Lecture notes for the course 88380:Quantum Theory in Applied Physics

Winter 08

April 8, 2008

Contents

1 Advanced topics in quantum mechanics 11.1 Review of the basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Coherent states and squeezing . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Transition probabilities and perturbation theory . . . . . . . . . . . . . . . . . 14

2 Interactions of matter and classical electromagnetic radiation 272.1 The quantum mechanics of charges and classical electromagnetic radiation . 272.2 Transitions induced by electromagnetic radiation . . . . . . . . . . . . . . . . 302.3 Gain in an active medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3 Quantum electrodynamics 363.1 The free quantum electromagnetic field . . . . . . . . . . . . . . . . . . . . . . 363.2 Interactions of charges and the quantum electromagnetic field . . . . . . . . . 41

1 Advanced topics in quantum mechanics

1.1 Review of the basics

a. States & observables The basic objects in quantum mechanics are states and observables.States are normalized vectors in Hilbert space, and are labeled using Dirac bra-ket notation|ψ〉. Hilbert space is just a complex inner product space, which can be infinite-dimensional,with the additional property of completeness (which always holds when the inner productspace is finite-dimensional). The inner product between vectors |ψ〉 and |φ〉 is denoted by〈φ |ψ 〉 = 〈ψ |φ 〉∗. When 〈φ |ψ 〉 = 0, |ψ〉 and |φ〉 are termed orthogonal. The inner productof a vector with itself is its ‘length’ squared 〈ψ |ψ 〉 = ‖ψ‖2 ≥ 0 and is positive unless |ψ〉 isthe zero vector. Each vector |ψ〉 has a dual ‘bra’ vector 〈ψ| = (|ψ〉)†, i.e. a linear functionalwhose action on kets is defined by the inner product.

Observables, i.e. measurable physical quantities, are self-adjoint linear operators in theHilbert space. A linear operator A is just a mapping of vectors to vectors, which preserveslinearity A(a |ψ〉 + b |φ〉) = aA |ψ〉 + bA |φ〉. It has a natural action on bra vectors. Theadjoint of the action of an operator on a bra defines another linear operator, A† throughA† |psi〉 ≡ (〈ψ| A)†. If A |ψ〉 = A† |ψ〉 for all ψ then A is Hermitian. The eigenvalues

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of a Hermitian operators are all real, and the eigenvectors belonging to different eigen-values are mutually orthogonal. When the eigenvectors of a Hermitian operator span theHilbert space (i.e. they form a basis), the operator is called self-adjoint. Again, there isa difference between Hermitian and self-adjoint operators only when the Hilbert space isinfinite-dimensional.1

Exercise 1.1 1. Show that if A |a〉 = a |a〉 then 〈a| A† = a∗ 〈a| .

2. Show that (A†)† = A

b. Measurement In contrast with classical mechanics, the kinematics of a quantum systemdoes not present us with an obvious experimental interpretation. Therefore, we have to adda recipe for translating the mathematical to physical content. The orthodox interpretationis called the Copenhagen school, and it states:

1. The outcome of a measurement of an observable O is equal to one of its eigenvalues

2. If the state of the system is an eigenvector |o〉 of O then the outcome of the measure-ment is o

3. If O is to be measured for a general state |ψ〉, it should be expanded in a basis ofeignevectors, |ψ〉 = a1 |o1〉 + a2 |o2〉 + · · · The probability of obtaining outcome on is|an|2. When O has continuous spectrum, the sum is replaced by an integral, andyields a probability density. The previous item is of course a special case of this.

4. As a result of a measurement whose result was o, the system state becomes |o〉, en-suring that repeated measurements yield the same results. This process is called‘collapse’.2

A careful scrutiny of these measurement rules shows that the phase of the states doesnot have any measurable consequences. Only the relative phase between two states has aphysical sense (quantum interference).

The coefficients an = 〈on |ψ 〉 are called probability amplitudes. Since the eigenvectors ofnoncommuting operators are in general different, noncommuting observables cannot ingeneral be measured simultaneously with arbitrary precision. This is Heisenberg’s uncer-tainty principle. It follows from the measurement rules that the expectation value of anobservable is 〈O〉 = 〈ψ|O |ψ〉. We also define the uncertainty (standard deviation) in O by

∆O =√〈O2〉 − 〈O〉2.

Example 1.1 The operator x is the observable corresponding to the x-position of a quantumparticle. It has continuous spectrum ranging from −∞ to ∞, and its (generelized) eigenvec-tors are normalized by 〈x′ |x 〉 = δ(x − x′). The x-dependent probability amplitude 〈x |ψ 〉is called the wave function ψ(x) of the state: The probability density of finding the particleat x is |ψ(x)|2. The components of position operator~r commute, and therefore they can bediagonalized simultaneously, with eigenvectors |~r〉; the 3D wave function is ψ(~r) = 〈~r |ψ 〉

1Mathematical subtleties regarding the continuous spectrum and unbounded operators are beyond thescope of these notes.

2A ‘complete’ measurement as described here is called a von-Neumann measurement. Later on, a formal-ism was developed to deal with more general types measurements, which is fully consistent with the aboverules

2

Example 1.2 The commuting components of the operator ~p are the observables corre-sponding to the components of the canonical momentum. The spectral properties of ~pare identical with those of ~x, and it has similarly an eigenstate basis |~p〉. Heisenberg postu-lated that the components of the position and momentum satisfy the canonical commutationrelations (CCM)

[ri, pj] = ihδij , (1.1)

where h = h/(2π) is Planck’s constant and δ is the usual Kronecker delta. The CCMare inspired by the classical Poisson brackets, and serve as an example of quantization (seebelow) The CCM imply the uncertainty relations ∆x∆px ≥ h/2 by using the positivityof⟨(

λ(x− 〈x〉)− i(p− 〈p〉))(λ(x− 〈x〉)− i(p− 〈p〉))⟩

for any real λ. Equation (1.1) alsoimplies that ~p is the generator of translations - see below section ??

It is fair to say that the Copenhagen school interpretation of quantum mechanics iscontroversial. It postulates the existence of a classical observer, and gives no dynamicalexplanation of the measurment process and the wave function collapse.

c. Quantum interference Unlike classical states, quantum states can be combined in alinear superposition. Such superpositions are often characterized physically by interferencepatterns similar to those of wave mechanics. Feynman’s famous description of the two-slitexperiment serves as an example (see figure 1).3 The probability density of observingthe particle at a position x on the screen, assuming the quantum state is an equal weightsuperposition of the single slit wavefunction is

| 〈x| |( |ψ〉+ |φ〉√

2

)|2 =

12| 〈x |ψ 〉 |2 +

12| 〈x |φ 〉 |2 + Re 〈x |φ 〉∗ 〈x |ψ 〉 . (1.2)

The last term in this equation is the interference term which distinguishes the quantumfrom the classical case; in the latter you sum probabilities while in the former you addamplitudes. As Feynman pointed out, when a measuring device capable of distinguishingthrough which slit the particle went is introduced, then the probabilities add incoherently,as in classical physics.

d. Representations In performing concrete calculation it is often useful to work in a spe-cific basis—that is choosing a representation. Vectors are represented using their compo-nents ψr = 〈r |ψ 〉, and operators by their ‘matrix elements’ Ars = 〈r| A |s〉 or their action(Aψ)r = ∑s Arsψs. The index can be continuous (in which case the sum is replaced by anintegral) discrete or both. The transformation between different representations is unitary.

Example 1.3 (Position & momentum) The momentum wave function ψ(~p) = 〈~p |ψ 〉 is re-lated to the position wave function by Fourier transform

ψ(~p) =∫ dp

h3/2 ψ(~r)ei~r·~p

h , (1.3)

because 〈~p |~r 〉 = h−3/2ei~r·~p

h , which follows from the CCM up to a constant phase. Positionand momentum are position-represented by

(~rψ)(~r) =~rψ(~r) , (~pψ)(~r) =hi∇ψ(~r) , (1.4)

3Actually carrying out the experiment is not very easy and was only achieved in the 70s

3

Figure 1: Feynman’s qualitative demonstration of the phenomenon of quantum interference via a two-slit(gedanken) experiment. In all cases a quantum particle source is placed on the left, an opaque screen withone or more open slits in the middle, and sensitive detector screen on the right. The result of a schematicmeasurment histogram after many particles have been produced is shown to the right of the detection screen.(i) Only the top slit is open. A broad histogram is measured. (ii) A similar histogram is measured whenonly the bottom slit is open. (iii) Both slits are open: Since probability amplitudes, rather than probabilitiesare added, an interference pattern is produced. (iv) A light source is placed between the two screens, whichenables the experimenter to determine through which slit the particle has passed. The light source destroysthe coherence of the quantum dynamics, so that probabilities rather than amplitudes are added, as in classicalphysics, and no interference pattern is produced.

and momentum-represented by

(~rψ)(~p) = − hi∇ψ(~p) , (~pψ)(~p) = ~pψ(~p) . (1.5)

Remark: It is worth pointing out that it is usually more efficient to work without choosinga basis if possible.

e. Symmetries The assignment of Hilbert space vectors to physical states involves a largedegree of arbitrariness, since only transition probabilities | 〈φ |ψ 〉 |2 are measurable. ByWigner’s theorem (one of them) it follows that transition probabilities don’t change undera different assignment of vectors to states if and only if each state |ψ〉 is replaced by |ψ′〉 =U |ψ〉 where U is either linear and unitary or antilinear and antiunitary 4. It should bestressed that |ψ〉 and |ψ′〉 are different labellings of the same state—the transformationto the primed states is passive because no actual change has taken place. Observablestransform passively by O → O′ = UOU† so that measurement results are the same in bothschemes.

In contrast with passive transformations, active transformations involve an actual changein the system, like shifting it in space. Some operations are special: They don’t change tran-sition amplitudes, i.e., they are symmetries of the system. By Wigner’s theorem again, the

4The latter possibility will not be covered in these notes

4

states in the original and transformed are related by a (anti)unitrary transformation. Forinstance there is a family of shift operators U~r such that |ψ〉~r = U~r |ψ〉 is the state obtainedafter shifting the original system a distance ~r. The U~r operators form an abelian group:U~rU~r′ = U~r+~r′ , U0 = 1, U−~r = U−1

~r = U†~r . Therefore we can write U~r = e−

ih~r·P, where P is

the system momentum, the generator of translations.

Exercise 1.2

1. Show that e−ih apx |x〉 is an x-eigenvector with eigenvalue x + a. This means we can

define |x + a〉 = e−ih apx |x〉, fixing the phase of 〈px |x 〉.

2. The previous item shows that ~p is indeed the momentum of a single particle system.Show that momentum of an n-particle system is ~P = ~p1 +~p2 + · · ·+~pn.

Unlike in passive transformation, the observables are not transformed in an active trans-formation along with the states—the measurement results are shifted along with the sys-tem. Nevertheless, it’s possible to make a further passive transformation so that the observ-ables are transformed rather than the states. Namely, taking U = U†

~r we get |ψ′〉~r = |ψ〉 andO′

~r = U†~r OU~r. Borrowing terminology from dynamics, the former approach can be called

the Schrodinger picture of the transformation, while the latter is the Heisenberg picture.Note the active transformation of observables (in the Heisenberg picture) is the inverse ofthe passive transformation. This is also true in classical physics.

Example 1.4 (Rotations and angular momentum) Rotation about an axis n is generated byn ·~J, where J is the system angular momentum. The individual components of J do notcommute because rotations about different axes do not commute. The angular momen-tum of a single particle is ~j = ~ +~s, [~ ,~s] = 0. The orbital angular momentum ~ = ~r × ~pis the quantization of the classical angular momentum; no ordering issues arise, sincedifferent components of~r and ~p are multiplied. The spin angular momentum~s is a (finite-dimensional) irreducible representation of the rotations Lie algebra; ~s2 = hs(s + 1), is con-stant and the particle is spin s, s half-integer. The angular momentum of a system of a fewparticles is the sum of the individual angular momenta.

Exercise 1.3

1. Calculate the result of active and passive rotations by an angle θ about the z axis onthe vectors ~x and ~p. Hint: The result is the same in classical and quantum mechanics.

2. Show by explicit calculation that the active transformation is realized by e−ih θ`z .

f. The density matrix and mixed states It is possible to use Hilbert space operators in-stead of vectors to label states. The operator associated with |ψ〉 is the density matrixρ|ψ〉 = |ψ〉 〈ψ|. It contains less information than |ψ〉 because it disregards the vector’s abso-lute phase, but as already explained, the phase does not carry physical information; in fact,all the physical information is contained in the density matrix as transition probabilitiesfrom state 1 to 2 are given by Tr ρ1ρ2, so that the probability of measuring the value o forthe observable O is 〈o| ρ |o〉. The expectation value of an observable is Tr ρO.

When working with pure states density matrices are less convenient than state vectors,because, as they are quadratic in the state, they cannot be easily used to create linearsuperpositions. On the other hand, they are very useful in describing situation where thestate is not known with certainty, since then probabilities are added rather amplitudes. If

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the the probability that the system is in state |s〉 is ps then the density matrix of the systemis ρ = ∑s psρ|s〉. (As usual the index s may also be continuous, in which case the sum isreplaced by an integral.) When it is necessary to distinguish the two, a partially knownstate is called a mixed state, and a state known with certainty is called a pure state. Thedensity matrix contains the measurement probabilities for mixed states by taking trace, asfor pure states. The essential properties of ρ are: a) ρ is positive and bounded from aboveby 1, and b) Tr ρ = 1 5. It follows that ρ2 ≤ ρ. When the the inequality is saturated, ρ is aprojection and the state is pure. As in classical physics we can quantify the degree of ourignorance about the system by its entropy. The entropy of a state with density matrix ρ is−Tr ρ log ρ.

Exercise 1.4 Show that the entropy of a state is always nonnegative and that it is 0 if andonly if it the state is pure. Hint: Since it is positive, ρ can be diagonalized by a unitarymatrix. Show that the state is pure if and only if the diagonalized density matrix has asingle nonzero element.

Example 1.5 (Density matrix of a spin 1/2)

1. The density matrix of a system in the state α |↑〉+ β |↓〉 is ρ|ψ〉 = |α|2 |↑〉 〈↑|+ α∗β |↓〉 〈↑|+αβ∗ |↑〉 〈↓|+ |β|2 |↓〉 〈↓|, |α|2 + |β|2 = 1.

2. The density matrix of a system known to be either in state |↑〉 with probability p orin state |↓〉 with probability 1− p is ρp = p |↑〉 〈↑|+ (1− p) |↓〉 〈↓|. It is a mixed stateunless p = 0 or q = 0, and its entropy is −p log p − q log q. The entropy is maximalfor the ‘maximally mixed state’ with p = q = 1/2.

Note that pure state density matrices are characterized by nonzero off-diagonal elements.They are often called coherence matrix elements.

Exercise 1.5 Show that the state ρp defined in the second part of the last example withp = 1/2 is indeed maximally mixed, in the sense that it has the largest possible entropy ofa spin 1/2 system.

Exercise 1.6 The density matrix contains all the necessary information to predict the resultof any measurement performed on a mixed state system. However, it does not determineuniquely a probability distribution of pure states, as follows from this exercise. Showthat the density matrix of a system known to be with equal probability in either of thestates 1√

2(|↑〉 ± |↓〉) (these are eigenstates of sx if |↑〉 , |↓〉 are eigenstates of sz) is that of the

maximally mixed state.

g. Composite systems and entanglement The Hilbert space of a system composed of twosubsystems is the tensor product of the two sub-Hilbert spaces; that is if subsystem A hasthe set |a〉 as a basis and subsystem B has |b〉 (where a and b are index variables,taking discrete, continuous, or both values) then the set |ab〉 ≡ |a〉 ⊗ |b〉, is a basis of thecomposite system Hilbert space.6

Example 1.6 The simplest example of a composite system is a two (distinguishable) spin 12

system. Its 4-dimensional Hilbert space is spanned by |↑↑〉 , |↑↓〉 |↓↑〉 |↓↓〉.5The additional requirement of complete positivity, needed to ensure that ρ arises as the density matrix of

a subsystem of a system in a pure state, will not be discussed here6It possible to give a basis independent definition for the tensor product space, as well as a precise defini-

tion of the tensor product of two vectors. As usual such mathematical niceties will be ignored here.

6

It should be stressed that not all the states in a product space are themselves productstates. States that cannot be expressed as a tensor product of a subsystem A state and asubsystem B state are called entangled. When a system is an entangled state, no definitepure state can be assigned to the subsystems, but mixed state can, by taking a partial traceof the density matrix ρ of the composite system; for example, the density matrix ρA ofsubsystem A is given by ρA

aa′ = ∑b ρab,a′b. This can also be written as ρA = TrB ρ. Thisoperation is often performed when A is an open system and B is its environment.

Evidently, a system is in a product state if and only if the subsystem states are pure.This fact motivates measuring the degree of entanglement by the entropy of the subsystemstates—the entanglement entropy.7

Example 1.7 The density matrix of the ‘left’ subsystem for a two spin-12 system in the state

(|↑↑〉+ |↑↓〉)/√

2 is ρL = |↑〉 〈↑|, that is evidently the density matrix of a pure state. Hence,the original state is separable. In fact, (|↑↑〉+ |↑↓〉)/

√2 = |↑〉 ⊗ (|↑〉+ |↓〉)/

√2.

Exercise 1.7 Show that the singlet state 1√2(|↑↓〉 − |↓↑〉) in a two spin 1/2 system is en-

tangled, calculate its entanglement entropy, and show that its entanglement entropy ismaximal for this system. It is therefore a maximally entangled state.

The notion of a partial trace can be used also when the system state is not necessarilypure, and defines a consistent scheme for system and subsystem states.

Exercise 1.8 1. Show that ρA = TrB ρ has the necessary properties of a density matrix,i.e. 0 ≤ ρ ≤ 1, Tr ρ = 1, if ρ has them.

2. Show that the measurement results of an operator OA which acts only in the A sub-space is the same whether ρ or ρA is used.

The notion of entanglement can also be generalized to the case where the compositesystem is in a mixed state by defining a state to be reducible if its density matrix is a linearcombination of pure product state density matrices, and entangled otherwise. However,determining whether a system in a mixed state is entangled or not is hard.

h. Dynamics & energy Although isolated systems are invariant under time translation asmuch as under space translation, in nonrelativistic quantum mechanics the result of timetranslation is a state in a different Hilbert space rather than a different state in the sameHilbert space. The fact that time evolution is generated by a unitary operator U(t′, t) musttherefore be added as another postulate. Defining the Hamiltonian H(t) = 1

ih ∂tUU†, theinfinitesimal version of this statement is the Schrodinger equation

ihd |ψ〉

dt= H |ψ〉 . (1.6)

When the system is invariant in time, H is constant, and is the observable associatedwith the energy. It follows that the stationary states of the system are energy eigenstates|ε〉 of the Hamiltonian. When the Hamiltonian is diagonalized, the time evolution of anarbitrary state is ‘solved’ by expanding it eigenstates:

|ψ〉 = ∑ε

aε |ε〉 =⇒ |ψt〉 = ∑ε

ase−ih εt |εs〉 (1.7)

7It can be shown that the entanglement entropy is independent of which subsystem is used

7

Example 1.8 (Free particle) The Hamiltonian of a mass m free particle with H = p2

2m . Hcommutes with the momentum observable, hence its eigenstates are momentum eigen-

states |~p〉 and |ψt〉 =∫

dp |~p〉 〈~p |ψ 〉 e−ih

p22m t.

Example 1.9 The Hamiltonian of a system of n particles interacting by a static potential is

H = p21

2m1+ · · ·+ p2

n2mn

+ U(~r1, . . . ,~rn) where U is some reasonable function of 3n variables;mi is the mass of particle i. This is an example of ‘quantization’—an educated guess of thequantum Hamiltonian based on knowledge of the classical one, although the final wordis always obtained by experimental verification. In this particular case the quantizationis easy, because no ordering issues arise, and consists of simply substituting quantumoperators for the classical observables.8

Exercise 1.9 Show that the velocity observable is v = pm .

Example 1.10 The density matrix of a system in equilibrium with a heat bath at temper-ature T is ρ = Z−1e−H/T (choosing units such that Boltzmann’s constant is one). It is amixed state unless T = 0.

Exercise 1.10 Calculate the density matrix of a two-state system with a Hamiltonian ε↑ |↑〉 〈↑|+ε↓ |↓〉 〈↓| at temperature T and its entropy.

i. Particle in an external potential Consider two particles coupled by a static potential,one of which can be considered classical (the environment). Then the Hamiltonian for the‘system’ particle becomes H = p2

2m + Ut(~r), where the effective one-variable potential isobtained by substituting the classical environment coordinate in the two-variable potential.If the environment is also static, the system is subject to a static external field with potentialU.

In a real system, when the particle is far enough from the center of force, the forceand the potential decrease to zero (the value of the potential at infinity is conventional).Then, states with positive energy are unbound, i.e. scattering states, and H has continuousspectrum going from zero to infintity, and there may be negative energy bound states,forming the discrete spectrum of H. See figure 2.

Example 1.11 (Harmonic oscillator) The quadratic potential, U(x) = 12 mω2x2 is a good

approximation for a (generic) attractive potential near its minimum. It will also play acentral role in the quantization of the electromagnetic field. Define the lowering operator

a =√

mω2h x + i 1√

2mhωp and the raising operator a†. Then [a, a†] = 1 and H = hω(a†a + 1/2).

The ground state of the Hamiltonian is a zero eigenstate of a, a |0〉. Its phase is conven-

tionally defined so that its wave function is real. The excited states are |n〉 = (a†)n√

n!|0〉. The

integer n is the eigenvalue of the number operator N = a†a, and the energy eigenvalueis εn = (n + 1

2)hω. The 12 hω term is called zero-point energy, and its existence is a man-

ifestation of the uncertainty principle, or ‘quantum fluctuations’. The energy states wave

functions are 〈n |ψ 〉 = cnHn(√

mωh x)e−

mω2h x2

, where Hn are Hermite polynomials and cn arenormalization constants.

8Actually, there is no fundamental force which has the static potential form; however, it is quite a goodapproximation for a very large class of systems, such as the interaction of nuclei and electrons in atoms andmolecules

8

Figure 2: The spectrum of the Hamiltonian of a spinless particle in an attractive finite range potential. Inthree dimensions the number of discrete levels can be zero.

1.2 Coherent states and squeezing

The focus of this section is on quantum systems whose states are not necessarily eigen-states or a superposition of a small number of eigen-states of any observable, in particularthe energy. On the other hand they have the property of being phase-space localized, inthe sense that they have well-defined position and momentum up to small uncertainties.The position representation of such states are wave packets, and they often arise as a con-sequence of the interaction of the quantum system with classical, macroscopic systems.

a. External driving and Coherent states Suppose a particle lying in the ground state of aharmonic potential is subjected to external driving by a uniform time-dependent force f (t),starting from t = 0. We would like to know its state |ψt〉 at later times. The Hamiltonianfor the forced Harmonic oscillator is

H =p2

2m+

mω2

2q2 − f (t)q = hω(a†a + 1/2) + hφ(t)(a + a†) , (1.8)

where φ(t) = −(2mhω)−1/2 f (t). Note that the Hamiltonian is time-dependent. TheHeisenberg equation of motion for the lowering operator a is

a =ih[H, a]− iωa− iφ(t) ; (1.9)

it is a linear ordinary differential equation whose solution is

a(t) = ae−iωt − i∫ t

0dt′φ(t′)e−iω(t−t′) . (1.10)

The fact that a |ψ0〉 = 0 implies, using a(t) = eih Htae−

ih Ht, that a |ψt〉 = α |ψ0〉 (where

α = −i∫ t

0 dt′φ(t′)e−iω(t−t′)), that is, |ψt〉 is a coherent state.

b. Coherent states It is convenient to define the coherent states by |α〉 = D(α) |0〉, wherethe displacement operator D(α) = eαa†−α∗a is a unitary operator depending on a complexparameter. It acts by displacement on the operator a, D(−α)aD(α) = a + α 9, which impliesthe first property of coherent states:

9as can be proved using the identity eABe−A = B + [A, B] + [A,[A,B]]2! + · · ·

9

1. Coherent states are eigenstates of the lowering operator: a |α〉 = α |α〉. In particularthe ground state |0〉 is a coherent state.

Exercise 1.11 Show that a† has no eigenstates.

2. Let Q and P be the quadrature operators of a, that is, a = Q+iP√2

, (Q, P Hermitian).Q and P are non-dimensionalizd canonical variables proportional to q and p respec-tively, [Q, P] = i. Then 〈Q〉+ i 〈P〉 =

√2α in a coherent state.

3. Coherent states are minimum uncertaianty states: ∆Q∆P = 1/2 in a coherent state.

4. The coherent states form a non-orthogonal set. By the Baker-Campbell-Hasdorff iden-

tity10 D(α)D(β) = ei Im αβ∗D(α + β) and D(α) = e−|α|2

2 eαa†e−α∗a. Hence 〈β |α 〉 =

ei Im α∗β 〈0|D(α− β) |0〉 = e−12 |α|2+αβ∗− 1

2 |β|2 , and | 〈β |α 〉 |2 = e−|α−β|2

5. 〈n |α 〉 = 〈0| an√

n!|α〉 = e−

|α|22 αn√

n!. Hence the distribution of quanta in a coherent state

is Poissonian, pn = e−|α|2 |α|2n

n! , with mean number |α|2.

6. The coherent states form a spanning set, since

1π

∫d2α |α〉 〈α| = 1

π ∑nm

∫d2α |n〉 〈m| (α∗)n

√n!

αm√

m!e−|α|

2= ∑

n|n〉 〈n| = 1 , (1.11)

where d2α is a shorthand for d Re αd Im α. The main step is the middle equality whichfollows from

Exercise 1.12 Show that∫

d2α(α∗)mαne−|α|2= n!δnm.

The coherent states form in fact an overcomplete set since nontrivial linear combina-tions can be formed which are equal to zero, for example, by expressing one of thecoherent states in terms of the others via the resolution of the identity (1.11).

These properties, especially the minimum uncertainty property, and the fact that exter-nal driving excites coherent states, has earned them the label ‘classical states’. This namecan be further motivated by considering ac =

√ha under external driving. In the classical

limit ac becomes a classical (complex) variable αc, which is identical with the coherent statelabel (times

√h).

Exercise 1.13 Calculate the wave function of a coherent state. Hint: The easiest method isto use the fact that |α〉 is an eigenstate of the lowering operator. This property determinesthe wave function up to a phase.

Exercise 1.14 (External driving: The phase) We can now retake the problem of the exter-nally driven harmonic oscillator. It was shown that |ψt〉 = eiχt |αt〉, where α = −i

∫ t0 dt′φ(t′)e−iω(t−t′);

it is left to calculate the phase χt.

1. Show that ∂α |α〉 = (a† − α∗2 ) |α〉, and ∂α∗ |α〉 = − α

2 |α〉.

10The BKH identity states that eAeB = eA+B+ 12 [AB ]+higher order commutators. The higher order terms are non-

trivial, but vanish in the cases considered here.

10

2. Show that the Schrodinger equation implies that α∂α |α〉+ α∗∂α∗ |α〉+ iχ |α〉 = −iω(a†a +12) |α〉 − iφ(a + a†) |α〉.

3. Substitute the result of question 1 in that of question 2 to obtain(iχ − (iωα + iφ) +

φ Im α)|α〉 =

(−iω(αa† + 1

2)− iφ(α + a†))|α〉 .

4. Show that the terms in the last equation proportional to a† |α〉 and to (real) |α〉 aretrivially satisfied, and that the other terms imply that

χt = −12 ωt−

∫ t

0φt′ Re αt′dt′ . (1.12)

c. Poisson process and shot noise In this section we pause to consider the physical sig-nificance of coherent states. It will be explained in the last part of these notes that a freemode of the quantum electromagnetic field is a harmonic oscillator. Therefore, when elec-tromagnetic radiation is generated by classical currents, its normal modes are in coherentstates—this includes laser light, for which these states were considered by Glauber in theearly ’60s.

Suppose then that we have a source of quasi-monochromatic light that emits pulses ofduration τ in a coherent state |α〉 of an electromagnetic field mode (see figure ??) towarda perfect detector; this is well-defined as long as τ is much longer than an optical cycle11. According to the properties of a coherent state, the number of photons detected will bePoisson-distributed with mean photon number n = |α|2.

Consider now a random source of classical particles that releases pulses of duration τwith an average of n particles in such a way that the events of particle arrival are inde-pendent, like the fall of raindrops. In this case the probability that the detector observes aparticle during an interval dt is dp = n dt

τ . The pulse consists of N = τdt such intervals, and

the probability that the pulse contains n particles is given by the binomial distribution

p(n) =(

Nn

)(nN

)n (1− n

N

)N−n∼ nn

n!e−n . (1.13)

The last equality, obtained in the limit dt τ shows that the distribution of counter eventsis Poissonian, and the random arrival sequence is called a Poisson process. This purelyclassical phenomenon of current fluctuations due to discreteness of current carriers is calledshot noise and is often used to calculate charge. In our context it shows that in a coherentstate photons are distributed like independent classical particles—another motivation forusing the ‘classical’ label for coherent states.

Exercise 1.15 (Number variance)

1. Show that for a Poisson distribution Var n = n.

2. Calculate n and Var n for a harmonic oscillator in a ‘thermal state’ with temperatureT, i.e. in mixed state with ρ = Z−1 exp(−H/T), where H is the harmonic oscillatorHamiltonian. Show that Var n ≥ n with equality only if T = 0.

We can draw two conclusions from this exercise. Firstly, the relative strength of numberfluctuations in a Poisson process decreases like the square root of the mean number, ascould be expected on the basis of the law of large numbers. This result explains on one

11The phase of α is actually oscillating optically, but this does not affect the arguments

11

hand why macroscopic electric current or coherent light seems continuous, and on theother hand shows that highly excited coherent states are supported by a large number ofenergy states. Secondly, it is rather easy to generate photon distributions which are super-Poissonian, that is they have a larger number standard deviation than a coherent state withthe same mean number. Creating sub-Poissonian states is more tricky, and is related to thesubject of the next section.

d. Squeezed states Consider again the quadrature uncertainties ∆Q and ∆P. We saw thatcoherent states are minimal uncertainty states since the values ∆Q = ∆P = 1√

2saturate the

Heisenberg rule ∆Q∆P ≥ 12 . Obviously, however, one may improve on the coherent state

uncertainty values of a given quadrature, if one is willing to accept a larger uncertainty inthe other quadrature. States exhibiting sub-coherent-state uncertainties in a quadrature aretermed squeezed.

Example 1.12 A trivial example is the position eigenstate |q〉 where ∆Q = 0 and ∆P = ∞.

Exercise 1.16 Show that energy states |n〉 are not squeezed. 12

Naturally there is no need to restrict ourselves to Q and P; one can squeeze any quadratureQθ = 1√

2(eiθa + h.c.) (where h.c. stands for Hermitian conjugate), of which Q = Q0 and

P = Qπ/2 are special cases.

e. Squeezing by optical nonlinearities An important scientific and technological use oflight is precision measurement, and the limitation to accuracy posed by the quantum un-certainty principle play an important role there. For this reason interest has been raisedin the production use and detection of squeezed states of light, with the idea of using thereduced noise level of the squeezed quadrature to reduce measurement errors. As could beguessed on the basis of the results shown above, sources that produce light from classicalcurrents, and this includes lasers, are limited by the noise of coherent states, or ‘shot noise’.

As will be explained in the last part of these lectures, the operations of traditionallinear optics, affected by mirrors, beam splitters and interferometers, are unable to reduceuncertainties. The way to achieve squeezed states of light is to use nonlinear optics, wherethe optical properties of the medium depend on the intensity of the light. For example, theprocess of degenerate parametric down-conversion, in which one photon of frequency 2ω isconverted into two identical photon of frequency ω is well-modelled by the Hamiltonian13

HDPDC = hωa†a + 2hωb†b + g((a†)2b + a2b†) , (1.14)

where a and b are the (mutually commuting) annihilation operators for the field modes offrequency ω and 2ω, respectively, and g is the effective coupling coefficient.

The blue light is the ‘pump’ of the process. It is high-intensity, and may be approxi-mately considered classical, replacing b by βe−2iωt. Next consider the Heisenberg equationfor a,

a = −iωa− 2igβa†e−2iωt . (1.15)

It is most easily analyzed by substituting a(t) = a(t)e−iωt. Consider for simplicity the caseof β pure imaginary with positive imaginary part. Then equation (1.15) implies that the

12To be precise: They are not quadrature-squeezed, but they are amplitude-squeezed13In practice one has to achieve ‘phase-matching’ for the interaction to be resonant, and hence effective.

This is assumed in the model Hamiltonian.

12

quadrature Pt = Pe−2g|β|t. Supposing the a mode is initially in the ground (i.e. vacuum)state it follows that 〈Pt〉 = 〈P〉 = 0, and

∆P2t =

⟨P2

t

⟩=⟨

P2⟩

e−2g|β|t =12

e−2g|β|t , (1.16)

that is, the photon field is exponentially squeezed. It should be remarked, though, thatimperfections make the squeezing much harder to achieve in practice than on paper.

f. Ideal squeezed states Evidently, the dynamical squeezing in the previous example wasgenerated by the terms proportional to a2 and (a†)2 in the Hamiltonian. This observationmotivates the definition of the unitary squeeze operator S(z) = e

12 (z∗a2−z(a†)2). Its action on

the lowering operator is S(z)aS(z)† = S(z)aS(−z) = cosh ra + eiθ sinh ra†, where z = reiφ

is the polar decomposition.Consider for example the ‘squeezed ground state’ |0; z〉 = S(z) |0〉. The expectation of

the quadratures vanishes, 〈0; z| a |0; z〉 = 〈0| (a cosh r− a†eiφ sinh r) |0〉 = 0. The quadraturevariance is then

∆Q2θ =

⟨Q2

θ

⟩=

12〈0; z|

(cosh reiθ − ei(φ−θ) sinh r)a + (cosh re−iθ − e−i(φ−θ) sinh r)a†

)2|0; z〉

=12(cosh 2r− cos(2θ − φ) sinh 2r) . (1.17)

Plotted as a function of θ, then, the quadrature standard deviation describes an ellipse, withsemi-major axis along θ = φ

2 + π2 of length er/

√2 (see figure 3). The length of the semi-

minor axis (along θ = φ2 ) is e−r/

√2, which is always less than 1/sqrt2. The states |0; z〉 are

therefore indeed squeezed, exponentially in the modulus of z. When z is real ∆Q∆P = 12 ,

and the squeezed ground state is also a minimal uncertainty squeezed state; otherwise thesqueezed ground state attains minimum uncerainty in the φ

2 , φ2 + π

2 . Such states are calledideal squeezed state.

Figure 3: The ellipse, viewed as a representation of a function in polar coordinates, gives the standarddeviation of the quadrature Qθ as a function the angle θ for an ideal squeezed state.

Example 1.13 The quadrature expectations and variances of the state |α; z〉 = D(α) |0; z〉are 〈Q〉 + i 〈P〉 =

√2α, and the uncertainties are the the same as for |0; z〉. Hence the

error ellipse for|α; z〉 is obtained from that of |0; z〉 by shifting by√

2α, and they are all

13

ideal squeezed states, except the coherent states (z = 0), where the error ellipse becomesan error circle. This example illustrates a general principle: D(α) shifts the observableexpectation values of a state without changing theor uncertainties

Exercise 1.17 Show that S(z)D(α) |0〉 is an ideal squeezed state |α; z〉 with the same z.Calculate α.

Exercise 1.18 Calculate the quadrature expectations and variances of the state S(z) |n = 1〉(squeezed energy state). For which values of z are these states squeezed? Are they mini-mum uncertainty states?

In spite of its name, the squeezed ground state is an excited state. The expected num-ber of excitation is 〈0; z| a†a |0; z〉 = sinh2 r, proportional to the degree of squeezing. Wehave seen that the distribution of number of quanta in a coherent state is Poissonian, cor-responding to a process of independent events. The variance of the number of quanta⟨(a†a)2⟩− ⟨(a†a)

⟩2 = 2 sinh2 r cosh2 r in the squeezed ground state, which is always largerthan

⟨a†a⟩, so that the number statistics is super-Poissonian.

It follows that squeezing and sub-Poissonian statistics are independent concepts. Theyare related, however, and their relation can be deduced from the error ellipse, as shown inthe next exercise.

Exercise 1.19 Calculate the expectation and variance in the number of energy quanta in theideal squeezed state |α; z〉. Show that sub-Poissonian distributions are possible as well assuper-Poissonian ones. Show that the smallest number variance for a given |α| and |z| isachieved when 2 arg α = arg z, and that ∆n

n can be arbitrarily small (but positive of course),for sufficiently large α and z. Note: zero variance is achieved, obviously, only for energystates.

Exercise 1.20 We have seen that all the states |α; z〉, z real saturate the the Heisenberguncertainty relation ∆Q∆P ≥ 1

2 . Show the converse is also true. Hint: Show that a minimaluncertainty state must be an eigenstate of λQ + iP for some real λ.

1.3 Transition probabilities and perturbation theory

In quantum mechanics we are often interested in the following situation: The Hamiltonianis expressible as a sum of two terms H = H0 + V, where H0 is some reference Hamiltonianwhose spectrum and eigenstates are assumed to be known, sometimes called the ‘freeHamiltonian’, and a possibly time-dependent ‘interaction’ V. Suppose that at some pointthe system is in an eigenstate |i〉 of H0 with energy εi and ask what is the probability thatat some later time it is in a state | f 〉 with energy ε f . In this section we will address thisproblem for cases where the interaction is in some sense weak.

a. Transitions between two discrete states This question is most often posed in the con-text of scattering problems or decay of an unstable state, but first we will study this questionin the simplest case where the spectrum of H0 consists of two states only. No real physicalsystem has only two levels, but it often happens that only two discrete states of H0 arephysically important. This can happen because the H0 has discrete spin or orbital degen-eracy, for example, or when two specific states of H0 are resonant with V (see below), andall other transitions are nonresonant.

Suppose then that H0 = hωi |i〉 〈i| + hω f | f 〉 〈 f |, and consider first the special case〈i|V | f 〉 = 〈 f |V |i〉∗ = hΩei(νt+φ). The problem depends on three frequencies, the level

14

spacing ω (the absolute energy values have no physical meaning, as usual), the interactionfrequency ν, and the Rabi frequency Ω, which measures the strength of the interaction.

Exercise 1.21 Show that Hamiltonian of the spin degree of freedom of a spin 12 particle

with magnetic moment µ in a constant magnetic field Bzz in the z direction, and circularlypolarized electromagnetic radiation with magnetic field amplitude Br and freqeuncey ωrpropagating in the z-direction has the above form. Identify ω, ν and Ω in terms of thephysical parameters.

The Schrodinger equation for |ψt〉 = α(t) |i〉+ β(t) | f 〉 is equivalent to

α = −i(ωiα + Ωei(νt+φ)β

)β = −i

(Ωe−i(νt+φ)α + ω f β

),

(1.18)

with α(0) = 1, β(0) = 0.Equations (1.18) may seem hard to solve because the Hamiltonian H is time dependent,

but the simple transformation α = αei2 (νt+φ), β = βe−

i2 (νt+φ), transforms the Scrodinger

equation to the time-independent form

ddt

(αβ

)= −i

ω + δ Ω

Ω ω − δ

(αβ

)≡ M

(αβ

), (1.19)

where ω = 12(ωi + ω f ) and δ = 1

2(ν−ω) is the detuning.

Exercise 1.22 Show that

eMt = e−iωt

(cos Ωt− i δ

Ωsin Ωt −i Ω

Ωsin Ωt

−i ΩΩ

sin Ωt cos Ωt + i δΩ

sin Ωt

), (1.20)

where Ω =√

Ω2 + δ2. Hint: You may represent M using Pauli matrices.

It follows then, using the initial conditions, that the probability to find the system in state| f 〉 after time t is

|β(t)|2 =Ω2

Ω2 + δ2 sin2(√

Ω2 + δ2t) . (1.21)

i.e., the probability oscillates harmonically with frequency 2√

Ω2 + δ2, and reaches themaximal value Ω2

Ω2+δ2 . This phenomenon is named Rabi oscillations, after Rabi who usedit to measure the magnetic moment of nuclei. The technique is called nuclear magneticresonance (NMR), and won Rabi the Nobel prize.

The possibility to derive a simple exact solution to a time-dependent problem is theexception rather than the rule.

Exercise 1.23 ∗ Using the magnetic resonance realization of exercise ??, show that that thetransformation α, β → α, β, that mapped the problem to a stationary one, is equivalent to atransformation to a reference frame rotating in the xy plane with frequency ν, in which theradiation field is stationary.

In general it is not possible to transform the problem to a time-independent one. A simplealternative arises when the interaction is weak in the sense that Ω ω, a common situ-ation. Let then, 〈i|V | f 〉 = 〈 f |V |i〉∗ = hΩ(t), with an arbitrary small Ω(t) instead of the

15

specific exponential form. It is then reasonable to assume that α(t) ∼ e−iωit. Using this inthe equation for β gives

β = −iω f β− iΩ(t)e−iωit =⇒ β(t) = −ie−iω f t∫ t

0dt′Ω(t′)e−iωt′ . (1.22)

This reasoning is valid in most cases, but it obviously does not reproduce correctly theexact solution (1.21) when δ . Ω. This is not surprising, since β(t) is not small there andthe basic assumption leading to equation (1.22) breaks down. A more careful examinationof equation (1.22) shows that when ν is close to ω, the integrand can become very large astime increases. This is the familiar phenomenon of resonance, that is not at all limited toquantum systems: Weak external forces can have large effects if they vary on the systemsnatural cycle.

When the interaction has the exponential form, it is evident from equation (1.21) thatnear resonance the dependence on ω is contained only in the detuning δ, and this is thescale to which Ω has to be compared rather than ω. The saturation parameter s = (Ω

δ )2 isthen the only dimensionless parameter. When s is small, the interaction is non-resonant,validating equation (1.22), the frequency of the Rabi oscillation is close to δ, and the maxi-mum transition probability is close to s. As s increases, the interaction becomes resonant,equation (1.22) ceases to hold, the frequency of oscillations tends to Ω, and the maximumtransfer probability approaches one. One then says that the transition is saturated.

When resonant effects are present, they overshadow weak non-resonant effects. For thisreason, for example, terms with a e+iνt time-dependence in the interaction matrix element,ν ≈ ω are neglected with respect to the resonant e−iνt. For historical reasons, such terms inthe Hamiltonian are called counter-rotating, and their omitting is called the rotating waveapproximation.

b. Transitions to a continuum and the S matrix The ideas of the previous section can beapplied to transition between three or more well-separated discrete states, but when thefinal state is part of a continuum some basic new phenomena arise. If we let the interactionact for a time interval tI , we can expect that the energy of the final state should be definedwith an uncertainty of ∼ h

t I . For discrete levels the uncertainty is not noticeable if tI islarge enough, but when | f 〉 belongs to a continuum it means that the transition is alwaystowards a band of states rather than a single one — which is reasonable since continuumeigenkets are not physical anyhow. An important consequence is the loss of coherence inthe target state: Since it has different energy components whose amplitudes oscillate atdifferent frequencies, their phases become randomized after time T, possibly making thetransition virtually irreversible. This is the reason that unitary quantum mechanics canlead to irreversible processes like radioactive decay.

These complications make the analysis of transition to a continuum more subtle. Thefirst step is to replace the time interval t of observation by an infinite time interval inwhich the interaction is only ‘switched on’ during an interval of time t. In other words wemultiply V by a switch function i(tI)(t) of width tI whose Fourier tranform is 2πδ(tI)(ω)of width 1/tI (see figure 4); as tI → ∞ δ(tI) approaches an actual delta function. When aspecific choice is needed, it is convenient to use a smooth rather than a step switch function,since the Fourier transform of the latter decays very slowly.

Once the interaction is switched on and off, then, the transition problem can phrasedas that of obtaining | f 〉 when t → ∞ from |i〉 prepared in |i〉 at t → −∞. The transitionprobability in question is therefore given by

limti→−∞t f→∞

| 〈 f |U(t f , ti) |i〉 |2 . (1.23)

16

Figure 4: A typical switch function i(tI)(t) used to adiabatically ‘turn’ the interaction on and off, and itsFourier tranform 2πδ(tI)(ω).

In contrast, the transition amplitude does not tend a limit as times to infinity, because of theH0 time dependence. This fact motivates the definition of

S f i = limti→−∞t f→∞

eih (ε f−εi)t 〈 f |U(t f , ti) |i〉 = lim

ti→−∞t f→∞

〈 f | eih H0tU(t f , ti)e−

ih H0t |i〉 . (1.24)

The last expression can be turned into an operator expression

S = limti→−∞t f→∞

eih H0tU(t f , ti)e−

ih H0t ; (1.25)

this operator is called the S matrix.S is the large time limit of U, the presentation of the time evolution in the interaction

picture. In general a state |ψ〉 evolves in the interaction picture as |ψt〉 = eih H0te−

ih Ht |ψ〉,

and an operator as A(t) = eih H0t Ae−

ih H0t. In a sense, it stands as an intermediate between

the Schrodinger and Heisenberg pictures, but it lacks their fundamental standing. Its mainuse is in the context of scattering-like problems and perturbation theory.

In light of the previous section the problem of finding the transition probability boilsdown to the calculation of a matrix element of the S matrix. In almost all cases it is impos-sible to calculate an exact closed form expression for S, and one must resort to approxima-tions. By far the most common approximation is based on assuming that the interaction Vis proportional to a small parameter (which will not be displayed explicitly) and calculat-ing S as a power series in this small parameter. Naturally this method has its limitations,the foremost of which is that V is not necessarily small, but a sophisticated application ofperturbation theory can at least mitigate these deficiencies.

The common starting point for perturbation theory is the integral form of the Schrodingerequation for U, ih ∂

∂tU(t) = V(t)U(t), namely

U(t) = 1− ih

∫ t

ti

dt′V(t′)U(t′) , (1.26)

which also implements the initial condition U(ti = 1) (the paramteric dependence of U onthe initial time ti is suppressed). Note that V is time-dependent, since it’s in the interaction

17

picture, even if V is not. The integral equation can now be iterated to yield the infiniteDyson series

U(t) = 1− ih

∫ t

ti

dt1V(t1) +( i

h

)2 ∫ t

ti

dt1

∫ t1

ti

dt2V(t1)V(t2) + · · · (1.27)

The nth order term in the series contains n factors of V ordered in increasing time fromright to left. It can be expressed more concisely using the time ordering meta-operator T,which acts to order the operators in the above mentioned order:

U(n)(t) =(− i

h

)n 1n!

T∫ t

ti

dt1 · · · dtnV(t1) · · · V(tn) . (1.28)

The entire Dyson series can be written formally as a time-ordered exponential

U(t) = T exp∫ t

ti

dt′V(t′) . (1.29)

The S matrix is obtained by letting ti → −∞, t → ∞.

c. Perturbation theory and Fermi’s golden rule To this point no approximation has beenmade and no assumption other than the Dyson series converges. Perturbation theory con-sists of approximating the series by a finite truncation. The calculation of a low-orderapproximation reduces to the calculation of an integral. The leading term is just the trivialS(0) = 1. The lowest order nontrivial term S(1) is linear in V.

We have seen in the case of transitions between discrete states that weak interactionscan have significant effects when they are resonant. This is also true for transition to a con-tinuum, so we will assume that the interaction matrix elements vary exponentially in timewith frequency ν; this includes of course the case ν = 0 of a stationary interaction. Takinginto account also the switching we make the replacement 〈 f |V(t) |i〉 → 〈 f |V |i〉 itI (t)e−iνt,where the new V is now stationary. The lowest-order approximation for the transitionamplitude is then

〈 f | S(1) |i〉 = − ih

∫dteiωte−iνti(tI)(t) 〈 f |V |i〉 (1.30)

(where again hω = ε f − εi). Since the V matrix element is time-independent, the timeintegration simply yields the Fourier transform of i, i.e.,

〈 f | S(1) |i〉 = −2πih〈 f |V |i〉 δ(tI)(ω − ν) . (1.31)

and the transition probability is

| 〈 f | S(1) |i〉 |2 = (2π

h)2| 〈 f |V |i〉 |2(δ(tI)(ω − ν))2 . (1.32)

Exercise 1.24 The perturbative argument leading to (1.32) is the same as the one leading to(??). Find the correct choice for Ω(t) in equation (??), so that the two calculations agree.

Our purpose is to use equation (1.32) to calculate transition probabilities as tI → ∞. AstI → ∞ δ(tI) tends to a Dirac delta function, but this does not mean that (δ(tI))2 → δ2, sincethe delta ‘function’ is only defined via its convolution with smooth functions. To see what(δ(tI))2 tends when tI is large consider therefore the integral I =

∫f (ω)(δ(tI))2(ω)dω =

18

Figure 5: The level scheme (left) and Feynman diagram (right) of a first-order resonant absorption process.

(2π)−1∫

f ∗ i(tI)(t)i(tI)(t)dt, where f and its Fourier transform are local and smooth func-tions and ∗ means convolution. Now, as tI → ∞, f ∗ i(tI) tends to the constant functionf (0), so that I ∼ tI

2π when tI is large, and therefore δ(tI)(ω)2 ∼tI→∞tI2π δ(ω).

As explained, since | f 〉 belongs to the continuous spectrum, for a finite tI the probabilitythat result of the transition is preciseely | f 〉 is, strictly speaking, equal to zero. Thereforewe should calculate the transition probability to a box of volume d f around | f 〉,

p|i〉→| f 〉 = 〈i| S† | f 〉 g( f ) 〈 f | S |i〉 d f , (1.33)

where | f 〉 g( f ) 〈 f | d f is the projection operator on the final state box. The density of statesfactor g( f ) is included to allow for non-canonical normalization 〈 f | f ′ 〉 = g( f )−1δ( f − f ′).The index f may stand for quantum numbers other than the energy, and when we want tomake this explicit we’ll write f = ε f , f .

Example 1.14 The density of states of a free particle in one dimension is obtained from〈ε |ε′ 〉 =

⟨p =

√2mε

∣∣∣p′ = √2mε

⟩= δ(

√2mε−

√2mε′) =

(m2ε

)−1/2δ(ε− ε′), whence g(ε) =(m

2ε

)1/2. Note that in one dimension a free particle is labeled by its energy and its movementdirection, so the result is expressed more accurately as g(ε, right moving) = g(ε, left moving) =(m

2ε

)1/2.

The concept of density of states can be understood when the states of the continuousspectrum are thought of as a limit of discrete states | f 〉V obtained by putting the systemin a large box. The the projection on the states in a box in f space is ∑ | f ′〉V V 〈 f ′|, wherethe sum is over states with f ′ in the box. If d f small then matrix elements (of smoothobservables) for different states in the box are close to each other, and we can approximatethe projection operator by n | f 〉V V 〈 f | = gV( f ) | f 〉V V 〈 f | d f , where n is the number ofstates of the box, and gV( f ) is the number of states in the box divided by its volume,i.e. the density of states. One would like to define g( f ) as the limit of gV( f ) as V → ∞,but gV diverges as V → ∞, so that the a finite limit is reached moving a divergent factorproportional to V from gV to | f 〉V obtaining the continuum versions g to | f 〉.

Exercise 1.25 Calculate the density of states of a free particle in three dimensions as afunction of energy and momentum solid angle Ω, first in the continuum, and then for aparticle in a cubic box of side L.

Combining equations (1.32) and equation (1.33), we see that first of all when tI getlarge, the transition probability increases linearly with it, so its reasonable to talk about a

19

transition rate wi f , rather than a transition probability. Secondly, the energy window of thetarget state becomes smaller, and at some point cannot be considered constant in the boxd f = dε f d f , so we have to integrate with respect to energy to get the actual transition rate.This yields finally

wi f = d f∫

dε f g(ε f )2π

h2 | 〈 f |V |i〉 |2δ(tI)(ω − ν) =2π

h| 〈 f |V |i〉 |2g(ε f = εi + hν)d f . (1.34)

This result is known as Fermi’s golden rule. It gives in principle a differential rate, whichreduces to a total rate if there are no continuous quantum numbers in the final state otherthan the energy.

Example 1.15 (The Born approximation) Consider a beam of particles with initial momen-tum ~pi scattered by a constant static potential V (which could stand for some static heavytarget). Using the Fermi golden rule and the density of states of free particles it followsthat the rate of scattering into a solid angle element dΩ around momentum ~p f (with

|pi| = |p f | = p) is 2πh |Vi f |2mpdΩ, with Vi f =

∫ drh3 V(~r)e

ih (~p f−~pi)·r. In scattering it is cus-

tomary to measure the scattering rate per incoming flux, given by density× speed = pmh3 ,

which yields an approximation for the differential scattering cross section

dσ

dΩ= (2π)4h2m2|Vi f |2 . (1.35)

Despite its simplicity the golden rule is quite useful. However, one should be awarethat it relies on several assumptions, beside the obvious weakness of the interaction, thatlimit its applicability. First, although we have taken the large time limit, it is clear that theformula cannot be strictly true for arbitrarily long times, since at some point we’ll reach aprobability larger than one! A simple escape from this problem can be raised, though: Asthe probability is transferred to | f 〉, the probability of |i〉 must accordingly diminish, so,assuming that there are no back transitions, one can say that the actual rate is the one givenby (??) times the actual probability of |i〉. This is actually correct, but to show it we’ll needthe methods of section ?? that effectively sum an infinite number of terms from the Dysonseries.

The golden rule cannot hold for very short times either, since we know that then thetransition probability is proportional to t2

I rather than tI (see equation (??)). The reason isthat then the energy uncertainty h/tI is so large that one cannot view the density of statesin the energy window as constant anymore. On the other hand, when the interactions arequite weak, the transition probability in this time interval is anyhow very small, so that onedoes not need to worry about it.

d. Higher orders and Feynman diagrams An obvious direction to overcome the limita-tions of the golden rule, is to go further in the Dyson series. We’ll see that this cannot curethe basic problems, and even exposes a new one. Still, higher order terms are useful notonly as an improvement on the leading term, when the latter is qualitatively correct, butcan sometimes cause transitions that are not allowed in the leading approximation.

Using a resolution of the identity with eigenstates of H0, the second order term in theDyson series for S is written as

〈 f | S(2) |i〉 = ∑s

( ih

)2 ∫ ∞

−∞dt1

∫ t1

−∞dt2 〈 f |V |s〉 〈s|V |i〉 i(tI)(t1)i(tI)(t2)ei(ω f−ωs−ν)t1ei(ωs−ν−ωi)t2 ,

(1.36)

20

Figure 6: The level scheme (left) and Feynman diagram (right) of a resonant double-absorption process

where as usual the sum over s includes integration over states in the continuous spectrum.If ν is positive, the amplitude describes a double-absorption transition.

Before proceeding further with the analysis of (1.36), it is good to recast in the form aFeynman diagram, figure 6. It describes the amplitude term as a punctuated free evolution:The system evolves from the early past in state |i〉, until time t2, where as a result of theinteraction it passes to state |s〉 until time t1, whereupon it makes a transition to | f 〉, andcontinues in this state until late future. The diagram is interpreted according to the (real-time) Feynman rules: Assign an incoming line of state |s〉 ending at t a factor of e−iωst, anoutgoing line starting at t a factor of eiωst, and an internal line (propagator) from t to t′ afactor eiωs(t−t′); assign an interaction vertex a matrix element between ingoing and outgoingstates; sum over all internal line states and integrate over all interaction times. The last stepis an example of Feynman’s ‘sum over histories’ concept. Feynman then dictates that oneshould also some over all diagrams.

Feynman diagrams are a convenient tool to represent terms in the perturbation series forS; naturally the 1st order term calculated above is represented by a single vertex diagram,see figure 5. Once the relevant Feynman diagrams have been drawn, the correspondingterm follows by applying mechanically the Feynman rules (although the actual calculationof the integrals can be a nontrivial task). On the other hand, while the Feynman diagramslend a vivid physical sense to the terms of the perturbation series one should keep in mindthat many physical processes are not well-described by perturbation theory. Calculation-ally, Feynman diagrams become really useful in more complicated perturbation theories,including loops (probably will not be touched upon in these notes).

As in the calculation of the 1st order amplitude, the matrix elements factor out, and weare left with a time integral of known function. However, the integral is made more com-plicated by the constraint t1 > t2, which can be expressed also by writing a step functionθ(t1 − t2) in the integrand, and letting both time integrations run from −∞ to ∞.

Exercise 1.26 Calculate the Fourier transform of the step function:

θ(t) = − limη→0+

∫ ∞

∞

dω

2πie−iωt

ω + iη. (1.37)

Note: We will use the custom of letting ‘η’ stand for an infinitesimal positive quantity, andusually omit the ‘lim’.

21

The amplitude is then reexpressed as

〈 f | S(2)I |i〉 = ∑

s

( ih

)2〈 f |V |s〉 〈s|V |i〉

∫ dω

2πi1

ω + iη

×∫

dt1dt2i(tI)(t1)i(tI)(t2)ei(ω f−ωs−ν)t1e−iω(t1−t2)ei(ωs−ωi−ν)t2 , (1.38)

which after carrying out the time integrals becomes

〈 f | S(2)I |i〉 = −∑

s

(2πih

)2〈 f |V1 |s〉 〈s|V2 |i〉

∫ dω

2πi1

ω + iη

× δ(tI)(ω f −ωs −ω − ν)δ(tI)(ωs + ω −ωi − ν) . (1.39)

The δ(tI) functions limit the frequency integration to an interval of width ∼ htI

aroundzero; if the function 1

ω+ih changes slowly enough in the interval, i.e. if ω f − ωs − ν ∼ν + ωi −ωs h

tI, we replace ω by the constant ωi −ωs + ν, getting

〈 f | S(2) |i〉 = −2πi ∑s

〈 f |V1 |s〉 〈s|V2 |i〉εi − εs + hν2 + iη

δ(tI) ∗ δ(tI)(ε f − εi − 2hν) , (1.40)

where the star stands for convolution. The convolution of δ(tI) with itself is also a narrowselection function, which converges to the Dirac δ when tI → ∞.

We can now formulate the energy domain Feynman rules. As before, each vertex isassigned a corresponding matrix element Vrs, and each internal line with state |s〉 carryingenergy ε a factor 1

ε−εs. Energy is conserved in each vertex including the contribution of the

frequency of the interaction, but note that the energy of the internal line is independent ofits state. In quantum field theory one says that the states in the internal line are ‘off themass shell’ or ‘virtual’. The whole expression is multiplied by −2πi(δ(tI))∗n(ε f − εi − nhν),where (δ(tI))∗n is the nth order convolution of δ(tI) with itself.

The translation of the transition amplitude to a transition rate is entirely similar tothe one used above to derive the golden rule. In general one obtains in the nth or-der an expression for the T matrix (transition matrix) amplitude defined by 〈 f | S(n) |i〉 =−2πi 〈 f | T(n) |i〉 δ(ε f − εi + nhν), and the transition rate generated by this process is then2πh | 〈 f | T(n) |i〉 |2g(ε f ). Caution: As usual when contribution of several processes are com-

bined, amplitudes should be summed rather than probabilities or rates.

Exercise 1.27 Calculate the differential scattering cross section in a static potential V tosecond order.

As warned above, the 2nd order approximation to the transition rate suffers from thesame short-time and long-time limitations as the leading approximation, while addinganother problem when ωs ∼ ωi + ν = ω f − ν, i.e., when there is a resonant intermediatelevel. Obviously in this case we should also consider transition to the intermediate level.In the second part of the course we’ll discuss a general scheme to handle this situation,while in the last section of this part we’ll consider the case where the transition to theintermediate level can be handled in perturbation theory.

e. The resolvent (Green’s function) In the last two sections of this part we will examinewhat happens to a discrete level coupled by the interaction to a continuum. I.e. we willassume that the eigenstates |n, ε〉, depend on a discrete label n and a continuous label ε,

22

and that the ‘unperturbed energy’ is εn + ε (see figure 7). For the sake of simplicity wewill assume that the interaction is stationary, so it could be internal (example: coupling ofcharged matter and electromagnetic radiation). We know from the previous analysis thatwhen there are discrete levels below |n〉, it may decay irreversibly, transferring its energyto the continuum. Our purpose here is to study this process: It will be shown that thedecay rate is approximately exponential, and furthermore that the energy level is shiftedand broadened.

23

Figure 7: The level scheme of a two-degree-of-freedom system. Without interaction (top panel), one degreeof freedom is associated with discrete levels, and the other with a continuous spectrum. The spectrum of thetotal system (still without interaction, middle panel), consists of discrete levels embedded in the continuum.The discrete levels are unstable with respect to a perturbation which couples the types of degree of freedom,whose presence turns the spectrum to a pure continuum (bottom panel). The former presence of the discretelevels is felt by the sharp increase of the density of states

24

The resolvent of the Hamiltonian (or of any other operator for that matter) is G(z) =1

z−H . G is an operator valued function of the complex variable z. Evidently G has singu-larities on the spectrum of H, but since H is self-adjoint, G(z) is a nice analytic functionoff the real axis. There are many reasons to study the properties of an operator through

its resolvent; we will use the fact that θ(t)e−ih Ht = − 1

2πi

∞∫−∞

dεe−ih εtG(ε + iη − H), which

follows from the Fourier transform of the step function formula (1.37). Another importantproperty of the resolvent is that its imaginary part gives the density of states.

We will be interested in the matrix elements of G in a small set of states H0 eigentstates|1〉 , . . . , |n〉, and therefore focus our attention on PGP, where P = ∑j |j〉 〈j| is the orthogonalprojection on the subspace spanned by our states of interest. Our aim is to develop aperturbation scheme for this operator. Define also the complementary projection operatorQ = 1− P.

Exercise 1.28 1. An orthogonal projection P is by definition an self-adjoint idempotentoperator; i.e., P = P† = P2. Show that the operator P defined above is indeed anorthogonal projection.

2. Show that Tr P = n. It is called the rank of the projection.

3. Show that Q is also a projection and that QP = PQ = 0.

4. Show that P and Q commute with H0

The first step in approximating PGP is to multiply the definition (z − H)G(z) = 1 byQ and P on the left and the right, respectively, which after a resolution of the identity1 = P2 + Q2 gives QG(z)P = Q

z−QHQ VPG(z)P. Next we multiply (z − H)G(z) = 1 byP on both sides, resolve the identity in the same manner, and use the expression derivedQG(z)P to get PG(z)P = P 1

z−P(H0+R(z)) P, where the level shift operator

R(z) = V + VQ

z−QHQV . (1.41)

A comparison of the expressions of G(z) and PG(z)P shows that R(z) acts as (z-dependent)interaction restricted to the P-range subspace. Unlike the full Hilbert space, the subspaceis an open system, and R(z) is in general non-Hermitian. It is naturally expanded pertur-batively by iteration

R(z) = V + V1

z− H0QV + V

1z− H0

QV1

z− H0QV + · (1.42)

f. Broadening and decay of an unstable state We will now specialize to the case wherethe subspace consists of a single state |s〉. Then PG(z)P = PGs(z), where Gs = 〈s|G |s〉 =

1z−εs−Rs(z) . As usual we will assume that the diagonal matrix elements of V vanish 14, sothat the leading nonzero term in Rs is

Rs(z) = ∑s′ 6=s

| 〈s|V |s′〉 |2z− εs′

. (1.43)

Our aim is to use G(z) to calculate the forward evolution operator. For this purpose weneed to find G for z values just above the real axis.

14For a stationary perturbation this can always be accomplished by absorbing the diagonal matrix elementsof V into H0

25

Exercise 1.29 Show that (for ε real) 1ε+iη = P1

ε − iπδ(ε), where P stands for ‘Cauchy prin-cipal part’, by (a) integrating with a test function and taking η → 0+, and (b) complexvariable methods.

In this case the real and imaginary parts of the diagonal R matrix element are Rs(ε + iη) =h∆s(ε) + i h

2 Γ(ε), where

∆s(ε) = P1h ∑

s′ 6=s

| 〈s|V |s′〉 |2ε− εs′

, Γs(ε) =2π

h ∑s′ 6=s

| 〈s|V∣∣s′⟩ |2δ(ε− εs′) . (1.44)

It follows that Γs(ε) = 2πh

∫ds′| 〈εs s|V |εs′〉 |2ρ(ε, s′), and in particular if ε does not belong

to the continuous spectrum Γs(ε) = 0. Note that Γs is always positive, ensuring causality,and allowing us to drop the iη term from the denominator.

The approximation for Gs obtained from (1.43) is

Gs(ε + iη) =1

ε− εs − h∆s(ε) + i h2 Γs(ε)

. (1.45)

Exercise 1.30

1. Show that our previous perturbation scheme based on the Dyson series is equivalentto approximating G by terms from its geometric series G = 1

z−H0+ 1

z−H0V 1

z−H0+ · · ·

2. Using Feynman diagrams, show that the last approximation obtained for Gs is a sumof an infinite number of terms from all orders of the Dyson series. It is accordinglycalled a partial resummation of the series.

When Γs is small (and it should be, since we assume that the interaction is weak), Gs(ε)changes very rapidly as ε changes from values below εs and above εs, describing a resonancein the transition amplitude as a consequence of the nearness of the metastable state. At thesame time, ∆s and Γs experience no special changes. They may therefore be approximatedas constant in the interval where Gs is appreciable. In this approximation

Gs(ε + iη) =1

ε− εs − h∆s + i h2 Γs

, (1.46)

where ∆s, Γs = ∆s(εs), Γs(εs).Effecting the inverse Laplace transform then gives

Us(t) = e−ih (εs+h∆s)te−

12 Γst . (1.47)

h∆s(z) indeed expresses the shift in the energy of level as a consequence of the interaction.It is the continuous spectrum analog of the shifting of a level in 2nd order perturbationtheory. In relativistic quantum electrodynamics this term is nknown as the Lamb shift, firstdiscovered experimentally, whose theoretical explanation (first by Bethe) was a major steptoward the theory of renormalization

On the other hand Γs gives the decay rate of the level as a result of its coupling to thecontinuum, a quantity that has no analog in discrete spectrum perturbation theory. Itconfirms the hand-waving arguments based on the golden-rule transition rate, namely thatas the initial level becomes depleted, it continues decaying at a constant relative rate, solvingthe large time problem. On the other hand, we still can’t use the equation (1.47) for veryshort times, since we know that the initial transition rate is quadratic rather than linear

26

in time. This inaccuracy is the result of approximating Γs by a constant, that would onlybe exact if the continuum were infinitely broad. In truth, the energy spectrum is alwaysbounded from below, and as a result equation (1.47) breaks down for sufficiently shorttimes. However, when the interaction is weak, the decay probability is anyhow very smallin these short times, and the validity of equation (1.47) is sufficient.

Finally, let us notice that equation (1.46) offers a way to improve also higher ordertransition probability perturbation theory, by replacing the internal line bare propagator

1ε+iη−εs

with Gs, avoiding the resonant divergence as long as V hΓs.

2 Interactions of matter and classical electromagnetic radia-tion

2.1 The quantum mechanics of charges and classical electromagnetic ra-diation

a. Brief review of classical electromagnetic theory Electromagnetism in a continuousmedium involves four vector fields, the electric field ~E, the electric displacement ~D, themagnetic induction ~B and the magnetic field ~H. They evolve according to Maxwell’s equa-tions (SI units will be used exclusively in these notes)

∇ · ~D = ρ ∇× ~E = −∂t~B

∇ · ~B = 0 ∇× ~H =~j + ∂tD(2.1)

where the charge density ρ and current density ~j act as sources. The medium polarization~P and magnetization ~M are defined by D = ε0E + P and B = µ0(H + M).

The Maxwell equation have to be supplemented by constitutive relations P = P[E]and M = M[H]. In this course we will assume that M = 0. The functional P[E] is ingeneral nonlinear and nonlocal in time. If the medium is linear we have by time translationinvariance P(t) =

∫ t−∞ χ(t − t′)E(t′)dt′, or in the frequency domain, P(ω) = χ(ω)E(ω),

where χ is the electric susceptibility tensor.

Exercise 2.1 Show that the Fourier transform f (ω) of a function f (t) supported on thepositive values obeys the Kramers-Kronig relation Im f (ω) = 2ω

π P∫ Re f (ω′)

ω2−ω′2 , Re f (ω) =

− 2πP∫

dω′ω′ Im f (ω′)ω2−ω′2 . Hint: You can Fourier transform the identity f (t) = θ(t) f (t).

Hence the real and imaginary parts of χ obey the Kramers-Kronig relations. One definesthe frequency-dependent dielectric constant ε(ω) = ε0 + χ(ω). Then Maxwell’s equationhave solutions of the type ~E(r, t) = 2 Re ~E0ei(k·r−ωt) where ω(k) = c

n k, cn is the phase velocity

of light in the medium and n = √εµ is the index of refraction. Note that the imaginary

part of ε corresponds to amplification or attenuation. In passive media only attenuation ispossible (see below)

Maxwell’s equations allow us to write the fundamental fields in terms of potentaials,E = −∇φ − ∂t A, B = ∇ × A. A crucial property of the potentials is gauge invariance,namely, the physical fields remain unchanged under a gauge transformation φ → φ + ∂tλ,A → A−∇λ for an arbitrary scalar function λ(~r, t).

Finally we have to prescribe the motion of charges in electromagnetic fields by theLorentz force eE + ev× B acting on a charge e moving with velocity v. The classical Hamil-

27

tonian which generates these equations of motion is

H =(~p− eA(~r))2

2m+ eφ(~r) . (2.2)

Note the canonical momentum p is not the usual kinetic momentum, rather p = mv + eA,so that p is gauge-dependent, and that H is not equal to the energy when A or φ aretime-dependent, even when E and B are.

b. Quantization and gauge transformations In this part we will mostly consider the elec-tromagnetic field as given, focusing on the dynamics of quantum charges. For this purposewe need to quantize the classical Hamiltonian (2.2). The quantization is not as obvious,because, since the classical Hamiltonian has terms that are products of momentum andfunctions of positions, there are many quantum Hamiltonians that have it as a classicallimit. Still the following points in the right direction:

Exercise 2.2

1. Show that the only self-adjoint quantum Hamiltonian that has (2.2) as its classicallimit that (a) has no explicitly h-dependent terms, and (b) contains only integral pow-ers of A and p is (2.2) with~r and ~p interpreted as quantum observables.

2. Give an example of a different quantum Hamiltonian that does not satisfy (b)

The final word comes from the lab, of course, and it confirms that (2.2) indeed describesthe dynamics of quantum charges in external electromagnetic fields.

Fixing the ordering, however, does not resolve completely the arbitrariness in H, sinceit depends on the gauge-dependent potentials. In classical mechanics we know that thedynamics is gauge-invariant because the Hamilton (i.e. Lorentz) equations depend only onthe fields E and B. In quantum mechanics different gauges can be shown to be equivalentphysically if they are connected by a unitary transformation (passive symmetry transfor-mation)

Exercise 2.3 Let U = eih eλ(~r,t). Show that rλ = UrU† = r, pλ = UpU† = p− e∇λ.

This exercise shows that U implements a gauge transformation with generator λ. Since|ψλ〉 = U |psi〉 it acts by multiplying the wave function of every particle by position- (andcharge-) dependent phase factor.

When the gauge transformation is static the Hamiltonian is simply transformed byletting p → pλ. When λ depends on time, the Hamiltonian is also time-dependent, and isno longer the energy observable, so its transformation becomes a little more complicated.The general transformation rule can be deduced from the calculation ∂tOλ = i

h [Hλ, O′],where Hλ = UHU† − ihUU†. Hλ that describes dynamics in the new gauge is indeedobtained from H by replacing the original potentials by the gauge-transformed ones.

It is interesting to make two remarks here: Firstly, it is possible to make a gauge transfor-mation Uλ on a system that is not in the presence of electromagnetic field. Then we wouldof course get a Hamiltonian with electromagnetic potentials that are ‘pure gauge’. Thepoint is that while this seems perverse in classical physics, it looks quite natural in quan-tum physics since we know that both gauges are legitimate, and motivating the existenceof gauge potentials. This is one of the cases where quantum physics helps us understandclassical phenomena. Secondly, it might seem that the quantum gauge transformationsmakes the potentials void of direct physical significance like in classical mechanics, but this

28

is false. It may happen that one cannot remove the ‘pure gauge’ potentials in field freeregions because of topological obstructions. In this case one can measure quantum phaseassociated with line integrals of the vector potential—the Aharonov-Bohm effect.

c. The dipole approximation Let us consider now a neutral system comprised of severalcharges interacting electrostatically (e.g. an atom or molecule) in the presence of classicalelectromagnetic radiation. The quantized Hamiltonian is then

H = ∑n

(p2n − en A(rn, t))2

2mn+ ∑

n 6=m

enem

4πε0|rn − rm|+ ∑

nenφ(rn, t) . (2.3)

We would like to consider the electrostatic system as the ‘free’ system, and the radiationas a perturbation. As a first step we will assume that the typical length scale of variationsin the radiation field (i.e. the wavelength) is much larger than the distances between thecharges. This approximation is very good for atoms interacting with visible light. Then wemay approximate the fields by their value at the position of the atom, which we may placeconveniently at the origin, and get

H = ∑n

(p2n − en A(0, t))2

2mn+ ∑

n 6=m

enem

4πε0|rn − rm|+ ∑

nenrn · ∇φ(0, t) . (2.4)

This the second term in a large wave length expansion of the Hamiltonian. The leadingterm (monopole), proportional to φ vanishes because of charge neutrality.

Exercise 2.4 Write down the monopole term for a charged system, and show that it can beeliminated by a gauge transformation with a constant gauge function λ.

In atomic physics one considers the static part of the Hamiltonian

H0 = ∑n

p2n

2mn+ ∑

n 6=m

enem

4πε0|rn − rm|(2.5)

as the free Hamiltonian, regarding the terms proportional to A and φ as interaction. Thisdecomposition can then be used as a starting point for perturbation theory for transitionsbetween atomic levels induced by the external fields. However, it is possible to simplifythe form of the interaction Hamiltonian by a gauge transformation. In effect, the vectorpotential can be entirely ‘gauged away’, by choosing the gauge function λ(r, t) = A(0, t) · r,after which the Hamiltonian becomes

H = ∑n

p2n

2mn+ ∑

n 6=m

enem

4πε0|rn − rm|− ~d · ~E(0, t) , (2.6)

where ~d = ∑n en~rn is the system dipole moment, and E = −∂t A −∇φ is the electric field,constant in this approximation.

The dynamics of the Hamiltonians (2.4) and (2.6) are equivalent, as they are connectedby a unitary transformation, but the second form is more convenient since: (i) It dependsonly on the physical field E, and as a consequence the canonical momentum is equal to thephysical momentum. (ii) The interaction term is much simpler, as it contains a single term,linear in the electromagnetic field, rather than four, one of which is quadratic.

29

Exercise 2.5 The Hamiltonians (2.4) and (2.6) are both of the form H0 + V, where H0 isgiven by (2.5), but with different interaction Hamiltonians, yet both should give the sametransition amplitudes. Show that the first order perturbation theory transition amplitude〈 f |U(1)(ti, t f ) |i〉 is the same in both gauges. Hint: The H0 eigenstates are not the same inboth gauges. The gauge transformations of U(1) and the eigenstates cancel each other inthe final answer

2.2 Transitions induced by electromagnetic radiation

a. Stimulated emission and absorption in monochromatic radiation Consider now twolevels |1〉 and |2〉 of the atomic Hamiltonian H0, with energies ε1 and ε2, ε2 − ε1 = hω, andsuppose that 〈2| ~d · E |1〉 = hΩe−iνt, where ν ∼ ω and the interaction is not resonant withany other atomic transition. If E contains other frequencies, we resort to the rotating waveapproximation and neglect the non resonant terms in the interaction.

Exercise 2.6 Show that the matrix element has exactly this form when E is the electric fieldof circularly polarized light. Write down the selection rule relating the angular momentumeigenvalues of |1〉 and |2〉. Hint: Use the Wigner-Eckart theorem.

The electromagnetic field induces a transition from |1〉 to |2〉, whose rate according tothe golden rule is w12 = 2π

h | 〈2| ~d · ~E0 |1〉 |2g(ε1 + hν). w12 is the rate of absorption (orstimulated emission if ω < 0) of coherent radiation. Suppose now that (as usually happensin applications) the radiation is affecting a large population of atoms, of which n1 are instate |1〉 and n2 in |2〉, so that absorption/stimulated emission generates a population fluxw12n1 from |1〉 to |2〉. We know however that a necessary condition for the validity of thegolden rule is that the width Γ2 of the level |2〉 is significantly larger than the the transitionrate, so there is also a depletion rate Γ2n2 of level |2〉. Assuming the Lorentzian density ofstates of a narrow resonance g2(ε) = 1

πhΓ2/2

(hΓ2/2)2+(ε−ε2)we get the rate equation

n2 = −Γ2n2 +Ω2(Γ2/2)

(Γ2/2)2 + (2δ)2 n1 , (2.7)

where, as before δ = 12(ν− ω) is the detuning of the radiation from resonance. Similarly,

for n1

n1 = λ1 − Γ1n1 −Ω2(Γ2/2)

(Γ2/2)2 + (2δ)2 n1 , (2.8)

where λ1 is a flux of atoms to |1〉 from other sources, and we have allowed for a broadeningΓ1 Γ2 of |1〉. It has been assumed for simplicity that there is negligible spontaneoustransition rate between |1〉 and |2〉.

Exercise 2.7 Assume λ1 = Γ1 = 0, and consider the effect of a broad Gaussian pulse with

electric field E(t) = E0e−iνte−t2

2τ2 + non-resonant, with τ Γ−12 . Calculate the depletion of

the source state n1(+∞)− n1(−∞).

If the pumping λ1 is constant, the population in the source level tends to the steady statevalue

n1 =λ1

Γ1 + w12. (2.9)

This population then acts as a constant source term for n2, that also reaches a steady state

n2 =w12n1

Γ2. (2.10)

30

Note that the condition of validity Γ2 Ω of the golden rule implies that n2 n1 consis-tently with the physical assumptions.

b. The optical Bloch equations The transition dynamics derived in the last section isquite different from that of two discrete levels as obtained in section ??, and the sources ofthis difference have been discussed above: The golden rule describes transitions to a broadcontinuum, where dephasing makes the transition practically irreversible, while Bloch os-cillations occur when the levels are genuinely discrete. However, in practice the situationis often intermediate: The levels are sharp enough that coherent dynamics is possible, butdephasing and decay are not negligible either. We will not develop a first principles analy-sis of this case (it is possible though), but use a phenomenological combination of the twolimiting cases.

We’d like to reconsider the dynamics of the probability amplitudes, but this time forensemble of atoms as in the last section. In this case we have to study the density matrixρ. Following the convention in this field, we’ll let ρ stand for the population matrix, thatis the density matrix multiplied by the number of particles. It has the advantage that itsdiagonal elements are mean populations.

Equation (??) give the ideal Rabi dynamics of two coupled narrow states, while equation(??) gives the decay dynamics of levels. Making the phenomenological assumption that thetwo effects contribute additively to the Schrodinger equation we get

α = (−iω1 − 12 Γ1)α− iΩeiνtβ

)β = −iΩe−iνtα + (−iω2 − 1

2 Γ2)β)

,

where now hΩ = 〈1| ~d · ~E0 |2〉. These equations imply the dynamics of the populationmatrix ρ11 = N|α|2, ρ12 = ρ∗21 = Nαβ∗, ρ22 = N|β|2,

ρ11 = λ1 − Γ1ρ11 − 2Ω Im(e−iνtρ12) (2.11)

ρ22 = λ2 − Γ2ρ22 + 2Ω Im(e−iνtρ12) (2.12)

ρ12 = (iω − γ)ρ12 − iΩeiνt(ρ22 − ρ11) . (2.13)

where, as always, ω = ω2 − ω1, γ = 12(Γ1 + Γ2), and external flux terms λ1,2 have been

added in the spirit of the last section. Note that Tr ρ is not constant because of the couplingto levels other than |1〉 and |2〉.

The nondiagonal matrix elements of ρ express quantum correlations, so that γ thatmeasures the rate of their decay, is a quantum decoherence rate. It often happens that thedominant dephasing process is not spontaneous decay as assumed above, and then γ canbe significantly larger than 1

2(Γ1 + Γ2). 15 Accordingly we’ll consider γ a free parameterbounded from below by 1

2(Γ1 + Γ2). The inverse of γ is often labeled as T2, as comparedwith T1, the time scale for spontaneous transition between |1〉 and |2〉. Roughly speaking,processes that occur on time scales larger than T2 can be described approximately classi-cally.

c. The rate equations, stationary solutions, and saturation. While the diagonal popula-tion matrix elements are the level populations n1,2, the off-diagonal element is related to themedium polarization, defined as the overall dipole moment. Assuming, on the basis of par-ity invariance, that ~d has no diagonal matrix elements, ~P = 2 Re(〈2| ~d |1〉 peiνt), where the

15A common source of strong dephasing in gases is random Stark shofts caused by elastic collisions.

31

polarization amplitude p = e−iνtρ12. In terms of these variables the optical Bloch equationbecome

n1 = λ1 − Γ1n1 − 2Ω Im p , (2.14)n2 = λ2 − Γ2n2 + 2Ω Im p , (2.15)p = −(2iδ + γ)p− iΩ(n2 − n1) . (2.16)

Qualitatively speaking, these equations describe damped Rabi oscillations, but when thecoefficients are time-dependent, obtaining a general solution is hard.

The demand that the λ’s and Γ’s are zero from the last exercise is actually too strong.Since p is damped with time scale T2 = γ−1, which is often much smaller than the othertime scales in the system, in particular the rate of change of the parameters, then we canassume that p reaches a steady state value

p =iΩ(n2 − n1)

2iδ + γ, (2.17)

and upon substitution in (2.14–2.15) we get

n1 = λ1 − Γ1n1 +2Ω2γ

γ2 + (2δ)2 (n2 − n1) , (2.18)

n2 = λ2 − Γ2n2 −2Ω2γ

γ2 + (2δ)2 (n2 − n1) . (2.19)

These rate equations are remarkably similar to those of absorption/stimulated emission,the only difference being that γ replaces Γ2 in the transition rate w12. It follows then thelatter are valid if Γ2 Γ1 and is also the dominant dephasing mechanism. If the systemparameters change slightly on time scales Γ−1

1 , Γ−12 (or are constant), the populations also

reach a steady state. A simple calculation shows that the steady state values for the totalpopulation n, and the population difference r = n2 − n1 are

n =(Γ+ + 2w12)λ+ − Γ−λ−(Γ+ + 2w12)Γ+ − Γ2

−

r =Γ+λ− − Γ−λ+

(Γ+ + 2w12)Γ+ − Γ2−

,(2.20)

where Γ± = 12(Γ2 ± Γ1), λ± = λ2 ± λ1. These expressions have two limiting behaviors:

When w12 Γ+, the external radiation is too weak to produce a significant change in thestationary population n1 = λ1/Γ1, n2 = λ2/Γ2, obtained without interactions. In this casethe transition is unsaturated. When w12 Γ+ on the other hand, the populations are com-pletely mixed by the interaction so that n tends to λ+/Γ+ and r tends to zero, saturatingthe transition. If the dissipative coefficients are all of the same order of magnitude, thedegree of saturation is controlled by the saturation parameter s = Ω2

δ2+γ2 . This definitiongeneralizes the one obtained for ideal two-level dynamics. Saturation is achieved whens 1 i.e. when Ω is much larger than both γ and δ. This point will be retaken in section??.

Exercise 2.8 Consider again the response of the two level system to a Gaussian pulse as inexercise ??, with λ1 = λ2 = Γ1 = 0, and calculation the depletion of n1 for (a) Γ2 τ−1 γand (b) τ−1 Γ2, γ. Hint: In the case (b) the pulse dynamics is close to the ideal dynamics.

32

Exercise 2.9

1. Use the expressions for ~P in terms of p, to calculate the electric susceptibility χ(ν) insteady state when s 1 and show that it obeys the Kramers-Kronig relation.

2. The definition of electric susceptibility is sometimes modified phenomenologically toP(ν) = χph(ν)E0 for external field of the form we have been considering. Show thatχph does not satisfy Kramers-Kronig and explain why.

d. Interaction with incoherent broadband radiation The external radiation consideredso far has been coherent, meaning that it was either monochromatic, or compressed toa short pulse of duration inverse to its spectral width. Such radiation is coherent in thesense that there exists a well defined phase relation between the phase of electromagneticfield throughout the whole wave form. However, although this the case most useful forapplication, actual light sources are never ideally coherent, and after a certain coherencetime tc phase correlations decay fast, so that the phase relation between parts of the waveform separated by a time interval larger than tc is practically random. This is especiallytrue for traditional thermal light sources where tc is of the order of atomic time scales.

The incoherent counterpart of a short coherent pulse of duration τ is a stationaryelectromagnetic field with a coherence time τ. The mean intensity of this radiation isconstant, so its temporal structure is characterized by its autocorrelation function f (t) =

limT→∞1T

T/2∫−T/2

E(t′)E(t + t′)dt′. For a monochromatic wave f (t) is constant, and it is zero

for pulse wave, but for an incoherent source f (t) is a smooth function that decays to zerowhen t tc, and random phase contribution of E(t′)E(t′ + t) cancel in the integration. TheFourier transform of the autocorrelation function is the power spectrum f (ω), that gives

the ω contribution to the mean value of E2 =∞∫

−∞f (ω) dω

2π .

When a two-level system is exposed to incoherent radiation, the transition amplitudeobtained by properly modifying (??) is

〈2| S(1) |1〉 =ih

∫dteiωti(tI)(t) 〈2| ~d · E(t) |1〉 . (2.21)

The transition probability obtained by squaring the amplitude, and performing the timeintegrations (using the defining properties of the switch function) is

| 〈2| S(1) |1〉 |2 =| 〈2| ~d |1〉 |2

h2 f (ω)tI . (2.22)

As in transitions to a broad continuum, the transition probability grows linearly with theinteraction period, indicating a constant transition rate

w12 =| 〈2| ~d |1〉 |2

h2 f (ω) . (2.23)

The rate is the product of a material factor |〈2|~d|1〉|2h2 with the radiation power at the reso-

nant frequency ω. The material constant is close related to the Einstein stimulated emis-sion/absorption coefficient B, to be touched upon again below in part 3. Usually it is as-sumed that the polarization as well as the phase is random, and in that case w12 is smallerby a factor of 3.

33

The main qualitative difference between the transition rate in incoherent radiation andthe golden rule is that spectral density of the radiation replaces the energy density of thetarget state. This is because of the implicit assumption made in deriving (2.23), that theradiation is spectrally so broad that the target state can be approximately treated as discrete.

Exercise 2.10

A model for incoherent radiation is E(t) =∫ ∞−∞ E(ω)e−iωt dω

2π , where the amplitudes E(ω)are independent centered random variables. Calculate the autocorrelation function andshow that 〈E(ω)∗E(ω′)〉 = 2π f (ω)δ(ω −ω′).

2.3 Gain in an active medium

a. Homogeneous broadening and gain saturation So far, in the interaction between clas-sical light and quantum matter, we have focused mainly on the state of matter. Strictlyspeaking, it is not possible to discuss the state of light in this framework, since by interact-ing with a quantum system it acquires the quantum nature itself. Nonetheless, one mayapproximate the back action of the quantum matter on the classical by first approximat-ing the quantum observables by their expectation values. This uncontrolled approximationis justified for the current purpose of studying the energy flow between classical radia-tion and an ensemble of atoms, but one should keep in mind that in some situations this‘semiclassical’ approximation can break down completely.16

Let us first recall that the electromagnetic energy density is U = 12(εE2 + µH2), and that

the power density absorbed by a polarized medium is W = ~E · ∂∂t

~P 17 The incoming beamenergy is usually expressed by its intensity (average energy flux) I = c

nU,where n is therefractive index at the appropriate frequency. I = 1

2 cnε0|E0|2, for a plane wave of amplitudeE0.

Using the definition of the polarization amplitude p and the Rabi frequency, and the factthat the optical frequency is much larger than the rate of change of p, the energy absorbedby the resonant medium is

W = −2Ω Im p(hν) , (2.24)i.e., 2Ω Im p photons per unit time. In the rate equation approximation this becomes

W =2Ω2γ

γ2 + (2δ)2 (n1 − n2)(hν) = w12(n1 − n2)hν . (2.25)

As pointed out above, on time scales larger than T2, the two-level dynamics can be under-stood on classical terms. Here the process is w12(n2 − n1) transition per unit time, whoseconsequence is the absorption of the same number of photons from the incoming beam.

It is clear that (supposing ε2 > ε1) amplification will occur only if n2 > n1, i.e. there is apopulation inversion. As expected, population inversion will not occur due to the action ofthe resonant beam, and must be created by other means, modeled by the pumping termsλ, that are beyond our scope.

The action of the medium on the beam can be modeled by adding a rate equation forthe beam intensity. Using the proportionality I = aΩ2, a = 1

2 cnε0| 〈2| ~d |1〉 |2, we get

I = (g− l)I = (cn

1a

2γ

γ2 + (2δ)2 (n2 − n1)(hν)− l)I , (2.26)

16That is not to say that a systematic controlled semiclassical approximation cannot be carried out, but it isbeyond the present scope.

17Since we are studying the transfer of energy from the electromagnetic radiation to the atoms and viceversa, we include in the electromagnetic energy contributions from ‘background’ non-resonant term in theelectromagnetic energy, and only resonant terms in the energy transfer expression.

34

where l is a phenomenological loss coefficient, and g is the intesnity-dependent amplifiergain. Together with equations (2.18–2.19) it form a closed system of equation describing(approximately) the matter-beam system. A system with a population inversion can func-tion as an amplifier for a beam that is just passed through it, while if put in a cavity canactivate a laser.

If the beam time scale is slower than that of the matter, or if the system reaches a steadystate, we can use the steady state value r to calculate the gain coefficient, that can be writtenas

g =g0

1 + I/Is, (2.27)

where Is = aw12Γ is the saturation intensity, Γ = 2Γ1Γ2/(Γ1 + Γ2). I/Is has the samesignificance as the saturation parameter s. As long as I Is the medium is unsaturatedand the gain is close to the zero field level. When I Is, the medium is saturated, thepopulation inversion drops to zero and with it the gain.

Exercise 2.11 Show that I = 0 is always a solution of equation (2.26), but that it is stableonly if the zero field gain g0 < l. This is the lasing threshold. What is the stable solution ifg0 > `?

b. The spectral response ,inhomogoneous broadening and ‘hole burning’ The amplifi-cation equation (2.26) can be used in a slightly more general ‘pump-probe’ setup, wherein addition to the strong pump beam at frequency ν, one shines a weak probe beam at anearby frequency ν′. Because the probe is weak, the populations are determined only bythe saturation level of the pump, but gain coefficient of the probe depends on its frequency

gprobe =c

na2γ

γ2 + (ν′ −ω)2 (n2 − n1)(hν′) =c

na2γ

γ2 + (ν′ −ω)2 (hν′)ru

(1 + Ipump/Is), (2.28)

where ru is the population inversion n2 − n1 in the absence of the pump beam. Such again curve, which maintains its Lorentzian shape, while being uniformly suppressed whenthe pump saturation increases characterizes amplifiers with uniform gain properties. Suchamplifiers are said be homogeneously broadened.

In applications one frequently encounters gain media where the atoms (or molecules)in the population are not completely identical. Gas atoms at finite temperature are anexample, where because of the Doppler effect the effective frequency of the light ’seen’ bythe atom is shifted to ν(1− v‖

c ), where v‖ is the velocity of the atom in the direction of thelight beam. To a good approximation this is the same as having the level spacing of theatom shifted by hω

v‖c .

Now suppose that in the probability of observing a level spacing of ω is p(ω); forexample if the Doppler shifted gas can be approximated as an ideal gas, then p(ω) ∼e−

m2T (ω−ω0)2

, where m is the atom mass, T is the temperature, and ω0 is the nonshifted levelspacing. The rate equations should be now solved for each component of the populationseparately, giving frequency-dependent steady state values r(ω), and the gain coefficentbecomes

g = hνc

na

∫dωp(ω)

ru

1 + I/Is(ω)γ

γ2 + (ν−ω)2 . (2.29)

If the width of distribution p is larger than γ then the medium is said to be inhomoge-neously broadened, since the broadening comes about from having a heterogeneous popula-tion, rather than being a property of each member of the population. Then, since the main

35

contribution to the gain integral comes from |ω − ν| . γ, we can approximate p(ω) byp(ν) and perform the integral giving

g = hνc

nap(ν)

πru√1 + I/Is0

, (2.30)

Is0 = Is(ω = ν) is a function of the various decay rates and does not depend on ν (unlikethe homogeneous broadening saturation intensity). Note that the saturation of the gain isslower than with homogeneous broadening since, while saturating the atoms in the centerof the band, increasing the intensity if the incoming beam allows more atoms to participatein the amplification.

In pump probe experiments (2.30) gives the gain coefficient of the pump. The probegain is once more obtained by integrating over frequencies

gprobe =c

nahν′

∫dωp(ω)

2γ

γ2 + (ν′ −ω)2ru

(1 + Ipump/Is(ω)). (2.31)

The spectral distribution as well the saturation caused by the pump change only slightacross the resonance at ν′, so we can approximate the integral by

gprobe =c

nahν′p(ν′)2π

ru

(1 + Ipump/Is(ν′)). (2.32)

The situation in the inhomogeneously broadened medium is quite different from the homo-geneously broadened one: the gain is determined by the saturated population inversion.Hence, the probe gain is mostly unaffected by the presence of the pump, except for a ’hole’near the frequency of the pump see figure ??. When s(ν) 1, the width of the ’hole’, i.e.the range of frequencies where I & Is, is ∼ γ

√I/Is0.

3 Quantum electrodynamics

3.1 The free quantum electromagnetic field

a. Field quantization in a cavity It has been already pointed out in these notes thatmutual interaction between quantum matter and the electromagnetic field can only bedescribed correctly and consistently if the field is also quantized, i.e. its degrees of freedom(which are infinite in number) should be treated according to the principles of quantummechanics.

How should the electromagnetic field be analyzed quantum mechanically, though, isnot an obvious question, even supposing that classical electrodynamics and the quantummechanics of particles are perfectly understood. The traditional approach, canonical quan-tization, uses the classical equations of motion to identify canonically conjugate observ-ables, and then imposes canonical commutation relations on them. In spite of its success,this approach suffers from several drawbacks; in addition to being inelegant, and failing torespect the symmetries of the system, it is conceptually flawed, in that it uses dynamicalinformation to derive the kinematic observable algebra.

The most satisfactory approach to field quantization starts from the fact that any (spe-cial) relativistically invariant quantum system must be a quantized field with bosonic orfermionic commutation relations between creation and annihilation operator. Furthermorethis approach shows why massless spin 1 fields (like the electromagnetic field) must inter-act as a gauge field. These ideas are well-presented in S. Weinberg’s Quantum theory of fieldsbut are beyond the scope of these notes.

36

The approach taken here, which can be thought of a stripped down version of thecanonical quantization approach, is closest to the original approach adopted by Born Jordanand Heisenberg in the late 20’s. It is an intuitive approach based on the following reasoning:

• The equations of the free electromagnetic field are linear.

• Therefore it can be regarded (classically) as a set of independent normal-mode har-monic oscillators.

• Therefore, each mode should be quantized as an independent harmonic oscillator.

In this section we implement this approach for the electromagnetic field inside a perfectcavity, and in the next we replace the cavity by a periodic-boundary-condition ‘box’ as amodel for the field in free space.

Consider the electromagnetic field in an empty finite cavity with perfectly conductingwalls. Maxwell’s equations imply that the electric field is divergenceless and obeys thewave equation ∂2

t~E = c2∇2~E in the cavity with E‖ = 0 on the walls. If the shape of the

cavity is reasonable we can find a complete set of normalized divergenceless normal cavitymodes ~Ek(~r), k = 1, 2, · · · , with vanishing tangential component on the boundary, which are

eigenfunctions of the (vector) Laplacian with (negative) eigenvalues −ω2k

c2 . This motivatesthe first kinematic quantization postulate: With each electromagnetic normal mode there is anassociated pair of annihilation and creation operators ak, a†

k satisfying the usual relation [ak, a†k ] = 1.

At this point it is not clear what relations an and a†n have to observable quantities. In

order to proceed we need further information on the free cavity dynamics: A normal modeconsists of oscillations between a magnetic field proportional to the eigenfunctions Bn andthe electric field proportional to En. The functions are related by Bn = −∇×En

ωn, and Bn

has zero normal component on the boundary. With this sign convention the magnetic fieldprecedes the electric field by a π

2 phase in the oscillations, like q precedes p, and we there-fore make the second kinematic quantization postulate: The electromagnetic field associatedwith cavity mode k is Ek = (ia†

k − iak)Ek(~r) and Bk = (a†k + ak)Bk(~r). The electromagnetic

field observables are then E = ∑k Ek and B = ∑n Bn. These statements still do not fullydetermine E and B, since the eigenmodes are only defined up to a constant. Here weuse the correspondence principle for the Hamiltonian of the kth mode Hk = hωk(a†a + 1

2),that should be equal to the quantization of the classical field energy

∫dV( ε0

2 E2k + 1

2µ0B2

k),where the integration is on the cavity interior. This last postulate sets the normal modenormalization to ε0

∫|Ek|2dV = 1

µ0

∫|Bk|2dV = hωk

2 .As long as only free fields are studied it is enough to quantize E and B, but as seen

above, the interaction terms in the Hamiltonian depend on the potentials rather than thefields. For the quantization of the potentials φ, A we must choose a gauge. Such a gaugefixing is necessary because only true degrees of freedom can be quantized. Here we choosethe Coulomb (or radiation) gauge that requires ∇ · A = 0, which fixes the potentials upto a constant. For the free fields this choice implies that φ = 0 and An = (a†

k + ak)Ak(~r),where Ak = Ek

ωk. It is called the Coulomb gauge, since in the presence of charges, it makes

φ equal to the instantaneous (rather than the retarded) Coulomb potential of the charges.Naturally, if the system is also under the influence of external (classical) electromagneticfields, these can be represented in any gauge. The most serious drawback of the Coulombgauge, that it is not relativistically invariant, will not concern us here.

b. Field quantization in free space The quantization of the electromagnetic field in freespace can now be regarded as a special case of the cavity quantization, where the ‘cavity’ is

37

a cube of volume V = L3, and the electromagnetic field obeys periodic boundary conditionson its faces (nothing in the previous subsection depends on the particular form of boundaryconditions used). The normal modes of the periodic cube ‘cavity’ are of the form E(c)

~k~σ(r) =√

hωkε0V cos(~k ·~r)~σ and E(s)

~k~σ(r) =

√hωkε0V sin(~k ·~r)~σ, where the wave vectors k have components

which integer multiples of 2πL , ωk = ck and the two polarizations ~σ are normalized basis

vectors of the transverse plane,~k ·~σ = 0.These statements define the free space quantization of the electromagnetic field by tak-

ing L → ∞, but it is more convenient to express the quantum fields using pure travel-ling wave modes. For this purpose we note that the modes for two opposite wave vec-tors ±~k (and the same polarization) are actually the same, so we can define travellingwave modes E~k~σ(r) = (E(c)

~k,~σ(r) + iE(s)

~k,~σ(r))/

√2, and corresponding annihilation operators

a~k~σ = (a(c)~k~σ− ia(s)

~k,~σ)/√

2. In terms of these the quantum electric field becomes

E(~r) = ∑~k,~σ

√hωk

2ε0L3 i(a(~k~σ)†e−i~k·~r~σ∗ − a~k,~σei~k·~r~σ) . (3.1)

If the polarization vectors are real, as in the original definition, then (3.1) represents theelectric field in the linear polarization basis, but we are free to transform the polarization

vectors unitarily(

~σ1~σ2

)−→ U

(~σ1~σ2

), and the expression (3.1) remains valid provided that(

a(k,~σ1)a(k,~σ2)

)−→ U†

(a(k,~σ1)a(k,~σ2)

). If the transformed polarization vectors are not real then the

electric field is expressed in terms of an elliptic (or circular) polarization basis.A similar calculation can be performed for the magnetic field ~B and vector potential ~A:

B(~r) = ∑~k~σ

√h

2ωkε0L3 i(a†~k~σ

e−i~k·~r~σ∗ − a~k~σei~k·~r~σ)×~k , (3.2)

A(~r) = −∑~k~σ

√h

2ωkε0L3 (a†~k~σ

e−i~k·~r~σ∗ + a~k~σei~k·~r~σ) (3.3)

c. The Fock space It is still left to construct the Hilbert state space. This is done as fora set of non-interacting harmonic oscillators, except that the number of modes is infinite.Energy consideration lead us to postulate the existence of vacuum state |vac〉 annihilatedby all the annihilation operators. The basis of the state space is obtained by all possibleapplications of a finite number of creation operators to |vac〉

|n1k1, n2k2, . . . , nsks〉 = (a†k1

)n1(a†k2

)n2 · · · (a†ks)ns√

n1!n2! · · · ns! |vac〉 . (3.4)

A state space obtained from such basis states is called a Fock space, and the basis statesare also called Fock states. The Fock states are eigenstates of all the number operatorsnk = a†

k ak with eigenvalue nk, and therefore also of the total number N = ∑k ak. They arealso free field energy eigenstates, but are not energy states in the presence of charges.

It follows that stationary states of the free electromagnetic field modes are characterizedby a finite number of excitations of energy hωk that are called photons, and a state with Nphotons behaves in many senses like an ordinary quantum mechanical state of N particles.The analogy is not perfect though. In particular, because photons are massless, it is im-possible to define a photon wave function. The fact that the creation operators commute

38

implies that |1k, 1k′〉 = |1k′, 1k〉 so that states are symmetric with respect to interchangeof photons—that is, that photons are bosons. Note that although photon energies for aparticular mode are discrete, there are free space photons of arbitrarily small frequency, sothat energy spectrum of the electromagnetic field in free space is continuous.

Fock states can be classified either by the photon number or (for states other than thevacuum) according to the number of excited modes. General electromagnetic states areformed by linear combination of Fock states. We can identify special classes of such su-perpositions. (1) Single mode states: |ψk〉 are states where only one mode of the electro-magnetic field, k is excited. They are of course superpositions of single mode Fock states|ψk〉 = ∑n ψn |nk〉, and ak′ |ψk〉 = 0 for k 6= k′. The coherent states and squeezed statesstudied in the first part of these notes were single mode states. One can similarly definetwo-mode and higher mode states. (2) Single photon states: |ψ 1-ph〉 These states are lin-ear combinations of single photon Fock states, |ψ 1-ph〉 = ∑k ψk |1k〉. They are no longereigenstates of individual number operators (and are not energy eigenstates), but are eigen-states of the total photon number, with eigenvalue 1. Similarly one can define two-photonstates three-photon states etc. The only states that are both single-mode and have a definitenumber of photons are single mode Fock states. General states, of course, can have bothan indefinite number of photons and involve an arbitrarily large number of modes.

Example 3.1 As free space states with a well defined wave number the single-photon basisstates

∣∣∣1~k~σ⟩ are completely delocalized. However, we can form one-photon wave packets—one photon states |ψ 1-ph〉 where the electromagnetic field is confined to a finite volume.

The (free space) state with ψ~k=kz,~σ=x = ce− (k−k0)2

2(δk)2 +ikz0 , and other ψ~k,~σ = 0 describes a photonpropagating in the z direction that is located near z = z0, with uncertainty of 1/δk. It is notlocalized in the transverse direction, though.

Exercise 3.1 Show that |c|2 ∼√

2πδkL for δkL 1.

Example 3.2 The state |α〉 = e∑k αka†k−α∗k ak |vac〉 = ∏k eαka†

k−α∗k ak |vac〉 is a multimode coherentstate. It is not single mode, and does not have a well-defined photon number. On the otherhand, it has well-defined electric and magnetic field values with minimal uncertainties.

Exercise 3.2

1. Show that 〈ψ 1-ph| ~E |ψ 1-ph〉 = 〈ψ 1-ph| ~E |ψ 1-ph〉 = 0 for any ψk.

2. Calculate 〈α| ~E |α〉 for α~k=kz,~σ=x = e− (k−k0)2

2(δk)2 +ikz0 .

Exercise 3.3 The function ψE(r) = 〈vac| E |ψ〉 has some properties that are desirable for aone-photon wave function.

1. Show that ψE(r) = 0 for unless ψ has a nonzero amplitude to be in a single-photonstate.

2. Calculate ψE(~r) for a given |ψ 1-ph〉, and in particular for the one in example 3.1

39

d. The Hamiltonian and dynamics So far we have focused on kinematic quantization,i.e. the field commutation relation and the state space structure. It is still left to makethe dynamical quantization, i.e. quantize the Hamiltonian. Since we have already useddynamical information it seems the only reasonable choice seems to assume that the elec-tromagnetic field Hamiltonian is a sum of (noninteracting) normal mode HamiltoniansH′ = ∑k hωk(a†

k ak + 12). It also fits nicely with the correspondence principle, since H′ =∫

dV( ε02 E2 + 1

2µ0B2). However, here we hit a snag: the energy H′ is always infinite because

of the infinite number of zero point energies 12 hωk. It is the first of the many infinities

which plague quantum field theory, most of which will not concern us here. A lot ofresearch in quantum field theory has gone into making sense out of expressions that areinfinite on the face of it. The main principle in dealing with the infinities is to rememberthat no real physical quantity is infinite, and that the infinities are simply a problem in ourmodeling. In our case the solution is very easy: since energy is anyhow defined up to aconstant, we simply declare that the Hamiltonian is actually H = ∑k hωka†

k ak. This sets thevacuum energy to zero, and the energy of all Fock space states finite, without changing thephysics on any finite mode truncation of the electromagnetic field. In terms of the quan-tum fields the ‘renormalized’ Hamiltonian is H =

∫dV( ε0

2 : E(~r)2 : + 12µ0

: B(~r)2 :), where:: denotes normal ordering, that is a reordering of all the creation and annihilation operatorssuch that all creation operators are to the left of all annihilation operators. For example: a†a :=: aa† := a†a. Note that everything commutes between the normal order signs.18

This renormalized Hamiltonian H is compatible with the correspondence principle, sincethe difference between O and : O : disappears in the classical limit where everything com-mutes.

Example 3.3 It follows from the field form of H that the energy density operator is ε02 :

E(~r)2 : + 12µ0

: B(~r)2 :. The electric energy density of a one-photon state is

〈ψ 1-ph| ε0

2: E(~r)2 : |ψ 1-ph〉 =

14 ∑

p|u~σ(r)|2 , (3.5)

where u~σ(~r) = 1V ∑~k hωkei~k·~r.

As mentioned above Fock basis states are energy eigenstates with

H |n1k1, n2k2, . . . , nsks〉 = (n1hωk1 + n2hωk2 + · · ·+ nshωks) |n1k1, n2k2, . . . , nsks〉 (3.6)

Since H is quadratic in ak, a†k , the Heisenberg equations of motion for the free field are

linear and quite easy to solve:

ak =1ih

[ak, H] = −iωk ⇒ ak(t) = ake−iωkt , (3.7)

as for harmonic oscillators. This result can be used to find the time evolution of free fieldstates.

Example 3.4

1. The one-photon wave packet of example 3.1 evolves into a wave packet of the sameform but with z0 → z0 + ct (assuming k0 δk). That is the photon propagates in thepositive z direction with the speed of light.

18Normal order has a slightly different definition for fermions. We will not need it here.

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2. The coherent state |α〉 evolves into |αt〉, where α~k,~σ,t = e−iωktα~k,~σ,t.

Exercise 3.4 (Vacuum fluctuations) We know that zero-point energy in harmonic oscilla-tors is a consequence of the uncertainty principle. Sweeping it under the rug by redefiningthe zero energy point of course does not make them go away. Define the smeared electricfield operator Ea(r) = 1

a3

∫dVE(r′)dr′, where the integration is on a cube of side a centered

on ~r. Calculate 〈vac| Ea(r)2 |vac〉 in the free space vacuum for L → ∞. The existence ofelectromagnetic field fluctuations in the vacuum implies that there are random currents inconductors that are not connected to any voltage or current source. They are the quantumanalog of the Nyquist currents that are generated by thermal fluctuations in warm con-ductors, and like them, they cannot generate energy. Note that 〈vac| Ea(r)2 |vac〉 has thecurious property that it diverges when a tends to 0, another appearance of a quantum fieldinfinity. This result implies that E(r) is not an observable quantity.

Exercise 3.5 (Momentum)

1. Show that the operator ~P = ∑~k,~σ h~ka†~k,~σ

a~k,~σ is the generator of translations in the sense

that [~E(~r), ~P] = −ih~E′(~r) and therefore eih~r0·~P~E(~r)e−

ih~r0·~P = ~E(~r −~r0). It follows that a

(~k,~σ) photon carries momentum h~k.

2. Show that ~P = 1ε0

∫dV : ~E(~r)× ~B(~r).

3.2 Interactions of charges and the quantum electromagnetic field

a. The Hamiltonian and the dipole approximation For the last time in these notes weneed to quantize. Equation (2.3) gives the Hamiltonian for quantum charges and a classicalelectromagentic field, and we know how to quantize the electromagnetic field. These sug-gest a quantization for a system where there both quantum charges and quantum fields,by simply adding the two together

H = ∑n

(~pn − en ~A(rn))2

2mn+ ∑

n 6=m

enem

4πε0|rn − rm|+ ∑

khωka†

k ak . (3.8)

This is the full Hamiltonian of nonrelativistic quantum electrodynamics in the coulombgauge. Note that the particle part has a different sense here than in part 2, because ~Ahere is a quantum field, whose precise meaning is given by equation (3.3). As explainedabove, in the Coulomb gauge we quantize only the vector potential ~A as a dynamicalobject. The scalar potential φ(r) = ∑n

en4πε0|r−rn| is quantized becuase the charge positions

are quantized, but is not a quantum field, and has no dynamics of its own.We are going to use the Hamiltonian (3.8) to study the spontaneous decay of an atom

from an excited state to a lower energy state while emitting electromagnetic radiation.As with classical radiation, the large wave length implies that the (quantum) field is nearlyconstant on the scale of the atom. We will therefore again make the dipole approximation—the leading term in the long-wavelength approximation, replace ~A(~r) by ~A(0).

b. Spontaneous emission and radiative lifetime The process of spontaneous emissionis similar to stimulated emission. An atom (or molecule, etc.) in state |2〉 of the atomicHamiltonian makes a transition a transition to a lower energy state |1〉 while emittingradiation of frequency ω = 1

h (ε2− ε1). Unlike the stimulated emission process, the radiativetransition occurs without any ambient electromagnetic radiation, or to put it otherwise, the

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system interacts with the electromagnetic vacuum, and therefore the process cannot bedescribed using classical electromagnetic radiation.

Since we include the electromagnetic field in our quantum system, the transition shouldbe written as |2; vac〉 −→

∣∣∣1;~k,~σ⟩

for some single-photon basis state with wave vector~k andpolarization ~σ (see Fig. ??). We will use again the golden rule to calculate the transitionrate. The initial and final states are eigenstates of

H0 = ∑n

~p2n

2mn+ ∑

n 6=m

enem

4πε0|rn − rm|+ ∑

khωka†

k ak , (3.9)

and the transition is caused by the interaction

V = ∑n

2en~pn · ~A + (en A)2

2mn. (3.10)

The second term in the interaction is quadratic in the charges, and therefore weak andnegligible in the dipole approximation. The matrix element of the interaction is therefore

M =⟨

1;~k~σ∣∣∣ ∑

n~k′~σ′

√h

2ωk′ε0Ven

mn(a†

~k′~σ′~pn ·~σ∗ + a~k′~σ′~pn ·~σ′) |2; vac〉 . (3.11)

As before we would like to express the transition amplitude in terms of the fields

rather than the potential. For this purpose note that pn = ihmn[rn, Hc], where Hc ∑n~p2

n2mn

+∑n 6=m

enem4πε0|rn−rm| is the charge system Hamiltonian, so that 〈1|~pn |2〉 = mn

ih 〈1| [~rn, Hc] |2〉 =mnih (ε1 − ε2). Using this result we get

M = iω

√h

2ωkε0V〈2| ~d ·~σ∗ |1〉 , (3.12)

where ~d = ∑ en~rn is again the electric dipole moment operator. Using the golden rule(1.34) we get that the differential rate to decay and emit a photon with wave vector ~k andpolarization ~σ is

dW|2;vac〉→|1;~k,~σ〉dΩ

=2π

h|M|2 dg

dΩ(ωk = ω, Ωk) . (3.13)

The electromagnetic field is isotropic, so that dgdΩ is in fact independent of Ω. This equation

also implies the obvious result that spontaneous emission can only occur if ε2 > ε1.

Exercise 3.6 Show that the photon density of states, i.e. the number of states in a ‘box’ ofenergy extent dε and solid angle extent dΩ divided by dεdΩ is dg

dΩ = L3

(2π)3ε2

h3c3 . No

The rate of spontaneous emission per unit solid angle and unit time is therefore

dWdΩ

=1

8πε0

ω3

hc3 | 〈2| ~d ·~σ |1〉 |2 . (3.14)

The total transition rate by definition the radiative decay rate Γ21 of |2〉 to |1〉 is obtainedby summing over the polarizations, and integrating over the angle of the emitted radiation,and the result is

Γ =1

3πε0

ω3

hc3 | 〈1| ~d |2〉 |2 . (3.15)

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The spontaneous emission decay rate is also known as the Einstein A coefficient. Aswith stimulated emission in incoherent radiation, it is a product of a material factor—thesquare of a dipole matrix element—and a field factor. Einstein expressed the stimulatedemission rate in terms of the electromagnetic spectral energy density E(ω). For unpolarizedlight, equation (??) becomes then w21,stimulated = BE(ω) with

B =π

31h2 | 〈1| ~d |2〉 |

2 . (3.16)

Hence we get the universal ratio AB = 1

π2hω3

c3 derived by Einstein on statistical physicsgrounds.

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