Electromagnetic radiation - Wiley€¦ · 6 ELECTROMAGNETIC RADIATION Table 1.1 (continued) Year Discoverer Discovery 1893 Friedrich Pockels Linear electro-optic (Pockels) effect

1 Electromagnetic radiation

James Clerk Maxwell Carl Friedrich Gauss

Andre Marie Ampere Charles Augustin de Coulomb Michael Faraday

Light and its interaction with matter play crucialroles in the dynamics of the universe. Our interestand fascination with them dates back to antiquity.The opening sentences of the Bible attest tothis fact:

In the beginning God created the heaven and theearth. . . . And God said, ‘Let there be light’, andthere was light. And God saw the light, that it wasgood, and he divided the light from the darkness.(Genesis, 1:1–3)

Light and Matter Yehuda Band 2006 John Wiley & Sons, Ltd

COPYRIG

HTED M

ATERIAL

2 ELECTROMAGNETIC RADIATION

In the context of modern scientific thought, thesefirst sentences of the Bible can perhaps be inter-preted as the creation of vacuum (heaven), matter(earth) and light.

Light and matter interact by virtue of matterbeing composed of charged particles. Electromag-netic radiation accelerates charged particles, andaccelerating charges are a source of electromag-netic radiation. The laws of electromagnetism (see

Appendix B for an enumeration of the seven laws)are embodied in Maxwell’s equations, named inhonor of James Clerk Maxwell, who first formu-lated these equations in 1860, the constitutive rela-tions and the Lorentz force law (see Appendix B).These provide the full classical description oflight and its interaction with matter. Maxwell’sequations can be written as:

∇ · D = 4πρ {∇ · D = ρ} (Coulomb’s law) (1.1)

∇ · B = 0 {∇ · B = 0} (Gauss’ law–magnetic) (1.2)

∇ × E = −1

c∂tB {∇ × E = −∂tB} (Faraday’s law) (1.3)

∇ × H =(

4π

cJ + 1

c∂tD)

{∇ × H = (J + ∂tD)} (Ampere’s law) (1.4)

Here D is the displacement field, E is the electricfield, B is the magnetic induction, H is themagnetic field, ρ is the charge density and J isthe current density. The constitutive relations arephenomenological relations that can be derived

by making assumptions about the nature of theinteraction of light and matter (see Chapters 3and 4). In most cases these relations are a goodapproximation of reality. The constitutive relationstake the form:

D(r, t) = ε E(r, t) = E(r, t) + 4πP(r, t) {D = ε E = ε0 E + P} (1.5)

H(r, t) = µ−1 B(r, t) = B(r, t) − 4πM(r, t) {H = µ−1B = µ0−1B − M} (1.6)

J(r, t) = σE(r, t) {J = σE} (Ohm’s law) (1.7)

Here ε and µ are constants known as the mediumpermittivity and permeability, respectively, σ is theconductivity, P is the polarization field and M isthe polarization field. If the medium is a vacuum,

ε = 1, µ = 1, σ = 0{ε = ε0 = 107

4πc2(C2/N/m2)

= 8.854187 × 10−12 (C2/N/m2),

µ = µ0 = 4π × 10−7(m kg/C2), σ = 0

}

and P and M vanish. The (non-relativistic) Lorentzforce law for a particle with charge q located atposition r at time t is given by:

F(r, t) = q

[E(r, t) + v(r)

c× B(r, t)

]{F = q[E + v(r) × B]} (Lorentz force law) (1.8)

LIGHT IN VACUUM 3

The reader need not have prior experience nordeep familiarity with the manipulation of theseequations to use this book. We shall proceedin a step-by-step manner without assuming suchfamiliarity. Furthermore, readers who feel theyare not sufficiently familiar with the conceptsand basics of electromagnetism can review thismaterial in Appendix B. Moreover, Appendix Acontains a tutorial on vector analysis for readerswho feel they might need to brush up on vectors,and divergence and curl notation. Appendix Ccontains a review of quantum mechanics andAppendix D a review of perturbation theory forreaders without a background in quantum theory,since some elements of quantum theory andperturbation theory will be used in certain sectionsof this book (primarily Chapters 5, 6 and 9).Maxwell’s equations, the Lorentz force law andthe constitutive relations are discussed at lengthduring the course of our study.

The parts of Eqs (1.1)–(1.8) in curly bracketsare in SI units (System International, i.e. MKSAunits), while those not in curly brackets arein Gaussian units. These systems of units arediscussed in Appendix B. We shall use bothsystems of units in most of the first chapter.In the other chapters, Gaussian units are usedpreferentially (but SI units are often also provided).

Before beginning our study of the interactionand the propagation of electromagnetic waves, inthis introductory chapter we briefly review thehistory of the discoveries comprising the bodyof our knowledge of the interaction of light andmatter. We then examine the character of lightpropagating in vacuum, and consider the natureof matter as a source of light.

1.1 Brief history of the interactionof light and matter

It follows from the fact that matter interacts withlight, that optical instruments can be developedto affect the propagation of light, hence the earlyinterest in the interaction of light and matter. Thescientific study of the laws of interaction of lightand matter were initiated even before the riseto power of the Greek empire. Euclid describes

the law of reflection in his book Catapropics(300 BCE). Positive lenses were developed asearly as 424 BCE (mentioned by Aristophanesin his play The Clouds). Magnifying glasseswere used before the start of the first century.Modern optical instruments, such as the telescope,the microscope and spectacles, were developedalmost immediately following improvements inglass-making at the beginning of the seventeenthcentury. After hearing about the invention of therefracting telescope, Galileo Galilei (1564–1642)built his own telescope, grinding the lenses byhand. He was thereby able to discover the moons ofJupiter and Saturn’s rings in 1610. The compoundmicroscope was invented about the same time.In 1784 Benjamin Franklin invented bifocals, andin 1827 Sir George Biddell Airy invented thecylindrical lens to correct astigmatism.

Soon thereafter, the field of optics began todevelop rapidly, with important discoveries occur-ring at a fast rate. Table 1.1 presents the high-lights of these discoveries from the 1600s tothe beginning of the twentieth century. Progressin optics continued to advance at a breakneckpace in the twentieth century. Some of the mostimportant developments are listed in Table 1.2.Additional important twentieth-century discoveriesand achievements in optics and related fieldsabound; Tables 1.1 and 1.2 are by no means com-plete. Moreover, not all those involved in theachievements have been noted due of lack of space.Nevertheless, the tables testify to the wealth andbreath of the discoveries in these areas of sciencerelated to the interaction of light and matter.

1.2 Light in vacuum

Electromagnetic radiation propagates in vacuum,as well as in matter. In this sense it is of coursedifferent from sound waves, which require a mate-rial medium in which to propagate. The propaga-tion of light in vacuum is easier to describe thanthe propagation of light in matter. In matter, lightis continuously absorbed and re-emitted, therebynecessitating the incorporation of the interactionof the light and matter in order to describe thepropagation. This interaction can be quite compli-cated to describe. It can be tensorial and nonlinear


Table 1.1 Highlights of important discoveries in optics and the interaction of light and matter through thenineteenth century

Year Discoverer Discovery

1611 Johannes Kepler Total internal reflection

1611 Johannes Kepler First-order optics for thin lens (positive and negative) telescopes

1621 Willebrord Snell Law of refraction of light at dielectric interfaces (Snell’s law)

1637 Rene Descartes Formulation of the law of refraction in terms of sines; theory ofrainbows

1657 Pierre de Fermat Formulation of principle of least time; re-derivation of law ofrefraction from this principle

∼1660 Francesco Maria Grimaldi Diffraction of light within the shadow of a rod illuminated by asmall source

1665 Robert Hooke Wave theory of light–observation and explanation of coloredinterference patterns generated by thin films

∼1670 Isaac Newton Dispersion; corpuscular theory of light; chromatic aberration

∼1670 Christian Huygens Double refraction (in calcite); polarization of light; wave theory oflight (Huygens’ principle)

1676 Dane Olaf Romer First measurement of the speed of light (2.14 × 108 m/s) byobserving successive eclipses of Jupiter’s moons

∼1730 Leonhard Euler Wave theory used to predict the construction of an achromatic lens

1800 William Herschel Infrared radiation described as long-wavelength electromagneticwaves

1801 Thomas Young Principle of interference used to explain the colored fringes of thinfilms

∼1801 Augustin Jean Fresnel Wave front propagation, description of partial refraction andreflection from interface

1808 Etienne Louis Malus Polarization of light upon reflection

1811 Dominique F.J. Arago,Jean Baptiste Biot

Optical activity discovered (first in quartz then in some liquids)

∼1812 Augustin Jean Fresnel Transverse polarization character of light postulated

∼1812 Josef Fraunhofer Diffraction gratings; fraunhofer lines in solar spectrum

LIGHT IN VACUUM 5

Table 1.1 (continued )


1820 Hans Christian Oersted Magnetic needle aligns perpendicular to a current-carrying wire(connection between electricity and magnetism)

∼1821 Andre Marie Ampere Electric current source of magnetic field (Ampere’s law)

1827 Georg Simon Ohm Current proportional to voltage (Ohm’s law)

∼1830 Carl Friedrich Gauss Potential theory (Gauss’s law)

1831 Michael Faraday Law of electromagnetic induction (Faraday’s law)

1842 Christian Doppler Doppler effect (observed frequency shift of a moving light source)

1845 Michael Faraday Faraday effect (rotation of polarization plane of light in the presenceof a magnetic field along the direction of propagation in matter)

1849 Armand Hippolyte Louis Fizeau First terrestrial measurement of the speed of light (3.15 × 108 m/s)

1850 Jean Bernard Leon Foucault Measurement of the speed of light in water (less than that in air)

∼1860 James Clerk Maxwell Maxwell’s equations formulated; unified optics andelectrodynamics

1864 August Toepler Schlieren method (for observing variations in index of refraction)

1870 John William Strutt (LordRayleigh)

Rayleigh scattering of light, energy density emitted by blackbody(Rayleigh–Jeans formula)

1875 John Kerr Quadratic electro-optic effect (Kerr effect) discovered, wherebyrefractive index changes are proportional to the square of electricfield

1879 Thomas Alva Edison Invention of the first electrical light bulb

1880 Pierre and Jacques Curie Piezoelectric effect (electrical polarization of a material uponapplication of mechanical stress)

1880 Alexander Graham Bell Acousto-optic detection of light

1882 Gustav Kirchhoff Diffraction theory formulated on a sound mathematical basis

1886 Albert A. Michelson,Edwin W. Morley

Experimentally shown that the speed of light is constant even whenmeasured in a moving reference frame

1888 Heinrich Rudolf Hertz Discovery, generation and detection of long-wavelength electro-magnetic waves (radiowaves); photo-emission (photoelectric effect)

∼1890 Ernst Abbe Diffraction theory of optical images

(continued overleaf )




1893 Friedrich Pockels Linear electro-optic (Pockels) effect discovered, whereby refractiveindex changes are proportional to electric field in some crystals

1895 Nikola Tesla,Guglielmo Marconi

Radio-frequency communication invented

1895 Pierre Curie Temperature dependence of paramagnetism discovered

1895 Wilhelm Konrad Rontgen Discovery of X-rays

1896 Pieter Zeeman Observation of magnetic field splitting of spectra

1896 Henrik A. Lorentz Explanation of normal Zeeman effect (splitting of spectral lines ina magnetic field)

in character, and details of the structure of thematter within which the light propagates are oftennecessary to properly account for the interaction.Consideration of light propagation in vacuum isa good starting point from which to embark onthe study of the propagation and interaction oflight and matter. The general principles of prop-agation in vacuum are as follows: (a) light travelsat a constant speed, the speed of light in vacuum,c = 2.99792458 × 1010 cm/s; (b) light has polar-ization properties and therefore must be describedin terms of vector fields; (c) light behaves asa wave (diffraction and interference phenomenacan occur) and therefore the vector fields obey awave equation; (d) the principle of superpositionapplies, i.e. if two optical waves [electromagneticfields {E1(r, t), B1(r, t)} and {E2(r, t), B2(r, t)}]are present simultaneously in vacuum, the result-ing optical wave [resultant electromagnetic field{E(r, t), B(r, t)}] is the coherent superpositionof the two electromagnetic fields {E(r, t) =E1(r, t) + E2(r, t), B(r, t) = B1(r, t) + B2(r, t)};hence the wave equation is linear. These principlesare embodied in Maxwell’s equations as applied tovacuum, as we shall see below.

1.2.1 The electromagnetic spectrum

Electromagnetic radiation at a desired frequency isproduced by driving charged particles to oscillate

at this frequency. The oscillating charged particlesgenerate the desired electromagnetic waves byradiating at this frequency. The frequency ofelectromagnetic radiation, ν, is often expressedin units of cycles per second (1 Hz = 1 cycleper second), or in units of radians per sec-ond or the angular frequency, ω (there are 2π

radians per cycle, so 1 Hz is equivalent to 2π

rad/s, i.e. ω = 2πν). However, other units of fre-quency are sometimes used. Two commonly usedunits for the frequency of electromagnetic radi-ation are inverse centimeters (cm−1) and elec-tron volts (eV). The former is an appropriateunit for frequency because of the relationshipof frequency and the wavelength of light, ν =cλ−1 (λ−1 can have the units cm−1 and, sinceν is proportional to λ−1, this accounts for theuse of cm−1 as a unit of frequency). The lat-ter unit derives from the relationship of the fre-quency and energy of a ‘photon’ (the quan-tum of light energy), ν = E/h, where h is thePlanck constant (h = 6.62606876 × 10−27 erg s =6.62606876 × 10−34 J s). Clearly, the angular fre-quency ω is given by ω = E/h, where h isPlanck’s constant divided by 2π . Since the energyE can be measured in units of electron volts (eV),ν can be expressed in terms of eV.

The conversion of frequency (energy) from oneof these units to another is given as follows:

LIGHT IN VACUUM 7

Table 1.2 Important twentieth-century discoveries in optics and the interaction of light and matter


1900 Max K.E.L. Planck Blackbody radiation law; quanta of energy in absorption–emission

1900 Jules Henri Poincare Postulate of relativity

1905 Albert Einstein Special theory of relativity; photoelectric effect explained

1905 Paul Langevin Statistical explanation of the temperature variation ofparamagnetism

1907 Pierre-Ernest Weiss Internal field theory of the Curie–Weiss law for ferromagnetism

∼1910 Max von Laue,William L. Bragg,William H. Bragg

Diffraction of X-rays from crystals (and diffraction of light fromother periodic structures, e.g. those created by sound); X-rayspectrometer developed

1913 Niels Henrik DavidBohr, ArnoldSommerfeld

Bohr-Sommerfeld semi-classical theory of the quantization ofenergy levels

1915 Albert Einstein General theory of relativity–effect of gravity on light

1921 Joseph Valasek Ferroelectric effect (spontaneous polarization of a material)

1921 Leon Brillouin Prediction of Brillouin scattering in liquids (scattering of light bysound waves)

1921–1922 Wolfgang Pauli, OttoStern, WaltherGerlach

Suggestion and demonstration of atomic magnetic moments thatgive rise to magnetic phenomena

1923 Arthur Holly Compton Compton effect (scattering of photons by free electrons)

1924 Louis-Victor Debroglie Wave particle duality

19251926

Werner Heisenberg,Erwin Schrodinger

The matrix form of quantum mechanics developed; wave natureof matter described via the Schrodinger equation

1926 Max Born Statistical interpretation of wave mechanics

1927 Werner Heisenberg The uncertainty principal formulated

1927 Paul Adrien MauriceDirac

Quantum theory of radiation

1928 ChandrasekharaVenkata Raman

Raman scattering–inelastic light scattering observed in liquids





1931 Maria Goppert-Mayer Two-photon absorption cross section calculation

1932 Peter Debye,Francis W. Sears

Observation of Brillouin scattering

1934 Fritz Zernike Phase contrast microscopy

1934 Pavel A. Cherenkov,Ilya M. Frank,Igor E. Tamm

Discovery and the interpretation of the Cherenkov effect

1946 Felix Bloch,Edward M. Purcell,Y.K. Zavoysky

Magnetic resonance phenomena

1947 Julian S. Schwinger,Richard P. Feynman,Sin-itiro Tomonaga

Quantum electrodynamics

1947 Dennis Gabor Invention of holography

1947 Willis E. Lamb Fine structure in the spectrum of the hydrogen atom explained

1949 Norman F. Ramsey Optical double resonance; development of the hydrogen maser

∼1950 Hannes Alfven Magneto-hydrodynamics and electrodynamics of plasmas

1952 Alfred Kastler Double resonance and optical pumping

1955 Abraham C.S. vanHeel,N.S. Kapany

Fiber optics (cladding layers improve transmission characteristics)

1955 J. Weber,G.P. Gordon,J.K. Zeiger,C.H. Townes

Microwave amplification by stimulated emission in NH3 maser

1956 R. Hanbury Brown,R.Q. Twiss

Photon correlations and anti-correlations

1957 Rudolf L. Mossbauer Resonance absorption of γ -rays in condensed matter

1958 Arthur L. Schawlow,Charles H. Townes

Optical light amplification by stimulated emission of radiation

∼1959 Nicolay G. Basov,Aleksandr M.Prokhorov

Principle of light amplification by stimulated emission of radiation

LIGHT IN VACUUM 9



1959 Gordon Gould Invention of the laser

1960 Theodor H. Maiman Optical maser – the ruby laser

1961 A. Javan,W.R. Bennett,D.R. Herriot

cw He–Ne laser

1961 Peter Franken Observation of second harmonic generation

1962 NicholaasBloembergen

Many nonlinear optical interactions in matter explained

1962 Eric J. Woodbury,Won K. Ng

Stimulated Raman scattering observed

1962 GE,IBM,Lincoln Lab teams

Semiconductor injection lasers

1962–1964 G.A. Askaryan,R.Y. Chiao,E. Garmire,C.H. Townes

Self-focusing of light

1963 Zhores I. Alferov,Herbert Kroemer

Principle of the heterostructure laser

1964 R.Y. Chiao,C.H. Townes,B.P. Stoicheff

Stimulated Brillouin scattering observed

1965 N.A. Kurnit,I.D. Abella,S.R. Hartmann

Photon echoes observed

1965 Arno A. Penzias,Robert W. Wilson

Discovery of 3◦ microwave background radiation

1966 Arthur Ashkin,F.S. Chen

Photorefractive effect discovered

1968 Veniamin S. Letokhov Trapping of atoms by a far-off-resonance dipole force predicted





1970 F.P. Kapron,D.B. Keck,R.D. Mauer

Low loss (≤20 dB/km) silica (SiO2) glass fibers produced

1970 I. Hayashi Room temperature semiconductor laser operation

1971 T.W. Hansch,Christian Borde

Two-photon Doppler free spectroscopy

1972 R.G. Brewer,R.L. Shumaker

Optical freeinduction decay

1972 B. Ya Zel’dovich Optical phase conjugation (wave front reversal) achieved usingstimulated Brillouin scattering

1972 Stephen E. Harris Third harmonic generation in gases

1973 Akira Hasegawa,F.D. Tappert

Optical solitons predicted and suggested for fiber opticcommunications

1973 Paul Lauterbur Magnetic resonance imaging with magnetic field gradientsdemonstrated

1975 T.W. Hansch,A.L. Schawlow,D. Wineland,H. Dehmelt

Laser cooling of atoms by Doppler cooling predicted

1977 R.W. Hellwarth,A. Yariv,D. Pepper,D.M. Bloom,G.C. Bjorklund

Four-wave mixing observed and explained

1977 Peter Mansfield Magnetic resonance imaging speed-up using echo planar imaging

1978 J. Bjorkholm,R. Freeman,A. Ashkin,D. Pearson

Dipole light force acting on atoms demonstrated

∼1979 Erich P. Ippen,Charles V. Shank

Ultrashort laser sources developed

1980 L.F. Mollenauer,R.H. Stolen,J.P. Gordon

Observation of optical solitons in fibers

LIGHT IN VACUUM 11



1980 Klaus von Klitzing Observation of quantization of the Hall conductance in a two-dimensional electron gas at low temperature (used for accuratemeasurement of the fine structure constant)

1985 R.E. Slusher Observation of squeezed photon states

1986 S. Chu,J. Bjorkholm,A. Cable,A. Ashkin

Atom trapping using radiation pressure

1987 Steven Chu,David E. Pritchard

Magneto-optical traps for atoms

1987 D. Kroekel,N. Halas,G. Guiliani,D. Grischkowsky

Dark solitons observed

1988 C. Cohen-Tannoudji,W.D. Phillips,S. Chu

Laser cooling of atoms using optical molasses and Sisyphus cooling

∼1990 A. Lewis,D. Pohl,E. Betzig

Near-field optics developed

∼1992 Arthur Ashkin,Steven Chu

Optical tweezers manipulate microscopic and macroscopic objects

1997 W. Ketterle,W.D. Phillips,T.W. Hansch,E.A. Cornell,C. Wieman

Atom lasers (coherent matter waves) demonstrated usingBose–Einstein condensate sources

1999 T.W. Hansch,J.L. Hall,and colleagues

Frequency comb locking for measuring optical frequencies

1999 Lene V. Hau,Stephen E. Harris

Light slowed to the speed of a bicycle


X Hz (s−1) = X

2.99792 × 1010cm−1

= X

2.99792 × 1010 · 8065.545eV

(1.9)i.e. unity can be written in the form

2.99792 × 1010 Hz

1 cm−1 ,1 eV

8065.545 cm−1 ,

2.417986 × 1014 Hz

1 eV(1.10)

To obtain the wavelength λ corresponding to agiven frequency, ν, we make use of the equation

λ = c/ν (1.11)

If the frequency ν is given in units of cm−1,we need only invert the numerical value ofthe frequency (in units of cm−1) to obtain thewavelength in cm. For example, a frequencyof 1000 cm−1 corresponds to a wavelength of10−3 cm (10 µm). The frequency in units of s−1

is obtained from the frequency in units of cm−1

by multiplying by c in units of cm/s, ν (s−1) =ν (cm−1) × c (cm/s).

The electromagnetic spectrum is (artificially)divided into various regimes according to fre-quency: radio-frequency, microwave, infrared, vis-ible, ultraviolet, X-ray, and γ -ray. The wavelengthand frequency of these regimes are presentedin Table 1.3. Figure 1.1 schematically representsthe electromagnetic spectrum divided into thesedivisions and indicates frequencies within theseregimes. We shall consider each of these regimesin turn.

RadiowavesIn 1887, Heinrich Hertz generated radio-frequencyradiation using an oscillatory discharge across a

Table 1.3 Wavelength and frequency regimes of theelectromagnetic spectrum

Regime λ (m or asindicated)

ν (Hz)

Radiowave 3 × 109 to 0.3 ∼1–109

Microwave 30 cm to 3 mm 109 –1012

Infrared 3 mm to 780 nm 1012 to 4 × 1014

Visible 780–390 nm 4 × 1014 to 8 × 1014

Ultraviolet 390–3 nm 8 × 1014 to 1017

X-ray 3 × 10−9 to 1017 –1020

3 × 10−12

γ -Ray <3 × 10−12 >1020

spark gap (two conducting surfaces separated by agap across which a dielectric breakdown currentcan flow), focused the radiation, determined itspolarization, reflected and refracted it, caused itto interfere and measured its wavelength (∼1 m).Radiowave frequencies are defined as the regionof the spectrum from a few Hz up to about109 Hz (0.3 m). This region of the spectrum isused heavily for communications, and includesair navigation bands, AM, ham and police bands,shortwave, FM, VHF-TV, UHF-TV and radar,in order of increasing frequency. The AM radioband is around 1 MHz, while the FM band isaround 100 MHz. Television bands lie somewhatabove 100 MHz (greater bandwidth is necessaryfor TV bands to carry the greater informationcontent of TV as compared with radio). Radarwaves have frequencies in the vicinity of 1 GHz(a gigahertz is 109 Hz). Radar is used routinely forthe ranging of objects (e.g. airplanes) and velocitydetermination (as in radar traps on the roads). Thelatter is determined by the Doppler shift in thedetected frequencies of the scattered radar waves(ν = νv/c where ν is the original frequency, v isthe velocity normal to the radar wave propagationdirection and c is the velocity of light).

radio

1−109 109−1012 1012−4 × 1014 8 × 1014−1017

Visible

1017−1020 1020 up

microwave IR UV X-ray g-ray

Figure 1.1 The electromagnetic spectrum divided into regimes and frequencies within these regimes. Frequency isin units of Hz (i.e., cycles/s). (For a color reproduction of this figure, see page 1 of the color plate section.)

LIGHT IN VACUUM 13

Scientific measurements in this region are preva-lent. Astronomers use the 21 cm (1420 MHz)hyperfine transition line of hydrogen [1S1/2(F =1 → F = 0)] to detect interstellar hydrogen. TheF = 1 and F = 0 total angular momentum statesdiffer in energy because the nucleus of the hydro-gen atom (the proton) possesses a spin angularmomentum (i) which interacts with the spin of theelectron (s), and the interaction energy is differentdepending on the relative alignment of these spins,just as the interaction energy of two bar magnetsdepends on their alignment. Figure 1.2 schemati-cally shows the spins of the electron and protonin the excited hyperfine 1S1/2F = 1 and groundhyperfine 1S1/2F = 0 energy levels of H, andexplains why the F = 1 state is higher in energythan the F = 0 state in terms of interaction of barmagnets. More will be said about hyperfine inter-actions in Chapters 4 and 6. The 21 cm hydrogen

line lies on the border of the radiowave andmicrowave regimes. Hydrogen masers (microwaveamplification by stimulated emission of radiation)that lase on the 21 cm hydrogen line have beendeveloped. These devices are used as frequencystandards in astrophysics applications.

Low-frequency radiowave emission is also gen-erated in the interstellar media and in hydrogencloud regions due to e + p collisions which accel-erate the electrons and therefore emit radiation(e + e collisions do not generate radiation becausethe net second derivative with respect to time ofthe dipole moment vanishes). This process is calledBremsstrahlung (braking radiation, i.e. radiationfrom decelerating charged particles). Note that pro-cesses that involve an acceleration (not a deceler-ation) are also often termed Bremsstrahlung.

Radiowaves are produced and transmitted usingan antenna. Electrons in the transmitter antenna

F = 1 (excited) state

Magnetic moments(and analogy) Spins

e−

p

se

i = sp

me

mp

F = 0 (ground) state

21 cm hyperfinetransition

Electron magnetic moment me = −2 (eh/2mec) se

Proton magnetic moment mp = 2.79 (eh/2Mpc) sp

s = 1/2, ms = 1/2

i = 1/2, mi = 1/2

Spins

p

N

N

S

Se

Se

Sp Sp

e− e−

p p

Magnetic moments(and analogy)

S

N

S

N

S

mee−

se

s = 1/2, ms = −1/2

mpi = sp

i = 1/2, mi = 1/2

Figure 1.2 Hyperfine F = 1 and F = 0 states of the 1S1/2 manifold of hydrogen and a magnetic analogy to explainwhy F = 0 is the ground state. The spins and magnetic moments of the electron and proton are shown schematicallyin both F = 1 and F = 0 states. The magnetic moments are not drawn to scale; the magnetic moment of the protonis smaller than that of the electron by a factor of about 1000. The F = 1 state shown is the stretched state withMF = 1. If no external magnetic field is present, the three F = 1 states (with MF = 1, 0, and −1) are degenerate inenergy. The F = 0 state is lower in energy that the F = 1 states by 1420 MHz.


are driven to oscillate at a desired radio-frequencyand the accelerating electrons radiate electro-magnetic waves at this frequency. Informationis attached to the carrier wave by modulatingthe electromagnetic radiation; amplitude mod-ulation or frequency modulation methods areused (Sections 1.2.5 and 1.2.6). Radiowaves aredetected and their frequency determined in modernradio receivers using the superheterodyne detec-tion principle. The radio signal is picked up inthe receiver antenna and the signal at frequency ν1

is heterodyned (i.e. mixed) with the output of anoscillator whose frequency ν2 is tuned (by turning aknob on the radio) so that the difference frequencyν1 − ν2 is kept constant. This difference frequencyis around 455 kHz (kilohertz) for AM receivers,10.7 MHz for FM receivers and 38 MHz for tele-vision. The oscillator frequency is tuned by varyingthe capacitance or inductance of interleaved sep-arated metal plates or moving a ferrite core in acopper wire coil.

MicrowavesThe microwave frequency regime ranges fromabout 1 GHz to about 1 THz (terahertz is 1012 Hz),i.e., wavelengths of 30 cm to about 3 mm. Cel-lular telephone communications and Wi-Fi Wire-less local area network communications are in themicrowave regime. Most molecular rotational tran-sitions lie in the microwave regime. Microwaveovens operate at a frequency of 2.45 GHz; this issignificantly less than the water rotational transi-tion absorption frequencies in the 3–8 × 1011 Hzregime. The frequency shift from the absorptionpeaks of water is to insure that the microwaveradiation is not absorbed by the first layer ofwater it encounters, thereby allowing more evenheating. The frequency used in microwave ovensexcite the shifted and broadened transitions ofwater molecules that are hydrogen bonded to otherwater molecules and to other molecules. Atomicfine structure transition frequencies also lie in themicrowave regime. For example, the fine structuretransition of Na, 3p 2P3/2 to 3p 2P1/2, has a fre-quency of 17.2 cm−1 (516 GHz). For other alkaliatoms with one electron in a P state, the fine struc-ture splitting FSS(Z, n) can be estimated by the

formula,

FSS(Z, n)

FSS(Na) = 17.2 cm−1∼ Z4/n3

114/33(1.12)

where Z is the nuclear charge, and n is theprincipal quantum number of the atomic state.

Problem 1.1(a) Estimate the fine structure splitting of theCs 6p 2P excited state.

Answer: FSS(Z = 55, n = 6) ≈ (55/11)4/

(6/3)3 × 17.2 cm−1 ≈ 1344 cm−1.

(b) The actual 6p 2P3/2 to 6p 2P1/2 spinorbit splitting is 554.11 cm−1 (1.66 × 1013 Hz).What are the sources of error that give rise tothe discrepancy?

Answer: The Z and n dependence given inEq. (1.12) is obtained from hydrogenic atomconsiderations. Screening of the actual wave-function can modify the hydrogenic approxima-tion significantly.

The Cs clock frequency is exactly 9.192631770 ×109 Hz (roughly 0.3066 cm−1) and is in themicrowave regime. This transition is presently thestandard of frequency and time (actually the secondis defined as ‘9 192 631 770 periods of the radia-tion corresponding to the transition between the twohyperfine levels of the ground state of the caesium133 atom’). It corresponds to the ground electronicstate hyperfine transition 6s 2S1/2 (F = 4) to 6s2S1/2 (F = 3) of caesium (the nuclear spin of 133

55 Csis 7/2 so the total spin, F , can be 7/2 + 1/2 = 4 and7/2 − 1/2 = 3). Chapter 6 will discuss both finestructure and hyperfine interactions.

Most microwave sources make use of electronguns and resonant cavities to produce microwaveradiation. A klystron is an electron tube inwhich thermionically produced electrons are accel-erated in a potential of several hundred volts.It has two resonant cavities separated by a driftregion through which the electron beam passes.A microwave field in the first cavity bunches theelectrons into groups due to the varying voltage

LIGHT IN VACUUM 15

of the microwave field which speeds up someelectrons and slows down others. The bunchedelectrons then pass through the second cavity andthere they amplify microwave photons. Part ofthe output signal is fed back into the first cavity,thereby providing positive feedback and sustainingthe oscillation. Klystrons can efficiently producehigh power radiation in a frequency range fromabout 1 to 500 GHz with power in the MW range.The frequency bandwidth can be narrow becausethe (second) cavity selects only a narrow range offrequencies that are resonant. The size of the cavitydetermines the resonant frequency. Figure 1.3(a)is a schematic diagram of a klystron. A mag-netron is another high-power narrow-bandwidthtube used to generate microwaves. The magnetronwas invented at MIT during World War II as asource for intercept aerial (high frequency) radarand it quickly revolutionized the field of radarduring the war. Magnetrons are typically higherefficiency devices than klystrons. They are usedas the source of radiation in microwave ovens.The name is due to the role of a magnetic fieldin the microwave generation process. Electronsare produced by a d.c. electric field at the cath-ode which is located on the axis of a multi-cavityresonant cylindrical cavity structure. The mag-netic field along the axis of the cylinder togetherwith a radial electric field causes the electrons toorbit in a radius about the cylindrical axis withan azimuthal drift velocity. The microwave fieldin the cavities picks up energy from the elec-trons that orbit in the cavities. Figure 1.3(b) is aschematic diagram of a magnetron. Another typeof microwave source which has extremely widebandwidth is the travelling-wave tube. A widebandwidth is desirable in applications where it isuseful to shift the frequency over a wide range,for example, to avoid jamming. The traveling-wave tube also has an electron gun and a tubeinto which a microwave signal is fed. However,a helical coil is built into the tube, which lowersthe resonant frequency as a function of positiondownstream. An external magnet along the tubekeeps the electron beam focused as the energyfrom the electrons is transferred to the microwavefield.

Microwave radiation can be detected usingextensions of superheterodyne receiver techniquesused in the radiowave frequency range using point-contact crystal diode rectifiers. A crystal rectifierconsists of a fine wire in point contact with ablock of semiconducting material such as siliconor germanium. The contact resistance is greaterwhen the voltage is in one direction than when itis in the reverse direction. When a positive voltageis applied to the anode, electrons emitted fromthe heated cathode flow to the anode, whereas,if a negative voltage is applied, electrons cannotpropagate from the cathode and no current flows.Hence, electrons flow from the cathode to theanode when a positive voltage is applied to theanode. If an alternating voltage is applied, currentflows only during the time when the anode ispositive. Thus, an alternating voltage is rectified,i.e. converted to d.c. Bolometers are also used todetect microwave radiation. A bolometer consistsof a small length of metal wire which is heatedby the presence of a microwave signal, resultingin a change in resistance. Sometimes carbon orgermanium are used as resistive elements. Thechange in resistance is detected as a voltagesignal by passing a small current through thebolometer and the total absorbed power can bedetermined. The response time is of millisecondorder and the devices can be used from theinfrared to the millimeter wavelength range. Thedevices need to be cooled to liquid heliumtemperatures for high sensitivity. An importantadvantage of bolometer over a crystal diodemicrowave detection is the very low noise at lowfrequencies. For more information on microwavesources and detectors, see, for example, Townesand Schawlow [1].

Infrared

This frequency regime ranges from about 1012 Hzto about 4 × 1014 Hz. It overlaps the microwaveregion and runs into the optical region. The‘optical’ telephone cables that are now replacingcopper wire cables are infrared carrying fibersand carry either 2.4 × 1014 Hz (1.3 µm), or 2.0 ×1014 Hz (1.5 µm) radiation pulses. The latterfrequency is at the minimum of the scattering loss


(a)

Highvoltage

Photocathode

Glass envelope

Incoming (c)light

Electrons

Dynodes

Output electron

pulse

Anode

vsvs

+−

Cathode

Bunchingcavity

Outputcavity

rfinput

rfoutput

CollectorGride− beam

(b)

−

− −

−

−

−−

−

−

+

+

+ +

+

+

++

electrons

cathode

anode

cavities

dc magnetic field B

electrons

Blowup

rf field

cavities

dc magnetic field B

−−

−

ac current I

++

+

Figure 1.3 Schematic diagram of (a) a klystron, (b) a magnetron and (c) a photomultiplier tube. Reproducedcourtesy of HyperPhysics, copyright Rod Nave, Georgia State University

of these fibers. The infrared regime correspondsto the frequency range of vibrational excitationof most molecules. For example, one very highfrequency vibrational degree of freedom in organicmolecules is the C–H stretching mode, which hasfrequency ∼3000 cm−1 (9 × 1013 Hz, 3.333 µm).

Room temperature corresponds to an energyof 1/40 eV, 200 cm−1, or roughly 5 × 1012 Hz.Matter in thermal equilibrium (with its ownradiation field) radiates electromagnetic energyaccording to the Planck radiation law (see Section1.2.1). This radiation has intensities which peak

LIGHT IN VACUUM 17

at frequencies ν near ν = kT /h. Hence, thespace around us is filled with infrared radiationemanating from matter on our Earth.

Intense infrared sources include klystrons (seethe previous section), gas lasers, e.g. the CO2

laser which can be made to lase on the vibrationaltransitions near 10.6 or 9.6 µm, solid-state dopedcrystals, e.g. Nd+3 –YAG (yttrium aluminium gar-net, Y3Al5O12), which lases at 1.064 µm, andinfrared solid-state semiconductor diode lasers,e.g. GaAs–GaAlAs heterojunction diodes, whichcan be made to lase over a wavelength from0.69 to 0.85 µm. The term ‘laser’ is an abbre-viation for ‘light by amplified stimulated emis-sion of radiation’, a means of making very brightlight sources. There are many examples of gas,solid-state doped crystal and semiconductor diodelasers which are commonly used in the infraredrange (see Chapter 6). Moreover, infrared gen-eration by mixing visible lasers of frequencies,say ω1 and ω2, to generate infrared light at fre-quency ω3 in a nonlinear mixing crystal is acommonly used technique called difference fre-quency generation (ω3 = ω1 − ω2). Furthermore,light from an existing visible frequency laser withfrequency ω1 can be shifted in frequency by a pro-cesses known as stimulated Raman scattering toproduce light at infrared frequencies, ω3 = ω1 −ωv, where ωv is a vibrational frequency of themedium in which the stimulated Raman scatter-ing occurs.

Infrared radiation can be detected and the powerof the radiation determined using a bolometer, apyroelectric detector and a Golay cell. Bolometerswere discussed in the previous section. Pyroelectricdetectors measure the change in the capacitance ofa circuit containing a ferroelectric crystal due tothe change of the dipole moment of the crystalupon heating (see Section 2.5.4). The radiationis chopped at a given frequency (typically kHzrange) for determination of the power. The Golaycell detector contains an air space between twofilms, one of which is an absorbing film thatheats the air space upon absorption of incidentradiation and the other is a thin mirror. The airspace contains xenon gas. The gas heated by thetemperature increase of the absorbing film deflects

the mirror, and the deflection is measured by aphotocell which monitors the light reflected off themirror. The radiation is chopped at about 10 Hz,which is the response time of the instrument. Bothpyroelectric and Golay detectors can operate atroom temperature, but are not as sensitive nor asfast as cooled bolometer detectors.

Visible light

The optical frequency range is from about 4 × 1014

to about 8 × 1014 Hz or 1.6–3.2 eV (wavelengthsfrom 780 to 390 nm). The middle of the opti-cal range corresponds to a frequency of about6 × 1014 Hz (2.5 eV or 20 000 cm−1), which cor-responds to a wavelength of 500 nm. The low-est electronic excitations of many atoms lie inthis regime (the sodium yellow line is ∼590 nm).The wavelength sensitivity of the human eye isroughly Gaussian and is centered at about 555 µm.Wavelengths below 390 (ultraviolet) and above780 (infrared) cannot be seen by the human eye.Sources of visible light include matter heatedto high temperatures, such as incandescent lightbulbs, arc lamps, fire (as in a wood campfire), flu-orescent light bulbs and, more recently, lasers.

Visible light can be detected with great sensi-tivity using a photomultiplier tube (PMT). Theyare available in various frequency ranges fromthe infrared through the ultraviolet, 110–1100 nm.PMTs can measure intensities weaker than thosethat can be seen by the human eye (the eye candetect a light pulse consisting of about 10 photons)and are used in detecting very low intensity lightsignals. They are often used in fluorescence spec-troscopy applications where the number of photonsemitted is very small. They produce photoelectronsat the photocathode and amplify the number ofelectrons produced at the photocathode to yield ameasurable current proportional to the number ofphotons detected. The response time is typicallyof the order of 1–10 ns. PMTs consist of a pho-tocathode and a series of anode electrodes calleddynodes that are held at increasingly higher volt-ages. The incident photons impinge on the surfaceof the cathode and cause the emission of electronsby the photoelectric effect. The quantum efficiency,defined as the number of photoelectrons ejected at


the photocathode divided by the number of inci-dent photons, is of the order of 1–25%. Theseelectrons from the cathode are then accelerated tothe second plate, which is maintained at a positivepotential relative to the cathode and knocks outmore electrons by secondary emission. The sec-ondary electrons are then attracted to the next dyn-ode and the process repeats until the last plate, thecollector or anode, is reached. The amplification ofthe number of electrons originally produced at thecathode (determined by the quantum efficiency ofthe cathode) can be 106 –108. Figure 1.3(c) showsa schematic representation of a PMT.

For sufficiently intense light sources, photodi-odes and photovoltaic detectors can be used fordetection. Photodiode detectors are not as sen-sitive as PMTs but they are small, robust andrelatively inexpensive. Photodiodes emit electronsupon absorption of radiation and the current isproportional to the photon flux. When photonsin the right frequency range strike a semiconduc-tor, they can promote electrons from the valenceband to the conduction band, creating electronand hole pairs. The concentration of these elec-tron–hole pairs depends on the amount of lightstriking the semiconductor, making the semicon-ductor suitable as an optical detector. In photodi-odes, a voltage bias is applied across the semi-conductor, and the voltage induces a current ofelectrons and holes. The concentration of the pho-toinduced electron–hole pairs determines the cur-rent through semiconductor. Photovoltaic detectorscontain a p–n junction (see Section 6.5.2) thatcauses the electron–hole pairs to separate to pro-duce a measurable voltage.

Photodiodes can be arranged in linear arrayson an integrated circuit chip. Such an arrange-ment can be used for spectroscopy if the arrayis placed at the image plane of a spectrom-eter to allow a range of wavelengths to bedetected simultaneously. They are available with512, 1024 or 2048 elements. Another type of arraythat can be used to measure light is a charge-coupled device (CCD). CCDs are integrated-circuit chips which contain an array of capac-itors that store charge. Incident light createselectron–hole pairs and the charges accumulate

and are measured in a given time interval. CCDscan be more sensitive than photodiode arraysand are therefore used for measurement in lowlight conditions.

Ultraviolet

This frequency range is from 8 × 1014 to about1017 Hz (390 to about 3 nm). Ultraviolet radia-tion is emitted by electronically excited atoms andmolecules in gaseous, liquid and solid-state mate-rials. This radiation is detectable by photocells andfluorescent screens. Ultraviolet photons that areless energetic than the Lyman limit {13.6 eV or91.2 nm, i.e. the limit as n → ∞ of transitionsfrom high n principal quantum number to n = 1[hν ≈ 0.5 × α2mec

2(1 − 1/n2)] where α ≈ 1/137is the fine structure constant; see Appendix C}can escape absorption in the interstellar medium,which contains hydrogen atoms. Observation ofsuch light has given us much insight into thecosmos. Copious amounts of UV radiation bom-bard our planet, most of it emanating from the Sun,and the ozone layer protects us from this harmfulUV radiation. Overexposure to ultraviolet radia-tion can cause damage to skin tissue and growthof cancerous tissue.

Continuous emission in the ultraviolet can beobtained by producing an electrical or microwavedischarge in high-pressure hydrogen gas or inertgases (xenon or argon). Another method of produc-ing ultraviolet emission is by high-intensity laser-produced plasmas. A high-intensity laser beam,e.g. from a Q-switched Nd–YAG laser, on a rareearth target can produce quasi-continuous radiationfrom about 200 to 4 nm. An increasingly popularsource of ultraviolet radiation for scientific studyis Bremsstrahlung radiation from accelerated elec-trons in a synchrotron accelerator or storage ringemitted perpendicular to the orbital radius of theelectron motion. The central frequency emitted isproportional to E3/r , where E is the energy of theelectrons and r is the radius of the orbit, and thetotal power varies as E4.

X-raysX-rays were discovered by Wilhelm ConradRontgen (1845–1923) in 1895. Their frequency

LIGHT IN VACUUM 19

range is from 1017 to 1020 Hz (3 × 10−9 to 3 ×10−12 m). This frequency range corresponds toenergies of about 1 to 100 keV. One way thatX-rays originate is from inner electron transi-tions in atoms and molecules. X-rays are artifi-cially produced in X-ray tubes by bombardmentof metals by highly accelerated electrons. Theyare also generated when very high-energy chargedparticles accelerate and radiate (Bremsstrahlung).From astrophysical observations, it is believed thatquasars (quasi-stellar radio sources which usuallyalso emit X-rays and infrared) and quasi-stellarobjects (similar to quasars but lacking strong radioemission) provide most of the hard X-ray back-ground (i.e. the higher range of X-ray energies).Soft X-rays are uniformly present and appear tobe associated with the interstellar medium of ourown galaxy.

Problem 1.2(a) To what velocities must electrons be accel-erated to produce 5 keV X-ray radiation bybombarding metals? (Do the calculation non-relativistically and relativistically).

Answer:

mev2/2 = 5 keV, so

v =

√√√√√2 × 5 × 103 eV × 1.6 × 10−12 erg

eV9.1 × 10−28 g

= 4.2 × 109 cm/s

An accurate estimate requires a relativis-tic calculation, (γ − 1)mec

2 = E, where γ =1/√

1 − v2/c2

(b) What voltage is necessary to accelerateelectrons to these speeds?

Answer: V = 5000 V.

γ -Raysγ -Rays can emanate from nuclear transitions.These high-energy photons have frequenciesof ≥1020 Hz (≤3 × 10−10 cm wavelength). One

example of a nuclear reaction that takes placein the Sun and produces high-energy γ -rays isthe collision of a deuteron with a proton to formhelium 3 and a γ -ray photon:

21H + 1

1H → 32He + 0

0γ (1.13)

Here 21H denotes the deuteron and 1

1H denotesthe proton (the subscript gives the charge and thesuperscript gives the nucleon number).

Problem 1.3What is minimum value of the energy ofthe γ -ray in reaction (1.13)? The atomic massexcesses, M , of the nuclei involved areM(1

1H) = 7.288 MeV, M(21H) = 13.136

MeV, M(32He) = 14.931 MeV, where M =

(M − AMamu)c2. Here one atomic mass unit is

given by Mamu = 1.6605 × 10−27 kg.

Answer: 5.493 MeV (1.3281 × 1021 Hz)

Another nuclear reaction responsible for produc-tion of γ -rays involves the collision of a relativis-tically moving proton and a slow proton (foundin profundity in the interstellar medium). One ofthese protons can be excited to a high-energy statewhich decays to a proton and a π0 meson. Theπ0 in turn decays into two gamma rays in about10−16 s. The reactions involved are:

p (relativistic) + p → p∗ + p (1.14)

p∗ → π0 + p (1.15)

π0 → γ + γ (1.16)

γ -Rays can be artificially made in acceleratorsand can be detected in crystal spectrometers,pair spectrometers, lithium drifted germaniumspectrometers and scintillation counters. Cosmicray showers can degrade the photon energy ofhigh-energy γ -rays as they enter the atmosphereby a process of pair creation and re-radiation

γ → e + e (1.17)

e → e + γ (1.18)

e → e + γ (1.19)


The original photon energy is soon divided amongmany particles.

Astrophysical mechanisms for the production ofγ -rays also include synchrotron radiation (radi-ation emitted by a charged particle spiraling ina magnetic field) emitted by relativistic elec-trons, and Compton scattering (the collision ofa relativistic electron with Lorentz factor γ =1/√

1 − v2/c2 and a photon of frequency ν, result-ing in a loss of energy of the electron, and aphoton with frequency roughly equal to γ 2ν).Bremsstrahlung of ultrarelativistic electrons alsoyields γ -rays. The observed spectrum of high-frequency radiation is a power law spectrum ofthe form ν−α with α in the range 0.5–1. Chargedparticles giving rise to synchrotron radiation with apower law ν−α have an energy distribution givenby E−σ where σ is related to α by the relationα = (σ − 1)/2 (see Section 3.4.2). Hence, if syn-chrotron radiation gives rise to the high-frequencyradiation, we infer that the electron energy spectrashould be a power law with σ = 2–3. For com-parison, it is interesting to note that the observedspectrum of cosmic rays, high-energy particlesbombarding the earth that include electrons, pro-tons and α particles but also heavy nuclei, is of theform E−σ with σ = 2.7, up to 1019 eV energies.This value of σ corresponds to α = 0.8.

Blackbody radiation

When matter is in thermal equilibrium withthe electromagnetic radiation surrounding it, theradiation emitted by the body is completelydetermined in terms of the temperature of the body.Such matter is called ‘blackbody matter’, and theradiation is called ‘blackbody radiation’, becausethe requirement that the body be in equilibriumwith its own radiation field requires that the bodybe opaque. The law for the radiation intensity ofsuch bodies was derived by Max Planck in 1900.Planck’s constant, h, was first introduced in thetheory of thermal radiation. Let uω(T ) dω be themean energy per unit volume in the frequencyrange between ω and ω + dω for photons of bothpolarizations. Planck’s blackbody radiation law isgiven by:

uω(T ) = h

π2c3

ω3

exp(hω/kBT ) − 1(1.20)

Defining the dimensionless parameter

x = hω

kBT

we can write

uω(T ) = kB3T 3

π2c3h2 f (x)

where the function f (x) takes the form:

f (x) = x3

exp(x) − 1

A plot of the function f (x) vs x is shownin Figure 1.4(a). The low frequency (x 1)

dependence of the Planck energy density is ω2,as is clear from expanding the exponent in thedenominator of Eq. (1.20), whereas the high-frequency dependence is exponential, uω(T ) ≈exp(−hω/kBT ) for x � 1. The high-frequencybehavior (the exponential tail) is the Wien tail,and the low-frequency limit of the blackbodyspectrum, uω(T ) ≈ ω2kBT /(π2c3) for x 1, isthe Rayleigh–Jeans limit. The number density ofphotons of frequency ω is given by Nω(T ) =uω(T )/(hω). The total energy density in allfrequencies is given by:

u(T ) =∫ ∞

0uω(T ) dω

= h

π2c3

∫ ∞

0

ω3 dω

exp(hω/kBT ) − 1= 4σ

cT 4

(1.21)where the integral has been analytically evaluated.The Stefan–Boltzmann constant, σ , is given by

σ = π2kB4

60h3c2= 5.6697 × 10−5 erg/s/cm2/K4

(1.22)

(in SI units 5.6697 × 10−8 W m−2 K−4) andthe Boltzmann constant has numerical value

LIGHT IN VACUUM 21

2500(b)

(a)

2250

WRC spectrum2000

1750

1500

1250

Sola

r sp

ectr

al e

nerg

y di

stri

butio

n

1000

750

500

250

00 0.4 0.8

Wavelength (µm)

1.2 1.6 2.0

1.5

1

0.5

00 2 4 6

x = hw/kBT

f(x)

= x

3 /[ex

p(x)

−1]

8 10 12

Assuming sun as ablackbody at 5777 K

(2.82, 1.42)

(Planck's blackbodyradiation law)

Figure 1.4 (a) Planck’s blackbody radiation law,f (x) = x3/(exp(x) − 1), where x = hω/kBT . Theenergy density per unit frequency is given by uω(T ) =[(kB

3T 3)/(π2c3h2)]f (x). (b) Measured spectral energydistribution of the light from the sun, F(λ) =f (x) dx/dλ versus λ (see discussion just beforeProblem 1.12 for the relation of the distribution asa function of ω and of λ) and comparison with abackbody distribution at T = 5777 K (reproduced fromIqbal (1984), with permission from Elsevier).

kB = 1.3806503 × 10−16 erg/K (in SI units,1.3806503 × 10−23 J K−1). Equation (1.21), is theStefan–Boltzmann law of radiation; the T 4 tem-perature dependence of the total energy emittedand the numerical value agree with experimentalobservations. The energy per unit time per unitarea (i.e. the intensity) emitted by the blackbodyis given by:

I (T ) = c

4u(T ) = σT 4 (1.23)

The factor of 1/4 will be derived in Problem 1.5(d)by comparing the power emitted per unit areaand per unit frequency of a blackbody with theenergy density uω(T ). The frequency at whichthe intensity, Iω(T ) = cuω(T )/4, is a maximumis determined numerically to be:

ωmax = 2.82 kBT /h (1.24)

Lord Rayleigh (John William Strutt) and SirJames Jeans were the first to suggest in around1870 that the equipartion principle, whereby eachdegree of freedom of a system has a mean energyequal to kBT /2, be applied to the electromagneticenergy of a cavity. The differential number ofmodes of the electromagnetic field per unit fre-quency per unit volume is given by

N(ω) dω = 21

(2π)3d3k

= 24π

(2π)3k2 dk = 8π

(2π)3c3ω2 dω

There are two polarization states of a photon, thusthe factor of 2 in the number of modes. Hencethe number of modes per unit angular frequency isgiven by

gω = 1

π2c3ω2

(a more detailed account of this counting argumentis given in Section 1.2.7). Assigning a meanenergy per unit frequency interval of kBT , weobtain the Rayleigh–Jeans limit:

uω(T ) = gωkBT = ω2kBT /(π2c3) for x 1

This yields the energy per unit time per unitarea per angular frequency interval (the intensityper angular frequency interval) known as theRayleigh–Jeans formula:

Iω(T ) = c

4uω(T ) = ω2kBT /(4π2c2) for x 1


The ω2 dependence of the blackbody radiationspectral distribution agrees well with experimentat low frequency, but it fails badly at high fre-quency. This contradiction is called the ‘ultravioletcatastrophe of classical radiation theory’ (since itleads to an infinite integrated intensity as well asan infinite number of high-frequency photons). Thequantum theory of light is required to obtain thecorrect frequency dependence. The correct ‘tail’ ofthe distribution (the high-frequency dependence) isobtained by replacing the Boltzmann distributionof statistical mechanics, which yields the meanenergy of kBT per angular frequency interval, withthe Bose–Einstein distribution (the Planck distri-bution for the special case of photons), whichyields the occupation number per angular fre-quency interval

〈nω〉 = 1

exp(hω/kBT ) − 1

and therefore the mean energy per angular fre-quency interval

〈hωnω〉 = hω

exp(hω/kBT ) − 1

rather than kBT . Now, the number of modesfield per unit frequency per unit volume yieldsEq. (1.20).

All radiation impinging on blackbodies is, bydefinition of a blackbody, absorbed. Since theblackbody is in thermal equilibrium with its radia-tion field, it emits as much radiation as it absorbs.What about non-blackbodies? Some of the radi-ation impinging on such a body is absorbedand some is reflected. Thus, the reflection prob-ability of light at frequency ω and wavevec-tor k, R(ω, k), must somehow be involved inthe modification of the blackbody radiation lawfor non-blackbodies. The probability of transmis-sion into the body is given by T (ω, k) = 1 −R(ω, k).

Additional discussion of the Planck distributioncan be found in Landau and Lifshitz, [3] andReif [4].

Problem 1.4(a) Derive the Planck blackbody radiation lawusing the fact that the absorption is proportionalto the photon number density nk,α , whileemission is proportional to (nk,α + 1).∗This is a more advanced problem and requiressome knowledge of quantization of the radiationfield (see Section 1.2.11 and Chapter 5) andquantum mechanics. See also Sakurai [5].

Answer: consider the reversible process

A ↔ hω + B

wherein thermal equilibrium is established, i.e.

N(B)wabs = N(A)wem

where N (A) and N (B) are the populationdensity of states A and B, and wabs and wem

are the rates for absorption and emission. Sincethe populations of states A and B are inequilibrium,

N(B)

N(A)= exp(−EB/kBT )

exp(−EA/kBT )= exp(hω/kBT )

From the quantum theory of radiation (Sec-tion 1.2.10 and Chapter 5):

wabs ≈ (nk,α)|〈A|[exp(ik · r)p

+ p exp(ik · r)]|B〉|2,wem ≈ (nk,α + 1)|〈B|[exp(−ik · r)p

+ p exp(−ik · r)|A〉|2,

but from microscopic reversibility, 〈A| exp(ik · r)p|B〉 = 〈B|p exp(−ik · r)|A〉. Substitu-ting into the thermal equilibrium expression,

exp(hω/kBT )(nk,α) = (nk,α + 1)

and rearranging, we find

nk,α = 1

exp(hω/kBT ) − 1

LIGHT IN VACUUM 23

The energy density per unit angular frequencyinterval is given by:

uω(T ) dω = 2 · hωnk,α

1

(2π)34πk2 dk

= h

π2c3

ω3

exp(hω/kBT ) − 1dω

(b) How would you formulate a modificationof the blackbody radiation law for bodiesthat reflect with reflection coefficient R(ω)

and internally reflect with reflection coefficientR(ω).

Answer: Modify the thermal equilibriumexpression given in part (a) to read

N(B)wabs[1 − R(ω)] = N(A)wem[1 − R(ω)]

where R(ω) is the reflectivity at frequency ω

and R(ω) is the reflectivity at frequency ω

from within.

Problem 1.5(a) Derive the value of the constant 2.82in Eq. (1.24) by differentiation of f (x) withrespect to x.

(b) Show how the Wien limit and theRayleigh–Jeans limit are obtained fromEq. (1.20).

(c) What is the temperature dependence ofthe average frequency of blackbody radiation?

Hint: Defining the number density of photonsof frequency ω,

Nω(T ) = uω(T )/(hω)

= 1

π2c3

ω2

exp(hω/kBT ) − 1

the average frequency is given by

ωav =

∫ ∞

0ωNω(T ) dω∫ ∞

0Nω(T ) dω

=

∫ ∞

0

ω3 dω

exp(hω/kBT ) − 1∫ ∞

0

ω2 dω

exp(hω/kBT ) − 1

= constant kBT /h (1.25)

Find the constant. Note that the denominatorin Eq. (1.25) can be evaluated in terms of theRiemann zeta function; its numerical value canbe found in Abramowitz and Stegun [6].

Answer: ωav = 2.7012 kBT /h.(d) Derive the factor c/4 in the first part ofEq. (1.23) by comparing the power emitted perunit area and per unit frequency of a black-body with the energy density. (See Reif [4],Section 9.13–9.15.)

(e) Determine the density of photons at agiven temperature.

Answer:

N(T ) =∫ ∞

0Nω(T ) dω

= 1

π2c3

∫ ∞

0

ω2 dω

exp(hω/kT ) − 1

= 60.4(kBT /hc)3

How could you have predicted the temperaturedependence from dimensional considerations?

(f) What is the density of photons at T =2.7 K (the temperature of the microwave black-body background radiation)?

Answer: 400 cm−3.

Problem 1.6(a) What color is lampblack heated to temper-atures of 300, 8600 and 11 000 K?

Answer: black, yellow and blue.(b) Why is there a difference between the colorof lampblack and stainless steel at 300 K?


The Sun and stars as light sources

The source of energy of the Sun is the nuclearreactions that occur in its interior. Hans Betheand others established that the most importantnuclear reactions in stars similar to our Sun arefusion reactions which produce helium nuclei bythe fusion of protons, 4 p → He. In the process,γ -rays, neutrinos and positrons are also created.Since the mass of helium is 3.97 mp, an amountof energy equal to (4 − 3.97) mpc

2 = 0.03 mpc2

is liberated.

Problem 1.7If the Sun continues to shine with its presentluminosity, L = 3.9 × 1033 erg/s, and the massof the Sun is 2.0 × 1033 g, how long will theSun shine for?

Answer: Mc2 =Lt , so t =1.5 × 1013 years.Actually, the mass deficit in burning 4 p →He is only 0.7% of the original mass 4 mp,so t = 0.007 × 1.5 × 1013 = 1 × 1011 years.Moreover, after the Sun has used up about 13%of its hydrogen, its luminosity will change sig-nificantly [2].

The principal thermonuclear reactions occurringin the Sun that are responsible for its energyproduction are as follows:

11H + 1

1H → 21H + 0

1e + 00ν (1.26)

21H + 1

1H → 32He + 0

0γ (1.27)32He + 3

2He → 42He + 1

1H + 11H (1.28)

The net reaction is given by

4 11H → 4

2He + 2 01e + 2 0

0ν + 2 00γ (1.29)

Here the symbol nmX indicates n nucleons (bary-

ons), nuclear charge of m, and species X.

Problem 1.8Demonstrate that Eq. (1.29) is the net reactionof the reactions (1.26–1.28).

The first reaction, reaction (1.26), is slow fortwo reasons. First, the protons must get closeenough to react. This requires that they have suf-ficient energy to overcome the Coulomb poten-tial energy barrier of height e2/2rp, where rp isthe radius of the proton (∼1 fermi = 10−13 cm).Secondly, during the collision of the protons, aweak interaction must take place (neutrino pro-duction is a weak interaction process). The rela-tively slow rate of this reaction assures our Sun along lifetime.

Problem 1.9Calculate the tunneling probability for twoprotons to reach a separation of 2rp (so that theyoverlap) at a temperature T . Give a numericalvalue for the tunneling probability.

Answer:

exp

(−2π

h

∫ 2rp

rT

|p(r)| dr

)

where

p(r) =√

2m(kT /2 − e2/r)

and rT is the radius at which p(r) = 0.

The temperature of the sun can be deter-mined from: (a) measurement of the solar constant,defined as the energy which impinges perpendicu-larly per unit time per unit area outside the Earth’satmosphere; (b) the measured distance of the Earthfrom the Sun, res = 1.5 × 1013 cm; (c) the angu-lar diameter of the Sun, θ = 9.28 × 10−3 rad;and (d) the assumption of a blackbody intensity,

LIGHT IN VACUUM 25

I (T ) = σT 4 [2]. The measured solar constant isf = 1.36 × 106 erg/s/cm2. The total power radi-ated by the Sun is given by the intensity of theSun times the area of the Sun. This quantity, thesolar luminosity, is the product of the solar con-stant and the area 4πres

2. The temperature of thesurface of the sun determined in this way is 5800 K(see Problem 1.10).

Problem 1.10If the temperature of the surface of the Sun is5800 K, and the angular diameter of the Sun isθ = 9.28 × 10−3 rad, what is its luminosity?

Answer: L = surface area of the Sun ×I (T ) = 4πrs

2 × σ T 4. The radius of the sun isgiven by rs ≈ resθ/2 = 6.96 × 1010 cm.

Problem 1.11Carry out the numerical calculation of thetemperature of the Sun using the above method.

Hint: L = f 4πres2 = Surface area of Sun ×

I (T ) = 4πrs2 × σT 4. Thus, T 4 =f res

2/(σrs2).

Table 1.4 Wavelength and frequency ranges forcolors in the visible spectrum

Color λ (nm) ν (1012 Hz)

Red 780–622 384–482Orange 622–597 482–503Yellow 597–577 503–520Green 577–492 520–610Blue 492–455 610–659Violet 455–390 659–769

One might think that the Sun’s frequency withmaximum intensity, νmax, must be yellow (sincethe Sun appears to be yellowish-white), i.e. fromTable 1.4, νmax ≈ 5.1 × 1014 Hz (corresponding to

a wavelength λ ≈ 590 nm). However, if thisvalue of νmax is substituted into Eq. (1.24), T =hνmax/(kB × 2.82), we obtain a solar temperatureof 8627 K, rather higher than the temperatureof 5800 K determined above. Moreover, if wecalculate νmax from Eq. (1.24) using a temperatureof 5800 K, we obtain a frequency νmax ≈ 3.4 ×1014 Hz (corresponding to a wavelength in theinfrared near 880 nm!). A partial reason for thisdiscrepancy could be the fact that the Sun is nota perfect blackbody. By looking at the spectrumof the Sun, Figure 1.4(b), small discrepanciesfrom a pure blackbody spectrum can be seen.At wavelengths shorter than the peak of theintensity there is considerably more intensity thanthe clearly evident blackbody contribution to thespectrum. This suggests that the Sun is not auniform temperature and that its surface is notfully in equilibrium with its radiation field. Athigher resolution, absorption lines appear in thesolar spectrum. These absorption lines originatefrom the absorption of the light generated in theinterior of the Sun by matter in the outermostregions of the Sun’s atmosphere, which is coolerthan the matter in the interior. Furthermore, ourcolor vision does not accurately determine the peakintensity of light with a wide spectrum because ofthe way our eye determines colors. We can easilybe fooled in our color vision by background colorsand other factors. We should also keep in mind thatthe spectrum of the Sun as it reaches us is partiallyaffected by the atmospheric scattering of light. Thisscattering is responsible for the Sun’s red color atsunrise and sunset. However, this would make theactual color of the sun bluer than what we see andwould therefore raise the 8627 K estimate, makingthe discrepancy with the 5800 K value even worse.Another point which can be considered in light ofthe apparent conundrum raised in this paragraphis that the peak wavelength, λmax, of a blackbodydistribution curve plotted as P(λ) vs λ is differentfrom the wavelength corresponding to c/νmax,the corresponding peak of the Planck distributionvs frequency curve. This can be understood


as follows:

P(λ) dλ = P(ω) dω

hence

P(λ) = P(ω)dω

dλ= P(ω)

2πc

λ2

Therefore,

dP(λ)

dλ=

d

[P(ω)

ω2

2πc

]dω

dω

dλ=[

dP(ω)

dω+ 2πc

λ3

]

Problem 1.12(a) Calculate the dimensionless quantity, η,defined by the relation

λmax = η−1 hc

kBT

To do so, numerically calculate the value of

x ≡ hc

λkBT

for which f (x) = x2x3/[exp(x) − 1] is maxi-mum. Compare η with 2.82.

Answer: x∗ = 4.96, i.e., η = 4.96.(b) Determine λmax for T = 5800 K.

Answer: 500 nm (this is much more in linewith our expectation, 590 nm, than 880 nm!).

1.2.2 Wave equation in vacuum

Propagation of light (i.e. electromagnetic waves)through vacuum is described by the source freeMaxwell’s equations (see Appendix B for a dis-cussion of Maxwell’s equations):

∇ · E = 0 {∇ · E = 0} (Coulomb’s law) (1.30)


∇ × E = −1


∇ × B = 1

c∂tE {∇ × B = µ0 ε0∂tE} (Ampere’s law) (1.33)

The propagation equation for light in vacuum iseasily derived by taking the curl of Eq. (1.32) andusing Eq. (1.33) to re-express the right hand side(RHS) of the resulting equation to obtain:

∇ × ∇ × E = −c−2∂t2E

{∇ × ∇ × E = −µ0ε0∂t2E} (1.34)

Noting the mathematical identity,

∇ × ∇ × E = −∇2E + ∇(∇ · E) (1.35)

and using Eq. (1.30) to show that the second termon the RHS of Eq. (1.35) vanishes, we reduceEq. (1.34) to the form {note that µ0 ε0 = c−2 in

SI units}:

∇2E − c−2∂2t E = 0

{∇2E − µ0 ε0∂t2E = 0} (1.36)

This equation is the wave equation (its solutionis a wave, or a superposition of waves, aswill be shown below) for the electric field withpropagation velocity c.

Taking the curl of Eq. (1.33) and usingEq. (1.32) and Eq. (1.30) in a manner similar tothat used to obtain Eq. (1.36), we obtain the waveequation for the magnetic induction vector, B:

LIGHT IN VACUUM 27

∇2B − c−2∂t2B = 0

{∇2B − µ0 ε0∂t2B = 0} (1.37)

As can be easily demonstrated by substitution,A cos(k · r − ωt) is a solution to Eq. (1.36) [orEq. (1.37)], where ω is the angular frequencyand k the wavevector. The magnitude of k isgiven by k = ω/c [as otherwise A cos(k · r − ωt)

cannot be a solution to the wave equation] andits direction k is arbitrary. The wavelength λ ofthis periodic wave is inversely proportional to theangular frequency, and is given by λ = 2πc/ω, i.e.k = 2π/λ = ω/c. A convenient form of solutionto Eq. (1.36) [and Eq. (1.37)] corresponding toangular frequency ω and wavevector k is givenby the exponential form, A exp[i(k · r − ωt)]. Thisis a complex function; since the electric (andmagnetic) field is real, it is given by the real partof A exp[i(k · r − ωt)]. A is a constant vector (itdoes not depend on r or t). In vacuum, the vectorA must be perpendicular to k, i.e. k · A = 0, since∇ · E = 0 (or ∇ · B = 0). The general solution toEq. (1.36) [and Eq. (1.37)] is given by an arbitrarysuperposition of harmonic functions of the formA exp[i(k · r − ωt)]:

E(r, t) =∫

dω

∫dk A(k) exp[i(k · r − ωt)]

+ c.c. (1.38)

where A(k) is an arbitrary vector function of thevector k, provided k · A(k) = 0 and c.c. standsfor complex conjugate. Since the frequency ω inEq. (1.38) equals kc, the integral over ω can beremoved as the value of ω is specified by the valuesof kx , ky and kz. Alternatively, the integral over themagnitude of k can be removed, and Eq. (1.38) canbe rewritten as:

E(r, t) =∫

dω

∫d2k A(ω, k)

× exp[iω(k · r/c − t)] + c.c. (1.39)

Here d2k indicates the differential over the angularpart of the wavevector (d2k = d�, where this is

the differential solid angle), with the length ofthe k vector given by ω/c. The electric (andmagnetic) fields are directly related to the forceon a charged particle, and therefore they are realfields. Therefore, the electric field is given by thereal part of the complex exponential forms on theRHS of Eq. (1.38) or (1.39). If we choose to writeEq. (1.39) in the form:

E(r, t) =∫

dω

∫d2k A(ω, k)

× exp[iω(k · r/c − t)]

the reality of the electric field imposes the condi-tion A(−ω, k) = A∗(ω, k). (See Problem 1.14.)

To use the analogy of Richard P. Feynman,the propagation of electromagnetic waves is sim-ilar to the propagation of water waves on thesurface of a pool. The waves of different fre-quency do not disturb one another and in factpropagate through one another without affectingeach other. That is, these waves propagate inde-pendently in vacuum. The waves emanating froma radio station can be detected without interfer-ence from the radio waves of a different radiostation, or the short waves transmitted from tele-vision stations or the γ -rays impinging on theEarth from sources in outer space. They driveelectrons up and down, much like waves in apool drive particles on the surface of the pool upand down.

Problem 1.13The solution to the wave equation can bewritten in the form:

E(r, t) =∫

dω

∫dk {A(k) exp[i(k · r − ωt)]

+ C(k) exp[−i(k · r + ωt)]}

Is the second term on the RHS really necessary?

Hint: Consider the integral over the angularvariables,

∫d2k, in Eq. (1.39). The integral

contains k and −k.


Problem 1.14(a) The electromagnetic field is real, as isclear from the fact that the force on a testparticle is

F(r, t) = q

[E(r, t) + v(r, t)

c× B(r, t)

]{F(r, t) = q(E + v × B) in SI units}

where q is the charge of the test particle andv is its velocity. Find what restriction thisimposes on the coefficients A(ω, k) by writingthe electric field in terms of non-negativefrequencies and using the restriction that theelectric field must be real.

Answer: Break up the integral∫

dω inEq. (1.39) into

∫∞0 dω + ∫ 0

−∞ dω and changevariables from ω to −ω in the second termto obtain:

E(r, t) =∫ ∞

0dω

∫d2k {A(ω, k)

× exp[iω(k · r/c − t)] + A(−ω, k)

× exp[−iω(k · r/c − t)]} (1.40)

Since exp[−iω(k · r/c − t)] in the second termis the complex conjugate of exp[iω(k · r/c −t)], and since the first and second terms mustbe complex conjugates for each frequency, itfollows that A(−ω, k) must be the complexconjugate of A(ω, k),

A(−ω, k) = A∗(ω, k)

(b) Show that, if A(ω, k) is real, the electricfield can be written as:

E(r, t) = 2∫ ∞

0dω

∫d2k A(ω, k)

× cos[ω(k · r/c − t)]

Answer: Because of the condition andEq. (1.40), the electric field can be writtenas

E(r, t) =∫ ∞

0dω

∫d2k {A(ω, k)

× exp[iω(k · r/c − t)] + A∗(ω, k)

× exp[−iω(k · r/c − t)]} (1.41)

However, if A(ω, k) is real, this can be writtenas:

E(r, t) = 2∫ ∞

0dω

∫d2k A(ω, k)

× cos[ω(k · r/c − t)] (1.42)

(c) [For readers who already know about fre-quency modulation.] Can A(ω, k) in Eq. (1.39)or (1.40) always be taken real?

Hint: Consider frequency modulation (FM).

Problem 1.15(a) Write an expression for the magnetic fieldgiven expression (1.39) for the electric field.

Hint: Use Faraday’s law, and write B(r, t) interms of its Fourier decomposition in the form:

B(r, t) =∫

dω

∫d2k B(ω, k)

× exp[i(ωk · r/c − ωt)]

Answer:

B(r, t) =∫

dω

∫d2k k × A(ω, k)

× exp[i(ωk · r/c − ωt)]

(b) Write an expression for the magnetic fieldgiven the electric field written in terms of asuperposition of positive frequency waves.

Answer: Using the form given inEq. (1.41), we find:

B(r, t) =∫ ∞

0dω

∫d2k k × {A(ω, k)

LIGHT IN VACUUM 29

× exp[iω(k · r/c − t)] + A∗(ω, k)

× exp[−iω(k · r/c − t)]}

(c) For A(ω, k) is real, write this expressionin terms of cosines of ω(k · r/c − t).

Answer:

B(r, t) = 2∫ ∞

0dω

∫d2k k × A(ω, k)

× cos[ω(k · r/c − t)]

Problem 1.16(a) For a monochromatic plane wave, what isthe magnitude of the magnetic field in gauss,given the magnitude of the electric field instatcoulombs/cm2? What can you say aboutthe relative magnitudes of the magnetic andelectric fields for a superposition of monochro-matic waves?

Hint: Take E(r, t) = E0 cos[ω(k · r/c − t)].B(r, t) must also be of the same func-tional form, i.e. B(r, t) = B0 cos[ω(k · r/c −t)]. Substitute these forms into Eq. (1.32), anddetermine B0.

(b) Carry out the same calculation using SIunits.

(c) Show that the amplitude vector A(ω, k)

in Eq. (1.39) must be perpendicular to k invacuum.

Hint: Use Coulomb’s law.

(d) Why is the term ‘transverse field’ isused in describing the electromagnetic radia-tion field?

Hint: Substitute E(r, t) = E0 cos[ω(k · r/c −t)] and B(r, t) = B0 cos[ω(k · r/c − t)] intoFaraday’s and Ampere’s laws and determinethe geometric relation between k, E0 and B0.‘Transverse’ means perpendicular to the direc-tion of propagation.

Problem 1.17Draw the electric and magnetic field vectors ofa single frequency wave of angular frequencyω = 2πc/λ propagating in the z direction, withthe electric vector pointing in the x direction,as a function of z for three different times,t = λ/(8c), t = λ/(4c), and t = λ/(2c).

Hint: see the figure below.

λ

λ

E

E

B

By

y

x

z

z

Polarizationdirection, x

Direction ofpropagation

Initialinstant

A shorttime tlater

ct

In a homogeneous isotropic medium with refrac-tive index n, the basic propagation equation for theelectric field E is given by:

∇2E − n2

c2

∂2

∂t2E = 0 (1.43)

This wave equation can be derived using Fara-day’s and Ampere’s laws, together with the consti-tutive relations, B = µH and D = E(x, t) + 4πP,to obtain

∇ × ∇ × E + µ

c2

∂2

∂t2E = −4πµ

c2

∂2

∂t2P

×{∇ × ∇E + µ/µ0

c2

∂2

∂t2E = −µ/µ0

ε0c2

∂2

∂t2P}

(1.44)


The steps in obtaining Eq. (1.39) are: take the curl

of ∇ × E = −1

c∂tB, substitute B = µH into the

resulting equation, use

∇ × H =(

4π

cJ + 1

c∂tD)

in the resulting equation, and substitute D = E +4πP in the resulting equation. Then, substituting

P = χE = ε − 1

4πE = n2/µ − 1

4πE

on the RHS of Eq. (1.44), one obtains ∇ × ∇ ×E + n2

c2

∂2

∂t2E = 0. Using Eq. (1.35) and further,

using Eq. (1.30) to show that the second termon the RHS of Eq. (1.35) vanishes, one finallyobtains Eq. (1.43). We shall say much more aboutthe refractive index n, and the susceptibility χ inChapters 2 and 3.

For a single frequency field, E(r, t) = E(r)exp(−iωt) + c.c., the spatial function E(r) satis-fies the Helmholtz equation:

∇2E + k2E = 0 (1.45)

with k2 = (n2ω2)/c2.

1.2.3 Propagation of one componentin one dimension

Let us consider the propagation of one componentof an electromagnetic wave in one spatial dimen-sion. The wave equation for the x component ofthe electromagnetic wave propagating along the z

axis is given by (we use the common shorthandnotation ∂xu = ∂u/∂x , ∂tu = ∂u/∂t , etc.):

∂z2Ex(z, t) − c−2∂t

2Ex(z, t) = 0 (1.46)

The general solution to this equation is:

Ex(z, t) = F(z − ct) + G(z + ct) (1.47)

where F and G are any arbitrary functions oftheir arguments. A pulse of light travelling in thepositive z direction can be written in the form:

Ex(z, t) = F(z − ct) (1.48)

An electric field with this form has no componentswhich travel with velocity −c. For this case thefield can be expressed in terms of a superpositionof waves travelling with different wavevectors k

all having velocity c:

F(z, t) =∫ ∞

−∞dkf (k) exp[ik(z − ct)] + c.c.

(1.49)

More generally, for an arbitrary pulse havingcomponents with velocity c and −c, we have

E(z, t) =∫ ∞

−∞dk{f (k) exp[ik(z − ct)]

+ g(k) exp[ik(z + ct)]} + c.c.(1.50)

i.e.

E(z, t) =∫ ∞


+ f ∗(k) exp[−ik(z − ct)]

+ g(k) exp[ik(z + ct)]

+ g∗(k) exp[−ik(z + ct)]} (1.51)

Since E is the x component of the electromagneticfield, and the electromagnetic field represents theforce (per unit charge) on a static test charge, E

must be real. It is for this reason we have includedthe complex conjugate terms in Eqs (1.49), (1.50)and (1.51). Without loss of generality, we canalways take f (−k) = f ∗(k) and g(−k) = g∗(k).With this condition, it is not necessary to includethe complex conjugate terms in Eqs (1.49), (1.50)and (1.51). Hence, Eq. (1.51) can be rewritten inthe form:

E(z, t) = 2∫ ∞

0dk{fe(k) exp[ik(z − ct)]

+ ge(k) exp[ik(z + ct)]} (1.52)

Here fe(k) = f (k) + f (−k) = f (k) + f ∗(k) andge(k) = g(k) + g(−k) = g(k) + g∗(k) are realeven functions of k. Equation (1.52) has theappealing property of involving only positivefrequencies ω = kc. Physically, we tend to think oflight having only positive frequencies. Sometimes,a factor of 1/2 multiplies the RHS of Eq. (1.49)

LIGHT IN VACUUM 31

and (1.50). This convention is quite common [itwas used in the discussion following Eq. (1.39)].With this convention, the factor of 2 in Eq. (1.52)would be cancelled.

Problem 1.18(a) Why is it permissible to take f (−k) =f ∗(k) and g(−k) = g∗(k) without loss ofgenerality?

(b) Demonstrate that Eq. (1.52) follows fromEq. (1.51) using the above conditions.

Electromagnetic propagation in vacuum is sim-ply a linear superposition of the propagation of dif-ferent frequencies, and this superposition does notchange as a function of time. In matter, this neednot be the case. Dispersion of the medium (i.e.the frequency dependence of the refractive indexof the medium) causes different frequency compo-nents to travel at different speeds. Moreover, in adissipative medium, the radiation is attenuated as ittravels through the medium, and different frequen-cies may be attenuated at different rates. Moreover,nonlinear propagation effects may also occur.

In vacuum, let us take an initial light pulse,characterized by E(z, 0), and see what thispulse evolves into at later times. Let us writeEq. (1.50) as:

E(z, t) =∫ ∞


+ g(k) exp[ik(z + ct)]} (1.53)

where the real part of the field is understood inEq. (1.53) and in what follows. At t = 0, we have

E(z, 0) =∫ ∞

−∞dk {f (k) + g(k)} exp[ikz] (1.54)

Now, from Eq. (1.53) we also have:

∂

∂tE(z, t)|t=0

=∫ ∞

−∞dk{ikf (k) + ikg(k)} exp[ikz] (1.55)

∂

∂zE(z, t)|t=0

=∫ ∞

−∞dk{−ickf (k) + ickg(k)} exp[ikz]

(1.56)Multiplying the last equation by i/c, and addingto −i multiplied by the previous equation, weconclude that:

− i∂

∂zE(z, 0) + i

c

∂

∂tE(z, t)|t=0

=∫ ∞

−∞dk2kf (k) exp[ikz] (1.57)

Similarly, we can easily show that:

− i∂

∂zE(z, 0) − i

c

∂

∂tE(z, t)|t=0

=∫ ∞

−∞dk2kg(k) exp[ikz] (1.58)

Therefore, from the Fourier theorem, f (k) andg(k) can be extracted:

f (k) = (4πk)−1∫ ∞

−∞dz

{−i

∂

∂zE(z, 0)

+ i

c

∂

∂tE(z, t)|t=0

}exp[−ikz] (1.59)

g(k) = (4πk)−1∫ ∞

−∞dz

{−i

∂

∂zE(z, 0)

− i

c

∂

∂tE(z, t)|t=0

}exp[−ikz] (1.60)

The temporal dependence of the field is thenobtained by substituting these expressions intoEq. (1.53). If g(k) = 0, then

∂

∂zE(z, 0) = −1

c

∂

∂tE(z, 0) (1.61)

and

f (k) = (2πk)−1∫ ∞

−∞dz

{−i

∂

∂zE(z, 0)

}

× exp[−ikz] = (2π)−1∫ ∞

−∞dzE(z, 0)

× exp[−ikz] (1.62)


Note that E(z, t) can be written in terms of anintegral over frequency:

E(z, t) =∫ ∞

−∞dω{f (ω) exp[iω(z/c − t)]

+ g(ω) exp[iω(z/c + t)]} (1.63)

where

f (ω) = (4πω/c)−1∫ ∞

−∞dz

{−i

∂

∂zE(z, 0)

+ i

c

∂

∂tE(z, t)|t=0

}exp[−iωz/c]

(1.64)

g(ω) = (4πω/c)−1∫ ∞

−∞dz

{−i

∂

∂zE(z, 0)

− i

c

∂

∂tE(z, t)|t=0

}exp[−iωz/c]

(1.65)

The slowly varying envelopeIt is often convenient to write the electric field ofa light pulse propagating in the z direction in theform:

E(z, t) = A(z, t) exp[ik(z − ct]) + c.c. (1.66)

The (generally) complex function A(z, t) is theslowly varying envelope of the electric field.Substitution of Eq. (1.66) for F(z, t) into the waveequation yields equations for the amplitude A(z, t):

∂z2A(z, t) + 2ik∂zA(z, t) − c−2[∂t

2A(z, t)

− 2ikc∂tA(z, t)] = 0

If the amplitude A(z, t) varies slowly in time andspace, i.e.,

|∂z2A(z, t)| 2π |∂zA(z, t)|/λ and |∂t

2A(z, t)| ω|∂zA(z, t)| (1.67)

we obtain the following equation for the slowlyvarying envelope A(z, t):

∂zA(z, t) + c−1∂tA(z, t) = 0 (1.68)

This approximation (i.e. dropping the secondderivative terms of the amplitude), called theslowly varying envelope approximation, is mostuseful not for propagation in vacuum, wherethe general solution to the wave equation isknown, but rather in matter where a source termappears on the RHS of the wave equation (thissource term is proportional to the polarization andis often proportional to the electric field). Theslowly varying envelope approximation is usefulbecause it alleviates the necessity for propagatingthe electric field over distances that are smallcompared with the wavelength or times smallcompared with the optical period. The envelopeof the field varies substantially over distancesand times that are very large compared with thewavelength and period. Therefore, solution forthe envelope of the field can be performed muchmore economically. In any case, it is important tounderstand the nature of this approximation, evenfor propagation in a vacuum. The general solutionto Eq. (1.68) is A(z, t) = f (z − ct), where f isan arbitrary function of its argument. However,we know that f (·) is a slowly varying functionof its argument when the slowly varying envelopeapproximation is valid, and in particular we knowthat, over a temporal duration of 2π/ω andover a spatial region 2πc/ω, the function f isnearly constant. For example, the electric field ofpulse of light with a Gaussian envelope can bewritten as:

E(z, t) = A0 exp

[− (z − ct)2

2σ 2

]× exp[ik(z − ct)] + c.c. (1.69)

where the slowly varying envelope

A(z, t) = A0 exp

[− (z − ct)2

2σ 2

]

varies slowly over a temporal duration of theperiod 2π/ω and over a spatial region of the wave-length 2πc/ω provided the width of the Gaus-sian, σ , is large compared with the wavelength2πc/ω.

LIGHT IN VACUUM 33

Problem 1.19The form of the electric field in Eq. (1.69) iscertainly a solution of the full wave equation,since it is a function of z − ct . Write a moregeneral expression for the field which satisfiesthe initial condition

E(z, 0) = A0 exp

[− z2

2σ 2

]exp[ikz] + c.c.

and satisfies the wave equation.

Hint: Consider a component of the wave whichpropagates in the −z direction as well as a com-ponent which propagates in the +z direction.

The intensity of the field given in Eq. (1.66)is given by [see Section 1.2.7, Eq. (1.120) andProblem 1.26]:

I (z, t) = nc

2π|A(z, t)|2{

I (z) = 2n

(ε0

µ0

)1/2

|A(z, t)|2}

(1.70)

Here has been included the factor n, indicatingthe index of refraction of the medium (seeSection 2.3); for vacuum, n = 1. In Gaussian units,I has units of erg/cm2/s and A (i.e. the fieldstrength) has units of statvolts/cm. In SI units, I

has units of J m−2 s−1 and A (i.e. the field strength)has units of V m−1. The intensity is the energy fluxin units of energy per unit second per unit area. InSection 1.2.7 the intensity will be related to thePoynting vector whose magnitude is the intensity.

It is easy to show, using Eq. (1.68), that theequation of motion for intensity I (z, t) in vacuum,where no absorption, emission and dispersion(frequency dependent velocity) are present, isgiven by:

∂t I (z, t) + c∂zI (z, t) = 0. (1.71)

To do so, we simply multiply Eq. (1.68) byA∗(z, t), take the complex conjugate of Eq. (1.68)

and multiply it by A(z, t), add the two resultingequations and note that ∂t |A(z, t)|2 = ∂tA

∗(z, t)A(z, t) + A∗(z, t)∂tA(z, t), and similarly for ∂z.

It is often convenient to define a time likevariable, τ = t − z/c, and a space like variableξ = z, to describe a pulse of light propagatingalong the z-axis with velocity c, and to write thefield in terms of these two variables instead of thevariables t and z, i.e.

E(z, t) = G(ξ, τ) (1.72)

What does the wave equation, ∂z2E(z, t) − c−2∂t

2

E(z, t) = 0, look like in terms of the variables ξ

and τ? Let us calculate ∂z and ∂t in terms of ∂τ

and ∂ξ to answer this question:

∂z = ∂τ

∂τ

∂z+ ∂ξ

∂ξ

∂z= ∂τ

−1

c+ ∂ξ

∂t = ∂τ

∂τ

∂t+ ∂ξ

∂ξ

∂t= ∂τ (1.73)

Squaring these expressions for ∂z and ∂t andsubstituting the resulting equations into the waveequation, we obtain:

(−2c−1∂ξ∂τ + ∂ξ2)G(ξ, τ ) = 0 (1.74)

Clearly no simplification results in writing thewave equation in terms of these variables. How-ever, let us now consider the slowly varying enve-lope A(z, t) in terms of the variables τ = t − z/c

and ξ = z. What does the slowly varying envelopeapproximation equation,

∂zA(z, t) + c−1∂tA(z, t) = 0 (1.75)

look like in terms of the variables τ and ξ =z? Substituting Eqs (1.73) into the slow varyingenvelope approximation (1.75) yields

∂ξA(ξ, τ ) = 0 (1.76)

This is a much simpler propagation equation thanEq. (1.75). It is an ordinary differential equation in


ξ , with τ appearing as a parameter. The solutionto Eq. (1.76) is of course

A(ξ, τ ) = A(τ) = constant as a function of ξ

(1.77)

This substitution of variables will also prove to bemost useful when the RHS of the wave equation(and therefore the RHS of the slowly varyingenvelope approximation equation) is not zero.

For light propagating in the −z direction, it isconvenient to write the electric field for a pulse oflight propagating in the z direction in the form:

E(z, t) = C(z, t) exp[ik(z + ct)] + c.c. (1.78)

C(z, t) satisfies the slowly varying envelopeapproximation equation:

∂zC(z, t) − c−1∂tC(z, t) = 0 (1.79)

The general solution to this equation is

C(z, t) = f (z + ct) (1.80)

We can consider the slowly varying envelopeapproximation in terms of the variables τ = t +z/c and ξ = z. The equation for the slowly varyingenvelope approximation using these variables isobtained with the aid of the equations:

∂z = ∂τ

∂τ

∂z+ ∂ξ

∂ξ

∂z= ∂τ

1

c+ ∂ξ

∂t = ∂τ

∂τ

∂t+ ∂ξ

∂ξ

∂t= ∂τ (1.81)

and we obtain(∂τ

1

c+ ∂ξ

)C(ξ, τ ) − c−1∂τC(ξ, τ ) = 0

i.e.∂ξC(ξ, τ ) = 0 (1.82)

1.2.4 Phase and group velocityof a light pulse

For an arbitrary pulse of light travelling in thepositive z direction, one can define two velocities,

the group velocity and phase velocity of the pulse.The electric field of the light pulse, E(z, t), can bewritten in terms of these velocities in the form:

E(z, t) = C(z − vgt) exp[−iω(t − z/vp)] + c.c.(1.83)

The phase velocity, vp, is the velocity of thetrajectory of points for which the phase remainsconstant, i.e. for z/t = vp the phase ω(t − z/vp)

in Eq. (1.83) remains zero. The group velocity ofthe pulse is the velocity at which the amplitudeof the pulse remains constant, i.e. the velocity atwhich the peak of the pulse moves through themedium.

Let us consider a light pulse with centralfrequency ω0,

E(z, t) = A(z, t)

= exp[−iω0(t − z/c)] + c.c.(1.84)

For example a pulse with a Gaussian envelope,

A(τ) = A0 exp[−(τ − t0)2/2σ 2] (1.85)

has an electric field which takes the form

E(z, t) = A0 exp[−(τ − t0)2/2σ 2] exp[−iω0τ ]

+ c.c. (1.86)

where τ = t − z/c, or in terms of z and t ,

E(z, t) = A0 exp[−(t − z/c − t0)2/2σ 2]

× exp[−iω0(t − z/c)] + c.c. (1.87)

By comparing Eq. (1.87) with the form inEq. (1.83) we immediately see that the phasevelocity is given by c, and the group velocity isalso c.

Problem 1.20For a Gaussian distribution in frequency

E(0, ω) = A0

√2πσ 2 exp

[−1

2σ 2(ω − ω0)

2

]

LIGHT IN VACUUM 35

show that the Fourier transform of this functionat z = 0 is

E(0, t) = A0 exp[−t2/2σ 2] exp[−iω0t]

The definitions of phase and group velocity aremost useful in describing pulse propagation ina dispersive medium, i.e. a medium with indexof refraction that is frequency dependent, n(ω).They are also important in crystals where theindex of refraction is not isotropic, but dependson the propagation direction of the light field (andits polarization) relative to the crystal axes. Weshall pause here to compare pulse propagation invacuum with that of the spreading of a pulse asit propagates in a dispersive medium which ispresent in a region z ∈ [0, L]. We can consider,for example, an incident pulse which for z < 0propagates in vacuum and takes the form:

E(z, t) = (2π)−1∫ ∞

−∞dωE(ω) exp{i[ωz/c − ωt]}

(1.88)

which, at z = 0, the boundary of a dispersivemedium, is of Gaussian form,

E(0, t) = (2π)−1∫ ∞

−∞dωE(ω) exp[−iωt]

(1.89)

Propagation for z > 0 is through the dispersivemedium with index of refraction n(ω). The timeand space dependence of the pulse as it propagatesthrough this medium is given by:

E(z, t) = (2π)−1∫ ∞

−∞dωE(ω)

× exp{i[n(ω)ωz/c − ωt]}= (2π)−1

∫ ∞

−∞dωE(ω)

× exp{i[k(ω)z − ωt]} (1.90)

where the wavevector k(ω) is given by

k(ω) = n(ω)ω/c (1.91)

Upon expanding this integral about ω = ω0 andretaining only the linear term in (ω − ω0) weobtain

E(z, t) = exp{i[k(ω0)z − ω0t]}(2π)−1

×∫ ∞

−∞dωE(ω)

× exp{−i[t − dk(ω0)/dω z](ω − ω0)}= exp{i[n(ω0)ω0z/c − ω0t]}

× E(0, t − dk(ω)/dω z)

× exp[iω0(t − dk(ω)/dω z)]. (1.92)

The first term in Eq. (1.92) is the plane waveat the central frequency, and the product of thesecond and third terms is the envelope. Uponcomparing (1.92) with Eq. (1.83), we determinethat phase and group velocities are given by:

vp = ω0/k(ω0) = c/n(ω0) (1.93)

vg = 1

dk(ω0)/dω= c

n(ω0) + dn

dω(ω0)ω0

(1.94)

The expression for the group velocity can beunderstood heuristically as follows: the phase ofthe wave is φ = [k(ω)z − ωt] = [n(ω)ωz/c − ωt]and the requirement for the group velocity isthat there be no change in the phase with fre-quency, dφ/dω = 0. Using the expression φ(ω) =n(ω)ωz/c − ωt , and taking the derivative withrespect to ω we find dn/dω ωz/c + nz/c − t = 0,which, when written in the form z = vgt , yieldsthe expression in Eq. (1.93) for vg.

We shall see in Chapter 2 that the energy of apulse with central frequency ω0 propagates withthis group velocity.

Problem 1.21(a) For E(ω) of the previous example the formof the field for arbitrary z becomes:

E(z, t) = (2π)−1√

2πσ 2

∫ ∞

−∞dω


× exp

[−1

2σ 2(ω − ω0)

2

]× exp[in(ω)ωz/c − iωt]

The integrand is sharply peaked about ω0 byvirtue of the Gaussian term. Show that whenthe integrand is expanded about ω0 the fieldcan be written as:

E(z, t) = (2π)−1√

2πσ 2

× exp[in(ω0)ω0z/c − iω0t]∫ ∞

−∞dω exp

[−1

2σ 2(ω − ω0)

2

]

× exp

{i

[dn

dω(ω0)ω0z/c

+ n(ω0)z/c − t

](ω − ω0)

}

Show that the integral over frequency can beperformed analytically, to obtain:

E(z, t) = exp

(−{[

n(ω0) + dn

dω(ω0)ω0

]

× z/c − t

}2

/σ 2

)exp[in(ω0)ω0z/c − iω0t]

using the formula [see Gradshteyn andRyzhik [7], Eq. (3.323.2)]:∫ ∞

−∞dt exp(±qt − p2t2)

= exp

(q2

4p2

)√π

p2[Re(p) > 0]

(b) Derive the second equation in Eq. (1.92)by showing that:

(2π)−1∫ ∞

−∞dωE(ω)

× exp{−i[t − dk(ω)/dωz](ω − ω0)}= E(0, t − dk(ω)/dωz)

× exp[iω0(t − dk(ω)/dωz)]

Quite generally, the group and phase velocitiesof a pulse of light propagating through a dispersivemedium in three dimensions can be obtained ifthe dispersion relation, ω(k), i.e. the frequency vswave vector, is known. The general superpositionof plane waves in a dispersive medium is given by:

E(r, t) = (2π)−3∫ ∞

−∞d3kE(k)

× exp{i[k · r − ω(k)t]} (1.95)

where the pulse shape at t = 0 is given by

E(r, t = 0) = (2π)−3∫ ∞

−∞d3kE(k) exp{ik · r}

(1.96)

Expanding the exponential in Eq. (1.95) aboutthe value k0 for which E(k) peaks, i.e. where∇kE(k0) = 0, we find

E(r, t) = (2π)−3 exp{i[k0 · r − ω(k0)t]}×∫ ∞

−∞d3kE(k)

× exp{i(k − k0) · [r − ∇kω(k0)t]}E(r, t) = exp{i[k0 · r − ω(k0)t]}

× {E[r − ∇kω(k0)t], t = 0}× exp{−ik0 · [r − ∇kω(k0)t]} (1.97)

where {E[r − ∇kω(k0)t], t = 0} exp{−ik0 · [r −∇kω(k0)t]} is the slowly varying envelope. Thus,comparing with Eq. (1.83) we find that the phaseand group velocities are given by:

vp = k0ω(k0)/|k0| (1.98)

vg = ∇kω(k0) (1.99)

In the one-dimensional case, these equationsare identical to Eqs (1.93) and (1.94). Generally,however, the dispersion relation, ω(k), is notdirectly available for propagation in crystals orother non-isotropic media. Instead, the indexof refraction as a function of frequency andwavevector direction is the quantity which can be

LIGHT IN VACUUM 37

directly obtained either from experimental data (orfrom calculation).

In non-isotropic dispersive media, such as mostcrystals or matter in the presence of external staticfields, there are two indexes of refraction functionscorresponding to two different orthogonal polar-izations for every value of ω and s, n(ω, s), wheres is a unit vector indicating the direction of thewavevector k. In a medium with an index of refrac-tion which is a function of frequency and directionof the wavevector,

k = (ω/c)n(ω, s)s (1.100)

The phase and group velocities of a pulse (orwavepacket) of light with central frequency ω anddirection s can be determined as follows. The phasevelocity of a pulse of light with central frequencyω in a medium with index of refraction n(ω, s)is given by Eq. (1.98) as vp = sc/n(ω, s). For thegroup velocity we need to evaluate the gradient,∇kω(k). Taking the magnitude of both sides ofEq. (1.100) and rearranging, we find:

ωn(ω, s) = kc (1.101)

The gradient of Eq. (1.101) with respect tok yields

∇kωn(ω, s) + ω∇kn(ω, s) = cs (1.102)

and this equation can be used to determinethe group velocity ∇kω. Depending upon thetype of crystal symmetry, either one or bothof the polarizations has an index of refractionwhich depends on the direction of propagations. We first consider the case of a uniaxial crys-tal where one index of refraction correspond-ing one of the two possible polarization statesis independent of direction and the other indexof refraction depends upon frequency and theangle θ between the propagation direction s andthe optical axis of the crystal (see Section 2.4.2).The gradient ∇kn(ω, s) appearing in Eq. (1.102)can be evaluated by noting that ∇kn(ω, s) =

[∂n(ω, θ)/∂ω∇kω + ∂n(ω, θ)/dθ∇kθ ]. HenceEq. (1.102) can be written as:

∇kω[n(ω, θ) + ω∂n(ω, θ)/∂ω]

= −ω∂n(ω, θ)/dθ∇kθ + cs (1.103)

We can now evaluate ∇kθ by noting thatk = nω/c (sin θ cos φ, sin θ sin φ, cos θ), hencedk/dθ = nω/c (cos θ cos φ, cos θ sin φ, sin θ) =nω/ceθ and the inverse of this quantity is∇kθ . Substituting the expression for ∇kθ intoEq. (1.103) yields:

∇kω[n(ω, s) + ω∂n(ω, s)/∂ω]

= −c/n dn(ω, θ)/dθeθ + cs (1.104)

and upon rearranging we finally obtain

vg = ∇kω(k, s) = c

[n(ω, s) + ω∂n(ω, s)/∂ω]

×[

s − dn(ω, θ)/dθ

neθ

](1.105)

This will be elaborated in Section 2.4.2.

Problem 1.22Find an expression for the group velocity in abiaxial medium where both indices of refractioncorresponding to the two possible polarizationsdepend on n(ω, θ, φ).

Answer:

∇kωn(ω, θ, φ) + ω[∂n(ω, θ, φ)/∂ω∇kω

+ ∂n(ω, θ, φ)/∂θ∇kθ

+ ∂n(ω, θ, φ)/∂φ∇kφ] = cs

∇kω[n(ω, θ, φ) + ω∂n(ω, θ, φ)/∂ω]

= −ω∂n(ω, θ, φ)/∂θ∇kθ

− ω∂n(ω, θ, φ)/∂φ∇kφ

+ cs∂n(ω, θ, φ)/∂θ∇kθ

+ ∂n(ω, θ, φ)/∂φ∇kφ


k = nω/c(sin θ cos φ, sin θ sin φ, cos θ)

∂k/∂θ = nω/c(cos θ cos φ, cos θ sin φ, sin θ)

= nω/c eθ

∂k/∂φ = nω/c(− sin θ sin φ, sin θ cos φ, 0)

= sin θnω/c eφ

ω[∇θn(ω, θ)∇kθ + dn(ω, θ, φ)/dφ∇kφ]

= c/n dn(ω, θ, φ)/dθ eθ

+ c

n sin θdn(ω, θ, φ)/dφ eφ

∇kω[n(ω, s) + ω∂n(ω, s)/∂ω] = cs

−[c/n∂n(ω, θ, φ)/∂θ eθ

+ c

n sin θ∂n(ω, θ, φ)/∂φ eφ

]

1.2.5 Amplitude modulation

Amplitude and frequency modulation of electro-magnetic waves is important for communicationstechnologies; AM and FM radio bands are clearlymarked on our radio receivers. These concepts areintroduced here.

The discussion of amplitude and frequencymodulation (and the discussion of phase andgroup velocities above) is presented in termsof a scalar wave field, E(z, t). However, theseconcepts carry over to a vector field in threespatial dimensions, E(r, t), in an almost trivialway. When the propagation of light in matter istreated in Chapter 2, we shall deal with phase andgroup velocities and the amplitude and frequencymodulation of the three (spatial) dimensionalvector waves.

The electric vector for amplitude modulatedlight can be written in the form:

E(z, t) = A(z, t) exp[ik(z − ct)] + c.c.

with amplitude A(z, t) being a real function thatis modulated, i.e. varies with time (and space),

but the frequency kc appearing in the phase doesnot vary.

For example, consider the electric field of a laseremitting light at several cavity mode frequencies.The dependence of the electric field, E(τ), whereτ = t − z/c, can then be written as:

E(τ) = exp(−iω0τ)A(τ) + c.c. = exp(−iω0τ)

×n∑

j=−n

Aj exp[i(jτ + φj )] + c.c.

(1.106)where we have defined the cavity mode fre-quencies as ωj ≡ ω0 + j, is the mode fre-quency spacing of the cavity, and φj are thephase shifts for the different modes. Let the inputbeam contain three equal intensity modes, i.e.n = 1, with equal amplitudes, A−1 = A0 = A1 =1/3, and equal phases φ−1 = φ0 = φ1 = 0. Theslowly varying envelope is then given by:

A(τ) = [1 + 2 cos(τ)]/3 (1.107)

This is an amplitude-modulated field. In fact, anyform of this sort with arbitrary n and arbitrary realAj is an amplitude modulated field, provided thephases φj are equal.

1.2.6 Frequency and phasemodulation

Frequency- or phase-modulated light has an elec-tric vector given in the form:

E(z, t) = A(z, t) exp[ik(z − ct)]) + c.c. (1.108)

with complex amplitude A(z, t) given by A(z, t) =A0 exp[ig(z − ct)], where g is a function ofits argument (usually a periodic function ofits argument). Thus, the phase and thereforethe effective frequency vary with time, but themagnitude of the electric field is constant.

How does one make such light? When a singlefrequency field propagates through a materialwhose index of refraction varies with time, theoutput field is frequency- (or phase-) modulated.An electro-optic phase modulator is a device

LIGHT IN VACUUM 39

whose index of refraction changes with time due toa time-varying electrical potential impressed acrossthe device. A piezoelectric crystal (e.g. the quartzcrystal in your wristwatch) subjected to a time-varying electrical potential along the direction ofpropagation of the light is often used to make aphase modulator. The resulting electric field hasthe form:

E(τ) = E0{exp(−iω0τ) exp[iF0 cos(τ)] + c.c.}(1.109)

The intensity of the field is constant with time. TheFourier transform of this field is:

E(ω) = E0

∞∑m=−∞

δ(ω − ω0 + m)imJm(F0)

(1.110)

1

0.5

0

−0.50 1 2 3

J n(X

)

X4 5 6 7

J0(x)

J1(x)J2(x) J3(x)

J4(x)

Figure 1.5 Bessel functions Jm(x) vs x for m = 0, 1, 2and 3. The Bessel functions of negative orders are givenby J−m(x) = (−1)mJm(x).

Figure 1.5 shows a plot of the Bessel func-tions Jm(x) vs x for m = 0, 1, 2 and 3. Fornegative orders, the Bessel functions are givenby J−m(x) = (−1)mJm(x). If you have accessto a fast Fourier transform computer subroutine,numerically calculate and plot the power spectrumfor E for various values of F0 and .

Problem 1.23(a) Derive the analytic expression for thespectrum of a phase modulated electric field

using the integral representation of the Besselfunction:

Jm(x) = 1

2πim

∫ 2π

0dϕ exp[ix cos(ϕ) + imϕ]

to derive the expression

exp[ix cos(θ)] =∞∑

m=−∞im exp[imθ ]Jm(x)

and then use this expression in Eq. (1.109).(b) Show that the expressions in (a) can be

transformed into:

Jm(x) = 1

2π

∫ 2π

0dϕ exp[ix sin(ϕ) − mϕ]

exp[ix sin(θ)] =∞∑

m=−∞exp[imθ ]Jm(x)

respectively.

Hint: Make the substitution θ = ϑ − π/2 inpart (a).

(c) Show that Jacobi–Anger identity exp[ixcos(θ)] =∑∞

m=−∞ im exp[imθ ]Jm(x) that youderived in part (a) can also be obtained in termsof the generating function:

exp[x

2(t − 1/t)

]=

∞∑m=−∞

tmJm(x)

Hint: let t = exp[i(θ + π/2)].

It is easy to write a form for an electric field witha linear frequency chirp, i.e. light with frequencychanging linearly with time,

ω(t) = ω0 + Ct (1.111)

For a field with constant amplitude and a linearchirp, the electric field can be written as:

E(t) = E0{exp(−iω0t) exp[−iCt2] + c.c.}(1.112)

Clearly, the frequency of this field is linearlyvarying with time in the form given by (1.111).


We can consider the more general case of a pulseof light whose frequency is linearly chirped. AGaussian pulse form with a linear frequency chirpcan be written as:

E(t) = E0{exp(−iω0t)

× exp(−t2/2σ 2 − iCt2) + c.c.} (1.113)

Figure 1.6 shows the amplitude and the instanta-neous frequency, ω(t), of such a pulse. The Fouriertransform of this field is easily calculated usingthe expression given in Problem 1.21. Thus theFourier transform of Eq. (1.113) is:

E(ω) = E0

∫ ∞

−∞dt exp(−iω0t)

× exp(−t2/2σ 2 − iCt2) exp(iωt) + c.c.

= E0 exp

[−1

2(σ 2 + i2C−1)(ω − ω0)

2

]

×√

2π(σ 2 + i2C−1) + c.c. (1.114)

Problem 1.24Derive this expression for the Fourier transformof a Gaussian pulse with a linear frequencychirp given the expression in Problem 1.21from Gradshteyn and Ryzhik [7].

Pulse shaping and characterizationTechniques to shape short optical pulses arereadily available today. By ‘shape a pulse’ wemean control the amplitude and phase of anoptical pulse. The concept of pulse shaping wasfirst introduced in Weiner et al. [8]. The inputinto pulse shapers is lasers that provide shortpulses with a given spectral bandwidth and centralwavelength in the visible and infrared regions

Amplitude(a)

Timetp

Frequency w(t)

w(t) = w0 + (dw/dt) (t − tp)

Timetp

Grating Grating

Inputpulse

Output pulse

Lens

(b)

LensPhase and Amplitude mask

f f f f

Figure 1.6 (a) Amplitude and instantaneous frequency ω(t) of a pulse whose electric field is given by

E(t) = E0 exp

[− i

2

dω

dt(t − tp)

2 − (t − tp)2

2σ 2

]exp(−iω0t) + c.c.

Here the center of the pulse occurs at t = tp and at this time, ω(tp) = ω0. (b) Schematic arrangement of a pulseshaper for short optical pulses.

LIGHT IN VACUUM 41

of the spectrum (see Chapter 7). For example,one of the most common short pulse lasersused today is the titanium–sapphire laser. Thecentral wavelength and the spectral bandwidthof such lasers can be readily controlled, andpulses of desired waveforms can be formed toprovide input pulses to a pulse shaper device. Apulse shaper modifies the optical field originatingfrom the laser by filtering its spatially dispersedfrequency components. It is composed of (1) apair of diffraction gratings, (2) a pair of lensesand (3) a spatial filter, sometimes called a mask,that applies amplitude and phase modificationsto different spectral components of the pulse[see Figure 1.6(b)]. The first grating and lensspatially separate the different spectral componentsof the pulse in such a fashion that different ω

components of the light are imaged on differentspatial positions of the mask. The mask canchange the amplitude and/or the phase of thedifferent ω components of the pulse. The secondlens and grating combine the spectrally filteredcomponents and produce the shaped pulse. Thegratings and the mask are in the focal planeof lenses. Programmable pulse shapers usingcomputer-controlled liquid crystal or acousto-optic spatial light modulators as masks can bemade; these devices can even be controlled usinglearning algorithms to self-optimize specific pulsecharacteristics.

Short optical pulses can be characterized usinga technique called ‘frequency-resolved opticalgating’, or FROG, by interacting one or morepulses in a nonlinear medium. The technique wasfirst pioneered by Rick Trebino and colleagues inthe 1990s. One pulse forms a ‘gate’ that lets a timeslice of the other pulse pass to a spectrometer. Thissignal pulse is spectrally resolved and recorded asa function of the delay between the input pulse andthe gate. This record is called a ‘FROG trace’; itis a plot of signal intensity vs both frequency andtime. The temporal and spectral profile of the inputpulse (intensity and phase) is obtained from theFROG trace using two-dimensional phase-retrievalmethods.

1.2.7 Energy, momentum and angularmomentum of electromagneticwaves

We now develop expressions for the energy,momentum and angular momentum densities ofan electromagnetic field. These are fundamentalquantities for describing the physics of matter, andthey are just a fundamental for electromagneticfields.

In this section we retain D and B as distinctfrom E and H. However, if we explicitly considerall the charge and current density and the full(microscopic) electric and magnetic fields arisingfrom these particles (without carrying out averagesto obtain the macroscopic mean electric andmagnetic fields), then D and B can be replacedby E and H in what follows. The currentdensity J(r, t) should be interpreted as the full(microscopic) current density and ρ(r, t) as thefull (microscopic) charge density, i.e. the chargeand current densities are given by the expressions:

ρ(r, t) =∑

1

qiδ[r − r(t)]

andj(r, t) =

∑1

qiviδ[r − r(t)]

The work performed on a particle experiencing aforce F(r) is given by the expression

U =∫

dr · F(r) =∫

dtv(t) · F(t) (1.115)

and therefore the work per unit time, or power, isgiven by dU/dt = v(r, t) · F(r, t). For a collectionof charged particles with charge density ρ(r, t)and current density J(r, t) = ρ(r, t)v(r, t), thisexpression can be written as:

dU/dt =∫

d3rJ(r, t) ·[

E(r, t) + v(r, t)c

× B(r, t)]

=∫

d3rJ(r, t) · E(r, t)


{dU/dt =

∫d3rJ(r, t) · [E(r, t)

+ v(r, t) × B(r, t)]

=∫

d3rJ(r, t) · E(r, t) in SI units

}(1.116)

Problem 1.25(a) Transform the sum over particles in theexpression for the rate of change of energy,

dU/dt =∑

1

vi · Fi

=∑

1

qiv(ri ) ·[

E(ri ) + v(ri )

c× B(ri )

]

into the integral over space given in the firstpart of Eq. (1.116).

(b) Why does the term originating from themagnetic force not contribute to the rate ofchange of energy?

Using Ampere’s law, Eq. (1.116) can be writ-ten as:

dU/dt = 1

4π

∫d3r(c∇ × H − ∂tD) · E

{dU/dt =

∫d3r(∇ × H − ∂tD) · E

}(1.117)

which can be rewritten using the identity ∇ ·(E × H) = H · (∇ × E) − E · (∇ × H) as

dU/dt = 1

4π

∫d3r[−c∇ · (E × H)

+ cH · (∇ × E) − ∂tD · E]{dU/dt =

∫d3r[−∇ · (E × H)

+ H · (∇ × E) − ∂tD · E] in SI units

}

Now, using Faraday’s law to rewrite the secondterm, we obtain

dU/dt =∫

d3rJ ·E = 1

4π

∫d3r[−c∇ · (E×H)

− H · ∂tB − ∂tD · E]{dU/dt =

∫d3r[−∇ · (E × H)

− H · ∂tB − ∂tD · E] in SI units

}(1.118)

Assuming the material permittivity and permeabil-ity are constant in time, the last two terms can bewritten as −1

2 ∂t (H · B + D · E) to finally obtain theexpression:∫

d3r[

J · E + 1

8π∂t (H · B + D · E)

+ c

4π∇ · (E × H)

]= 0

{∫d3r[

J · E + 1

2∂t (H · B

+ D · E) + ∇ · (E × H)

]= 0

}(1.119)

The first term is the electromagnetic power densitydue to the interaction of the charged matterwith the electromagnetic field, the second termis the power density of the electromagnetic field,and the last term is the power density escapingthe volume element due to the propagation ofthe electromagnetic field out of the volume. Weconclude that the energy density of the radiationfield, u(r, t), is given by:

u(r, t) = 1

8π[D(r, t) · E(r, t)+H(r, t) · B(r, t)]{

u = 1

2[D · E + H · B]

}(1.120)

and the energy flux (whose magnitude is called theintensity I (r, t) of the radiation field), in units ofenergy per unit second per unit area, is given bythe Poynting vector S,

LIGHT IN VACUUM 43

S(r, t) = c

4πE(r, t) × H(r, t)

{S = E × H} (1.121)

For the microscopic field case, the integrand ofEq. (1.119) can be written as:

∂

∂t

(E2

8π+ B2

8π

)+ ∇ · S = −J · E

{∂

∂t

(E2

2+ B2

2

)+ ∇ · S = −J · E

}(1.122)

This is the Poynting theorem, which states thatthe time rate of change of the electromagneticfield energy density plus the divergence of theelectromagnetic flux S equals the negative of therate of work done per unit volume on the matterby the electromagnetic field. It is directly relatedto the radiation pressure of the electromagneticfield. We shall see in Section 3.5 that light canimpart momentum to a particle when the light isabsorbed by the particle. This will be used fortrapping and cooling of atoms, and for opticaltweezers for holding particles. The law that therate of change of energy density dissipated equalsJ · E, i.e. du/dt = −J · E, is Joule’s law.

Problem 1.26(a) Show that I = c/8π |E(r, t)|2 and in a non-magnetic material with index of refraction n

I = nc/8π |E(r, t)|2(b) For an electric field of the form E(r, t) =

A(r, t) exp(−iωt) + c.c., show that the inten-sity in vacuum is given by I = c/2π |A(r, t)|2and in a non-magnetic material with index ofrefraction n, I = (nc/2π)|A(r, t)|2 when oneneglects the contribution coming from ∂A/∂t

relative to ωA.

Problem 1.27For those familiar with fluid mechanics, relateEq. (1.119) to the continuity equation for thedescription of fluids.

The energy flux of the field is directly relatedto the momentum of the field. A light waveimpinging on a body exerts a pressure on the bodyupon which it is incident. The light pressure wasdiscovered by Maxwell. Similarly, emission of alight wave by a body imparts a recoil equal andopposite to the momentum carried by the lightwave. We shall now show that the momentumdensity of the field is given in vacuum by:

g(r, t) = S(r, t)/c2 = 1

4πcE(r, t) × H(r, t){

g = ε0µ0E × H = 1

c2E × H

}

In matter, the momentum density is given (as weshall soon see) by:

g(r, t) = 1

4πcD(r, t) × B(r, t)

{g = D × B} (1.123)

The rate of change of the momentum exertedon a charged body having charge density ρ

and current density J in the presence of anelectromagnetic field is given by:

dPdt

=∫

d3r[ρE(r, t) + J(r, t)

c× B(r, t)

]{

dPdt

=∫

d3r[ρE(r, t) + J(r, t) × B(r, t)]}

(1.124)We can use Coulomb’s law and Ampere’s law,

ρ = ∇ · D/4π (1.125)

J = 1

4π(c∇ × H − ∂tD) (1.126)

to obtain the expression

dPdt

= 1

4π

∫d3r[∇·D E(r)

+ (c∇ × H − ∂tD)

c× B(r)

](1.127)


Substituting the relationship

−∂tD × B = −∂t [D × B] + D × ∂tB

into this equation, we obtain

dPdt

= 1

4π

∫d3r[∇·D E(r) + ∇ × H × B(r)

+−1

c{∂t [D × B] + D × ∂tB}

]d

dt

(P + 1

4πc

∫d3r[D × B]

)

= 1

4π

∫d3r [∇·D E(r) + ∇ × H

× B(r) + D × −1

c∂tB]

(1.128)

The ith component of the right-hand side (RHS)of this equation can be written as:∫

d3r∂

∂xj

[1

4π

{εEiEj + µHiHj

−1

2(εE · E + µB · B)δij

}](1.129)

Thus, Eq. (1.128) can be written as:

d

dt

(P + 1

4πc

∫d3r[D × B]

)i

=∫

d3r∂

∂xj

Tij

(1.130)where the Maxwell stress tensor Tij is defined as

Tij = 1

4π

[εEiEj + µHiHj

−1

2(εE · E + µB · B)δij

](1.131)

Application of the divergence theorem to the vol-ume integral on the right hand side of Eq. (1.130)gives:

d

dt

(P + 1

4πc

∫d3r[D × B]

)i

=∫

dS Tijnj

(1.132)

where n is the outward normal to the closedsurface. Thus, we identify the momentum densityof the electromagnetic field as the term

g(r, t) = 1

4πcD(r, t) × B(r, t)

appearing on the left-hand side (LHS) of Eq.(1.132). I should point out that the momentumdensity, g, is sometimes called the Minkowskimomentum density, whereas the density

1

4πcE(r, t) × H(r, t)

is called the Abraham density [9]. The momentumdensity, g, is the quantity obtained upon usingthe volume Lorentz force law, Eq. (1.124), for theforce on a particle in a medium.

Problem 1.28(a) Prove that the RHS of Eq. (1.128) can bewritten in the form of Eq. (1.129).

(b) Write Eqs (1.125)–(1.132) using SI units.(c) Why does D × B appear on the RHS of

Eq. (1.132), and not E × H? What does thisimply about the direction of the momentum of aplane wave relative to the vector of the wave ina non-magnetic non-isotropic medium whereinD = εE and B = µH with dielectric constant ε

that is a tensor?

Answer: the momentum of the light field isnot along the direction of the Poynting vectorE × H.

The vectors {B, D, k} form an orthogonaltriad, and the vectors {B, E, S = E × H} forman orthogonal triad (see Figure 2.14). FromFigure 2.14 it is clear that D × B is in the direc-tion of the wavevector k, i.e. the momentumof the light wave is along the wavevector. Thedirection of the energy flow is at an angle rel-ative to the direction of the momentum flow ina non-isotropic medium.

Reviewing our results, we obtained the follow-ing expressions for the energy and momentum of

LIGHT IN VACUUM 45

the electromagnetic field:

U = 1

8π

∫(ε|E|2 + µ|H|2) d3x

{U = 1

2

∫(ε|E|2 + µ|H|2) d3x

}(1.133)

P = 1

4πc

∫(D × B) d3x

{P =

∫(D × B) d3x

}(1.134)

The expressions for the energy and the momen-tum of the electromagnetic field can be written interms of the vector and scalar potentials A andφ using the expressions for the fields in terms ofthese potentials,

B = ∇ × A, E = −∇φ − 1

c

∂A∂t

(1.135)

It is particularly important to do so when a quan-tum description of the field, and the interaction ofthe field with matter, is considered. In what fol-lows, and for quantizing the electromagnetic field(Section 1.2.11), it is most convenient to use theradiation gauge in which the transversality con-dition, ∇ · A = 0, is satisfied [see Appendix B,Eq. (B38)]. That is, we choose the scalar function� such that the scalar potential vanishes and

B = ∇ × A, E = −1

c

∂A∂t

In this gauge, the vector potential can be expandedin plane waves (each wavevector k and polariza-tion α can be thought of as a mode of the electro-magnetic field, and the field is expanded in termsof these modes):

A(r, t) = 1

V 1/2

∑k

2∑α=1

[ckα(t) e(α) eik·r

+ e(α)∗ckα(t)∗ e−ik·r] (1.136)

where V is the volume of the region containingthe electromagnetic field, and where the Fourier

coefficients ck,α(t) have the time dependencegiven by

ckα(t) = ckα e−iωt (1.137)

and therefore

dckα(t)

dt= −iωckα e−iωt

since A satisfies the wave equation in vacuum[see Problem 1.29(a)]. The unit vectors e(α) dependon the propagation direction k, and satisfy thetransversality condition k · e(α) = 0. For a givenk, we can choose e(1) and e(2) such that (e(1),e(2), k/|k|) form a right-handed set of mutuallyorthogonal unit vectors. The Hamiltonian andmomentum of the field are given in terms of thevector potential by:

U = 1

8π

∫(|(1/c)∂A/∂t |2 + |∇ × A|2) d3x

{U = 1

2

∫[ε0|(1/c)∂A/∂t |2

+µ0||∇ × A|2] d3x

}(1.138)

P = −1

4πc2

∫∂A∂t

× ∇ × A d3x

{P = −1

c2

∫∂A∂t

× ∇ × A d3x

}(1.139)

The angular momentum of the electromagneticwave, L, can be written as:

L =∫

r × g(r, t) d3x

= 1

4πc

∫r × (D × B) d3x

{L =∫

r × (D × B) d3r} (1.140)

This is the angular momentum of the field relativeto the origin. In vacuum, Eq. (1.140) reduces to

L = 1

4πc

∫r × (E × B) d3x


The angular momentum can be broken up into apiece which can be identified with the ‘spin’ ofthe photons and the ‘orbital’ angular momentumof the photons. Furthermore, expressions for theangular momentum in terms of the plane waveexpansion of the vector potential in the radiationgauge can be easily derived. We shall not pause todo so here.

Problem 1.29(a) Prove that, in the radiation gauge, thevector potential satisfies the wave equation byfollowing the treatment of Section 1.3.2 withthe charge and current density set to zero andusing the transversality condition instead of theLoretz gauge condition.

(b) Explicitly show that Eq. (1.136) satis-fies the wave equation if Eq. (1.137) is satis-fied.

(c) Derive the following expressions for theenergy and momentum of the field:

U = 1

2π

∑k

∑α

2(ω/c)2ck,α(t)∗ck,α(t)

(1.141)

P = K ′∑k

∑α

ωkck,α(t)∗ck,α(t) (1.142)

Determine the constant K ′.

Answer: we need to calculate ∂A/∂t and∇ × A. Differentiating Eq. (1.136) with respectto time, we obtain:

∂A∂t

= 1

V 1/2

∑k

∑α

[∂ckα(t)

∂te(α) eik·r

+ e(α)∗ ∂ckα(t)∗

∂te−ik·r

]

= i

V 1/2

∑k

∑α

ω[−ckα(t)e(α) eik·r

+ e(α)∗ckα(t)∗ e−ik·r]

Taking the curl of Eq. (1.136), we find

B = ∇ × A = 1

V 1/2

∑k

∑α

ik × [ck,α(t)e(α)

× eik·r − e(α)∗ck,α(t)∗ e−ik·r] (1.143)

We can now substitute these formulas intothe expressions for the energy and momen-tum of the electromagnetic field, Eqs (1.138)and (1.139).

(d) Derive the expressions for Eqs (1.130)–(1.140) in SI units.

(e) Show that the transverse electric field canbe written as

E = −1

c

∂A∂t

= i

cV 1/2

∑k

∑α

ω[ckα(t)e(α)

× eik·r − e(α)∗ckα(t)∗ e−ik·r] (1.144)

The electromagnetic field will be quantized inSection 1.2.11, and this quantization yields explicitquantum mechanical expressions for the Fouriercoefficients, ck,α(t). Substituting these expressionsinto the equations developed in this section forthe energy, momentum and angular momentumof the radiation field yields quantum expressionsfor the energy, momentum and angular momentumof the radiation field in terms of the quantum oflight, called the photon. Before proceeding with thequantization of the radiation field, we discuss thepolarization of light and its coherence properties.

Modes of the radiation field and cavitymodesWe already indicated just before Eq. (1.136) thateach wavevector k and polarization α can bethought of as a mode of the electromagnetic field,and that the field can be expanded in terms ofthese modes. It is convenient for many applicationsto know the number of modes available perunit volume per unit energy. Moreover, sincewe will be dealing with cavities (e.g. lasersare cavities containing a gain medium), it isuseful to also consider the number of modes

LIGHT IN VACUUM 47

of electromagnetic cavities. Since electromagneticcavities can be considered as boxes with boundaryconditions corresponding to the field vanishing(or nearly vanishing) at the boundaries of thecavity, we explicitly consider this case too. Itis not an absolute necessity to consider thistype of boundary condition; in some applications(e.g. in solid-state physics when dealing withcrystal lattices) it is more convenient to considerperiodic boundary conditions. However, the natureof the counting of modes is very similar in thesetwo cases.

We consider a region with dimension Lx ,Ly and Lz, where, as a special example, thesedimensions could all be equal. For periodicboundary conditions, the wavevectors of the fieldmodes of this region (which can be written as e(α)

eik·r) are given by:

k = 2π

(nx

Lx

x + ny

Ly

y + nz

Lz

z)

(1.145)

where nx , ny and nz are integers. The magnitudesof these wavevectors are:

|k| = 2π

(nx

2

Lx2 + ny

2

Ly2 + nz

2

Lz2

)1/2

and the allowed frequencies are

ω = c|k| = 2πc

(nx

2

Lx2 + ny

2

Ly2 + nz

2

Lz2

)1/2

(1.146)

For each allowed wavevector k, there are twopolarizations e(α) perpendicular to k that areallowed (a parallel component is not allowed since∇ · A = 0).

We now calculate the number of modes [i.e. thenumber of integers (nx , ny , nz)] that correspondto frequencies ω2 = c2k2 ≤ ωm

2. The number oflattice points (nx , ny , nz) with

(2πc)2

(nx

2

Lx2 + ny

2

Ly2 + nz

2

Lz2

)≤ ωm

2

is given by the volume of the sphere with(dimensionless) radius

R(ωm) =[( ωm

2πc

)3LxLyLz

]1/3

Multiplying this (dimensionless) volume, (4π/3)

R(ωm)3, by 2 for the two different polarizationsthat are allowed, we obtain the number of modesN(ωm) with c2k2 ≤ ωm

2:

N(ωm) = 24π

3

( ωm

2πc

)3LxLyLz (1.147)

The number of modes per unit volume withinthe frequency interval dω is therefore given bydividing the above equation by the volume LxLyLz

and taking the derivative with respect to angularfrequency:

n(ω) dω = 8π

(2π)3c3ω2 dω (1.148)

The quantity n(ω) is the spectral mode densityof the cavity, or the density of states per unitfrequency.

Let us now consider standing wave modes.These modes, which vanish at the boundaryof the cavity, comprise standing waves (timespolarization factors) and are called ‘cavity modes’.The modes can be written for a rectangularcavity as e(α) sin(kxx) sin(kyy) sin(kzz), with kj =πnj/Lj , where nj are positive integers. Theallowed frequencies are:

ω = c|k| = πc

(nx

2

Lx2 + ny

2

Ly2 + nz

2

Lz2

)1/2

The number N(ωm) of modes with frequency ω2 =c2k2 ≤ ωm

2 is then given by twice (for the twopolarizations) the number of positive lattice points(nx , ny , nz) with

(πc)2

(nx

2

Lx2 + ny

2

Ly2 + nz

2

Lz2

)≤ ωm

2


which in turn is given by the volume of thespherical section with (dimensionless) radius

R(ωm) =[(ωm

πc

)3LxLyLz

]1/3

with x, y, z > 0:

N(ωm) = 21

8

4π

3

(ωm

πc

)3LxLyLz

This is the same result as in Eq. (1.147). The spec-tral mode density is again given by Eq. (1.148).

The modes of an electromagnetic field inside acavity are classified not only by their frequency,but also by their field distribution. Hence themodes of a cavity are classified by their longitudi-nal character, which specifies their frequency, andby their transverse character, which specifies thefield distribution inside the cavity. We shall takeup this classification in Section 7.5.

Problem 1.30(a) Differentiate Eq. (1.147) with respect tofrequency to obtain the number of modes infrequency interval ω to ω + dω.

Answer:

dN(ω) = 2 V

(1

2πc

)3

4πω2 dω

where V = LxLyLz.(b) Show that the total number of photon

states with momentum spread d3k around k isgiven by

2V

(2π)3d3k

We have considered the density of states of pho-tons in vacuum or in an empty cavity. It is pos-sible to modify the density of photon states byusing artificial structures fabricated in an opticalmaterial with unit cells whose dimensions are com-parable to the optical wavelength. Such structuresare called photonic crystals. If the artificial struc-ture has appropriate symmetry and properties, itcan exhibit a photonic band gap, i.e. a range of

frequencies in which photons cannot propagate.The bandgap is analogous to electron bandgapsin semiconductors (Section 6.5.3). Photonic crys-tals can be made by nanofabrication of a struc-ture which has three-dimensional (or two- or one-dimensional) periodicity in the dielectric constant(as compared with the three-dimensional period-icity in the potential for electrons in semiconduc-tors). These structures can make good reflectorssince photons with energies within the bandgap arenot allowed to propagate inside the photonic crys-tal and are therefore reflected. Photonic crystalsare interesting for optoelectronic devices becausethey make it possible for photons to be con-trolled and confined in small structures (on a wave-length scale).

The above mode counting argument can also beused for massive particles (not only for photons).For non-interacting, identical mass particles in abox, the Hamiltonian

H =∑

i

pi2

2m

has eigenvalues

En =∑

i

h2π2

2m

(nxi

2

Lx2 + nyi

2

Ly2 + nzi

2

Lz2

)

Therefore, the same considerations as describedabove apply. The number of quantum states forone particle with energy ≤ E is given by:

N(E) = 4π

3

(2mE

h2π2

)3/2

LxLyLz

For N particles

NN(E) = C3N

[(2mE

h2π2

)3/2

LxLyLz

]N

where

Cj = 2πj/2

�(j/2)

is a dimensionless constant (see the URL http://mathworld.wolfram.com/Hypersphere.html). The

LIGHT IN VACUUM 49

density of states per unit energy is given bydN(E)/dE and is proportional to E3N/2−1V N .

Gaussian beams in a homogeneousmediumWaves with wavefront normals that make smallangles with the average propagation direction(taken here as the z-axis) are called ‘paraxialwaves’. Gaussian beams are a special case ofparaxial waves. The intensity of a Gaussian beamis concentrated in a small cylinder surroundingthe beam axis. The radius of the cylinder changesslowly (compared with the wavelength of the lightcomprising the beam) with propagation of thebeam in the medium. Let us consider a beam oflight of a single frequency ω,

E(r, t) =∫

dk A(k) exp[i(k · r − ωt)] + c.c.

propagating in a homogeneous medium withrefractive index n. A single frequency paraxialGaussian beam has an electric field of the form

E(r, t) = E(r) exp(−iωt) + c.c.

with the complex scalar amplitude E(r) given bythe product of a plane wave eikz modulated by aslowly varying (complex) envelope (SVE) A(r),

E(r) = A(r) exp(ikz) (1.149)

The SVE A(r) is assumed to be almost constantover a region of size λ, thus the wave can be locallyviewed as a plane wave with paraxial rays aswavefront normals. The complex scalar amplitudeE(x, y, z), satisfies the scalar wave equation [theHelmholtz equation, see Eq. (1.26)]:

∇2E + k2E = 0 (1.150)

In vacuum (a dispersive medium with refrac-tive index n) k = 2π/λ (k = 2πn/λ). Below weexplicitly treat the case of vacuum, but to obtainthe case of a dispersive medium with index n, we

need only replace k by nk and λ by λ/n every-where where k or λ appears in this section. Sub-stituting Eq. (1.149) into Eq. (1.150) we find thatthe SVE A(r) satisfies the equation:

∇⊥2A + 2ik ∂A/∂z + ∂2A/∂z2 = 0 (1.151)

where ∇⊥2 = ∂2/∂x2 + ∂2/∂y2. The slow vari-ation of the SVE A implies that |∂2A/∂z2| k ∂A/∂z, hence the second derivative term inEq. (1.151) can be neglected and we obtain:

∇⊥2A + 2ik ∂A/∂z = 0 (1.152)

This equation is the paraxial Helmholtz equation,or simply the paraxial wave equation. Note thesimilarity of this equation to the time-dependentSchrodinger equation with z replaced by t . Asimple solution to Eq. (1.152) is given by:

A(x, y, z) = A0

q(z)exp

[ikr2

2q(z)

](1.153)

where q(z) is a monic linear function of the formq(z) = z + ξ , and r2 = x2 + y2. Direct substitu-tion of Eq. (1.153) into Eq. (1.152) verifies thatthis form is indeed a solution. If ξ is taken to beimaginary,

q(z) = z − iz0

with real z0, the argument of the exponentialbecomes

ikr2

2q(z)= ikr2

2

1

z − iz0

= ikr2

2

[z

z2 + z02

+ iz0

z2 + z02

](1.154)

The real part of this argument [the second termin the last equality of Eq. (1.152)] determines thebeam waist, while the imaginary part determinesthe phase of the wave and hence the wavefrontcurvature. The beam waist, w0, is defined by theGaussian form exp(−r2/w0

2) for the electric fieldamplitude as the width at which the intensitydrops by a factor of 1/e at the focus, z = 0


[see Eq. (1.159)]. By comparing this form withEq. (1.153) we find:

w02 = 2z0

k⇒ z0 = πw0

2

λ(1.155)

The parameter z0 is the Rayleigh range (moreover,the quantity 2z0 is sometimes called the ‘confocalparameter’). It is convenient to write q(z)−1 interms of the z-dependent radius of curvature R(z)

and the beam radius w(z) as:

1

q(z)= 1

R(z)+ i

λ

πw(z)2(1.156)

where

R(z) = z[1 + (z0/z)2] (1.157)

w(z) = w0[1 + (z/z0)2]1/2 (1.158)

The complex scalar amplitude E(r) can be writtenusing the definitions just introduced as follows:

E(x, y, z) = A0w0

w(z)exp[i tan−1(z/z0)]

× exp(ikz) exp

[i

kr2

2R(z)

]

× exp

[ −r2

w(z)2

](1.159)

The phase tan−1(z/z0) appearing in the exponentof Eq. (1.159) is the phase of the factor q(z)−1

appearing on the RHS of Eq. (1.153). It issometimes called the ‘Guoy phase’. It changes signupon going through the focus. Figure 1.7 showsthe profile of a Gaussian beam E(x, y, z) andgraphically indicates the Rayleigh range z0, thebeam waist w0, the z-dependent beam radius w(z),the z-dependent radius of curvature R(z), and theangular beam spread θ , defined as:

θ = limz→∞

w(z)

z= λ

πw0(1.160)

R(z)

w0w0

w(z)2z0

zq

z0 = pw02/l

w2(z) = w02[1+(z/z0)2]

q = l/(pw0)

√2

Figure 1.7 Profile of a Gaussian beam in a homoge-neous medium. Beam waist, w0, Rayleigh range, z0, andangular beam spread, θ , are shown.

The optical intensity is given by I (r, z) =|E(r, z)|2.

Higher-order modes, whose intensity is notmaximum on the beam axis, are also paraxialsolutions to the wave equation in vacuum or ina homogeneous dispersive medium. Their form isgiven by a product of the Gaussian form developedhere, Eq. (1.157), and Laguerre (for cylindricalsymmetry) or Hermite (for Cartesian coordinates)polynomials, as discussed in Section 7.5.2.

1.2.8 Polarized light

The electric field of monochromatic light propa-gating in the z-direction can be written as:

Ex(z, t) = i E0x cos(kz − ωt + φx)

Ey(z, t) = j E0y cos(kz − ωt + φy)

i.e.

E(z, t) = Ex(z, t) + Ey(z, t)

= i E0x cos(kz − ωt + φx)

+ j E0y cos(kz − ωt + φy) (1.161)

If the phases are functions of space and time (orfunctions of the variable z − ct), i.e.

φx = φx(z − ct) φy = φy(z − ct) (1.162)

LIGHT IN VACUUM 51

or the amplitudes E0x and E0y are functions ofspace and time, the field is not monochromatic withsingle frequency ω, since the Fourier transformof the field yields additional frequency compo-nents. We shall therefore restrict ourselves hereto phases and amplitudes that are constant (oralmost constant for quasi-monochromatic radia-tion). Equation (1.161) can be written in terms ofexponentials:

E(z, t) = Re{i E0x exp[i(kz − ωt + φx)]

+ j E0y exp[i(kz − ωt + φy)]} (1.163)

= Re{[i E0x + j E0y exp i(φy − φx)]

× exp i(kz − ωt + φx)} (1.164)

Linear and circular polarizationThe monochromatic field is linearly polarizedwhen φx = φy = φ. The electric field of the lin-early polarized monochromatic field is given by:

E(z, t) = (i E0x + j E0y) cos(kz − ωt + φ)

(1.165)

The trajectory swept out by the electric field at agiven z as a function of time is a straight line withslope E0y/E0x .

The monochromatic field is said to be circularlypolarized if (a) the phase difference of the x andy components of the electric field is a multipleof π /2,

φy − φx = 2mπ ± π/2 (1.166)

and (b) the amplitudes of the x and y componentsare equal, E0x = E0y = E0. The electric field ofa monochromatic left (right) circularly polarizedfield takes the form:

E(z, t) = E0[i cos(kz − ωt + φ)

+ (±)j sin(kz − ωt + φ)] (1.167)

The direction of polarization at a given value ofz is periodically varying as a function of time.Let us plot this field at z = 0 as a functionof time for various times. At t =φ/ω, E(0, t =φ/ω)=E0 i, at t = (φ + π/4)/ω, E[0, t = (φ +

E

y

x

5

6

1

2

3

Left circularly polarized light

4

E

y

x

3

2

1

6

5

Right circularly polarized light

4

Figure 1.8 Direction of polarization as a function oftime of a right and left circular polarized field. The lightis propagating in the z-direction (out of the plane of thepaper).

π/4)/ω]=E0/√

2 [i − (±)j], at t = (φ + π/2)/ω,E = E0[−(±)j], at t = (φ + π)/ω, E = −E0 i, att = (φ + 3π/2)/ω, E = −E0/

√2 [i − (±)j], and

at time t = (φ + 7π/4)/ω, E = E0/√

2 [−i −(±)j]. These six points are designated as 1, 2, 3,4, 5 and 6 in Figure 1.8, which plots these fieldsfor both left and right circular polarization, look-ing at the field with the wavevector out of theplane of the paper (along the z-direction). Thefield E0[i cos(kz − ωt) + j sin(kz − ωt)] is left cir-cular polarized and the arrows in Figure 1.8 moveclockwise as a function of increasing time, whereasE0[i cos(kz − ωt) − j sin(kz − ωt)] is right circu-larly polarized and the arrows move counterclock-wise (in accordance with the right-hand rule whenone puts one’s right thumb in the direction of the


propagation direction and one’s fingers show thedirection of the polarization direction). Right circu-larly polarized light has positive helicity (the pro-jection of the angular momentum along the wavevector is positive).

Elliptical polarizationLinear and circular polarized light are special casesof the general form:

E(z, t) = i E0x cos(kz − ωt + φx)

+ j E0y cos(kz − ωt + φy) (1.168)

with E0x �= E0y . This monochromatic wave is saidto be elliptically polarized because the electric fieldvector at a given position z maps out an ellipse asa function of time as is evident from the followingconsiderations:

Ex(z, t)/E0x

= cos(kz − ωt + φx) (1.169)

Ey(z, t)/E0y

= cos(kz − ωt + φx) cos(φy − φx)

− sin(kz − ωt + φx) sin(φy − φx)

= Ex(z, t)/E0x cos(φy − φx)

− {1 − [Ex(z, t)/E0x]2}1/2 sin(φy − φx)

(1.170)Therefore, after some manipulation we find:

[Ey(z, t)/E0y − Ex(z, t)/E0x cos(φy − φx)]2

= {1 − [Ex(z, t)/E0x]2} sin2(φy − φx)

×[

Ey

E0y

]2

− 2Ey

E0y

Ex

E0x

cos(φy − φx)

+[

Ex

E0x

]2

= sin2(φy − φx) (1.171)

This is an equation of an ellipse:

(x ′/a)2 + (y ′/b)2 = 1 (1.172)

where the rotation of coordinates

x ′ = x cos α + y sin α (1.173)

y ′ = −x sin α + y cos α (1.174)

has been made, as can be easily seen by comparingthe expression

x2(cos2 α/a2 + sin2 α/b2)

+ 2xy sin α cos α(1/a2 − 1/b2)

+ y2(sin2 α/a2 + cos2 α/b2) = 1 (1.175)

with Eq. (1.170) using the identifications, x ↔ Ex ,y ↔ Ey , a ↔ E0x , and b ↔ E0y . The angle α

made by the principal axis of the ellipse with thex axis is given by:

tan 2α = 2E0xE0y cos(φy − φx)

E0x2 − E0y

2 (1.176)

Clearly, if cos(φy − φx) = 0, then α = 0, and theequation of the ellipse reduces to

[Ey(z, t)/E0y]2 + [Ex(z, t)/E0x]2 = 1 (1.177)

If cos(φy − φx) = 1, i.e. φy − φx is a multiple of2π , the equation of the ellipse degenerates to

Ey(z, t)/E0y = Ex(z, t)/E0x (1.178)

and similarly for odd multiples of π ,

Ey(z, t)/E0y = −Ex(z, t)/E0x (1.179)

These last two cases correspond to linearly polar-ized light. Figure 1.9 shows the trajectory of theelectric field or an elliptically polarized field.

It is easy to specify an optical element whichchanges the polarization of an incident light wave.Such a device is known as a retarder or waveplate. All one needs is a material whose index ofrefraction is different for x polarized light than fory polarized light. Such materials are birefringent

LIGHT IN VACUUM 53

Ey

E0y

E0x

Exa

Figure 1.9 Elliptically polarized light.

(see Section 2.4 for a discussion of birefringentmaterials). An incident electric field of the form:

Ei(z, t) = i E0x cos(kz − ωt) + j E0y cos(kz − ωt)

(1.180)

will, upon traversing a length l of such a material(and emerging from it), take the form

Et (z, t) = i E0x cos(kz − ωt + φx)

+ j E0y cos(kz − ωt + φy) (1.181)

where phases of the transmitted field are

φx = (nx − 1)kl φy = (ny − 1)kl (1.182)

and the relative phase factor φ = (φy − φx)

between the i and i components of the field isgiven by

φ = (φy − φx) = (ny − nx)kl (1.183)

If φ is zero for the initial field, as assumed inEq. (1.180), one can turn this linearly polarizedfield into an arbitrary elliptical polarized fieldgiven a proper length, l, of such a material.

Problem 1.31A quarter wave plate is a retarder that can turnlinearly polarized light into circular polarizedlight.

(a) What value of φ = (φy − φx) is neces-sary to make a quarter wave plate?

(b) Suppose ny = 1.5443 and nx = 1.5534(the ordinary and extraordinary indices ofrefraction of quartz for λ = 589.3 nm). Whatminimum length of material would be needed tomake a quarter wave plate? What other lengthsare possible?

(c) What angle must the polarization of theincident linear polarized light make with the x-axis of the quarter wave plate to turn it intocircularly polarized light?

(d) What lengths of material are necessaryfor a half wave plate?

A wave plate imparts a fixed phase difference(φy − φx) on an incident polarized field. A deviceto impart a variable retardance on an incidentpolarized field is a compensator. This can beaccomplished by adjusting the total thickness, d1,of two wedges of birefringent material whose opticaxes are parallel to one another and perpendicularto the propagation direction of the field. Thesetwo wedges lie on top of a block of fixedthickness, d2, of the same material but with theoptic axis perpendicular to the optic axis of thewedges and also perpendicular to the propagationdirection of the field (see Figure 1.10). This deviceis known as a ‘Soleil compensator’. The phasedifference imparted by the wedges is φ1 = (ny −nx)kd1 and the phase difference imparted by theblock is φ2 = (nx − ny)kd2, so the total phasedifference equals

φ = φ1 + φ2 = (ny − nx)kd1

+ (nx − ny)kd2 = (ny − nx)k(d1 − d2)

(1.184)Since it is variable, the retardance is adjustable.

Babinet compensatorSoleil compensator

45k

kE 45

E

Figure 1.10 Soleil and Babinet compensators.


Also shown in Figure 1.10 is a Babinet com-pensator. Here the two wedges are cut so that theoptic axes in the wedges are mutually perpendic-ular and both are perpendicular to the propagationdirection.

Problem 1.32If quasi-monochromatic light is incident oncrossed polarizers containing a Babinet com-pensator placed between them at an angle of45◦ to their pass planes, what is the expectedpattern that will be observed?

Answer: a series of fringes with black whenφ = 0 and light when φ = π/2.

Unpolarized lightLight from an ordinary light source is not polarized.In a light bulb, for example, the excited state ofthe atoms that radiate by spontaneous emission tothe ground state are oriented in random directionsrelative to one another. Therefore, the emitted radi-ation, which is the sum of the radiation from each ofthe excited atoms, will have polarization which israndom. If the light from the source is dispersedin frequency (for example, by putting it througha prism and selecting only a narrow frequencyrange), the resulting quasi-monochromatic radiationis unpolarized. The electric vector of such light canbe expressed in terms of a phase factor (φy − φx)

which varies very quickly and randomly with timecompared with the time of observation of the field,and amplitudes E0x and E0y whose magnitude alsovaries randomly. Figure 1.11 shows the variation ofthe phase with time over times scales that are longcompared with the period 2π/ω (otherwise the radi-ation would not be quasi-monochromatic), but shortcompared with observation times.

Another way to represent the electric vector ofsuch light is to write it as a sum of electric vectorsover many sources:

E(z, t) =N∑n

{i Enx(z, t) cos[kz − ωt + φnx(z, t)]

+ j Eny(z, t) cos[kz − ωt + φny(z, t)]}(1.185)

φy − φx

π

−π

t

Figure 1.11 Variation of the phase difference φy − φx

of the field with time for unpolarized light.

where the spatial and temporal dependence ofthe amplitudes and phases are slowly varyingcompared with the wavelength and period of thefield. The sum over n of the x component of thefield, averaged over a time that is large comparedwith the period of the field, equals the sum over n

of the y component of the field:

N∑n

〈Enx(z, t)〉 =N∑n

〈Eny(z, t)〉 (1.186)

Moreover, the average over time of the product ofthe phase factors

〈φmi(z, t)φnj (z, t)〉 = 0 for

× m �= n = 1, . . . , N and i, j = x, y, z

(1.187)vanish because of the lack of correlation betweenatom n and atom k. Furthermore, the amplitudesfrom different atoms are uncorrelated:⟨

N∑m�=n

N∑n

Emx(z, t)Enx(z, t)

⟩

=N∑

m�=n

N∑n

〈Emx(z, t)〉〈Enx(z, t)〉 (1.188)

and the phases from different atoms are also uncor-related. Therefore, in determining the intensity ofthe emitted radiation, cross terms involving differ-ent atoms drop out,

〈I (z, t)〉 = c

4π〈E(z, t) · E(z, t)〉

LIGHT IN VACUUM 55

{ε0c

2〈E(z, t) · E(z, t)〉 in SI units

}

= c

4π

⟨N∑m

N∑n

{Emx(z, t)Enx(z, t)

× cos[kz − ωt + φmx(z, t)]

× cos[kz − ωt + φnx(z, t)]

+ Emy(z, t)Eny(z, t)

cos[kz − ωt + φmy(z, t)]

× cos[kz − ωt + φny(z, t)]}⟩

= c

4π

N∑n

〈{Enx(z, t)2 cos2[kz − ωt

+ φnx(z, t)] + Eny(z, t)2

× cos2[kz − ωt + φny(z, t)]}〉 (1.189)

Note that a monochromatic field cannot be unpo-larized; it must have some (in general, elliptical)polarization. We can say that a monochromaticfield is fully polarized. See Born [10] for a com-plete discussion of unpolarized light.

Jones matricesFor determining the effect of polarizers, retardersand other polarization changing optical elementson coherent radiation, it is often convenient towrite the electric field vector of monochromaticradiation in the form:

E(z, t) = Re{[i E0x + j E0y exp i(φy − φx)]

× exp i(kz − ωt + φx)}= Re

{[E0x

E0y exp i(φy − φx)

]

× exp i(kz − ωt + φx)

}(1.190)

In what follows we shall drop the explicit referenceto taking the real part, as well as the exponentialexp i(kz − ωt + φx). Thus, Eq. (1.190) becomes:

E(z, t) =[

E0x


](1.191)

Problem 1.33Write the electric field for linearly polarizedlight and left and right circularly polarized lightin this fashion.

Answer: (E0x

E0y

)

and

E0

(1±i

)

with the + sign for left and the − sign for right.

The intensity of the field is proportional toE(z, t) · E(z, t) averaged over a time that islong compared with the period of the field [theproportionality constant is c/(4π) in Gaussianunits; see Eq. (1.121) for the Poynting vector].Taking the average over a period we obtain

〈E0x(z, t)2〉 + 〈E0y(z, t)

2〉= 1

2[E0x(z, t)2 + E0y(z, t)2]

This can be written as:

1

2[E0xE0y exp i(φy − φx)]

∗

×(

E0x


)(1.192)

What is the effect of passing a quasi-monochro-matic field through a polarizer with the passplane axis along x? The polarizer removes they component of the field. Such an operationcan be represented by a matrix (called a ‘Jonesmatrix’ after R. Clark Jones [11]) operating on theincident electric field vector and turning it into thetransmitted electric field vector:

Et = A Ei (1.193)

or written out explicitly,(Etx

Ety

)=(

a11 a12

a21 a22

)(Eix

Eiy

)


Etx = a11Eix + a12Eiy

Ety = a21Eix + a22Eiy (1.194)

The matrix representing a polarizer with axis alongx is simply

A =(

1 00 0

)

(linear polarizer along x), and the matrix for apolarizer along y is

A =(

0 00 1

)

(linear polarizer along y). Thus, the operation ofthe x and y polarizers can be written as:

Et =(

E0x

0

)=(

1 00 0

)(E0x

E0y

)× (linear polarizer along x) (1.195)

Et =(

0E0y

)=(

0 00 1

)(E0x

E0y

)× (linear polarizer along y) (1.196)

If the beam passes through a polaroid with its passplane at an angle of ±45◦ to the x-axis, the Jonesmatrix is given by:

A = 1

2

(1 ±1

±1 1

)

For a polarizer with its pass plane at an arbitraryangle θ to the x-axis,

Aθ =(

cos2 θ cos θ sin θ

cos θ sin θ sin2 θ

)(1.197)

Thus, the x component of the transmitted fieldis obtained by adding the x component of theincident field multiplied by cos2 θ (one factor offor the projection of the x component of theincident field on the pass plane and one factor forprojection of the pass plane of the x axis) to they component of the incident field multiplied bycos θ sin θ , and similarly for the y component ofthe transmitted field. This can be represented by

rotating the original Jones matrix, A, by an angle θ

using the rotation matrix R(θ), to obtain the Jonesmatrix, Aθ , representing the rotated polarizer,

Aθ = R(−θ)A R(θ) (1.198)

where

R(θ) =(

cos θ sin θ

− sin θ cos θ

)(1.199)

Problem 1.34Prove that Eq. (1.197) for the Jones matrix for apolaroid at an arbitrary angle is obtained usingthis formula with A given by(

1 00 0

)

How is the operation of a retarder representedin Jones algebra? A retarder multiplies the x and y

components of the field by phase factors exp(ix)

and exp(iy), respectively. Thus,

[E0x exp(ix)

E0y exp iφ exp(iy)

]

=[

exp(ix) 00 exp(iy)

](E0x

E0y exp iφ

)(wave-retarder) (1.200)

Here φ is the relative phase shift between the x

and y components of the incident electric field.It is customary to remove the common phasefactor and write the Jones matrix for the retarderin terms of the difference of the phases betweenthe fast and slow axes of the retarder: exp(i) =exp i(y − x)[

E0x

E0y exp iφ exp(i)

]

=[

1 00 exp(i)

](E0x

E0y exp iφ

)(1.201)

LIGHT IN VACUUM 57

For example, for a quarter-wave plate with the fastaxis along x:

(E0x

iE0y exp iφ

)=(

1 00 i

)(E0x

E0y exp iφ

)(quarter-wave plate – fast axis along x) (1.202)

whereas for the fast axis along y,

(E0x

−iE0y exp iφ

)=(

1 00 −i

)(E0x

E0y exp iφ

)(quarter-wave plate – fast axis along y) (1.203)

If the axis of the retarder is rotated by an angle θ ,the rotation matrix R(θ) can be used to obtain thematrix representing the rotated retarder. Thus, for aquarter-wave plate with the fast axis at an angle θ

relative to the x-axis, we obtain the Jones matrix:

A =[

cos2 θ + i sin2 θ cos θ sin θ(1 − i)

cos θ sin θ(1 − i) + i cos2 θ + sin2 θ

](1.204)

Finally, we point out that the operation of acompound device made of a series of polarizingelements 1, 2, . . . , n can be represented as aproduct of Jones matrices:

Et = An . . . A2 A1 Ei (1.205)

Problem 1.35(a) Work out the Jones matrix for a half-waveplate with the fast axis at an angle θ relative tothe x-axis.

(b) Determine the Jones matrix formed by alinear polarizer along a 45◦ angle between x andy, a quarter-wave polarizer with fast axis alongx, and a polarizer with pass plane along y. Letthis matrix operate on the electric field vectorfor unpolarized light. What is the polarizationof the transmitted light?

Answer:

i/2

(0 01 1

)

is the Jones matrix. Therefore, only the y

component of the electric field will remain. Thephase of the y component will vary randomly.

Stokes parametersThe description of the polarization properties ofincoherent, or partially coherent radiation cannotbe carried out in terms of Jones algebra; theStokes vector representation of the light field isnecessary for this description. For coherent light,the electric vector moves periodically around anellipse (or its degenerate limit of a circle or astraight line). For incoherent or partially coherentlight, the electric vector trajectory can be extremelyirregular and chaotic, and the time-dependence(and spatial dependence) of the phases of the lightfield may also be so. We take up this descriptionnow. For monochromatic light, we define the fourquantities, S0, S1, S2 and S3, called the Stokesvector of the electric field:

S0 = E0x2 + E0y

2

S1 = E0x2 − E0y

2

S2 = 2E0xE0y cos(φy − φx)

S3 = 2E0xE0y sin(φy − φx) (1.206)

For quasi-monochromatic radiation which is coher-ent, we can write the electric field as:

E(z, t) = iE0x(z, t) cos[kz − ωt + φx(z, t)]

+ jE0y(z, t) cos[kz − ωt + φy(z, t)](1.207)

where E0x(z, t), E0y(z, t), φx(z, t) and φy(z, t) areslowly varying functions of time. We use a similardefinition of the Stokes parameters for partiallycoherent light:

S0 = 〈E0x(z, t)2〉 + 〈E0y(z, t)

2〉S1 = 〈E0x(z, t)

2〉 − 〈E0y(z, t)2〉


S2 = 2〈E0xE0y cos(φy − φx)〉S3 = 2〈E0xE0y sin(φy − φx)〉 (1.208)

where the brackets indicate a temporal averageover many periods of the light. For linearlypolarized light, with polarization along x and y

respectively, the Stokes parameters are

S = I

1±1

00

(1.209)

whereas polarization at an θ angle relative to thex-axis gives

S = I

cos2 θ + sin2 θ

cos2 θ − sin2 θ

2 cos θ sin θ

0

= I

1cos2 θ − sin2 θ

2 cos θ sin θ

0

(1.210)Right and left circularly polarized light is given by

S = I

100

±1

(1.211)

For monochromatic radiation of arbitrary ellip-tical polarization (i.e. for fully polarized light),

S02 = S1

2 + S22 + S3

2 (1.212)

since E0x , E0y and cos(φy − φx) are constantwith time and therefore the average is simply thequantity within the bracket. Therefore, the aboveequation can be reduced to the form:

(〈E0x2〉 + 〈E0y

2〉)2

= (〈E0x2〉 − 〈E0y

2〉)2 + 2(2〈E0xE0y〉)2

(1.213)However, for unpolarized light, and more generallyfor partially incoherent radiation (see below), the

above equation is not satisfied. The Stokes vectorfor unpolarized light is

S = I

1000

(1.214)

where we have taken the liberty of normalizing theStokes vector to intensity I .

The degree of polarization, P , of a monochro-matic light field is defined as:

P =√

S12 + S2

2 + S32

S02 (1.215)

If P vanishes, the light is unpolarized, whereas ifP equals unity, the light is completely polarized.Note that as long as P remains unity, S1, S2 andS3 can take only any values between −1 and 1 andthe light remains completely polarized.

The Stokes parameters for any field can beindependently determined by making four separateexperimental measurements. We assume that theintensity of a beam of light can be measured usinga photocell. The six measurements are as follows:

(1) A polaroid sheet is placed in the beam with itspass plane parallel to the x-axis. The intensityis measured and is proportional to 〈E0x

2〉.(2) The polaroid is rotated so its pass plane is

parallel to the y-axis. The intensity is measuredand is proportional to 〈E0y

2〉. The first twoStokes parameters are then given by the sumand difference of these measurements,

S0 = 〈E0x(z, t)2〉 + 〈E0y(z, t)

2〉,S1 = 〈E0x(z, t)

2〉 − 〈E0y(z, t)2〉

(3) A polaroid is placed in the beam with its passplane at 45◦ to the x-axis and lying in the firstand third quadrants. The intensity is measured

LIGHT IN VACUUM 59

and is proportional to

〈|E0x cos(π/4) + E0y exp i(φ) sin(π/4)|2〉

=[

1

2〈E0x

2 + E0y2 + 2E0xE0y cos(φ)〉

]

= 1

2(S0 + S2)

Knowing S0 from the previous two measure-ments, S2 is obtained from this measurement.

(4) We now place a quarter-wave plate with itsfast axis along the x-axis and a polaroid withits pass plane at 45◦ to the x-axis and lying inthe first and third quadrants in the beam. Theaction of the quarter-wave plate and polarizerproduces an electric field given by:

1

2

(1 11 1

)(1 00 i

)(E0x

E0y exp iφ

)

= 1

2

(1 11 1

)(E0x

iE0y exp iφ

)

= 1

2

(E0x + iE0y exp iφ

E0x + iE0y exp iφ

)(1.216)

so the intensity is proportional to

[1

2〈E0x

2 + E0y2 + 2E0xE0y sin(φ)〉

]

= 1

2(S0 + S3)

Knowing S0 from the first two measurements,S3 is obtained from this measurement.

Mueller matrices

Since the Jones calculus cannot be used for inco-herent, or partially coherent, radiation, another cal-culus based on Mueller matrices has been devel-oped. The action of various polarizing elementscan be represented by a matrix, the Mueller matrix,which multiplies the incident Stokes vector toobtain the transmitted Stokes vector, St = BSi , or

explicitly:

S0t

S1t

S2t

S3t

=

b11 b12 b13 b14

b21 b22 b23 b24

b31 b32 b33 b34

b41 b42 b43 b44

S0i

S1i

S2i

S3i

(1.217)The Mueller matrix representing a polarizer withaxis along x is simply

1

2

1 1 0 01 1 0 00 0 0 00 0 0 0

(1.218)

For a polaroid with its pass plane at an arbitraryangle θ with the x-axis,

1

2

1 cos 2θ sin 2θ 0cos 2θ cos2 2θ cos 2θ sin 2θ 0sin 2θ cos 2θ sin 2θ sin2 2θ 0

0 0 0 0

(1.219)This can be represented by rotating the originalMueller matrix, B, by an angle θ using the rotationmatrix R(θ),

A(θ) = R(−θ)B R(θ) (1.220)

where

R(θ) =

1 0 0 00 cos 2θ sin 2θ 00 − sin 2θ cos 2θ 00 0 0 0

(1.221)

Problem 1.36Prove that Eq. (1.219) for the Mueller matrixis obtained for a polaroid at an arbitrary angleby using Eq. (1.220) with Eq. (1.221) for R(θ)

and B given by Eq. (1.218).


A quarter-wave plate with the fast axis along x isgiven by:

1 0 0 00 1 0 00 0 0 10 0 −1 0

(1.222)

Problem 1.37What is the Mueller matrix for a quarter-waveplate with the fast axis at an angle θ relativeto the x-axis? What is the Mueller matrix fora half-wave plate with the fast axis along thex-axis? What is the Mueller matrix for a half-wave plate with the fast axis at an angle θ

relative to the x-axis?

A linear retarder with retardation along the x-axis is given by a Mueller matrix:

1 0 0 00 1 0 00 0 cos sin

0 0 − sin cos

(1.223)

Operation of a compound device made from aseries of polarizing elements 1, 2, . . . , n can berepresented as a product of Mueller matrices, i.e.

St = Bn . . . B2B1Si (1.224)

1.2.9 Diffraction

The intensity distribution of light in regions nearan illuminated aperture in an opaque screen is notadequately described by geometric optics or anycorpuscular (particle-like) theory of light. Opti-cal ray tracing of the light from the light sourcethrough the aperture does describe the light inten-sity near the aperture within several wavelengths ofthe aperture. Waves appear to bend around opaqueobjects so that there is not a distinct shadow ofthe object but rather the shadow of the objectis accompanied by diffraction fringes. The inten-sity distribution near an aperture has the form ofdark and bright bands of light, diffraction fringes.

This inadequacy of geometrical optics is partic-ularly evident if the dimensions of the apertureare comparable to the wavelength of the light, asthen the intensity pattern far behind the opaquescreen also contains diffraction fringes. Diffrac-tion was first accurately described by FrancescoMaria Grimaldi (ca 1660). Christian Huygens,Augustin Fresnel and Gustav Kirchhoff, amongothers, developed the wave theory of diffractionphenomena. Diffraction problems are very difficultto solve exactly, but approximate solution methodshave been developed, e.g. methods describing theelectric field in the far field (or far zone), far fromthe aperture, are easy to use. Moreover, numericalmethods for propagating light wavepackets havebeen developed that treat diffraction exactly, andthese methods allow the determination of the fieldin the near-field as well as in the far field. As anillustration of diffraction and interference patternsas they appear in the far field (far away from theobject giving rise to these patterns), Figure 1.12shows the diffraction of light from a rectangularand a circular aperture. Consider the diffraction ofa single-frequency beam of light from an orificein a opaque screen. The diffraction of light from arectangular aperture shown in Figure 1.12(a) cor-responds to the intensity pattern on a screen alarge distance, z, away from the opaque screencontaining the rectangular aperture with center atx = 0 and y = 0. The intensity is approximatelygiven by:

I (x, y, z)

= I (0)k2LxLy

16π2z2

(sin(kLxx/2z)

(kLxx/2z)

)2(

sin2(kLyy/2z)

(kLyy/2z)2

)2

.

(1.225)where k is the wavenumber of the light, Lx andLy are the dimensions of the rectangular aperture,and I (0) is the intensity at the aperture. Theintensity pattern on a screen a distance z from anopaque screen containing the circular aperture withcenter x = 0 and y = 0 is approximately given bythe form:

I (x, y, z) = I (0)2πa2k2

16π2z2

[2J1(kar/z)

kar/z

]2

(1.226)

LIGHT IN VACUUM 61

(a)

(b)

Figure 1.12 (a) Diffraction pattern for the intensity oflight passing through a square aperture. (b) Diffractionpattern for a circular aperture.

where a is the radius of the circular aper-ture, r = √

(x2 + y2), and J1 is the Bessel func-tion of the first kind of order unity. We shallexplain the nature of these intensity patterns inwhat follows.

According to Huygens’ principle of optics,every point on the wave front of the lightin the aperture of an aperture gives rise tospherical waves, and the light field behind theaperture can be regarded as the envelope ofthese wavelets originating from the points in theaperture. However, to account for diffraction, theHuygens principle needs to be supplemented bythe postulate that secondary wavelets of the opticalfield interfere in the sense of the superpositionof the amplitude and relative phases of thewavelets. This combination of Huygens principlewith Fresnel’s principle of interference is the

Huygens–Fresnel principle, and it does accountfor diffraction phenomena. The Huygens–Fresnelprinciple states that every point of a wavefrontis the source of a secondary spherical waveletand that the amplitude of the optical field at anypoint beyond the wavefront is the superpositionof the amplitude and relative phases of all thesewavelets.

The general mathematical methods developed byKirchhoff for diffraction theory can be approxi-mated when the position of viewing the light fieldis sufficiently far from the aperture. If the planeof observation is sufficiently far away, the lightimage of the aperture is such that the apertureis easily recognized, but it becomes increasinglymore structured with diffraction fringes as the dis-tance from the aperture increases. In this regimeof distance the Fresnel or near-field diffractionapproximation is valid. At a large distance, thelight image spreads out and loses its resemblanceto the actual aperture. Moving the plane of obser-vation further away results only in an increasedsize of the diffraction pattern but no change inshape. This is the region where the Fraunhofer orfar-field diffraction approximation is valid. Herewe shall consider scalar diffraction theory, i.e.diffraction theory for waves obeying the scalarwave equation, Eq. (1.227). Generalization to vec-tor diffraction theory to account for the polarizationcharacteristics of diffraction is important; we men-tion it briefly but we shall not pause to developit here. The interested reader is referred to Jack-son [13].

Kirchhoff and Fresnel diffraction theory relatethe complex amplitude at a certain location to thatat another location, e.g. that at a screen to that at adiffracting aperture, through a Fourier transform.This leads to the important method of Fourieroptics in general, sub-topics of which includeFourier transform filtering and Fourier transformspectroscopy. For a full treatment of Fourier optics,see Goodman [14].

Kirchhoff diffraction theory

We consider a monochromatic field of frequencyω for which the wave equation (the Helmholtz


equation) in vacuum is:

(∇2 + k2)U(r) = 0 (1.227)

with k = ω/c [see Eq. (1.45)]. We seek a solutionto this equation behind an opaque screen withan aperture having surface S with light sourceof angular frequency ω to the left of the screenand movable detector to the right of the screen.Kirchhoff diffraction theory makes use of Green’stheorem (after George Green, who developed it in1824) using the divergence theorem:

∫V

d3x ∇ · A =∫

S

da n · A (1.228)

where A is an arbitrary vector field. Letting A =φ∇ψ and applying the identities

∇ · (φ∇ψ) = φ∇2ψ + ∇φ · ∇ψ (1.229)

(φ∇ψ) · n = φ ∂ψ/∂n (1.230)

to Eq. (1.228) results in the expression

∫V

d3x(φ∇2ψ + ∇φ · ∇ψ) =∫

S

da φ ∂ψ/∂n

(1.231)

Writing Eq. (1.231) with φ and ψ interchanged,and subtracting, we obtain Green’s theorem:

∫V

d3x(φ∇2ψ − ψ∇2φ)

=∫

S

da(φ ∂ψ/∂n − ψ ∂φ/∂n) (1.232)

The wave equation, Eq. (1.227), can be convertedinto an integral equation by choosing ψ to be theGreen function, G(r, r′) (see Green’s functions inAppendix A), which satisfies the equation:

(∇2 + k2)G(r, r′) = δ(r − r′) (1.233)

and, choosing φ = U , which satisfies the waveequation, Eq. (1.227), we find that Eq. (1.232)

becomes∫V

d3x(U∇2G − G∇2U)

=∫

S

da(U ∂G/∂n − G ∂U/∂n) (1.234)

Using Eqs (1.227) and (1.233), and letting r′ bewithin the volume V , Eq. (1.234) becomes

U(r′) =∫

S

da[U(r)∂G(r, r′)/∂n

− G(r, r′)∂U(r)/∂n]

or, upon rewriting and making use of the fact thatG(r, r′) = G(r′, r),

U(r) =∫

S

da′[U(r′) n′ · ∇′G(r, r′)

− G(r, r′) n′ · ∇′U(r′)] (1.235)

The Green functions satisfying Eq. (1.233) are:

G(±)(r, r′) = − exp(±ikR)

4πR, R = |r − r′|

(1.236)

Using the Green function G(+)(r, r′), Eq. (1.235)becomes:

U(r) = 1

4π

∫S

da′ exp(±ikR)

R

× n′ ·[ik

(1 + i

kR

)RR

U(r′) + ∇′U(r′)]

(1.237)The integration over S involves two parts: (1) overthe aperture and the screen and (2) over the surfaceat infinity to the right of the screen. Since U(r′)behaves as 1/r ′ as r ′ goes to infinity, the secondintegral vanishes. Equation (1.237) is the Kirchhoffdiffraction formula.

In order to use the Kirchhoff diffraction formula,both U and ∂U/∂n must be known on the screenand the aperture. In general, they are not known,so the Kirchhoff approximation for these quantitiesis made: (1) U and ∂U/∂n vanish on the screen,and (2) U and ∂U/∂n in the aperture equal the

LIGHT IN VACUUM 63

incident wave values as if the screen were absent.These assumptions are only approximations. Infact, there is a theorem that states that any solution,U , to the wave equation for which U and ∂U/∂n

vanish on any finite surface, vanishes everywhere,i.e. U ≡ 0. This contradicts approximation (2).The mathematical contradiction can be removedby assuming that U is not zero on the screenbut becomes vanishingly small on the screen,and then using the proper Green function, G(D),which satisfies the Dirichlet boundary conditions,G(D)(r, r′) = 0 on S, i.e.

G(D)(r, r′) = − exp(ikR)

4πR− − exp(ikR′)

4πR′(1.238)

where R = |r − r′| and R′ = |r − r′′| with r′′being the mirror image of r′ in the screen. Then thegeneralized Kirchhoff diffraction formula becomesU(r) = ∫

Sda′ ∂G(r, r′)/∂n′U(r′) or, explicitly,

U(r) = ik

4π

∫S

da′ exp(ikR)

R

×(

1 + i

kR

)n′ · R

RU(r′)

(1.239)

Thus, we make use of the generalized Kirchhoffapproximation: (1) U becomes vanishingly smallon the screen, and (2) U in the aperture equals theincident wave value as if the screen were absent.In Eq. (1.239), the quantity n′ · R/R is simplygiven by:

n′ · R/R = cos θ

where θ is the angle between the wavevector ofthe light and the normal to the surface S. It shouldbe noted that, if the light impinging on the aperturecomes from a point source a long-distance from theaperture, then at the aperture we can approximateU(r′) = exp(ikr ′)/r ′, where r ′ is the distance fromthe source to the aperture.

There are three length scales involved inthe Kirchhoff diffraction formula: the size ofthe aperture, a, the distance from the aperture to

the observation point, x, and the wavelength, λ.

R = |r| − n · r′ + 1

2|r|× [r′2 − (n · r′)2] + · · · , for |r| � a

(1.240)where n = r/|r| is the unit vector in the obser-vation direction. When ka2/|r| 1, the quadraticterms in x ′ and y ′ that appear in exp(ikR) [i.e. thefactors resulting from the third term on the RHSof Eq. (1.240)] can be neglected in Eq. (1.239)and this limit is called the ‘Fraunhofer diffractionlimit’. Substitution of the expansion of exp(ikR)

into Eq. (1.239), keeping the quadratic terms inr′, yields Fresnel diffraction results, whereas keep-ing only terms up to first order yields Fraunhoferdiffraction results.

Upon substituting Eq. (1.240) into the exponen-tial inside the integral on the RHS of Eq. (1.239),and assuming that the quantity n′·(R/R) = cos δ isslowly varying with x ′ and y ′ so it can be taken outof the integral, we obtain the following expressionfor |r| � a:

U(r) = −ik cos δ exp(ik|r|)4π |r|

∫S

dx ′ dy ′

× exp

{−ikn · r′ + ik

2|r|

× [r′2 − (n · r′)2]

}U(r′) (1.241)

Noting that for z � x, y, |r| = z + (x2 + y2)/

2z3, and expanding the dot product n · r′ thatappears inside the exponential on the RHS ofEq. (1.241) as:

n · r′ = (xx ′ + yy ′)/z − (xx ′ + yy ′)

× (x ′2 + y ′2)/2z3 + . . . , (1.242)

we find

U(r) = −ik cos δ exp(ik|r|)4πz

∫S

dx ′ dy ′

exp

{− ik(xx ′ + yy ′)/z


+ ik

2z[(x ′2 + y ′2) − (xx ′ + yy ′)2/2z2]

}U(r′)

(1.243)When ka2/|r| ≈ 1 the quadratic terms in x ′and y ′ appearing in the exponential on theRHS of Eq. (1.243) cannot be neglected, andmost are retained. Equation (1.243) is the FresnelFraunhofer diffraction formula. When ka2/|r| 1, the quadratic terms can be neglected and weobtain the Fraunhofer diffraction formula:

U(r) = −ik cos δ exp(ik|r|)4πz

×∫

S

dx ′ dy ′ exp{−ik(xx ′ + yy ′)/z} U(r′)(1.244)

This method is the basis for Fourier optics.

Fraunhofer diffractionWe now consider the rectangular aperture in theFraunhofer diffraction limit. Let us take U(r′) =U0 on the rectangular aperture:

U(r) = −ik cos δ U0 exp(ik|r|)4πz

×∫ Lx/2

−Lx/2dx ′ exp{−ikxx ′/z}

∫ Ly/2

−Ly/2dy ′

× exp{−ikyy ′/z} (1.245)

Performing the integrals yields:

U(r) = −ik cos δ U0 exp(ik|r|)4πz

× LxLy

sin(kxLx/2z)

kxLx/2z

sin(kyLy/2z)

kyLy/2z(1.246)

Squaring the field to obtain the intensity, anddefining a constant I (0) for the intensity near thecenter of the aperture, we find

I (r)

= I (0)

[kLxLy

4πz

sin(kxLx/2z)

kxLx/2z

sin(kyLy/2z)

kyLy/2z

]2

(1.247)

Figure 1.12(a) shows the intensity obtained as aresult of diffraction from a rectangular aperture.The RHS of Eq. (1.247) involves the square ofthe function sinc(x) ≡ [sin(x)/x]. This function isplotted in Figure 1.13.

1.0

0.80

0.60

[sin

(x)/

x]2

0.40

0.20

0.0

−8 −6 −4 −2 0

x

2 4 6 8

Figure 1.13 Plot of the square of the functionsinc(x) ≡ sin(x)/x.

Carrying out a similar Fraunhofer diffractioncalculation for the circular aperture, we obtain:

U(r) = −ik cos δU0 exp(ik|r|)4πz

∫ a

0ρ dρ

×∫ 2π

0dθ exp{−ikρ(x cos θ + y sin θ)/z}

(1.248)and defining the distance r = √

(x2 + y2) andangle χ so that x = r cos χ , y = r sin χ , rewrit-ing the factor exp{−ikρ(x cos θ + y sin θ)/z} =exp{−ikρr cos(θ − χ)/z} in the integral over θ onthe RHS of Eq. (1.248), and noting that axial sym-metry demands that the result of the integration beindependent of the angle χ , we find:∫ 2π

0dθ exp{−ikρ(x cos θ + y sin θ)/z}

=∫ 2π

0dθ exp{−ikρr cos(θ − χ)/z}

= 2πJ0(kρr/z) (1.249)

Substituting this into the remaining integral over ron the RHS of Eq. (1.248), we obtain:

2π

∫ a

0dρ J0(kρr/z)ρ

LIGHT IN VACUUM 65

= 2π(z/kr)2[(kar/z)J0(kar/z)]

= 2π(az/kr) J1(kar/z)

so Eq. (1.248) takes the form

U(r) = −ik cos δU0 exp(ik|r|)2z

× a2(z/kar)J1(kar/z) (1.250)

Again defining an intensity I (0) near the center ofthe aperture, we obtain

I (r) = I (0)

[ka2

2z

J1(kar/z)

kar/z

]2

(1.251)

This formula was first derived by G.B. Airy in1835. The disc patterns shown in Figure 1.12(b)are called Airy discs after him. The first minimumof intensity occurs at about kar/z = 3.833. Theintegrated intensity of Eq. (1.251) falls off with z

as 1/z2. Figure 1.5 shows a plot of J1(x) vs x.

Fresnel and near-field diffractionIn the Fraunhofer limit, the diffracting system istaken as small and the point of observation is takenas very distant. The near-field region is one wherethe point of observation is at a distance comparablewith the size of the diffracting system. Theintermediate regime between Fraunhofer and near-field limits is called the Fresnel diffraction regime.As discussed at the beginning of the previoussection, the expression in Eq. (1.241), obtainedassuming that n′ · (R/R) = cos δ is slowly varyingwith x ′ and y ′, and |r| � a, can be approximatedfurther by expanding the dot product n · r′ as inEq. (1.242) to obtain Eq. (1.243). Taking U(r′) =U0, we obtain from Eq. (1.243):

U(r) = U0−ik cos δ exp(ik|r|)

4πz

×∫

S

dx ′ dy ′ exp

{− ik(xx ′ + yy ′)/z

+ ik

2z[(x ′2 + y ′2) − (xx ′ + yy ′)2/2z2]

}(1.252)

In problems involving Fresnel diffraction froma slit and from a straight edge (a semi-infiniteopaque plane bounded by a sharp straight edge),the quadratic form appearing in the exponential ofEq. (1.252) can be messaged into forms involvingthe integrals:

C(w) =∫ w

0dζ cos

[π2

ζ 2],

S(w) =∫ w

0dζ sin

[π2

ζ 2]

(1.253)

where C(w) and S(w) are know as Fresnel’sintegrals (see Born and Wolfe [10], Section 8.7).Figure 1.14 plots the Fresnel integrals and Cornu’s

−0.80

−0.60

−0.40

−0.20

0.0

0.20

0.40

0.60

0.80

−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

C(x)

[C(−∞), S(−∞)]

[C(∞), S(∞)]

C(∞) = 1/2, S(∞) = 1/2

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 0.5 1 1.5 2 2.5 3 3.5

C(x

), S

(x)

S(x)

x

S(x)

C(x)

(a)

(b)

Figure 1.14 (a) Fresnel integrals, C(x) and S(x) vs x.(b) Cornu’s spiral, C(x) vs S(x).


spiral. Using these integrals, one can show that theintensity behind a straight edge is given by:

I (x) = I01

2{[1/2 + C(w)]2 + [1/2 + S(w)]2}

(1.254)

where w = (2/λr)1/2 x, where x is the distancefrom the edge and r = z is the distance from theplane of the opaque screen. Figure 1.15 plots theintensity behind the screen.

1.6

1.4

1.2

1.0

0.80

0.60

0.40

0.20

0.0−1 0 1

w = (2/lr)1/2x

I(x)

/I0

2 3

I(x)/I0 = {[1/2 + C(w)]2 + [1/2 + S(w)]2}2

Figure 1.15 Integral I (x)/I0 vs w = (2/λr)1/2 x.

Problem 1.38Derive Eq. (1.254).

Hint: note that C(∞) = S(∞) = 1/2, and that∫ 0−∞ dζ cos[(π/2)ζ 2] = C(∞).

The diffraction theory presented here (Fresneland Fraunhofer diffraction) is for scattering oflight off objects that are large compared with thewavelength of light. One can, however, imaginean aperture with small size compared with thewavelength of light. Experimentally, it is knownthat light can get through such an object (evento the far field, i.e. even to very large distancesbehind such apertures). The field behind a circularaperture in an opaque screen, for the case when theradius of the circular aperture is small comparedwith the wavelength of impinging light, a < λ, wasfirst solved by H. Bethe and then modified by C.J.

Bouwkamp [15]. The solution took into accountthe vector properties of the electromagnetic fieldimpinging on the aperture.

Consider the fields behind an extremely smallorifice in an opaque screen. In a vacuum, theelectric and magnetic fields can be expanded asa Fourier series in the form:

E(r, t) = (2π)−3∫ ∞

−∞dω

∫ ∞

−∞dkx

∫ ∞

−∞dky E(k, ω) exp{i[k · r − ωt]} (1.255)

where in the exponential and in E(k, ω) onesubstitutes the vacuum dispersion relation

kz = (ω2/c2 − kx2 − ky

2)1/2 (1.256)

The sum over kx and ky is over all values,even values such that the quantity in the squareroot on the RHS of Eq. (1.256) is negative.For such values, kz is imaginary; these ‘modes’of the electromagnetic field are closed modesand decay as they propagate away from theorifice. The smaller the aperture of the orifice,the more such modes with large values of kx

and ky are needed to represent the amplitudeE(r, t) in terms of a Fourier expansion. Theseevanescent modes can have significant amplitudenear the hole, but the intensity due to these modesdecreases exponentially with increasing distancefrom the hole.

The Kirchhoff diffraction formula, Eq. (1.237)is valid for each component of the electric andmagnetic fields, E(r) and B(r), at frequency ω =kc. The corresponding vector formulas expressingE(r) and B(r) in terms of surface integrals areuseful for obtaining approximate expressions fordoing vector diffraction theory, and for derivingformal expressions (such as the optical theoremrelating the total scattering of optical waves tothe forward scattering amplitude). The readerinterested in these topics is encouraged to read theappropriate sections of Jackson [13].

1.2.10 Interference

A field composed of two or more optical wavespresent in the same region of space is a

LIGHT IN VACUUM 67

superposition (sum) of these waves, since thewave equation for light is linear and the super-position principle holds. Thus, if we have twocomplex monochromatic amplitudes with fre-quency ω, U1(r, t) = U1(r) e−iωt and U2(r, t) =U1(r) e−iωt , the complex amplitude for the totalfield is given by:

U(r, t) = U(r) e−iωt = U1(r, t) + U2(r, t)

= [U1(r) + U2(r)] e−iωt (1.257)

and the intensity of the total wave is

I (r) = |U(r)|2 = |U1(r) + U2(r)|2= |U1|2 + |U2|2 + U1U2

∗ + U1∗U2 (1.258)

Substituting U1(r) = I1(r)1/2 exp[iφ1(r)], andU2(r) = I2(r)1/2 exp[iφ2(r)], where φ1 and φ2 arethe phases of the waves, we obtain:

I (r) = I1(r) + I2(r)

+ 2[I1(r)I2(r)]1/2 cos φ(r) (1.259)

withφ(r) = φ2(r) − φ1(r) (1.260)

Thus, the intensity of the sum of the fields is not thesum of the intensities. The last term in Eq. (1.259),due to the interference of the two waves, canbe constructive or destructive depending upon thephase φ. For φ = 2mπ with m an integer, theinterference is constructive. For φ = (2m + 1)π ,the interference is destructive. If I1 = I2 = I0, I =4I0 for the former, and I = 0 for the latter.

For two plane waves of the same frequency,propagating in different directions, U1(r) = I1

1/2

exp(ik1 · r + φ1) and U2(r) = I21/2 exp(ik2 · r +

φ2) (where φ1 and φ2 are not now functions ofposition since we assume the two waves are planewaves), we find:

I (r) = I1 + I2 + 2(I1I2)1/2 cos(k · r + φ)

(1.261)

with φ = φ2 − φ1 and

k = k2 − k1 = 2π

λ2 sin

(θ

2

)e (1.262)

Thus, the intensity is periodic with period

p = 2π/|k| = λ/

[2 sin

(θ

2

)]

(see Figure 1.16 for the determination of thelength |k|). If we consider two equal intensitywaves, one wave propagating along the z-axisand the other in the x –z plane, so that U1(r) =I0

1/2 exp(ikz) and U2(r) = I01/2 exp[ik(cos θz +

sin θx)], at the z = 0 plane, we find

I = 2I0[1 + cos(k sin θ x)] (1.263)

so the period in the z plane is p = 2π/(k sin θ) =λ/ sin θ .

|k1| = |k2| = k

∆k = k2 − k1

|k2 − k1| = 2k sin (θ/2)

k2

k1

θ

Figure 1.16 Length |k| = 2k sin(θ/2) where k =k2 − k1 and k = |k2| = |k1|.

When N monochromatic waves of the samefrequency and intensity are present, and the phasedifferences between successive waves is constant(so φj = jφ), we can take Uj = I0

1/2 exp(iφj ) =I0

1/2 zj , j = 1, . . . , N , with z = exp(iφ). The totalfield is given by the sum:

U =N∑

j=1

Uj = I01/2

N∑j=1

zj = I01/2 z

1 − zN

1 − z

(1.264)


and the intensity is given by

I = |U |2 = I0

∣∣∣∣1 − zN

1 − z

∣∣∣∣2

= I0

∣∣∣∣z−N/2(zN/2 − z−N/2)

z−1/2(z1/2 − z−1/2)

∣∣∣∣2

= I0sin2 Nφ/2

sin2 φ/2(1.265)

We will need to use this result in considering thecase of a Fabry–Perot interferometer in the nextsection.

Problem 1.39(a) Show that

S ≡N∑

j=1

zj = z + z2 + z3 · · · + zN = z1 − zN

1 − z

by taking the difference of S and zS = z2 +z3 · · · + zN + zN+1.

(b) Show that

∞∑j=0

zj = 1

1 − z

(c) Calculate a closed form expression for

N∑j=−N

zj

Problem 1.40(a) Consider two equal intensity sphericalwaves eminating from points located at(a/2,0,0) and (−a/2,0,0). Obtain an expressionfor the intensity at a screen at z = d [i.e. deter-mine the intensity I (x, y, d)], by taking thefield at the z = d plane to be in the form:

U(r) = U1(r) + U2(r)

= −ik exp[ik|r − x(a/2)|]4π |r − x(a/2)|

+ −ik exp[ik|r − x(−a/2)|]4π |r + x(a/2)|

and then substitute |r ± x(a/2)| = d in thedenominator.

Answer: I (x, y, d) = 2I0{1 + cos[2π cos(2πθx/λ)]}, where θ = a/d .

(b) What is the nature of the periodicity ofthe bright stripes in the z = d plane?

Answer: the period in x is p = λ/θ = λd/a.(c) Determine the intensity reduction for

increasing x by substituting |r ± x(a/2)| = d +[(x ± a/2)2 + y2]/2d3 in the denominators ofthe expression for u(r).

InterferometersInterferometers are optical instruments that split alight beam or pulse into two waves using a beam-splitter, delay one wave relative to the other andrecombine the waves so that they interfere. Thereare many types of interferometers. Four importantexamples are the Fabry–Perot, the Michelson, theMach–Zehnder and the Sagnac interferometers.We shall consider the Fabry–Perot interferometerat some length in the next section. The Michelson,Mach–Zehnder and Sagnac interferometers areschematically drawn in Figure 1.17. The Michel-son interferometer has the initial wave split by abeamsplitter (labelled BS1). The two waves enterseparate adjustable length arms, are retro-reflectedand then are recombined in the output (another rep-resentation of a Michelson interferometer is shownin Figure 1.18). The Mach–Zehnder interferome-ter has an arrangement with two mirrors and twobeam splitters. The advantage of this arrangementis that, if a dispersive object is contained in one ofthe arms of the interferometer, the light in that armtraverses the body only once (it does not repassthrough the body). The Sagnac interferometer hasa ring geometry arrangement, with light traversingthe ring in both directions. It is used to measureacceleration (e.g. rotation) because an accelerationchanges the fringe pattern obtained by the inter-ferometer; this is the Sagnac effect, named afterGeorges Sagnac, who used it in 1913 to searchfor ‘ether drift’ in a rotating coordinate frame.

LIGHT IN VACUUM 69

M2

M1

BS1

Input

Output

(b)

(a)

Input

M1 M2

BS2BS1

Output

(c)

Input

M1 M2

BS1

Output

M3

Figure 1.17 Schematic diagrams of (a) Michelsoninterferometer, (b) Mach–Zehnder interferometer and(c) Sagnac interferometer.

M1

L1

L2

Output

Light wave

M2BSf

x

Figure 1.18 Michelson interferometer setup.

Many other kinds of interferometers exist, includ-ing Fizeau, Lummer–Gehrke and Twyman–Green,to name a few, but we will not be able to describethese here.

Fabry–Perot interferometerA Fabry–Perot interferometer (sometimes alsocalled a Fabry–Perot etalon or a Fabry–Perot res-onator) consists of two parallel optical surfaces,M1 and M2, between which a light wave can prop-agate. The reflectivities of the surfaces (mirrors)are taken to be R1 and R2, and the surfaces aretaken to be a distance L apart. We take the mediumbetween the surfaces to have index of refraction n,and we allow for absorption loss (with absorptioncoefficient α) within the medium. Consider a wavewhich propagates inside the medium at an angleθ relative to the surface normal (see Figure 1.19).Multiple reflections occur at the surfaces, and inter-ference of the fields that are multiply reflectedoccurs at the surfaces. Upon making one roundtrip between the surfaces, the electric field Ei ismodified to become:

Ei+1 = r exp(iϕ)Ei (1.266)

L

R1M1

M2

n, αR2

T2 = 1 − R2

T ′1 = 1 − R1

θ

Figure 1.19 Schematic drawing of a Fabry–Perotinterferometer of thickness L with mirror reflectivitiesR1 and R2 having index of refraction n and absorp-tion coefficient α. The wave inside the resonator prop-agates at an angle θ relative to the surface normal.The transmissions T ′

1 and T2 are given in terms of thereflectivities.

where

r = (R1R2)1/2 exp[−2(α/2)L/ cos θ ]

= (R1R2)1/2 exp(−αL/ cos θ) (1.267)


φ = 2π

λ2nL cos θ + φR

= 2π(ν/c)2nL cos θ + φR (1.268)

Here φR is the additional phase due to the(complex) reflection and transmission amplitudes.The total field E inside the resonator is given by:

E = E1 + E2 + E3 + · · · = E1

1 − r exp(iφ)

(1.269)since S = 1 + x + x2 + x3 + · · · = 1/(1 − x). Inour case,

x = r exp(iφ) (1.270)

Therefore the intensity inside the cavity isgiven by:

I = I1

[(1 − r cos φ)2 + r2 sin φ2]

= I1/(1 − r)2

[1 + (2F/π)2 sin2(φ/2)](1.271)

where we have defined the finesse of the resonator,F , as

F = πr1/2

(1 − r)(1.272)

Problem 1.41Carry out the algebra leading to Eq. (1.271).

The intensity in Eq. (1.271) is a periodic func-tion of the phase φ, defined in Eq. (1.268), withperiod 2π . If F is large, I is sharply peaked at val-ues φm = 2πm, with full width at half maximumδφ = 2 × 2 arcsin(π/2F) ≈ 2π/F . The maximumvalue of the intensity I is given by:

Imax = I1/(1 − r)2 (1.273)

and occurs for φm = 2πm, and the minimum isImin = Imax/[1 + (2F/π)2], which occurs at the

midpoints between the φm. In terms of Imax,Eq. (1.271) can be written as:

I = Imax

[1 + (2F/π)2 sin2(φ/2)]

The spacing between different frequency maximais, using the definition of φ in Eq. (1.268),given by:

ν = c

2nL cos θ(1.274)

and the full width at half maximum is

δν = (2/π) arcsin(π/2F)ν ≈ ν

F(1.275)

The spacing between frequency maxima, ν, isinversely proportional to the cavity length, and theresonance linewidth, δν, is inversely proportionalto the finesse and the cavity length. ν is calledthe free spectral range of the Fabry–Perot for areason to be specified below. The intensity as afunction of frequency can be written as:

I (ν) = Imax

[1 + (2F/π)2 sin2(πν/ν)](1.276)

Figure 1.20 plots the intensity inside a Fabry–Perot resonator vs frequency.

1.4

1.2

I = Imax/[1 + (2F/π)2sin2(πν/∆ν)]

∆ν = 1 GHz, F = 10δν = ∆ν/F

1.0

0.80

0.60

0.40

Inte

nsity

(ar

bitr

ary

units

)

0.20

0.0

0.5 1 2 3 41.5 2.5

Frequency (GHz)

3.5 4.5

∆ν

δν

Figure 1.20 Intensity inside a Fabry–Perot resonatorvs frequency.

The losses of the Fabry–Perot resonator are dueto the non-unity reflection of the mirrors (i.e. the

LIGHT IN VACUUM 71

output of the resonator), and the absorption of themedium inside the cavity. The losses are quantifiedby the parameter r2:

r2 = (R1R2) exp(−2αL/ cos θ) (1.277)

This attenuation factor can be written in terms ofan effective cavity loss coefficient, αeff, given by

r2 = exp(−2αeffL) (1.278)

where

αeff ≡ α/ cos θ + 1

2Lln[|R1R2|−1/2] (1.279)

The finesse is given in terms of the effective cavityloss coefficient by the expression

F = π exp(−αeffL/2)

1 − exp(−αeffL)(1.280)

The finesse decreases with increasing loss. Forsmall loss, αeffL 1,

F ≈ π

αeffL(1.281)

i.e. the finesse is inversely proportional to the lossαeffL, and the finesse is very high.

The losses of the Fabry–Perot resonator result ina finite lifetime of the photons in the cavity. Thislifetime is manifested in the finite width of themodes of the resonator. The relationship betweenthe width of the modes and the photon lifetime isunderstood by rewriting the width of the modes δν

given in Eq. (1.275) as follows:

δν ≈ ν

F≈ c/(2nL cos θ)

π/(αeffL)= cαeff

2πn cos θ(1.282)

The quantity (c/n)αeff has the units of an inversetime and from the time–frequency uncertaintyrelation we understand this equation to meanthat the photon lifetime, τph, in the Fabry–Perotresonator [the electric field decays as exp(−t/2τph)

inside the cavity] is given by:

τph = 1

(c/n)αeff(1.283)

so that Eq. (1.282) can be rewritten as

δν = 1

2πτph(1.284)

It is useful to relate the photon lifetime to thequality factor, Q, of a Fabry–Perot resonator. Thequality factor, Q, is defined as:

Q = 2π × stored energy

energy loss per cycle(1.285)

The energy stored inside the cavity is given byhν times the number of photons in the cavity(equal to NphV , where Nph is the photon densityin the cavity and V is the cavity volume), and theenergy loss per cycle is the stored energy times(1/ν)/τph. Hence,

Q = 2π × NphV hν

NphV hν (1/ν)/τph= 2πντph (1.286)

The Q can also be written in terms of thefinesse as:

Q = ν

νF (1.287)

Since the laser frequency, ν, is typically very muchlarger than the free spectral range, ν, Q � F .

In addition to the applicability of this descriptionof the Fabry–Perot resonator to a laser cavityfor the purpose of understanding cavity modes,this treatment describes the use of a Fabry–Perotresonator as a frequency filter and a spectrumanalyzer. For these purposes, it is important tocalculate the relation of the incident intensity, Ii,and the transmitted intensity, It, to the total internalintensity, I . We have the following relations:

I1 = T ′1Ii = (1 − R1)Ii (1.288)

It = T2I = (1 − R2)I (1.289)

from which we can obtain the relation

It = Tmax

[1 + (2F/π)2 sin2(πν/ν)]Ii (1.290)


where

Tmax = (1 − R1)(1 − R2)

[1 − (R1R2)1/2 exp(−αL/ cos θ)]2

(1.291)Hence, the transmittance of a Fabry–Perot, T (ν) ≡It/Ii, has the same frequency dependence as theinternal intensity. A Fabry–Perot filter or spectrumanalyzer is tuned by changing the free spectralrange, ν, either by changing the length, L, or bychanging the angle, θ . If the spectral width of theincident light is narrower than ν, the spectrumof the light can be resolved by the Fabry–Perotresonator without ambiguity, hence the name freespectral range for ν.

In most configurations of a Fabry–Perot inter-ferometer, light from an extended source with fre-quency bandwidth greater than the free spectralrange of the Fabry–Perot will form narrow brightcircular fringes surrounded by dark regions in thefocal plane of the exit lens of the interferometer.The area enclosed by the bright fringes will haveequal incremental area, i.e. the diameters of thecircles are given by

Dm2 = 4cf 2

n2Lν0(m − ζ )

where f is the focal length of the exit lens, ν0

is the central frequency of the light source andζ is a constant dependent upon the phases of thereflection amplitudes of the mirrors forming theFabry–Perot interferometer.

1.2.11 Temporal and spatialcoherence

We mentioned in Section 1.2.8 that natural lighteminating from a heated object, e.g. the Sun, iscomposed of a superposition of radiation from alarge number of excited atoms or molecules thatradiate independently into different frequencies,directions and phases. However, even light ofa single frequency and direction can be maderandom in direction and in phase by scatteringoff media with randomness, e.g. a rough surface.

The degree of random fluctuations of the light isrelated to the coherence of the light: the morerandom the fluctuations, the less coherent thelight. One can measure the temporal and spatialcoherence of light independently. Both temporaland spatial coherence affect the ability to obtaininterference fringes with light, e.g. in an interfero-meter.

Let us begin our discussion of coherence byconsidering a scalar field U(r, t) of the formU(r, t) = A(r, t) exp{i[k · r − ωt]}. The intensityof this field is given by I (r, t) = |U(r, t)|2. It isconvenient to define the average over all time ofthe product U ∗(r1, t)U(r2, t + τ),

G(r1r2, τ )

= 〈U ∗(r1, t)U(r2, t + τ)〉

= limT →∞

1

2T

∫ T

−T

dtU ∗(r1, t)U(r2, t + τ)

(1.292)

This average over time, indicated by the brackets〈· · ·〉, is the mutual coherence function, as denotedby G(r1, r2, τ ) where τ is the time differencebetween the temporal arguments of the two fieldsappearing in the average. If the spatial argumentsr1, r2 are taken to be identical, r1 = r2, we oftenwrite G(r1, r1, τ ) simply as G(τ) and drop thespatial arguments, G(τ) = 〈U ∗(t)U(t + τ)〉. Thisfunction is called the temporal coherence function.One can also consider the mutual coherencefunction at zero time difference,

G(r1, r2, 0) = 〈U ∗(r1, t)U(r2, t)〉 (1.293)

This quantity is the mutual intensity, and itis sometimes denoted G(r1, r2). The averageintensity of the field at position r, 〈I (r)〉 isgiven by G(r,r). The normalized form of themutual coherence function, g(r1, r2, τ ), is thecomplex degree of coherence or complex coherencefunction, and is given by:

LIGHT IN VACUUM 73

g(r1, r2, τ )

= 〈U ∗(r1, t)U(r2, t + τ)〉{[〈U ∗(r1, t)U(r1, t)〉

× 〈U ∗(r2, t)U(r2, t)〉]1/2

}

= G(r1, r2, τ )

[G(r1, r1, 0)G(r2, r2, 0)]1/2(1.294)

The complex degree of coherence is the cross-correlation of U ∗(r1, t) and U(r2, t + τ). Themagnitude of the complex degree of coherence isbounded as follows:

0 ≤ |g(r1, r2, τ )| ≤ 1 (1.295)

Note that g(r, r, 0) = 1. The quantity g(r, r, τ )

characterizes the temporal coherence of light andg(r1, r2, 0) characterizes its spatial coherence.

The coherence time, τc, is a measure of the timeat which the magnitude of the complex degreeof coherence, g(r1, r1, τ ), drops from unity toa smaller number. The coherence time is oftendefined as:

τc =∫ ∞

−∞dτ |g(r1, r1, τ )|2 (1.296)

Clearly, the coherence time for a monochromaticfield is infinite since |g(r1, r1, τ )|2 = 1 for all r1.The coherence distance can be defined as lc = cτc.

Problem 1.42(a) Suppose that the complex degree of coher-ence, g(τ), is exponential. Show that it shouldbe written as g(τ) = exp(−|τ |/τc), i.e. provethat Eq. (1.296) follows from this exponentialform. What is the value of g(τc)?

(b) Suppose that the complex degree ofcoherence, g(τ), is Gaussian. Show that itshould be written as g(τ) = exp(−πτ 2/2τc

2),i.e. prove that Eq. (1.296) follows from thisGausian form. What is the value of g(τc)?

(c) Consider the amplitude U(t)=exp(−iωt)

for −τ/2 < t < τ/2 and U(t) = 0 otherwise.Calculate g(τ) and τc.

(d) For an amplitude U(t) = exp(−iωt) exp(−t2/2τ 2), calculate g(τ) and τc.

Another possible definition of τc is given by:

τc2 =

∫∞−∞ dττ 2|g(r1, r1, τ )|2∫∞

−∞ dτ |g(r1, r1, τ )|2 (1.297)

This definition, and that in Eq. (1.296), do notin general yield identical coherence times for allpulses, but their results are usually very similar.

The mutual coherence function G(r1, r2, τ )

gives the spatial correlation between the fieldat r1 and r2 at time delay τ . The Fouriertransform of G(r1, r2, τ ) is the cross-spectraldensity, G(r1, r2, ω), and is given by:

G(r1, r2, ω)

= 1

(2π)1/2

∫ ∞

−∞dτ exp(−iωτ)G(r1, r2, τ )

(1.298)This quantity at r1 = r2, G(r1, r1, ω), can beshown to equal the spectral power density,|U(r1, ω)|2,

G(r1, r1, ω) = |U(r1, ω)|2 (1.299)

This relation is known as the Wiener–Khinchintheorem.

Problem 1.43(a) Prove the Wiener–Khinchin theorem.

(b) Show that G(r1, r2, ω) is the Fouriertransform of the mutual coherence function

G(r1, r2, ω)

= 1

(2π)1/2

∫ ∞

−∞dτ exp(−iωτ)

× 〈U ∗(r1, t)U(r2, t + τ)〉


It is sometimes more useful to consider theone-sided spectral power density which doesnot distinguish between positive and negativefrequencies, and is defined as:

P(r1, r1, ω) = G(r1, r1, ω) + G(r1, r1, −ω)

[= 2S(r1, r1, ω) if G(r1, r1, τ ) is real] (1.300)

The spectral width, ω = 2πν, can be definedas the full width at half maximum of P(r1, r1, ω).

Problem 1.44(a) Consider a general plane wave pulsemoving in the +z direction of the formU(r, t) = A(t − z/c) exp[iω0(t − z/c)]. Showthat G(r1, r2, τ ) = GA[τ − (z2 − z1)/c] exp{iω0[τ − (z2 − z1)/c]}, where GA(τ ) = 〈A∗(t)A(t + τ)〉.

(b) For a general spherical pulse ofthe form U(r, t) = [A(t − r/c)/r] exp[iω0(t −r/c)], show that

G(r1, r2, τ )

= 1

r1r2GA[τ − (r2 − r1)/c]

× exp{iω0[τ − (r2 − r1)/c]}

Let us consider the interference of a par-tially coherent light field. When a partially coher-ent plane wave, U(r, t) = A(t − z/c) exp[iω0(t −z/c)], is introduced into a Michelson interferome-ter (composed of a beamsplitter and two opticalarms through which the split beams propagate,see Figure 1.18) having a difference in the lengthsof the arms of L and hence a time differencebetween the arms given by τ = L/c, the outputof the interferometer is given by:

Uout(t) = U(t) + U(t + τ) (1.301)

The average intensity is then given by:

Iout = 〈|U(t) + U(t + τ)|2〉= 〈|U(t)|2〉 + 〈|U(t + τ)|2〉

+ 〈U ∗(t)U(t + τ)〉+ 〈U ∗(t + τ)U(t)〉

= G(0) + G(0) + 2Re[G(τ)]

= 2G(0){1 + Re[g(τ)]} (1.302)

If we write g(τ) = |g(τ)| exp[iφ(τ )], we find

Iout = 2G(0)[1 + |g(τ)| cos φ(τ)] (1.303)

For a fully coherent field, |g(τ)| = 1 and φ(τ) =ω0τ , but for a partially coherent field, |g(τ)| < 1.This shows that the nature of the interferencefringes of a field is (a) governed by the complexdegree of coherence, and (b) the interference isreduced for a field that is partially coherentrelative to a fully coherent field. The modulationdepth, or contrast of the interference pattern, isthe modulation transfer function (MTF) and isgiven by

MTF = Iout;max − Iout;min

Iout;max + Iout;min= max[|g(τ)|] (1.304)

Thus, the maximum contrast is given in termsof the complex degree of coherence, g(τ), bymax |g(τ)|, and this is a direct measure of thetemporal coherence of the light.

Spatial coherence of light can be measuredby a Young-type interference measurement (recallthe Young double-slit experiment) carried out byintroducing two small holes into an impenetrablescreen located at r1 and r2 relative to the originof coordinates located at a point of measurementbehind the screen. These holes are then thesources of two waves behind the screen, eachof the form Ui(r, t) = Ui(t − |r − ri |/c)/|r − ri|.The measurement of the average intensity at theorigin yields:

〈|U1(0, t) + U2(0, t)|2〉

=⟨∣∣∣∣U1(t − |r1|/c)

r1+ U2(t − |r2|/c)

r2

∣∣∣∣2⟩

≈ 1

z2〈|U(t − |r1|/c) + U(t − |r2|/c)|2〉

LIGHT IN VACUUM 75

= 1

z2(I1 + I2 + 2Re{G[r1, r2, τ (r1, r2)]})

(1.305)where

τ(r1, r2) = (|r1| − |r2|)/c

≈ [(x12 + y1

2) − (x22 + y2

2)]

zc

and z is the perpendicular distance to the screenfrom the measurement point. We have assumedthat z is larger than the distance between thetwo holes. If the measurement point is equidistantto the two holes, (x1

2 + y12) − (x2

2 + y22) = 0,

and the average measurement intensity is given interms of Re[G(r1, r2, 0)]. Otherwise, τ(r1, r2) ≈xθ/c, where x is the distance between theholes and θ is the angle made by the line joiningthe measurement point to the position on the screenbetween the two apertures and the normal to thescreen, and the average measurement intensity isgiven in terms of Re[G(r1, r2, xθ/c)].

The above treatment is adequate for describingthe coherence of a polarized light source, sincethen only one polarization state needs to be consid-ered. The coherence of a partially polarized source(with radiation field propagating along the z-axis)requires the generalization of the mutual coherencefunction to the mutual coherence matrix:

G(r1, r2, τ ) =[

Gxx(r1, r2, τ )Gxy(r1, r2, τ )

Gyx(r1, r2, τ )Gyy(r1, r2, τ )

](1.306)

where Gij (r1, r2, τ ) = 〈Ui∗(r1, t)Uj (r2, t + τ)〉

and i and j denote the i and j components ofthe complex wavefunctions.

One can define quantum coherence functionsin terms of expectation values of the quantumfield operators defined in the next section. It isparticularly important to do so when consideringhigher order coherence functions, as is necessaryto explain the Hanbury–Brown and Twiss exper-iment, first carried out in 1956, which measuredtwo-photon interference from light sources. As aconsequence of the correlation of photons presentin partially coherent light, fluctuations of the lightfield are such that photons in such fields prefer to

bunch (there is a tendency of photons to arrive inpairs); this phenomenon is called photon bunch-ing. Space limitations prevent us from presentingthese topics fully [16,17]. A discussion of quan-tum coherence of light fields will, however, bepresented in Chapter 9, which also contains a shortdiscussion of bunching (preferential arrival of pho-tons in pairs) and antibunching of photons.

1.2.12 Photons: quantization of theelectromagnetic field

Photons are degrees of freedom of the radiationfield. The concept of a photon originates fromthe second quantization picture of the radiationfield. The population of a particular mode of acavity by no photons, one photon or N photonscorresponds to having zero, one or N excitationsin this mode. We now briefly consider the secondquantized picture of the radiation field.

The time dependence of the radiation fieldresembles the dynamic behavior of a collection ofharmonic oscillators. This is easy to see from thedifferential equation for the Fourier coefficients ofthe field, ck,α(t),

d2ckα(t)

dt2= −ωk

2ckα (1.307)

and from Eq. (1.144), which expresses the fieldenergy in terms of the Fourier coefficients, ck,α(t),as a sum of the energies of the individual harmonicoscillators,

U = 2

2π

∑k

∑α

(ωk/c)2ck,α(t)∗ck,α(t)

{in SI units, U = 1

ε0

∑k

∑α

× (ωk/c)2ck,α(t)∗ck,α(t)

}

Let us define the following real-time-dependentquantities:

Qkα =√

4π

c(ckα + ck,α

∗) (1.308)


Pkα = −iωk√

4π

c(ckα − ck,α

∗) (1.309)

In terms of these quantities the Hamiltonian (i.e.the energy) for the radiation field takes the form:

H = 2∑

k

∑α

(ω/c)2

[c(ωkQkα − iPkα)

2ω

]

×[c(ωkQkα + iPkα)

2ω

]

=∑

k

∑α

1

2(Pkα

2 + ωk2Qkα

2)

(1.310)From this expression, it is clear that the radiationfield can be regarded as a collection of indepen-dent harmonic oscillators for each k and α (i.e.for each mode of the radiation field labelled bythe indices k and α) with momentum like vari-able Pkα and position like variable Qkα . Around1870 Lord Rayleigh and J.H. Jeans obtained anexpression for the spectral energy distribution ofthe radiation field by assigning an average energyof kBT for each oscillator. Their expression agreedwith observations of the density of photons at lowfrequencies for high-temperature bodies (the ω2

power law at low frequencies), but failed at highfrequencies. The derivation of the correct black-body law for radiation was derived by Max Planckin 1900, when he postulated that the energy ofeach oscillator is a multiple of hω. In 1927 P.A.M.Dirac took this postulate one step further and pro-posed that each oscillator of the radiation field bequantized, just like the quantum mechanical treat-ment of the one-dimensional harmonic oscillator.Thus, Dirac introduced the quantization of the radi-ation field. A particularly good discussion of thequantization of the electromagnetic field is givenby Sakurai [5], and we will follow that discussionhere. A simple treatment of the one-dimensionalharmonic oscillator in quantum mechanics is pre-sented in Appendix C. The discussion below of thequantization of the electromagnetic field followsthat of the harmonic oscillator in Appendix C.7.

The quantization of the electromagnetic fieldproceeds by making the quantities Qkα and

Pkα operators. The ‘second quantization’ of theelectromagnetic field can be implemented utilizingthe commutation relations of the operators Qkα

and Pkα (in a fashion similar to the commutationrelation used for first quantization):

[Qkα, Pkα] = ihδk,k′δα,α′ (1.311)

[Qkα, Qkα] = 0 (1.312)

[Pkα, Pkα] = 0 (1.313)

The linear combinations of Qkα and Pkα given by

ak,α = 1√2hω

(ωkQkα + iPkα) (1.314)

ak,a† = 1√

2hω(ωkQkα − iPkα) (1.315)

satisfy the commutation relations

[ak,α, ak,α†] = δk,k′δα,α′ (1.316)

[ak,α, ak,α] = 0 (1.317)

[ak,α†, ak,α

†] = 0 (1.318)

The Hamiltonian can be written in terms of theoperators ak,α(t) and ak,α

†(t) as:

H =∑

k

∑α

1

2(Pkα

2 + ωk2Qkα

2)

= 1

2

∑k

∑α

hωk(ak,α†ak,α + ak,αak,α

†)

(1.319)Note the care in which the operators ak,α andak,α

† have been ordered, since they are nownon-commuting operators. The time-dependenceof ak,α(t) can be obtained from the Heisenbergequation of motion for ak,α(t). The Heisenbergequation of motion for ak,α(t) is:

d

dtak,α(t) = 1

ih[H, ak,α(t)]

= −iωk ak,α(t) (1.320)

LIGHT IN VACUUM 77

and hence the time-dependence of the operatorak,α(t) is harmonic,

ak,α(t) = ak,α e−iωkt (1.321)

The operators ak,α(t) and ak,α†(t) are related to

the coefficients ckα(t) and ck,α(t)∗, but now theseFourier coefficients are no longer considered as c-numbers but as operators, i.e.

ck,α(t) → c(2πh/ωk)1/2ak,α(t)

{ck,α(t) → c [h/(2ε0ωk)]1/2ak,α(t)} (1.322)

Using the commutation properties of the operators,it is easy to rewrite Eq. (1.319) as:

H =∑

k

∑α

hωk

(ak,α

†ak,α + 1

2

)(1.323)

It is useful to define the number operator for modek and α,

Nk,α = ak,α†ak,α (1.324)

The number operator is Hermitian. The eigenvec-tors |nk,α〉 of the number operator Nk,α thereforehave real eigenvalues:

Nk,α|nk,α〉 = nk,α|nk,α〉 (1.325)

The number nk,α represents the number of photons(light particles) of wavevector k, frequency ω

and polarization α in the state |nk,α〉; nk,α isan eigenvalue of the number operator Nk,α and|nk,α〉 is an eigenvector of this operator. From thecommutation relations of the operators ak,α andak,α

†, one can show that:

ak,α|nk,α〉 = (nk,α)1/2 |(n − 1)k,α〉 (1.326)

a†k,α|nk,α〉 = (nk,α + 1)1/2|(n + 1)k,α〉 (1.327)

Therefore, the operators ak,α and ak,α† are annihi-

lation (lowering) and creation (raising) operatorsfor the eigenvectors of the number operator Nk,α .The lowest value of nk,α is 0, and the loweringoperator ak,α annihilates the state |0k,α〉. The full

state of the radiation field can be written in terms oflinear combinations of the product of eigenvectors:

|nk1,α1, nk2,α2 , nk3,α3 , nk4,α4 , . . . 〉= |nk1,α1〉|nk2,α2〉|nk3,α3〉|nk4,α4〉 . . .(1.328)

Applying the Hamiltonian to this state, we obtain:

H |nk1,α1 , nk2,α2, nk3,α3, . . . 〉

=[∑

k

∑α

hωk

(nk,α + 1

2

)]

× |nk1,α1, nk2,α2, nk3,α3, . . . 〉Since the absolute energy scale is arbitrary,we can subtract the energy hω/2 for everydegree of freedom, yielding the Hamiltonian(energy) operator:

H =∑

k

∑α

hωk Nk,α (1.329)

and therefore,

H |nk1,α1 , nk2,α2 , nk3,α3 , . . . 〉

=[∑

k

∑α

hωknk,α

]

× |nk1,α1 , nk2,α2 , nk3,α3 , . . . 〉 (1.330)

Hence, the energy of one quantum of light atfrequency ω is E = hω. If n photons are presentin this degree of freedom, the total energy in thisdegree of freedom is nhω. If photons of frequencyω have density N , the energy density of thesephotons is Nhω.

It is easy to establish using Eq. (1.142)and (1.322) that the momentum operator for theelectromagnetic field is:

P =∑

k

∑α

hk(

Nk,α + 1

2

)(1.331)

The 12 that appears in Eq. (1.331) can be dropped

because, in the sum over k, the hk term cancelsthe −hk term. Hence,


P =∑

k

∑α

hk Nk,α (1.332)

The momentum of state |nk1,α1 , nk2,α2 , nk3,α3 , . . . 〉is therefore given by

∑k

∑α hk nk,α .

For the description of the quantum theory ofradiation, it is important to obtain an expressionfor the vector potential A(r, t). Using Eq. (1.136)and (1.319), we obtain the expression

A(r, t) = c∑

k

√2πh

V ω

2∑α=1

× [ak,α(t) e(α) eik·r + e(α)∗ akα(t)† e−ik·r]

and upon using Eq. (1.322) we finally obtain:

A(r, t) = c∑

k

√2πh

V ω

2∑α=1

[ak,α(0) e(α)

× eik·r−iωt + e(α)∗ akα(0)† e−ik·r+iωt ]{A(r, t) =

∑k

√h

2ε0V ω

2∑α=1

[ak,α(0)e(α) eik·r−iωt

+ e(α)∗akα(0)† e−ik·r+iωt ]

}

(1.333)

We have dropped the subscript k on thefrequency. We will use this expression in thederivation of the rate for spontaneous emissionof radiation in Chapter 5. Note that the operatorfor the vector potential A(r, t) is Hermitian, theparameters r and t (the space time coordinates ofthe photon) appear as parameters upon which thequantized vector potential depends, and the vectorpotential contains a lowering operator for thephotons multiplied by e(α) eik·r−iωt and a raisingoperator for the photons multiplied by e−ik·r+iωt .The expressions for the electric and magnetic fieldsin terms of the raising and lowering operators areas follows:

E(r, t) =∑

k

∑α

i

√2πhω

V[ak,α(0) e(α) eik·r−iωt

− e(α)∗akα(0)† e−ik·r+iωt ]{E(r, t) =

∑k

∑α

i

√hω

2ε0V[ak,α(0) e(α) eik·r−iωt

− e(α)∗akα(0)† e−ik·r+iωt ]

}

(1.334)

and similarly for B(r, t), using Eq. (1.135): B =∇ × A.

Problem 1.45(a) Derive the expression for the transverseelectric field in terms of the raising andlowering operators akα(0)† and akα(0).

Hint: using the expressions in Problem 1.29,substitute ck,α(t) = c(2πh/ω)1/2ak,α(0) e−iωt .

(b) Obtain the expression for the electric fieldin SI units.

Hint: use Eq. (1.333) and write E in terms ofthe vector potential.

(c) Obtain expressions for the magnetic fieldin terms of the raising and lowering operatorsin Gaussian and SI units.

The quantization of the radiation field changesour whole conception of the nature of electro-magnetic radiation for weak fields. Fluctuationof the radiation field is built into the quan-tum formalism. We can calculate the quantityE ≡ (〈E2〉 − 〈E〉2)1/2, which is the root meansquare deviation of the electric field strength,where the expectation is for the quantum stateof the field, |nk1,α1 , nk2,α2 , nk3,α3, . . . 〉. The quan-tity E is non-zero due to the commutationrelation [ak,α(0), ak,α

†(0)] = δk,k′δα,α′ . The quan-tity E is non-zero even for the vacuum state|0k1,α1 , 0k2,α2 , 0k3,α3 , . . . 〉. Vacuum fluctuations andzero-pint energy present conceptual problems inquantum field theory. Since there are an infinite

MATTER–SOURCE OF LIGHT 79

number of modes, each with a zero-point energy,the total zero-point energy is infinite. Yet, noobservable consequences arise from this diver-gence. In any case, there are observable effects ofthe vacuum fluctuations. The Dirac theory for thehydrogen atom spectrum predicts a degeneracy inthe energy levels of the 22S1/2 and 22P1/2 states. In1947, Lamb and Retherford used microwave spec-troscopy techniques to show that the 22S1/2 stateis higher in energy than the 22P1/2 state by about1 GHz. The discrepancy was explained by HansBethe in terms of the interaction of the bound elec-tron with the vacuum fluctuations. This shift in theenergy levels of these states is the Lamb shift.

1.3 Matter–source of lightSo far we have dealt with the propagation of radi-ation through vacuum, and developed the math-ematical description of electromagnetic radiationfield in vacuum, but how are electromagneticwaves generated? The one-line classical answerto this question is, through oscillating (accelerat-ing) charges which account for the source termsin Coulomb’s law, Eq. (1.1), and Ampere’s law,Eq. (1.4). These time-dependent charge distribu-tions and currents produce the spontaneous andstimulated emission from (a) excited bound stateswhich decay to energetically lower lying states byvirtue of the fundamental interaction of light andmatter, and (b) from free charges that lose energyby emission of photons to the radiation field. In thepresent section we discuss the emission of elec-tromagnetic radiation by oscillating charges andcurrents within the context of Maxwell’s equations.We describe the classical interaction of chargedparticles with the electromagnetic field, in the formof electric and magnetic dipoles, quadrupoles, etc.Further elaboration is provided in Chapter 3, wherewe describe the fundamental interaction of lightwith matter, and the electromagnetic fields ema-nating from matter that result due to the sourceterms in Maxwell’s equations.

1.3.1 Classical expressions for thecharge density and current

Charges and currents are the sources of electric andmagnetic fields. The two Maxwell equations which

contain the source of the fields are Coulomb’s lawand Ampere’s law, Eqs (1.1) and (1.4). The chargedistribution due to a collection of point chargedparticles at positions ri (t) is given by:

ρ(r, t) =∑

1

qiδ[r − ri (t)] (1.335)

and the current is given in terms of the chargesand their velocities vi (t) by the expression

J(r, t) =∑

1

qivi (t) δ[r − ri (t)] (1.336)

The current and charge distribution are relatedthrough the continuity condition,

∇ · J + ∂tρ = 0 (1.337)

which is a statement of charge conservation.Integrating the continuity equation over a volumeelement dr, and using Gauss’s theorem, yieldsthe relation

∂t

∫dr ρ(r, t) = −

∫dS n · J

The physical interpretation of this equation is thatthe rate of change of charge within a volume equalsthe negative of the integral of the projection of thecurrent on the surface normal through the surface.

For macroscopic neutral matter, the microscopicdetermination of the charge and current density isnot feasible because of the large number of parti-cles. However, for the macroscopic description ofthe electromagnetic fields, it is sufficient to obtainthe average charge and current densities of thematter in order to determine the macroscopic elec-tromagnetic fields. As we shall see (Sections 4.3and 4.4), for macroscopic neutral (i.e. not plasma)matter, the macroscopic (average) charge and cur-rent density can be written quite generally in theform:

ρ = −∇ · P (1.338)

J = ∂tP + c∇ × M {J = ∂tP + ∇ × M}(1.339)


where P is the macroscopic polarization vectorand M is the macroscopic magnetization vector.Maxwell’s equations, Eqs (1.1)–(1.4) are for themacroscopic electric and magnetic fields, and thesource terms for them are the average macroscopiccharge and current density. Equations (1.338)–(1.339) for the charge and current density arisingfrom neutral matter are the source terms for themacroscopic electric and magnetic fields.

1.3.2 The wave equation with sourceterms: Lienard–Wiechertpotentials

Propagation of light through vacuum is describedby the source free Maxwell’s equations,Eqs (1.30)–(1.33). If charged point-particles arepresent, the source terms must be added, andMaxwell’s equations become:

∇ · E = 4πρ {∇ · E = ρ/ε0} (Coulomb’s law) (1.340)


∇ × E = −1


∇ × B =(

4π

cJ + 1

c∂tE)

{∇ × B = µ0(j + ε0∂tE)} (Ampere’s law) (1.343)

where the sources are given by

ρ(r, t) =∑

1

qiδ[r − r(t)] (1.344)

J(r, t) =∑

1

qiviδ[r − r(t)] (1.345)

The propagation equations in vacuum for thelight originating from the sources can be derivedby taking the curl of Eq. (1.342) and usingEq. (1.343) to re-express the RHS of the resultingequation. We obtain:

∇ × ∇ × E + c−2 ∂t2E = −4π

c∂tJ

{∇ × ∇ × E + µ0 ε0∂t2E = −µ0 ∂tJ} (1.346)

Noting that

∇ × ∇ × E = −∇2E + ∇(∇ · E)

and using ∇ · E = 0 in vacuum (and noting thatµ0 ε0 = c−2 in SI units), Eq. (1.346) reduces to

∇2E − c−2 ∂t2E = −4π∇ρ + 4π

c2∂tJ

{∇2E − c−2 ∂t2E = −∇ρ − µ0∂tJ} (1.347)

This equation is a wave equation with propagationvelocity c that contains source terms. Taking thecurl of Eq. (1.342) and using Eq. (1.343) and ∇ ·E = 0 in a manner similar to that used to obtainEq. (1.347), we obtain the driven wave equationfor B,

∇2B − c−2 ∂t2B = −4π

c∇ × J

{∇2B − c−2 ∂t2B = −µ∇ × J} (1.348)

Problem 1.46(a) Derive wave equations for E and B ina dielectric medium containing free movingcharges starting from Maxwell’s equations andthe first two constitutive equations, Eq. (1.5)–(1.6), D = ε E, H = µ−1 B.

Answer: take the curl of Eq. (1.342),and use the two constitutive equations andEq. (1.343) in the resulting equation. Take thecurl of Eq. (1.343), and use the two constitutiveequations and Eq. (1.342) in the resultingequation. Now use the identity for ∇ × ∇ × Eand Eqs (1.340)–(1.341) to obtain:


∇2E − εµ

c2∂t

2E = 4π∇ρ + 4πµ

c2∂tJ

{∇2E − εµ ∂t2E = ε−1∇ρ + µ∂tJ}

∇2B − εµ

c2∂t

2B = −4πµ

c∇ × J

{∇2B − εµ ∂t2B = −µ∇ × J}

(b) Can one now use the last consti-tutive equation, J = σE and the equation−∇(∇ · P) = ∇ρ to further reduce these waveequations?

Hint: think about double counting.

These wave equations for E and B are oftenused in a different guise. The electromagneticfields generated from charged particles can be mosteasily written in terms of the vector and scalarpotentials A and φ, in terms of which the fieldsare given by (see Appendix B):

B = ∇ × A (1.349)

E = −∇φ − 1

c∂tA

{E = −∇φ − ∂tA, in SI units} (1.350)

Moreover, quantum mechanically, the vector andscalar potentials (rather than the E and B fields)are directly used in, for example, writing theSchrodinger equation, and therefore these poten-tials take on an important role quantum mechani-cally. However, the electromagnetic fields do notuniquely define the vector and scalar potentials Aand φ. As described in Appendix B, the transfor-mations

A → A′ = A + ∇� (1.351)

φ → φ′ = φ − 1

c∂t�

{φ → φ′ = φ − ∂t�, in SI units} (1.352)

with an arbitrary scalar function �(r, t), a gaugefunction, leave the electric and magnetic fieldsunchanged. For radiation problems it is convenient

to choose the scalar gauge function �(r, t) so thatA and φ satisfy the Lorentz gauge,

∇· A + 1

c∂tφ = 0

{∇ · A + ∂tφ = 0, in SI units} (1.353)

Substituting Eqs (1.349)–(1.350) for E and B intoEqs (1.347)–(1.348), we obtain

∇ × ∇ × A + c−2 ∂t2A = 4π

cJ − 1

c∇(∂tφ)

(1.354)and

−∇2φ − 1

c∂t (∇· A) = 4πρ (1.355)

Using the Lorentz gauge condition, Eqs (1.354)–(1.355) can be rewritten as:

∇2A − c−2 ∂t2A = −4π

cJ

{∇2A − c−2 ∂t2A = −µ0J, in SI units}

(1.356)

∇2φ − c−2 ∂t2φ = −4πρ{

∇2φ − c−2 ∂t2φ = − ρ

ε0, in SI units

}(1.357)

One way to obtain the solutions to these equationsis in terms of the Green’s function, G(r, t, r′, t ′),which satisfies the delta function inhomoge-neous equation:

[∇2 − c−2 ∂t2] G(r, t, r′, t ′)

= −4π δ(r − r′) δ(t − t ′) (1.358)

The retarded Green function (that specifies asolution at position r and time t that depends ona signal originating at time t ′ = t − |r − r′|/c) isgiven by

G(−)(r, t, r′, t ′) = δ[t ′ − (t − |r − r′|/c)]|r − r′|

(1.359)


The retarded solution specifies a solution atposition r and time t that depends on a signaloriginating at time t ′ = t − |r − r′|/c that is earlierthan time t , whereas the unphysical advancedsolution,

G(+)(r, t ; r′, t ′) = δ[t ′ − (t + |r − r′|/c)]|r − r′|

depends on a signal originating at time t ′ = t +|r − r′|/c that is later than time t . In terms ofGreen’s function, G(−)(r, t, r′, t ′), the vector andscalar potentials A and φ are given by:

A(r, t) = c−1∫

dt ′∫

dr′ G(−)(r, t, r′, t ′) J(r′, t ′)

(1.360)

φ(r, t) =∫

dt ′∫

dr′ G(−)(r, t, r′, t ′) ρ(r′, t ′)

(1.361)Hence, Eqs (1.356)–(1.357) for A and φ have theparticular solution:

A(r, t) = c−1∫

dt ′∫

dr′δ(t ′ − [t − |r − r′|/c])/

R J(r′, t ′) = c−1∫

dr′ J(r′, t − |R|/c)/R (1.362)

φ(r, t) =∫

dt ′∫

dr′δ(t ′ − [t − |r − r′|/c])/

R ρ(r′, t ′) =∫

dr′ ρ(r′, t − |R|/c)/R (1.363)

where R = r − r′.To better understand the significance of these

relations, let us take as an example the case ofradiation from a moving point charged particlewith charge q following a trajectory r(t). Then,

J(r′, t ′) = q v(t ′) δ[r′ − r(t ′)] (1.364)

ρ(r′, t ′) = q δ[r′ − r(t ′)] (1.365)

The vector and scalar potentials resulting fromthe moving point charge particle are the Lien-ard–Wiechert potentials. A simple way to derive

these potentials is to use the following way of writ-ing the charge and current densities:

J(r′, t ′) = q v(t ′) δ[r′ − r(t ′)]

= q

∫dτ v(τ )δ[r′ − r(τ )] δ(τ − t ′)

(1.366)

ρ(r′, t ′) = q δ[r′ − r(t ′)]

= q

∫dτ δ[r′ − r(τ )] δ(τ − t ′)

(1.367)When these forms for the charge and currentdensity are substituted into Eqs (1.280)–(1.281),we obtain, upon integration over r′,

A(r, t) = qc−1∫

dτv(τ )δ(τ − t ′)

|r − r(τ )| (1.368)

φ(r, t) = q

∫dτ

δ(τ − t ′)|r − r(τ )| (1.369)

where t ′ is given by the retarded time,

t ′ = t − R(τ)/c (1.370)

and we have defined R(τ ) = r − r(τ ). Equa-tions (1.368)–(1.369) show that the retarded Lien-ard–Wiechert potentials are given by an integralover the history of the particle in which only theinstant on the light cone that reaches the locationin space–time (r, t) contributes. Upon making atransformation of variables from τ to τ ′,

τ ′ = τ − t ′ = τ − t + R(τ)/c

Eqs (1.366)–(1.367) become

A(r, t) = qc−1∫

dτ ′ v(τ )δ(τ ′)R(τ) − R(τ ) · v(τ )/c

(1.371)

φ(r, t) = q

∫dτ ′ δ(τ ′)

R(τ) − R(τ ) · v(τ )/c

(1.372)In obtaining Eqs (1.371)–(1.372) we have used therelations:

dτ ′ = dτ − dt ′ = dτ

(1 + 1

c

dR(τ)

dτ

)


where the expression

dR(τ)

dτ= −R(τ ) · v(τ )

R(τ)

for dR(τ)/dτ can be obtained by differentiatingR(τ)2 = R(τ ) · R(τ ) with respect to τ and notingthat

v(τ ) = dr(τ )

dτ

Consequently,

dτ

R(τ)= dτ ′

R(τ) − R(τ ) · v(τ )/c

Integrating Eqs (1.371)–(1.372) over τ ′ by settingτ ′ = 0, we find

A(r, t) =[

qv(τ )/c

R(τ) − R(τ ) · v(τ )/c

]ret{

A(r, t) = µ0

[qv(τ )

R(τ ) − R(τ ) · v(τ )/c

]ret

}(1.373)

φ(r, t) =[

q

R(τ) − R(τ ) · v(τ )/c

]ret{

1

ε0

[q

R(τ) − R(τ ) · v(τ )/c

]ret

}(1.374)

where the subscript ‘ret’ indicates that the quantityinside the square brackets is to be evaluated at theretarded time τ :

τ = t − R(τ)/c (1.375)

Here, t is the time of observation of the fieldsat position r, τ is the time at which the particleat position r′ [i.e. r′ = r′(τ )] emitted the photonsresponsible for the fields at their observation pointat time t , and R(τ) is the distance at time τ

between the observation point r and the positionof the particle at time τ , r′(τ ).

Clearly, when v = 0, the Lienard–Wiechertpotentials reduce to φ = q/R and A = 0. Inlater chapters (primarily Chapter 3, Section 3.4)

we shall use the Lienard–Wiechert potentials toderive the emission characteristics of acceleratingcharged particles. To do so, we must be able todifferentiate φ and A with respect to t and r, sincethe electromagnetic fields are given by B = ∇ ×A, E = −∇φ − 1

c∂tA. To calculate the required

derivatives the following relations are useful:

∂R(τ)

∂t

= dR(τ)

dτ

∂τ

∂t= −R(τ ) · v(τ )

R(τ)

[1 − ∂R(τ)

∂t/c

]

so

∂R(τ)

∂t=

−R(τ ) · v(τ )

R(τ)

1 − R(τ ) · v(τ )/c

R(τ)

(1.376)

∂τ

∂t= 1

1 − R(τ ) · v(τ )/c

R(τ)

(1.377)

and

∇τ = −∇R(τ)/c = −[

dR(τ)

dτ∇τ + R(τ )

R(τ)

]/c

(1.378)so

∇τ = R(τ )

cR(τ)

[1 − R(τ ) · v(τ )/c

R(τ)

] (1.379)

Using these relations, we finally obtain:

E =

{q{(1 − v2/c2)(R − vR/c)

+ c−2 R × [(R − vR/c) × dv/dτ ]}}

[R(τ) − R(τ ) · v(τ )/c]3

(1.380)

H = R(τ ) × ER(τ)

(1.381)

where the numerator of Eq. (1.380) is evaluated attime τ . In the remainder of this section we shall


consider only the case of a charged particle movingwith constant velocity [i.e. dv(τ )/dτ = 0].

Problem 1.47(a) Derive the equation for dR(τ)/dτ in termsof R(τ ) and v(τ ) and Eqs (1.376) and (1.377).

(b) Do the algebra leading to Eqs (1.380)and (1.381).

The first term in the electric field given inEq. (1.380) depends only on the velocity, but noton the acceleration; it varies at large distancesas 1/R2. This term can be called the ‘Coulomb’part of the electromagnetic field. The second termdepends on acceleration, and varies as 1/R. Thedenominator is given by [R(τ){1 − [R(τ )/R(τ)] ·v(τ )/c}]3. The factor κ ≡ [1 − R(τ ) · v(τ )/c] canbe significantly different from unity if β ≡ v/c

is close to unity and the angle between the unitvector R(τ ) and the velocity is small. In this case,κ = 1 − β cos θ(τ ), and κ can become very closeto zero, hence the fields can become large.

The Coulomb part of the electromagnetic fieldgenerated by a particle is the only contributionto the electromagnetic field when the particle hasvanishing acceleration. This field can be writtenin terms of the instantaneous configuration, i.e.in terms of the vectors R(t) and v(t), where t

is the time of measurement of the field, not thetime of emission of the radiation from the chargedparticle. In Problem 1.48 you will demonstrate thatthe fields are given by:

E = q(1 − β2)R(t)

R(t)3[1 − |R(t) × v(t)|/c]3/2(1.382)

H = (v/c) × E (1.383)

The electric field points radially away from theinstantaneous position of the (positively) chargedparticle (for a negatively charged particle thesign is opposite). The factor [1 − |R(t) × v(t)|/c]appearing in the denominator of the expressionfor the electric field is the factor κ ≡ [1 − R(τ ) ·v(τ )/c] [see Problem 1.48(f)]. This factor ‘beams’the field in the direction of the velocity if theparticle is relativistic, since κ = 1 − β cos θ(τ )

becomes very small when β ≈ 1 and θ(τ ) ≈ 0.More about the properties of the radiation frommoving charged particles will be presented inSection 3.4.

Problem 1.48(a) For the case of a particle with constantvelocity, show that the vector R(t) = R(τ ) −(t − τ)v = R(τ ) − vR(τ)/c is the vector dis-tance at time t from the charge to the observa-tion distance by constructing a vector diagramcontaining the vectors R(t), R(τ ) and (t − τ)v.

Hint: use the definition of the retarded time inEq. (1.376).

(b) Calculate the length of R(t) in terms ofthe angle θ(τ ) between R(τ ) and v(τ ).

Hint: take the dot product of R(t) with itself.

Answer: R2(t) = R2(τ )[1 + β2 − 2β cosθ(τ )].

(c) From the vector diagram, find a relation-ship between θ(τ ) and the angle θ(t) betweenR(t) and v(τ ) [which equals v(t) by ourassumption of constant velocity].

Answer:

cos θ(τ ) = (t − τ)v + R(t) cos θ(t)

R(τ)

= β + R(t) cos θ(t)/R(τ)

(d) Eliminate cos θ(τ ) in favor of cos θ(t)

in the expression for the length of R(t) foundin (b).

(e) Use the expression in (d) to derive thefollowing relationship by solving the quadraticequation for R(τ):

R(τ) = R(t)

(1 − β2){β cos θ(t)

+ [1 − β2 sin2 θ(t)]1/2}Hint: use the identity cos2 θ(t) − 1 = sin2 θ(t).

(f) Obtain the expression for the electric fieldof a particle for the case of constant velocity.


Hint: use the expression

R(τ) − R(τ ) · v(τ )/c

= R(τ)(1 − β cos θ)

= R(τ)(1 − β2) − R(t)β cos θ(t)

= R(t)[1 − β2 sin2 θ(t)]1/2.

Answer:

E = q(1 − β2)R(t)

R(t)3[1 − β2 sin2 θ(t)]3/2

(g) Show that the magnetic field is given byH = (v/c) × E, by substituting R(τ ) = R(t) +vR(τ)/c into the expression

H = R(τ ) × ER(τ)

In Section 3.4 we will continue our study ofthe radiation emitted by accelerating relativisticcharged particles, using Eqs (1.380) and (1.381),to calculate the fields generated by synchrotron andother radiation emission processes.

Isaac Newton Max Planck

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