2 ‚t2

c ‚t 0

2

1‚

∇2 ’ 2 2 B = ’µ0 ∇ — j , (B.23)

c ‚t

and the ¬rst-order equations for the transverse vector potential A can be combined

to yield the wave equation

1 ‚2

A = ’µ0 j⊥ ,

∇2 ’ (B.24)

c2 ‚t2

where j⊥ is the transverse part (see Section 2.1.1-B) of the current.

B.3.1 Propagation in the vacuum

Since the B ¬eld and the transverse part of the E ¬eld are derived from the vector

potential, we can concentrate on the wave equation (B.24) for the vector potential. In

the vacuum case (j = 0), A satis¬es

1 ‚2

∇2 ’ A (r, t) = 0 , (B.25)

c2 ‚t2

and the transversality condition ∇ · A = 0.

Classical electrodynamics

The general solution of the wave equation can be obtained by a four-dimensional

Fourier transform, which yields

ω2

’k + 2 A (k, ω) = 0 .

2

(B.26)

c

The solution of the last equation is

A (k, ω) = A(+) (k) 2πδ (ω ’ ck) + A(’) (k) 2πδ (ω + ck) , (B.27)

and the reality of A (r, t) requires

—

A(+) (k) = A(’) (’k) . (B.28)

The inverse transform yields the general solution in (r, t)-space as

A (r, t) = A(+) (r, t) + A(’) (r, t) , (B.29)

where

—

d3 k

A 3A (k) ei(k·r’ωk t) = A(’) (r, t)

(+) (+)

(r, t) = , (B.30)

(2π)

and ωk = ck.

The relation between the E and B ¬elds and the vector potential can be used to

express them in the same way. In k-space

E (+) (k) = iωk A(+) (k) , B(+) (k) = ik — A(+) (k) , (B.31)

and in (r, t)-space

E (r, t) = E (+) (r, t) + E (’) (r, t) ,

(B.32)

B (r, t) = B(+) (r, t) + B(’) (r, t) ,

where

—

d3 k

3 iωk A (k) ei(k·r’ωk t) = E (’) (r, t)

E (+) (+)

(r, t) = ,

(2π)

(B.33)

—

d3 k

B(+) (r, t) = 3 ik — A (k) ei(k·r’ωk t) = B(’) (r, t)

(+)

.

(2π)

B.3.2 Linear and circular polarization

The forms in eqns (B.32) and (B.33) are valid for any real vector solutions of the wave

equation, but we are only interested in transverse ¬elds, e.g. A (r, t) should satisfy

∇ · A (r, t) = 0, as well as the wave equation. In k-space the transversality condition,

k · A(+) (k) = 0, requires A(+) (k) to lie in the plane perpendicular to the direction of

the k-vector. We choose two unit vectors e1 (k) and e2 (k) such that e1 (k) , e2 (k) , k

form a right-handed coordinate system, where k = k/k is the unit vector along the

propagation direction. The unit vectors e1 and e2 are called polarization vectors.

Wave equations

Since an arbitrary vector can be expanded in the basis e1 (k) , e2 (k) , k , the three

unit vectors satisfy the completeness relation

esi esj + ki kj = δij , (B.34)

s

as well as the conditions

k · es (k) = 0 , (B.35)

es (k) · es (k) = δss , (B.36)

e1 — e2 = k (et cycl) , (B.37)

where s, s = 1, 2. The vector A(+) (k) can therefore be expanded as

A(+) (k) = A(+) (k) es (k) , (B.38)

s

s

(+)

where the polarization components As (k) are de¬ned by

A(+) (k) = es (k) · A(+) (k) . (B.39)

s

The general transverse solution of the wave equation is therefore given by eqn (B.29)

with

d3 k

A (r, t) = A(+) (k) es (k) ei(k·r’ωk t) .

(+)

(B.40)

s

3

(2π) s

Each plane-wave contribution to the solution of the wave equation, say for E and

B, has the form

E = Re (E1 e1 + E2 e2 ) ei(k·r’ωk t) , (B.41)

1

B= k—E, (B.42)

c

where E1 and E2 are complex scalar amplitudes, and e1 and e2 are real polarization

vectors k. If the phases of E1 and E2 are equal, the ¬eld is linearly polarized. If the

phases are di¬erent, the ¬eld is elliptically polarized. In general, the time-averaged

Poynting vector is

1 1

Re E — B— =

0 2 2

|E1 | + |E2 |

S= k, (B.43)

µ0 2 µ0

so the intensity is given by

0 2 2

I = |S| = c |E1 | + |E2 | . (B.44)

2

If the two phases di¬er by 90—¦ , then

E1 e1 + E2 e2 = E0 (e1 ± ie2 ) , (B.45)

Classical electrodynamics

where E0 is real. The ¬eld is then said to be circularly polarized. In this case it is

useful to introduce the complex unit vectors

1

es = √ (e1 + ise2 ) , (B.46)

2

where s = ±1. The complex vectors satisfy the (hermitian) orthogonality relation

e— · es = δss , (B.47)

s

and the completeness relation

esi e— + ki kj = δij . (B.48)

sj

s

The transversality, orthogonality, and completeness properties of the linear and circular

polarization vectors are both incorporated in the relations

k · es (k) = 0 (transversality) ,

e— (k) · es (k) = δss (orthonormality) ,

s

(B.49)

esi (k) e— (k) = δij ’ ki kj (completeness) .

sj

s

Note that the completeness relation can also be written as

esi (k) e— (k) = ∆⊥ (k) , (B.50)

sj ij

s

where ∆⊥ (k) is the Fourier transform of the transverse delta function. The general

ij

solution (B.40) has the same form as for linear polarizations, but the polarization

component is now given by

A(+) (k) = e— (k) · A(+) (k) . (B.51)

s s

In addition to eqn (B.49), the circular polarization vectors satisfy

k — es = ’ises , (B.52)

es — e— = ’isδss k , (B.53)

s

where s, s = ±1.

The linear polarization basis for a given k-vector is not uniquely de¬ned, since

a new basis de¬ned by a rotation around the k-direction also forms a right-handed

coordinate system. It is therefore useful to consider the transformation properties of the

polarization basis. Let ‘ be the rotation angle around k; then the linear polarization

vectors transform by

e1 = e1 cos ‘ + e2 sin ‘ ,

(B.54)

e2 = ’e1 sin ‘ + e2 cos ‘ ,

which implies

es = e1 + ise2 = e’is‘ es . (B.55)

When viewed at a ¬xed point in space by an observer facing into the propagation

direction of the wave (toward the source), the unit vector e+ (e’ ) describes a phasor

Wave equations

rotating counterclockwise (clockwise). In the traditional terminology of optics and

spectroscopy, e+ (e’ ) is said to be left (right) circularly polarized. In the ¬elds of

quantum electronics and laser physics, the observer is assumed to be facing along the

propagation direction (away from the source), so the sense of rotation is reversed.

In this convention e+ (e’ ) is said to be right (left) circularly polarized. In more

modern language e+ (e’ ) is said to have positive (negative) helicity (Jackson, 1999,

Sec. 7.2).

For a plane wave with propagation vector k, there are two amplitudes Es (k), where

for circular (linear) polarization s = ±1 (s = 1, 2). The general vacuum solution can be

expressed as a superposition of plane waves. In this context it is customary to change

the notation by setting

ωk

Es (k) = 2i ±s (k) , (B.56)

20

2

where the is introduced only to guarantee that |±s (k)| is a density in k-space,

i.e. the new amplitude ±s (k) has dimensions L3/2 . This yields the Fourier integral

expansion

d3 k ωk

E (+) (r, t) = i ±s (k) es (k) ei(k·r’ωk t) . (B.57)

3 20

(2π) s

The Fourier integral representation is often replaced by a discrete Fourier series:

ωk

E (+) (r, t) = ±ks eks ei(k·r’ωk t) , (B.58)

2 0V

ks

√

where eks = es (k), ±ks = ±s (k) / V , and V is the volume of the imaginary cube

used to de¬ne the discrete Fourier series.

B.3.3 Spatial inversion and time reversal

Maxwell™s equations are invariant under the discrete transformations

r ’ ’ r (spatial inversion or parity transformation) (B.59)

and

t ’ ’t (time reversal) , (B.60)

as well as all continuous Lorentz transformations (Jackson, 1999, Sec. 6.10). The phys-

ical meaning of spatial inversion is as follows. If a system of charges and ¬elds evolves

from an initial to a ¬nal state, then the spatially-inverted initial state will evolve

into the spatially-inverted ¬nal state. Time-reversal invariance means that the time-

reversed ¬nal state will evolve into the time-reversed initial state.

For any physical quantity X, let X ’ X P and X ’ X T denote the transforma-

tions for spatial inversion and time reversal respectively. The invariance of Maxwell™s

equations under spatial inversion is achieved by the transformation rules

ρP (r, t) = ρ (’r, t) , jP (r, t) = ’j (’r, t) , (B.61)

Classical electrodynamics

E P (r, t) = ’E (’r, t) , (B.62)

BP (r, t) = B (’r, t) . (B.63)

Thus the current density and the electric ¬eld have odd parity, and the charge density

and the magnetic ¬eld have even parity. Vectors with odd parity are called polar

vectors, and those with even parity are called axial vectors, so E is a polar vector and

B is an axial vector.

Time-reversal invariance is guaranteed by

ρT (r, t) = ρ (r, ’t) , jT (r, t) = ’j (r, ’t) , (B.64)

E T (r, t) = E (r, ’t) , (B.65)

BT (r, t) = ’B (r, ’t) . (B.66)

As a consequence of these rules, the energy density and Poynting vector satisfy

uP (r, t) = u (’r, t) , SP (r, t) = ’S (’r, t) ,

(B.67)

uT (r, t) = u (r, ’t) , ST (r, t) = ’S (r, ’t) .

For many applications, e.g. to scattering problems, it is more useful to work out

the transformation laws for the amplitudes in a plane-wave expansion of the ¬eld. We

begin by using eqn (B.58) to express the two sides of eqn (B.62) as

ωk P

E P (r, t) = ±ks eks ei(k·r’ωk t) + CC

i (B.68)

2 0V

ks

and

ωk

’E (’r, t) = ’ ±ks eks ei(’k·r’ωk t) + CC .

i (B.69)

2 0V

ks

Changing k to ’k in the last result and equating the coe¬cients of corresponding

plane waves yields

±P eks = ’ ±’k,s e’k,s . (B.70)

ks

s s

In order to proceed, we need to relate the polarization vectors for k and ’k. As

a shorthand notation, set es = eks , es = e’k,s , and k = ’k. The vectors es lie in

the same plane as the vectors es , so they can be expressed as linear combinations of

e1 and e2 . After imposing the conditions (B.35)“(B.37) on the basis {e1 , e2 , k }, the

relation between the two basis sets must have the form

e1 = e1 cos ‘ + e2 sin ‘ ,

(B.71)

e2 = e1 sin ‘ ’ e2 cos ‘ .

The transformation matrices in eqns (B.54) and (B.71) represent proper and improper

rotations respectively. The improper rotation in eqn (B.71) can be expressed as the

product of a proper rotation and a re¬‚ection through some line in the plane orthogonal

Planar cavity

to k. Since the polarization basis can be freely chosen, it is convenient to establish a

convention by setting ‘ = 0, i.e.

e’k,1 = ek1 , e’k,2 = ’ek2 . (B.72)

For the circular polarization basis, with s = ±, this rule takes the equivalent forms

e’k,s = e— ,

ks

(B.73)

e’k,’s = eks .

The transformation law derived by applying this rule to eqn (B.70) is

±P = ’±’k,’s (s = ±) , (B.74)

ks

which relates the amplitude for a given wavevector and circular polarization to the

amplitude for the opposite wavevector and opposite circular polarization. For the linear

polarization basis the corresponding result is

±P = ’±’k,1 ,

k1

(B.75)

±P = ±’k,2 .

k2

Turning next to time reversal, we express the right side of eqn (B.65) as

ωk ωk — — ’i(k·r+ωk t)

E (r, ’t) = ±ks eks ei(k·r+ωk t) ’

i i ±ee , (B.76)

2 0 V ks ks

2 0V

ks ks

and again change the summation variable by k ’ ’k. This is to be compared to the

expansion for E T (r, t), which is given by eqn (B.68) with ±P replaced by ±T . The

ks ks

result is

±— e—

±T eks = ’ ’k,s ’k,s . (B.77)

ks

s s

The circular polarization vectors satisfy e— —

’k,s = ek,’s = ek,s , so the transformation

law in this basis is

±T = ’±— ’k,s . (B.78)

ks

Thus for time reversal the amplitude for (k, s) is related to the conjugate of the

amplitude for (’k, s). The wavevector is reversed, but the circular polarization is

unchanged. For the linear basis the result is

±T = ’±—

’k,1 , (B.79)

k1

±T = ±—

’k,2 . (B.80)

k2

B.4 Planar cavity

A limiting case of the rectangular cavity discussed in Section 2.1.1 is the planar cavity,

L. In most applications, only the limit L ’ ∞ will

with L1 = L2 = L and L3 = d

be relevant, so the only physically meaningful boundary conditions are those at the

¼ Classical electrodynamics

planes z = 0 and z = d. Periodic boundary conditions can be used at the other faces

of the cavity. Thus the ansatz for the solution is E = eiq·r U (z), where q = (kx , ky )

is the transverse part of the wavevector k. Inserting this into eqns (2.11), (2.1), and

(2.13) leads to the mode functions E qns . For n = 0 there is only one polarization:

1

E q0 = √ eiq·r uz . (B.81)

2d

L

For n 1 there are two polarizations, the P polarization in the (q, uz )-plane and the

orthogonal S polarization along uz — q:

2 n»qn q

E qn1 = eiq·r sin (kz z) q + i cos (kz z) uz , (B.82)

L2 d 2d kz

2 iq·r

E qn2 = sin (kz z) uz — q ,

e (B.83)