−iH ,

withH† = H,

so that|out〉 = U |in〉 = e−iH |in〉 . (6.11)

If we assume low reflectivity (ε 1), then H = H (ε) can be expanded in a Taylorseries with respect to ε, i.e.,

H (ε) = H0 + EH ′ +O(E2).

Plugging H (ε) in (6.11) and Taylor expanding of the exponential we obtain:

|out〉 = e−iH |in〉 = e−i[H0+EH′] |in〉

≈(

1− iEH ′)|in〉 = |k1〉 − iEH ′ |k1〉

∝ |k1〉+ E |k2〉 . (6.12)

Note thatH ′ |k1〉 ∝ |k2〉

which leads to〈k2| H ′ |k1〉 = c− number (6.13)

How can we find an expression for H ′? Given Eq. (6.13), we can say that

H ′ = a†2a1 + Q,

with〈k2| Q |k1〉 = 0.

From(H ′)

=(H ′)†, follows that

a†2a1 + Q = a†

1a2 + Q†,

and therefore it is not difficult to prove that

Q = a†1a2.

Therefore we obtain:H ′ = a†

1a2 + a†2a1, (6.14)

62


which in general readsH ′ = g∗a†

1a2 + ga†2a1. (6.15)

The Beam Splitter operator can be then written as follows:

U = eiE(g∗a†

1a2+ga†2a1). (6.16)

Now we will prove, that the matrix representation of the beam splitter operator Uintroduced above is given by B, therefore we introduce:

x =

(a1

a2

), (6.17)

and build the following scalar product

(x,Mx) = x†Mx

= m11a†1a1 +m12a

†1a2 +m21a

†2a1 +m22a

†2a2 = B, (6.18)

with M = M †. Let us now assume to introduce an operator U (λ) such that

U (λ) = eiλB. (6.19)

In this case U (1) = eiB and calculate

UxU † = eiλBxe−iλB ≡ f (λ) .

By using the Heisenberg equation we can see how f (λ) evolves with respect to λ,i.e.,

idf

dλ= i[B, f (λ)

], (6.20)

with f (0) = x. Calculating the RHS gives[B, f (λ)

]=

[B, eiλBxe−iλB

]= BeiλBxe−iλB − eiλBxe−iλBB

= eiλB[B, x

]e−iλB (6.21)

(note that B and e±iλB commute). Computing the matrix element of the commu-tator, with B = mij a

†i aj gives[B, x

]=

[mij a

†i aj, ak

]= mij

[a†i aj, ak

]= mij

a†i [aj, ak] +

[a†i , ak

]aj

= −mijδikaj = −mkj aj = (Mx)k . (6.22)

63


By then generalizing the previous result we get:[B, f (λ)

]= −eiλBMxe−iλB

= −MeiλBxe−iλB

= −Mf (λ) (6.23)

(note that M is a c-number matrix), i.e., it is not an operator anymore. Eq. (6.20)now becomes

df

dλ= −iMf ,

whose solution is given byf (λ) = e−iλMx. (6.24)

Observe that U †xU can be then represented in two equivalent ways, i.e.,

U †xU = ei(g∗a†

1a2+ga†2a1)xe−i(ga

†2a1+g∗a†

1a2), (6.25a)

andU †xU = eiλBxe−iλB = eimij a

†i ajMe−imij aia

†j . (6.25b)

Comparing these two equivalent representations, i.e., the RHS of (6.25a) with theRHS of (6.25b) we get

M =

(0 g∗

g 0

). (6.26)

Exponentiating M we have the following result

e−iM =∞∑k=0

(−i)k

k!Mk =

(cos |g| −i g∗|g| sin |g|

−i g|g| sin |g| cos |g|

). (6.27)

If the phases g|g| and

g∗

|g| are properly chosen, then

(cos |g| −i g∗|g| sin |g|

−i g|g| sin |g| cos |g|

)→

(t ir

ir t

),

thereforet = cos |g| , (6.28a)

andr = −ieiφg sin |g| . (6.28b)

64


6.3 Classical VS Quantum Beam Splitter

6.3.1 Classical BS

We consider a classical beam splitter, where we have a field in only one arm. There-fore the beam splitter equation reads(

a4

a3

)=

(t ir

ir t

)(a1

0

). (6.29)

The incoming and outgoing energy then are given by

Ein = |v1|2 , (6.30a)

and

Eout = |a3|2 + |a4|2 = |irv1|2 + |ta1|2

=(|r|2 + |t|2

)︸︷︷︸=1

|v1|2 = |a1|2 . (6.30b)

As one can expect the energy conservation is fulfilled in this case.

6.3.2 Quantum BS

As in the case of a classical beam splitter we want to study the case of a quantumbeam splitter where we have a field only in one input. The beam splitter equationfor the quantum case then reads(

a4

a3

)=

(t ir

ir t

)(a1

0

). (6.31)

Since the beam splitter is a passive device, the nature of the incoming radiation mustnot be changed. Therefore, if the bosonic commutation relations (see Eq. (6.32))hold for the incoming field, they must also hold for the outgoing fields.[

ai, a†j

]= δij, (6.32a)

[ai, aj] = 0 =[a†i , a

†j

]. (6.32b)

In order to see if above statement is valid we need to compute[a3, a

†3

](where we

expect result=1),[a4, a

†4

](=1) and

[a3, a

†4

](=0)

[a3, a

†3

]=[ira1,−ir∗a†

1

]= |r|2

[a1, a

†1

]= |r|2 6= 1 (6.33a)

65


[a4, a

†4

]=[ta1, t

∗a†1

]= |t|2

[a1, a

†1

]= |t|2 6= 1 (6.33b)[

a3, a†4

]=[ira1,−ir∗a†

1

]= irt∗

[a1, a

†1

]= irt∗ 6= 0 (6.33c)

Why are this results different from our expectations? In order to understand whythe results are not the expected ones let us consider one little (subtle) detail (seeFig. 6.3):

Figure 6.3: Comparison classical vs quantum BS

As we see from Fig. 6.3 we need to account for the presence of the quantumvacuum in the second arm of the BS. We indicate this by assigning an operator a0

on the other input port of the BS. In this case Eq. (6.31) becomes(a4

a3

)=

(t ir

ir t

)(a1

a0

), (6.34)

if we now calculate for example[a3, a

†3

], we get

[a3, a

†3

]=

[ira1 + ta0,−ir∗a†

1 + t∗a†0

]= |r|2

[a1, a

†1

]+ irt∗

[a1, a

†0

]− itr∗

[a0, a

†1

]+ |t|2

[a0, a

†0

]= |r|2 + |t|2 = 1,

which is the expected result. The other commutator follows analogously.

66


6.4 Effects of Vacuum in BS: Coherent State Input

We now want to ask ourselves if taking into account the presence of the vacuum statein the unused port of the BS has some sort of consequence or not. To this aim, letus first consider the case of a coherent state as input, and assume, for simplicity, toconsider a 50% BS, i.e., t = r = 1/

√2. The input state is then given by (remember

displacement operator, see 3.52):

|in〉 = |α〉1 |0〉2 = D (α) |0〉1 |0〉2 = eαa†−α∗a |0〉1 |0〉2 . (6.35)

And, according to Eq. (6.34)(a4

a3

)=

1√2

(1 i

i 1

)(a1

a2

).

Since we want to calculate how the input state gets transformed by the BS, it ismore useful to express the input operators a1 and a2 as a function of the outputoperators (a3, a4). This is obtained by inverting the above equation, thus obtaining(

a1

a2

)=

1√2

(1 −i−i 1

)(a4

a3

), (6.36)

we then have that the input state |in〉 tronasforms to the following output state:

|out〉 = D′ (α) |0〉3 |0〉4 = eα√2(a†

4+ia†3)− α

∗√2

(a4−ia3) |0〉3 |0〉4 , (6.37)

Moreover, since a4 and a3 commute

eα√2(a†

4+ia†3)− α

∗√2

(a4−ia3)= exp

[iαa†

3√2− iα∗a†

3√2

]exp

[αa†

4√2− α∗a†

4√2

](6.38)

= D3

(iα√

2

)D4

(α√2

). (6.39)

As a result we have that the input coherent state is split (50%) into two coherentstates. The average number of photons per output port is therefore 50% of theinitial one, since coherent states contain the quantum noise, this corresponds to theclassical case

67


6.5 Effects of Vacuum in BS: Single Photon Input

We now consider the case of a single photon in a Fock state impinging on a 50% BS.Again, r = t = 1√

2, and we have

|1〉1 |0〉2 = a†1 |0〉1 |0〉2

BS−−→ 1√2

(a†

4 + ia†3

)|0〉3 |0〉4

=1√2

(i |1〉3 |0〉4 + |0〉3 |1〉4) .

We have therefore:

|1〉1 |0〉2 →1√2

(i |1〉3 |0〉4 + |0〉3 |1〉4) . (6.40)

This means that the output state of the BS is, because of the impact of the vacuumstate in the second port of the input, an entangled state, where we can not tellanymore, where the photon is.

6.6 Hong-Ou-Mandel Effect [9]

Consider a 50% BS with two photons as input, one single photon from the field infirst arm and one single photon from the vacuum in second port. What will be ouroutput? Let us have a look at the possibilities how two photons would behave theclassical case:

Figure 6.4: Classical prediction of the behaviour of two photons in a BS

In order to check if our expectations are correct, let us do some QM calculations.

68


First of all let us rewrite the input operators as a function of the output ones, namely(a1

a2

)=

1√2

(1 −i−i 1

)(a4

a3,

)

and similarly, for the hermite conjugate(a†

1

a†2

)=

1√2

(1 i

i 1

)(a†

4

a†3

). (6.41)

Since the incoming field is given by:

|in〉 = |1〉1 |1〉2 = a†1a

†2 |0〉 , (6.42)

the outgoing field is then given by

|out〉 =

[1√2

(a†

4 + ia†3

)] [ 1√2

(ia†

4 + a†3

)]|0〉

=1

2

[i(a†

4

)2

+ a†4a

†3 − a

†3a

†4 + i

(a†

3

)2]|0〉

=1√2

i(a†

4

)2

√2

+ i

(a†

3

)2

√2

|0〉=

1√2

(|2〉4 |0〉3 + |0〉4 |2〉3) . (6.43)

Figure 6.5: Schematic sketch of a HOM interferometer

Fig. 6.5 shows schematically how a HOM interferometer works. Here a nonlinearprocess (PDC-cell) is used to generate single photon states. The BS works as seen

69

Quantum Theory of Photodetection

above. If we slightly move the BS, we can introduce a delay between the two counts.The detected signal is illustrated in Fig. 6.6.

Figure 6.6: Coincidence rate detected by a HOM interferometer depending on thedelay between the photons

The coincidence rate can be written as follows

Rcoincidence ≈ 1− exp[−∆ω2 (τs − τi)2] , (6.44)

where −∆ω2 is the bandwidth of the photons (non-monochromatic signal/idler) andτs−τi is proportional to the distance difference that the two photons travelled. If τs−τi τcorr we obtain a maximum, i.e. the photons are not impinging simultaneouslyon the BS, which shows classical behaviour (simultaneous clicks possible). Butfor τs − τi = 0 we obtain that Rcoincidence = 0, i.e. the photons are impingingsimultaneously on the BS, therefore we observe bunching (no simultaneous clickpossible) and hence, the coincidence is zero, as can be seen.

7 Quantum Theory of Photodetection

7.1 Direct Detection

Number of photons that arrive at the detector during the integration time t is givenby

M (t, τ) =

t+τ∫t

dt′a† (t′) a (t′) (7.1)

just like the time averaged classical intensity (average over integration time). Let usassume, that the light is nearly monochromatic and that T < ∞. This is the ideal

70


case. In reality we have no detector with a 100% conversion efficiency (photons mayfail to trigger ionization events, dead time masks subsequent ionization events).Dead time is the time that the photo-tube needs to recover, therefore we havea certain quantum efficiency η < 1. Therefore we can model a real (imperfect/inefficient) photo-detector as shown in fig. 7.1.

Figure 7.1: Schematic sketch of the model of a real detector

Fig. 7.1 shows the actual model of a real detector. The BS models the inefficiencyand v (t) accounts for the quantum noise. The transmittance and reflectivity of theBS are given by:

T =√η (7.2a)

R = i√

1− η (7.2b)

• The output on arm 3 is lost (detector inefficiency)

• the output on arm 4 enters a perfect detector

• v (t) is a continuous vacuum

We have:d (t) =

√ηa† (t) + i

√1− ηv (t) (7.3)

71


The photo-count operator M is now defined as

MD (t, τ) =

t+τ∫t

dt′d† (t′) d (t′) . (7.4)

Mean Photocount:

〈m〉 ≡⟨MD (t, τ)

⟩=

⟨ t+τ∫t

dt′d† (t′) d (t′)

⟩

=

⟨ t+τ∫t

dt′[√

ηa† (t′)− i√

1− ηv† (t′)] [√

ηa (t′) + i√

1− ηv (t′)]⟩

= η⟨M (t, τ)

⟩+ i√η (1− η)

⟨ t+τ∫t

dt′a† (t′) v (t′)

⟩

− i√η (1− η)

⟨ t+τ∫t

dt′v† (t′) a (t′)

⟩+ (1− η)

⟨ t+τ∫t

dt′v† (t′) v (t′)

⟩

and therefore:〈m〉 ≡ η

⟨M (t, τ)

⟩(7.5)

Second Moment:

〈m (m− 1)〉 =

t+τ∫t

dt′t+τ∫t

dt′′⟨d† (t′) d† (t′′) d (t′′) d (t′)

⟩= . . .

= η2⟨M (t, τ)

[M (t, τ)− 1

]⟩and therefore

〈m (m− 1)〉 = η2⟨M (t, τ)

[M (t, τ)− 1

]⟩(7.6)

We can calculate now the variance photocount:

〈∆m〉2 = η2⟨

∆M (t, τ)2⟩

+ η (1− η)⟨M (t, τ)

⟩(7.7)

The first term of the variance photocount expresses the variance of the integratedphoton number of the light beam. The second term expresses the partition noise,which is based on a random selection of a fraction η of the incident photons by theimperfect photo-detector.Note: with direct detection we are only able to retrieve properties pertaining tothe incident intensity. No phase information can be retrieved.

72


7.2 Homodyne Detection

Homodyne detection is a method for measuring the electric field operator (actu-ally its quadratic operator) and therefore allows phase measurement (particularlyimportant for squeezed light because its nonclassical properties are encoded in thephase).

Figure 7.2: Sketch of a balanced homodyne detector

We have a 50% BS, i.e. |R| = |T | = 1√2and φR − φT = π

2. The signal that we

want to measure is in arm 1 of the BS. If we place photodetectors in both outputarms, then we speak of balanced homodyne detection.We measure the difference of the two output signals, during the integration time.This corresponds to:

M− (t, τ) =

t+τ∫t

dt′[a†

3 (t′) a3 (t′)− a†4 (t′) a4 (t′)

](7.8)

in terms of the output signal this means that(a4

a3

)=

1√2

(1 −i−i 1

)(a

aL

)(7.9)

where:

•

a†3 (t′) a3 (t′) =

1

2

[ia† (t′) + a†

L (t′)]

[−ia (t′) + aL (t′)]

=1

2

a† (t′) a (t′) + ia† (t′) aL (t′)− ia†L (t′) a (t′) +

a†L (t′) aL (t′)

73


•

a†4 (t′) a4 (t′) =

1

2

[a† (t′) + ia†

L (t′)]

[a (t′)− iaL (t′)]

=1

2

a† (t′) a (t′)− ia† (t′) aL (t′) + ia†L (t′) a (t′) +

a†L (t′) aL (t′)

Therefore:

M− (t, τ) = i

t+τ∫t

dt′[a† (t′) aL (t′)− a†

L (t′) a (t′)]

(7.10)

Although the detector measures photocount, the presence of the local oscillatormakes it possible to actually measure the field!Note: If we use an inefficient detector:

MH (t, τ) =

t+τ∫t

dt′[d†

3 (t′) d3 (t′)− d†4 (t′) d4 (t′)

]. (7.11)

Therefore the mean homodyne difference photocount is given as

〈m−〉 =⟨MH (t, τ)

⟩= iη

t+τ∫t

dt′⟨a† (t′) aL (t′)− a†

L (t′) a (t′)⟩

(7.12)

and the variance is given as

〈∆m−〉2 = η2∆⟨MH (t, τ)2

⟩+ η (1− η)

t+τ∫t

dt′⟨a†L (t′) aL (t′) + a† (t′) a (t′)

⟩.

(7.13)Let us take a specific example:Local oscillator: Single-mode coherent state:

αL (t) =√FLe

−iωLt+iϑL (7.14)

where ωL is chosen in such a way, that ωSignal = ωL (homodyne). Since a (t) |α〉 =

α (t) |α〉 we obtain:

〈m−〉 = iη√FL

t+τ∫t

dt′⟨a† (t) e−iωLt+iϑL − a (t) eiωLt−iϑL

⟩(7.15)

Let us define:

EH (χ, t, τ) =1

2√τ

t+τ∫t

dt′[a† (t) e−iωLt+iϑL − a (t) eiωLt−iϑL

](7.16)

74


the homodyne electric field operator, where χ = ϑL + π2, so that:

〈m−〉 = 2η√FLT

⟨EH (χ, t, τ)

⟩(7.17a)

〈∆m−〉2 = ηFLT

4η⟨

∆EH (χ, t, τ)2⟩

+ 1− η

(7.17b)

And since substantially E ∼ x1 we can conclude that with the homodyne detectionwe are able to measure the phase space of the field and therefore its phase properties.

Example 1

Input: single mode coherent state α (t) =√Fe−iωLt+iϑL and in this case:

S ≡⟨EH (χ, t, τ)

⟩=√FT cos (χ− ϑ) (7.18)

andN ≡

⟨∆EH (χ, t, τ)2

⟩=

1

4(7.19)

With this we obtain the signal to noise ratio:

SNR =S2

N=

⟨E2H

⟩⟨

∆E2H

⟩ = 4FT cos2 (χ− ϑ) (7.20)

For an inefficient detector: SNRH = ηSNR

Example 2

Input: Squeezed stateWith EH ∼ x1 follows that

⟨EH

⟩= 0 for a squeezed state and therefore 〈m− = 0〉.

For continuous squeezed state we have that:

⟨a† (ω) a (ω′)

⟩= sinh2 [s (ω)] δ (ω − ω′) (7.21)

with the photon flux1

2π

∫dω sinh2 [s (ω)] = F (7.22)

and〈a (ω) a (ω′)〉 = −1

2eiϑ(ω) sinh [2s (ω)] δ (ω + ω′ − ωp) (7.23)

Note: Here we used the following representation of a continuous coherent state:

ϕ (ω) = s (ω) eiϑ(ω)

therefore:

75

Degree of Coherence and Quantum Optics

• we need to Fourier transform the operators because the properties of squeezedlight are best caught in the frequency domain

• Assumption: T long enough that s (ω) and ϑ (ω) vary insignificantly over 1T

After some calculations follows⟨∆EH (χ, t, τ)2

⟩=

1

4

e2s(ωL) sin2

[χ− 1

2ϑ (ωL)

]+ e−2s(ωL) cos2

[χ− 1

2ϑ (ωL)

](7.24)

and we can show that0 ≤

⟨∆EH (χ, t, τ)2

⟩≤ 1

4(7.25)

and therefore0 ≤ (∆m−)2 < ηF2T, (7.26)

where F2 F . This means that this noise derives only by the beating of the strongcoherent field (of the local oscillator) with the noise in the signal.Note: Shot noise of the local oscillator automatically cancels out if we use balancedhomodyne detection.

8 Degree of Coherence and Quantum Optics

8.1 First Order Degree of Coherence

Figure 8.1: Representation of a Mach-Zehnder interferometer (taken from [1])

• tk = t− zkcalready contains eventual delay coming from different path lengths

• BS → (r, t)

Intensity on output E4 (t) averaged over an oscillation cycle:

I4 =1

2ε0c |r|2 |t|2

|E1 (t1)|2 + |E2 (t2)|2 + 2< [E∗1 (t1)E2 (t2)]

(8.1)

76


where E1,2 are the fields entering the second BS (E1 → path 1, E2 → path 2). Wemoreover need to average the result over the detector integration time T (normallyassumed to be much larger than the source’s coherence time). DEF:

〈E (t)〉 =1

T

T∫0

dtE (t) (8.2)

The intensity now becomes:

〈I4 (t)〉 =1

2ε0c |r|2 |t|2

⟨|E1 (t1)|2

⟩+⟨|E2 (t2)|2

⟩+ 2< [〈E∗1 (t1)E2 (t2)〉]

(8.3)

the third term is responsible for interference and it is related to the first- orderdegree of coherence.First-Order Degree Of Coherence

g(1) (τ) =〈E∗ (t)E (t+ τ)〉〈E∗ (t)E (t)〉

(8.4)

with g(1) (−τ) = g(1) (τ) and τ = z1−z2c

. It is basically related to the fringe visibility.By collecting 〈I (t)〉 = 〈E∗ (t)E (t)〉 we have:

〈I4 (t)〉 =1

2ε0c |rt|2 〈I (t)〉

1 + <

[g(1) (τ)

](8.5)

Example: Collision Broadened Source

We consider a source that emits radiation characterized by random phase jumps dueto the collisions between the atoms of the source (see fig. 8.2).

Figure 8.2: Phase jumps caused by collisions (taken from [1])

In this case even if each atom emits only the same frequency, the phase of theemitted field is different (assumed to be random). The total emitted electric field is

77


given byE (t) = E0e

−iω0t∑k

e−iϕk(t) (8.6)

Then if we calculate g(1) (τ) with this definition by recalling that the correlationfunction that we gat for the phase

⟨ei[ϕk(t+τ)−ϕk(t)]

⟩is proportional to the probability

that the atom has a period of free flight longer than τ given by:

p (τ) =1

τ0exp

[τ

τ0

](8.7)

with τ0 ≡ τc being the source’s coherent time. We get at the end

g(1)CB (τ) = exp

[−iω0τ −

|τ |τc

](8.8)

Figure 8.3: Degrees of first order coherence. Lorentzian shape: collision broadenedsource. Gaussian shape: Doppler broadened source

Example: Doppler Broadened Source

In this case all atoms emit coherently but the emission frequency is slightly differentdue to Doppler broadening, therefore

E (t) = E0

∑k

exp [−iωkt+ iϕk (t)] (8.9)

78


noting, that ϕk for k 6= j averages to zero (randomly distributed), after some calcu-lations we arrive at the final result:

g(1)DB (τ) = exp

[−iω0τ −

π

2

(τ

τc

)2]

(8.10)

Message:

• g(1) (0) = 1

•∣∣g(1) (τ)

∣∣ =

1 fully coherent

0 fully incoherent

∈ ]0, 1[ partially coherent

8.2 Quantum Theory of g(1) (τ)

Note: We consider single mode fieldsThe field operator is given by

E (z, t) = E+ (z, t) + E− (z, t) =i

2

[ae−i(ωt−kz) − a†ei(ωt−kz)

](8.11)

we can generalize g(1) (τ) for two fields at different space-time points (z1, t1) and(z2, t2) write it as:

g(1) (z1, t1; z2, t2) =〈E∗ (z1, t1)E (z2, t2)〉√⟨|E (z1, t1)|2

⟩ ⟨|E (z2, t2)|2

⟩ . (8.12)

If we want to write its quantum version, we just need to replace the fields with theiroperator versions.

Intensity operator:I ∼ a†a → for the electric field this means a† ∼ E−,a ∼ E+ and thereforeI ∼ E−E+

Operator Orderingwe always use normal ordering (a† on the left and a on the right). The reasonwe do this is because with this the vacuum energy becomes zero

we consider light beams that are stationary (fluctuations of the field do net varyin time and so g(1) (τ) does noet depend on τ)

Therefore:

g(1) (τ) =

⟨E−T (t) E+

T (t+ τ)⟩

⟨E−T (t) E+

T (t)⟩ (8.13)

and we can calculate the intensity as for the MZ case.

79


Single Mode Field

Substitute E± and calculate

g(1) (τ) = ei(χ1−χ2) (8.14)

with χ = ωt− kzMessage: We are not really able to extract new information about the nature ofa quantum state by simply looking at its first order degree of coherence. Thereforeany single mode field is first order coherent for all pairs of space-time points.

8.3 Second Order Degree of Coherence

Instead of measuring correlations in the field we now measure intensity correlations.For simplicity we assume simple polarization and that the intensity readings aretaken at a fixed pint in space and at a fixed pint in space and at a fixed delay τ .

g(2) (τ) =

⟨I (t) I (t+ τ)

⟩I2

=〈E∗ (t)E∗ (t+ τ)E (t)E (t+ τ)〉

〈E∗ (t)E (t)〉2(8.15)

Properties:

• g(2) (−τ) = g(2) (τ)

• g(2) (0) ≥ 1

Proof: Let us consider two intensity measurements one at t1 and the other one att2, so that the Cauchy-Schwarz inequality is fulfilled by

2I (t1) I (t2) ≤ I (t1)2 + I (t2)2 . (8.16)

Generalization: 1

N

∑k

I (tk)

2

≤ 1

N

∑k

I (tk)2 (8.17)

this means that for the statistical average we obtain:

I2 =⟨I (t)

⟩2 ≤⟨I (t)2⟩ (8.18)

and therefore with

g(2) (0) =

⟨I (t) I (t)

⟩I2

≥ 〈I2〉I2

= 1 (8.19)

we obtain g(2) (0) ≥ 1. q.e.d.In particular it can be demonstrated that there is no upper band

1 ≤ g(2) (0) ≤ ∞ (8.20)

80


with similar arrangements we can also prove that

0 ≤ g(2) (τ) ≤ ∞, τ 6= 0 (8.21a)

andg(2) (τ) ≤ g(2) (0) (8.21b)

Figure 8.4: Degree of second order coherence for various light sources. Dashed line:coherent source. Lorentzian shape: collision broadened. Gaussian shape: dopplerbroadened (taken from [1])

How to measure g(2) (τ)? → Hanbury-Brown-Twiss Interferometer

Figure 8.5: Schematic illustration of the Hanbury-Brown-Twiss experimental setting(taken from [2])

81


8.4 Quantum Theory of g(2) (τ)

Analogously to the case of g(1) (τ) we can easily quantize the expression of g(2) (τ)

given above:

g(2) (τ) =

⟨E−T (t) E−T (t+ τ) E+

T (t+ τ) E+T (t)

⟩⟨E−T (t) E+

T (t)⟩2 . (8.22)

For a single mode field, and for the case of zero delay, i.e., τ = 0, this reduces to thefollowing expression

g(2)SM (0) =

⟨a†a†aa

⟩〈a†a〉2

, (8.23)

which we call single-mode second order coherence function. It is now clear thatg(2) (τ) will have different outcomes depending on the quantum state consideredbecause

⟨a†a†aa

⟩gives different results for different quantum states. Let us rewrite

g(2) (0) in a more easy way as a function of 〈n〉 and (∆n)2 as follows:

•⟨a†a⟩

= 〈n〉 ≡ 〈n〉

•⟨a†a†aa

⟩=⟨a†(aa† − 1

)a⟩

=⟨a†aa†a− a†a

⟩= 〈n (n+ 1)〉

Therefore:

g(2) (0) =〈n2〉 − 〈n〉〈n〉2

=〈n2〉 − 〈n〉2 + 〈n〉2 − 〈n〉

〈n〉2= 1 +

(∆n)2 − 〈n〉〈n〉2

and so we obtain

g(2) (0) = 1 +(∆n)2 − 〈n〉〈n〉2

. (8.24)

g(2) (0) is therefore able to distinguish different quantum states!

8.4.1 Number States

• 〈n〉 = n

• (∆n)2 = 0

it followsg

(2)N (τ) = 1− 1

n, (8.25)

where n ≥ 1. Note:

• Here g(2)N (0) = 1− 1

n, which is in contrast to the classical case, where g(2) (0) ≥

1!

• if |n〉 ≡ |0〉 then g(2) (0) is undefined.

82


8.4.2 Coherent States

• 〈n〉 = |α|2

• (∆n)2 = |α|2

and so:g

(2)C (0) = 1 (8.26)

this gives a classical result (coherent states are the most classical states among thequantum states...).

8.4.3 Chaotic Light

We did not treat chaotic light during this course. Anyway it can be characterized bya photon mean number 〈n〉 and a variance (∆n)2 = 〈n〉2 + 〈n2〉 and a distributionproperty hat is essentially the Bose-Einstein statistics

pn =〈n〉n

(1 + 〈n〉)1+n (8.27)

So we obtain|ψ〉 =

∑pn |n〉〈n| . (8.28)

With this:g(2) (0) = 2 (8.29)

it is also in accordance with the classical result (cfr. both collision/Doppler broad-ened).

8.4.4 Squeezed Vacuum

• 〈n〉 = sinh2 (s)

• (∆n)2 = 2 〈n〉 (〈n〉+ 1)

so thatg(2) (0) = 3 +

1

〈n〉(8.30)

Note: When 〈n〉 = 0 then g(2) (0) 6= 1, so it does not give the coherent state limit.Care must be taken in how the vacuum limit is approached.

8.5 Comparison and Nonclassical Light

g(2) (τ) Classical: • 1 ≤ g(2) (0) ≤ ∞, τ = 0

• 0 ≤ g(2) (τ) ≤ ∞, τ 6= 0

83


g(2) (τ) Single Mode:

1− 1

〈n〉≤ g(2) (τ) ≤ ∞

for 〈n〉 ≥ 1 and ∀τ ∈ R

If we compare these ranges, we can see that for the case of quantum (single mode)light we can access also the ... below ..., which is not possible in the classical case,where g(2) (0) ≥ 1, i.e.

1− 1

〈n〉≤ g(2) (0) ≤ 1 (8.31)

for 〈n〉 ≥ 1. Note: If light lies in this range, then it is called nonlcassical light.

Figure 8.6: Comparison second order coherence of different light forms (taken from[1])

Why are the classical and quantum g(2) (τ) different? Consider measuringintensity/photon number twice in succession:

⟨I2⟩

this leaves the value of I unchanged forthe second measurement and thereforeI2 is to be averaged

〈n (n− 1)〉

It appears explicit the fact that therehas been a first measurement followedby a second one, i.e. the measurementinterferes with the measured system.While the first measurement counts nphotons, the second one (affected bythe first one) counts only (n− 1) pho-tons.

84

Elements of Quantum Nonlinear Optics

9 Elements of Quantum Nonlinear Optics

In the first part of this chapter we will introduce the nonlinear susceptibility, whichis responsible for nonlinear processes. Furthermore the quantum optical expressionsfor the electric and magnetic field operators in the presence of a medium will beintroduced. We will start with the classic formulation of the Maxwell equationsin presence of a medium and derive the wave equation. At this point we will in-troduce the nonlinear polarization of the medium which is caused by the nonlinearsusceptibility. In the second part we will treat nonlinear effects like second harmonicgeneration (SHG) and parametric up/down conversion.

9.1 Nonlinear Wave Equation

In order to derive the wave equation w need to start with the Maxwell equations ina medium:

∇× E (r, t) = −∂B (r, t)

∂t, (9.1a)

∇ ·D (r, t) = ρfree, (9.1b)

∇ ·B (r, t) = 0, (9.1c)

∇×H (r, t) =1

c2

∂D (r, t)

∂t+ j. (9.1d)

We assume to consider a material for which the free charges ρfree = 0 and thecurrent j = 0. Moreover, let us assume to consider a non-magnetic material, i.e.B = µ0H. In this case we also have that (see e.g. [3])

D (r, t) = ε0E (r, t) + P (r, t) , (9.2)

where P (r, t) is the polarization vector, which accounts for the effect of th materialon the propagation of the electromagnetic field inside the medium.In the following we will drop the dependence of the fields, but we will keep in mindthat all fields depend on r and t if not mentioned explicitly. We obtain the waveequation by applying the curl operator two times resulting in:

∇×∇× E = ∇ (∇ · E)−∇2E = −µ0∂2D

∂t2. (9.3)

In general the first term on the r.h.s. of this equation, i.e. ∇ (∇ · E), is nonzero,because in a material ∇ · D = 0 but ∇ · E 6= 0. However, for practical casesin nonlinear optics we can treat this term as being very small, i.e. ∇ · E ≈ 0.Furthermore with Eq. (9.2) we obtain:

µ0∂2D

∂t2=

1

c2

∂2E

∂t2+ µ0

∂2P

∂t2. (9.4)

85


Substituting this result into Eq. (9.3) and keeping in mind that ∇ · E ≈ 0 leads to(∇2 − 1

c2

∂2

∂t2

)E = µ0

∂2P

∂t2. (9.5)

For the case of a nonlinear medium the electric polarization P (z, t) consists of twocontributions

P = PL + PNL, (9.6)

where PL = ε0χ (ω)E is the linear part of the polarization, which is responsible forthe refractive index. PNL =

∑k χ

(k) (ω)Ek is the nonlinear part of the polarization,which is related to anharmonic terms in the Lorentz model (see [11]), e.g.

χ(2) (ω)→ PNL = χ(2)E · E.

As it can be seen, the second order nonlinear polarization is proportional to theproduct of the square of the field, with the proportionality constant being the secondorder nonlinear susceptibility χ(2)(ω).For third order processes, the nonlinear polarization can be analogously written asfollows

χ(3) (ω)→ PNL = χ(3)E · E · E,

where χ(3)(ω) is the third order nolinear susceptibility.Substituting these expressions in Eq. (9.5) leads to(

∇2 − n (ω)2

c2

∂2

∂t2

)E = µ0

∂2PNL

∂t2, (9.7)

where we have defined the refractive index

n (ω)2 = 1 + χ (ω) . (9.8)

Normally the electric susceptibility χ (ω) ∈ C, and therefore

n (ω) = n1 (ω) + in2 (ω) , (9.9)

where n1 (ω) describes dispersion and n2 (ω) describes absorption. For simplicitywe assume n2 (ω) = 0. This is a good approximation as long as we are consider-ing a frequency range where the considered material appear transparten or quasi-transparent.To consider a simple enough system, let us assume an electromagnetic field whichpropagates along the z-axis, so that ∇2 → ∂2

∂z2. Moreover, let us take the positive

86


frequency part of the Fourier expansion of the field E (z, t) in plane waves, i.e.

E (z, t) =

∞∫0

dω E (z, ω) e−i[ωt−k(ω)z], (9.10)

with k (ω) = k0n (ω). Analogously for the nonlinear polarization we obtain

PNL (z, t) =

∞∫0

dω PNL (z, ω) e−iωt. (9.11)

Here we took the positive frequency part instead of the full Fourier transform becausethis would provide an easier way to write the quantum counterparts of these fields,looking back at the electric field operator, which only contains the positive frequencypart, i.e. E+ (z, t).Applying ∂2

∂z2on E (z, t) gives

∂2E

∂z2=

∂2E (z, ω)

∂z2+ 2ik (ω)

∂E (z, ω)

∂z− k2 (ω) E (z, ω) . (9.12)

where we used the slow varying envelope approximation (SVEA), namely we as-sumed

∂2E

∂z2 k

∂E

∂z. (9.13)

Substituting (9.11) and (9.12) in (9.5) leads to the expression

2ik (ω)∂E

∂z− k2 (ω) E − n (ω)2 ω2

c2E = −µ0ω

2PNLe−ik(ω)z (9.14)

where

k (ω)2 = k2n (ω)2 =ω2n (ω)2

c2(9.15)

was used. This can be written in the form

∂E

∂z=

iω

2ε0n (ω)PNLe

−ik(ω)z, (9.16)

where we used that

− µ0ω2

2ik (ω)= i

µ0ω2

2ωn· c = i

µ0ω

2µ0ε0n= i

ω

2ε0n.

9.2 Quantization of the Electromagnetic Field in Dielectric

Media

Since Eq. (9.16) is valid in a nonlinear medium, we need to find the correct quan-tization rule for the electromagnetic field in a dielectric medium in order to solve

87


it. In the beginning of this course we have already treated the case of quantizingthe e.m. field in free space. Furthermore, we already found an expression for thequantized e.m. field for the case of the presence of an atom. Here we have assumedthat the atomic medium is sufficiently diluted so that the quantized field can retainits free-space form.For the case of the propagation of the electromagnetic field in a dielectric medium,it is not possible anymore to make the approximation of dilute gas (as in the caseof a single atom interacting with the field), as the atoms in a dielectric medium areorganized in a compacted crystalline structure.To find an easy way to approach the problem of field quantization in a dielectric,we therefore need to make some assumptions, in order to be able to use the samequantization method we used for the case of free space. First of all, assume that theoptical nonlinearities of the medium do not significantly affect the quantization pro-cedure, and thus only the linear effects are included in the determination of the fieldoperators. Furthermore we assume no absorption, i.e. n2 (ω) = 0. This restricts theanalysis to the range of frequencies of which the medium appears transparent. Thedefinitions of the field operator are then only valid in this frequency interval. Thelast assumption that we make, is that the nonlinear effects are not significant ef-fected by the beam configurations. This means we can employ 1D-continuous modefield operators. Therefore we will consider propagation in z-direction and the polar-ization of the e.m. field is such that E ‖ x and B ‖ y. The quantization procedureis essentially similar to the free-space quantization, i.e. first we calculate the fieldHamiltonian H and express it as a sum of harmonic oscillators. We obtain from thiscalculations the following electric field operator:

E+ (z, t) =1√2π

∞∫0

dω E+ (ω) exp −i [ωt− k (ω) z] , (9.17)

where E+ (ω) is the Fourier transformed electric field operator and k (ω) is the k-vector in the medium, which accounts for medium dispersion through n (ω). Thisresult can be written in terms of a (ω) as follows:

E+ (z, t) = i

∞∫0

dω f (ω) a (ω) exp −i [ωt− k (ω) z] , (9.18)

where the normalization factor f (ω) is to be determined.The displacement operator D+ (z, t) in a dispersive material is obtained by meansof its classical definition in a dispersive (linear) medium. The classical displacementoperator is given by

D (z, t) = ε0εRE (z, t) = ε0n2E (z, t) , (9.19)

88


from which we obtain the nonlinear form as follows

D+ (z, t) = iε0

∞∫0

dω f (ω)n (ω)2 a (ω) exp −i [ωt− k (ω) z] . (9.20)

From Maxwell’s equations we obtain the magnetic field operator as follows

B+ (z, t) = i

∞∫0

dω f (ω)n (ω)

ca (ω) exp −i [ωt− k (ω) z] . (9.21)

To determine f (ω) we make use of the quantum counterpart of the continuity equa-tion for the electromagnetic field. In the classical case the energy-continuity equationreads

∂

∂t〈W (z, t)〉 =

∂

∂z〈S (z, t)〉 (9.22)

where 〈W (z, t)〉 is the time averaged energy density and 〈S (z, t)〉 the time aver-aged Poynting vector (see [3]). From which we obtain the similar quantum opticalexpression of the continuity equation given by

∂W (z, t)

∂t= −∂S

∂z, (9.23)

where the operators W and S are assumed to be normal-ordered.

Energy Density in a Medium

The energy density is given by

W (z, t) = E ·D + B ·H = E (z, t)D (z, t) +1

µ0

B (z, t)B (z, t) . (9.24)

Therefore we obtain the time derivative

∂W

∂t= E

∂D

∂t+∂E

∂tD +

1

µ0

B∂B

∂t+

1

µ0

∂B

∂tB. (9.25)

Hence, we can write the time derivative of the energy operator as follows

∂W (z, t)

∂t= E− (z, t)

∂D+ (z, t)

∂t+∂D− (z, t)

∂tE+ (z, t)

+1

µ0

[B− (z, t)

∂B+ (z, t)

∂t+∂B− (z, t)

∂tB+ (z, t)

]. (9.26)

In the general case, it is not possible to retrieve an expression for W (z, t) from(9.26), because the r.h.s. is not a total time derivative and cannot be written likeone. This is due to the presence of the medium. However, if we first insert the

89


expressions of the field operators defined above and then look for an expression forW (z, t), then Eq. (9.26) can be integrated and leads to the following result:

W (z, t) = ε0c

∞∫0

dω

∞∫0

dω′ f (ω) f (ω′)k (ω)− k (ω′)

ω − ω′[n (ω) + n (ω′)]

× a† (ω) a (ω′) ei(ω−ω′)t−[k(ω)−k(ω′)]z, (9.27)

and we assume f (ω) ∈ R. Now we can obtain the explicit expression of the normal-ization factor f (ω) by imposing that the z-integrated energy density can be writtenas a collection of harmonic oscillators, i.e.,

A

∞∫−∞

dz W (z, t) =

∞∫0

dω ~ωa† (ω) a (ω) , (9.28)

where A is the space volume. Therefore we obtain

f (ω) =

√~ω

4πε0cAn (ω). (9.29)

Poynting Vector

Analogously to the classical formulation, we define the Poynting vector operator asfollows

S (z, t) = ε0c2E− (z, t) B+ (z, t) + B− (z, t) E+ (z, t)

. (9.30)

Substituting the expression of the field operators gives

S (z, t) =~

4πA

∞∫0

dω

∞∫0

dω′

√ωω′

n (ω)n (ω′)[n (ω) + n (ω′)]× (9.31)

×a† (ω) a (ω′) ei(ω−ω′)t−[k(ω)−k(ω′)]z. (9.32)

If we calculate the total energy that flows through a plane of constant z we have

A

∞∫−∞

dt S (z, t) =

∫dω ~ωa† (ω) a (ω) , (9.33)

since the continuity equation must be valid. Moreover, the total energy flow overthe entire length of the z-axis at a given time t is given instead by

A

∞∫−∞

dz S (z, t) =

∫dω ~ωva (ω) a† (ω) a (ω) , (9.34)

90


where va (ω) denotes the group velocity of the medium. It weights each frequencycomponent of the energy flow.Note: For a quasi-monochromatic beam, i.e. a beam with narrow spectrum distri-bution centred at ω0, the Poynting vector is given as

S (z, t) = 2ε0cn (ω0) E− (z, t) E+ (z, t) =

~ω0

Aa† (tR) a (tR) , (9.35)

where tR = t− z/va (ω0) is the retarded time and va(ω) is the group velocity of thefield inside the dielectric medium.

9.3 Second Harmonic Generation

In this part of our introduction we want to have a closer look at the second harmonicgeneration in a χ(2) medium. Therefore consider an incident, narrow-band signal atfrequency ωp, as it is shown in Fig. 9.1. Furthermore, SHG means that two photonsat the frequency ωp are simultaneously annihilated in order to create one photon at2ωp.

Figure 9.1: Schematic sketch of the SHG in a χ(2) medium, where ωp is the frequencyof the pump.

In what follows, we employ the so-called “non-depleted pump approximation”,which implies that the input (pump) beam, initially represented by a very brightcoherent state, remains almost unchanged after the interaction with the nonlinearcrystal. This means, for exmaple, that is the pump beam was originally containing109 photons, the SHG interaction maybe converts 50 pump photons into SHG pho-tons, thus leaving 109− 50 ' 109 photons in the pump beam after the process. Thepositive frequency part of the nonlinear polarization is then given by

P(+)NL (z, t) =

ε0

2π

∞∫0

dω′∞∫

0

dω′′ χ(2) (ω′ + ω′′) E(+) (ω′) E(+) (ω′′) (9.36)

× exp −i [(ω′ + ω′′) t− [k (ω′) + k (ω′′)] z] , (9.37)

91


where ω′ and ω′′ are the frequencies of the incident light. This nonlinear polarizationacts as a source for the second-harmonic signal, according to

∂ Ê (z, ω)

∂z=

iω

2ε0n (ω)ˆPNL (z, ω) e−ik(ω)z. (9.38)

Substituting (9.18) in the expression for the nonlinear polarization leads to

P(+)NL (z, t) = − ~

4πcA

∞∫0

dω′∞∫

0

dω′′

√ω′ω′′

n (ω′)n (ω′′)χ(2) (ω′ + ω′′) a (ω′) a (ω′′)

× exp −i [(ω′ + ω′′) t− [k (ω′) + k (ω′′)] z] , (9.39)

where A is the space volume.The component of the nonlinear polarization oscillating at the frequency ω is thenobtained by Fourier transform of the expression above, according to

PNL (z, t) =1√2π

∞∫0

dω PNL (z, ω) e−iωt. (9.40)

Comparing the expressions above leads to

ˆPNL (z, ω) = − ~√2πcA

∞∫0

dω′∞∫

0

dω′′

√ω′ω′′

n (ω′)n (ω′′)χ(2) (ω′ + ω′′) a (ω′) a (ω′′)

× exp i [k (ω′) + k (ω′′)] z δ (ω − ω′ − ω′′) . (9.41)

With this we can solve Eq.(9.16) with respect to Ê(+) (z, ω)

Ê(+) (L, ω) =iω

2ε0n (ω)

L∫0

dz ˆP(+)NL (z, ω) e−ik(ω)z. (9.42)

Using the Fourier relation we can then reconstruct the time-dependent electric fieldoperator, i.e.

E(+) (L, t) =i√

2π2ε0c

∞∫0

dω

L∫0

dzω

n (ω)P

(+)NL (z, ω) e−i[ωt−k(ω)z]. (9.43)

We can now calculate the intensity of the second harmonic signal as follows:⟨I (L, t)

⟩=

1

2ε0cn (ω)

⟨E(−) (L, t) E(+) (L, t)

⟩. (9.44)

92


In order to get the full expression, let us first calculate E(+) (L, t)

E(+) (L, t) =i√

2π2ε0c

∞∫0

dω

L∫0

dzω

n (ω)P

(+)NL (z, ω) e−i[ωt+k(ω)z]

= − i~4πε0c2A

L∫0

dz

∞∫0

dω

∞∫0

dω′∞∫

0

dω′′ω

n (ω)e−i[ωt+k(ω)z]

×

√ω′ω′′

n (ω′)n (ω′′)χ(2) (ω′ + ω′′) ei[k(ω′)+k(ω′′)]za (ω′) a (ω′′)

× δ (ω − ω′ − ω′′)

= − i~4πε0c2A

L∫0

dz

∞∫0

dω′∞∫

0

dω′′ω′ + ω′′

n (ω′ + ω′′)e−i[(ω

′+ω′′)t+k(ω′+ω′′)z]

×

√ω′ω′′

n (ω′)n (ω′′)χ(2) (ω′ + ω′′) ei[k(ω′)+k(ω′′)]za (ω′) a (ω′′)

If we assume that the pump beam can be considered having a narrow-band spectrumcentered around the frequency ωp (quasi-monochromatic approximation), then wecan set ω′ ' ωp and ω′′ ' ωp in the quantities appearing inside the integral above,except for the operators and time-dependent exponential. By doing so we obtainthe following result:

E(+) (L, t) =

(− i~

8πε0c2A

) L∫0

dz2ωp

n (2ωp)e−i[k(2ωp)+2k(ωp)]z

× ωpn (ωp)

χ(2) (2ωp)

∞∫0

dω′a (ω′) e−iω′t

∞∫0

dω′′a (ω′′) e−iω′′t. (9.45)

By noting that the integrals in ω′ and ω′′ are esesntially the Fourier transform of thetime dependent operators a(t), we can then rewrite the previous equation as follows

E(+) (L, t) =

(−

i~ω2p

2πε0c2A

)1

n (2ωp)n (ωp)χ(2) (2ωp) a (t) a (t)

L∫0

dz e−i[k(2ωp)+2k(ωp)]z.

(9.46)We obtain a equivalent result for E(+) (L, t), which substituted in the equation for

93


the intensity of the second harmonic reads as follows:⟨ISHG (L, t)

⟩= 2ε0cn (2ωp)

⟨E− (L, t) E+ (L, t)

⟩= 2ε0cn (2ωp)

(−

i~ω2p

4πε0c2A

)(i~ω2

p

4πε0c2A

)×

∣∣χ(2) (2ωp)∣∣2 [ 1

n (2ωp)n (ωp)

]2

×

∣∣∣∣∣∣L∫

0

dze−i[k(2ωp)+2k(ωp)]z

∣∣∣∣∣∣2 ⟨a† (t) a† (t) a (t) a (t)

⟩. (9.47)

Defining ∆k = k (2ωp)− 2k (ωp) we have∣∣∣∣∣∣L∫

0

dz e−i∆kz

∣∣∣∣∣∣2

=

L∫0

dz ei∆kzL∫

0

dζ e−i∆kζ

=

(i

∆k

)(−i∆k

)[e−i∆kz

]L0

[e−i∆kζ

]L0

=1

∆k2

[e−i∆kL − 1

] [ei∆kL − 1

]=

(1

∆k

)2 1− e−i∆kL − ei∆kL + 1

=

2

∆k2[1− cos (∆kL)]

=2

∆k2

[1− 1 + 2 sin2

(∆kL

2

)]=

sin2 (∆kL/2)

(∆kL/2)2

Summarizing everything leads to the expression for the time averaged second har-monic intensity

⟨ISHG (L, t)

⟩=

( ~2ω4p

8π2ε0c3A2

)[|χ(2)(2ωp)|2

n (2ωp)n (ωp)2

]sin2 (∆kL/2)

(∆kL/2)2

× χ(2) (2ωp)⟨a† (t) a† (t) a (t) a (t)

⟩. (9.48)

If we consider a continuous wave at the frequency ωp we can write⟨a† (t) a† (t) a (t) a (t)

⟩as a second order correlation function g(2)

p,p (0), i.e.

⟨a† (t) a† (t) a (t) a (t)

⟩= g(2)

p,p (0)⟨a† (t) a (t)

⟩2. (9.49)

The second harmonic signal is proportional to g(2) (0) and to the square of the meanphoton flux of the incident pump beam. This is the case, because, as already men-tioned SHG is the process of simultaneous two photon absorption at the frequency

94


ωp. Furthermore, g(2) (τ) is a measure for two photon absorption. Therefore, ifg(2) (τ) of the pump is zero, we will not obtain second harmonic generation, e.g.consider a single photon state |1〉 as pump. There is no second photon to absorband therefore ISHG = 0 and also g(2) (τ) = 0 for |1〉, since g(2) (0) = 1−1/n for Fockstates.Moreover, for a thermal source we obtain

g(2)THERMAL (0) = 2 6= g

(2)COH (0) ,

since the availability of photon pairs in thermal light is greater than in coherentlight. This leads to the fact, that I(thermal)

SHG /I(coh)SHG = 2.

In anti-bunched light (e.g. Fock states), we have that:

g(2)A.B. (0) < g

(2)COH (0) ,

therefore, anti-bunched light will generate less SHG than coherent light.

Phase Matching

For phase matching we need to consider the L-dependent term of the intensity, i.e.

sin2 (∆kL/2)

(∆kL/2)2 .

If ∆k = 0 this term becomes one, and the second harmonic intensity ISHG reachesits maximum. This means that for ∆k = 0 we obtain

k (2ωp) = 2k (ωp) , (9.50)

which is the phase matching condition. This corresponds to the conservation ofmomentum in passing from ωp to 2ωp.

95


9.4 Parametric Down Conversion

Figure 9.2: Schematic sketch of the quantum optical parametric down conversion ina χ(2) medium, where ωp is the frequency of the pump, ωi is the idler frequency andωs the signal frequency.

The nonlinear polarization is given by

P(+)NL (z, t) =

ε0

2π

∞∫0

dω′∞∫

0

dω′′ χ(2) (ω′′ − ω′) E(−) (z, ω′) E(+) (ω′′)

× exp i [(ω′ − ω′′) t− [k (ω′)− k (ω′′)] z] , (9.51)

where ω′ and ω′′ are the frequencies of the signal and the pump, respectively. If wetake the Fourier transform of the nonlinear polarization and evaluate it at the idlerfrequency ωi = ωp − ω we have:

ˆP(+)NL (z, ωp − ω) =

ε0√2π

∞∫0

dω′∞∫

0

dω′′ χ(2) (ω′′ − ω′) E(−) (z, ω′) E(+) (ω′′)

× e−i[k(ω′)−k(ω′′)]zδ (ωp − ω + ω′ − ω′′) . (9.52)

For our calculations we will assume that the pump is a bright coherent state, namely

a (ω′′) |αp〉 = αp (ω′′) |αp〉 . (9.53)

Furthermore, we assume that we have quasi monochromatic light, i.e., we have anarrow band around ωp which we express as

αp (ω′′) =√

2πFpeiϑpδ (ω′′ − ωp) , (9.54)

where Fp is the mean photon flux and ϑp is the pump phase. The phase matchingcondition is given by

∆k = k (ωp)− k (ω)− k (ωp − ω) , (9.55)

96


where we have for ∆k 6= 0 the non-degenerate and for ∆k = 0 the degenerate para-metric down conversion, respectively.Since signal and idler have different frequencies, we define two different field opera-tors (they are, in fact, two different field modes):

E(+) (z, ω) = i

√~ω

2ε0cAn (ω)az (ω) , (9.56a)

as electric field operator of the signal and

E(+) (z, ωp − ω) = i

√~ (ωp − ω)

2ε0cAn (ωp − ω)bz (ω) , (9.56b)

is the electric field operator of the idler. They both obey (9.16), i.e.

∂E(+) (z, ωp − ω)

∂z=

ωp − ωn (ωp − ω)

√~ωpFp

8ε0c3An (ωp)χ(2) (ωp − ω) E(−) (z, ω)

× exp i [ϑp + (k (ωp)− k (ω)− k (ωp − ω)) z] , (9.57a)

and

∂E(−) (z, ωp)

∂z= − ω

n (ω)

√~ωpFp

8ε0c3An (ωp)χ(2)∗ (ω) E(+) (z, ωp − ω)

× exp −i [(k (ωp)− k (ω)− k (ωp − ω)) z] . (9.57b)

This set of coupled wave equations can be written with respect to a and b. Afterintroducing

s (ω) eiϑ(ω) = −

√~ωpω (ωp − ω)Fp

8ε0c3An (ωp)n (ω)n (ωp − ω)χ(2) (ωp − ω) eiϑpL (9.58)

we obtain∂bz (ωp − ω)

∂z= −s (ω)

Leiϑ(ω)a†

z (ω) , (9.59a)

and∂az (ω)

∂z= −s (ω)

Le−iϑ(ω)bz (ωp − ω) . (9.59b)

whose solution is given by

a†L (ω) = a†

0 (ω) cosh [s (ω)]− b0 (ωp − ω) e−iϑ(ω) sinh [s (ω)] , (9.60a)

and

bL (ωp − ω) = b0 (ωp − ω) cosh [s (ω)]− a†0 (ω) eiϑ(ω) sinh [s (ω)] , (9.60b)

97

Appendix A: The Harmonic Oscillator

for the case of non-degenerate parametric down conversion.For the degenerate case we have the same theory as before, just with ˆ

a ≡ b. There-fore, the solution of the coupled equations becomes

a†L (ω) = a†

0 (ω) cosh [s (ω)]− a0 (ωp − ω) eiϑ(ω) sinh [s (ω)] , (9.61)

which corresponds to a single mode squeezed state.This means, that PDC and degenerate PDC are natural sources of squeezed light(and they possess all the non-classical properties of squeezed light). Therefore wecan write

|ψ〉 PDC−−−→ |ξ (ω)〉 , (9.62)

withξ (ω) = s (ω) eiϑ(ω). (9.63)

10 Appendix A: The Harmonic Oscillator

The Hamiltonian of a harmonic oscillator is given by

H =p2

2m+

mω2

2x2. (10.1)

To solve the problem, we introduce the so called creation (a†) and annihilation (a)operators, defined as follows:

x =

√~

2mω

(a+ a†) , (10.2a)

p = i

√mω~

2

(a† − a

). (10.2b)

Substituting this into Eq. (10.1), we obtain the so-called second quantized form ofthe harmonic oscillator Hamiltonian, i.e.,

H =~ω2

(a†a+ aa†) (10.3)

with the commutation rules

[a, a†] = aa† − a†a = 1 (10.4a)

[a, a] = 0 =[a†, a†] . (10.4b)

Using the commutation rule (10.4), the Hamiltonian of the harmonic oscillator canbe rewritten as follows:

H = ~ω(a†a+

1

2

). (10.5)

98

Appendix B: The Quantization of a Paraxial Electromagnetic Field

If |n〉 is the state of the harmonic oscillator with n quanta of energy on it, then theHamiltonian operator written above acts on such a state in the following way:

H |n〉 = n~ω |n〉 . (10.6)

11 Appendix B: The Quantization of a Paraxial Elec-

tromagnetic Field

The quantization procedure described in the first chapters of these notes has beencarried for the electromagnetic field in free space, with the aid of a fictitious cavity,whose main aim was to introduce a suitable (and regular) set of modes upon whichthe general electromagnetic field could be expanded. In carrying out that quanti-zation procedure, however, no attention at all was put on the spatial properties ofthe field itself, as the main result of the quantization of the electromagnetic fieldwas the introduction of Fock states, i.e. energy eigenstates associated to the field.The analysis carried out throughout all these notes, therefore, only focused on theenergy-eigenstates description of the quantized field, and the spatial structure of thefield itself has not been taken into account, assuming that the field could always berepresented by a plane wave in space.

In this last chapter, however, we intend to overcome this limitation, by discussinghow can we quantize a paraxial electromagnetic field, and how can we introduce fieldoperators, which are able to create, for example, a single photon in a given spatialmode (for example, in a Hermite-Gauss or Laguerre-Gauss mode). To do that,however, we need to reconsider the problem of field quantization, putting particularattention to the fact that the quantized field must be a solution of the paraxialequation, and therefore the corespondent field state must be defined in a suitablychosen Hilbert space, which guarantees the feasibility of the quantization procedureand at the same time ensures the field to be a solution of the paraxial equations.

The problem of paraxial quantization was first solved by Deutsch and Garrisonin 1991 [12] and can be found (in a slightly more didactical version) in Chapter 6of Ref. [13]. However, the approach presented in Ref. [14] is much more clear andeasier to adapt to the scope of these notes. Therefore, the procedure of quantizationof a paraxial light beam will be explained in this chapter taking Ref. [14] as a mainguidance. The results presented here, however, are in complete agreement with thework by Deutsch and Garrison presented in Ref. [12].

11.1 From the Vector Potential to the Paraxial Operator

As a starting point for our analysis, let us consider the expression of the positive fre-quency part of the quantized vector potential in the Coulomb gauge, for a continuous

99


mode field, i.e.,

A(+)(r, t) =2∑

λ=1

∫d3k

√~

16π3cε0|k|e(λ)(k)aλ(k)ei(k·r−c|k|t). (11.1)

To correctly carry out the quantization procedure for a paraxial beam, we use a tricksimilar to the one used for the general case in chapter 1, i.e., we introduce a longi-tudinal quantization length L, so that the longitudinal component kz of the field’swave vector can be discretized as follows: ζn = (2π/L)n, where n ≥ 0 selects onlyforward propagating waves. By doing this, the kz-integration in the above equationreduces to a summation over the allowed values of n, and only the integration overthe transverse momentum kT = kxx + kyy remains. Therefore∫

d3k → 2π

L

∑n

∫d2kT ,

where the prefecture 2π/L has been inserted to correctly account for normalization.Together with this substitution, we must also redefine de annihilation operator ap-pearing in Eq. (11.1) as follows

aλ(k)→√

L

2πaλ(k, n),

in such a way that the bosonic commutation relation

[aλ(k, n), a†

µ(k′,m)]

= δλµδnmδ(k− k′) (11.2)

continues to be valid. After these transformations, Eq. (11.1) can be then rewrittenas follows:

A(+)(r, t) =∑λ,n

∫d2kT

√~

8π2cε0|k|Le(λ)(k)aλ(kT, n)ei[(kt+ζnz)·r−c|k|t]. (11.3)

Our aim is to be able to write the expression of the vector potential operator in sucha way that it could be interpreted as a solution of the paraxial wave equation. To dothat, we shoudlrecall that a paraxial field can be always written as the product ofan envelope function ψ(x, y, z), which varies slowly with z0 and a plane wave termexp [i(k0z − ω0t)], which accounts for the propagation along the z direction and theharmonic time dependence of the field itself [13]. A closer inspection to Eq. (11.3),however, reveals that we do not have such exponential factor.

However, if we introduce the following, trivial identity

1 =L2π

∫ L/2π0

dk0ei(k0z−ω0t)

ei(k0z−ω0t),

100


into Eq. (11.3), we can rewrite it in the following form:

A(+)(r, t) =L2π

∫ L/2π0

dk0e−i(k0z−ω0t) ψ(r, t), (11.4)

where

ψ(r, t) =∑λ,n

∫d2kT

√~

8π2cε0|k|Le(λ)(k)aλ(kT, n)ei[(k−k0z)−(c|k|−ω0)t], (11.5)

and k = kT + ζnz.Written in this form, Eq. (11.4) is an exact solution of the paraxial wave equa-

tion, if the field operator ψ(r, t) is itself an exact solution of the paraxial equation,i.e., (

2ik0∂

∂z+∇2

T

)ψ(r, t) = 0. (11.6)

Substituting Eq. (11.5) into the above equation brings to the following dispersionrelation:

ζnk0

= 1− |kT |2

2k20

, (11.7)

which is the usual paraxial dispersion relation [15]. It is worth noticing that theabove relation constitutes a constraint on the allowed valued of ζn that a paraxialwave can take, as a function of the transverse wave vector kT . This, in our case,constitutes a constraint that defines a two dimensional sub-space in the k-space,where Eq.(11.4) is at the same time an exact solution of the paraxial equation(because of the way we defined ψ(r, t)) and also an exact solution of the waveequation, a condition which is necessary for being able to write the quantized form ofthe vector potential as given by Eq. (11.1). This 2D sub-space is the paraxial Hilbertspace introduced in Ref. [12]. In what follows, we will try to rewrite Eq. (11.4) ina form which automatically accounts for this constraint, thus making not necessaryto introduce the paraxial Hilbert space explicitly, thus making the quantizationprocedure easier to understand.

11.2 The Paraxial Dispersion Relation

To implement a form of the paraxial operator ψ(r, t) which automatically accountsfor the paraxial constraint, we first need to elaborate a bit more eq. (11.7). To dothat, let us rename the transverse wave vector as q = kt and let us define q = |kT |.Moreover, let us introduce the following quantity

θ√

2 =q

k0

, (11.8)

101


which is a generalization of the well known divergence angle of a Gaussian beam[16]. By then introducing the angle φ formed by the propagation direction z andthe field k-vector k, it is not difficult to prove that

tanφ =q

k · z=

q

ζn.

Using this result and Eq.(11.8) we can rewrite the dispersion relation (11.7) in thefollowing, dimensional form:

θ√

2 = − cotφ+√

2 + cot2 φ. (11.9)

The above equation is exact and links the beam divergence θ√

2 with the angleformed by the beam with the propagation axis. For the paraxial case, φ 1, anda Taylor expansion of the above relation brings to the following result

θ√

2 ' φ− φ3

6+O(φ5). (11.10)

With this approximation, the dispersion relation (11.7) becomes

ζn ' k0(1− θ2). (11.11)

We can use this result to write the k-vector and the polarization vectors appear-ing in Eq. (11.5) as θ-dependent quantities, in such a way that it becomes veryeasy to isolate their paraxial contribution. In fact, in analogy with paraxial gaus-sian beams [17], the purely paraxial terms are the one independent on θ, and theterms which depend from θ2 represent the lowest order corrections to the paraxialapproximation.

Using Eq. (11.10) we can then write the following, approximated, quantities:

k→ k ' qq + k0(1− θ2)z, (11.12a)

e(2)(k)→ e(2)(q, θ) ' z× q, (11.12b)

e(1)(k)→ e(1)(q, θ) ' e(1)(q, θ)× k. (11.12c)

Moreover, Eq. (11.11) needs to be compared with the longitudinal quantizationcondition ζn = (2π/L)n that we assumed at the beginning of our analysis. Since theallowed longitudinal k-vectors are only the ones which have the form ζn = (2π/L)n,the paraxial dispersion relation given by Eq. (11.11) forces the integer n to acquirea precise value, i.e.

n ≡ n(θ) =

[k0L

2π

(1− θ2

)]I.P

, (11.13)

where the subscript I.P. stands for “Integer Part”.This result is exactly what we were looking for. Among all the possible value

102


of ζn deriving from the introduction of a longitudinal quantization length L, theconstraint imposed by ψ(r, t) being an exact solution of the paraxial equation selectsonly one single value of ζn, namely the one corresponding to n = n(θ). If we nowinsert artificially a condition of the kind δn,n(θ) in the expression of ψ(r, t) givenby Eq. (11.5), we therefore have a field operator which automatically accounts forthe paraxial constraint (11.10). The final expression of the paraxial field operatorψ(r, t) is then given by

ψ(r, t)∑λ,n

∫d2kT

√~

8π2cε0|k|Le(λ)(k)aλ(kT, n)δn,n(θ)e

i[(k−k0z)−(c|k|−ω0)t]. (11.14)

11.3 Back to Continuous Mode Operators

Now that we have obtained our desired result of writing the expression of a fieldoperator, which is automatically defined (through the paraxial constraint δn,n(θ))in a suitable paraxial Hilbert space, we can restore the continuous character of thelongitudinal wave vector kz. We have in fact disctretized the longitudinal componentof the wave vector only for convenience, in order to deal with well defined quantities(i.e., in order to avoid singularities and problems with infinities).

If we then make the following substitutions, where now kz has been replacedwith cω: ∑

n

→ L

2πc

∫dω,

aλ(q, n) →√

2πc

Laλ(q, ω), (11.15)

and we take the continuos limit of δn,n(θ) as

δn,n(θ) →2πc

Lδ(ω − ck0(1− θ2)), (11.16)

we can rewrite the expression of the vector potential operator (11.1) as follows:

A(+)(r, t) =∑λ

∫dω e−iω(t−z/c)

√~

16π3ε0cω

×∫

d2q E (λ)(q, ω, z, t)aλ(q, ω)ei[q·x−q2cz/(2ω)], (11.17)

where x = xx + yy and

E (λ)(q, ω, z, t) = e(λ)(q, θ)

(1 + θ2

√1 + θ4

)1/2

e−iωt(√

1+θ4−1). (11.18)

One can easily prove that Eq. (11.17) represents a bona fide vector potential, whichfor t = 0 is an exact solution of the paraxial equation and for t ≥ 0 is an exact

103


solution of the wave equation. The vector potential operator (11.17) is thereforedefined in the paraxial Hilbert space defined by the paraxial constraint (11.10).

11.4 The Maxwell-Paraxial Modes

Equation (11.17) is already a good form of paraxial vector potential operator. How-ever, the annihilation operator that appears there is written in the frequency space,as it annihilates a photon with a specific transverse momentum q. To be able tomanage the creation and annihilation of photons in spatial modes of the field, how-ever, we will need to rewrite Eq. (11.17) in such a way that it contains operatorsdefined in the representation space spanned by x To do that, let us introduce thefollowing Fourier transform for the operators aλ(q, ω)

aλ(x, ω) =1

2π

∫d2q aλ(q, ω)eiq·x. (11.19)

It is left to the reader to prove that aλ(x, ω) obeys the bosonic commutation rule

[aλ(x, ω), a†

µ(y,Ω)]

= δλµδ(ω − Ω)δ(x− y). (11.20)

Substituting Eq. (11.19) into Eq. (11.17) and calling k0 = ω/c gives the following,final, result:

A(+)(r, t) =∑λ

∫dω

ei−ω(t−z/c)√4πε0cω/~

A(+)(x, z, ω, t), (11.21)

whereA(+)(x, z, ω, t) =

∫d2x′F(λ)(x′, z,x, ω, t)aλ(x

′, ω), (11.22)

with F(λ)(x′, z,x, ω, t) are the so-called Maxwell-paraxial modes, whose explicit ex-pression is given by

F(λ)(x′, z,x, ω, t) =1

(2π)2

∫d2q E (λ)(q, ω, z, t)ei[q·(x−x

′)−q2cz/(2ω)] (11.23)

Let us comment this result. First of all, notice that the Maxwell-paraxial modesF(λ)(x′, z,x, ω, t) represent the field at a plane z at time t generated by a point sourcelocated at x′ in the transverse plane z = 0, oscillating at the paraxial frequency ω.From this perspective, F(λ)(x′, z,x, ω, t) can be interpreted as the quantum coun-terpart of the Huygens-Fresnel diffraction integral. Moreover, if the paraxial ap-proximation ids valid, i.e, if the integration in Eq. (11.23) is limited to the domainCω = q : q ω/c, the Maxwell-paraxial modes reduce to

P(λ)(x′, z,x, ω, t) =

∫Cω

d2q

(2π)2e(λ)(q)ei[q·(x−x

′)−q2cz/(2ω)], (11.24)

which is nothing but the paraxial propagator, i.e. the Fresnel diffraction integral.

104


Moreover, the Maxwell-paraxial modes are quasi-orthogonal, in the sense that∫d2x F(λ)(x′, z,x, ω, t)F(µ)(x′′, z,x, ω, t) = δλµ

×∫

d2q

(2π)2

(1 + θ2

√1 + θ4

)1/2

eiq·(x′−x′′). (11.25)

Notice that these modes become fully orthogonal in the paraxial regime.

11.5 The Paraxial States of the Electromagnetic Field

Let us now concentrate on Eq. (11.22) and its meaning. According to Eq. (11.22),the Fourier component at frequency ω of the paraxial vector potential operatorA(+)(r, t) is given as the superposition of all the annihilation operators at the po-sition x′, weighted by the corresponding Maxwell-paraxial mode F(λ)(x′, z,x, ω, t).This means, from a very general point of view, that if we want to create a photonin a certain paraxial mode given by F(λ)(x′, z,x, ω, t) (or, equivalently, its paraxialponterpart P(λ)(x′, z,x, ω, t)), we need to construct a field operator, which is givenby the superposition of al the field operators that create a photon in the point ofspace x′, each of them weighted by the value of the Maxwell-paraxial mode at thatspatial point.

If we then introduce the following single photon paraxial state

|x, ω, λ〉 = a†λ(x, ω)|0〉, (11.26)

then〈0|A(+)(r, t)|x, ω, λ〉 ' F(λ)(x′, z,x, ω, t)e−iω(t−z/c), (11.27)

and therefore the Maxwell-paraxial modes can be interpreted as the single photonwave function. The single photon state defined in Eq. (11.26) can also be used tofind the spatial representation of any state of the electromagnetic field. Given infact a general state |ψ〉 of the electromagnetic state, the projection onto |x, ω, λ〉gives in fact

〈x, ω, λ|ψ〉 = ψλ(x, ω), (11.28)

which is an exact solution of the paraxial equation.

11.6 An Example of Paraxial Mode Operators

As it is well known [16], the fundamental solution of the paraxial equation is givenin the form of a Gaussian beam. The form of the higher order solutions, however,depend on the symmetry of the problem. For the case of cartesian symmetry, theyare given by the Hermite-Gauss modes, while for cylindrical symmetry, they aregiven by Laguerre-Gauss modes [16]. In both cases, the modes are defined by two

105

REFERENCES

integer indices n and m. Moreover, these modes represent an orthonormal andcomplete basis in cartesian and cylindrical coordinates, respectively. Without anyloss of generality, then we can define ψnm(x, ω) to be one of such paraxial eigenmodes.Any more general paraxial field can be then represented as a suitable superpositionof these modes.

Let us then define the state of the field corresponding in having a single photonin one of these modes as follows:

|n,m, ω, λ〉 =

∫d2xψnm(x, ω)|x, ω, λ〉, (11.29)

in such a way that

〈y,Ω, µ|n,m, ω, λ〉 = ψnm(y, ω)δλµδ(ω − Ω). (11.30)

Analogously to what we did in Eq. (11.26), we can define the mode operators asfollows:

|n,m, ω, λ〉 = a†nmλ(ω)|0〉. (11.31)

It is then not difficult to prove that the following relations hold:

anmλ(ω) =

∫d2xψ∗nm(x, ω)aλ(x, ω), (11.32a)

a†nmλ(ω) =

∫d2xψnm(x, ω)a†

λ(x, ω), (11.32b)

for the mode operators and

aλ(x, ω) =∑n,m

ψnm(x, ω)anmλ(ω), (11.33a)

a†λ(x, ω) =

∑n,m

ψ∗nm(x, ω)a†nmλ(ω), (11.33b)

(11.33c)

for the correspondent spatial operators.

12 Bibliography

References

[1] R. Loudon, “The Quantum Theory of Light” , Third Edition (Oxford, 1997).

[2] C. Gerry and P. Knight, “Introductory Quantum Optics” (Cambridge, 2004).

[3] J. D. Jackson, “Classical Electrodynamics”, Third Edition (Wiley, 1998).

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REFERENCES

[4] B. Mehmani and A. Aiello, Eur. J. Phys. 33, 1367 (2012).

[5] S. Gasiorowicz, “Quantum Physics", First Edition (Wiley, 1974).

[6] A. Fetter and J. Walecka, “Quantum Theory of Many-Particle Systems”(Dover, 2003).

[7] C. Kittel, “Quantum theory of solids” (Wiley, 1987).

[8] V. Mukhanov and S. Winitzki, “Introduction to Quantum Effects in Gravity”(Cambdrige, 2007).

[9] C. K. Hong, Z. Y. Ou, and L. Mandel, Phys. Rev. Lett. 59, 2044 (1987).

[10] H. Walther, B. T. H. Varcoe, B. G. Englert and T. Becker, Rep. Prog. Phys.69, 1325 (2006).

[11] R. W. Boyd, “Nonlinear Optics”, Third Edition (Academic Press, 2008).

[12] I. H. Deutsch and J. C. Garrison, Phys. Rev. A 43, 2498 (1991).

[13] J. C. Garrison and R. Y. Chiao, “Quantum Optics" (Oxford, 2008).

[14] A. Aiello and J. P. Woerdman, Phys. Rev. A 72, 060101(R) (2005).

[15] L. Mandel and E. Wolf, “Optical Coherence and Quantum Optics” (Cambridge,1995).

[16] O. Svelto, “Principles of Lasers”, Fifth Edition (Springer, 2010).

[17] M. Lax, W. H. Louisell, and W. B. McKnight, Phys. Rev. A 11, 1365 (1975).

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