2. Classical dynamics and electrodynamics

2-1

2. Classical dynamics and electrodynamics

Interactions in chemistry are almost exclusively electromagnetic in nature, with gravity, weakand strong interactions seldom of interest. Classical electrodynamics or its quantized versionsare therefore required to understand molecular interactions in detail. The standard text inclassical electrodynamics is Jackson's Classical Electrodynamics.

Similarly, classical mechanics can be of help. Often our intuition is better in the classicaldomain, and we can guess what an answer might look like, even though classical dynamics willnot provide an exact solution. Even for those with a perfect "quantum intuition", classicaldynamics still provides an alternate and useful viewpoint. Ironically, most physical chemistsknow quantum mechanics much better than classical mechanics, although their intuition islargely based on classical dynamics. The following pages provide a condensed review of thesubject, while appendix B contains a more extensive discussion. A short and excellent text onthe subject is Landau and Lifshitz, Classical Mechanics, while a more exhaustive and wordytreatment is provided by H. Goldstein, Classical Mechanics.

2.1 A brief review of classical dynamics

2.1.1 Newton's, Lagrange's and Hamilton's equations of motion

We start by motivating the Lagrangian from Newton's equation of motion:

miai = Fi ⇒ middt

˙ x i = −∂V∂xi

; if K =12i

∑ mi ˙ x i2

⇒ddt

( ∂∂ ˙ x i

K) = −∂V∂xi

2-1

⇒

ddt

( ∂∂˙ x i

L) − ∂∂xi

L = 0

where L = K − V is the Lagrangian. Although this is derived here in cartesian coordinates, moregeneral considerations (Appendix B) show that this holds even when L ≠ K-V and coordinatesare not cartesian. Note that L is a function of the coordinates and velocities. Usually, L does notexplicitly depend on time (only implicitly via q and ˙ q ). To solve the dynamics, we need to findall "constants of the motion", of which there are at most n for n coordinates (for n degrees offreedom). From eq. 2-1

if ddt

∂∂ ˙ q i

L⎧ ⎨ ⎩

⎫ ⎬ ⎭

= 0 ⇒

∂∂ ˙ q i

L is a constant of the motion: 2-2

One constant of the motion is always guaranteed: In one dimension, what is the value of

ddt

( ˙ q ∂L∂˙ q

− L) = ? = ˙ ̇ q ∂L∂ ˙ q

+ ˙ q ddt

(∂L∂˙ q

) − ddt

L

= ˙ ̇ q ∂L

∂ ˙ q + ˙ q ∂L

∂q−∂L∂q

dqdt

−∂L∂ ˙ q

d ˙ q dt

−∂L∂t

2-3

=0 if L is not explicitly time-dependent and ∂L/∂t is zero.

2-2

In cartesian coordinates specifically, we find that

∂L∂˙ x

=∂K∂˙ x

=∂˙ x ( 12

m ˙ x 2 ) = m˙ x = p

⇒ ˙ x ∂L

∂˙ x − L = ˙ x p − K + V (x) = p2

m−

p2

2m+V = K + V = E = H 2-4

Therefore, the sum of kinetic and potential energies, or total energy, is always a conservedquantity if the Lagrangian does not explicitly depend on time. Again, these equations hold inany coordinate system:

H(q, p) = ˙ q i∑ pi − L where pi =

∂L∂˙ q i

2-5

where pi is the momentum conjugate to qi. It follows directly from eq. 2-2 that if a coordinate qidoes not appear in the Lagrangian (or Hamiltonian), its conjugate momentum pi is conserved.For example, for a free particle H = p2/2mis independent of x, therefore p must be constant (andof course E is constant).

The Hamiltonian and Lagrangian are related by the Legendre transform 2-5, which also playsa role in thermodynamics. They depend on different variables: L on q and ˙ q , and H on q and p.To obtain H from L, one

i) computes pi according to equation 1-5, then solves this for the ˙ q i as a function of the pi and qi.

ii) computes H(qi,pi) according to equation 2-5 by using i) to eliminate ˙ q i.Due to the properties of Legendre transforms, one cannot write H as a function of q and p or L asa function of q and ˙ q : no consistent equations of motions can be obtained if this is attempted(see Callen, Thermodynamics).

The quantum Hamiltonian can usually be obtained from the correspondence principle by thesubstitution

H(q, p)→ ˆ H ( ˆ q , ˆ p ) = ˆ H ( ˆ q ,

i∂∂ ˆ q

) , 2-6

barring a commutation problem (see appendix A-8).What are the equations of motion in terms of H rather than L? The perfect differential of H

can be written as (using only one dimension for simplicity)

dH =

∂H∂q

dq + ∂H∂p

dp; also, dH = d( ˙ q p − L ) = d˙ q p + ˙ q dp −∂L∂q

dq −∂L∂˙ q

d ˙ q .

= ˙ q dp −ddt

(∂L∂˙ q

)dq

= ˙ q dp − ˙ p dq2-7

⇒ ˙ q =

∂H∂p

, ˙ p = −∂H∂q

(symplectic form analogous to ˙ Ψ i = −HΨr / ˙ Ψ r = + HΨi /

) 2-8

Eqs. 2-8 are Hamilton's equations of motion. We now have two ways of solving the problem ofclassical motion:

-n second order Lagrange equations in n variables qi in eq. 2-1-2n first order Hamilton equations in 2n variables qi, pi in eq. 2-8

2-3

Area conserved

p

q

q(0),p(0)

q(t),p(t)

q

p

ρ

ρ(0)ρ(t)

Fig. 2-1. Top shows a 2-D phase spacecorresponding to one degree of free-dom. Velocity and position can be readoff simultaneously and trajectoriesnever cross. Bottom shows evolutionof a classical density function. Its shapemay change, but the area is always 1,since the number of trajectories is a conserved quantity.

The solution is complete with n constants of the motions of which one, the energy, is alwaysguaranteed for a time-independent Lagrangian or Hamiltonian. It may be difficult to findtheremaining n-1 in practice, but sometimes they do not even exist in principle: if the actualnumber of constants of the motion is less than n, the motion will be chaotic, and nearby orbitswill diverge from one another rapidly and unpredictably. In fact, this is the case for mostsystems of many degrees of freedom, and the old adage "given initial position and velocity, theorbit can be computed for all future" needs to be amended "given initial position and velocitywith infinite accuracy...". Of course, infinite accuracy is never available, and so classicalmechanics is in practice rarely deterministic.

The energy is independent of the motion, and the time dependence of q and p is given byequations 2-8. How does an arbitrary dynamical variable F(q,p,t) depend on time? The answeris provided by the Poisson bracket. Consider the time derivative of the function F,

dFdt

=∂F∂qi

dqidt

+∂F∂pi

dpidt

+∂F∂ti

∑

=∂F∂qi

∂H∂pi

−∂F∂pi

∂H∂qi

+∂F∂ti

∑ or

dFdt

= [F,H]P + ∂F∂t

2-9

The final equation is the Liouville equation of motion, and the quantity []P defined by the middleand final equations is the Poisson bracket of F with H. If F does not explicitly depend on time,and [F,H]P=0, then F is a constant of the motion.

By the correspondence principle, if F → ˆ F and H → ˆ H then dˆ F / dt = 1

i [ˆ F ,H] + ∂ ˆ F / ∂t.The latter equation is the Heisenberg equation of motion, which was the first fully correctversion of nonrelativistic quantum mechanics, as developed by Heisenberg in 1925.

2.1.2 Motion in phase spaceThe space of the 2n coordinates and momenta of Hamilton's equations of motion is called the

classical phase space. For one degree of freedom, this space is two-dimensional. For example,the small amplitude oscillations of a pendulum appear as an ellipse (fig. 2-2); the pendulum is

2-4

-2π 2π

V(θ)

θ

L

θ

Vibration Nearly free rotation

Fig. 2-2. Potential energy and phase spacediagram of the pendulum. At the anglesmarked by a dashed line lies a potentialenergy barrier. Below it, motion resem-bles that of a harmonic oscillator; above it,motion resembles that of a free rotor.

discussed below in more detail. Consider a distribution of trajectories in phase-space with adistribution function ρ(qi , pi ,t) ≥ 0 . Such trajectories can never cross: two trajectories with onepoint in common have the same initial conditions, and subject to the same Hamiltonian, musthave all points in common. Such trajectories also cannot disappear or appear out of nothing(conservation of particles), so ρ must satisfy

.. . dqidpiρ(qi , pi ,t) =1∫∫∫∫ 2-10at all times. ρ gives the probability of finding a set of initial conditions. Since it is constant atall times, it follows that

ddt

. .. dqidpi∫∫ ρ = 0⇒ .. . dqidpi∫∫ ∑ ( ∂ρ∂qi

dqidt

+∂ρ∂pi

dpidt

+∂ρ∂ t) = 0 2-11

or, since this can only hold in general if the integrand dρ/dt is identically zero,∂ρ∂t

= −[ρ, H]P 2-12

For example, if we start a swarm of particles with a normal distribution of x and p values aboutx=p=0, ρ would start out as a gaussian function, but no matter what shape it evolved into under

2-5

eq. 2-12, the integral of ρ over all phase space must remain constant. ρ(qi(t = 0), pi(t = 0))specifies the initial condition of the distribution of trajectories.

Eq. 2-12 is also referred to as the Liouville equation of motion; although it superficiallyresembles eq. 2-9, it is in fact very different. The dynamical variable F in 2-9 is usually notexplicitly dependent on time, but only via x(t) and p(t); the Poisson bracket gives its total timederivative. In contrast, ρ has no explicit time dependence, since its total integral over phasespace is constant; in 2-12, the Poisson bracket yields minus the partial time derivative of ρ at agiven position (p,q) in phase space, since the trajectory distribution may change locally subject tooverall trajectory conservation. Finally, it should be noted that the quantum mechanical quantitycorresponding to ρ is the density matrix ˆ ρ in section 3.

As another particular example let ρ(t = 0) =1 / V over a phase space volume ∫ ∫V i∏ dpidqi ; that

is, ρ represents a "pillbox" in phase space: constant in some are ΔqΔp phase space volume, zeroelsewhere. Then it follows from eq. 2-11 that

dqi∫∫ (t = 0)dpi(t = 0) = dqi∫∫ (t)dpi(t) . 2-13Thus classical propagation in phase space is an incompressible flow. A bundle of trajectorieswhich occupies a certain volume in phase space will always occupy the same volume. However,the initially smooth volume may become highly interdigitated and spread "arbitrarily narrowtendrils" throughout phase space, thereby occupying a point arbitrarily close (but not necessarilyON) any chosen point in phase space. This is known as classical chaos. The structure ofHamilton's equation's 2-8 is highly conducive to producing chaotic trajectories. In matrix form,we could rewrite equation 2-8 as

˙ q ˙ p

⎛ ⎝ ⎜

⎞ ⎠ ⎟

~=

∂ / ∂q∂ / ∂p⎛ ⎝ ⎜

⎞ ⎠ ⎟

~0 −11 0⎛ ⎝ ⎜

⎞ ⎠ ⎟

H 00 H

⎛ ⎝ ⎜

⎞ ⎠ ⎟ . 2-8a

The middle matrix has unit determinant (is area-preserving, as noted above). It belongs to a classof matrices known as "symplectic". Although symplectic transformations preserve area, theyinduce shear and folding of trajectories in phase space, and this causes chaos.

2.1.3 Cyclic motions and action-angle variablesMany important types of motions are periodic, for example vibrations and rotations. In

quantum mechanics, these periodic motions lead to discrete energy levels labeled by quantumnumbers. In the case of vibrations, discrete quantum numbers arise because the wavefunction isbounded on both sides by a potential; in the case of rotations, discrete quantum numbers arisebecause of circular boundary conditions. Do quantum numbers have a classical analog?

We consider explicitly the example of the pendulum Lagrangian and Hamiltonian to applysome of the ideas introduced in i) and ii). The potential surface and phase space structure of thependulum are shown in figure 2-2: at low energies, it has the topology of a harmonic oscillator;at high energies, that of a free rotor (as the pendulum swings over the top).ex:

L =

12mr2 ˙ Θ 2 + a cosϕ hindered rotor/pendulum 2-14

∂L∂ ˙ Θ

= mr2 ˙ ϕ = Lϕ , the momentum conjugate to ϕ : the angular momentum 2-15

H = ˙ ϕ Lϕ − L =

L2

2mr2 − a cosϕ = K + V . 2-16

In one dimension, H = H(p,q) = E(p,q)→ p(q, E) (usually 2m(E − V)) : we can always solvefor the momentum in terms of E and q. We now define a new momentum (see fig. 2-3)

2-6

p'= I = 12π

p(q, E)dq = I(E)∫ . 2-17

Fig. 2-3. Integration contour for action variables lies at constant energy as shown by the arrow. Integrationis over a complete period of the motion.

E=H(p,q)

p

q

To this definition also corresponds a new coordinate. The old phase space coordinates (q,p) andthe new phase space coordinates (Θ,I) are related by a canonical transformation (canonicaltransformations are described in appendix B)

q→ Θ(p,q), p→ I(p,q) . 2-18

As indicated in 2-17, we can solve E=E(I), which does not depend on coordinateΘ (or on q,which has been integrated out) only on I; according to Hamilton's equations, we then have

∂H∂I

= ˙ Θ = ω(I)

∂H∂Θ

= −˙ I = 0 2-19

Therefore, I is a constant of the motion, and ˙ Θ depends only on I since H depends only on I.⇒Θ = Θ0 + ˙ Θ (I)t = Θ0 + ωt 2-20

The variable I is called an action variable; it is constant, like an angular momentum under freerotation. The variable Θ is an angle variable, since it evolves linearly in time like an angle underfree rotation. In fact, if we let a=0 in eq. 2-14, we have

H =L2

2mr2I = 12π

LdΘ = L;∫∂H∂L

=Lmr2

= ω(L)→ L =mωr2 2-21

=mr2 ˙ Θ so I = L is just the standard angular momentum for the free rotor system.

Consider as a final example the central force problem in two dimensions (e.g. rotation of aclassical electron under Coulomb attraction about a nucleus):

F = −

∂V(r)dr

; K =12µ( ˙ x 2 + ˙ y 2 )⇒ L =

12µ( ˙ x 2 + ˙ y 2 ) − V(r ) 2-22

First, let us do a coordinate transformation to polar coordinates:

x = r cosΘ→ ˙ x = −r ˙ Θ sinΘ + ˙ r cosΘ

y = rsinΘ → ˙ y = r ˙ Θ cosΘ + ˙ r sinΘ2-23

2-7

From this follows the transformation of the Lagrangian to polar coordinates:

⇒ ˙ x 2 + ˙ y 2 = r2 ˙ Θ 2 + ˙ r 2

⇒ L =12

µ(r 2 ˙ Θ 2 + ˙ r 2 ) − V(r )

⇒ pΘ = LΘ = ∂L∂ ˙ Θ

= µr2 ˙ Θ = µr2ω

2-24

pr =

∂L∂˙ r

= µ˙ r 2-25

pΘ and pr are the angular and radial momenta conjugate to Θ and r, and ω = ˙ Θ is the angularvelocity. Since the Lagrangian does not depend on Θ explicitly, it follows that LΘ is a constantof the motion: angular momentum is conserved. The same is not true for the r degree offreedom, since it follows from Lagrange's equation that

ddt

(µ˙ r ) − µr ˙ Θ 2 +∂V∂r

= 0⇒ µ˙ ̇ r = µr ˙ Θ 2 −∂V∂r

, so r is not constant. 2-26

In fact, the motion in the radial direction might follow an elliptical path for the CoulombHamiltonian. We can now derive the Hamiltonian and Hamilton's equations from theLagrangian:

H(r,pr,Θ, LΘ ) = ˙ q i pi − L = ˙ r ∑ pr + ˙ Θ LΘ −

12

µ(˙ r 2Θ 2 + ˙ r 2 ) + V(r) 2-27

=LΘ2

2µr2+pr2

2µ+V(r) 2-28

The kinetic energy is clearly separated into a rotational part (which depends on r) and avibrational part, which is independent of Θ. Inserting into the four Hamilton's equations ofmotion, one obtains

1) ˙ r = pr

µ 2) ˙ Θ = LΘ

µr2

3) ˙ p r =LΘ

2

2µr3 −∂V∂r

4) ˙ L Θ = 0 which implies LΘ = const.2-29

We can verify that LΘ is indeed a constant of the motion by evaluating its Poisson bracket: dLΘ

dt=∂LΘ

∂Θ∂H∂LΘ

−∂LΘ

∂LΘ

∂H∂Θ

+∂LΘ

∂r∂H∂pr

−∂LΘ

∂pr

∂H∂r

= −∂H∂Θ

= 0 so ˙ L Θ = [LΘ,H] = 0 2-30

Consider the action angle variables for this problem (fig. 2-3). In one dimension, we had anangle and a constant action, so the motion followed a circle or ring. In two dimensions, a similarresult is obtained, but for two degrees of freedom the cyclical motions move on two rings.Topologically, motion on two circles in 4 phase space coordinates corresponds to a torus in 4-D.Often, the motion is visualized as a torus in 3-D as shown in the figure, but this is only avisualization aid and should not be taken too literally. For the action variables I1 and I2,

I1 =1

2πLΘdΘ∫ = LΘ; ∂H

∂I1=ω(I1) =ω (LΘ) = LΘ

µr 2 or LΘ = µr 2ω ; 2-31

I2 =12π

dr 2µ E − V(r ) − LΘ2

2µr 2( )∫ 2-32

2-8

p1

q1

p2

q2

I1

Θ1

I2

Θ2

I1

I2

Fig. 2-4. Motion of the 2-D rotor in 4-D phase space (or ofa pair or harmonic oscillators). Each action variable cor-responds to a circle (a "1-torus") in phase space, and themotion goes around the circle clockwise. Simultaneousmotion around two circles corresponds to motion on a "4-torus", which is often depicted symbolically in threedimensions by a donut as shown at the bottom.

If we let V(r) be a harmonic potential, we obtain in the special case of L=0

V(r) = 12µ

ω 2r2 ⇒ I2 =1π

2µE (1 − ( µω' 2

2E−

2Eµω ' 2

+ 2Eµω ' 2

∫ r) 2 )1/ 2dr 2-33

= 1π

2µE (1 − cos2π

0

∫ u )12−sinu

µω2

2E

⎛

⎝ ⎜

⎞

⎠ ⎟ du

= 1π2Eω

sin20

π

∫ ΘdΘ =1π2Eω(u2− sin 2u

0

π)

= Eω

⇒∂E∂I2

= ω .

The variable substitution cosu = r

√(µω2/2E) was used in the second line.Therefore, E = ωI2 = ωn if we let I2 = n. This closely resembles the harmonic oscillator

equation and indicates that by the correspondence principle I → n . Quantum-mechanically, Evib = ω (n +1) if the correct zero point energy factor for a 2-D degenerate oscillator isintroduced (which of course cannot be predicted by classical theory). It holds generally thataction variables become "quantum numbers times Planck's constant", while the angle variablesbecome the time evolution phase factors of the wavefuntion. For one-dimensional potentials, theWentzel-Kramers-Brillouin approximation provides a rigorous semiclassical formulation of eq.2-17.

2-9

2.2 A brief review of classical electrodynamics

Before we begin with Maxwell's equations, here is a summary of some vector identities (seealso appendix 2):

∇ ×∇a = 0∇⋅∇ × a = 0

∇ ×∇ × a = −∇a2 +∇(∇⋅ a)∇⋅ (a × b) = b ⋅ (∇ × a) − a ⋅(∇× b)v = ∇ × f − ∇χ for any v

2-34a-e

2.2.1 Maxwell's equationsThe speed of light is given by c ≡{εµ}−1/ 2 ≈ ε −1/ 2 where ε is the dielectric constant and µ is

the magnetic permeability. In terms of charge and current densities ρ and j, Maxwell's equationsare

∇⋅E = 4πρ 2-35∇⋅B = 0 2-36

∇ × E = −1c∂B∂t

2-37

∇ × B =1c∂E∂t

+4πcj 2-38

For example, a point charge e at position r0 would have a charge density eδ(r-r0). Combiningthe time derivative of 2-35 and the divergence of 2-38, one obtains the continuity equation

∇⋅ j + ∂ρ∂t

= 0 or jA∫ ⋅ d ˆ a = −

∂∂t

ρdVV∫ , 2-39

the latter using Gauss' law in Appendix A. The Maxwell equations therefore require chargeconservation, since the current flowing out of a closed surface must equal the loss of chargedensity in the volume contained within that surface. Expanding ρ in a multipole series,

ρ = ρ0 − ∇ ⋅P+.. . 2-40where ρ0 is the monopole charge density and P is the dipole moment; this type of Taylorexpansion is motivated in section 9 for the magnetic density ρM = −∇⋅M , and the reason is thesame here: the terms in the series fall off ever more rapidly as 1/r (monopole), 1/r2 (dipole), 1/r3(quadrupole), etc. Using the continuity equation 2-39,

j = j0 +∂P∂ t

+... 2-41where j0 is the dc current, ∂P/∂t is the oscillating dipole current, etc.

V

A

jρ

Fig. 2-5. Relationship between vectorflow field out of a surface and scalardensity contained within that surface.

2-10

In terms of these multipole charges and currents, the Maxwell equations can be rewritten todipole order as

∇⋅ (E + 4πP) = 4πρ0 2-42∇⋅B = 0 2-43

∇ × E = −1c∂B∂t

2-44

∇ × B =1c∂∂t(E + 4πP) + 4π

cj0 2-45

Here, any electric dipole density is explicitly taken into account. The quantity E + 4πP is oftenreferred to as the electric displacement D.

2.2.2 Vector and scalar potentials

Since ∇⋅B = 0⇒ B = ∇ ×A , where A is a "vector potential". 2-46

From 2-37⇒ ∇× (E +1c∂A∂t

) = 0 if P = 0

⇒ E +1c∂A∂t

= −∇ϕ ,

where ϕ is a scalar potential (hence ∇2ϕ = 0 , Poisson's equation), or

E = −∇ϕ −1c∂A∂t

. 2-47

Eqs. 2-46 and 2-47 express the 6 components of B and E in terms of only four quantities A andϕ. Under static conditions, ϕ is just the familiar electrostatic (e.g. Coulomb) potential.

2.2.3 Gauge invariance

Since ∇ ×∇χ = 0 for any χ it follows that B is invariant if we let A→A' = A +∇χ

Since E = −∇ϕ −1c∂A∂t

→ E = −∇ϕ −1c∂A∂t

+∇1c∂∂t

χ ⇒ E = −∇(ϕ +1c∂χ∂t) − 1

c∂A'∂t

2-48

ϕ 'Thus, the substitutions

A' = A +∇χ 2-49

ϕ '= ϕ −1c∂χ∂t

2-50

leave E and B unchanged. This gives us the freedom to place an additional constraint on A orϕ , i.e. to impose a gauge, since E and B are gauge-invariant; the Coulomb and Lorentz gaugesare common; we will use the Coulomb gauge since it is most convenient for nonrelativisticdiscussion of radiation fields. Can we pick

∇⋅A = 0 ? 2-51Let ∇⋅A ≠ 0; ∇⋅A'= ∇⋅ (A +∇χ ) = 0 is desired.⇒ A +∇χ =∇ × f for some f, since ∇⋅∇ × f = 0 always.⇒ A = ∇ × f -∇χ ; any arbitrary vector can be written like this, so we can indeed choose∇⋅A =0.

2-11

2.2.4 Wave equations

We can write wave equations both for E and B or A andϕ . The solution of such equationsclearly shows that electromagnetic fields can propagate at the speed of light.

Example 1: wave equation for E with ρ0=0 and j0=0, but P not necessarily zero:

From 2-44 ⇒ ∇×∇ × E = −1c∂∂ t

∇ × B or − ∇2E +∇(∇⋅E) = −1c∂∂t∇ × B . 2-52

From 2-45 ⇒ −1c∂∂t∇ × B = −

1c2

∂ 2

∂ 2t(E + 4πP) 2-53

Combining these two equations, ∇ ×∇ × E +1c2

∂ 2

∂2tE = −

4πc2

∂2

∂ 2tP . 2-54

This is an equation for an electric field wave E(r,t) emitted or absorbed by a dipole density P.

Example 2: wave equation for A, with ρ0=0, j0=0 and P=0 (free space):

From eq. 2-45 ∇ × B =1c∂E∂t

⇒ ∇×∇ ×A = −1c∂∂t

∇⋅ ϕ −1c2

∂ 2A∂ 2t

2-55

Far away from source charges (free space)∇⋅ ϕ →0, and

−∇2 ⋅A +∇(∇⋅A) = − 1c2

∂ 2A∂ 2t

2-56

Making use of the Coulomb gauge, we finally obtain

∇2A −1c2

∂ 2A∂2t

= 0 . 2-57Far from charges, only A counts, ϕ ≈ 0. The solutions of this equation, such as the one below forplane wave boundary conditions, can be used to calculate E and B.

2.2.5 Plane wave solutions and their properties in free space

Consider the AnsatzA(r,t)=A0cos(k ⋅r − ωt + ϕ0 ) =A0e

ik⋅r−iωt +A *0 e− ik⋅r+iωt 2-58

for the vector potential. Inserting into the wave equation 2-57,−k2A +

1c2

ω 2A = 0 . 2-59

Fig. 2-6. Plane wave vector quantities.

A,E

k

B

2-12

E

qEv Fig. 2-7: Projection of the electric field

onto the particle trajectory needed to calculate the change in the energy U.

Trajectory

This is satisfied if |k|= k =ω / c . Since k .r should repeat after r has cycled through onewavelength, k=2π/λ. Similarly, ω = 2πν, and therefore νλ = c: the waves propagate at a speedc ≅ E . Evaluating E and B,

E ≅ −1c∂A∂t

=iωcA0e

ik⋅r−iωt + c.c 2-60

B =∇ × A = ik ×A0eik⋅r−iωt + c.c 2-61

From 2-60, A is clearly parallel to E, while from 2-61, B is perpendicular to both k and A. Fig.2-4 shows the direction of the relevant vectors. They correspond to a single-frequency,transverse electromagnetic wave with two independent polarizations λ =1,2 (k defines a planeperpendicular to it which contains E and can be described by two vectors, fig. 2-6).

2.2.6 Energy density of the field: the radiation Hamiltonian

To apply this in quantum-mechanics we need to derive the Hamiltonian for the radiation, so thatHtot=Hmol+Hfield+Hint. 2-62

From eq. 2-38c4π(∇ × B −

1c∂E∂t) = j⇒ 1

4π(cE ⋅ ∇ × B − E ⋅

∂E∂t) = j ⋅E

⇒14π(cB ⋅ ∇× E − c∇ ⋅(E × B) − E ⋅ ∂E

∂t) = j ⋅E

⇒14π(−B ⋅ ∂B

∂t− E ⋅ ∂E

∂t− c∇ ⋅(E × B)) = j ⋅E

⇒∂u∂t

+∇⋅S = − j ⋅E 2-63

whereu = 1

8π(B2 + E2 ) and S = c

4πE × B . 2-64

In the above derivation, first E. was added to both sides, then a vector identity was employed,and finally eq. 2-37. Eq. 2-63 is another continuity equation, which states that the integratedchange in electromagnetic energy density (u) inside a closed surface equals the losses throughthe surface due to radiation (S) and current flow (j). Units are j = charge per unit area and unittime ;u = energy per unit volume; S = energy per unit area per unit time. That u is indeed an energydensity can be seen as follows. Consider a particle of charge q (fig. 2-7) moving through apotential difference ΔV; its energy change ΔU is

ΔU = qΔV = −qEvΔx

⇒ ΔUΔt

= −qEvΔxΔt

or ∂U∂t

= −qv ⋅E2-65

2-13

For many small charges we can write instead−∑ qivi ⋅Ei = −V∫ j ⋅Edr3 . The total energy U is

the integral over the energy density in a volume, U = ∫dr3u, and therefore 2-65 yields∂u∂t

= − j ⋅E . 2-66

This is the same as 2-63, except that here we only took charge currents into account as sources ofenergy loss; the radiative loss S is included in the full derivation in 2-63.

2.2.7 Total energy of an unperturbed electromagnetic field

We now derive the classical Hamiltonian for an electromagnetic field. Using 2-64 and 2-47U = udv = 1

8πv∫ dr3∫ (E2 + B2) = 1

8πdr3∫ ( 1

c2˙ A 2+|∇ × A|2 ) 2-67

if the field is measured far enough from the charges so ϕ≈0. For simplicity, we will assume thatour molecules and radiation are contained in a conducting cubic box V=L3 much larger than λor molecular size. Expanding A in terms of plane waves 2-58,

A = Ak,λk,λ∑ (t)e ik⋅r +A *k,λ i

(t)e− ik⋅r 2-68

where |kj|=njπ / L and λ =1 or 2 (two polarizations perpendicular to k). The restriction on thevalues of kx, ky and kz arises from the cubic box, as multiples of the wavelength must fit withinthe box: the largest allowed wavelength is λ max = 2L. The time dependence e±iωt is included inAk ,λ (t). From this follows

U =1

8πdr3∫ { 1

c2 | ˙ A k,λk,λ∑ eik ⋅r + ˙ A *k,λ e

−ik⋅r |2 + |k,λ∑ ik × A k,λe

ik⋅r + ik × A *k,λ e− ik⋅r |2 }

=18π2V{ 1

c2| ˙ A k ,λ

k,λ∑ |2 + k 2 | Ak ,λ

k ,λ∑ |2 } 2-69

The simplification in the second step arises because e±2ik⋅r cross-terms average to zero over theintegration volume; also Ak ,1 ⋅Ak ,2=0 since the two polarization states are perpendicular so thesummation over λ remains a single sum. Finally,

. U =V4π{ ω 2

c2|Ak,λ

k,λ∑ |2 + k2 |Ak ,λ

k ,λ∑ |2}

=V2π

k2k,λ∑ Ak ,λA *k,λ 2-70

since k=ω/c in free space. Defining

ak,λ =Ak ,λ{Vkhc}1/2

eq. 2-70 can be rewritten as

U = ω

k ,λ∑ a *k,λ ak ,λ 2-71

This is the classical energy of the radiation field far away from the charges that generated it, interms of vector potentials for each wavevector and polarization.Note that

Ek,λ =

iωcAk,λ + c.c = iω

c{2πc

Vk}1/ 2 ak,λ ˆ e

k,λ−

iωc

{2πcVk

}1/ 2 a *k,λ ˆ e k,λ

. 2-72

2-14

2.2.8 Charge-field interaction energy

An important equation that must be added to Maxwell's equations is the force exerted on acharge by electromagnetic fields:

F = qE +qcv × B 2-73

This can be used to derive the Lagrangian for the particle in an electromagnetic field (homework#1), which yields (using only one charge for simplicity)

L =

12mv2 − qϕ +

qc

v ⋅A ⇒ p =∂L∂˙ r

= mv +qc

A ⇒ v =pm−qmc

A . 2-74

Note that the conjugate momentum is not simply mv anymore. Evaluating the Hamiltonian byLegendre transform,

H = v.p-L=p ⋅ ( pm

−qmcA) − 1

2m( p

m−qmcA)2 + qϕ −

qc( pm−qmcA) ⋅A 2-75

=p2

m−12p2

m−12q2

mc2⋅A2 + qϕ −

qmcp ⋅A +

q2

mc2⋅A2 2-76

The potential ϕ is taken to be the potential of the charge q, or ϕ=V. This accounts for most ofthe potential, although in the presence of a radiation field ϕ cannot necessarily be rigorouslyinterpreted as the potential of the isolated charge q. However, any residual potential due to faraway charges which may have been involved in creating the field is small, so V is the dominantcontribution to the potential. The total Hamiltonian for particle and radiation from eq. 2-76 istherefore

Htot =1

2m(p − q

cA)2 + V(r) + Hfield

= 12mpmech

2 +V (r) + Hfield

2-77

where Hfield ~ A2 as derived in 2-71, V(r) accounts for the potential energy of the charge q dueto nearby charges, and the coupling to the radiation field is given purely by the term qA/c, sincethe contribution of the radiation to the potential is assumed to be negligible. pmech is the usualmechanical momentum mv.