ADVANCED QUANTUM MECHANICS - ULisboaADVANCED QUANTUM MECHANICS Filipe R. Joaquim Complements of Quantum Mechanics MEFT – 4th year, 1st Sem. 2018/2019

ADVANCED QUANTUM

MECHANICS

Filipe R. Joaquim

Complements of Quantum Mechanics

MEFT – 4th year, 1st Sem.

2018/2019

Table of Contents

Page

1 Second quantisation formalism 11.1 First quantisation: single-particle system . . . . . . . . . . . . . . . . . . . 11.2 First quantisation: many-particle systems . . . . . . . . . . . . . . . . . . . 4

1.2.1 Systems of identical particles . . . . . . . . . . . . . . . . . . . . . 61.3 Second quantisation formalism . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3.1 Occupation number representation and Fock spaces . . . . . . . . . 91.3.2 Bosonic creation and annihilation opperators . . . . . . . . . . . . . 111.3.3 Fermionic creation and annihilation opperators . . . . . . . . . . . . 14

1.4 Second-quantised operators formalism . . . . . . . . . . . . . . . . . . . . 151.4.1 One-particle operators . . . . . . . . . . . . . . . . . . . . . . . . . 151.4.2 Many-particle operators . . . . . . . . . . . . . . . . . . . . . . . . 181.4.3 Change of basis and field operators . . . . . . . . . . . . . . . . . . 201.4.4 Momentum representation, spin and field equations . . . . . . . . . 24

1.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2 Interaction of radiation with matter 292.1 Recap of Maxwell equations . . . . . . . . . . . . . . . . . . . . . . . . . . 292.2 Quantization of the electromagnetic field . . . . . . . . . . . . . . . . . . . 32

2.2.1 Hamiltonian description of the classical electromagnetic field . . . . . 332.2.2 Quantization procedure . . . . . . . . . . . . . . . . . . . . . . . . 34

2.3 States and field operators of the electromagnetic field . . . . . . . . . . . . 362.3.1 Coherent states of the electromagnetic field . . . . . . . . . . . . . 39

2.4 Interaction of matter with the quantised electromagnetic field . . . . . . . . 402.4.1 Emission of radiation from an atom . . . . . . . . . . . . . . . . . . 432.4.2 Einstein coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . 462.4.3 Photon absorption . . . . . . . . . . . . . . . . . . . . . . . . . . 472.4.4 Planck distribution formula . . . . . . . . . . . . . . . . . . . . . 48

2.5 Matter+radiation in second quantisation . . . . . . . . . . . . . . . . . . . 49

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TABLE OF CONTENTS

2.5.1 Example: non-relativistic Bremsstrahlung . . . . . . . . . . . . . . 512.6 Supplement 1: The infinite volume limit . . . . . . . . . . . . . . . . . . . . 572.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3 Relativistic quantum mechanics 613.1 Minkowski spacetime and Lorentz transformations: basics . . . . . . . . . . 623.2 The Klein-Gordon equation . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.2.1 The free particle Klein-Gordon equation . . . . . . . . . . . . . . 653.2.2 Non-relativistic limit of the Klein-Gordon equation . . . . . . . . 673.2.3 Klein-Gordon equation with electromagnetic field . . . . . . . . . 683.2.4 Non-relativistic limit of the Klein-Gordon equation with electro-

magnetic field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693.2.5 Gauge invariance of the Klein-Gordon equation . . . . . . . . . . 703.2.6 Box normalization of Klein-Gordon free particle solutions . . . . . 713.2.7 Klein Gordon equation with Coulomb potential (the pionic atom) 72

3.3 The Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 753.3.1 Dirac equation in covariant form . . . . . . . . . . . . . . . . . . 773.3.2 Covariance of the Dirac equation . . . . . . . . . . . . . . . . . . 78

3.4 S(Λ) for proper Lorentz transformations . . . . . . . . . . . . . . . . . . . 803.4.1 Boosts along an arbitrary direction . . . . . . . . . . . . . . . . . 823.4.2 Spatial rotations around an arbitrary axis . . . . . . . . . . . . . 84

3.5 S(Λ) for improper Lorentz transformations: spatial reflection and time reversal 853.6 The spin operator ~S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873.7 The free-particle solutions of the Dirac equation . . . . . . . . . . . . . . . 883.8 Helicity and spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 893.9 The E<0 solutions of the Dirac equation: the "hole theory" . . . . . . . . . 903.10 Charge conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 923.11 Non-relativistic limit of the Dirac equation . . . . . . . . . . . . . . . . . . 933.12 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

Bibliography 99

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1Second quantisation formalism

“Quantum theory provides us with a striking illustration of the fact that we can fullyunderstand a connection though we can only speak of it in images and parables.”

Werner Heisenberg

Quantum theory provides the most complete description of Nature at the microscopiclevel. From elementary particle physics to astrophysical objects, a plethora of phenomenahave been understood thanks to a quantum description of matter and energy. Historically,quantum physics has been first applied in the quantization of particle motion, in which physicalobjects are described by wave functions. On the other hand, the surrounding environment likethe electromagnetic and gravitational fields, is treated classically (hence the name quantummechanics). This first step in the construction of the quantum theory is usually called firstquantisation, as opposed to second quantisation, in which the environment is also quantisedand the wave function interpreted as being dynamic and subject to quantization. In short,we take the wave function of a quantised system and transform it into an operator. Beforedescribing the method of second quantisation and some applications, we will review some keyaspects of quantum mechanics covered in the courses Quantum Mechanics I and II.

1.1 First quantisation: single-particle system

The quantum state (or one of the states) of a particle (or system) is described by the wavefunction ψ(r, t) which provides the mathematical description of the particle’s De Brogliewave. Therefore, when addressing quantum effects, |ψ(r, t)|2 is somehow connected with theintensity of the wave related to those quantum effects. This intensity of the wave at a givenpoint in space and instant of time will be proportional to the probability of finding, at thatpoint and at that time, the material particle to which the wave corresponds. In 1927 Borninterpreted |ψ(r, t)|2 as the probability density and |ψ(r, t)|2d3r as the probability, dP (r, t),of finding a particle at time t in the volume element d3r located between r and r + dr.Therefore:

dP (r, t) = |ψ(r, t)|2d3r . (1.1)

1

CHAPTER 1. SECOND QUANTISATION FORMALISM

For a n-dimensional system, |ψ(r, t)|2 has dimensions of [L]−n. Obviously, since we are certainthat the particle is somewhere in space, integrating the above equation over the entire spacemust return 1, i.e. ∫

all spacedP (r, t) =

∫all space

|ψ(r, t)|2d3r = 1 . (1.2)

We have seen in previous courses that the possible states (more precisely, the purestates) of a quantum mechanical system are represented by unit vectors (called state vectors)belonging to a complex separable Hilbert space H, known as the state space, well definedup to a complex number of norm 1 (the phase factor). The exact nature of this Hilbertspace depends on the system1; for example, the position and momentum states for a singlenon-relativistic spin zero particle is the space of all square-integrable functions, while thestates for the spin of a single electron are unit elements of the two-dimensional complexHilbert space spanned by the spinors |↑〉 and |↓〉.

Given two state vectors |φ〉 and |ψ〉 describing configurations of a quantum system, theirinner product is a complex number known as probability amplitude 〈φ|ψ〉. Upon an idealmeasurement of a particular observable of the quantum system, the probability the state|φ〉 collapses to a particular eigenstate |ψ〉 is given by |〈φ|ψ〉|2. In particular, the possibleoutcomes of the measurement of a certain quantity A are just the (real) eigenvalues of thecorresponding quantum mechanical operator A, which have to be real. Since we are dealingwith physical observables, quantum mechanical operators must be self-adjoint (or Hermitian).These features of quantum mechanics summarise the postulates of quantum mechanics, whichhave been presented in the course Quantum Mechanics I (and can be found in any textbook).

Two important state vectors are |r〉 and |p〉, which are the eigenvectors of the corre-sponding position and momentum operators, r and p,

r|r′〉 = r′|r′〉 , p|p′〉 = p′|p′〉 . (1.3)

Notice that the vectors |r〉 and |p〉 do not belong to the Hilbert space H since their normis not finite. Still, they span the whole Hilbert space. This can be seen recalling the closurerelations in the case of continuos state basis∫

d3r|r〉〈r| = 1 ,∫d3p|p〉〈p| = 1 . (1.4)

Using these relations, one can show that the wave function describing a quantum system in astate |ψ〉 in the position and momentum configuration is given by:

ψ(r) = 〈r|ψ〉 , ψ(p) = 〈p|ψ〉 . (1.5)

In physical terms, |r〉 describes the state of a localised particle at the point r and |p〉corresponds to a state of definite momentum p. In particular, we have:

〈r|r′〉 = δ(3)(r − r′) , 〈p|p′〉 = δ(3)(p− p′) , 〈r|p〉 =

(1

2π~

)3/2

eip.r/~ . (1.6)

In configuration space, the operators r and p act as:

〈r|r|ψ〉 = rψ(r) , 〈r|p|ψ〉 = −i~∇ψ(r) , (1.7)meaning that we can write r = r and p = −i~∇.

1The mathematical properties of Hilbert spaces were mentioned in the Quantum Mechanics II course.

2

1.1. FIRST QUANTISATION: SINGLE-PARTICLE SYSTEM

Exercise 1.1. Prove relations (1.5),(1.6) and (1.7) .

So, in quantum mechanics it’s everything about the wave function ψ(r, t)! But how do wecompute it? In some sense, if ψ describes the state of our quantum system, we must find someequation of motion analogous to Newton’s second law or, the equivalent Euler-Lagrange andHamilton’s equations. Contrarily to what happens in classical mechanics where the equationsof motion predict the behaviour of a system giving its configuration at any time, in QM weask what is the probability of finding the system in a given configuration among all possibleones. The quantum analogue of Newton’s law is of course the Schrödinger equation

i~∂ψ(r, t)

∂t= Hψ(r, t) , (1.8)

where H is the Hamiltonian operator containing the information about the particle’s totalenergy (kinetic plus potential). In case H does not depend explicitly on time, we can definethe time-independent Schrödinger equation HΨ(r) = EΨ(r), with E constant (the possiblevalues of E define the energy spectrum). This equation describes the stationary states of thesystem (remember that the wave function is still time-dependent, but in a trivial way). TheSchrödinger equation, although has been presented as a postulate, can be actually derivedfrom powerful symmetry principles [1] or by using Lagrangian formulation. Like Newton’ssecond law (F = ma), the Schrödinger equation can be defined differently by using, forinstance, the Heisenberg’s matrix mechanics or Feynman’s path integral formalisms (pathintegrals will be covered later on).

Considering a particle of mass m in a potential V (r), the corresponding Hamiltonian is

H =p2

2m+ V (r) , (1.9)

for which the time-independent Schrödinger equation reads

H|φα〉 = Eα|φα〉 , (1.10)

where |φα〉 (also represented shortly by just |α〉), denote the eigenstate basis of thesystem spanned by the states with well-defined energy. In the coordinate representation theabove equation has the form[

− ~2

2m∇2 + V (r)

]φα(r) = Eαφα(r) , (1.11)

In case the particle has internal degrees of freedom like spin s, isospin I, etc., the statevector |α〉 has several components depending on the possible values of the internal degrees offreedom. Therefore, we can define the general states |r, ξ〉 as being the state of the particlelocalised at r, with internal degrees of freedom denoted generically by ξ (these can be spin,isospin, etc.). These states span the Hilbert space of the particle, for which the closure relation∑

i

∫d3r|r, ξi〉〈r, ξi| = 1 , (1.12)

holds. The projections of different states obey

〈r′, ξi|r, ξj〉 = δi,jδ(3)(r′ − r) . (1.13)

which are the well-known normalisation rules.

3


1.2 First quantisation: many-particle systems

Let us now generalise what we have seen in the previous section to the case of N -particlequantum systems. The Hilbert space of HN is the space of complex (square-integrable)functions defined in theN -particle configuration space. The wave-function ΨN (r1, r2, ..., rN , t)is now interpreted as the probability amplitude of finding the N particles at the positionsr1, r2, ..., rN at a time t. In other words, the probability dPN(r1, r2, ..., rN , t) of finding thesystem in a configuration in which each particle is localised in a region d3ri around ri is givenby

dPN(r1, r2, ..., rN , t) = |ΨN(r1, r2, ..., rN , t)|2d3r1d3r2...d

3rN . (1.14)

Obviously,∫all space

dPN(r1, r2, ..., rN , t) =

∫all space

|ψN(r1, r2, ..., rN , t)|2d3r1d3r2...d

3rN = 1 .

(1.15)The above wave functions define HN , which is nothing but the tensor product of thesingle-particle Hilbert spaces Hi, namely HN = H1 ⊗H2 ⊗ ...⊗HN . If |α(k)

nk 〉 denotes anorthonormal basis spanningHk, then the state of the system corresponding to the configurationin which each particle k is in the state |α(k)

nk 〉 is

|α(1)n1, α(2)

n2, .., α(N)

nN〉 = |α(1)

n1〉 ⊗ |α(2)

n2〉 ⊗ ...⊗ |α(N)

nN〉 , (1.16)

to which corresponds the wave function:

ψα(1)n1,α

(2)n2,..,α

(N)nN

(r1, r2, ..., rN) = 〈r1, r2, ..., rN |α(1)n1, α(2)

n2, .., α(N)

nN〉 (1.17)

= (〈r1| ⊗ 〈r2| ⊗ ...⊗ 〈rN |)(|α(1)n1〉 ⊗ |α(2)

n2〉 ⊗ ...⊗ |α(N)

nN〉). (1.18)

In the same way as for the single-particle quantum system, the projections among states are

〈α(1)

n′1, α

(2)

n′2, .., α

(N)

n′N|α(1)n1, α(2)

n2, .., α(N)

nN〉 = 〈α(1)

n′1|α(1)n1〉〈α(2)

n′2|α(2)n2〉...〈α(N)

n′N|α(N)nN〉 , (1.19)

and the closure relation∑ni

|α(1)n1, α(2)

n2, .., α(N)

nN〉〈α(1)

n1, α(2)

n2, .., α(N)

nN| = 1N , (1.20)

where 1N is the unit operator defined in HN .The dynamics of a N -particle system is attained by generalising the dynamics of a single

particle. Namely, the time-dependent Schrödinger is now expressed as:

i~∂

∂tΨN(r1, r2, ..., rN , t) = HΨN(r1, r2, ..., rN , t) (1.21)

being the Hamiltonian of the system

H = −∑j

~2

2mj

∇2j + V (r1, r2, ..., rN , t). (1.22)

The potential V (r1, r2, ..., rN , t) accounts for all types of interactions, either internal orexternal, among the particles of the system and with external agents. The commutation

4

1.2. FIRST QUANTISATION: MANY-PARTICLE SYSTEMS

relations for each particle of the system remain valid, while operators of different particlescommute. For instance, [xj, pxk ] = i~δj,k, while [xj, xk] = [pxk , pxj ] = 0. For time-independentpotentials, the solutions of (1.21) are stationary states described by ΨN(r1, r2, ..., rN , t) =ψN(r1, r2, ..., rN)e−iEt/~, where ψN(r1, r2, ..., rN) is the solution of the time-independentSchrödinger equation:[

−∑j

~2

2mj

∇2j + V (r1, r2, ..., rN)

]ψN(r1, r2, ..., rN) = EψN(r1, r2, ..., rN) , (1.23)

with E being the total energy of the system. Notice that the above equation is a many-bodydifferential equation which cannot, in general, be separated into a set of N single-particledifferential equations. This can only be done in the case of no mutual interactions among theparticles of the system. Also remember that, for a system of N noninteracting distinguishableparticles, one can write the Hamiltonian as

H =N∑j=1

Hi =N∑j=1

[− ~2

2mj

∇2j + V (ξj)

], (1.24)

where we have now written the potential in terms of ξj , the general set of degrees of freedomξj ≡ (rj, sj, Ij, ...). Notice that the Hamiltonians Hi of different particles commute, i.e.[Hk, Hj] = 0, since [xj, xk] = [pxk , pxj ] = 0. The Schrödinger equation Hψ(ξ1, ξ2, ...ξN) =Eψ(ξ1, ξ2, ...ξN) separates into N independent equations:[

− ~2

2mj

∇2j + V (ξj)

]ψ(ξj) = Ejψ(ξj) , (1.25)

where Ei is the energy of each particle and E =∑

j Ej is the total energy of the system.The wave function of the system is therefore

ψN(ξ1, ξ2, ...ξN) = ψ(ξ1)ψ(ξ2)...ψ(ξN) =N∏i=1

ψ(ξi) . (1.26)

The above formalism can be particularised for the case in which the state of each particle isdescribed by a set of quantum numbers ni, leading to, for instance,

ψni,n2,...,nN (ξ1, ξ2, ...ξN) = ψn1(ξ1)ψn2(ξ2)...ψnN (ξN) =N∏i=1

ψni(ξi) , (1.27)

which, in Dirac notation, could be shortly expressed as |n1;n2; ...;nN〉 = |n1〉|n2〉...|nN〉.For example, the ground state of a system of three-distinguishable spin 1/2 noninteractingparticles with sz = 3/2 confined to a one dimensional infinite square potential would be|1, ↑; 1, ↑; 1, ↑〉 = |1, ↑〉|1, ↑〉|1, ↑〉, in the basis |ni, szi〉.

When generalising the single-particle quantum formalism for N-particle systems theproperties of stationary states remain unchanged. In particular, probability densities |ψN |2,probability currents j and the expectation values of (time-independent) operators:

〈Ψ|O|Ψ〉 = 〈ψ|O|ψ〉 =

∫d3r1

∫d3r2...

∫ψ∗(r1, r2, ..., rN)Oψ(r1, r2, ..., rN)d3rN ,

(1.28)are conserved.

5


1.2.1 Systems of identical particles

In physics one is often led to study systems of many identical particles, i.e. particles for whichall intrinsic properties are the same (mass, electric charge, spin, etc.). In classical physics,identical-particle trajectories can (in principle) be followed. In contrast, quantum identicalparticles with uncertain position and momentum cannot be tracked and, therefore, are trulyindistinguishable. Such property of identical quantum particles leads to interesting phenomenafor which there is no analog in classical physics. In particular, indistinguishability introducesthe concept of exchange symmetry, and to particle classification into bosons and fermions.

Let us consider a system of N -identical particles in which the state of each particle isdescribed by a set of parameters ξi = (ri, si, ...), where ri is the position vector and si arespin coordinates. As we have seen in QMII, particles can have other intrinsic degrees offreedom like colour, isospin, etc. All of them are included in ξi. With this notation, the wavefunction of the system will be denoted as ψN(ξ1, ξ2, ...ξN) (to keep it as general as possible)and quantum indistinguishability implies that:

|ψN(ξ1, .., ξk, ..., ξj, ..., ξN)|2 = |Pkjψ(ξ1, .., ξk, ..., ξj, ..., ξN)|2 = |ψ(ξ1, .., ξj, ..., ξk, ..., ξN)|2 ,(1.29)

where Pkj is the permutation (or transposition) operator which exchanges the coordinates ξkof particle k with those of particle j, i.e. ξj. In a few words, this means that the probabilityof finding the system in a configuration where particle j and k are interchanged is the same.Notice that when we talk about interchanging the two particles we are actually referringto something more physical than just the mathematical permutation operation in the wavefunction. In fact, permuting two particles really means taking them around each other incontinuous paths in the system’s configuration space. Interestingly, it can be shown thatin a 3-dimensional space P 2

kj = 1 (interchanging twice is the same as doing nothing) andtherefore2

ψN(ξ1, .., ξk, ...ξj, ..., ξN) = ±ψ(ξ1, .., ξj, ...ξk, ..., ξN) . (1.30)

The plus (minus) sign corresponds to a symmetric (antisymmetric) wave function. Fromexperiment we know that there are in Nature two types of particles: bosons and fermions.The wave function of a system of identical bosons (fermions) is completely symmetric(antisymmetric). Fermions like electrons, muons and quarks have half-integral spin, whereasbosons like, for instance, the Higgs particle, the pion and the photon have integral spin. Theconnection spin-symmetry follows from the spin-statistics theorem (SST) which can be provedin the more general framework of quantum field theory. For now, just accept that

The wave function of N identical bosons (fermions) is completely symmetric (antisymmetric)under permutations of any two particles.

The main concern now is to construct the wave function of a N -particle system with thesymmetry properties imposed by the spin-statistics theorem. In QMII, we have worked outseveral examples of quantum systems of identical bosons and fermions and studied someimportant consequences of the spin-statistics theorem. For fermions, the SST leads to thePauli principle which settles the starting point for the understanding of atomic and condensedmatter physics and, of course, chemistry. Moreover, the same principle plays a crucial role in

2Interestingly, this is not always the case. Wanna learn more? Google the word "anyon" or have a look at[7].

6

1.2. FIRST QUANTISATION: MANY-PARTICLE SYSTEMS

the study of stars (degeneracy pressure in white dwarfs and neutron stars) and high-energyscattering processes. For bosons, the spin-statistics connection turns out to be fundamentalto understand laser light, Planck’s radiation law, superfluidity and, of course, Bose-Einsteincondensation. Here I will just recap some aspects which will be useful later.

Let’s start by considering a system with N identical particles. The Hilbert space of eachparticle is spanned by the single-particle states |i〉 = |1〉, |2〉, .... The basis states for theN -particle system are then

|i1, i2, ..., iN〉 = |i1〉|i2〉...|ik〉...|iN〉 , (1.31)

where |ik〉 denotes the state of particle k. If the set |i〉 is complete and orthonormal, then thestates defined above represent a complete orthonormal system in the N -particle state space.In order to define the completely symmetric or antisymmetric basis states, we have somehowto contemplate all permutations of N particles, i.e. we have to apply all elements of thepermutation group SN which are N ! in total. Notice that, any permutation can be achieved bysuccessive permutations of two particles (transpositions). For example, |bca〉 can be obtainedfrom |abc〉 by |bca〉 = P12P13|abc〉 = P12|cba〉 = |bca〉. This is an even permutation (becauseit is made of an even number of transpositions). We can define a symmetrizer (S+) andantisymmetrizer (S−) operator which picks up |i1, i2, ..., iN〉 and delivers the N -particle statewith the desire symmetry properties. Namely3,

S±|i1, i2, ..., iN〉 =1√N !

∑P

(±1)pP |i1〉|i2〉...|ik〉...|iN〉 (1.32)

=1√N !

∑P

(±1)p|iP1〉|iP2〉...|ik〉...|iPN 〉 , (1.33)

where the sum is for all permutations P = (P1, P2, ..., PN) of SN , and p = 1 (−1) foreven (odd) permutations (this permutations should be simply interpreted as in the |abc〉case illustrated above). The symmetry/antisymmetry constraints imposed on a system of Nidentical bosons or fermions imply corresponding restrictions on its Hilbert space HN . One cansay that the wave function ψN of HN belongs to the Hilbert space of N bosons (fermions) SN(AN) if it is symmetric (antisymmetric) under permutations of particles. Formally speaking,the operators S± project HN onto SN and AN , i.e.

SN = S+HN , AN = S−HN . (1.34)

An important property of S+ is that, if any state is occupied by more than one particlethen S+|i1, i2, ..., iN〉 is no longer normalised. Let us assume that each state |k〉 is occupiednk times (this is called the multiplicity of the state). Notice that S+|i1, i2, ..., iN〉 containsN ! terms from which only N !/(n1!n2!...) are different (this is the number of permutationsof N objects with some repeated ones). Therefore, each of the different terms will appearn1!n2!... times. Given this, the norm of the state S+|i1, i2, ..., iN〉 will be

|S+|i1, i2, ..., iN〉|2 = 〈i1, i2, ..., iN |S†+S+|i1, i2, ..., iN〉 =1

N !(n1!n2!...)2 N !

n1!n2!...= n1!n2!...nN ! (1.35)

3We will use the notation |i1, i2, ..., iN 〉B,F ≡ S±|i1, i2, ..., iN 〉 is also used to express the state vectorwith the required symmetry properties.

7


Therefore, the normalised Bose states are:

|i1, i2, ..., iN〉B =1√

n1!n2!...S+|i1, i2, ..., iN〉 =

1√N !n1!n2!...

∑P

P |i1, i2, ..., iN〉 . (1.36)

Obviously, if nk > 1 for any k, then S−|i1, i2, ..., iN〉 = 0, and the wave function cannotbe antisymmetrized. In other words, the Pauli exclusion principle is a consequence of thespin-statistics theorem, i.e. in a system of N identical fermions there can be no particlesin the same state 4. A practical way to antisymmetrise the wave function of a system is bycomputing the Slater determinants

|i1, i2, ..., iN〉B,F =1√

n1!n2!...S±|i1, i2, ..., iN〉 =

1√N !n1!n2!...

∣∣∣∣∣∣∣∣∣|i1〉1 |i2〉1 ... |iN〉1|i1〉2 |i2〉2 ... |iN〉2...

......

...|i1〉N |i2〉N ... |iN〉N

∣∣∣∣∣∣∣∣∣±

,

(1.37)where the ± sign indicates weather one has to take the usual permutation sign in thecalculation of the determinant. Obviously, in case the system has a state k with multiplicitynk > 1, the real determinant vanishes and the wave function cannot be antisymmetrised.

Exercise 1.2. Show that S2±|i1, i2, ..., iN〉 =

√N ! S±|i1, i2, ..., iN〉.

1.3 Second quantisation formalism

Historically quantum mechanics was first applied to particles. The state of a system isdescribed by the wave function and observables by operators acting on those states. On theother hand, interactions between particles and fields have been first treated classically. Secondquantisation extends quantum mechanics to fields opening the window to quantum fieldtheory. The method is important for the study of many-particle systems in which the numberof particles varies due, for instance, to reactions occurring in the system. One example isthe photon gas where photons are continuously emitted and absorbed by matter, being theirnumber not fixed. Similar situations can be encountered in condensed matter physics andhigh-energy physics where particles are constantly annihilated and created in collisions. Theexistence of such phenomena calls for the development of a quantum formalism which allowsthe description of systems with an undetermined number of particles which interact throughinteractions which violate particle number. Such quantum formalism is second quantisation.In some sense, the wave function which described a certain particle, acquires now a dynamicalcharacter. Notice however that by introducing second quantisation we are not actually invokingnew physical laws or principles, namely there is only one quantisation, that which considersparticles as quanta. In the course of condensed matter physics you will see many applicationsof second quantisation (superconductivity, superfluidity, etc.), here we will use it to introducequantum fields. From the practical point of view, second quantisation naturally accountsfor the symmetrisation and antisymmetrisation of quantum states. The violation of particlenumber is due to the use of creation and annihilation operators.

4Here "in the same state" means with the same quantum numbers.

8

1.3. SECOND QUANTISATION FORMALISM

1.3.1 Occupation number representation and Fock spaces

The physics of many-particle systems is formulated in terms of the so-called occupationnumber representation. The starting point is the notion of indistinguishability introduced in theprevious sections and the Slater determinants of Eq. (1.37) constructed from single-particlestates. Such determinants form the Hilbert spaces of N -particle systems of fermions or bosons,namely AN and SN , respectively.

Let us start by considering the one-particle state basis |1〉, |2〉, ..., which spans all possibleconfiguration states of a single particle, which are nothing but all possible solutions for theSchrödinger equation for the system. However, most of these solutions are redundant since fora system of identical particles only symmetric and anti-symmetric occur in Nature. Moreover,due to particle indistinguishability, the many-particle wave function obeying the symmetrisationpostulates is unique for each set of occupation numbers, i.e. the number of particles whichoccupy each state or level |1〉, |2〉, |3〉, .... In the occupation number representation, the basisstates are obtained by listing the occupation nk of each state |k〉, namely

|n1, n2, n3, ...〉 ,∑k

nk = N , (1.38)

which is the state with n1 particles in state |1〉, n2 particles in state |2〉, and so on. Obviously,the sum of all occupation numbers must be equal to the total number of particles N . Itis therefore natural to define occupation number operators nk which count the number ofparticles occupying the state |k〉 for a certain occupation number representation,

nk|n1, n2, ..., nk, ...〉 = nk|n1, n2, ..., nk, ...〉 . (1.39)

Obviously, nk = 0, 1 (due to antisimmetrization) for fermions, while for bosons nk = 0, 1, 2, ....

Exercise 1.3. Find the occupation basis states for a system of three non-interacting iden-tical bosons subject to a potential characterised by the one-particle states |1〉, |2〉, |3〉.Repeat the exercise for fermions.

The question now is whether the state |n1, n2, ...〉 fully describes the system. The answeris yes. The abstract vector |n1, n2, ...〉 indicates that there are n1 particles in state |1〉, n2

particles in state |2〉, etc. Then, the state of the system is just:

|i1, i2, ..., iN〉B,F =1√

n1!n2!...S±|i1, i2, ..., iN〉 ≡ |n1, n2, ...〉 . (1.40)

In practice, by knowing |n1, n2, ...〉 one can compute the symmetrised (or antisymmetrised)state by just computing the Slater determinant of Eq. (1.37) where we put the first n1 columnswith all particles in state |1〉, then n2 columns with all particles in state |2〉, and so on. Forinstance, suppose a system of 3 identical bosons is described by the occupation-number statevector |2, 0, 0, 1, 0, 0, ...〉. The (symmetric) state of the system |ψ〉S in terms of single-particle

9


N Fermion occupation number basis states |n1, n2, n3, ...〉0 |0, 0, 0, 0, ...〉1 |1, 0, 0, 0, ...〉, |0, 1, 0, 0, ...〉, |0, 0, 1, 0, ...〉 , ...2 |1, 1, 0, 0, ...〉, |0, 1, 1, 0, ...〉, |1, 0, 1, 0, ...〉 , ......

......

...N Boson occupation number basis states |n1, n2, n3, ...〉0 |0, 0, 0, 0, ...〉1 |1, 0, 0, 0, ...〉, |0, 1, 0, 0, ...〉, |0, 1, 0, 0, ...〉 , ...2 |2, 0, 0, 0, ...〉, |0, 2, 0, 0, ...〉, |1, 1, 0, 0, ...〉 , |1, 0, 1, 0, ...〉 , ......

......

...

Table 1.1: Some occupation number state basis elements.

states will be given by:

|ψ〉S =1√

3!2!1!

∣∣∣∣∣∣|1〉1 |1〉1 |4〉1|1〉2 |1〉2 |4〉2|1〉3 |1〉3 |4〉3

∣∣∣∣∣∣+

=1√12

( 2|1〉1|1〉2|4〉3 + 2 |1〉1|4〉2|1〉3 + 2 |4〉1|1〉2|1〉3)

=1√3

( |114〉+ |141〉+ |411〉) , (1.41)

which is conveniently normalised.

Exercise 1.4. Repeat the previous example for a system of three fermions. Do you needto do any calculation? Find the state of the system in terms of single-particle states forthree fermions in the occupation-number state |1, 0, 0, 1, 0, 1, 0, ...〉.

The space spanned by the occupation number basis with arbitrary particle number isdenoted by Fock space F and is given by

F = F0 ⊕F1 ⊕F2 ⊕ ... , FN = span

|n1, n2, ...〉

∣∣∣∣∣∑k

nk = N

. (1.42)

Therefore, the Fock space is spanned by all states |n1, n2, ...〉 without restriction in the totalnumber of particles. In other words, the Fock space is nothing but an Hilbert space containingall vectors which describe a system of arbitrary number of states, including that with noparticles at all, the vacuum |0〉 = |0, 0, 0, ...〉. In Table 1.1 some basis states for bosons andfermions are shown in the occupation number representation.

In conclusion, the Fock space is obtained by combining the occupation number states|n1, n2, ...〉 with arbitrary number of particles N = 0, 1, 2, 3, ... (N = 0 correspond to thevacuum), i.e. with no restrictions on

∑k nk. States with different occupation numbers are

orthogonal〈n1, n2, ...|n′1, n′2, ...〉 = δn1,n′1

δn2,n′2... , (1.43)

10


which means that, for instance, states with different number of particles do not overlap. InFock space, the closure relation is∑

n1,n2,...

|n1, n2, ...〉〈n1, n2, ...| = 1 . (1.44)

By now you may wonder why we have introduced this new type of space F which describesystems with arbitrary number of particles. In fact, the operators we are used to (position,momentum, energy,...) do not alter the number of particles, they just act on each HilbertsubspaceHk. The reason why Vladimir Fock came up with the idea of this kind of mathematicalobjects is that the number of particles is not conserved in Nature. In fact, an electron anda positron can annihilate into two photons or a Higgs boson can decay into two photons(although this is a more complicated process). Therefore, perhaps writing the usual operatorsin terms of others which change the number of particles may turn out to be useful in theanalysis of systems which would be otherwise intractable. This is what we will do in thefollowing sections.

1.3.2 Bosonic creation and annihilation opperators

We now take the first step in the direction of connecting first and second quantisation byintroducing a new kind of operators which act as vehicles to move among states with differentoccupation numbers, i.e. with different number of particles. For that, let us define the bosoniccreation operator (BCO) b†i which raises the occupation number ni by one unit. Then, takingadvantage of the occupation number representation,

b†i |n1, n2, ..., ni−1, ni, ni+1, ...〉 = Bi+|n1, n2, ..., ni−1, ni + 1, ni+1, ...〉 , (1.45)

where Bi+ is a normalisation constant to be determined later. It is straightforward to see that

for b†i , the only nonvanishing matrix elements are:

〈n1, n2, ..., ni−1, ni + 1, ni+1, ...| b†i |n1, n2, ..., ni−1, ni, ni+1, ...〉 . (1.46)

The operator adjoint to b†i can be found by complex conjugation since

〈n1, n2, ..., ni−1, ni + 1, ni+1, ...| b†i |n1, n2, ..., ni−1, ni, ni+1, ...〉∗

= 〈n1, n2, ..., ni−1, ni, ni+1, ...| (b†i )† |n1, n2, ..., ni−1, ni + 1, ni+1, ...〉 . (1.47)

Thus, one can define the bosonic annihilation operator (BAO) bi = (b†i)† which acts on

occupation number states and lowers the occupation of state |i〉 by one unit, i.e.

bi |n1, n2, ..., ni−1, ni, ni+1, ...〉 = Bi−|n1, n2, ..., ni−1, ni − 1, ni+1, ...〉 , (1.48)

where again Bi− is a normalisation constant. If ni = 0, then

bi |n1, n2, ..., ni−1, 0, ni+1, ...〉 = 0 , (1.49)

because the occupation of a state must be ni ≥ 0 (in the above equation the 0 in theright-hand side is not the vacuum state).

11


The BCOs and BAOs are therefore the fundamental operators when working in theframework of occupation number representation. Their action in the Hilbert spaces HN is:

b†i : HN −→ HN+1 , bi : HN −→ HN−1 . (1.50)

Therefore, upon acting on the one-particle states |i1, i2, ..., iN〉 we have:

b†j : |i1, i2, .., iN〉 −→ |i1, i2, .., iN〉|ij〉N+1 ,

b†k : |i1, i2, .., iN〉 −→ |i1, i2, .., iN〉|ik〉N+1, (1.51)

which means that b†j (b†k) creates a particle in state |ij〉 (|ik〉), with label N+1. It is important

to infer some properties of b†j and bj. First, we note that since identical bosons occupysymmetric states, then b†j and b

†k commute

[b†j, b†k] = 0 =⇒ b†j b

†k = b†kb

†j , (1.52)

This can be proved by using relations (1.51) which give

b†j b†k : |i1, i2, .., iN〉 −→ |ψ〉jk ≡ |i1, i2, .., iN〉|ik〉N+1|ij〉N+2 ,

b†k b†j : |i1, i2, .., iN〉 −→ |ψ〉kj ≡ |i1, i2, .., iN〉|ij〉N+1|ik〉N+2. (1.53)

Since |ψ〉kj = Pjk|ψ〉jk and the state of the system must be symmetric under permutations oftwo particles (remember we are dealing with bosons) we conclude that b†j b

†k = b†k b

†j . Following

the same procedure, or just by taking the Hermitian conjugate of b†j b†k = b†k b

†j, we can prove

that:[bj, bk] = 0 =⇒ bj bk = bkbj . (1.54)

Exercise 1.5. Prove that [bj, b†k] = δjk.

The commutation relations for the BCOs and BAOs are then:

[b†j, b†j] = 0 , [bj, bj] = 0 , [bj, b

†k] = δj,k . (1.55)

By definition the operators b† and b are not Hermitian. Instead, b†b and bb† are. In fact, wewill now show that b†j bj is the occupation number operator nj defined in Eq. (1.39), i.e. theoperator which counts the number of particles nj in state |ij〉. From (1.55) it can be shownthat

[b†j bj, bj] = −bj , [b†j bj, b†j] = b†j . (1.56)

Exercise 1.6. Prove the commutation relations (1.56).

In addition, for any |ψ〉 = |n1, n2, ..., nj, ...〉 ∈ F (with nj 6= 0) we have that

〈ψ|b†j bj|ψ〉 = |Bj−|2 ≥ 0 , (1.57)

12


since this in nothing more than ||bj|ψ〉||2, which is always positive. Let us now define theproperties of bj|ψ〉 by writing:

(b†j bj)bj|ψ〉 =Eq. (1.56)

(bj b†j − 1)bj|ψ〉 =

Eq. (1.57)(|Bj−|2 − 1)(bj|ψ〉) . (1.58)

This means that when bj acts on a eigenstate of b†j bj gives an eigenstate of b†j bj with eigenvaluereduced by one unit. By applying repeatedly bj to |ψ〉 one eventually encounters zero atsome point, otherwise we could find a negative eigenvalue, which contradicts (1.57) sinceb†j bj|ψ〉 = |Bj

−|2|ψ〉. So, zero is also an eigenvalue of b†j bj . By using the second commutationrule of (1.56), one can prove that

(b†j bj)b†j|ψ〉 =

Eq. (1.56)b†j(b

†j bj + 1)|ψ〉 =

Eq. (1.57)(|Bj−|2 + 1)(b†j|ψ〉) , (1.59)

which shows that b†j|ψ〉 is an eigenstate of b†j bj with eigenvalue increased by one unit. Theabove two results show that b†j bj is the occupation number operator nj defined in Eq. (1.39),i.e.

nj|n1, n2, ..., nj, ...〉 = b†j bj|n1, n2, ..., nj, ...〉 = nj|n1, n2, ..., nj, ...〉 . (1.60)

From this we also conclude that, since 〈ψ|b†j bj|ψ〉 = |Bj−|2 = nj, one has Bj

− =√nj (apart

from a phase factor which we consider to be 1), leading to [see Eq. (1.48)]:

bj |n1, n2, ..., nj, ...〉 =√nj |n1, n2, ..., nj − 1, ...〉 . (1.61)

In a similar way, the relation

b†j |n1, n2, ..., nj, ...〉 =√nj + 1 |n1, n2, ..., nj + 1, ...〉 . (1.62)

can be proved for the BCOs (Exercise 1.7). It is important to stress that although the abovediscussion has been carried out for a single mode j, all the above results can be generalisedfor the total system since BAOs and BCOs for different modes commute. In particular, atotal number particle operator N which counts the total number of particles N =

∑j nj can

be defined as

N |n1, n2, ...〉 =∑j

nj|n1, n2, ...〉 =

[∑j

(b†j bj)

]|n1, n2, ...〉 =

(∑j

nj

)|n1, n2, ...〉

= N |n1, n2, ...〉 . (1.63)

Exercise 1.7. Prove (1.62).

Since (b†j)nj | 0〉 =

√nj! |nj〉, we can now define the relation which allows us to construct

any state |n1, n2, ...〉 ∈ F starting from the vacuum state |0〉 ≡ |0, 0, ...〉, namely:

|n1, n2, ...〉 =(b†1)n1(b†2)n2 ...√

n1!n2!...|0〉 . (1.64)

Therefore, the connection between first and second quantised boson states is now clear, i.e.

|i1, i2, ..., iN〉B =1√

n1!n2!...S+|i1, i2, ..., iN〉 ≡ |n1, n2, ...〉 =

(b†1)n1(b†2)n2 ...√n1!n2!...

|0〉 . (1.65)

13


1.3.3 Fermionic creation and annihilation opperators

We now define the fermionic creation c† and annihilation operators c (which from now on willbe shortly referred as FCOs and FAOs, respectively). Acting on the occupation number states|n1, n2, ..., nj, ...〉 the FCOs give:

c†j|n1, n2, ..., nj, ...〉 = Cj+|n1, n2, ..., nj + 1, ...〉 ,

cj|n1, n2, ..., nj, ...〉 = Cj−|n1, n2, ..., nj − 1, ...〉 . (1.66)

Since we are dealing with fermions, some properties of the normalisation constants Cjcan already be inferred. Namely, since for a system of N identical fermions nj = 0, 1 weimmediately conclude that if nj = 1, then Cj

+ = 0 (since each state can be occupied at mostby one particle). Obviously, if nj = 0 then Cj

− = 0 because an unoccupied state cannot befurther emptied. Thus

c†j|..., 1j, ...〉 = 0 , cj|..., 0j, ...〉 = 0 . (1.67)Contrarily to what happened for bosons, the order of the occupied states has a meaning sincecreating particle 1 in state j and particle 2 in state k is not the same than creating particle 1in state k and after particle 2 in state j. In fact, due to antisymmetry of identical fermionstates |nk = 1, nj = 1〉 = −|nj = 1, nk = 1〉. Therefore, in general

|..., nj = 1, ..., nk = 1, ...〉 = −|..., nk = 1, ..., nj = 1, ...〉 . (1.68)

Notice that this operation corresponds to switching the column of the state |ik〉 with that ofstate |ij〉 in the Slater determinant | |− of Eq. (1.37), giving the desired minus sign. Therefore,this means that, for j 6= k:

c†j c†k = −c†kc

†j =⇒ c†j, c

†k = 0 , (1.69)

i.e. FCOs anti commute. One can arrive at the same conclusion by working with single-particlestates, as done in Eq. (1.51). By Hermitian conjugation we immediately conclude that ckand cj anticommute. A general vector state in the occupation number state representation isgiven by:

|n1, n2, ...〉 = (c†1)n1(c†2)n2 ...|0〉 , nk = 0, 1 . (1.70)With this, we can see how a FCO acts on a general state |n1, n2, ...〉, namely:

c†j|..., nj, ...〉 = (1− nj)(−1)n<j |..., nj + 1, ...〉 , n<j =

∑k<j

nk . (1.71)

Obviously, if level |ij〉 is already occupied the result must be zero (this is why the factor nj−1

appears). The phase factor (−1)n<j stems from the number of anticommutations needed to

bring c†j to position j in Eq. (1.71).


The above relation sets the value of the normalisation constant Cj+ = (1 − nj)(−1)n

<j in

(1.66). As for the action of the FAO cj, it can be shown that (Exercise 1.9)

cj|..., nj, ...〉 = nj(−1)n<j |..., nj − 1, ...〉 (1.72)

which sets Cj− = nj(−1)n

<j in (1.66).

14

1.4. SECOND-QUANTISED OPERATORS FORMALISM


By using (1.71) and (1.72) we arrive at the following relations:

cj c†j|..., nj, ...〉 = (1− nj)|..., nj, ...〉 ,

c†j cj|..., nj, ...〉 = nj|..., nj, ...〉 ,ck, c†j = δkj . (1.73)

Remember that for fermions nj = 0, 1. The second relation above shows that the occupationnumber operator nj = c†j cj also for fermions (although its eigenvalues can be nj = 0, 1).

Exercise 1.10. Prove relations (1.73).

As we have done for the bosonic case in Eq. (1.65), we can now establish the connection

|i1, i2, ..., iN〉F = S−|i1, i2, ..., iN〉 = c†i1 c†i2...c†iN |0〉 . (1.74)

between first and second quantised fermion states.

1.4 Second-quantised operators formalism

In second quantisation we can express any operator in terms of the fundamental creation andannihilation operators. The way to go from first to second quantised operators is based onthe relations between one-particle state and occupation number representations defined inthe previous section.

1.4.1 One-particle operators

We start by considering a general operator O in a Hilbert space SN or AN (for N boson orfermion system, respectively). Since indistinguishability implies that any physical quantity isinvariant under permutation of particles. Therefore, for any permutation P

〈i1, i2, ..., iN |O|i′1, i′2, ..., i′N〉 = 〈iP1 , iP2 , ..., iPN |O|i′P1, i′P2

, ..., i′PN 〉 . (1.75)

If the action of an operator W on a state |i1, i2, ..., iN〉 belonging to HN is the sum of actingan operator wk to each particle state |ik〉, then this operator is a one-body operator and:

W =N∑k=1

wk , W |i1, i2, ..., iN〉 =N∑k=1

wk |i1, i2, .., ik, ..., iN〉 . (1.76)

In the above equation, wk only acts on the state of particle k, i.e. |ik〉 and, therefore, onecan show that for non-orthogonal states the matrix elements are given by:

〈i′1, i′2, ..., i′k, ..., i′N |W |i1, i2, ..., ik, ..., iN〉 =N∑k=1

〈i′k|wk|ik〉 . (1.77)

The above relation shows that the properties of W depend solely on its matrix elements inthe Hilbert space H of each particle.

15


Exercise 1.11. Prove relation (1.77).

We now address the question of how to write one-body operators in second quantisedlanguage. For that we simplify the notation and just express the matrix elements of theone-particle operator wk by:

[wk]ij = 〈i|wk|j〉 = wij . (1.78)

Since we are dealing with a system of identical particles we have soften the notation by justwriting [wk]ij ≡ wij. With this, we can define the operator wk and W =

∑k=1,...,N wk as

wk =∑i,j

wij|i〉k〈j|k , W =∑i,j

wij

N∑k=1

|i〉k〈j|k . (1.79)

What we have just obtained is the representation of the one-particle operator in the one-particle state basis |i〉. Our goal is now to express the same operator in such a way that itcan act on states |n1, n2, ...〉. In other words, we are looking for a representation of W interms of creation and annihilation operators.

Let us first consider the bosonic case and apply W to a state |n1, n2, ...〉. From Eq. (1.79),the relevant term to consider is then

N∑k=1

|i〉k〈j|k|n1, n2, ...〉 ≡1√

n1!n2!...

N∑k=1

|i〉k〈j|kS+|i1, i2, ..., iN〉 . (1.80)

Consider for now i 6= j. The first thing to notice is that∑N

k=1 |i〉k〈j|k is a symmetric operatorand, therefore, S+ commutes with it. This allows us to write

N∑k=1

|i〉k〈j|k|n1, n2, ...〉 =1√

n1!n2!...S+

N∑k=1

|i〉k〈j|k|i1, i2, ..., iN〉 . (1.81)

In the above equation, each time ik = j then ik must be replaced by i. This can be easilyseen in the following way: suppose ik = j, then

|i〉k〈j|k |i1, i2, ik−1, j, ik+1, ..., iN〉 = |i〉k|i1, i2, ik−1〉(〈j|j〉k)︸︷︷︸=1

|ik+1, ..., iN〉

= |i1, i2, ik−1, i, ik+1, ..., iN〉 . (1.82)

Therefore, for each of these occurrences the occupation number nj (ni) decreases (increases)by one unit. Symmetrisation of each of these configurations yields to the occupation numbervector: √

n1!...(ni + 1)!...(nj − 1)!... |n1, ..., ni + 1, ..., nj − 1, ...〉 . (1.83)

This result will appear for each occurrence of j, i.e. nj times. Summing everything up, wethen get

S+

N∑k=1

|i〉k〈j|k|i1, i2, ..., iN〉 = nj

√n1!...(ni + 1)!...(nj − 1)!... |n1, ..., ni + 1, ..., nj−1, ...〉 .

(1.84)

16


We then have from Eqs. (1.81) and (1.84)N∑k=1

|i〉k〈j|k|n1, n2, ...〉 =√nj√ni + 1 |n1, ..., ni + 1, ..., nj − 1, ..〉

= b†i bj|1, ..., ni, ..., nj, ...〉 . (1.85)

where the properties for BCOs and BAOs given in Eqs. (1.61) and (1.62) have been used. Ifi = j, after summing in k in Eq. (1.82), the state |i〉 will appear ni times and it is easy tosee that

N∑k=1

|i〉k〈i|k|n1, n2, ...〉 = ni|1, ..., ni, ...〉 = b†i bi|1, ..., ni, ...〉 . (1.86)

Putting everything together, we conclude thatN∑k=1

|i〉k〈j|k = b†i bj . (1.87)

Let us now analyse the case of fermions. For the state |i1, i2, ..., iN〉 we assume i1 < i2 <... < iN . The equivalent of Eq. (1.81) is

N∑k=1

|i〉k〈j|k|n1, n2, ...〉 ≡N∑k=1

|i〉k〈j|kS−|i1, i2, ..., iN〉 = S−

N∑k=1

|i〉k〈j|k|i1, i2, ..., iN〉 .

(1.88)The difference with respect to the boson case is that the state |j〉 can appear only once in|i1, i2, ..., iN〉. If this happen, then it will be replaced by |i〉, i.e. we will have |i1, i2, ..., i, ..., iN〉.Obviously, antisimmetrization will only give a nonzero result if the state |i〉 is not initiallypresent in |i1, i2, ..., iN〉. So, the only possible nontrivial configuration is ni = 0, nj = 1.Therefore, summing in k we have

S−

N∑k=1

|i〉k〈j|k|i1, i2, ..., iN〉 = S−|i1, i2, ..., iN〉j→i . (1.89)

In the above equation, j → i means that |ij〉 is replaced by |ii〉. When doing this, we alterthe order of the states initially postulated. In order to bring |i〉 to the right position, wemust perform the appropriate number of permutations. in particular, if i ≤ j the number ofpermutations is n<j +n<i , where the notation of Eq. (1.71) has been used. On the other hand,for i > j, the number of permutations is n<j + n<i − 1. This can be shown to yield

N∑k=1

|i〉k〈j|k = c†i cj . (1.90)

for fermions. Inserting this result and that of Eq. (1.87) into Eq. (1.79), one can write

W =∑k

wk =∑ij

wij a†i aj , wij = 〈i|w|j〉 , (1.91)

where a ≡ b (c) for bosons (fermions). The matrix elements wij are calculated as done infirst quantisation. The graphical representation of a one-particle operator in second-quantisedlanguage is shown in Fig. 1.1 on the left. The operator first annihilates the state |j〉 and thencreates the state |i〉 with an amplitude given by the matrix element wij.

17


Figure 1.1: Graphical representation of one- and two-particle operators (left and rightdiagrams, respectively).

Exercise 1.12. Consider a system of non-interarcting particles described by single-particlestates |i〉, which are eingenstates of the Hamiltonian H0 with energies Ei. Show that thesecond quantised form of H0 is: H0 =

∑iEini.

1.4.2 Many-particle operators

We now define many-body operators T for which the action on the N -particle state|i1, i2, ..., iN〉 can be represented as the sum of operators tαβ acting in sets of more than oneparticle. We start by two-body operators which are sum of operators acting on all distinctpairs of particles (αβ), i.e.

T |i1, i2, ..., iN〉 =∑α<β

tαβ|i1, i2, ..., iN〉 , (1.92)

where it should be clear that tαβ only acts on particles α and β. In order to ensure summationover distinct pairs only we have imposed the constraint k < j. Since T must obey thesymmetry property given in Eq. (1.75), we conclude that tαβ = tβα must be verified. Withthis, we can rewrite (1.92) as

T |i1, i2, ..., iN〉 =1

2

∑α 6=β

tαβ|i1, i2, ..., iN〉 , (1.93)

where the factor 1/2 takes care of the fact that we have double the number of each termby relaxing the constraint α < β to α 6= β. Similarly to what we have done for one-bodyoperators, we can show that for non orthogonal states

〈i′1, i′2, ..., i′N |T |i1, i2, ..., iN〉 =1

2

∑k 6=j

〈i′ki′j|t|ikij〉 , (1.94)

which shows that the properties of T are determined by its matrix elements 〈i′ki′j|t|ikij〉 inthe 2-particle hilbert space H2.

18


Exercise 1.13. Prove relation (1.94) and generalise it for a potential which can bewritten as a sum of n-particle potentials Vi1,i2,...,in acting in all possible n-particle subsetsof the system.

We now write T in the one-particle eigenstate basis. Namely, we have:

T =1

2

∑α 6=β

tαβ =1

2

∑α 6=β

∑ijkl

tijkl|i〉α|j〉β〈k|α〈l|β

=1

2

∑ijkl

tijkl∑α 6=β

|i〉α|j〉β〈k|α〈l|β , tijkl ≡ 〈ij|t|kl〉 . (1.95)

The matrix elements tijkl are computed as usual. We now concentrate on the last sum of theabove equation. We start by noting that∑

α 6=β

|i〉α|j〉β〈k|α〈l|β =∑α,β

|i〉α|j〉β〈k|α〈l|β −∑α=β

|i〉α|j〉α〈k|α〈l|α

=∑α

|i〉α〈k|α∑β

|j〉β〈l|β −∑α

|i〉α〈l|αδkj (1.96)

Using now the results of Eqs. (1.87) and (1.90), one has∑α

|i〉α〈k|α = a†i ak ,∑α

|j〉α〈l|α = a†j al ,∑α

|i〉α〈l|αδkj = a†iδkj al . (1.97)

One therefore concludes that:∑α 6=β

|i〉α|j〉β〈k|α〈l|β = a†i a†j alak , (1.98)

being the result the same for bosons and fermions. Once more, a ≡ b (c) for bosons (fermions).

Exercise 1.14. Using (1.97) and using the (anti)commutation relations for (fermionic)bosonic creation and annihilation operators, prove (1.98).

Replacing now (1.98) into (1.95),we get the final result:

T =1

2

∑α 6=β

tαβ =1

2

∑ijkl

tijkla†i a†j alak , tijkl ≡ 〈ij|t|kl〉 , (1.99)

which is the general form for two-particle operators in second-quantised language. Thegraphical representation of a two-particle operator in second-quantised language is shown inFig. 1.1 on the right. The operator first annihilates the states |k〉 and |l〉 and then createsthe states |j〉 and |i〉 with an amplitude given by the matrix element tijkl.

19


1.4.3 Change of basis and field operators

Up to now, we have developed our second-quantised formalism using the single-particle states|α〉, for which the wave functions are ψα(r) ≡ 〈r|α〉 where |α〉 = a†α|0〉. It is howeveruseful to consider alternative basis like, for instance, |r〉 and find the relations betweenoperators in the two basis, in particular aα and aα. Therefore, we will now see how to changethe representation of our operators from one basis to the other within the second quantisationformalism. For that, we consider a different single-particle basis |α〉 described by thewavefunctions ψα(r) = 〈r|α〉, where |α〉 = a†α|0〉. The closure relation for both single-particlebasis allow us to write: ∑

α

|α〉〈α| =∑α

|α〉〈α| = 1 (1.100)

We can now write

a†α|0〉 = |α〉 =∑α

〈α|α〉|α〉 =∑α

〈α|α〉 a†α|0〉 ⇒ a†α =∑α

〈α|α〉 a†α . (1.101)

The last relation and its hermitian conjugate determine that the creation and annihilationoperators in two distinct single-particle basis |α〉 and |α〉 are related by

a†α =∑α

〈α|α〉 a†α , aα =∑α

〈α|α〉 aα . (1.102)

Exercise 1.15. .a) Obtain the transformation relation between the wave functions ψα(r) and ψα(r).b) Show that the basis transformations Uαα ≡ 〈α|α〉 are unitary transformations.c) Show that the creation and annihilation operators in the new basis satisfy the samecommutation relations.c) Show that the basis transformations leave the number of particles unchanged.

One particularly interesting basis is that of the position eigenstates |r〉. It is important toobtain the representation of creation and annihilation operators in this basis. These are usuallycalled field operators and are denoted by ψ†(r), ψ(r). The form of these operators can beobtained by using relations (1.102), namely:

ψ†(r) ≡ a†r =∑α

〈α|r〉 a†α =∑α

ψ∗α(r) a†α

ψ(r) ≡ ar =∑α

〈r|α〉 aα =∑α

ψα(r) aα .

(1.103)

(1.104)

The field operators are nothing but the sum of all ways of adding (or removing) a particle tothe system through any of the basis sates |α〉. Suppose that α ≡ p, then

ψ†(r) =∑p

〈p|r〉 a†p =∑p

ψ∗p(r) a†p (1.105)

ψ(r) =∑p

〈r|p〉 ap =∑p

ψp(r) ap . (1.106)

20


If one considers plane waves in a box with volume V , sides Li with i = x, y, z with periodicboundary conditions in each coordinate, then we have the momentum eigenstates

ψp(r) =eip.r/~√V

, pi =2~πniLi

(ni = 0,±1,±2, ...) , (1.107)

and, we get for the field operators ψ†(r), ψ(r)

ψ†(r) =∑p

e−ip.r/~√V

a†p , ψ(r) =∑p

eip.r/~√V

ap . (1.108)

The operator ψ†(r) adds a particle to the system in a superposition of momentum stateswith amplitude eip.r/~/

√V . In order to prove that we must prove that |r〉 = ψ†(r)|0〉. The

proof is quite simple

ψ†(r)|0〉 =∑p

e−ip.r/~√V

a†p|0〉 =∑p

e−ip.r/~√V|p〉 =

∑p

|p〉〈p|r〉 = |r〉 . (1.109)

The commutation relations for field operators are easy to obtain from the ones of the creationand annihilation operators using the basis transformation properties (1.103) and (1.104).

[ψ(r), ψ(r′)]± = 0 , [ψ†(r), ψ†(r′)]± = 0 , [ψ(r), ψ†(r′)]± = δ(r − r′) , (1.110)

where we have used for simplicity the notation [ ]+ ≡ and [ ]− ≡ [ ]. The com-mutation relations for the field operators show that there is no interference if we destroy orcreate two bosons or fermions in any point in space. Still, creating one and then destroying itat the same point is not the same to destroying it and then creating it. In some sense, theconcept of quantum field operator expresses the wave/particle duality in quantum physics.On the one hand ψ†(r), ψ(r) are defined as fields, i.e. as a kind of waves, but on the otherhand they obey the particle commutation relations.

Exercise 1.16. Obtain the (anti)commutation relations given in (1.110).

It is often useful to transform between the position and momentum representations, i.e.to perform a Fourier transformation. Given the field operators ψ†(r), ψ(r), the creation andannihilation operators in momentum space are given by:

a†q =1√V

∫eiq.r/~ψ†(r)d3r , aq =

1√V

∫e−iq.r/~ψ(r)d3r , (1.111)

which can be proved by multiplying Eq. (1.108) by e±iq.r/~ and integrating over r.We can now see how operators are written in terms of field operators. For that we write

Eq. (1.91) as

W =∑ij

∫ψ∗i (r) w ψj(r)a†i aj d

3r , (1.112)

21


which, after considering the definition of field operator given in (1.104), leads to

W =

∫ψ†(r) w ψ(r) d3r , (1.113)

We can do the same for the two-particle operators, taking care of the fact that the states|ij〉 mean that of the two particles one is in state |i〉 and the other in state |j〉. Therefore,different space coordinate labels must be used for the field operators. It can be shown that:

T =1

2

∫ψ†(r)ψ†(r′) t ψ(r′)ψ(r)d3rd3r′ , (1.114)

in terms of the field operators.We are now in the position of defining some operators in second quantised form. To

simplify the notation, we will sometimes in this section speak of wave number k as beingmomentum p, remembering that the relation between them is p = ~k.

• Momentum operator - P

We consider the states |i〉 as being the one-particle states with momentum ~k, namely |k〉.Using Eq. (1.91), we can write

P =∑k

pk =∑k,k′

〈k|p |k′〉a†kak′ =∑k

~ka†kak (1.115)

Exercise 1.17. Obtain the result (1.115) using the representation of the momentumoperator in terms of field operators [use Eq. (1.112)].

In terms of field operators we have:

P =

∫ψ†(r) p ψ(r) d3r = −i~

∫ψ†(r)∇ ψ(r) d3r , (1.116)

• Kinetic energy - K

Similarly to what we have done for the momentum operator P , for the kinetic energy wehave:

K =∑k

~2k2

2ma†kak (1.117)

In terms of field operators, K can be written taken into account its form in configurationspace and Eq. (1.112). Namely, one has

K = − ~2

2m

∫ψ†(r)∇2 ψ(r) d3r =

~2

2m

∫∇ψ†(r)∇ ψ(r) d3r , (1.118)

where the last equality has been obtained integrating by parts and considering that the wavefunction where the momentum operator is acting decreases sufficiently fast to zero in such away that the surface term in the integration can be neglected.

22


Exercise 1.18. Use Eq. (1.118) and the wave functions given in (1.107) to demonstrateEq. (1.117).

• Total number operator - N , and density operator n(r)

As you remember from QMI, the first-quantised form of the density operator is n(r) =δ(3)(r − r′). We can now write the number density operator in second quantisation usingEq. (1.91).

n(r) =∑ij

∫ψ∗i (r

′) δ(3)(r − r′)ψj(r′)a†i aj d3r′ =∑ij

ψ∗i (r)a†i ψj(r)aj = ψ†(r)ψ(r) .

(1.119)

Notice that the particle density operator in terms of field operators looks like the probability offinding a particle in the state ψ(r). However, remember that ψ(r) is an operator, while ψ(r)is a complex function. This correspondence is responsible for the term second quantisationsince the operators in the second-quantised language are obtained by replacing ψ(r) by ψ(r)in the single particle densities.

With the number density operator obtained in Eq. (1.119) we can define the total numberoperator which is

N =

∫n(r)d3r =

∫ψ†(r)ψ(r)d3r . (1.120)

• Single and two-particle potential - U , V

Consider the one particle potential U(r) like, for instance, the Coulomb potential felt byan electron due to the presence of a nucleus. According to Eq. (1.91), the correspondingsecond-quantised operator will be given by:

U =∑i,j

∫a†iψ

∗i (r)U(r)ψj(r)ajd

3r, (1.121)

which, in terms of field operators takes the form:

U =

∫ψ†(r)U(r)ψ(r)d3r . (1.122)

For a two-particle potential V (r, r′), we use once more Eq. (1.99) to define the correspondingoperator in terms of creation and annihilation operators.

V =1

2

∑i,j,k,m

∫ψ∗i (r)ψ∗j (r

′)V (r, r′)ψk(r)ψm(r′)a†i a†j amakd

3rd3r′ . (1.123)

The corresponding representation using field operators is straightforward from Eq. (1.124)

V =1

2

∫ψ†(r)ψ†(r′)V (r, r′) ψ(r′)ψ(r)d3rd3r′ , (1.124)

23


With the above results we can now write the second-quantised form of the Hamiltonianof a many-particle quantum system in terms of field operators:

H =

∫ [~2

2m∇ψ†(r)∇ ψ(r) + ψ†(r)U(r)ψ(r)

]d3r

+1

2

∫ψ†(r)ψ†(r′)V (r, r′) ψ(r′)ψ(r)d3rd3r′ , (1.125)

where we have assumed that there are only one and two-particle interactions involved.

1.4.4 Momentum representation, spin and field equations

Within the context of second quantisation, the momentum representation is frequently usedto express operators in terms of the creation and annihilation operators ak and a†k. For theone-particle potential we have, using the momentum eigenfunctions normalised in a box, thefollowing matrix elements

〈k′|U |k〉 =

∫ψ∗k′(r)U(r)ψk(r)d3r =

1

V

∫e−i(k

′−k).rU(r)d3r ≡ 1

VUk′−k , (1.126)

where Uk′−k denotes the Fourier transform of U defined as: Ak =∫e−ik.rA(r)d3r with the

corresponding inverse: A(r) =∑kAke

ik.r/V .For the two particle potential V (r − r′) we proceed in the same way to arrive to the

result:

〈p′,k′|V |p,k〉 =1

V 2

∫e−ip

′.re−ik′.r′V (r − r′)eik.r′eip.rd3rd3r′ . (1.127)

Using the inverse Fourier transform, we can further write:

〈p′,k′|V |p,k〉 =1

V 3

∑q

Vq

∫d3r

∫e−ip

′.r−ik′.r′+iq.(r−r′)+ik.r′+ip.rd3r′

=1

V

∑q

Vq

∫1

Vei(q−p

′+p).rd3r

∫1

Vei(k−k

′−q).r′d3r′

=1

V

∑q

Vqδp′,q+p δk′,k−q , (1.128)

where in the last step of the calculation the integral relation:∫V

ei(k−k′).r d3r = V δkk′ , (1.129)

has been used.We can now use the results of equation Eqs. (1.117), (1.126) and (1.128) together with

Eqs. (1.112) and (1.124) to write the Hamiltonian in the form

H =∑k

~2k2

2ma†kak +

1

V

∑k′,k

Uk′−ka†k′ ak +

1

2V

∑q,p,k

Vqa†p+qa

†k−qakap , (1.130)

24


Figure 1.2: Graphical representation of kinetic (left diagram) one- and two-particle spin-independent potential operators (center and right diagrams, respectively) in the momentumrepresentation.

The interpretation of the interaction terms above can be understood diagrammatically asshown in Fig. (1.2).

Up to now we have not included spin in our definitions of the creation and annihilationoperators. In second-quantised formalism, spin is just another quantum number that has tobe included to label states. For instance, in the momentum representation we now define thecommutation relations:

[ak,σ, ak′,σ′ ]± = [a†k,σ, a†k′,σ′ ]± = 0 , [ak,σ, a

†k′,σ′ ]± = δk,k′δσ,σ′ , (1.131)

where σ, σ′ denote spin states. The generalisation of the total Hamiltonian operator (1.130)to include spin degrees of freedom is straightforward since the only aspect to be taken intoaccount is the sum in all possible states. Therefore, for a totally spin-independent Hamiltonianwe have

H =∑k,σ

~2k2

2ma†k,σak,σ +

1

V

∑k′,k,σ

Uk′−ka†k′,σak,σ +

1

2V

∑q,p,k,σ,σ′

Vqa†p+q,σa

†k−q,σ′ ak,σ′ ap,σ ,

(1.132)

As for the field operators, ψ†(r) and ψ(r) are now replaced by ψ†σ(r) and ψσ(r) denoting theoperators which create or annihilate a particle at a given point r is a spin state σ. Therefore,

ψ†σ(r) ≡ a†r,σ =∑α

〈α|r, σ〉 a†α =∑α

ψ∗α(r, σ) a†α

ψσ(r) ≡ ar,σ =∑α

〈r, σ|α〉 aα =∑α

ψα(r, σ) aα ,

(1.133)

(1.134)

which are the generalisation of Eqs. (1.103) and (1.104) to the case in which spin is considered.Notice that in the above expression, ψα(r, σ) is the σ component of the wavefunction (spinor)for the particle in state α. After the inclusion of spin, new (anti)commutation relations forfield operators should be defined, namely

[ψσ(r), ψσ′(r′)]± = 0 , [ψ†σ(r), ψ†σ′(r

′)]± = 0 , [ψσ(r), ψ†σ′(r′)]± = δσ,σ′δ(r − r′) ,

(1.135)

25


which generalise those in (1.110).We will now define the field equations for ψσ(r, t), i.e. the Schrödinger equation which

determines the evolution of the field operators. For that, we recall what we have seen in MQIIabout the representation of a time-dependent operator O(t) in the Heisenberg representation,namely

O(t) = eiHt/~Oe−iHt/~ , i~∂O(t)

∂t= −[H, O(t)] = −[H(t), O(t)] , (1.136)

where in the last equality we have used the fact that H(t) = eiHt/~He−iHt/~ = H. To derivethe equations of motion, we must be able to compute commutators and anticommutatorsat an arbitrary time t. This is easier since if [A, B]± = C, then [A(t), B(t)]± = C(t) (youcan prove this as an exercise). This means that the operator evolution in the Heisenbergrepresentation conserves the (anti)commutation relations for operators. We will also needthe useful relation [AB,C]− = A[B,C]± ∓ [A,C]±B, where in the right-hand side of thisequation the upper (lower) sign corresponds to the fermion (boson) case.

Our goal is therefore to determine the explicit form of the Schrödinger equation for ψ(r)with the Hamiltonian given in Eq. (1.125), which including spin (but still with spin-independentinteractions) is

H =∑σ

∫ [~2

2m

∫∇ψ†σ(r)∇ ψσ(r)

]d3r︸︷︷︸

K

+∑σ

∫ [ψ†σ(r)U(r)ψσ(r)

]d3r︸︷︷︸

U

+1

2

∑σ,σ′

∫ψ†σ(r)ψ†σ′(r

′)V (r, r′) ψσ′(r′)ψσ(r)d3rd3r′︸︷︷︸

V

. (1.137)

Let us start from the Heisenberg equation for ψσ(r, t)

i~∂ψσ(r, t)

∂t= −[H, ψσ(r, t)] . (1.138)

Considering the Hamiltonian (1.137) and making use of the commutation relations for thefield operators given in (1.135), and of the commutator properties given in the previousparagraph, we obtain:

−[K, ψσ(r, t)] = − ~2

2m∇2ψσ(r, t) ,

−[U , ψσ(r, t)] = U(r)ψσ(r, t) ,

−[V , ψσ(r, t)] =∑σ′

∫ψ†σ′(r

′, t)V (r, r′)ψσ′(r′, t)d3r′ψσ(r, t) . (1.139)

Exercise 1.19. Compute the commutators given in (1.139).

Plugging (1.139) into the Heisenberg equation (1.138) we finally obtain

i~∂ψσ(r, t)

∂t=

[− ~2

2m∇2 + U(r)

]ψσ(r, t) +

∑σ′

∫ψ†σ′(r

′, t)V (r, r′)ψσ′(r′, t)d3r′ψσ(r, t) ,

(1.140)

which looks like the Schrödinger equation in first quantisation, except for the last term.26

1.5. PROBLEMS

1.5 Problems

Problem 1.1. Applications of commutation rules:

(a) Use commutation rules BCOs and FBOs to prove that [a†kaj, am] = −δmkaj is valid forboth fermions and bosons.

(b) Given the Hamiltonian of non-interacting particles described by one-particle states |αi〉with energies Ei, show that

dakdt

= −iEk~ak.

(c) For an ideal boson gas of spinless particles compute the matrix elements 〈ψ| bi1 bi2 b†i3b†i4|ψ〉,

where |ψ〉 = |..., ni1 , ..., ni2 , ..., ni3 , ..., ni4 , ...〉.Answer: 〈ψ| bi1 bi2 b

†i3b†i4 |ψ〉 = (δi1i3δi2i4 + δi1i4δi2i3)[(1 + ni1)(1 + ni2)− 1/2ni1(1 + ni2)] .

(d) Compute a similar matrix element 〈ψ| c†i1 ci2 c†i3ci4|ψ〉 for fermions.

Answer: 〈ψ| ci1 ci2 c†i3c†i4|ψ〉 = δi1i2δi3i4ni1ni3 + δi1i4δi2i3ni1ni3(1− ni3) .

Problem 1.2. Consider a system of N noninteracting spinless bosons in a three dimensionalbox of volume volume V = L3 at zero temperature.

(a) Obtain the one-particle momentum eigenfunctions using periodic boundary conditions.

Answer: ψk = eik.r/√V , ki = 2πni/L

(b) Describe the ground state of the system |ψ0〉 in both the one-particle state and occupationnumber representations.

Answer: |ψ0〉 = |0〉1|0〉2 ...|0〉N ; , |ψ0〉 = |N, 0, 0, ...〉

(c)Write down the local density operator n(r) = ψ†(r)ψ(r) in terms of annihilation (creation)operators bk (b†k) of particles with momentum p = ~k.

Answer: n(r) = 1V

∑k1,k2

ei(k2−k1).r b†k2 bk1

(d) Compute the average density of bosons n = 〈n(r)〉 in the ground state, as well as thenumber of particles in a sub volume v < V of the system.

Answer: n = N/V , Nv = Nv/V

(e) Calculate fluctuations of the number of particles ∆Nv in volume v by evaluating

(∆Nv)2 = 〈ψ0|N2

v |ψ0〉 − 〈ψ0|Nv|ψ0〉2 ,

where Nv =∫vn(r)d3r.

Answer: (∆Nv)2 = Nv(1− v/V )/V

27


Problem 1.3. Show that the Fourier transform of the density operator n(r) is given bynq =

∑p a†pap+q.

Problem 1.4. Find the eigenstate |ψ〉 of the annihilation operator such that b|ψ〉 = α|ψ〉.

Answer: |ψ〉 = e−|α|2/2∑∞

n=0αn√n!|n〉

Problem 1.5. Consider a system ofN spinless bosons enclosed in a volume V . The interactionbetween pairs of these particles is described by a short-range repulsive potential U(r1−r2) ≥ 0.Compute the ground state energy of the system using first order perturbation theory.

Answer: E0 ' 12U0

N2

Vwith U0 =

∫U(r)d3r.

Problem 1.6. Repeat the previous problem for a system of N spin 1/2 fermions interactingthrough the same type of spin-independent potential obeying kFR0 1, where R0 is thepotential range and kF the Fermi momentum.

Answer: E0 ' E(0)0 + E

(1)0 = 3

10

(3π2NV

)3/2~2mN + 1

4U0

N2

Vwith U0 =

∫U(r)d3r.

Problem 1.7. Find the second-quantised form of the spin operator S =∑N

j=1 sj for asystem of N electrons which, besides spin can be characterised by a set of quantum numbersα. Remember that sj = ~σ/2, where σ are the Pauli matrices.

Answer: S = ~2

∑α

[c†α↓cα↑ + c†α↑cα↓, i(c

†α↓cα↑ − c

†α↑cα↓), c

†α↑cα↑ − c

†α↓cα↓

].

Problem 1.8. Consider a system where particle interactions are described by the contactpotential V = g

2

∑i 6=j δ(ri − rj). Express V in second quantised form.

Answer: V = 12V

∑q

∑k,σ

∑k′σ′ a

†k+q,σa

†k′−q,σ′ ak′,σ′ ak,σ.

Problem 1.9. Field and number operators

(a) Show that the field operators ψσ(r) and the total particle number operator N satisfy thecommutation relations:

[N , ψσ(r)] = −ψσ(r) , [N , ψ†σ(r)] = ψ†σ(r)

for both bosons and fermions. The relation [AB,C]± = A[B,C]±∓ [A,C]±B may be useful.

(b) Consider the operator ψσ(r, θ) = eiNθψσ(r)e−iNθ. Show that ψσ(r, θ) = e−iθψσ(r) andψ†σ(r, θ) = eiθψ†σ(r)

(c) Show that any second-quantised operator A which represents and observable is invariantunder the transformation ψσ(r)→ ψσ(r, θ) and ψ†σ(r)→ ψ†σ(r, θ). Compute [A, N ].

28

Ch

ap

te

r

2Interaction of radiation with matter

“Changing electric fields produce magnetic fields, and changing magnetic fields produceelectric fields. Thus the fields can animate one another in turn, giving birth to

self-reproducing disturbances that travel at the speed of light. Ever since Maxwell, weunderstand that these disturbances are what light is.”

Frank Wilczek, The Lightness of Being (2008)

Up to know we have treated the electromagnetic field and its interaction with matter(charged particles) in a classical way, i.e. electric and magnetic fields were treated as externalagents not subjected to the process of quantization. However, we know from the Planck-Einstein relation that the quanta of the electromagnetic field are discrete particle with energy~ω: the photons. In classical electrodynamics this cannot be explained by just taking theclassical field equations and their solutions, i.e. the Maxwell equations. The first attemptto quantize the e.m. field was done in 1927 by Dirac in his work "The quantum theory ofemission and the absortion of radiation". In particular, he was able to describe spontaneousand induced emission in a natural way (rememeber that we have done this in QMII but notfrom first principles). This was a triumph for Dirac’s quantum theory of radiation and astarting point to generalise quantisation also to matter.

In this chapter we aim at finding a quantum representation of the electromagnetic fieldwhich will account for the existence of photons. This will be done through the process ofquantization. After this we will be able to study quantum processes where not only matterparticles are created and annihilated but also photons. But before doing, let us briefly reviewMaxwell equations and their properties.

2.1 Recap of Maxwell equations

As you have seen in the courses of electromagnetism and classical electrodynamics, theclassical electromagnetic field is described by the Maxwell equations (in Gaussian units):

29

CHAPTER 2. INTERACTION OF RADIATION WITH MATTER

∇×E +1

c

∂B

∂t= 0, (2.1)

∇.D = 4πρ , (2.2)

∇×H − 1

c

∂D

∂t=

4π

cj , (2.3)

∇.B = 0 , (2.4)

where D = (1 + 4πχe)E, H = (1 + 4πχm)B for linear media (χe and χm are the electricand magnetic susceptibilities, respectively). Taking the divergence of Eq. (2.3) and usingEq. (2.2) together with the mathematical relation ∇.(∇× V ) = 0 one obtains:

∇.j +∂ρ

∂t= 0 , (2.5)

which is nothing but the continuity equation for the charge and current densities ρ and j,respectively, describing the transport of the (conserved) electric charge.

The electric and magnetic fields can be written by means of auxiliary potentials ϕ (scalar)and A (vector) as:

E = −1

c

∂A

∂t−∇ϕ , B = ∇×A , (2.6)

which automatically obey Eqs. (2.1) and (2.4). As you know, the above definitions are notunique in the sense that E and B remain invariant under the "gauge transformations"

A→ A′ = A+ ∇χ , ϕ→ ϕ′ = ϕ− 1

c

∂χ

∂t, (2.7)

where χ is a scalar function. If we now replace (2.6) in (2.2) and (2.3), one easily obtains:

∇× (∇×A) +1

c∇∂ϕ

∂t+

1

c2

∂2A

∂t2=

4π

cj , (2.8)

1

c

∂(∇.A)

∂t+ ∇2ϕ = −4πρ. (2.9)

If we now use ∇× (∇×A) = ∇(∇.A)−∇2A 1 and use the gauge transformation freedom(2.7), we can simplify the above equations. In particular, chosing the Lorentz gauge for which

∇.A′ +1

c

∂ϕ′

∂t= 0 , (2.10)

Eqs. (2.8) and (2.9) read:

∇2A′ − 1

c2

∂2A′

∂t2= −4π

cj , (2.11)

∇2ϕ′ − 1

c2

∂2ϕ′

∂t2= −4πρ. (2.12)

1Here, ∇2A represents a vector with components ∇2Ai.

30

2.1. RECAP OF MAXWELL EQUATIONS

Exercise 2.1. .a) Show that E and B remain invariant under (2.7).b) Prove Eqs. (2.8) and (2.9).c) Find the second order order differential equations which allows you to find χ for thecase of the Lorentz gauge.

Let us now consider Maxwell equations in empty space (vacuum), i.e. ρ = 0 and j = 0.In this case, those equations reduce to:

∇×E = −1

c

∂B

∂t, (2.13)

∇.E = 0 , (2.14)∇.B = 0 . (2.15)

∇×B =1

c

∂E

∂t, (2.16)

It can be shown that, in this case, we can choose a gauge function χ such that ϕ′ = 0 and∇.A′ = 0 (Coulomb gauge) which automatically fulfill Eqs. (2.13)-(2.15). In turn, Eq. (2.16)reads

∇2A′ − 1

c2

∂2A′

∂t2= 0. (2.17)

Therefore, from now on we will consider the vacuum case in the Coulomb gauge, for which:

∇2A(r, t)− 1

c2

∂2A(r, t)

∂t2= 0 , ∇.A(r, t) = 0 , ϕ = 0 , (2.18)

where we have dropped the primes for simplicity.

Exercise 2.2. Show that the Maxwell equations are consistent in the Coulomb gaugeonly for ρ = 0 and ∇.j = 0.

We can now ask what are the solutions A(r, t) of Eqs. (2.18). It is not difficult to convinceyourself that the general solution of the first equation in (2.18) is of the form:

A(r, t) = 2εA0 cos(k.r − ωt+ α) = A0ei(k.r−ωt) +A∗0e

−i(k.r−ωt)

= A0ei(k.r−ωt) + c.c. , (2.19)

where "c.c." stands for complex conjugate and A0 = A0εeiα, with ε being the polarization

vector. Imposing the above solution to first equation in (2.18) implies ω = |k|c = kc, whilethe condition ∇.A(r, t) = 0 automatically leads to k.A(r, t) = 0 or k.ε = 0, i.e. the wavevector k is transverse to the polarization vector ε: k⊥A(r, t). It is also commonly said thatA0 is transverse. With the help of Eq. (2.19) one can now compute E(r, t) and B(r, t), inparticular it can be shown that:

E(r, t) = −2kεA0 sin(k.r − ωt+ α) (2.20)B(r, t) = −2A0(k × ε) sin(k.r − ωt+ α) . (2.21)

31


Exercise 2.3. Show Eqs. (2.20) and (2.21).

2.2 Quantization of the electromagnetic field

In order to reach a quantum description of the electromagnetic field, let us consider a finitebox of volume V = L3 and study the radiation field confined inside it. in addition, we imposeperiodic boundary conditions in such a way that the classical field A:

A(0, y, z, t) = A(L, y, z, t) ,

A(x, 0, z, t) = A(x, L, z, t) ,

A(x, y, 0, t) = A(x, y, L, t) . (2.22)

These conditions basically mean that the normal modes allowed for the electromagnetic fieldinside this box are those with wavelengths which are multiples of L. These oscillate at thesame frequency ω and therefore we can write the solution of the wave equation (2.18) as

A(r, t) = A(r)e−iωt , (2.23)

which, after inserted in Eq. (2.18) leads to:(∇2 +

ω2

c2

)A(r) = 0 ⇒ Akλ(r) = Nk εkλ e

ik.r , (2.24)

where εkλ is the polarization vector for the normal mode with momentum k and Nk isa normalization constant which will be defined later. Taking into account the transversalcharacter of electromagnetic waves, one has

εkλ.k = 0 , εkλ.εkλ′ = δλλ′ , λ = 1, 2 , (2.25)

where εk1 and εk2 are two polarization vectors laying in the plane orthogonal to k. Pluggingthe solution (2.24) back into the wave equation, one obtains the frequency-momentumrelation:

|k|2 ≡ k2 = k2x + k2

y + k2z =

ω2k

c2. (2.26)

Since we are in the presence of an electromagnetic field confined to a box with boundaryconditions, the normal modes will be discretized (the passage to free infinite space can bealways done as described in Supplement 1). From the periodic boundary conditions (2.22),the discrete set of normal mode momenta allowed in our finite box is:

k = (kx, ky, kz) =2π

L(nx, ny, nz) , ni ∈ Z . (2.27)

The most general solution for Akλ(r) in Eq. (2.24) will be a superposition of all the abovenormal modes. Therefore, the vector potential A will be just a discrete Fourier decompositionof the form:

A(r, t) =∑k

∑λ

[akλ(t)Akλ(r) + a∗kλ(t)A∗kλ(r)] , (2.28)

32

2.2. QUANTIZATION OF THE ELECTROMAGNETIC FIELD

which, from Eq. (2.24) leads to

A(r, t) =∑k

∑λ

Nkεkλ[akλ(t)e

ik.r + a∗kλ(t)e−ik.r] . (2.29)

Written in this form, the solution for the vector potential is real in all space. By inserting thissolution into the wave equation (2.18) one obtains the equation of motion:

akλ(t) = −ω2kakλ(t) =⇒ akλ(t) = e−iωktakλ(0) . (2.30)

This same equation can be obtained through the definition of an Hamiltonian which correspondsto the total energy of the field.

2.2.1 Hamiltonian description of the classical electromagnetic field

Let us now calculate the total energy of the electromagnetic field confined in our box ofvolume V which, if you remember from your electromagnetism or classical electrodynamicscourses, is given by:

H =1

8π

∫V

(|E|2 + |B|2) dr . (2.31)

Given the definitions for E and B in Eq. (2.6), it can be shown that the energy can be solelywritten in terms of the vector potential A as:

H =1

8π

∫V

[1

c2

∂A

∂t.∂A∗

∂t+ (∇×A).(∇×A∗)

]dr . (2.32)

Inserting now the solution (2.29) into the above equation, one gets for the energy contributioncoming from E:

1

8πc2

∫V

∂A

∂t.∂A∗

∂t=∑k,λ

ω2k

8πc2N2kV[(akλa

∗kλ + a∗kλakλ) + (−1)λ(akλa−kλ + a∗kλa

∗−kλ)

],

(2.33)while that stemming from B is

1

8π

∫V

(∇×A).(∇×A∗) =∑k,λ

ω2k

8πc2N2kV[(akλa

∗kλ + a∗kλakλ)− (−1)λ(akλa−kλ + a∗kλa

∗−kλ)

].

(2.34)The above results were obtained taking into account that ωk and Nk are the same for ±k, andusing the relation εkλ.ε−kλ′ = −(−1)λδλλ′ . Moreover, we used the fact that the transversalitycondition εkλ.k = 0 is valid for both ±k, and also the relation (1.129).

Exercise 2.4. Show Eqs. (2.32)-(2.34).

Notice that the above two equation differ only from the sign of the second term. Therefore,their sum (the total energy) will be given by:

H =∑k,λ

ω2k

4πc2N2kV (akλa

∗kλ + a∗kλakλ) . (2.35)

33


This Hamiltonian can be written in more suggestive way choosing for the normalizationconstant:

N2k =

2π~c2

V ωk=⇒ H =

1

2

∑k,λ

~ωk(akλa∗kλ + a∗kλakλ) . (2.36)

Notice that the Fourier coefficients akλ are just amplitudes (complex numbers) which obviouslycommute with their conjugate. We have taken them as non-commuting objects since thesewill be treated later as operators upon quantization. In any case, let us for now treat them aswhat they are: complex numbers. In this case:

H =∑k,λ

~ωk a∗kλakλ , (2.37)

which indicates that, somehow, the energy contained in the electromagnetic field resemblesthe total energy of a system of classical uncoupled harmonic oscillators of amplitude |akλ(0)|2and energy ~ωk|akλ(0)|2. Notice that as soon as the akλ are treated as noncommuting objects,Eqs. (2.36) and (2.37) will not be identical anymore.

Let us appreciate the above fact in a clearer way by defining the generalized coordinatesand respective momenta:

Qkλ =

√~

2ωk(akλ + a∗kλ) , Pkλ = −i

√ωk~

2(akλ − a∗kλ) . (2.38)

In terms of these coordinates, the Hamiltonian of our system reads:

H =∑k,λ

(P 2kλ

2+

1

2ω2kQ

2kλ

), (2.39)

which is the Hamiltonian for a set of oscillators, each with frequency ωk. To this classicalHamiltonian functions correspond the following equations of motion:

Qkλ =∂H

∂Pkλ, Pkλ = − ∂H

∂Qkλ=⇒ Qkλ = Pkλ , Pkλ = −ω2

kQkλ , (2.40)

for the mode (k, λ).

2.2.2 Quantization procedure

We are now ready to quantize our system, i.e. the classical electromagnetic field. Sincewe have just seen that this can be interpreted as a set of infinite oscillators, the path toquantization would be to postulate that the transition to the quantum is achieved by replacingthe generalized coordinates Qkλ and Pkλ by the corresponding operators Qkλ and Pkλ in theHamiltonian (2.39) and the equations of motion (2.40), namely:

H =∑k,λ

(P 2kλ

2+

1

2ω2k Q

2kλ

), (2.41)

34

2.2. QUANTIZATION OF THE ELECTROMAGNETIC FIELD

and

˙Qkλ = Pkλ ,

˙Pkλ = −ω2

k Qkλ . (2.42)

This however is not sufficient, in the sense that compatibility with quantum mechanics has tobe insured. In other words, the above quantization procedure is only valid if the dynamics ofthe operators follows the quantum mechanics rules. To better understand this, let us considerthe Heisenberg equations for Qkλ and Pkλ:

i~∂Qkλ∂t

= [Qkλ, H] , i~∂Pkλ∂t

= [Pkλ, H] , (2.43)

which govern the time-dependence of those operators. If our quantization procedure is correct,then these equations should be identical to those in (2.40). In order to check this, we have tocompute the commutators [Qkλ, H] and [Pkλ, H]. Remember that same-type operators anddifferent-type operators related to distinct oscillators always commute, i.e.

[Qkλ, Qk′λ′ ] = 0 , [Pkλ, Pk′λ′ ] = 0 , [Qkλ, Pk′λ′ ]k 6=k′∨λ6=λ′ = 0 . , (2.44)

Using these commutation relations and the identity [A, B2] = [A, B]B + B[A, B]one canshow that:

i~∂Qkλ∂t

=1

2

[Qkλ, Pkλ]Pkλ + Pkλ[Qkλ, Pkλ]

, (2.45)

i~∂Pkλ∂t

=ω2k

2

[Pkλ, Qkλ]Qkλ + Qkλ[Pkλ, Qkλ]

. (2.46)

Comparing these equations with those in (2.42) we immediately realize that both sets ofrelations are equivalent if the commutation relation [Qkλ, Pkλ] = i~ is satisfied. Therefore,the quantization postulate makes all sense if the commutation relations between operators

[Qkλ, Qk′λ′ ] = 0 , [Pkλ, Pk′λ′ ] = 0 , [Qkλ, Pk′λ′ ] = i~ δkk′δλλ′ . , (2.47)

are valid.

Exercise 2.5. Prove Eqs. (2.45) and (2.46).

We now go back to Eqs. (2.38) and define their quantum analogues, namely

Qkλ =

√~

2ωk(akλ + a†kλ) , Pkλ = −i

√ωk~

2(akλ − a†kλ) , (2.48)

from which one gets

akλ =ωkQkλ + iPkλ√

2~ωk, a†kλ =

ωkQkλ − iPkλ√2~ωk

, (2.49)

35


which are adjoints of each other. Using this together with the commutation relations (2.47)one gets:

[akλ, a†k′λ′ ] = δkk′δλλ′ , [akλ, ak′λ′ ] = [a†kλ, a

†k′λ′ ] = 0 . (2.50)

But these are the commutation relations for bosonic creation and annihilation operators thatwe have obtained in Chapter 1. Therefore, the electromagnetic quanta (the photon) must bea boson! The fact that the photon has two polarization states could indicate that it has spin1/2. However, this is incompatible with what we have just seen. In fact, the photon has spin1 to which, in principle, correspond three polarization states with Sz = ±1, 0. However, dueto the transversal character of the electromagnetic field, the longitudinal polarization state ofthe photon (which corresponds to Sz = 0) is absent. Later on we will understand this betterwhen we will realize that for particles with zero mass there can be no longitudinally polarizedmodes (basically because there is no rest frame for particles moving at the speed of light).

With the help of Eqs. (2.41) and (2.48) we can now write the (second-quantized)Hamiltonian for the radiation field

H =∑k,λ

~ωk(a†kλakλ + 1/2) =∑k,λ

~ωk(nkλ + 1/2) ≡∑k,λ

Hkλ , Hkλ = ~ωk(nkλ + 1/2) ,

(2.51)where nkλ is the number operator defined in Chapter 1.

Exercise 2.6. Prove Eqs. (2.50)-(2.51).

2.3 States and field operators of the electromagneticfield

From what we have just seen in the previous section, it is clear that the quantization of theelectromagnetic field leads us to a system with an infinite number of harmonic oscillators(one oscillator corresponding to each degree of freedom, namely momentum and polarization)with no mutual interaction. Using the commutation relations (2.50), one can establish thefollowing commutation rules

[akλ, Hkλ] = akλ~ωk =⇒ [akλ, nkλ] = akλ ,

[a†kλ, Hkλ] = −a†kλ~ωk =⇒ [a†kλ, nkλ] = −a†kλ . (2.52)

Since the harmonic oscillators are independent, the eigenstates ψ of H is the product of theeigenstates of each Hkλ, namely:

|ψ〉 = |k1, λ1〉|k2, λ2〉...|kn, λn〉 , Hkλ|k, λ〉 = Ekλ|k, λ〉 , (2.53)

where Ekλ is the energy of the eigenstate |k, λ〉. Given this, we can now try to find outwhat is the physical significance of the operators akλ and a†kλ. For that, we start with thecommutation relations (2.52) to notice that

Hkλ a†kλ|k, λ〉 = (Ekλ + ~ωk) a†kλ |k, λ〉 , (2.54)

Hkλ akλ|k, λ〉 = (Ekλ − ~ωk) akλ |k, λ〉 . (2.55)

36

2.3. STATES AND FIELD OPERATORS OF THE ELECTROMAGNETIC FIELD

From this, we conclude that a†kλ |k, λ〉 and akλ |k, λ〉 are eigenstates of Hkλ with eigenvalues(Ekλ + ~ωk) and (Ekλ + ~ωk), respectively. Therefore, the operators a†kλ and akλ can beinterpreted as the creation and annihilation operators of a photon with momentum k andpolarization ελ.

In the occupation number representation developed in Chapter 1, the quantum stateof the electromagnetic field is a state in the bosonic Fock state spanned by basis vectorswhich correspond to states |nkλ〉 with certain occupation numbers nkλ (the number ofphotons with momentum k and polarization ελ). Therefore, in second-quantized language,an eigenstate of the electromagnetic field can be represented as

|nk1λ1 , nk1λ2 , nk2λ1 , nk2λ1 , ...〉 =∏

i=1,2,... ; j=1,2

(a†kiλj)nkiλj√

nkiλj !|0〉 , (2.56)

where |0〉 is the electromagnetic vacuum in which all modes have zero occupation numberand any state of the radiation field can be obtained by applying creation and annihilationoperators to it. In particular, the operator a†kλ (akλ) increases (decreases) the occupationnumber nk,λ by one unit as:

a†kλ|nkλ〉 =√nkλ + 1 |nkλ + 1〉 , akλ|nkλ〉 =

√nkλ |nkλ − 1〉 . (2.57)

It is now easy to see that the energy of the electromagnetic vacuum is given by:

〈0|H|0〉 = 〈0|∑kλ

(nkλ + 1/2)~ωk|0〉 =∑kλ

1/2~ωk =∞ ! (2.58)

The above result is telling us that the electromagnetic field configuration with no photons hasinfinite energy. This is due to the fact that each mode has a finite zero-point energy. This isactually not a problem in the sense that this is a background energy and we will be interestedin differences between energies before and after a given process. In fact, in Chapter ?? wewill shift the vacuum energy to zero.

Having quantised the radiation field Hamiltonian, we can now repeat the procedure forthe classical fields A, E and B. We start by defining the quantized version of the classicalfield A(r, t) given in (2.29) as

A(r, t) =∑k

∑λ=1,2

Nkεkλ

[akλ(t)e

ik.r + a†kλ(t)e−ik.r

], (2.59)

where we have replaced the Fourier coefficients ak,λ and a∗k,λ by the corresponding quantumoperators akλ and a†kλ. From this, we can now find the general expression for the quantisedelectric field E(r, t). For that, we use the Maxwell equation relating E(r, t) and A(r, t),and akλ(t) = −iωkakλ(t) to find:

E(r, t) = −1

c

∂A(r, t)

∂t= i∑k

∑λ=1,2

√2π~ωkL3

εkλ[akλ(t)e

ik.r − a∗kλ(t)e−ik.r]. (2.60)

Thus, the quantised version of the classical field E(r, t) is:

E(r, t) = i∑k

∑λ=1,2

√2π~ωkL3

εkλ

[akλ(t)e

ik.r − a†kλ(t)e−ik.r

]. (2.61)

37


We should however be cautious when doing the straightforward replacements akλ(t)→ akλ(t)and a∗kλ(t)→ a†kλ(t), since the above result for E(r, t) has to be compatible with the factthat due to the quantum nature of the system under study, the relation

E(r, t) = −1

c

∂A(r, t)

∂t= −1

c

(− i~

[A(r, t), H]

), (2.62)

must be fulfilled (if you have not noticed, we have used the Heisenberg equation for thetime evolution of the quantum operator A). In order to check this, we have to compute thecommutator [A(r, t), H] using Eqs. (2.51) and (2.59). Namely,

[A, H] = AH − HA =∑k,λ

∑k′,λ′

Nk~ωk′εkλ[akλ(t)e


](nk′λ′ + 1/2)

−∑k,λ

∑k′,λ′

Nk~ωk′εkλ(nk′λ′ + 1/2)[akλ(t)e


]. (2.63)

The terms proportional to the zero point energy of each radiation mode cancel in the abovecommutator and, therefore,

[A, H] =∑k,λ

∑k′λ′

Nk~ωk′εkλeik.r[akλ, nk′λ′ ] + e−ik.r[a†kλ, nk′λ′ ]

. (2.64)

The commutators [akλ, nk′λ′ ] and [a†kλ, nk′λ′ ] can be determined by using Eqs. (2.50) andthe operator relation [A, BC] = [A, B]C + B[A, C], leading to the result:

[akλ, nk′λ′ ] = [akλ, a†k′λ′ ak′λ′ ] = [akλ, a

†k′λ′ ]︸︷︷︸

=δkk′δλλ′

ak′λ′ + a†k′λ′ [akλ, ak′λ′ ]︸︷︷︸=0

= δkk′δλλ′ ak′λ′ ,

[a†kλ, nk′λ′ ] = − ([akλ, nk′λ′ ])† = −δkk′δλλ′ a†k′λ′ . (2.65)

Inserting the above results into (2.64) we finally get:

[A, H] =∑k,λ

∑k′λ′

Nk~ωk′εkλδkk′δλλ′[ak′λ′(t)e

ik.r − a†k′λ′(t)e−ik.r

](2.66)

=∑k,λ

Nk~ωkεkλ[akλ(t)e


], (2.67)

which, together with Eq. (2.62), leads to:

E(r, t) = i∑k

∑λ=1,2

√2π~ωkL3

εkλ

[akλ(t)e


]. (2.68)

Notice that this result coincides with (2.61), meaning that the correspondence akλ(t)→ akλ(t)and a∗kλ(t) → a†kλ(t) is compatible with the dynamical evolution of quantum operators(governed by the Heisenberg equation). The magnetic field operator can be obtained bytaking into account that B = ∇×A and, again, making the replacements akλ(t)→ akλ(t)

38

2.3. STATES AND FIELD OPERATORS OF THE ELECTROMAGNETIC FIELD

and a∗kλ(t)→ a†kλ(t). In the end, the field operators for the quantised electromagnetic fieldare

A(r, t) =∑k

∑λ=1,2

Nkεkλ

[akλ(t)e


], (2.69)

E(r, t) = i∑k

∑λ=1,2

√2π~ωkL3

εkλ

[akλ(t)e


], (2.70)

B(r, t) = i∑k

∑λ=1,2

Nk(k × εkλ)[akλ(t)e


]. (2.71)

2.3.1 Coherent states of the electromagnetic field

In the previous sections we have seen that the quantum states of the electromagnetic field arecharacterized by oscillation modes with momentum k and polarization ελ. Therefore, the stateof this "many-mode" field is the Fock state |..., nkλ, ..., nk′λ′ , ...〉 ≡ ...⊗|nkλ〉⊗...⊗|nk′λ′〉⊗...,meaning that each mode |k, λ〉 contains nkλ photons. Given that the quantisation of theelectromagnetic field has been consistently done by replacing the Fourier coefficients byoperators in the classical solution of the Maxwell wave equation, we would expect that, forinstance, the classical value of the electric field for a normal mode |k, λ〉 could be found bycomputing the expectation value 〈nkλ|E|nkλ〉. However, for a single normal mode of theelectromagnetic field,

〈nkλ|E|nkλ〉 = i

√2π~ωkL3

εkλ

[〈nkλ|akλ(t)|nkλ〉eik.r − 〈nkλ|a†kλ(t)|nkλ〉e

−ik.r]

= 0 ,

(2.72)since |nkλ〉 is not an eigenstate of the creation and annihilation operators akλ and akλ,respectively. So, we can immediately conclude that the classical state of the electromagneticfield does not have a fixed number of photons. If we compute the expectation values ofEE†/(8π), which denotes the electric-field contribution to the energy density [see Eq. (2.31)],we get

1

8π〈nkλ|EE†|nkλ〉 =

~ωk4L3

[〈nkλ|akλ(t)a†kλ(t) + a†kλ(t)akλ(t)|nkλ〉

]=

~ωk4L3

[〈nkλ|2a†kλ(t)akλ(t) + 1|nkλ〉

]=

~ωk2L3

(nkλ + 1/2) , (2.73)

meaning that, although 〈E〉 vanishes, the expectation value of the energy density is (apartfrom the vacuum energy term) half of the photon number times ~ωk divided by the volumeL3 (the other half comes from the magnetic term BB†). We may therefore ask ourselves howdoes a classical state of the electromagnetic field looks like. It is straightforward to noticethat such a state has to be an eigenstate of the annihilation operator akλ. But this state hasbeen found in Problem 1.5, namely the coherent (or Glauber) state |ψc〉 given by:

|ψc〉 = e−|α|2/2

∞∑nkλ=0

αnkλ

√nkλ!

|nkλ〉 ≡∑k,λ

bnkλ|nkλ〉 , bnkλ

= e−|α|2/2 αnkλ

√nkλ!

, (2.74)

39


obeys akλ|ψc〉 = α|ψc〉. The probability of finding nkλ photons in a Glauber state is just

|〈nkλ|ψc〉|2 = |bnkλ|2 = e−|α|

2 |α|2nkλ

nkλ!, (2.75)

from which we conclude that the number of photons nkλ is not fixed in a Glauber state2.Consequently, the expectation value of the creation and annihilation operators is nonzero. Inparticular:

〈ψc|akλ|ψc〉 =∑

nkλ,nk′λ′

bnkλb∗nk′λ′

〈nk′λ′|akλ|nkλ〉 =∑

nkλ,nk′λ′

bnkλb∗nk′λ′

√nkλ δnk′λ′ ,nkλ−1

=∑nk′λ′

bnk′λ′+1b∗nk′λ′

√nk′λ′ + 1 = αe−|α|

2∑nk′λ′

|α|2nk′λ′

nk′λ′ != α . (2.76)

Taking into account that 〈ψc|a†kλ|ψc〉 = (〈ψc|akλ|ψc〉)∗, one has 〈ψc|a†kλ|ψc〉 = α∗. We cannow use these results to find the expectation value

〈ψc|E|ψc〉 = i

√2π~ωkL3

εkλ

[〈ψc|akλ(t)|ψc〉eik.r − 〈ψc|a†kλ(t)|ψc〉e

−ik.r]

= i

√2π~ωkL3

εkλ[αeik.r − α∗e−ik.r

]= −2

√2π~ωkL3

εkλ|α| sin(k.r + δα) , δα = arg(α) . (2.77)

which looks like a classical wave with amplitude and phase set by |α| and arg(α), respectively.In fact, the second equation above looks like the classical solution (2.60) with akλ = α. Noticehowever that a Glauber state of the electromagnetic field is not equivalent to a classical fieldsince there are quantum fluctuations inherent to this state. In fact, by computing the meanquadratic deviation of the electric field (∆E)2, it can be shown that

(∆E)2 = 〈ψc|EE†|ψc〉 − |〈ψc|E|ψc〉|2 =2π~ωkL3

, (2.78)

which is nonzero. However, in the classical limit ~ → 0, the fluctuations vanish.

Exercise 2.7. Prove Eq. (2.78).

2.4 Interaction of matter with the quantised electro-magnetic field

Matter, as we know it, is made of particles with different mass and electric charge. Particleinteractions are often translated in terms of potentials V (ri, ξi) which may depend not only

2Obviously, it can be easily shown that all the above probabilities sum up to 1.

40

2.4. INTERACTION OF MATTER WITH THE QUANTISEDELECTROMAGNETIC FIELD

on particle position ri, mass mi and electric charge ei, but also on other degrees of freedom ξilike, for instance, spin. This approcah allows us to study more complicated systems like atoms,molecules or other bound states like atomic nuclei which can also be described by interactionpotentials. The most popular example of a oftenly used potential is the one describing theCoulomb repulsion among electrons VC(r1, r2, ...)

VC(r1, r2, ..., rN) =1

2

N∑i 6=j=1

eiej|ri − rj|

, (2.79)

In view of this potential description for interactions, the general form of the Hamiltonian of amany-particle (MP) system is

HMP =N∑i=1

p2i

2mi

+ V . (2.80)

Notice that, as mentioned before, the potential V describes any kind of interactions dependingon the particle degrees of freedom.

We want now to address the question of how to describe the quantum behaviour of amatter system which interacts with radiation. In previous courses of quantum mechanics thishas been done by considering the electromagnetic field as being classical and not subject toquantisation. With what we have studied in the previous sections, we are now able to describethe quantum behaviour of the system matter+radiation.

When dealing with the electromagnetic field, we have to ensure that our quantum theorymust remain invariant under the gauge transformations defined in (2.7). In physical terms, thismeans that eigenvalues, probabilities, cross sections and other observable quantities must begauge invariant. You have seen in your analytical mechanics course that this can be ensuredby performing in HMP the minimal substitution (or minimal coupling)

pi −→ pi −eicA(ri, t) . (2.81)

With this, we can finally write the Hamiltonian of a N-particle system which interacts withthe electromagnetic field using Eqs. (2.31), (2.80) and (2.81). In particular,

H =1

8π

∫V

(EE† + BB†)dr +N∑i=1

(pi −

eicA)(pi −

eicA)

2mi

+ V . (2.82)

Let us develop the second term in the above expression. We have:(pi −

eicA)(pi −

eicA)

= p2i −

eic

(Api + piA) +e2i

c2A2 . (2.83)

We may further simplify this by noting that we are dealing with an operator and, therefore,we should think about it as acting in a wavefunction ψ:

p.Aψ = (p.A)ψ + A.(pψ) ∝ (∇.A)ψ + A.(∇ψ) . (2.84)

Taking into account that ∇.A = 0 in the Coulomb gauge, we have p.A+ A.p = 2p.A andEq. (2.83) reduces to(

pi −eicA)(pi −

eicA)

= p2i − 2

eicpi.A+

e2i

c2A2 . (2.85)

41


Figure 2.1: Example of processes involving two photons. (Left) Raman (inelastic) scattering: aphoton is absorbed and a photon with different energy is emitted (both energy and momentumchange). (Right) Rayleigh (elastic) photon scattering: a photon is absorbed and a new photonwith the same energy is emitted (only the momentum direction changes).

Relacing this into Eq. (2.82) we have

H =1

8π

∫V

(EE† + BB†)dr︸︷︷︸Hrad

+N∑i=1

p2i

2mi

+ V︸︷︷︸HMP

+N∑i=1

(− eimic

pi.A+e2i

2mic2A2

)︸︷︷︸

Hint

, (2.86)

where Hrad denotes the Hamiltonian of the radiation field, HMP contains the kinetic andpotential energy of the N-particles with charge ei and mass mi, and Hint describes theinteraction of the particles with the electromagnetic field. We further split Hint into H

(1)int and

H(2)int with:

H(1)int = −

N∑i=1

eimic

pi.A , H(2)int =

N∑i=1

e2i

2mic2A2 . (2.87)

Using now the expression for the field operator (2.69) we can write the above interactionHamiltonians as:

H(1)int = −

N∑i=1

eimic

∑k,λ

√2π~c2

L3ωkpi.εkλ

[akλ(t)e


], (2.88)

and

H(2)int =

2π~L3

N∑i=1

e2i

2mi

∑k,λ

∑k′,λ′

εkλ.εk′λ′√ωkωk′

[akλak′λ′e

i(k+k′).r + akλa†k′λ′e

i(k−k′).r

+a†kλak′λ′ei(k′−k).r + a†kλa

†k′λ′e

−i(k+k′).r]. (2.89)

By looking at the above expressions we can immediately interpret the physical meaning ofthe interaction terms H(1)

int and H(2)int . Namely, H(1)

int contains one creation operator a†kλ and aannihilation operator akλ and, therefore, induces transitions where a photon is either emmitedor absorved. On the other hand, H(2)

int is responsible for processes where two photons areinvolved. For instance, the term a†kλak′λ′ corresponds to absorption and reemission, as ithappens in Raman scattering or elastic photon scattering (see Fig. 2.1).

42


To conclude this section we remark that we have obtained the Hamiltonian of the systemMP+radiation in such a way that

H = H0 + Hint , H0 = Hrad + HMP , Hint = H(1)int + H

(2)int , (2.90)

where H0 does not mix matter and radiation. Denoting the state vectors of H0 by |MP+ rad〉,we have that |MP+ rad〉 = |MP〉|rad〉. Here, |MP〉 is the state of the MP system (for instancean atom), and |rad〉 = |..., nk,λ, ...〉 is a state of the radiation field. In the following sections,we will treat these eigenstates of H0 as the ones to be used in perturbation theory, beingHint the perturbation.

2.4.1 Emission of radiation from an atom

Let us consider an atom in an excited state |ai〉 which decays to a lower-energy state |af〉. Inthis process, there is emission of a photon with momentum k and polarization λ. In terms ofinitial and final states of the whole system "matter+radiation field", we write:

|i〉 = |ai〉|..., nkλ, ...〉 , |f〉 = |af〉|..., nkλ + 1, ...〉 , (2.91)

notice that in the process nkλ → nkλ + 1 due to the emission of the photon.We have seen in QMII that the |i〉 → |f〉 transition probability per unit time will be

(Fermi golden rule):Pi→ft

=2π

~|Mfi|2δ(Ef − Ei) , (2.92)

where Mfi is the transition amplitude, and Ef − Ei. The transition amplitude is computedusing perturbation theory with the Hamiltonian

Mfi = 〈f |Hp|i〉+∑j

〈f |Hp|j〉〈j|Hp|i〉Ei − Ej

+ ... , (2.93)

where Hp is the perturbation Hamiltonian. In the present case, since we are dealing withone-photon absorption, we have that Hp = H

(1)int , where H

(1)int has been defined in (2.88).

Using the states (2.150)

Mfi = 〈af |〈..., nkλ + 1, ...|Hp|ai〉|..., nkλ, ...〉

= 〈af |〈nkλ + 1|

[− e

mc

∑k′,λ′

√2π~c2

L3ω′kp.εk′λ′

(ak′λ′ e

ik′.r + a†k′λ′ e−ik′.r

)]|ai〉|nkλ〉

= − e

mc

√2π~c2

L3ωk〈af |p.εkλ e−ik.r|ai〉〈nkλ + 1|a†kλ|nkλ〉

(2.94)

which leads to

Mfi = − e

mc

√2π~c2

L3ωk

√nkλ + 1 〈af | p.εkλ e−ik.r|ai〉 . (2.95)

43


Notice that only the term with k = k′ and λ = λ′ contributes to the sum. This result,together with Eq. (2.92) leads to

Pi→ft

=2π

~|Mfi|2δ(Ef − Ei)

=2π

~

( e

mc

)2 2π~c2

L3ωk(nkλ + 1) |〈af | p.εkλ e−ik.r|ai〉|2 δ(Eaf + ~ωk − Eai) , (2.96)

for the transition probability per unit time of an atom from state |ai〉 to |af〉 with emission ofa photon with energy ~ωk. Let us analyse this expression with more detail. Notice in particularthat in Pi→f/t there is a term proportional to nkλ, i.e. the number of photons present inthe initial state. This is stimulated emission, and the probability is higher the larger nkλ is.However, Pi→f/t 6= 0 even if nkλ = 0, this is spontaneous emission, i.e. emission stimulated bythe zero point oscillations of the electromagnetic field. These are purely quantum mechanicaloscillations. The ’1’ appearing in (nkλ + 1) has the same origin as the ’1/2’ in the zero pointenergy, i.e. the commutation relations between creation and anihilation operators.

We can use the above results to compute the lifetime of an atomic excited state (dueto spontaneous emission). For that, we set nkλ = 0 and sum over all possible momenta kand polarizations λ of the emmited photon. The lifetime of the excited state τi→f is then theinverse of Pi→f/t. Thus,(

1

τ

)i→f

=4π2e2

m2L3

∑k,λ

1

ωk|〈af | p.εkλ e−ik.r|ai〉|2 δ(Eaf + ~ωk − Eai) . (2.97)

We now let the volume of our box go to infinity L3 →∞. In Section 2.6 (Supplement 1), itis shown that this ammount to perform the replacement∑

k

−→ L3

(2π)3

∫dk , (2.98)

and, therefore,(1

τ

)i→f

=e2

2πm2

∑λ=1,2

∫1

ωk|〈af | p.εkλ e−ik.r|ai〉|2 δ(Eaf + ~ωk − Eai) dk . (2.99)

Let us now turn our attention to the term |〈af | p.εkλ e−ik.r|ai〉|2. The chosen configurationfor the polarization vectors is that represented in Fig. 2.2, for which∑

λ=1,2

|〈af | p.εkλ e−ik.r|ai〉|2 = |〈af | p.εk2 e−ik.r|ai〉|2 = |εk2.〈af | p e−ik.r|ai〉|2 . (2.100)

We now evaluate k.r for the characteristic scales and energies we are dealing with. For lightemmited from an atom we tipically have ~ω ∼ 10 eV. On the other hand, the typical dimensionsof an atom are of the order of the Bohr radius aB ∼ 5× 10−11 m. It is straightforward toprove that, in this case, kr ∼ ωaB/c ∼ 2.7× 10−3 1. Therefore:

e−ik.r = 1− ik.r − 1

2(k.r)2 + ... ' 1 , (2.101)

44


Figure 2.2: Configuration of polarization vectors used to compute de sum in Eq. (2.100).Here P ≡ 〈af | p|ai〉.

and Eq. (2.100) can be approximated by

|〈af | p.εk2 e−ik.r|ai〉|2 ' |εk2.〈af | p |ai〉|2 . (2.102)

This is the so called dipole approximation, which we have already seen in the QMII course.The remaining terms in the expansion would give rise to multipole terms, which we neglecthere. Looking at Fig. 2.2 we have

εk2.〈af | p |ai〉 = |εk2||〈af | p |ai〉| cos(90 − θ) = |〈af | p |ai〉| sin θ . (2.103)

Taking all this into account, we can write(1

τ

)i→f

=e2

2πm2

∫1

ωk|〈af | p|ai〉|2 sin2 θ δ(Eaf + ~ωk − Eai) dk . (2.104)

Using spherical coordinates dk = k2 sin θdkdθdφ = 2π(ω2k/c

3)dωk sin θdθ (the factor 2πcomes from integration in φ). Finally,(

1

τ

)i→f

=e2

2πm2

∫2πωkc3|〈af | p|ai〉|2δ(Eaf + ~ωk − Eai) dωk

∫ π

0

sin3 θdθ , (2.105)

which, after performing the integration leads to:(1

τ

)i→f

=4

3

e2ωfim2c3~

|〈af | p|ai〉|2 , ωfi =Eaf − Eai

~. . (2.106)

We now evaluate the matrix element

〈af | p|ai〉 = 〈af |mdr

dt|ai〉 = −im

~〈af | rHMP − HMPr|ai〉 = −im

~(Eaf − Eai)〈af |r|ai〉

= −imωfi〈af |r|ai〉 , (2.107)

so that we can write:(1

τ

)i→f

=4

3

e2ω3fi

~c3|〈af | r|ai〉|2 =

4

3

ω3fi

~c3|〈af | d|ai〉|2 , (2.108)

where d = e r is the dipole operator.

45


2.4.2 Einstein coefficients

Spontaneous emission has been observed from the beggining of the 20th century, either inchemical reactions or in the form of radioactivity. Still, by that time, this phneomenon wasonly understood statistically. In 1916, Einstein came up with a very simple proof of the relationbetween spontaneous emission and absoption, based on thermal equilibrium arguments (thishas been discussed in the QMII course). Notice that by that time nothing was known aboutthe quantisation of the electromagnetic field. Einstein’s proof starts by assuming a system(e.g. an atom) with two energy levels E1 and E2 > E1, in such a way that n1 (n2) atoms arein state E1 (E2). The probability of a 2 → 1 transition with emission of one photon withfrequency ω is called the coefficient of spontaneous emission A. The probability of a photonwith frequency ω to induce a 2→ 1 transition will be proportional to the stimulated emissioncoefficient B, multiplied by the number of photons with frequency ω present, i.e. the intensityI(ω) (for einstein this is the energy density). Therefore, the change in n2 will be given by:

dn2

dt= −[A+BI(ω)]n2 . (2.109)

On the other hand, absorption (transition 1→ 2) will decresase n1 by I(ω)B′n1, where B′ isthe coefficient for absoption. Since the total number of photons in the system is conserved,we have:

dn2

dt= −dn1

dt= −[A+BI(ω)]n2 + I(ω)B′n1 . (2.110)

By using thermal equilibrium arguments, we have seen in QMII that:

A

B=

~ω3

π2c3, B′ = B . (2.111)

We now use what we have learned about the quantised electromagnetic field to find thesame result. Treating the system matter+radiation in a quantum way, we have seen fromEqs. (2.92) and (2.95) that

P2→1

t=

4π2e2

~m2L3

∑k,λ

(nkλ + 1)

ωk|〈1| p.εkλ e−ik.r|2〉|2︸︷︷︸

|M0|2

δ(ω21 − ωk) . (2.112)

Taking the infinite volume limit (2.98), the change in n2 per unit time is:

dn2

dt= −P2→1

tn2 = − 4π2e2n2

~m2(2π)3

∑λ

∫(nkλ + 1)

ωk|M0|2δ(ω21 − ωk)dk . (2.113)

We now have to relate the number of photons nkλ with the intensity I(ωk). First we noticethat the number of photons with momentum between k and k + dk is given by

2︸︷︷︸λ=1,2

L3

(2π)3dk =

L3k2

π2dk =

ω2k

π2c3︸︷︷︸ρ(ωk)

L3dωk , (2.114)

where ρ(ωk) is the number of photons with frequency ωk per unit volume. The energy densityI(ωk) will be defined by:

I(ωk)dωk = (~ωk)︸︷︷︸Energy of each photon

Density of modes︷︸︸︷ρ(ωk) nωk︸︷︷︸

# of photons with ωk

dωk , (2.115)

46


from which we conclude that the number of photons nωk is related with the energy densityI(ωk) by:

nωk =I(ωk)

~ρ(ωk)ωk=π2c3I(ωk)

~ω3k

. (2.116)

Replacing this in (2.113) and comparing with (2.110) leads to

A =e2

2π~m2

∫1

ωk|M0|2δ(ω21 − ωk) dk , B =

π2c3

~ω321

A =⇒ A

B=

~ω321

π2c3, (2.117)

as obtained by Einstein.

2.4.3 Photon absorption

We now consider the case in which there is absorption of one photon with momentum kand polarization λ, causing the excitation of an atom from a state |ai〉 to a state |af〉 withEai < Eaf . In terms of the notation introduced in the previous section, the initial and finalstates of the system matter+radiation are represented by:

|i〉 = |ai〉|..., nkλ, ...〉 , |f〉 = |af〉|..., nkλ − 1, ...〉 . (2.118)

Notice that the occupation number of the mode |kλ〉 is reduced by one unit due to theabsorption of the photon. Following the same procedure as we did for the case of photonemission, the relevant interaction for one-photon absorption is H(1)

int given in (2.88). Therefore,the matrix element we are interested in is

Mfi = 〈af |〈..., nkλ − 1, ...|Hp|ai〉|..., nkλ, ...〉

= 〈af |〈nkλ − 1|

[− e

mc

∑k′,λ′

√2π~c2

L3ω′kp.εk′λ′

(ak′λ′ e

ik′.r + a†k′λ′ e−ik′.r

)]|ai〉|nkλ〉

= − e

mc

√2π~c2

L3ωk〈af |p.εkλ eik.r|ai〉〈nkλ − 1|akλ|nkλ〉

= − e

mc

√2π~c2

L3ωk

√nkλ 〈af |p.εkλ eik.r|ai〉 . (2.119)

Using now Eq. (2.92), we have for the transition probability per unit time:(Pi→ft

)abs.

=2π

~|Mfi|2δ(Ef − Ei)

=4π2e2

m2ωkL3nkλ |〈af | p.εkλ eik.r|ai〉|2 δ(Eai + ~ωk − Eaf ) . (2.120)

comparing this with (2.96) we conclude that:(Pi→ft

)abs.

=

(Pi→ft

)stim. em.

. (2.121)

47


We now compute the cross section σi→f (k, λ) for absortion of a photon with momentum kand polarization λ. This is defined as the probability of absoption per unit time, divided bythe flux of incoming photons jk,λ = nk,λc/L

3. We therefore have:

σi→f (k, λ) =(Pi→f/t)abs

jk,λ=

4π2e2

m2ωkc|〈af | p.εkλ eik.r|ai〉|2 δ(Eai + ~ωk − Eaf ) , (2.122)

where we have used the result in (2.120).

2.4.4 Planck distribution formula

In order to derive the Planck distribution formula for black-body radiation, we considera two-level atomic system in thermal equilibrium, with N1 (N2) atoms in level |1〉 (|2〉).Transitions |1〉 |2〉 occur with absorption and emission of photons from and to the radiationfield. Therefore, the time variation of N2 and N1 is given by

N2 = −N2

(P2→1

t

)emi.

+N1

(P1→2

t

)abs.

N1 = −N1

(P1→2

t

)abs.

+N2

(P2→1

t

)emi.

. (2.123)

Thermal equilibrium requires that N2 = N1 = 0, leading to

N2

N1

=(P1→2/t)abs

(P2→1/t)emi=

nk,λnk,λ + 1

. (2.124)

On the other hand, the probability distribution of states follow a Boltzmann distribution, suchthat:

N2

N1

=e−E2/(kBT )

e−E1/(kBT )= e−(E2−E1)/(kBT ) = e−∆E/(kBT ) . (2.125)

Comparing the two previous equations, we can see that

nk,λ =1

e~ωk/(kBT ) − 1, (2.126)

where we have used the fact that ∆E = ~ωk. We now define the energy for modes withmomentum between k and k + dk:

dErad = 2(~ωk)nk,λL3

(2π)3dk . (2.127)

Writing dk = k2dkdΩ = (ω2k/c

3)dωkdΩ, we can define the spectral function u(ωk) whichgives the energy density in the range of frequencies ωk and ωk + dωk

u(ωk) =

∫Ω

dErad

L3=

∫Ω

2~ωkL3

L3

(2π)3

ω2k

c3nk,λdΩ =

~ω3k

π2c3

1

e~ωk/(kBT ) − 1, (2.128)

which is the well-known Planck distribution for the spectral energy density of the black-bodyradiation for a temperature T .

48

2.5. MATTER+RADIATION IN SECOND QUANTISATION

2.5 Matter+radiation in second quantisation

in the previous section we have studied matter-radiation interaction by treating matter(typically electrons) as we have always done in first quantisation. We now follow an alternativeapproach where everything is second-quantised, i.e. we want to treat particles in terms of(fermionic) creation and annihilation operators b, b†.

To start, we consider once more the Hamiltonian (2.82) of a particle of mass m and chargee, interacting with the radiation field. As it was seen in Section 2.4, this can be divided inthree terms HMP, Hrad and Hint, defined in Eq. (2.86). Considering V as being a one-particleoperator, we can write the Hamiltonian in second-quantised form using field operators ψ andψ† defined in Eqs. (1.103) and (1.104). Namely, from what we have learned in Section 1.4.3we can write

HMP =

∫ψ†(r, t)

[− ~2

2m∇2 + V

]ψ(r, t)dr

Hint =

∫ψ†(r, t)

[− e

mcp.A+

e2

2mc2A2

]ψ(r, t)dr

Hrad =1

8π

∫V

(EE† + BB†)dr =∑k,λ

~ωk(nkλ + 1/2) , (2.129)

where:ψ†(r) =

∑α

ψ∗α(r) b†α , ψ(r) =∑α

ψα(r) bα . (2.130)

So, we can now write the Hamiltonian in terms of creation and anhilitation operators forphotons and matter particles. For H(1)

int we have then:

H(1)int =

∑α,α′

∑k,λ

εkλ.[M (1)(k, α, α′) b†αbα′ akλ +M (1)(−k, α, α′) b†αbα′ a

†kλ

]. (2.131)

where

M (1)(k, α, α′) =

√2π~c2

L3ωk

∫ψ∗α(r, t)

(ie~mc

eik.r∇)ψα′(r, t) dr . (2.132)

The above result has been obtained using (2.129), together with the definition of A given in(2.59) and the definition p = −i~∇. For the two-photon interaction term interaction H(2)

intwe have:

H(2)int =

∑α,α′

∑k1,λ1

∑k2,λ2

b†αbα′εk1λ1εk2λ2[M (2)(k1,k2, α, α

′) ak1λ1 ak2λ2

+ M (2)(k1,−k2, α, α′) ak1λ1 a

†k2λ2

+M (2)(−k1,k2, α, α′) a†k1λ1 ak2λ2

+M (2)(−k1,−k2, α, α′) a†k1λ1 a

†k2λ2

]. (2.133)

with

M (2)(k1,k2, α, α′) =

2π2~c2

L3√ωk1ωk2

∫ψ∗α(r, t)

e2

2mc2ei(k1+k2).rψα′(r, t) dr . (2.134)

49


Figure 2.3: Diagrammatic illustration of the 1-photon (diagrams a and b) and two-photon(diagrams c-f) processes described in Eqs. (2.131)-(2.134).

In fact, H(1)int and H(2)

int can be written in a more compact form:

H(1)int =

∑α,α′

∑k,λ

εkλ.[M (1)(k, α, α′) b†αbα′ akλ + H.c.

](2.135)

H(2)int =

∑α,α′

∑k1,λ1

∑k2,λ2

b†αbα′εk1λ1εk2λ2[M (2)(k1,k2, α, α

′) ak1λ1 ak2λ2

+ M (2)(k1,−k2, α, α′) ak1λ1 a

†k2λ2

+ H.c.], (2.136)

The interactions written in Eqs. (2.131) and (2.133) involve different combinations of creationoperators, corresponding to the processes illustrated in Fig. 2.3.

Exercise 2.8. Prove Eqs. (2.135) and (2.136).

The unperturbed Hamiltonian is, once more, H0 = HMP + Hrad. Since the whole hamil-tonian is now in second-quantised form, the eigenstates of H0 in the occupation numberrepresentation are |ψ〉 = |..., nα, ...〉|..., nkλ, ...〉, such that

H0|ψ〉 = Eψ|ψ〉 =

[∑α

nαEα +∑k,λ

~ωk(nkλ + 1/2)

]|ψ〉 . (2.137)

50


where nα is the number of (matter) particles in state |α〉 and nkλ the number of photonswith momentum k and polarization ελ.

2.5.1 Example: non-relativistic Bremsstrahlung

As an example of the formalism developed above, we study Bremsstrahlung (braking radiation)which is the emission of radiation by a charged particle when deflected by another chargedparticle, typically an electron by an atomic nucleus. Since the mass of a nucleus is muchlarger than that of an electron, we can neglect recoil effects and consider the nucleus at theorigin of the reference frame. The position of the electron with respect to that of the nucleuswill be denoted by r. Therefore, the (Coulomb) electron-nucleus interaction is:

V (r) = −Ze2

r, (2.138)

where Z is the atomic number of the nucleus. This corresponds to the Hamiltonian:

HV =

∫ψ†(r, t)V (r)ψ(r, t)dr . (2.139)

The states |α〉 of the (free) electron are now the states of definite momentum |p〉 to whichcorrespond the wavefunctions

ψp(r) = 〈r|p〉 =1√L3

eip.r/~ , (2.140)

and, therefore, the field operators for the electron are:

ψ(r, t) =∑q

bqψq(r) , ψ†(r, t) =∑q

b†qψ∗q(r) . (2.141)

As see in Chapter 1, the kinetic energy the operator is

Hkin =

∫ψ†(r, t)

(−~2∇2

2m

)ψ(r, t) dr . (2.142)

On the other hand, for the Coulomb potential operator we have:

HV =∑q,q′

∫V (r)

ei(q′−q).r

L3b†q bq′ dr =

∑q,q′

b†q bq′Vq′−q , (2.143)

where Eqs, (2.138)-(2.141) have been used. In the above equation, Vq′−q is the Fouriertransform of V

Vq′−q =

∫V (r)

ei(q′−q).r

L3dr (2.144)

In the particular case of the Coulomb potential, the Fourier transform is:

Vk = −Ze2

L3

∫eik.r

rdr = −Ze

2

L3limµ→0

eik.r−µr

rdr = −4πZe2

L3k2. (2.145)

51


Since we are interested in studying one-photon Bremsstrahlung, the relevant interactionHamiltonian is H(1)

int given in Eq. (2.131). Namely, the matrix elements we are interested inare

εkλ.M(1)(k, q, q′) =

√2π~c2

L3ωkεkλ.

∫ψ∗q(r, t)

(ie~mc

eik.r∇)ψq′(r, t) dr , (2.146)

which, taking into account the momentum eigenfuctions (2.140), take the form

εkλ.M(1)(k, q, q′) =

ie~mc

√2π~c2

L3ωkεkλ.

∫e−iq.r√L3

(eik.r∇

)( eiq′.r√L3

)dr

= − e~mc

√2π~c2

L3ωkεkλ.q

′∫ei(k+q′−q).r

L3dr

= − e~mc

√2π~c2

L3ωkεkλ.q

′ δq,q′+k . (2.147)

Replacing the above equation into (2.131), we get:

H(1)int = − e~

mc

∑k,λ

∑q

√2π~c2

L3ωkεkλ.q

(b†q+kbqakλ + b†q bq+ka

†kλ

), (2.148)

As shown in Problem 2.11, a free electron cannot emit or absorve photons. This is onlypossible in the presence of a surrounding medium (as it happens in Cherenkov radiation) orif there is a force center (like a nucleus) which is responsible for the (des)acceleration ofthe particle. In the process, the initial state of the system is an electron with momentumq1, and the final state is an electron with momentum q3 and a photon with momentum kand polarization λ. Therefore, at first order of perturbation theory, the matrix element tocompute is

M(1)fi = 〈f | HV + H

(1)int | i〉 , (2.149)

where HV is responsible for the interaction of the electron with the nucleus and H(1)int for the

photon emission. The initial and final states are expressed in the following way

|i〉 = |q1〉e|0〉R = b†q1|0〉e|0〉R , |f〉 = |q3〉e|1k,λ〉R = b†q3 a†kλ|0〉e|0〉R , (2.150)

where the index ′e′ refers to the electron state and ′R′ to the radiation field. It is easy tosee that 〈f |H(1)

int |i〉 = 0, since this would correspond to emission of a photon from a freeparticle. On the other hand, 〈f |HV |i〉 = 0 because HV does not create any photon (HV isonly responsible for the Coulomb interaction). We therefore conclude that Bremsstrahlungdoes not occur at first order in perturbation theory. We now look to what happens in secondorder. In this case:

M(2)fi =

∑I

〈f |HV + H(1)int |I〉〈I|HV + H

(1)int |i〉

Ei − EI, (2.151)

52


Figure 2.4: Diagrammatic illustration of the two contributions to one-photon Bremsstrahlunggiven in Eq. (2.151). In the left panel, the electron with momentum q1 is scattered by thenucleus (Coulomb interaction) and then there is emission of the photon with simultaneousscattering of the electron. In the right panel, the photon is emmited with electron scatteringand subsequent interaction with the nucleus.

where the sum is done in all intermediate states |I〉. These intermediate states correspond tothe state of the system obtained by interaction of either HV or H(1)

int in |i〉, i.e we have

|I1〉 = |q2〉e|0〉R = b†q2 |0〉e|0〉R , |I2〉 = |q′2〉e|1kλ〉R = b†q′2a†kλ|0〉e|0〉R . (2.152)

The matrix element (2.151) will now be

M(2)fi =

〈f |H(1)int |I1〉〈I1|HV |i〉Ei − EI1

+〈f |HV |I2〉〈I2|H(1)

int |i〉Ei − EI2

. (2.153)

The first (second) contribution to the process is represented in the left (right) panel of Fig. 2.4.Let us now compute each of the matrix elements entering in the above expression. Takinginto account that HV =

∑q,q′ Vq′−q b

†q b†q′ , where Vq′−q is the Fourier transform defined in

(2.144), Then,

〈I1|HV |i〉 = e 〈0| R〈0| bq2

[∑q,q′

Vq′−q b†q bq′

]b†q1|0〉e|0〉R =

∑q,q′

Vq′−q〈0| bq2 b†q bq′ b†q1|0〉

=∑q,q′

Vq′−q〈0| bq2 b†q(δq′,q1 − b†q1 bq′)|0〉 =∑q,q′

Vq′−q 〈0| bq2 b†q︸︷︷︸δq2,q

|0〉 δq′,q1

=∑q,q′

Vq′−q δq2,q δq′,q1 = Vq2−q1 = Vq3+k−q1 = − 4πZe2

L3(q3 + k − q1)2. (2.154)

In the same way, it can be shown that 〈I1|HV |i〉 = 〈f |HV |I2〉. We now compute the matrix

53


element 〈f |H(1)int |I1〉

〈f |H(1)int |I1〉 = 〈0| bq3 akλ

[− e~mc

∑k′,λ′,q

√2π~c2

L3ωk′εk′λ′ .q

(b†q+k′ bqak′λ′ + b†q bq+k′ a

†k′λ′

)]b†q2|0〉

= − e~mc

√2π~c2

L3

∑k′,λ′,q

1√ωk′

q.εk′λ′ 〈0| bq3 b†q bq+k′ b†q2|0〉〈0|akλa†k′λ′|0〉︸︷︷︸

δk′,kδλ′,λ

= − e~mc

√2π~c2

L3ωk

∑q

q.εkλ 〈0| bq3 b†q︸︷︷︸δq,q3

bq+kb†q2︸︷︷︸

δq2,q+k

|0〉

= − e~mc

√2π~c2

L3ωkq3.εkλ δq2,q3+k . (2.155)

In a similar way we can obtain

〈I2|H(1)int |i〉 = −e~

m

√2π~L3ωk

q1.εkλ δq′2,q1−k . (2.156)

Notice the kronecker deltas in Eqs. (2.155) and (2.156) which translate momentum conserva-tion (see Fig. 2.4).

Having computed the matrix elements entering in (2.151), we now have to deal with theenergy-difference denominators. In this sense, we notice that:

Ei =~2q2

1

2m, EI1 =

~2

2m(q3 + k)2 , EI2 =

~2

2m(q1 − k)2 + ~ωk , Ef =

~2q23

2m+ ~ωk ,

(2.157)

so that the energy differences are:

Ei − EI1 = ~ωk(

1− k.v3

kc− ~k

2mc

)Ei − EI2 = −~ωk

(1− k.v1

kc− ~k

2mc

). (2.158)

Using the above relations for the matrix elements and energy-differences, and the fact thatthe initial and final velocites of the electron are v1 = ~q1/m and v3 = ~q3/m, we can finallywrite:

M(2)fi =

4πZe2

L3(q3 + k − q1)2

√2πe2

L3~ω3k

εkλ.v3

1− k.v3

kc− ~k

2mc

− εkλ.v1

1− k.v1

kc− ~k

2mc

. (2.159)

We now make some approximations. First, we consider that the photon momentum is muchsmaller than that of the electron, i.e. ~k ~q = mv. On the other hand, since theelectrons are non-relativistic we have v/c 1, and thus v1/c, v3/c 1. This implies that~k mv mc. These relations imply that:

1− k.v3

kc− ~k

2mc' 1 , 1− k.v1

kc− ~k

2mc' 1 , q3 + k − q1 ' q3 − q1 . (2.160)

54


Figure 2.5: (left) Configuration for the initial and final electron momentum vectors v1 andv3 with v1 ' v3. (Right) Final electron and photon phase-space configuration.

Plugging this into Eq. (2.159) and defining ∆v = v3 − v1, we have

M(2)fi '

4π~2Ze2

L3m2|∆v|2

√2πe2

L3~ω3k

εkλ.∆v . (2.161)

With this matrix element, we can use Eq. (2.92) to compute the transition probability perunit time (Pif/t). The total cross section σ is obtained by summing these probabilities for allpossible final states, i.e,. for all possible momenta of the final electron and emmited photon,and dividing by the incoming electron current v1/L

3. Therefore, we can write:

σ =2π

~L3

v1

∑q3,k

|M(2)fi |

2δ

(~2q2

3

2m− ~2q2

1

2m

), (2.162)

We now pass to the infinite volume limit using relation (2.98) for both the final electron andemmited photon momentum summation

σ =2π

~L3

v1

[L3

(2π)3

]2 ∫|M(2)

fi |2δ

(~2q2

3

2m− ~2q2

1

2m

)dq3 dk , (2.163)

Taking into account that dq3 = q23 dq3 dΩe and dk = k2 dk dΩk, and using (2.161) for the

matrix element, we have

σ =e2

v1(2π)4~2

(4π~2Ze2

m2

)2 ∫q2

3 k2 |εkλ.∆v|2

|∆v|4 ω3k

2m

~2δ(q2

3 − q21) dq3 dΩe dk dΩk , (2.164)

where we have used the property δ(ax) = δ(x)/|a|. Considering that δ(q23 − q2

1) = [δ(q3 −q1) + δ(q3 + q1)]/(2q1), and since δ(q3 + q1) = 0 (because we are integrating for positive q3),we have δ(q2

3 − q21) = δ(q3 − q1)/(2q1). We have also considered that the photon momentum

is negligible with respect to the electron one. Therefore, v1 ' v3 (there is only change inthe direction of the electron momentum) and ∆v = 2v1 sin(θ/2) (see left panel in Fig. 2.5).Finally, we can write the cross section as

σ =Z2e4

m2v41

e2

16π2c3~

∫|εkλ.∆v|2

sin4(θ/2)

dωkωk

dΩe dΩk . (2.165)

55


Lookign at the above expression, we can define the differential cross section:

d3σ

dωk dΩe dΩk

=Z2e4

m2v41 sin4(θ/2)

e2 |εkλ.∆v|2

16π2c3~ωk=

(dσ

dΩe

)Rutherford

dP

dΩkdωk. (2.166)

The above expression shows that the Bremsstrahlung differential cross section can be writtenas the product of the Rutherford differential cross section for elastic electron scattering witha term which is nothing but the probability that a photon with momentum k and polarizationεkλ is emitted into dΩkdωk. One could now choose a coordinate system for the photon phasespace, in such a way that

k = k (sin θk cosφk, sin θk cosφk, cos θk) ,

εk1 = (cos θk cosφk, cos θk sinφk,− sin θk) ,

εk2 = (− sinφk, cosφk, 0) ,

dΩk = sin θkdθkdφk , (2.167)

and compute the total cross section by integrating (2.166).

56

2.6. SUPPLEMENT 1: THE INFINITE VOLUME LIMIT

2.6 Supplement 1: The infinite volume limit

In this chapter we have quantized the electromagnetic field confined in a cubic box of volumeL3. Here, we will see how to take the limit L3 →∞. In Section 2.2 we have seen that, in thefinite box, momentum is quantised as

k = (kx, ky, kz) =2π

L(nx, ny, nz) ≡

2π

L~n , ni ∈ Z. (2.168)

Therefore, in the discrete 3D lattice defined by the points (nx, ny, nz), each point representsa normal mode of the electromagnetic field defined by the vector n (see Fig. 2.6).

Figure 2.6: 3D lattice of the quantised discrete modes of the electromagnetic field.

Therefore, inside a lattice volume ∆nx∆ny∆nz we have ∆nx∆ny∆nz normal modes. Giventhat, according to the above quantisation, a variation ∆ni is related with ∆ki through therelation ∆ni = L∆ki/(2π), one has

∆nx∆ny∆nz =L3

(2π)3∆kx∆ky∆kz . (2.169)

This counts the number of modes falling into the range [ki, ki + ∆ki]. If we now take thelimit L→∞, we can see from the above expression that ∆ki → 0, and therefore we shouldmake the replacement ∆ki → dki. Thus,the number of modes is now

∆nx∆ny∆nz →L3

(2π)3dkxdkydkz =

L3

(2π)3dk . (2.170)

This means that, a sum over all discrete modes in the finite volume V , has to be replaced byan integral in the continuous variable ~k in infinite space. Namely, the transition to infinitevolume corresponds to performing the replacement∑

k

−→ L3

(2π)3

∫dk . (2.171)

57


We have also seen in the previous chapter that in a finite volume V the followingorthogonality relations are valid ∫

L3

eik.re−ik′.rdr = L3δkk′ . (2.172)

In infinite space (where momentum is a continuous variable) we have∫eik.re−ik

′.rdr = (2π)3δ(k − k′) . (2.173)

From the above two equations, we see that when passing from finite to infinite volume wemust perform the replacement:

δkk′ −→(2π)3

L3δ(k − k′) . (2.174)

With this, we immediately realize that if when passing to continuous modes k the creationand annihilation operators are replaced by:

akλ, a†kλ −→

(2π

L

)3/2

akλ,

(2π

L

)3/2

a†kλ , (2.175)

then the commutation relations (2.50) become

[akλ, a†k′λ′ ] = δλλ′δ(k − k′) , [akλ, ak′λ′ ] = [a†kλ, a

†k′λ′ ] = 0 , (2.176)

where we have used the result (2.174).

58

2.7. PROBLEMS

2.7 Problems

Problem 2.1. Starting from the classical expression for the momentum of the electromagneticfield

p =

∫V

E ×B4πc

,

show that the second quantized form of the momentum operator of the electromagnetic fieldcan be written as:

p =∑k,λ

~ka†k,λak,λ .

Problem 2.2. Show that for coherent states of the electromagnetic field which obeya|ψc〉 = α|ψc〉 (α is a complex number), the following relations hold

(a) 〈ψc|a†a|ψc〉 = |α|2 = 〈n〉 .(b) 〈ψc|aa†|ψc〉 = |α|2 + 1 = 〈n〉+ 1 .(c) 〈ψc|a†aa†a|ψc〉 = |α|4 + |α|2 = 〈n2〉 ..

Problem 2.3. Consider spontaneous emission corresponding to the decay of an atom fromthe level |2〉 to |1〉, with E2 > E1. Show that, in the dipole approximation, the energy radiatedby the atom per unit time can be written in the form:

E/t =4e2

3c3

∣∣∣∣⟨1

∣∣∣∣d2r

dt2

∣∣∣∣ 2⟩∣∣∣∣2 .Of which classical equation is this the quantum analogue?

Answer: This is the quantum analogue of the Larmor equation.

Problem 2.4. When using the dipole approximation one often needs to compute matrixelements of the position operator r between states |nlm〉 of, for instance, an electron inan atom. For this reason it is worth to verify whether there are selection rules for thesematrix elements. For this reason, show that dipole transitions only occur between states with∆l = ±1 and ∆m = 0,±1.

Problem 2.5. Show that the 2s state of the Hydrogen atom cannot decay through theinteraction p.A (do not use the dipole approximation and ignore spin). What if spin isconsidered?

Problem 2.6. Consider the photoelectric emission of an electron from Hydrogen in itsground state. Assuming that the incoming energy is very large, compute the differential crosssection for the emission of an electron into the solid angle Ωe. Use the dipole approximationand consider that the incoming radiation is polarised in the z direction and that the photonmomentum is parallel to the x axis.

Answer:dσ

dΩe

= 8a2

(e2

~c

)2q0

kcos2(θ′)f 2(q0a)

where a = ~2/(mee2) is the Bohr radius, q2

0 = 2me(E1s + ~ω)/~2, f(x) = 2x/(1 + x2)2 andθ′ is the angle which the electron momentum does with the z axis.

59


Problem 2.7. The magnetic moment of the electron is µ = eS/(mec). Therefore, besidesthe interactions of the electron with radiation of the type p.A and A.A, we must add theinteraction of the magnetic moment with the electromagnetic field of the type: H(3)

int = −µ.B.Show that this can be written as:

H(3)int = −µ.B = − ie~

2mec

∑k,λ

Nk σ.(k × εkλ)[akλe

ik.r − a†kλe−ik.r

],

where the notation for field operators is the one used in the lecture notes.

Problem 2.8. As you have seen in QMI, the interaction between the nuclear spin andthat of the electron originates the splitting of the Hydrogen ground state in two levelscorresponding to the two total spin states with S = 0, 1 (Hyperfine splitting). The emittedphoton corresponding to this transition has a wavelength of 21 cm. Use the results of theprevious problem to compute the lifetime of this transition. Chose for the calculation theinitial spin state S = 1.

Answer: τ =3λ30

8π3α3a20c' 3.4× 1014 s ' 107 years. Here a0 = ~2/(mee

2) ' 5.29× 10−11 m isthe Bohr radius, α = e2/(~c) = 1/137 is the fine-structure constant and λ0 = 21cm.

Problem 2.9. Consider the scattering of a very energetic photon by an electron, i.e. γ+e− →γ + e−. Compute the total cross section for unpolarised incoming photons in the limit wherethe change in the wavelength of the incoming radiation is neglected.

Answer: σtotal ' 8π3

e2

mec2' 0.66 b.

Problem 2.10. Consider the two-photon decay of the 2s decay of the Hydrogen atom.Estimate the value of the lifetime of this state (do not compute the exact matrix elements,make reasonable assumptions).

Answer: The estimate for an Hydrogen-like atom with atomic number Z is τ = 0.11Z6 (s).For Hydrogen one has τH = 0.114 s, which is not far from the value τH = 0.12 s that wewould have obtained with a full computation of the matrix elements.

Problem 2.11. Show that a free electron cannot absorb nor emit a photon.

60

Ch

ap

te

r

3Relativistic quantum mechanics

“A great deal of my work is just playing with equations and seeing what they give. I don’tsuppose that applies so much to other physicists; I think it’s a peculiarity of myself that I liketo play about with equations, just looking for beautiful mathematical relations which maybe

don’t have any physical meaning at all. Sometimes they do...”Paul Dirac, AHPQ Interview, 1902

Quantum mechanics, as developed around 1925, was an extremely successful theory,Nevertheless, it was not in agreement with Einstein (special) theory of relativity. In fact, withthe gradual experimental verification of special relativity, it was soon widely accepted thatthis theory provides a correct description of phenomena involving high velocities close to thespeed of light, i.e. v ' c. On the other hand, it also became clear that quantum mechanicshad to be reformulated if it aimed at describing physical processes at very short scales andhigh velocities. This is evident if we think about the famous Einstein’s energy-mass relationE = mc2, which basically states that energy and mass are equivalent and can be transformedinto each other. This is somehow in contradiction with the quantum formalism we have seenuntil now, in the sense that the Schrödinger equation predicts conservation of probabilitycurrents and, therefore, of the number of particles. In fact, the wavefunction of an electronalways describes an electron and cannot simply transform into something else. Moreover, somefundamental aspects like the existence of particle spin cannot be explained in the light of theold quantum theory. In this chapter we introduce and explore some fundamental aspects ofrelativistic quantum mechanics like the Klein-Gordon and Dirac equations, and the conceptof quantum mechanical spin. This will prepare the way for the formulation of relativisticquantum field theories, which will be introduced in the next chapter.

So, when is a particle relativistic? Relativity impacts when the velocity approaches thespeed of light c or, more intrinsically, when its energy is large compared to its rest massenergy, mc2. For instance, protons at the CERN Large Hadron Collider (LHC) are acceleratedto energies of 7 TeV (1 TeV= 103 GeV) which is much larger than their rest mass energy,0.94 GeV. Electrons at the CERN Large Electron-Positron collider (LEP) were acceleratedto even larger multiples of the electron rest energy (30 GeV compared to the 5× 10−4 GeV

61

CHAPTER 3. RELATIVISTIC QUANTUM MECHANICS

of their rest mass energy). In fact we do not have to go to such sophisticated machines tosee relativistic effects since, for instance, high-resolution electron microscopes use relativisticelectrons. Also, photons always travel at the speed of light - they are never non-relativistic.

Before diving into the realm of relativistic quantum mechanics, we will first review somebasic aspects about Lorentz transformations and calculus in the 4-dimensional Minkowskispacetime.

3.1 Minkowski spacetime and Lorentz transformations:basics

The Minkowski space or Minkowski spacetime is the result of combining three-dimensionalEuclidean space and time into a four-dimensional manifold where the interval between twoevents does not depend on the inertial frame of reference. In other words, this means thatspacetime distance is invariant. Historically, the concept of spacetime was first developed bymathematician Hermann Minkowski for Maxwell’s equations of electromagnetism. Later on,the mathematical structure of Minkowski spacetime was shown to be most useful to Einsteinin the generalization of special relativity to accelerated frames.

In special relativity an event in spacetime is expressed by a contravariant vector xµ(t)

xµ(t) =

x0(t)x1(t)x2(t)x3(t)

, x0(t) = ct , (x1, x2, x3) = r , (3.1)

where c is the speed of light. Alternatively, we can write xµ = (ct, r). Acording to thepostulates of special relativity, the distance between two events xµ1 and xµ2 in spacetime isgiven by:

∆x2 = c2(t1 − t2)2 − (x1 − x2)2 − (y1 − y2)2 − (z1 − z2)2 , (3.2)

and is independent of the inertial frame chosen. The invariance of the above quantity folowsfrom the invariance of c2t2− x2− y2− z2 under coordinate transformations between differentframes. This allows to define a bilinear form:

(xµ).(yν) = xµgµνyν = xµyµ = xµg

µνyν = (xµ).(yν) . (3.3)

The mathematical object g is the metric which, from what we have seen above is:

gµν = gµν =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

, gµαgαν = gµν = δµν . (3.4)

The metric can be used to lower or raise contravariant and covariant indices, respectively. Forexample:

xµ = gµνxν , T µνρ = gµαT νρ

α = gµαgνβT ραβ = gµαgνβgργTαβγ . (3.5)

Events (or vectors) in Minkowski space are classified as time, light or space-like, dependingon whether xµxµ is > 0, 0 or < 0.

62

3.1. MINKOWSKI SPACETIME AND LORENTZ TRANSFORMATIONS: BASICS

In Euclidean space, the set of transformations which leave the length of a 3-vector invariantare rotations. In a 4-dimensional space, the equivalent transformations which leave a 4-vectorlength invariant are Lorentz transformations. The fact that 3D rotations are implemented byorthogonal transformations imply that the Euclidean metric is trivial. In the case of Minkowskispacetime, the set of transformations which leave the length of 4-vectors invariant is theLorentz group. Under a Lorentz transformation (or more generally a Poincaré transformation)

xµ −→ x′µ = Λµνx

ν + aµ , (3.6)

which also includes a translation by the constant vector aµ. Under the above transformation,the distance between two quadrivectors must remain invariant, i.e.

(x′ − y′)µ(x′ − y′)µ = (x− y)µ(x− y)µ . (3.7)

This automatically implies a strict rule for the metric transformation under Lorentz transfor-mations, namely

(x′ − y′)µ(x′ − y′)µ = gµν(x′ − y′)ν(x′ − y′)µ = gµνΛ

νβ(x− y)βΛµ

α(x− y)α

= ΛµαgµνΛ

νβ(x− y)β(x− y)α = Λµ

αgµνΛνβg

βσ(x− y)σ(x− y)α ,

(3.8)

has to be equal to (x− y)α(x− y)α, implying ΛµαgµνΛ

νβg

βσ = δσα or

ΛµαgµνΛ

νβ = gαβ or ΛTgΛ = g . (3.9)

The last equation above (invariance of the metric in matricial form) is the equivalent ofRTR = 1 for 3D rotations in Euclidian space. Multiplying the above equation by [Λ−1]ασ([Λ−1] is the inverse Lorentz transformation) one has

Λµα[Λ−1]ασgµνΛ

νβ = [Λ−1]ασgαβ ⇒ δµσgµνΛ

νβ = [Λ−1]ασgαβ ⇒ gσνΛ

νβ = [Λ−1]ασgαβ ,

(3.10)which, in matrix form, corresponds to gΛ = [Λ−1]Tg. Multiplying Eq. (3.10) by gβµ we have

gσνΛνβg

βµ︸︷︷︸Λ µσ

= [Λ−1]ασ gαβ gβµ = [Λ−1]ασ δ

µα = [Λ−1]µσ , (3.11)

from which we conclude that:[Λ−1]µσ = Λ µ

σ , (3.12)

and[Λ−1]σµΛµ

γ = gµνΛνβg

βσΛµγ = δσγ . (3.13)

The transformation properties of xµ under Lorentz transformations can be found taking intoaccount that:

x′µ = gµνx′ν = gµνΛ

νβx

β + aµ = [Λ−1]αµ gαβxβ + aµ = xα[Λ−1]αµ + aµ , (3.14)

where in the third equality we have used Eq. (3.10).

63


The above results allow us to obtain the transformation properties for differential operators.First, notice that:

x′µ = Λµνx

ν −→ xν = [Λ−1]νµx′µ ⇒ ∂xν

∂x′µ= [Λ−1]νµ , (3.15)

which we can use to conclude that:∂

∂x′µ=

∂

∂xν∂xν

∂x′µ=

∂

∂xν[Λ−1]νµ ≡ ∂′µ . (3.16)

From te above equation we conclude that the derivative ∂/∂xµ ≡ ∂µ transforms as a covariantvector. In the same way it can be shown that ∂/∂xµ ≡ ∂µ transforms as a contravariantvector. Therefore, we can define:

∂µ =∂

∂xµ= (∂/∂x0, ∂/∂x1, ∂/∂x2, ∂/∂x3) = (∂t/c, ∂x, ∂y, ∂z) = (∂t/c,∇) . (3.17)

In the same way,∂µ = gµν∂ν = (∂t/c,−∇) . (3.18)

We can now define the d’Alembertian operator 2 = ∂µ∂µ which, from the definitions above is

2 = ∂µ∂µ =

1

c2

∂2

∂t2−∇2 . (3.19)

3.2 The Klein-Gordon equation

Historically, the first attempt to obtain a relativistic version of the Schrödinger equation wascarried out by applying the quantization rules to the relativistic energy-momentum invariant. Innon-relativistic quantum mechanics, the correspondence principle states that the momentumoperator is p = −i~∇, and the energy operator E = i~∂t. The Schrödinger equation hasbeen obtained by quantizing the classical Hamiltonian. To obtain the relativistic version of thatequation, one might apply the correspondence principle to the energy-momentum dispersionrelation p2 = (E/c)2 − p2 = m2c4. Namely:

E(p) = [m2c4 + p2c2]1/2 =⇒ i~ ∂tφ(x) = [m2c4 − ~2c2∇2]1/2φ(x) , (3.20)

where the x in the argument of φ(x) is to be taken as xµ. This first attempt to obtaina quantum relativistic equation poses some problems. In particular, we are faced with thedifficulty of computing a square root of an operator. The only way to define this is to expressthe square-root as a Taylor expansion

i~ ∂tφ =

[mc2 − ~2∇2

2m− ~4(∇2)2

8m3c2+ ...

]φ , (3.21)

meaning that an infinite number of boundary conditions is needed to express the time-dependence of the wave-function. Also, this equation does not treat time and space in thesame way (the equation is of first order in time derivatives and infinite order in spatialderivatives). Finally, the Schrödinger equation is not Lorentz covariant, i.e., taking intoaccount that ∂′µ = Λ ν

µ ∂ν , φ′(x′) = φ(x) (because φ is a scalar) and x′µ = Λµ

νxν , one can

show that

i~∂φ(x)

∂t=

~2∇2

2mφ(x) −→/ i~

∂φ′(x′)

∂t′=

~2∇′ 2

2mφ′(x′) , (3.22)

and, therefore, conclude that the Schrödinger equation does not have the same form in allrelativistic reference frames.

64

3.2. THE KLEIN-GORDON EQUATION

3.2.1 The free particle Klein-Gordon equation

A second approach, is to apply the correspondence principle directly to the relativistic invariantE2 = m2c4 + p2c2 giving

− ~2 ∂2φ(x)

∂t2= (m2c4 − c2~2∇2)φ(x) . (3.23)

Using the fact that pµ = i~(∂t/c,−∇), we can write the above equation in the form:

(pµpµ −m2c2)φ(x) = 0 or

(2 +

m2c2

~2

)φ(x) = 0 . (3.24)

where we have used the definition of the 2 operator given in (3.19). This equation is theKlein-Gordon equation for a free particle, named after the physicists Oskar Klein and WalterGordon, who in 1926 proposed it to describe relativistic electrons. Although it turns out thatthe Klein-Gordon equation does not describe the spinning electron, it correctly describesthe spinless pion, a composite particle of a quark-antiquark pair. Since the Higgs boson(discovered at CERN on July 4, 2012) is a spin-zero particle, it is the first elementary particlethat is described by the Klein-Gordon equation. Up to now, all experimental data indicatesthat the boson found at the CERN LHC is that of the Standard Model.

The Klein-Gordon equation was first considered by Schrödinger. In fact, that equation isfound in his notebooks from late 1925, and he appears to have written a manuscript applyingit to the hydrogen atom. Yet, the Klein-Gordon equation does not describe spin and, whenapplied to the Hydrogen atoms, it predicts incorrectly the hydrogen atom’s fine structure.Soon after the formulation of the Schrödinger equation, Vladimir Fock wrote an article in1926 about its generalization for the case of magnetic fields, and independently derived theKlein-Gordon equation.

It is straightforward to show that the Klein-Gordon equation is invariant under Lorentztransformations. Namely, in a rotated reference frame we would have:(

∂′µ∂′µ +

m2c2

~2

)φ′(x′) = 0 . (3.25)

Since φ(x) is a scalar function, it transforms trivially under the Lorentz transformation, i.e.φ′(x′) = φ(x). Obviously, the speed of light c and the Planck constant ~ is the same in anyrelativistic frame. The same happens with the rest mass of the particle m. We just have tocheck how ∂′µ∂

′µ transforms. From Eq. (3.16) we have seen that ∂′µ = ∂ν [Λ−1]νµ. From the

fact that xα = [Λ−1] µα x

′µ, we have

∂′µ =∂

∂x′µ=∂xα∂x′µ

∂

∂xα= [Λ−1] µ

α ∂α . (3.26)

With this,

∂′µ∂′µ = [Λ−1]νµ[Λ−1] µ

α ∂ν∂α = [Λ−1] µ

α Λ νµ ∂ν∂

α = δ να ∂ν∂

α = ∂α∂α . (3.27)

With this, we have shown that the operator 2 transforms as a scalar under Lorentz transfor-mations and, therefore, the Klein-Gordon equation is Lorentz invariant (or covariant).

65


Figure 3.1: Free particle spectrum for spinless relativistic particles.

We can now look at the Klein-Gordon equation

∇2φ(x)− 1

c2

∂2φ(x)

∂t2=m2c2

~2φ(x) . (3.28)

and obtain its solutions for free particles. Since the Klein-Gordon equation resembles with awave equation, the solutions are plane waves of the type

φ(r, t) = Ae−i(ωt−k.r) . (3.29)

Replacing this into the Klein-Gordon equation, we obtain the dispersion relation:

− |k|2 +ω2

c2=m2c2

~2. (3.30)

We can now compute energy and momentum by using:

〈p〉 = 〈φ| − i~∇|φ〉 = ~k , 〈E〉 = 〈φ| i~ ∂t|φ〉 = ~ω . (3.31)

Replacing this in (3.30) we get the classic relativistic equation:

E2 = m2c4 + p2c2 ⇒ E = ±√m2c4 + p2c2 . (3.32)

Solutions have both positive and negative eigenvalues for energy (see Fig. 3.1). Note thatin QM, one cannot drop solutions that seem to be unphysical-the full set of eigenstates isneeded. So, for a given momentum p there are two solutions for the Klein-Gordon equation,namely:

φ±(r, t) = A±e−i(±|E|t−p.r)/~ , (3.33)

where A± are normalization constants for the positive and negative energy solutions. Takinginto account that pµxµ = p0x0 − p.r, we can write the solutions as

φ(x) = Ae−ipµxµ/~ . (3.34)

66


We can now construct the conserved current associated to the Klein-Gordon equation. Innon-relativistic quantum mechanics, the time-dependence of the probability density ρ = ψ∗ψ(always positive) is related to the probability current j by the continuity equation:

∂ρ

∂t= −∇.j , j = − i~

2m[ψ∗(∇ψ)− (∇ψ∗)ψ] . (3.35)

Let us now try to obtain a similar equation for spinless relativistic particles described by theKlein-Gordon equation. We start with Eq. (3.25) and multiply it by φ(x)∗ on the left, andwith the conjugate of the same equation multiplied by φ on the right, i.e.

φ∗(∂µ∂µφ) +

m2c2

~2φ∗φ = 0

(∂µ∂µφ∗)φ+

m2c2

~2φ∗φ = 0 . (3.36)

We now subtract the above two equations to obtain

φ∗(∂µ∂µφ)− (∂µ∂

µφ∗)φ = [φ∗(∂µ∂µφ) + ∂µφ

∗∂µφ ]− [∂µφ∗∂µφ+ (∂µ∂

µφ∗)φ]

= ∂µ[φ∗(∂µφ)− (∂µφ∗)φ] = 0 . (3.37)

In covariant form, the above equation is

∂µjµ = 0 , jµ =

i~2m

(φ∗∂µφ− φ ∂µφ∗) =i~2m

φ∗←→∂ µφ , (3.38)

where we have used the notation a←→∂ b = a(∂µb)− (∂µa)b. The above equation reflects the

fact that for spinless relativistic particles there is a conserved jµ = (cρ, j), which is equivalentto the continuity equation (3.35). The conserved probability density ρ is now given by j0, i.e.

ρ =j0

c=

i~2mc2

(φ∗∂tφ− φ ∂tφ∗) , (3.39)

We can now use solutions (3.33) to compute the probability density for free Klein-Gordonparticles. Writing φ(x) = e±i|E|t/~u(r) we obtain:

ρ =i~

2mc2(φ∗∂tφ− φ ∂tφ∗) =

∓|E|mc2|u(r)|2 . (3.40)

The above equation shows that the probability density can be either positive or negative, whichdoes not make any sense if ρ is to be interpreted as a probability. This is a very unaestheticaspect of the Klein-Gordon equation. Notice that the reason why this happens is the fact thatthe Klein-Gordon equation is second order in time derivatives. Therefore, a relativistic equationinvolving order one time derivatives will, in principle, have a positive-definite probability density.Later, we will see what is the physical significance of these negative energy solutions.

3.2.2 Non-relativistic limit of the Klein-Gordon equation

It is interesting to study the non-relativistic limit of the Klein-Gordon equation. For that, wesomehow have to extract out the relativistic part of the wave-function φ(x) as

φ(r, t) = ϕ(r, t) exp

(−imc

2

~t

). (3.41)

67


where ϕ(r, t) is the wavefunction in the non-relativistic limit in which the energy is E ′ =E −mc2 with E ′ mc2. We therefore expect that∣∣∣∣∂ϕ(r, t)

∂t

∣∣∣∣ ∼ E ′ϕ(r, t) mc2ϕ(r, t) . (3.42)

In order to obtain the desired limit, we must replace (3.41) into the Klein-Gordon equation.Thus we have to compute the second order time-derivative of φ(r, t), namely:

∂φ

∂t=

(∂tϕ− i

mc2

~ϕ

)exp

(−imc

2

~t

)' −imc

2

~ϕ exp

(−imc

2

~t

),

∂2φ

∂t2' −

(2imc2

~∂tϕ+

m2c4

~2ϕ

)exp

(−imc

2

~t

). (3.43)

Replacing this into (3.28) we obtain

− 1

c2

(2imc2

~∂t +

m2c4

~2

)ϕ =

(∇2 − m2c2

~2

)ϕ , (3.44)

which implies

i~∂ϕ(r, t)

∂t= − ~2

2m∇2ϕ(r, t) , (3.45)

which is the Schrödinger equation for ϕ(r, t). Notice that this result further supports the factthat the Klein-Gordon equation describes spinless particles.

3.2.3 Klein-Gordon equation with electromagnetic field

We now consider a relativistic scalar particle with charge q, interacting with an electromagneticfield

Aµ = (A0,A) , Aµ = gµνAν = (A0,−A) . (3.46)

The interaction of charged particles with such fields can be obtained by performing theminimal coupling:

i~∂

∂t−→ i~

∂

∂t− qA0 , −i~∇ −→ −i~∇− q

cA . (3.47)

In covariant form this can be written as:

pµ −→ pµ − q

cAµ , pµ −→ pµ −

q

cAµ . (3.48)

Therefore, the Klein-Gordon equation is now

(pµpµ −m2c2)φ(x) = 0 −→

[(pµ −

q

cAµ

)(pµ − q

cAµ)−m2c2

]φ(x) = 0 . (3.49)

Taking into account that pµ = i~∂µ = i~(∂t/c,−∇), the above equation can be expressed as[(i~∂

∂t− qA0

)2

− c2(−i~∇− q

cA)2

−m2c4

]φ(x) = 0 . (3.50)

68


Let us now suppose that we have the above equation for a particular solution with positiveenergy E and momentum p, i.e φ+(r, t). So we have[(

i~∂

∂t− qA0

)2

− c2(−i~∇− q

cA)2

−m2c4

]φ+(r, t) = 0 . (3.51)

If we now conjugate the above equation and define φ∗+(r, t) as φC+(r, t), we have[(i~∂

∂t+ qA0

)2

− c2(−i~∇ +

q

cA)2

−m2c4

]φC+(r, t) = 0 , (3.52)

which the Klein-Gordon equation of a particle with negative energy, mass m and charge −qsubject to the electromagnetic field. These opens up the interpretation of the negative-energysolutions as belonging to anti-particles.

Proceeding in the same way as we did for the free particle, we can show that there is aconserved 4-current jµ given by

∂µjµ = 0 , jµ =

i~2m

φ∗←→∂ µφ− q

mcAµφ∗φ . (3.53)

Exercise 3.1. Prove Eq. (3.53).

From the above equation, we get

ρ(x) =i~

2mc2

(φ∗←→∂t φ

)− q

mc2A0φ∗φ . (3.54)

for the probability density ρ(x). Notice that, once more, ρ(x) is not positive definite.

3.2.4 Non-relativistic limit of the Klein-Gordon equation with electromagneticfield

We consider again the wavefunction written as in (3.41). Remembering that ϕ is related withthe non-relativistic character of the particle, and considering that all potentials are weak whencompared with the rest mass, we can write∣∣∣∣∂ϕ(r, t)

∂t

∣∣∣∣ ∼ E ′ϕ(r, t) mc2ϕ , |qA0ϕ| mc2ϕ . (3.55)

We now compute the term containing the time-derivatives in the Klein-Gordon (3.49). Namely,for the first order derivative we have:(

i~∂

∂t− qA0

)φ(r, t) = (i~ ∂tϕ− qA0ϕ+mc2ϕ) exp

(− i~mc2t

), (3.56)

69


and (i~∂

∂t− qA0

)2

φ(r, t) = (−~2 ∂2t ϕ− 2i~qA0∂tϕ+ 2i~mc2∂tϕ− i~qϕ∂tA0

+ q2A20ϕ− 2qA0mc

2ϕ+m2c4ϕ) exp

(− i~mc2t

). (3.57)

Keeping now only the order-one terms in the small parameters, we get:(i~∂

∂t− qA0

)2

φ(r, t) ' (m2c4 − 2qA0mc2 + 2i~mc2∂t − i~q∂tA0)ϕ exp

(− i~mc2

).

(3.58)

Replacing this into Eq. (3.50) and isolating the first time derivative of ϕ, one obtains

i~∂tϕ '[p2

2m− q

mcA.p+ qA0 +

i~q2mc

(∇.A) +i~q

2mc2∂tA0 +

q2

mc2A2

]ϕ , (3.59)

where we have used the fact that p = −i~∇. Taking into account the Lorentz gauge condition∂tA0/c+ ∇.A = 0, we arrive at the result

i~∂tϕ '[p2

2m− q

mcA.p+ qA0 +

q2

mc2A2

]ϕ , (3.60)

which is the Schrödinger equation for a particle of charge q interacting with the electromagneticfield (remember the results we have obtained in the previous chapter).

3.2.5 Gauge invariance of the Klein-Gordon equation

In Chapter 2 we have seen that Maxwell equations are invariant under gauge transformations

Aµ(x) −→ A′µ(x) = Aµ(x) + ∂µχ(x) , (3.61)

where χ(x) is a scalar function. The question now is what are the consequences of imposinggauge invariance for the Klein-Gordon equation. We start by writing it with the transformedfield, namely(

i~ ∂µ − q

cA′µ)(

i~ ∂µ −q

cA′µ

)φ(x)−m2c4φ(x) = 0

⇔ gµν(i~ ∂ν −

q

cA′ν

)(i~ ∂µ −

q

cA′µ

)φ(x)−m2c4φ(x) = 0

⇔ gµν(i~ ∂ν −

q

cAν −

q

c∂νχ)(

i~ ∂µ −q

cAµ −

q

c∂µχ

)φ(x)−m2c4φ(x) = 0 . (3.62)

This equation is equivalent(i~ ∂µ − q

cAµ)(

i~ ∂µ −q

cAµ

)φ′(x)−m2c4φ′(x) = 0 . (3.63)

It is straightforward to see that φ′(x) = exp[ieχ(x)/(~c)]φ(x). In other words, the gaugetransformation change the phase of the wavefunction. The phase factor is the same for all

70


states φ and, therefore, does not play any role in physical observables. In fact, we could havereasoned in the opposite way (and we will later on). Suppose we impose gauge invariance ofthe free-particle Klein-Gordon equation under the transformation

φ(x) −→ φ′(x) = exp[ie

~cχ(x)

]φ(x) . (3.64)

This is called a U(1) local (because χ depends on the spacetime coordinates) gauge trans-formation. From what we have seen above, it is easy to conclude that invariance under thistransformation requires the addition of a vector field Aµ as shown in Eq. (3.61), and thatwe have to replace the usual derivative by a covariant derivative Dµ = ∂µ + ieAµ/(~c).Invariance under local gauge transformations demands adding vector fields. This is the essenceof the formulation of more fundamental theories like the Standard Model of electroweakinteractions.

3.2.6 Box normalization of Klein-Gordon free particle solutions

We have seen that the free-particle solutions of the Klein-Gordon equation can be written as:

φ±(r, t) = A±e−i(±|E|t−p.r)/~ , (3.65)

to which correspond a probability density ρ±(x) = ±|E|φ∗±φ±/(mc2). We now normalizethese solutions in a 3-dimensional cubic box of volume V = L3 in such a way that momentumgets quantised. Namely:

pn =2π~Ln =

2π~L

(nx, ny, nz) =⇒ En = ±√p2nc

2 +m2c4 , (3.66)

with ni integer. Although we cannot identify ρ as a probability density, we can define a4-current density of charge j′µ = qjµ. With this definition,the charge density is ρ′± = qρ±. Ifwe now demand that the total charge is the volume integral of ρ′ we have:

± q =

∫V

ρ′±dr = ±q|En|mc2

|A±n |L3 . (3.67)

If we choose real normalization constants we get:

A±n =

√mc2

L3|En|=⇒ φ±n (r, t) =

√mc2

L3|En|exp

[− i~

(±|En|t− pn.r)

]. (3.68)

Notice that the normalization of positive and negative-energy solutions is the same, beingthe only difference the time exponential. We can now write the most general solution of thefree-particle Klein-Gordon equation

φ+(r, t) =∑n

anφ+n (r, t) =

∑n

√mc2

L3|En|exp

[− i~

(|En|t− pn.r)

]

φ−(r, t) =∑n

bnφ−n (r, t) =

∑n

√mc2

L3|En|exp

[i

~(|En|t+ pn.r)

](3.69)

(3.70)

71


We can now ask what happens for neutral particles with ρ′ = 0. From the charge densityexpression

ρ′ =iq~

2mc2(φ∗∂tφ− φ ∂tφ∗) , (3.71)

we conclude that out of φ± we can define a solution for a neutral particle φ0 with ρ′ = 0 ifφ0 = φ0∗. We can actually do this by defining

φ0n =

1√2

[φ+n (pn) + φ−n (−pn)

]=

√2mc2

L3|En|cos

(pn.r − |En|t

~

), (3.72)

for which ρ′ = 0 and j ′ = 0. We have just seen that the relativistic quantum theory bringsup new degrees of freedom (the charge degrees of freedom). In the case of spinless particles,we just saw that for a given momentum p we can construct three solutions corresponding to±1, 0 charges.

3.2.7 Klein Gordon equation with Coulomb potential (the pionic atom)

A pionic atom is an atom where one or more electrons are replaced by π− mesons (spin-0 particles with mass mπ made of a quark-antiquark pair. In this case ud). To createpionic atoms, first pions must be produced. This is done through proton-proton scattering:p+ p→ p+ p+ π + π−. Afterwards, the negative pions are filtered out from the beam andstopped in matter. Exotic (pionic) atoms are formed when the π− mesons are captured byordinary atoms, ejecting bounded electrons from their Bohr orbits. Therefore, we can writethe π− capture process as

π− + atom→ [(atom− e−) + π+] + e− .

The ejection of electrons from the atoms happens when the pion reach the velocity of electronsin the atom. The typical quantum number n for the π mesons will be n ∼

√mπ/me ' 17.

The process of π− capture includes the so-called Auger effect in which a π− in the atomfilling an inner-shell vacancy causes the emission of an Auger electron which carries off theenergy excess. Subsequent deexcitations occur either by Auger or X-ray emission. At the end,the physical system is an Hydrogen-like atom with the e− replaced by the π−. In the case ofHydrogen we have a pionic Hydrogen.

So we now want to study the bound state of a pion π− and nucleus with atomic numberZ (and charge Ze). The potencial is the Coulomb potencial V (r) such that

qA0 = V (r) = −Ze2

r= −Z~cα

r, α =

e2

~c, (3.73)

where α is the fine-structure constant. Notice that since there are no magnetic fields, wehave A = ~0.

So, we start with the Klein-Gordon equation (3.63) for the case of the pionic atom, namely[(pµ − q

cAµ)(

pµ −q

cAµ

)−m2

πc4]

Φ(x) = 0 . (3.74)

72


By using pµ = i~∂µ, we can write the above equation as[(i~ ∂µ − q

cAµ)(

i~ ∂µ −q

cAµ

)−m2

πc4]

Φ(x) = 0 . (3.75)

Which, for A = ~0 is [(i~c∂t −

q

cA0

)2

+ ~2∇2 −m2πc

2

]Φ(r, t) = 0

⇔

[i~ ∂t − V (r)]2 + ~2 c2∇2 −m2πc

4

Φ(r, t) = 0 . (3.76)

This is a separable equation for which a solution of the type Φ(x) = e−iEt/~ϕ(r) can bewritten. Replacing this in (3.76), we get

[E − V (r)]2 + ~2 c2∇2 −m2πc

4ϕ(r) = 0

⇔ −~2 c2 ∇2ϕ(r) =

[E − V (r)]2 −m2πc

4ϕ(r) (3.77)

Expressing the differential operator ∇2 in spherical coordinates we have

− ~2 c2

[1

r2

∂

∂r

(r2 ∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2

]ϕ(r)

=

[E − V (r)]2 −m2πc

4ϕ(r) (3.78)

This is an equation similar to the one you found for the Hydrogen atom. The solution allowsfor a speration of variables ϕ(r) = u(r)Y (θ, φ). Replacing this into the above equation gives:

− ~2 c2

[Y

r2

∂

∂r

(r2∂u

∂r

)+

u

r2 sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

u

r2 sin2 θ

∂2Y

∂φ2

]=

[E − V (r)]2 −m2πc

4uY . (3.79)

Multiplying the above equation by r2/(uY ) leads to

− ~2 c2

[1

u

∂

∂r

(r2∂u

∂r

)+

1

Y sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

Y sin2 θ

∂2Y

∂φ2

]=

[E − V (r)]2 −m2πc

4r2 . (3.80)

Putting together all terms which depend on r and on θ, φ, we have

~2 c2

u

∂

∂r

(r2∂u

∂r

)+

[E − V (r)]2 −m2πc

4r2

= −~2 c2

[1

Y sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

Y sin2 θ

∂2Y

∂φ2

]. (3.81)

The left-hand side is a function of r only, while the right-hand side is a function of θ and φ.So, both sides have to be equal to a constant λ(~c)2, namely[

1

r2

∂

∂r

(r2 ∂

∂r

)+

[E − V (r)]2 −m2

πc4

~2 c2− λ

r2

]u(r) = 0 (3.82)[

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂φ2+ λ

]Y (θ, φ) = 0 . (3.83)

73


The solutions of the last equation1 are the spherical harmonic Y`,m, with λ = `(`+ 1) andm = 0,±1, ...,±`. Then, Eq. (3.82) becomes[

− 1

r2

∂

∂r

(r2 ∂

∂r

)+` (`+ 1)

r2

]u(r) =

[E − V (r)]2 −m2

πc4

~ c2

u(r) . (3.84)

Writing now u(r) = R`(r)/r, and replacing V (r) by the expression (3.73) for the Coulombpotential, we can finally obtain the (relativistic) radial equation for the pionic atom:[

∂2

∂r2− ` (`+ 1)− (Zα)2

r2+

2EZα

~cr+E2 −m2

πc4

~2 c2

]R`(r) = 0 . (3.85)

We can now compare this equation with the one we would have obtained if we had used theSchrödinger equation, namely:[

∂2

∂r2− `′ (`′ + 1)

r2+

2mπcZα

~r+

2mπE′

~2

]R`(r) = 0 . (3.86)

The Bohr energy levels would then be:

E ′ = −mπ c2(Zα)2

2n2, n = 1, 2, ... . (3.87)

This spectrum is independent of the quantum numbers ` and m. By comparing Eqs. (3.85)and (3.86) we see that:

`′ (`′ + 1) = ` (`+ 1)− (Zα)2 =⇒ `′ = −1

2+

√(`+

1

2

)2

− (Zα)2 . (3.88)

Remember that in the Schrödinger case n − ` is related with the number of nodes of thewavefunction and, therefore, must be an integer. For the pionic atom n is replaced by n′ inEq. (3.87) and n′ − `′ must also be an integer. So, this means that if `′ → ` −∆`, thenn′ → n−∆` with ∆` = `− `′.

So, in order to use the Bohr energy levels to compute the Klein-Gordon spectrum we haveto perform the replacements:

n→ n−∆` , E ′ → E2 −m2π

2mπc2, α→ αE

mπc2. (3.89)

Using now (3.87) we get:

E2 −m2π

2mπc2= − (Zα)2E2

(n−∆`)2mπc2. (3.90)

Solving for E, we finally obtain:

En,` =mπ c

2√1 + (Zα)2

(n−∆`)2

=mπ c

2√1 +

[Zα

n−(`+ 12)+

√(`+ 1

2)2−(Zα)2

]2, n = 1, 2, ...; ` = 0, 1, ..., n− 1 .

(3.91)1See Griffiths, Chapter 4 for the solution of this equation.

74

3.3. THE DIRAC EQUATION

Notice that for Zα & 1/2 ⇒ Z & 68 there is no energy eigenvalue for the 1s state (E1,0

becomes imaginary). For Z ∼ 68 we have E1,0 ' −mπc2/√

2 and the binding energy Eb isEb = mπc

2 −E1,0 ' −40 MeV. Notice that for Zα > 1/2 we have to consider the finite sizeof the nucleus and the fact that pions interact strongly with the nucleus.

Since α ∼ 1/137, in general Zα 1. Therefore, we can expand En,` in powers of (Zα)2.Namely,

En,` = mπc2

1︸︷︷︸Rest energy

−(Zα)2

2n2︸︷︷︸Bohr term

−(Zα)4

2n3

(1

`+ 1/2− 3

4

)︸︷︷︸

leading relativistic correction

. (3.92)

This result is in agreement with experiment. However, it does not explain the energy spectrumof Hydrogen which, for instance, has a splitting of the (n = 2, ` = 1) level. Therefore, theKlein-Gordon equation is not satisfactory for spin 1/2 particles like the electron.

3.3 The Dirac equation

In the previous section we have obtained a relativistic equation for spin-0 particles which sufferfrom two problems. First, the existence of both positive and negative energy solutions, andsecond the fact that we are no longer able to interpret the probability density ρ as being thetime-component of the conserved 4-current jµ. This problem is connected with the fact thatthe Klein-Gordon equation is of second order in time derivatives. Therefore, it is somehowclear that in order to obtain a positive definite probability density, we must search for arelativistic equation which is of first-order in time derivatives. Moreover, if we want suchequation to treat space and time in the same way, it must be also of first order in spatialderivatives. In conclusion, we are looking for an equation of the type

i~∂ψ(x)

∂t= Hψ(x) , (3.93)

Notice that the Hamiltonian H must obey the relativistic energy-momentum relation H2 =c2p2+m2c4 which implies E2 = c2p2+m2c4. This, together with the fact that the Hamiltonianmust be of first order in spatial derivatives, leads us to the following form of the Hamiltonianwe are looking for

H = cα.p+ βmc2 . (3.94)

The first conclusion we can immediately take is that α and β cannot be a 3-vector and anumber, respectively. If this was the case, H2 would have crossed terms linear in p, whichare incompatible with the energy-momentum relation. Therefore αi and β must be matrices.Keeping this in mind, let us replace H in Eq. (3.93)

i~∂ψ(x)

∂t=

(−i~c

∑i

αi∂i + βmc2

)ψ(x) , (3.95)

75


where we have used the fact that the momentum operator is p = −i~∇. Taking into accountthat H2 = −h2∂2

t , we have

− ~2∂2ψ(x)

∂t2= i~

∂

∂t

(−i~c

∑i

αi∂i + βmc2

)ψ(x)

= −i~c∑j

αj∂j

(−i~c

∑i

αi∂iψ + βmc2ψ

)+ βmc2

(−i~c

∑i

αi∂iψ + βmc2ψ

)

= −~2c2∑i,j

αiαj + αjαi2

∂i∂jψ − i~mc3(αiβ + βαi)∂iψ + β2m2c4ψ

= −~2c2∑i,j

αi, αj2

∂i∂jψ − i~mc3αi, β∂iψ + β2m2c4ψ . (3.96)

So, in order for this equation to obey the energy-momentum relation we must have:

αi, αj = 2δij , αi, β = 0 , α2i = β2 = 1 . (3.97)

The relation α2i = β2 = 1 implies that the eigenvalues of αi and β are ±1. Moreover the

hermiticity condition of the Hamiltonian implies that α†i = αi and β = β†. On the other hand:

Tr(αi) = Tr(β2αi) = Tr(β βαi︸︷︷︸=−αiβ

) = −Tr(αi) =⇒ Tr(αi) = 0 . (3.98)

The same relation could be obtained for β, i.e. Tr(β) = 0. Since the trace is the sum of theeigenvalues which can only be ±1 we conclude that the dimension N of these matrices mustbe even. The minimal dimension would be N = 2. However, there are only three independentmatrices which obey the above relations: the Pauli matrices. But we actually need four. So,the next case is N = 4, for which we can actually find four independent matrices havingthe required properties. This means that the minimum dimension of the algebra (3.97) isN = 4. The choice of the αi and β matrices is not unique in the sense that there are severalequivalent representations like:

Dirac representation: αi =

(0 σiσi 0

), β =

(112 00 −112

)Weyl representation: αi =

(σi 00 −σi

), β =

(0 −112

−112 0

).

(3.99)

(3.100)

Notice that what we have just found has profound implications. Namely, the search for arelativistic equation with first-order time derivatives together with an Hamiltonian which obeysthe Einstein energy-momentum relation led us to an Hamiltonian which is a 4-dimensionalmatricial operator. Therefore, since this Hamiltonian acts on wavefunctions, ψ(x) cannot bea number anymore and the new relativistic equation we have obtained is written in the form

i~∂ψi(x)

∂t=

4∑j=1

[c(α.p)ij + βijmc

2]ψj(x) , ψ(x) =

ψ1(x)ψ2(x)ψ3(x)ψ4(x)

. (3.101)

76


This is the Dirac equation written in the canonical (or Hamilton) form. The wavefunctionψ has now four components and is usually called bispinor. In vectorial notation the aboveequation reads:

i~∂ψ(x)

∂t=[c(α.p) + β mc2

]ψ(x) . (3.102)

We now look at the continuity equation. Let us define the hermitian conjugate of thebispinor ψ as ψ†(x) = [ψ∗1(x), ψ∗2(x), ψ∗3(x), ψ∗4(x)]. With this we can write:

i~ψ†∂ψ(x)

∂t= −i~c ψ†α.(∇ψ) +mc2ψ†βψ

−i~∂ψ†(x)

∂tψ = i~c (∇ψ†).αψ +mc2ψ†βψ . (3.103)

where the first (second) equation corresponds to taking the (Hermitian conjugate of the)Dirac equation and multiply it on the left (right) by ψ† (ψ). The difference of the above twoequations leads to:

ψ†∂ψ(x)

∂t+∂ψ†(x)

∂tψ = −cψ†α.(∇ψ)− c(∇ψ†).αψ , (3.104)

and, therefore, we conclude that the above equation can be written in a compact way:

∂ρ(x)

∂t= −∇.j , ρ(x) = ψ†(x)ψ(x) , j = cψ†(x)αψ(x) , (3.105)

which is the continuity relation for the Dirac equation. Notice that ρ(x) = |ψ(x)|2 is alwayspositive definite and, thus, can be interpreted as being a probability density.

3.3.1 Dirac equation in covariant form

For future purposes it is convenient to write the Dirac equation in covariant form, i.e. in a4-vector language. For that, we multiply (3.102) by β/c on the left and use p = −i~∇ toobtain

β

c

[i~∂

∂t+ i~c

3∑k=1

αk∂

∂xk− βmc2

]ψ(x) = 0[

i~β∂

∂(ct)+

3∑k=1

(βαk)i~∂

∂xk−mc

]ψ(x) = 0 . (3.106)

Defining the Dirac γ matrices as:

γ0 = β , γk = βαk . (3.107)

we can write (3.106) as

i~(γ0 ∂

∂x0+ γ1 ∂

∂x1+ γ2 ∂

∂x2+ γ3 ∂

∂x3

)ψ(x)−mcψ(x) = 0

=⇒ (i~γµ∂µ −mc)ψ(x) = 0 . (3.108)

77


Using the Feynman-slash notation A/ = γµAµ, we can write the Dirac equation in covariantform as

(i~∂/−mc)ψ(x) = 0 or (p/−mc)ψ(x) = 0 . (3.109)

Taking into account the form of the α matrices given in (3.99), we have

γ0 =

(112 00 −112

), γk =

(0 σk−σk 0

)(3.110)

for the gamma matrices in the Dirac representation.From the properties (3.97) we can prove that

(γk)†γk = (βαk)†(βαk) = αkβ

2αk = (αk)2 = 114 =⇒ (γk)† = (γk)−1 , k = 1, 2, 3 .

(3.111)

On the other hand

(γk)† = (βαk)† = (αk)

†(β)† = αkβ = −βαk = −γk =⇒ (γk)† = −γk , k = 1, 2, 3 .

(3.112)

As for the square of a gamma matrix γk one has

(γk)2 = (βαk)2 = (βαk)(βαk) = −αkβ2αk = −(αk)

2 = −114

=⇒ (γk)2 = −114 , k = 1, 2, 3 . (3.113)

The matrix γ0 is unitary and Hermitian since γ0(γ0)† = β2 = 114 and (γ0)† = β† = β = γ0.So,

(γ0)† = (γ0)−1 , γ0 = (γ0)† . (3.114)

From what we have seen above, we conclude that the Dirac matrices obey the Clifford algebra:

γµ, γν = 2gµν . (3.115)

In particular, these relation implies that (γµ)2 = gµµ and (γµ)† = gµµγµ.

3.3.2 Covariance of the Dirac equation

A relativistic theory must be Lorentz covariant, i.e. all its equations have to be invariantunder transformations from one inertial frame to the other. We want now to impose Lorentzcovariance to the Dirac equation. The situation is sketched in Fig. 3.2. Suppose observersA and B are solidary with their inertial reference frames. coordinates for observer B are theLorentz transformed of A, namely x′µ = Λµ

νxν . Suppose now that for A the wavefuntion of a

relativistic Dirac system is ψ(x). We want now to find the wave function of the same systemfor observer B, i.e. we want to compute ψ′(x′). According to the principle of relativity, allbasic Physics equations must have the same form in all reference frames. Therefore:(

i~γ′µ∂′µ −mc)ψ′(x′) = 0 . (3.116)

78


Figure 3.2: Observer B is solidary with the transformed reference frame related with that ofobserver A by the Lorentz transformation x′µ = Λµ

νxν .

Notice that γ′µ, γ′ν = 2gµν since the metric is the same for both observers. Actually, allproperties of the Dirac matrices remain valid irrespective of the reference frame. It can beshown that all γ matrices γµ′ which satisfy the above properties are identical to γµ up toa unitary transformation U , i.e. U †γµ′U = γµ. Since unitary transformations do not modifyphysics, we can use the gamma matrices γµ instead of γµ′ (if you wish to know more detailsabout this just google "Pauli’s fundamental theorem"). Therefore, the Dirac equation forobserver B is: (

i~γµ∂′µ −mc)ψ′(x′) = 0 . (3.117)

Our goal is to find a transformation law which allows us to obtain ψ′(x′) by knowing ψ(x)and Λ. We start by noting that, since both the Dirac equation and the Lorentz transformationare linear in the coordinates, the wavefunction transformation is also linear. Namely:

ψ′(x′) = ψ′(Λx) = S(Λ)ψ(x) . (3.118)

S(Λ) is a 4×4 matrix which depends on the parameters of Λ (for example, boost or rotations)and acts on the four components of ψ. Obviously, these transformation must have an inverse.From the above equation

ψ(x) = S(Λ)−1ψ′(x′) = S−1(Λ)ψ′(Λx) . (3.119)

On the other hand, from the point of view of observer A,

ψ(x) = S(Λ−1)ψ′(x′) = S(Λ−1)ψ′(Λx) . (3.120)

Comparing the above two equations we conclude that

S−1(Λ) = S(Λ−1) . (3.121)

Let us then start with the Dirac equation for observer A: (i~γµ∂µ −mc)ψ(x) = 0. Takinginto account that, from Eq. (3.119), ψ(x) = S−1(Λ)ψ′(x′), we have[

i~γµS−1(Λ)∂µ −mcS−1(Λ)]ψ′(x′) = 0 . (3.122)

79


Multiplying this equation on the left by S(Λ) leads to[i~S(Λ)γµS−1(Λ)∂µ −mc

]ψ′(x′) = 0 , (3.123)

where we have used the fact that S(Λ)S−1(Λ) = 114. We now have to relate ∂µ with ∂′µ.Remember that

∂µ ≡∂

∂xµ=∂x′ν

∂xµ∂

∂x′ν= Λν

µ

∂

∂x′ν= Λν

µ∂′ν . (3.124)

Replacing this in Eq. (3.123) one gets[i~S(Λ)γµS−1(Λ)Λν

µ∂′ν −mc

]ψ′(x′) = 0 . (3.125)

Covariance demands that this equation must be equal to Eq. (3.118) and therefore

S(Λ)γµS−1(Λ) Λνµ = γν . (3.126)

Therefore, once we prove that for every Lorentz transformation Λ there exists the correspondingtransformation S(Λ) obeying the above equation, we have proven the covariance of the Diracequation. Consequently, for every Λ we should determine S(Λ).

We have seen that [Λ−1]βµΛµα = δβα which is equivalent to Λµ

αΛ βµ = δβα. From this, we

conclude that det(Λ2) = 1⇒ det(Λ) = ±1. Lorentz transformations with det(Λ) = 1 (−1)are called proper (improper). The proper ones can be obtained by an infinite number ofinfinitesimal transformations (rotations and boosts), while the improper ones can’t (spatialand time reversal).

3.4 S(Λ) for proper Lorentz transformations

In this section we will determine the spinor transformations S(Λ) defined in Eq. (3.118) forany type of Lorentz transformation Λ. For that, let us consider an infinitesimal proper Lorentztransformation which can be written as:

Λνµ = δνµ + ∆ωνµ , (3.127)

where ωνµ is expected to be small. Taking into account that:

[Λ−1] µν Λ σ

µ = δ σν ⇔ Λµ

νΛσµ = δ σ

ν ⇔ (δµν + ∆ωµν)(δσµ + ∆ω σ

µ )

' δµνδσµ + δµν∆ω

σµ + δ σ

µ ∆ωµν = δ σν + ∆ω σ

ν + ∆ωσν = δ σν

⇔ ∆ω σν + ∆ωσν = 0⇔ gµν(∆ω σ

ν + ∆ωσν) = 0

⇔ ∆ωµσ + ∆ωσµ = 0⇒ ∆ωµσ = −∆ωσµ , (3.128)

we conclude that ∆ω must be antisymmetric. Since Lorentz transformations are representedby 4× 4 matrices, we immediately conclude that there will be six independent non-vanishingparameters ∆ωµν . Each of these group parameters or rotation angles in Minkowski spacegenerates an infinitesimal Lorentz transformation. Let us consider the cases of ∆ω10 6= 0 and

80

3.4. S(Λ) FOR PROPER LORENTZ TRANSFORMATIONS

∆ω12 6= 0.

Example 1 (Boost in the x direction): ∆ω10 = −∆ω01 = −∆β

In this case we have ∆ω 01 = g1σ∆ωσ0 = g11∆ω

10 = −∆ω10 = ∆β and ∆ω01 = −∆β. For

i = 1, 2, 3, ∆ω 1i = 0. We can now look at the Lorentz transformations x′ ν = Λν

µxµ '

(δνµ + ∆ωνµ)xµ: x′0 = x0 + ∆ω0

1x1 = x0 −∆β x1

x′1 = ∆ω10x

0 + x1 = x1 −∆β x0

x′2 = x2

x′3 = x3

, (3.129)

from which we see that observer B moves along the positive axis relative to A with velocity∆v = c∆β.

Example 2 (Rotation around the z axis): ∆ω12 = ∆ϕ

Similarly to what has been done above, we get ∆ω12 = −∆ϕ = −∆ω21, allowing us to write∆ω2

1 = −∆ϕ. Using again x′ ν = Λνµx

µ ' (δνµ + ∆ωνµ)xµ, we can writex′0 = x0

x′1 = ∆ω12x

2 + x1 = x1 + ∆ϕx2

x′2 = ∆ω21x

1 + x1 = x2 −∆ϕx1

x′3 = x3

. (3.130)

We can now compare this with the transformation which corresponds to an infinitesimalrotation around the z axis by an angle θ = ∆ϕ. Namely,(

x′1

x′2

)=

(cos θ sin θ− sin θ cos θ

)(x1

x2

)'(

1 ∆ϕ−∆ϕ 1

)(x1

x2

)=

(x1 + ∆ϕx2

x2 −∆ϕx1

).

(3.131)

The above examples allow us to determine the generators of infinitesimal Lorentz transfor-mations just as it has been done in QMII when you have studied symmetries in QM. So, wenow have to determine the ∆ω for arbitrary proper Lorentz transformations Λ (boosts androtations) and the corresponding spinorial transformations S(Λ).

We want to determine the operator S(Λ) ≡ S(ωµν). This is done expanding S in powersof omegaµν and keeping only the linear terms in the infinitesimal generators ω. We canparametrise S as

S(∆ωµν) ' 11− i

4σµν∆ω

µν , S−1(∆ωµν) ' 11 +i

4σµν∆ω

µν , σµν = −σνµ . (3.132)

The third relation is a consequence of the relation S−1(Λ) = S(Λ−1) given in Eq. (3.121).

Exercise 3.2. Prove that σµν = −σνµ, as shown in Eq. (3.132).

81


In order to determine S(Λ) we must find σµν . For that, we insert (3.127) and (3.132) in thecovariance relation (3.126), leading to

(δνµ + ∆ωνµ)γµ =

(11− i

4σµν∆ω

µν

)γν(

11 +i

4σµν∆ω

µν

). (3.133)

From this equation we can further write:

∆ωνµγµ = − i

4∆ωαβ(σαβγ

ν − γνσαβ) , (3.134)

and show that:

[σαβ, γν ] = −2i(gναγβ − gνβγα) . (3.135)

As σαβ must be anti-symmetric and depend on the γ matrices (as shown by the aboverelation), the most natural choice is:

σαβ =i

2[γα, γβ]⇒ S(∆ωµν) = 11 +

1

8[γµ, γν ]∆ω

µν . (3.136)

The above relation defines S(Λ) for any infinitesimal infinitesimal Λ depending on the ∆ω’s.

Exercise 3.3. Prove relations (3.134) and (3.135), and show that σαβ as defined in(3.136) obeys (3.135).

It is now time to think about finite Lorentz transformations which can be achieved bysuccessive infinitesimal ones. To proceed, we make the following definition

∆ωνµ = ∆ωIνµ , (3.137)

where ∆ω is an infinitesimal parameter, while Iνµ is a 4× 4 matrix which defines the type oftransformation. We now have to find the Iνµ for all types of transformations. Let us first seethe case of boosts.

3.4.1 Boosts along an arbitrary direction

Going back to the case ∆ω10 = −∆ω01 = −∆β (boost along the x direction) we have

x′ν = (δνµ + ∆ωνµ)xµ = (δνµ + δν1∆ω10 δ

0µ + δν0∆ω0

1 δ1µ)xµ (3.138)

= δνµ −∆β(δν1δ0µ + δν0δ

1µ)xmu , (3.139)

which, comparing with (3.139) gives ∆ω = −∆β and:

(Ix)νµ = −(δν1δ

0µ + δν0δ

1µ) = −

0 1 0 01 0 0 00 0 0 00 0 0 0

⇒ (Ix)2 = 114 , (Ix)

3 = Ix , (3.140)

82

3.4. S(Λ) FOR PROPER LORENTZ TRANSFORMATIONS

Let us now consider successive infinitesimal Lorentz transformations with ∆ω = ω/N . Namely

x′ν = limN→∞

[(11 +

ω

N

)N]νµ

xµ = [ exp(ωIx)]νµ x

µ =

coshω − sinhω 0 0− sinhω coshω 0 0

0 0 1 00 0 0 1

xµ .

(3.141)We can now relate ω with the relative velocity i the following way:

tanhω =v

c≡ β , coshω =

1√1− β2

. (3.142)

This implies

x′0 =x0 − βx1√

1− β2, x′1 =

x1 − βx0√1− β2

, x′2 = x2 , x′3 = x3 . (3.143)

The above procedure can be repeated for any proper Lorentz transformation. In total, thereexist six Iνµ for the three boosts and three rotations (these are the generalization of thegenerators for the spatial rotations that you have seen in QMII). For the boosts along the xand y directions we would obtain

Iy = −

0 0 1 00 0 0 01 0 0 00 0 0 0

, Iz = −

0 0 0 10 0 0 00 0 0 01 0 0 0

, (3.144)

which implies

Λy =

coshω 0 − sinhω 0

0 1 0 0− sinhω 0 coshω 0

0 0 0 1

, Λz =

coshω 0 0 − sinhω

0 1 0 00 0 1 0

− sinhω 0 0 coshω

, (3.145)

Exercise 3.4. Prove the last equality of (3.141) and relations (3.142)-(3.145).

The above results allow us to write the finite transformation for a boost along an arbitrary di-rection defined by the velocity vector ~v = v(cos θ1, cos θ2, cos θ3) for which it is straightforwardto show that

I~v = −

0 cos θ1 cos θ2 cos θ3

cos θ1 0 0 0cos θ2 0 0 0cos θ3 0 0 0

. (3.146)

To define S(Λ~v) we must compute σµνIµν~v . Taking into account that cos θk = vk/v, and

using Eqs. (3.136) and (3.146), one has:

σµνIµν~v = −2i

~α.~v

v, (3.147)

83


where ~α = (α1, α2, α3), being αk the matrices defined in (3.97). By performing successiveinfinitesimal S transformations we get:

S(Λ~v) = limN→∞

[11− i

4σµν∆ω

µν

]N= exp

[− i

4ωσµνI

µν~v

]= exp

[−ω

2

~α.~v

v

]. (3.148)

Expanding the exponential in power series and taking into account that (~α.~v)2n = 114 and(~α.~v)2n+1 = (~v)2n~α.~v, it can be shown that

S(Λ~v) =

√√√√1

2

(1√

1− β2+ 1

)114 −

~α.~v

v

√√√√1

2

(1√

1− β2− 1

). (3.149)

Exercise 3.5. Prove the results (3.147)-(3.149).

3.4.2 Spatial rotations around an arbitrary axis

Going back to case of an infinitesimal rotation around the z axis by an angle ∆ϕ, we haveseen in (3.130) and (3.131) that

(Λ3)νµ = δνµ + ∆ϕ(I3)νµ , I3 =

0 0 0 00 0 1 00 −1 0 00 0 0 0

. (3.150)

If we now generalize for a rotation around an arbitrary axis ~u = (u1, u2, u3) we get

(Λ~u)νµ = δνµ + ∆ϕ(I~u)

νµ , I~u =

0 0 0 00 0 u3 −u2

0 −u3 0 u1

0 u2 −u1 0

. (3.151)


Having defined the generators for an arbitrary rotation, we have to compute σµν∆ωµν inorder to be able to define S(Λ~u) as in (3.132). It can be shown that,

σµν∆ωµν = −2(σ12u3 + σ31u2 + σ23u1) = −2 ~u . ~Σ , ~Σ =

(~σ 00 ~σ

). (3.152)

Therefore, applying successive infinitesimal rotations with angle ∆ϕ = ϕ/N and taking thelimit N →∞, we arrive at:

S(Λ~u) = exp

(i

2ϕ~u . ~Σ

). (3.153)

84

3.5. S(Λ) FOR IMPROPER LORENTZ TRANSFORMATIONS: SPATIALREFLECTION AND TIME REVERSAL

By using the properties:

(~u.~Σ)2n = 114 , (~u.~Σ)2n+1 = ~u . ~Σ , (3.154)

one finally gets:

S(Λ~u) = cos(ϕ/2)114 + i sin(ϕ/2)~u . ~Σ . (3.155)

Notice that for a 2π rotation the original state is not reached since S(Λ~u)2π = −114, whichis characteristic of half-integer spins. This means that physical quantities must be bilinearin the spinors and, therefore, of even order in the Dirac fields ψ(x). Only in this case thosequantities are invariant under 2π rotations.

Exercise 3.7. Prove (3.152), (3.153) and (3.155).

3.5 S(Λ) for improper Lorentz transformations: spa-tial reflection and time reversal

We want now to define S(Λ) when Λ is a Lorentz transformation which cannot be achievedcontinuously. This is the case of spatial reflection (parity) and time reversal.

• Spatial reflection (parity) - P

Under spatial reflection t′ = t and ~r ′ = −~r. Therefore, the corresponding Lorentz transforma-tion is:

(ΛP )νµ =

(1 00 −113

). (3.156)

We have now to find spinor transformation S(ΛP ) which we will denote by P . According tothe covariance relation (3.126),

(ΛP )νµγµ = P γνP−1 , (3.157)

which implies[γ0, P ] = 0 , γk, P = 0 , (k = 1, 2, 3) . (3.158)

Taking into account the properties of the γ matrices, we conclude that the simplest choicefor the parity spinor operator P is

P = eiφγ0 → ψ′(x) = eiφγ0ψ(t, ~r) , (3.159)

where φ is an arbitrary phase, which is usually taken to be zero.

Exercise 3.8. Prove (3.158) using (3.156) and (3.157).

85


• Time reversal - T

For time reversal, the transformation is t′ = −t , ~r ′ = ~r to which corresponds the spinortransformation S(ΛT ) ≡ T such that:

ψ′(~r ′, t′) = ψ′(~r,−t) = S(ΛT )ψ(~r, t) ≡ Tψ(~r, t) . (3.160)

Notice that ψ′(~r,−t) describes a Dirac particle propagating backwards in time. This is possibleif ψ′(~r,−t) also satisfies the Dirac equation. In order to find an explicit form fot T , we startwith the Dirac equation and apply T on the left

T i~ T−1T∂

∂tψ(~r, t) = T H(~r, t) T−1T ψ(~r, t) , (3.161)

which, taking into account (3.160) and that t′ = −t, is equivalent to:

− T i~ T−1 ∂

∂t′ψ′(~r, t′) = T H(~r, t) T−1 ψ′(~r, t′) . (3.162)

In order for the Dirac equation to hold also for ψ′(~r, t′), one must have:

i~∂

∂t′ψ′(~r, t′) = H(~r, t)ψ′(~r, t′) , (3.163)

and, comparing with Eq. (3.162), this is verified if one of the following conditions is verified:

(i) T i~ T−1 = i~ and T H(~r, t) T−1 = −H(~r, t′)

(ii) T i~ T−1 = −i~ and T H(~r, t) T−1 = −H(~r, t′)

The latter possibility (ii) is the only valid one since T H(~r, t) T−1 = −H(~r, t′)would changethe spectrum of time-independent Hamiltonians (think for instance on a free particle whichwould have negative energy after time reversal). Let us consider the example of the free-particleHamiltonian given in Eq. (3.102). The covariance condition under time reversal (ii) implies:

T[−i~ cα.∇+ β mc2

]T−1 = −i~ cα.∇+ β mc2 , (3.164)

which is satisfied for:T α T−1 = −α , T β T−1 = β . (3.165)

Notice that, since T iT−1 = −i, the operator T must include the conjugation operator K, i.e.T = T0K. Thus, Eqs. (3.165) lead to

T0α∗ T−1

0 = −α⇒ T0 γk∗ T−1

0 = −γk (k = 1, 2, 3) , (3.166)

T0 β T−10 = β ⇒ T0 γ

0 T−10 = γ0 . (3.167)

Since in the Dirac basis only α2 is (purely) imaginary, we are looking for an operator thatobeys

T0 α1 T−1

0 = −α1 , T0 α2 T−1

0 = α1 , T0 α3 T−1

0 = −α3 , T0 β T−10 = β . (3.168)

Taking into account the anticommutation relation (3.97), and taking into account that T0

commutes (anticommmutes) with α2 and β (α1 and α3), it is straightforward to concludethat

T0 = −iα1α3 = iγ1γ3 ⇒ T = −iα1α3 = iγ1γ3K . (3.169)

86

3.6. THE SPIN OPERATOR ~S

3.6 The spin operator ~S

We now give a symmetry argument for the appearance of spin in the context of the Diracequation. Under Lorentz transformations, spinors transform as

ψ′(x′) = S(Λ)ψ(x)⇒ ψ′(x) = S(Λ)ψ(Λ−1x) , (3.170)

which, for an infinitesimal rotation around the arbitrary axis ~u with an angle ∆ϕ leads to

ψ′(x′) = S(Λ∆ϕ)ψ(x)⇒ ψ′(x) = S(Λ∆ϕ)ψ(Λ−1∆ϕx) = S(Λ∆ϕ)ψ(Λ−∆ϕx) , (3.171)

where in the last equality we have considered the fact that the inverse of a rotation with anangle ∆ϕ corresponds to a rotation with an angle −∆ϕ. According with what we have foundin Eq. (3.153), we can write

S(Λ∆ϕ) ' 11 +i

2∆ϕ~u . ~Σ . (3.172)

On the other hand, from (3.151),

(Λ−∆ϕ)x = x−∆ϕ

0 0 0 00 0 u3 −u2

0 −u3 0 u1

0 u2 −u1 0

x =

x0

x1 + ∆ϕ(−u3x2 + u2x

3)x2 + ∆ϕ(u3x

1 − u1x3)

x3 + ∆ϕ(−u2x1 + u1x

2)

.

(3.173)We have now to find ψ(Λ−∆ϕx) to complete Eq. (3.171). For that, we Taylor expandψ(Λ−∆ϕx) around ∆ϕ = 0, leading to:

ψ(Λ−∆ϕx) '(

11 +i

~∆ϕ~u.~L

)ψ(x) , (3.174)

where ~L is the orbital angular momentum. Together with (3.172), we now have for Eq. (3.171)

ψ′(x) '[11 +

i

~∆ϕ~u.

(~L+

~2~Σ

)]ψ(x) . (3.175)


Therefore, we conclude that for a general rotation with an angle ϕ,

ψ′(x) = exp

(i

~ϕ~u. ~J

)ψ(x) , ~J = ~L+ ~S , ~S =

~2~Σ . (3.176)

From this result we see that an intrinsic angular momentum appears, which we identify withthe spin.

87


3.7 The free-particle solutions of the Dirac equation

Lets us start with the Dirac equation in form

i~∂ψ(x)

∂t=[c (α.p) + β mc2

]ψ(x) . (3.177)

Given that the Hamiltonian is time independent, we can write

ψ(~r, t) = Ψ(~r) e−iEt/~ , (3.178)

which after inserting in the Dirac equation leads to

EΨ(~r) =[c (α.p) + β mc2

]Ψ(~r) . (3.179)

Using (3.99) and defining the spinors Ψ(~r) as Ψ(~r) = [ϕ(~r), χ(~r)]T , where ϕ and χ aretwo-component spinors, one has

E

(ϕχ

)= c

(0 ~σ~σ 0

).p

(ϕχ

)+mc2

(11 00 −11

)(ϕχ

), (3.180)

implying E ϕ = c ~σ.pχ+mc2 ϕE χ = c ~σ.pϕ−mc2χ .

. (3.181)

If we now write(ϕχ

)=

(ϕ0

χ0

)ei~p.~r ⇒

(E −mc2)ϕ0 − c ~σ.~p χ0 = 0(E +mc2)χ0 − c ~σ.~p ϕ0 = 0

. (3.182)

It can be shown that this system has solutions if

E2 −m2c4 − c2p2 = 0⇒ E±p = ±c√p2 +m2c4 ≡ ±Ep . (3.183)

Exercise 3.10. Starting from (3.182) and using the property (~σ. ~A)(~σ. ~B) = i~σ.( ~A ×~B) + ~A. ~B prove (3.183).

If we define ϕ0 as

ϕ0

(u1

u2

)≡ u , u†u = 1⇒ u∗1u1 + u∗2u2 = 1 , (3.184)

we get from (3.182)

χ0 =c (~σ.~p)

mc2 + E±pu , (3.185)

and, the free-particle solutions of the Dirac equation can be written as

ψ~pλ(~r, t) =N

(2π~)3/2

uc (~σ.~p)

mc2 + λEpu

exp

[i(~p.~r − λEpt)

~

], (3.186)

88

3.8. HELICITY AND SPIN

where λ = 1 (λ = −1) correspond to the positive (negative) energy solutions with E = Ep(E = −Ep). The normalization constant N is such that the solutions are normalised a laDirac, i.e. ∫

ψ†~pλ(~r, t)ψ~p′λ′(~r, t) d~r = δλλ′δ(~p− ~p′) . (3.187)

Replacing in this relation the solutions (3.186), we get for the normalisation constant N

N =

√mc2 + λEp

2λEp, (3.188)

leading to

ψ~pλ(~r, t) =1

(2π~)3/2

√mc2 + λEp

2λEp

uc (~σ.~p)

mc2 + λEpu

exp

[i(~p.~r − λEpt)

~

]. (3.189)

Exercise 3.11. Prove (3.188) from (3.186) and (3.187).

It is straightforward to see that the spectrum of a free Dirac particle propagating withmomentum ~p obeys |E| ≥ mc2. Moreover, the solutions ψ~pλ(~r, t) are eigenfunctions of themomentum operator p, i.e.

pψ~pλ(~r, t) = ~pψ~pλ(~r, t) . (3.190)

For each ~p there are two solutions corresponding to λ = ±1 with energy E = ±Ep.

3.8 Helicity and spin

In general, and since [HD, ~S] 6= 0, the solutions of the Dirac equation found in the previoussection are not eigenfunctions of spin. Therefore, it is convenient to find an operator whichcommutes with the Dirac Hamiltonian so that its eigenstates are also eigenstates of HD. Thiswill allow us to determine the spinors u which appear in the solutions (3.189). For this, let usnote that the operator Σ.p commutes with HD, i.e.

Σ.p =

(~σ 00 ~σ

).p ,⇒ [HD, ~Σ] = [p, ~Σ] = 0 . (3.191)

This means that HD, Σ.p and HD can be diagonalised simultaneously.

Exercise 3.12. Prove the relations [HD, ~Σ] = [p, ~Σ] = 0.

Defining the helicity operator h as

h =~2

~Σ.p

|~p|=~S.p

|~p|, (3.192)

89


we have [HD, h] = 0. Since ~S = ~~Σ/2 is the generalisation of the spin vetor operator infour dimensions, the helicity is nothing more than the spin projection onto the momentumdirection. Consider, for instance, a Dirac particle moving in the z direction with momentum~p = (0, 0, p). In this case

h =~S.p

|~p|= ~Sz =

~2

1 0 0 00 −1 0 00 0 1 00 0 0 −1

, (3.193)

which has eigenvalues ±~/2 and eigenvectors(u+

0

),

(u−0

),

(0u+

),

(0u−

), u+ =

(10

), u− =

(01

). (3.194)

For this particle, the solutions (3.189) are

ψ↑p,λ(~r, t) =1

(2π~)3/2

√mc2 + λEp

2λEp

(

10

)pc

mc2 + λEp

(10

) exp

[i(pz − λEpt)

~

],

ψ↓p,λ(~r, t) =1

(2π~)3/2

√mc2 + λEp

2λEp

(

01

)−pc

mc2 + λEp

(01

) exp

[i(pz − λEpt)

~

],

(3.195)

where ↑ (↓) correspond to the solutions with sz = ~/2 (sz = −~/2).

3.9 The E<0 solutions of the Dirac equation: the"hole theory"

The interpretation of the negative energy solutions of the Dirac equation as being single-particle states of an electron leads to some problems. If we consider electrons with energy−mc2 < E < mc2 in an atom, then we it is hard to avoid the fact that theses electrons couldundergo a transition from the 1s state to states with E < −mc2. This could actually occurcontinuously and the electron could constantly radiate towards a −∞ energy (see Fig. 3.3.A),i.e. an electron would radiate an infinite amount of energy... Dirac proposed a solution forthis problem which, although not satisfactory from the fundamental point of view, it wasperhaps the first time that the vacuum was considered to be more than just empty space.This is called the Dirac hole theory.

The main idea behind the Dirac hole theory is that, somehow, one has to avoid thatan electron falls off into the states with E < 0. With this purpose, in 1930 Dirac proposedthat all negative energy states are occupied by electrons. The vacuum is, therefore, a statewithout positive-energy electrons (see Fig. 3.3.B). The existence of this continuum filled withelectrons (the Dirac sea) solves radiation catastrophe problem mentioned in the previousparagraph since, due to the exclusion principle and to the fact that the Dirac sea is fully

90

3.9. THE E<0 SOLUTIONS OF THE DIRAC EQUATION: THE "HOLETHEORY"

Figure 3.3: A: Electrons in bound states can radiate energy continuously by occupying lowerand lower energy states in the negative-energy continuum. B: In the Dirac sea all negativeenergy states in the continuum are occupied with negative-energy electrons. In this way, thepositive energy electrons cannot fall off and occupy states in the Dirac sea due to the Pauliexclusion principle. C+D: Pair creation. An electron in the Dirac sea can absorb a photon(C) and occupy a state in the positive energy continuum. This leaves a hole in the Dirac sea(D) which is interpreted as a positron (absence of a negative charge=presence of a positivecharge).

occupied, no electron can occupy a state there. On the other hand, an electron in the Diracsea (E < 0) can absorb radiation (a photon). If the energy of the photon Eγ is such thatEγ = ~ω > mc2, an electron of negative energy can be excited to a state of positive energy(see Fig. 3.3.C). In the end we will have a real electron and a hole in the Dirac sea, whichbehaves like a particle of positive charge e+ (absence of a negative charge). Thus, the hole isto be interpreted as the antiparticle of the electron, the positron. The creation of an electronand a positron by photons is called pair creation, being the threshold ~ω = 2mc2. When theelectron drops in the hole, we have pair annihilation. The energy balance for pair creation is:

~ω = Ee−,E>0 − Ee−,E<0 = c√p2 +mc2 − (−

√p′2 +mc2 ) ≡ Ee− + Ee+ . (3.196)

According to this we must give the positron a positive energy Ee+ =√p′2 +mc2 . As for the

momentum,

~k + p′e−,E<0 = pe−,E>0 ⇒ ~k = p′e−,E>0 − pe−,E<0 . (3.197)

If we want this to be equivalent to ~k = p′e− + pe+ , we must set pe+ = −p′e−,E<0, meaningthat the positron has opposite momentum to the e− and E > 0. Therefore, a missing electronwith E < 0 behaves like a positron with the same mass and opposite momentum.

The most important feature of the hole theory is that it provides a naive model for thevacuum as a free particle state. Also notice that it is many-body theory describing positivelyand negatively charged particles. In fact, an infinite number of electrons is needed to describethe Dirac sea, and the single-particle wavefunction interpretation is not adequate to accountfor pair creation and annihilation. Although the hole theory provides a simple interpretationfor negative-energy solutions, we cannot apply a similar idea to bosons since these do notobey the Pauli exclusion principle.

91


3.10 Charge conjugation

The hole theory has introduced the concept of antiparticles or positrons (for eletrons).Therefore, there is a new fundamental symmetry which states that for each particle there isan antiparticle: charge cinjugation (C). We now have to formulate this in a mathematical way.From what we have seen, there exists a one-to-one correspondence between E < 0 solutionsfor the electron e− and the E > 0 solutions for the positron e+.

In order to study the charge conjugation transformation C, let us start by writing theequation of a Dirac particle with charge q subject to an electromagnetic field Aµ. We consideronce more the minimal coupling (3.48) and perform the replacement in Eq. (3.102), i.e.

Electron with E < 0: (i~∂/− q/cA/−mc)ψ = 0 , (3.198)Positron with E > 0: (i~∂/+ q/cA/−mc)ψc = 0 . (3.199)

Our goal is now to find the transformation which relates ψ and ψc.The correponding operatormust change the relative sign between the terms i~∂/ and q/cA/. This is done by applyingconjugation. Taking into account that(

i~∂

∂xµ

)∗= −i~ ∂

∂xµ, A∗µ = Aµ . (3.200)

If we take the complex conjugate of Eq. (3.198) we have

[ (i~∂µ + q/cAµ) γµ∗ +mc]ψ∗ = 0 . (3.201)

We want to find the transformation U satisfying Eq. (3.199) such that ψc = Uψ∗. ApplyingU on the left of Eq. (3.201):

[ (i~∂µ + q/cAµ)Uγµ∗ +mcU ]ψ∗ = 0⇒ (3.202)[(i~∂µ + q/cAµ)Uγµ∗U−1 +mc

]Uψ∗︸︷︷︸ψc

= 0 , (3.203)

and comparing with Eq. (3.199) gives:

Uγµ∗U−1 = −γµ . (3.204)

Defining:

ψc = Uψ∗ ≡ Cγ0ψ∗ = C(ψ)T , ψ = ψ†γ0 , (3.205)

then Eq. (3.204) implies,

Cγ0γµ∗(Cγ0)−1 = Cγ0γµ∗γ0C−1 = −γµ . (3.206)

Taking into account the explicit representation of the γ matrices given in (3.110), whichimplies γ0γµ∗γ0 = −γµ, the above relation turns out to be equivalent to

C−1γµC = −γµT . (3.207)

It is easy to see that C commutes with γ1 and γ3, and anticommutes with γ0 and γ2. Asimple form for C is, thus,

C = iγ2γ0 = −C−1 = −C† = −CT . (3.208)

92

3.11. NON-RELATIVISTIC LIMIT OF THE DIRAC EQUATION

Exercise 3.13. Prove the properties of C given in Eq. (3.209).

The charged-conjugated state is

ψc = CψT = Cγ0ψ∗ = iγ2ψ∗ , (3.209)

where C is the charge conjugation matrix defined by (3.209) in a particular representation.

3.11 Non-relativistic limit of the Dirac equation

Let us recall the form of the Dirac equation for a particle of mass m and charge q under theaction of a field Aµ

i~∂ψ(x)

∂t= Hψ(x) , H = cα.

(p− q

cA)

︸︷︷︸π

+qA0 + β mc2 . (3.210)

Defining

ψ =

(ϕχ

), (3.211)

and taking into account the explicit form of the α and β matrices, Eq. (3.210) can be writtenas

i~∂

∂t

(ϕχ

)=

(c ~σ.πχc ~σ.πϕ

)+ qA0

(ϕχ

)+mc2

(ϕ−χ

). (3.212)

We now write:

ψ =

(ϕχ

)=

(ϕχ

)exp

(−i mc

2

~t

), (3.213)

where ϕ and χ are the nonrelativistic solutions (remember what we did when we looked atthe nonrelativistic limit of the Klein-Gordon equation). Using this in (3.213), we get

i~∂

∂t

(ϕχ

)=

(c ~σ.πχc~σ.πϕ

)+ eA0

(ϕχ

)− 2mc2

(0χ

). (3.214)

We now consider the (nonrelativistic) limit∣∣∣∣i~∂χ∂t∣∣∣∣ , |eA0χ| |mc2χ| , (3.215)

which, for the lower components of (3.213) implies

χ =~σ.π

2mcϕ ∼ v

2cϕ . (3.216)

It is common to call χ (ϕ) the small (large) components of ψ. Inserting χ as given aboveinto (3.214) we have for ϕ

i~∂ϕ

∂t=

(~σ.π)(~σ.π)

2mϕ+ eA0 ϕ . (3.217)

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Using the relation

(~σ.π)(~σ.π) =(p− e

cA)2

− e~c~σ. ~B , (3.218)

where ~B is the magnetic field. Using this in (3.217) we arrive at

i~∂ϕ

∂t=

[1

2m

(p− e

cA)2

− e~2mc

~σ. ~B + eA0

]ϕ , (3.219)

which is the Pauli equation. For weak magnetic fields we can further write

i~∂ϕ

∂t=

p2

2m− e

2mc~L− 2︸︷︷︸

g=2

e

2mc~S. ~B + eA0

ϕ . (3.220)

This shows that, the nonrelativistic limit of the Dirac equation ci+ontains an interaction termbetween the spin magnetic moment of the electron and the magnetic field with the rightg factor g = 2. Therefore, the Dirac equation predicts g = 2 for the electron, as observedexperimentally.

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3.12. PROBLEMS

3.12 Problems

Problem 3.1. Show that the Maxwell equations

∂µFµν =

4π

cjν ,

where F µν = ∂µAν − ∂νAµ is the electromagnetic tensor and jν is the current 4-vector, areinvariant under Lorentz transformations.

Problem 3.2. Consider a scalar particle under the effect of the potential:

A0 =

0 , z < 0U0 , z ≥ 0

.

Compute the transmission coefficient when the energy of the particle is E > V0 +mc2 withV0 = qU0.

Answer: T = 4r(1+r)2

with r =√

(E−V0)2−m2c4

E2−m2c4.

Problem 3.3. Compute the energy spectrum of a relativistic spin-zero particle with mass mand charge q, which is under the action of a uniform magnetic field:

~B = B0 ~ez .

Answer: En = ~c√k2z + m2c2

~2 + (2n+ 1) qB0

~c .

Problem 3.4. Prove the following properties of the γ matrices:

(a) a/a/ = a2

(b) Tr(a/b/) = 4 ab(c) γµa/γµ = −2a/(d) γµγνγµ = −2γν .

Problem 3.5. Show that under Lorentz transformations

(a) ψγ5ψ transforms as a pseudoscalar.(b) ψγµγ5ψ transforms as a pseudo-vector.(c) ψσµνψ transforms as a second-rank tensor.

Problem 3.6. Introduce the spinors:

ψL =1

2(1− γ5)ψ , ψR =

1

2(1 + γ5)ψ .

and derive the system of Dirac equations for ψL,R. Discuss in which conditions those equationsdecouple.

Problem 3.7. Show that:[~σ.(~p− q ~A

) ]2

=(~p− q ~A

)2

− q~σ. ~B .

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Problem 3.8. An electron with Sz = 1/2 and momentum ~p = p~ez meets a potential barrier:

− eA0 =

0 , z < 0V0 , z ≥ a

.

Determine the reflection and transmission coefficients of this barrier for the several energyregimes of the incoming electron.

Problem 3.9. Consider the Dirac Hamiltonian HD = ~α.~p+mβ for a free spin 1/2 relativisticparticle with mass m. Compute the commutators [HD, ~L] and [HD, ~S], where ~L and ~S are theorbital angular momentum and spin operators, respectively. What do you conclude regardingthe commutator [HD, ~L+ ~S]? Comment on the relevance of this result.

Notes: The spin operator is ~S = −i~α× ~α/4 with αi, αj = 2δij, (αi)2 = β2 = 14×4. For

the orbital angular momentum components one has [La, Lb] = iεabcLc, [La, pb] = iεabcpc.Also remember that (~a×~b)i = εijkajbk.

Problem 3.10. Compute the time derivative of the position operator r and of the kineticmomentum Π = p−qA/c for Dirac particles in an electromagnetic field. Compare the resultswith the corresponding classical expressions.

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Bibliography

[1] L. Ballentine, Quantum Mechanics: A Modern Development, World Scientific Pub-lishing Co., 1st ed., 1998.

[2] A. Fetter and J. Walecka, Quantum Theory of Many-particle systems, (McGraw-Hill, 2003).

[3] W. Greiner, Relativistic Quantum Mechanics: wave equations, (Springer verlag, 2003).

[4] F. Gross, Relativistic Quantum Mechanics and Field Theory, (John Wiley and Sons,1993).

[5] W. Nolting, Fundamentals of Many-body Physics: Principles and Methods, (Springer-Verlag, 2009).

[6] T. Ohlsson, Relativistic Quantum Physics: From Advanced Quantum Mechanics toIntroductory Quantum Field Theory, (Cambridge University Press, 2011).

[7] S. Rao, An Anyon primer, (1992).

[8] J. C. Romão, Introdução á Teoria do Campo, (IST, 2018).

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ADVANCED QUANTUM MECHANICS - ULisboaADVANCED QUANTUM MECHANICS Filipe R. Joaquim Complements of Quantum Mechanics MEFT – 4th year, 1st Sem. 2018/2019