Source Free Electromagnetic Fieldsgleeson/ElectricityMagnetismChapter6.pdf · Source Free...

Chapter 6

Source Free ElectromagneticFields

Maxwell’s equations, Equations 5.1, with the source terms removed are rel-atively simple. There are six temporal evolution equations

1c2

∂

∂tE(r, t) = ∇× B(r, t)

∂

∂tB(r, t) = −∇× E(r, t), (6.1)

and two constraint equations

∇ · E(r, t) = 0∇ · B(r, t) = 0. (6.2)

These constraints require that the source free case has only transverse so-lutions, ET and BT . We will not carry the designation of the transversenature of the field in the following but the reader should keep this conditionin mind.

This is a set of first order in time coupled field equations much like thestring, Equation 2.59, with the addition of constraints. Similar to that case,a pair of second order decoupled equations can be found and these are theusual wave equation for both the fields,

1c2

∂2

∂t2E(r, t)− ∇2 E(r, t) = 0 (6.3)

and1c2

∂2

∂t2B(r, t)− ∇2 B(r, t) = 0. (6.4)

87

88 CHAPTER 6. SOURCE FREE ELECTROMAGNETIC FIELDS

Thus, all the solutions of this field system are travelers with velocity ±c.This, of course, is the usual definition of radiation. There are complicationswell beyond those of a string because of the three spatial dimensions andthe vector nature of the fields but otherwise all the intuition developed therecarries over to this system, see Section 2.2. As with the string, the systemis hyperbolic and, as we will see shortly, other than the phase relationsbetween E and B, it behaves much like a string when described using thedisplacement/velocity or Hamiltonian description, Equation 2.59. It is alsoworth pointing out that the source free electromagnetic field system hasonly one dimensional parameter, the speed of light c. This means thatthis system has no intrinsic length or time scales but does have an intrinsicvelocity. Thus the length and time scales for any field configuration areconnected; a disturbance of length scale L has a time scale L

c.

There is a special quadratic form for the source free fields that providesspecial constraints and interpretation. The quantity

ρe (x, t) ≡12

E2 (x, t) + c2 B2 (x, t)

(6.5)

is locally conserved in the same sense as the charge, Equation 5.3. Whenthe time rate of change of ρe (x, t) is examined

12

∂

∂t

E2 (x, t) + c2 B2 (x, t)

=

∂

∂tE

· E − c2 B ·

∂

∂tB

=c2

∇× B

· E − c2 B ·

∇× E

= c2∇ ·

E × B

. (6.6)

Thus je (x, t) ≡ −c2

E (x, t)× B (x, t)

is a local flow current that balances

the change in the density1, ∂ρe∂t

+ ∇ ·je = 0. Once we develop the action forthese fields, we will derive this local density from the Noether’s constructionthat develops from time translation symmetry, see Sections A.5.4 and 6.3 .Obviously, other continuous symmetries will lead to other local conservationdensities. Equations 6.1 and 6.2 also enjoy rotational and time and spacetranslational symmetries.

1In the presence of sources this relation becomes

∂ρe

∂t+ ∇ ·je = −c

2µ0j (x, t) · E (x, t) . (6.7)

There is a source for this otherwise conserved density. In the usual interpretation, poweris transferred to the charges by the electric field, we can identify this density as the energydensity.

6.1. SOURCE FREE FIELDS IN K SPACE 89

There are other symmetries of this dynamic. The substitution

E(x, t) = c B(x, t)

B(x, t) = −E(x, t)

c(6.8)

leaves Equations 6.1 through 6.6 unchanged.In addition, Equations 6.3 and 6.4 are wave equations and, as in the

string example, Section 2.3.5, are unchanged by the linear coordinate trans-formation, the Lorentz transformations, that preserves the form c2t2−x2 =c2t2−x2, see Section 2.3.5 and G.3. This symmetry is true for Equations 6.3and 6.4 also for the electric and magnetic fields in each component and, sincethey linear, they remain true for any linear transformation of the fields. TheMaxwell dynamic in the form of Equations 6.1 and 6.2 is not symmetricwithout also making a linear transformation in the fields. This problem isreviewed in Section 8.2.

Similarly, Equations 6.1 and 6.2 also have other discrete symmetries.Spatial reflection, x → −x, also requires that E(x, t) = − E(−x, t) andB(x, t) = B(−x, t); that E(x, t) be a vector field and B(x, t) be a pseu-dovector field or better said a two form, an area, see Section 5.3 . Timereversal invariance

6.1 Source Free Fields in k Space

The kinematics and dynamics of the source free Maxwell fields is best under-stood when the spatial labels are transformed into Fourier or wave-numberspace, k space. The properties and the nature of these transfoms andthe closely related generalized functions are treated in some detail in Ap-pendix B. It must be emphasized that the examination of the Maxwellfield in k space is appropriate only to the free space field properties. Thebounded system will require special considerations and be treated later, seeChapter 9.

For the case of the field in unbounded space, these transforms are definedas

E(k, t) =1

√2π

3

∞

−∞d3r E(r, t)e−ik·r

E (r, t) =1

√2π

3

∞

−∞d3k E(k, t)eik·r (6.9)


with a similar set of definitions for the magnetic field and any other fieldvariable.

In k space, the Maxwell’s dynamical equations become1c2

∂

∂tE(k, t) = −ik × B(k, t)

∂

∂tB(k, t) = ik × E(k, t), (6.10)

and the constraint equations becomek · E(k, t) = 0k · B(r, t) = 0. (6.11)

Equations 6.11 require that the source free electromagnetic fields be trans-verse. This is consistent with the dynamical equations, Equations 6.10,since the longitudinal projector can be used on Equation 6.10 to maintainthe constraint for all times,

1c2

∂

∂tEL(k, t) = 0

∂

∂tBL(k, t) = 0. (6.12)

It is important to realize that since the dynamical equations, Equa-tions 6.10, involve only the transverse fields which are only two dimensional,see Equation 5.17, there are only four equations instead of six as the no-tation may indicate. Also these fields in k space couple only fields of thesame k and, thus, for each k this is a particularly simple mechanical system;the evolution of the field for each k develops independently. Expanding thetransverse fields in terms of the (1)(k) and (2)(k) basis,

ET (k, t) = ET

(1)(k, t)(1)(k) + ET

(2)(k, t)(2)(k), (6.13)

with a similar expansion for BT (k, t).The dynamic for this system is

1c2

∂ET

(1)

∂t(k, t) = ikBT

(2)(k, t)

∂BT

(2)

∂t(k, t) = ikET

(1)(k, t)

1c2

∂ET

(2)

∂t(k, t) = −ikBT

(1)(k, t)

∂BT

(1)

∂t(k, t) = −ikET

(2)(k, t). (6.14)

6.1. SOURCE FREE FIELDS IN K SPACE 91

Thus the four dynamical equations for the fields decouple even further intoa pair of coupled first order differential equations with the ET

(1)(k, t) andBT

(2)(k, t) coupled and ET

(2)(k, t) and BT

(1)(k, t) coupled and each pair indepen-dent of the other. Calling these the (1) and (2) field systems respectively.These two coupled one dimensional fields are the only non-trivial fields thatare possible for the source free case of the electromagnetic fields.

The general solutions of these equations are those of two independentsimple harmonic oscillators both with radian frequency ω = ±kc or

ET

(1)(k, t) = ET

(1)(k, 0) cos(kct) + icBT

(2)(k, 0) sin(kct) (6.15)

cBT

(2)(k, t) =−i

kc

∂

∂tE

T

(1)(k, t) (6.16)

= cBT

(2)(k, 0) cos(kct) + iET

(1)(k, 0) sin(kct), (6.17)

with a similar pair of solutions for the ET

(2)(k, t) and BT

(1)(k, t) coupled pair.The easiest way to obtain this second set of solutions is to interchange 1 and2 in the labels and replace i by −i in Equations 6.15 through 6.17.

A simple example serves to show the application of this result. Considera source free situation with an initial configuration that is a pulse of elec-tric field that is symmetric in a plane. For orientation, chose the plane ofsymmetry to be the x-y plane. The condition of symmetry implies that theinitial field configuration has spatial dependence only in the z coordinate,E(z, t = 0). Equation 6.9 then requires that

E(k, 0) =√

2πδ(kx)δ(ky) ∞

−∞dz e−ikzz E(z, 0).

Similarly, Maxwell’s equations dynamical equations, the transversality con-dition, and the fact that the kx = ky = 0 implies that E(kz, 0) and thus alsoE(z, 0) cannot be along the z axis. For simplicity, we can take the x axisalong the direction of the initial E field. Choosing this direction to be thedirection of (1). This then implies the e(1) = x and e(2) = y. Thus, we haveBT

(2)(k, 0) = ET

(2)(k, 0) = BT

(1)(k, 0) = 0. To be concrete, we can choose as our

pulse shape a gaussian, E(x, 0) = E0e− z2

2σ2 x with width σ. Since the Fouriertransform of a gaussian is a gaussian, the full solution in k-pace is

E(k, t) = ET

(1)(k, t)x = 2πE0σδ(kx)δ(ky)e−k2zσ2

2 cos(|kz|ct)x

c B(k, t) = cBT

(2)(k, t)y = i2πE0σδ(kx)δ(ky)e−k2zσ2

2 sin(|kz|ct)y,


where I have used the fact that δ(kx) and δ(ky) force k ≡

k2x + k2

y + k2z =

|kz|.An important point to note is that the kx and ky dependence is com-

pletely localized at kx = 0 and ky = 0 whereas the x and y dependence isuniform for all values. This inversion relation also obtains in the kz case.The non-zero values of the field in kz is in the range kz = ± a few ( 1

σ) whereas

the range of non-zero field in z is z = ± a few σ. The more spread in con-figuration space the tighter the field in k-space or, better said, there is anuncertainty like relation between the ranges, ∆kz ×∆z ≈ 1.

Fourier transforming back to configuration space the full time dependentsolution is

E(x, t) =E0

2

e−

(z−ct)2

2σ2 + e−(z+ct)2

2σ2

x.

The corresponding B field is

c B(x, t) =E0

2

e−

(z+ct)2

2σ2 − e−(z−ct)2

2σ2

y.

The initial field in E produces two travelers, a right and left traveler, in theone dimension available, z, in the same shape as the original pulse with halfamplitude.

This solution also has an interesting structure given its genesis. The

original, t = 0, fields are E(x, 0) = E0e− z2

2σ2 x and B(x, 0) = 0 which aretransverse and agree with our solution. Unstated but assumed is that giventhe fields at t = 0, we want to know the fields at later times, t ≥ 0. Thissolution though is valid for all t with the two traveling B(x, t) fields conve-niently canceling at t = 0. We found this solution by solving the first orderin time simple harmonic oscillator differential equation.

How general is this result? The intuition of the string system, Chap-ter 2.2, tells us that all solutions are travelers formed from the initial config-uration of the fields. This is a general result for any shape and exactly thepattern of development as worked in the stretched string, Section 2.2.1, orany (1, 1) dimensional hyperbolic system. Here the initial plane symmetricE is used to establish the direction of propagation and orientation of thetravelers. The associated traveling B field is zero in the initial configura-tion because of cancelation from the overlap of the two oppositely movingtravelers. This is a common feature of (1, 1) dimensional wave fields eventhough the relationship of the two fields, E and B, for the electromagneticfield and y and v, for the stretched string, are based on a different dynamic.

6.2. ANOTHER REPRESENTATION 93

We obtained this result because of the fact that we forced the systemto be basically a (1,1) system by forcing the plane wave symmetry on thesystem. Is this technique restricted to situations for which only the planewaves are appropriate? Can any configuration of starting transverse fieldsbe represented and solved?

One beauty of the Maxwell system is the linearity of the theory. In thiscase, superposition can allow for the construction of more complex cases. Inother words, is the most general source free field configuration composed oflinear combinations of these plane wave solutions. These are the radiationsolutions and they are composed of these plane wave travelers as their basicconstituents. In other words, the basic constituents of the classical radiationfield, like the photons of quantum field theory, are the plane wave solutionsof Equations 6.14. Because of the constraints in Maxwell’s equations, allsource free fields must be transverse including the initial configurations.These when transformed to k are the basis for the travelers that emerge.

In order to understand this idea and make it rigorous, we must constructa consistent complete set of plane wave solutions that spans the space oftransverse source free solutions to Maxwell’s equations. This will req

Let’s work an example. Start with aIf the only system that we had was the source free electromagnetic fields,

we can quickly

6.2 Another Representation

The Fourier transformation is not the only set of functions that are completein a three dimensional space-time. Any of these sets can be used to expandthe transverse fields. There is an especially simple set that is a natural forexpansion of the electromagnetic fields. This is the set of vector sphericalharmonics discussed in Appendix F. This is an expansion that is directlyrelated to the rotational properties of the fields. In addition as discussedin Appendix B.4, for most applications of Fourier transform techniques, theangular integration required for the Fourier transforms in three space treatsthe unit circle without transformation. This allows these techniques to applyto both the original field configurations and to the Fourier transformations.

The vector spherical harmonic expansions are based on transformations

eJM(i)(r, t) ≡

d2Ωr

Y (i)∗

JM(r) · E (r.t) (6.18)

E (r.t) ≡

JM(i)

eJM(i)(r, t)Y(i)JM

(r) (6.19)


with a similar set of equations for the magnetic field and the Y (i)∗

JM(r) are

the Hansen vector spherical harmonics, Equation F.90.In this representation, it is easy to isolate the transverse fields; the sum

in Equation 6.19 is restricted to the two cases, (i) = (e) or (i) = (m). Oneconclusion that is then immediately apparent is that there is no J = 0,rotationally unchanged, transverse modes. Since we already know that thetransverse fields are the traveling modes, there is no such thing as monopoleradiation.

We can find the source free evolution equations for the scalar functions,eJM(i)(r, t).

∂eJM(e)(r, t)∂t

=

d2ΩrY (e)∗

JM(r) ·

∂ E (r.t)

∂t

= c2

d2ΩrY (e)∗

JM(r) ·

∇× B (r.t)

=

J M (i)

c2

d2ΩrY (e)∗

JM(r) ·

∇×

bJ M (i)(r, t)Y

(i)J M (r)

=

J M (i)

c2

d2ΩrY (e)∗

JM(r)

·i1

2J + 1

d

dr+

2J + 1r

bJ M (i)(r, t)Y

(i)J M (r)

= ic2

2J + 1

d

dr+

2J + 1r

bJM(e)(r, t) (6.20)

and

∂bJM(e)(r, t)∂t

=

d2ΩrY (e)∗

JM(r) ·

∂ B (r.t)

∂t

= −i1

2J + 1

d

dr+

2J + 1r

eJM(e)(r, t). (6.21)

Similarly,

∂eJM(m)(r, t)∂t

=

d2ΩrY (m)∗

JM(r) ·

∂ E (r.t)

∂t

= ic2

2J + 1

d

dr+

2J + 1r

bJM(m)(r, t) (6.22)

6.3. SOURCE FREE FIELD ACTION 95

and

∂bJM(m)(r, t)∂t

=

d2ΩrY (m)∗

JM(r) ·

∂ B (r.t)

∂t

= −i1

2J + 1

d

dr+

2J + 1r

eJM(m)(r, t). (6.23)

From these it follows that

∂2eJM(i)(r, t)∂t2

− (6.24)

the (1, 1)

6.3 Source Free Field Action

In order to produce a mechanical description of these two field systems, weeliminate the B fields from the dynamical equations, Equations 6.14, whichmanifests the oscillator nature of the the solutions.

1c2

∂2ET

(1)

∂t2(k, t) = −k2

ET

(1)(k, t)

1c2

∂2ET

(2)

∂t2(k, t) = −k2

ET

(2).(k, t). (6.25)

In the modern approach to mechanics, an action which reproduces thedynamics is the basis for a full mechanical description. In the case of thesource free electromagnetic fields in k space each of the two degrees of free-dom are simple oscillators and the action is thus well known. In this casethough, there is still some ambiguity since the dynamic is linear homo-geneous in the fields and there is thus no means to set the scale of thefields E and B, i. e. the units of the field strengths. Consistent with ourfield theoretic approach, the scale of the fields is set by the dimensional re-quirements. The dynamic, Equations 6.14, does set the relative dimensions,BT

(2)(k, t) dim= T

LET

(1)(k, t) where in this equation T is a time dimension and L

is a length dimension. Remember that k has the dimensions of an inverselength, see Equation 6.9. In addition, the units of the basic source free equa-tions, require that a parameter of dimension L

Tbe identified for the traveler

solutions to make sense. In this regard the c is a required conversion factorthat transforms times into lengths.


A “natural” system of units for this case would be to assign a unit tothe field such that a quadratic form of the time derivatives of the field be aLagrangian or a part of one; for example

Lag ∼12

E2

for each k serving as a kinetic energy like term. On taking the variation ofthis form with the field to identify the dynamical equations, this constructionwill yield the required first order differential system. Since the action is

S =

dtLag,

the Lagrangian has dimension

Lagdim=

ML2

T 2, (6.26)

the dimension of a kinetic energy, which implies that the field E has thedimension

Edim=

ML2

12 . (6.27)

Using Equation 6.9, the dimensions of the electric field are then

Edim=

M12

L2. (6.28)

Diversion on Units

In the usual approach to electromagnetism, the dimension of E is givenfrom its definition drawn from Gauss’s Law, ∇ · E(x, t) = ρ(x,t)

and the

electric part of the Lorentz force equation, f(x, t) ≡ ρ(x, t) E(x, t). Notethat the subscript on the has been dropped. This will allow an analysisof unit systems beside the rationalized SI system. Assuming that thereis an independent unit for charge, Q, these equations imply that Q

2

dim=ML

3

T 2 . Actually, the definition of charge is the time integral of a current

or Qdim= iT or in the rationalized SI system the coulomb is an ampere

second. In this form the proper relationship is i2

dim= ML3

T 4 . This equationmakes it clear that the is a factor that is required to convert the chargeunits into mechanical units. A similar analysis of the requirement that∂ E

∂t(x, t) = c2∇× B(x, t)− c2µ0j(x, t) be dimensionally homogeneous yields


Bdim= µ i

Land combined with the magnetic portion of the Lorentz force yields

µi2dim= ML

T 2 . Again, the µ plays the role of converting the current units tomechanical units. Since we really treat currents only as moving charges,these two conversions must be related. There really is no independent unitfor the source in the dynamic of the magnetic field. This restriction ofcurrents to being moving charges relates them dimensionally but also colorsour view of the sources for the electromagnetic field and this has a significantimpact on the Galilean invariance of our equations. This is discussed inSection 8.2. Solving for i2, this then constrains the dimensional content ofeach of these correction factors to satisfy 1

µ

dim= L2

T 2 . Not only that but sincethe travelers of the source free system are light, the velocity squared mustbe the speed of light. It is this identification that allowed us to replace 1

µ

with c2 from the beginning in our form of Maxwell’s equations. With thisunderstanding of the dimensional content of constants of electromagnetism,we can discuss several options. In the CGS ESU system, is chosen to be 1implying that there are no independent units of charge. In this case, the µattached to the current source must be fit to guarantee that 1

µis the speed

of light in the CGS sytem, 3× 1010 cmsec . in the rationalized MKS SI system,

the permeability of the vacuum, µ0 is defined to be 4π × 10−7 N

A2 . The 4πpart of the definition is geometric. It is the area of the unit sphere in unitsin which the area of the unit square is 1. If area had units independentlyof length units, we would need a conversion factor for areas from L2. Forexample areas are in units of scrums. The unit circle is one scrum and theconversion from scrums to the other dimensional description of area, L2, inwhich the area of the unit square is 1. For instance, in the MKS system,this implies that 1 scrumMKS = 4π M2.

Equation 5.6, E ≡ limQ→0FQ

Qwhere FQ is the force on a small test charge

Q. Inserting the usual form of Coulomb’s Law, the dimension of Edim= ML

T 2Q.

Coulomb’s law then requires Q2

dim= ML3

T 2 where is a dimension that is setby the unit system for charges being used. In other words, is a conversionfactor for returning charge units to mechanical units. For example, in theelectrostatic system of units = 1 and there is no independent unit ofcharge. For the rationalized SI units which are the most commonly usedunits, a separate unit of charge is defined from the the ampere and time,the coulomb C. In this case, the the electric field has dimension ML

T 2Q. This

would be an inappropriate approach in a source free setting but if we want tobe consistent with the usual definitions we need to connect with the usualapproach. Our form of Maxwell’s equations, Equations 5.1, are those of


the usual SI form and with the two dimensional factors 0 and µ0 locatedon the two source terms only. In the original form of Maxwell’s equations,the c2s in Equations 5.1 were c2 ≡

10µ0

. Although motivated by concernsabout local causality, it was Maxwell’s great contribution to realize that thedisplacement current, the term, 1

µ0

∂ E

∂t(x, t) in Equations 5.1, was required for

charge conservation and, once incorporated, allowed in certain cases for thefield dynamic to be hyperbolic. The existence of this term in the equationsthen implies that there are travelers and these have the velocity

1

0µ0.

When he realized this combination of the unit terms had a numeric valueclose to that of the speed of light, he concluded that light was the travelersolutions of the electromagnetic field dynamic.

This “natural” system of units for the fields could have been arrive atby using the fact that all disturbances of the transverse source free field aretravelers we the scale and thus dimensions of E can be set by measuringthe energy absorbed into a perfectly absorbing boundary. It is interestingto speculate that, had Fresnel realized that the transverse vector amplitudethat he was dealing with when he developed his wave theory of light was anelectric phenomena, he would have defined a field strength unit that was aresult of the energy transmitted when the light was absorbed and given us aunit for the field strength based on mechanical units and that the definitionabove that comes from the almost contemporaneous work of Faraday inEngland with the longitudinal part of the field which leads to the definitionabove may have never been introduced.

6.3.1 Return to the Action for E&M

Regardless of the historical precedents, in order to fully develop the mechan-ical basis of the source free electromagnetic fields, we will have to identifythe action in detail.

From Equations 6.25, there are two oscillators for each k and thus theLagrangian for our source free system is

LSF (k, t) =

i=1,2

λ(k)

12

∂ET

∗

(i)

∂t(k, t)

∂ET

(i)

∂t(k, t)

−(kc)2

2

E

T∗

(i) (k, t)

ET

(i)(k, t)

, (6.29)

where λ(k) is the dimensional factor which brings the Lagrangian into energyunits and the variational degrees of freedom which produce the dynamic,


Equation 6.25, are ET

(i)(k, t) and ET∗

(i) (k, t) and the subscript SF indicatesthat this is for the source free case. It is important to note that, because ofthe reality of the E(x, t), in order to not double the degrees of freedom, thevariables ET

(i)(−k, t) = ET∗

(i) (k, t), see B.2, should not be counted twice; therange of k is only over the half sphere.

There are conjugate momenta,

p(1)(k.t) ≡∂L(k)

∂∂ET

(1)

∂t

=λ(k)

2∂ET

∗

(1)

∂t= −ikc2 λ(k)

2B

T∗

(2)(k, t)

p∗(1)(k.t) ≡∂L(k)

∂∂ET∗

(1)

∂t

=λ(k)

2∂ET

(1)

∂t= ikc2 λ(k)

2B

T

(2)(k, t)

p(2)(k.t) ≡∂L(k)

∂∂ET

(2)

∂t

=λ(k)

2∂ET

∗

(2)

∂t= ikc2 λ(k)

2B

T∗

(1)(k, t)

p∗(2)(k.t) ≡∂L(k)

∂∂ET∗

(2)

∂t

=λ(k)

2∂ET

(2)

∂t= −ikc2 λ(k)

2B

T

(1)(k, t). (6.30)

where the last substitution in each line uses Equation 6.14 and, in thissense of Lagrangian dynamics, provides a definition of the magnetic field,BT

(i)(k, t), and these terms within a multiplicative factor now playing theroll of the conjugate momenta. With these in hand, we can construct the kspace Hamiltonian for this system:

HSF (k, t) =

i=1,2

λ(k)2

(kc)2E

T∗

(i) (k, t)ET

(i)(k, t) + c2B

T∗

(i) (k, t)BT

(i)(k, t)

.

(6.31)The dynamical equations, Equation 6.14, are the resulting Hamilton’s equa-tions. Using the completeness of the e(i) for the transverse fields and thefact that all the fields in the source free case are transverse,

HSF (k, t) =λ(k)

2(kc)2

ET

∗(k, t) · ET (k, t) + c2 BT

∗(k, t) · BT (k, t).

(6.32)Adding in the longitudinal terms which are zero in the source free case towrite the full k space Hamiltonian,

HSF (k, t) =λ(k)

2(kc)2

E∗(k, t) · E(k, t) + c2 B∗(k, t) · B(k, t)

. (6.33)


Summing over all the degrees of freedom, the total Hamiltonian for thissystem is

HSF (t) = ∞

−∞d3k

λ(k)4

(kc)2

E∗(k, t) · E(k, t) + c2 B∗(k, t) · B(k, t)

(6.34)This Hamiltonian emerged from a Lagrangian that was time translation in-variant and thus there is a conserved quantity, the energy, see Apendix A.Although the Hamiltonian is still expressed in terms of the k space descrip-tion, we can see that the choice λ(k) = 0

(kc)2 will yield an especially simpleform. Later, we can show that 0 is the usual permittivity of the vacuum.Making this substitution and, using the Parseval Identity, Equation B.19,the Hamiltonian becomes

HSF (t) = ∞

−∞d3x

04

E(x, t) · E(x, t) + c2 B(x, t) · B(x, t)

. (6.35)

Thus we identify the source free energy density, ρeSF (x, t), as

ρeSF (x, t) ≡04

E(x, t) · E(x, t) + c2 B(x, t) · B(x, t)

(6.36)

and, from Equation 6.6, the local current density which balances energydensity changes is

jESF (x, t) ≡ −0c2

2

E (x, t)× B (x, t)

. (6.37)

In a similar fashion, this Hamiltonian and the corresponding action arespace translation symmetric.

6.4 A Theory of Photons

6.4.1 Introduction

Although our purpose is a classical description of electromagnetic phenom-ena, the construction of these solutions of the source free field based onsimple harmonic oscillators is a natural and straight forward vehicle for theconstruction of a quantum field theory of electromagnetic phenomena. Ifwe deal with the source free sector of the theory, we would be developing atheory of photons. The full theory of Quantum Electrodynamics is a theoryof electrons and photons.

6.4. A THEORY OF PHOTONS 101

The route to this formulation is to simply make the oscillators in Sec-tion 6.1 quantum oscillators. This is not the usual path to a quantum theoryof a field system. In that case, from the classical action of the fields, themechanical degrees of freedom are identified and the field version of thePoisson bracket is related to the commutation relations appropriate to thea quantum system, see Appendix 10. In our case, we have developed theaction for our fields from the identification of the intrinsic oscillator natureof the source free fields. It should be clear that these two routes to quanti-zation lead to the same construction of a quantum system. The route usedhere has the advantage of interpretational simplicity and much of the con-structional ambiguities that emerge in the direct field theory approach aremanaged by the power of the Fourier transform machinery. In the follow-ing subsection we review the properties of the simple quantum mechanicaloscillator, particular the relationship to the classical oscillator. Followingthat construction, we develop our theoretical construction of the photons.These are simple photons. There are no charged particles and constructionof the full quantum field theory requires the presence of the charges. Thiswill require the full machinery of quantum field theory and will be saved forlater development, see Section ??.

6.4.2 The Quantum Classical Oscillator

The quantum oscillator is defined by the Hamiltonian

H =p2

2m+ k

q2

2(6.38)

and the commutation relationships for the hermitian operators, q and p;[q, q] = [p, p] = 0 and [q, p] = i.

Defining the the non-Hermitean operator a =

mω

2 q + i p√2mω where

ω ≡

k

m. It is important to note that each term of a is dimensionless

and ω is dimensionally a frequency but we have no reason for interpretingit as an oscillator frequency. Granted, when Equation 6.38 is a classicalHamiltonian, ω is the radian frequency of the oscillator.

The commutation relations for the operator a and its adjoint become[a, a] = [a†, a†] = 0 and [a, a†] = 1. It follows that [a, a†a] = a and [a†, a†a] =−a†. Thus, the spectrum is a ladder spectrum with a moving down theladder and a† moving up where, since the Hamiltonian in Equation 6.38which is now expressed as

H = ω

a†a +

12

(6.39)


is positive definite, there is a lowest step on the ladder,

a|0>= 0. (6.40)

Thus the ladder states generated by

|n>≡a†n

√n!|0>, n = 0, 1, 2, · · · (6.41)

are the stationary states with energy En = ωn + 1

2

. The states |n> form

an orthonormal basis, <n|m>= δnm, are complete,∞

n=0 |n><n| = 1, andare the eigenstates of the operator a†a with eigenvalue n.

Using Equation 6.40, we can find the representation of these stationarystates in the coordinate or q representation. The ground state satisfies

0 =

dq <q|a|q><q|0>=

2mω

mω

q +∂

∂q

ψ0(q) (6.42)

whose solution is the well known Gaussian with a standard deviation of

mωand which is normalized to

∞−∞ ψ2

0(q)dq = 1,

ψ0(q) = 4

mω

πe−

mω q2

2 , (6.43)

and where ψ0(q) =<q|0>. The probability distribution in q, ψ∗0(q)ψ0(q),

has mean of zero and a standard deviation of

2mω

. Since the expectationvalue for the position in the ground state,

0 < q >0= 0, (6.44)

where the label 0 on the brackets indicates the state used for the evaluationof the expectation value and, since the variance2 of a Gaussian is the squareof the standard deviation,

0 << q >>0=0< q2 >0=

2mω. (6.45)

All the higher ladder states wave functions can be found from Equa-tion 6.41,

<q|n>≡ ψn(q) =1√

n!

mω

q −∂

∂q

n

2mω

n2

ψ0(q). (6.46)

2The variance of an operator A in the state |β> is defined as β << A >>β=β< A2

>β

− (β < A >β)2. Note that β << A >>β has the dimensional content of A2.


These are the generator equations for the Hermite functions.Another important state for our purposes is the state |α> which is the

eigenstate of the ladder operator a or a|α>= α|α> where α is an arbitrarycomplex number3. This state is generated by the displacement operator

D(α) ≡ e(αa†−α

∗a) (6.47)

operating on the ground state,

|α>= e(αa†−α

∗a)|0> . (6.48)

These states are normalized and are expanded in the ladder basis using theBaker-Campbell-Hausdorf expansion4 as

|α> = e(αa†)e−(α∗a)e−

|α|22 [a.a

†]|0>

= e−|α|2

2

n=∞

n=0

αn

√n!|n> . (6.49)

That this state is an eigenstate of the operator a with eigenvalue α canbe verified from the operator equations, D†(α) = D−1(α) = D(−α) andD†(α)aD(α) = a+α. These states are called coherent states. The coherentstates are an over-complete set and the completeness relation is

1π

|α><α|d2α = 1. (6.50)

Another important property of the coherent states is that the probability ofoccupation of the state |n> is

P (n, α) ≡ |<n|α>|2 =

e−|α|2 |α|2

n

n!(6.51)

which is a Poisson distribution with mean |α|2 and standard deviation |α|.Two special cases will be especially important to us, α real and α imag-

inary. For the case α = d real, the operator D(d) = e−i

q2

ωm pd in the qrepresentation becomes

Dq(d) = e−

q2ωm d

∂∂q . (6.52)

3Since the operator a is not Hermitian its eigenfunctions have complex eigenvalues.4As a special case, e

A+B = eA

eB

e−[A,B] 12 , if [A, [A, B]] = [B, [A, B]] = 0.


Since for any smooth function of q

eγ∂∂q f(q) =

∞

n=0

γn

n!∂n

∂qnf(q) = f(q + γ),

the operator Dq(d) translates any wave function in the q representation by

the amount −

2ωm

d. Similarly, if α is imaginary, α = id, D(id) = eid

q2mω

q.In the p representation, q is i ∂

∂pand thus

Dp(id) = e−√

2mωd∂∂p (6.53)

which translates any wave function in the p representation by the amount−√

2ωmd.Applying Equation 6.52 to the ground state in the q representation,

Equation 6.43,

ψq2ωm d

(q) ≡ e−

q2mω d

∂∂q ψ0(q) = 4

mω

πe−

mω

„q−√

2mω d

«2

2 , (6.54)

we now have a state that has minimum uncertainty and is a Gaussian withstandard deviation

mωand mean

qd ≡

2mω

d. (6.55)

The probability distribution of this state is also a Gaussian with the samemean but now with standard deviation

2mω.

This new state, Equation 6.54, is not stationary. The evolution is bestunderstood in terms of the the evolution of the underlying ladder states.Although the ladder states are stationary, their probability distributions aretime independent, their wave functions have a time dependence in the phaseand, since each ladder state has a different phase development, superposi-tions of these states do have an evolution through interference5. In otherwords, the time dependence of the displaced state can be found by addingthe time dependence to Equation 6.49,

ψqd(q, t) = e−d2

2

∞

n=0

dn

√n!

ψn(q, t)

5Alternatively, ψqd(q, t) = e

−i H

tψqd

(q, 0) which leads to Equation 6.56 since Hψn(q) =

(− 2

2m∂2

∂q2 + mω2 q2

2 )ψn(q) = ω`n + 1

2

´ψn(q)


= e−d2

2

∞

n=0

dn

√n!

e−iω(n+ 12)tψn(q) (6.56)

= e−iωt2 e−

|de−iωt|22

∞

n=0

de−iωt

n

√n!

ψn(q)

= e−iωt2 Dq(de−iωt)ψ0(q), (6.57)

The operator

D(de−iωt) = ede−iωt

a†−de

iωta

= e−i

nq2mω

qd sin(ωt)+q

2mω pd cos(ωt)

o

(6.58)

= e−i

q2mω

qd sin(ωt)e−i

q2

mω pd cos(ωt)e

1 d

2 sin(ωt) cos(ωt)[q,p],

(6.59)

where Equation 6.59 follows from Equation 6.58 using the Baker-Campbell-Hausdorf expansion. In the q representation Equation 6.59 becomes

Dq(de−iωt) = e−i

q2mω

qd sin(ωt)e−

q2mω d cos(ωt) ∂

∂q eid2 sin(ωt) cos(ωt)

= e−id sin(ωt)

nq2mω

q−d cos(ωt)o

e−qd cos(ωt) ∂∂q (6.60)

and thus our evolving wave function is

ψqd(q, t) = 4

mω

πe−i

nωt2 +d sin(ωt)

nq2mω

q−d cos(ωt)oo

e−mω

(q−qd cos(ωt))2

2 .

(6.61)The magnitude of Equation 6.61 is

ψ∗qd(q, t)ψqd(q, t) =

mω

πe−

mω (q−qd cos ωt)2 (6.62)

and thus the probability distribution is again a Gaussian but now withmean qd cos ωt and the same standard deviation as the original ground state,

2mω

. Also since the distribution in q is a Gaussian, the variance is thestandard deviation squared,

qd << q >>qd=0<< q >>0=

2mω(6.63)

The interpretation of this result is clear. The minimum uncertainty wavepacket originally at a distance qd ≡

2ωm

d from the origin moves without


changing shape in the same way a classical oscillator with radian frequencyω would. It is also important to note that, although this result may appearto be a classical limit to the quantum oscillator, it is an exact quantumresult; there were no large n approximation or d1 or d1 conditions. Itis true that any macroscopic classical system has its displacement from theorigin large compared with the variance of the the quantum ground stateand thus operationally qd

2mωwhich implies that d 1

2 .It is worthwhile to develop other features of this special state. The

expectation value for the energy which is independent of time is

qd < E >qd = ∞

−∞ψ∗qd

(q, t)Hψqd(q, t)dq (6.64)

= ∞

−∞ψ∗qd

(q, 0)−

12m

∂2

∂q2+

mω2

2q2

ψqd(q, 0)dq

= ∞

−∞ψ∗0(q, 0)

−

12m

∂2

∂q2+

mω2

2(q − qd)2

ψ0(q, 0)dq

= 0<

H −mω2qqd +

mω2

2q2d

>0

=12

ω +mω2

2q2d

(6.65)

where qd is the is displacement of the oscillator zero point. As expectedthis result is the zero point energy plus the classical energy. Similarly, theexpectation value of the energy squared is

qd < E2 > qd = ∞

−∞ψ∗qd

(q, t)H2ψqd(q, t)dq

= 0<

H −mω2qqd +

mω2

2q2d

2

>0

=

12

ω +mω2

2q2d

2

+mω2

20< q2 >0q

2d

(6.66)

or, although the expectation value of the energy is the classical value plusthe zero point energy, the variance of the energy is

qd << E >>qd=mω2

2q2d 0<< q >>0=

mω2

2q2d

2mω

, (6.67)

and is purely a quantum effect, proportional to which vanishes in the limit→ 0.


It is important to understand that this oscillator is not the most generalcase of classical motion possible. Since the classical oscillator equation issecond order in time, there is both a starting position and an initial velocitypossible. It should be obvious that we can obtain the case of the initialvelocity distribution with similar results to the above by doing all of theanalysis in the p-representation of the oscillator quantum mechanics. In thep representation, the ground state condition, Equation 6.40, is

i

mω

2

∂

∂p+

p

mω

ψ0 (p) = 0, (6.68)

and the ground state wave function6 which leads to a normalized probabilitydistribution is

ψ0 (p) =1

4√

mωπe−

p2

2mω , (6.69)

which has a mean of 0 and standard deviation of√

mω. Again, the as-sociated momentum probability distribution has a mean of 0 and standarddeviation of

mω

2 .Following the same development as before except in the p representation,

but using Equation 6.53 or 6.58 with t = − π

2ω,

ψpd (p) =1

4√

πmωe−

(p−pd)2

2mω (6.70)

where pd ≡√

2mω d which is a Gaussian with mean of pd and standarddeviation of

√mω. As before the probability distribution is a Gaussian

with mean pd and standard deviation

mω2 .

Using the appropriately weighted Fourier transform, the q representationwave function is

ψpd (q) =1√

2π

∞

−∞ei

pq

1

4√

πmωe−

(p−pd)2

2mω

dp√

= 4

mω

π eipdq

e−mωq2

2 . (6.71)

6Another approach to finding the wave function in the p representation would be theuse of Fourier transforms, Appendix B, but some care must be exercised since, although p

and q are conjugate variables in Lagrange mechanics, the appropriate variables for Fouriertransforms are dimensionally inverses of each other, Equation B.29. For example, if q isa length, p is a linear momentum in mechanics but in the Fourier transform case, theappropriate variable is the wave number, k ≡ p

, which is an inverse length. Also thetransforms are dimensional inverses but the modulus of the wave function squared is aprobability distribution and, thus, the wave function is normed to have dimension that isthe inverse square root of the distribution variable.


This is a Gaussian with mean 0 and standard deviation

mω

multipliedby a phase factor and normed so that the modulus squared is normed for adistribution in q.

The temporal evolution follows the same pattern as before.

ψpd (p, t) = e−iωt2 Dp

de−i(ωt−π

2 )

ψ0(p)

=1

4√

mωπe−

iωt2 e

−id sin ωt

“q2

mω p−d cos ωt

”

e−(p−pd cos ωt)2

2mω .

(6.72)

Again, a Gaussian with an extra phase term. The probability distributionchanges momentum in the same pattern as a classical oscillator with radianfrequency ω.

The evolving q representation wave function is found using the Fouriertransformation of Equation 6.72 and is

ψpd (q, t) = 4

mω

π e−iωt2 eipd cos ωt

q e−

mω2 (q− pd

mω sin ωt)2

. (6.73)

This wave function produces a probability distribution in q that is a Gaussianwhich has a standard deviation of

2mωand a mean of pd

mωsin ωt. It moves

in space in the same way a classical oscillator with radian frequency ω excitedby an initial velocity of pd

m.

Returning to the p representation, we find the expectation of the energyto be pd < E >pd=

12ω+ p

2d

2m, the zero point energy plus the classical energy.

The variance follows similarly to the development of Equation 6.67 and is

pd << E >>pd=1

m2p2

d 0<< p >>0=p2

d

m2

mω2

. (6.74)

We are now in a position to incorporate the most general initial condi-tions and allowing it to evolve. Consider the case of a displacement in boththe momentum and position of the ground state,

α = d1 + id2 ≡ Deiγ (6.75)

with D ≡

d21 + d2

2 and γ ≡ tan−1

d2d1

. The position displacement is

qd1 ≡

2ωm

d1 =

2ωm

D cos γ and the momentum displacement is pd2 ≡√

2mω d2 =√

2mω D sin γ.


The analysis of Equations 6.56 through 6.77 holds and leads to the re-placement of ωt by ωt− γ in Equation 6.60 to yield,

Dq

De−i(ωt−γ)

= e

−iD sin(ωt−γ)nq

2mω q−D cos(ωt−γ)

o

e−

q2ωm D cos(ωt−γ) ∂

∂q .

(6.76)The evolving wave function is

ψqd1pd2

(q, t) = 4

mω

πe−i

nωt2 +D sin(ωt−γ)

nq2mω

q−D cos(ωt−γ)oo

×e−mω

„

q−qd1cos(ωt)−

pd2mω sin(ωt)

«2

2 . (6.77)

As expected, the magnitude is a Gaussian with mean qd1 cos(ωt)+ pd2mω

sin(ωt)

and standard deviation

mω

. The probability distribution for this stateis the magnitude squared and thus has the same mean and the standarddeviation is

2mω.

To study the energy expectation and the variance it is easier to work inthe operator form of Equation 6.59 with the appropriate argument. Againusing the time independence of the expectation value of the energy andfollowing the development of Equation 6.64 through 6.65,

qd1pd2

< E >qd1pd2

= 0< ei

q2mω pD cos γ

e−i

q2mω

qD sin γeiD

2 cos γ sin γ

×

p2

2m+

mω2

2q2

ei

q2mω

qD sin γ

×e−i

q2mω pD cos γ

e−iD2 cos γ sin γ >0

= 0<

e−i

q2mω

qD sin γ p2

2mei

q2mω

qD sin γ

+mω2

2q2

>0

= 0<(p− pd2)

2

2m+

mω2

2(q − qd1)

2 >0

= 0< H −ppd2

m+

p2d2

2m−mω2qqd1 +

mω2

2q2d1

>0

=ω

2+

p2d2

2m+

mω2

2q2d1

(6.78)

Again, this is the zero point energy plus the classical energy.


Proceeding similarly to find the variance in the energy for this case, wefollow the same pattern as Equation 6.66. By putting all the quadraticdependence on the operators p and q in H, the expectation for the energysquared becomes

qd1pd2

< E2 > qd1pd2

=

12

ω +mω2

2q2d1

+p2

d2

2m

2

+mω2

20< q2 >0q

2d2

+ 0< p2 >0p2d2

m2+0< q, p+ >0ω

2qd1pd2 (6.79)

Where q, p+ is the anticommutator of q and p and is the only new featureof the energy squared expectation. It is straight forward to show that thevacuum expectation value of this operator is zero7. Thus the variance of theenergy is the sum of the variances of the separate p and q translates of theground state. As in the previous cases, this is a purely quantum effect andwould vanish in the limit goes to zero.

6.4.3 The Quantized Electromagnetic Field

This result will be helpful in interpreting the large amplitude fields that arethe basis of classical electromagnetism, see Section 6.1 and Equation 6.25; foreach k, these fields are oscillators with radian frequency kc. In a sense, we arenow reversing the development of Section 6.3 where we took the mechanicaloscillator nature of the source free fields to discover the appropriate action.Following the route to quantum mechanics of using a classical action toproduce a quantum theory.

7The operator qp = −i2 (a + a

†)(a− a†) = −i

2 (aa− a

†a† − 1) has the vacuum expec-

tation 0< qp >0 = i2 and pq is the adjoint of qp.

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