Mitin V.V., Sementsov D.I., Vagidov N.Z. Quantum mechanics for nanostructures

Подождите немного. Документ загружается.

B.1 Equations of an electromagnetic ﬁeld 343

where the energy density of the electric ﬁeld is

E · D =



(B.31)

and the energy density of the magnetic ﬁeld is

B · H =

2µµ

. (B.32)

Using the dependences of ﬁeld characteristics on coordinates, we can ﬁnd with the help

of Eq. (B.30) the total energy of an electromagnetic ﬁeld in any arbitrary volume.

Example B.1. Let us consider a sphere of radius R, which is uniformly charged with

the total charge Q. Find the energy of the electric ﬁeld inside the sphere and outside it,

considering that the ﬁeld intensity distribution in such a sphere is known:

E(r) =











, r ≥ R,

, r ≤ R.

(B.33)

Outside of the sphere the electric ﬁeld is equal to the electric ﬁeld of a point charge Q

placed at the center of the sphere. Inside of the sphere the ﬁeld at a distance r from its

center is deﬁned also as a ﬁeld of a point charge

= Q

, (B.34)

which is placed into the center of the sphere and is equal to a charge inside of a sphere of

radius r.

The distribution of the intensity of the electric ﬁeld, E(r), deﬁned by Eq. (B.33)is

shown in Fig. B.1.

Reasoning. Let us substitute Eq. (B.33) into the expression for the energy density of the

electric ﬁeld, Eq. (B.31):

(r) =











8π

, r ≥ R,

8π

, r ≤ R.

(B.35)

Let us calculate with the help of Eq. (B.30) the energy of the electric ﬁeld, W

, concentrated

outside of the sphere, and the energy W

inside of the sphere. Taking into account that in

the case of spherical symmetry the differential volume is deﬁned as

dV = 4πr

dr, (B.36)

344 Appendix B. Electromagnetic ﬁelds and waves

Figure B.1 Distributions of

the electric ﬁeld, E (r)

(dashed line), and density

of electric ﬁeld energy,

(r) (solid line), in the

case of an electrically

charged sphere.

for the above-mentioned energies we obtain the following expressions:



out

8π

dV =

8π



∞

4πr

dr =

, (B.37)



8π

dV =

8π



4πr

dr =

10R

. (B.38)

The ratio of total energies of the electric ﬁeld outside and inside of the sphere does not

depend on the radius, R, and is equal to

= 5. (B.39)

Example B.2. The equation that deﬁnes the law of charge conservation and establishes

at each point of space the relation between the density of charge, ρ(r), and the density of

current, j(r), can be obtained from Maxwell’s equations. Derive the charge-conservation

equation.

Reasoning. Let us perform the scalar multiplication of the left-hand and right-hand sides

of Eq. (B.16) by the operator ∇ and, in the ﬁrst term on the right-hand side of the equation

obtained, let us change the order of taking time and spatial derivatives:

∇ · (∇ × H) =

∂∇ · D

∂t

+ ∇ · j. (B.40)

In Eq. (B.40) the left-hand side is identically equal to zero:

∇ · (∇ × H) = (∇ × ∇) · H = 0, (B.41)

B.2 Electromagnetic waves 345

because ∇ × ∇ = 0. On the right-hand side of Eq. (B.40) let us replace ∇ · D by ρ, taking

into account Eq. (B.19). As a result we obtain the equation

∂ρ

∂t

+ ∇ · j = 0, (B.42)

which is called the continuity equation (there is an analogous equation in hydrodynamics,

where ρ is the density of liquid and j = ρv is the density ﬂux of liquid).

Let us choose an arbitrary point r, and encircle it by the closed surface S. Then let us

integrate Eq. (B.42) over the volume V which this surface encloses:

∂

∂t



ρ(r)dV =−



(∇ · j)dV. (B.43)

The integral on the left-hand side of Eq. (B.43) is equal to the total charge Q in the volume

V that is enclosed by the surface S. The integral over the volume, V , on the right-hand

side of Eq. (B.43) can be transformed into an integral over the closed surface, S:

=−

j · dS =−I. (B.44)

This equation represents the fact that the change of the charge inside of the closed surface

is deﬁned by the current through this surface.

B.2 Electromagnetic waves

B.2.1 Wave equations

It follows from Eqs. (B.1)and(B.2) that electric and magnetic ﬁelds can exist in a medium

as well as in vacuum. These ﬁelds can exist near charges and currents, as well as far from

them. Moreover, it follows from Maxwell’s equations, (B.16)–(B.19) that time-varying

ﬁelds can exist without any charges or currents. The change in time of the ﬂux, D,leadsto

the creation of the solenoidal magnetic ﬁeld, H , and the changing magnetic ﬂux density,

B, induces the solenoidal electric ﬁeld, E. The above-mentioned changes (disturbances),

which are called electromagnetic waves, propagate in a medium as well as in vacuum.

The wave character of electromagnetic ﬁeld propagation is built into Maxwell’s equa-

tions. To check this statement, let us consider a uniform and isotropic medium from which

free charges and currents are absent (ρ = 0andj = 0). Let us extract from the system

of equations (B.16)–(B.19) two equations with rotors and let us replace the ﬁeld ﬂux

densities by ﬁeld intensities, assuming that  and µ do not depend on time, t:

∇ × H = 



∂E

∂t

, (B.45)

∇ × E =−µ

∂H

∂t

. (B.46)

346 Appendix B. Electromagnetic ﬁelds and waves

Let us take the time derivative of both sides of Eq. (B.45):

∂

∂t

∇ × H = ∇ ×

∂H

∂t

= 



∂

∂t

, (B.47)

where we changed the order of time and spatial derivatives. Let us substitute into Eq. (B.47)

∂H/∂t from Eq. (B.46):

−

∇ × (∇ × E) = 



∂

∂t

. (B.48)

The rule of opening of a double vector product of any three vectors A, B,andC is the

following:

A × (B × C) = B · (A · C) − C · (A · B). (B.49)

Using this rule, we can carry out the following transformations:

∇ × (∇ × E) = ∇ · (∇ · E) −∇

E =−∇

E. (B.50)

Here we took into account that, according to Eq. (B.19), for the case considered here

( = constant and ρ = 0) the following equality has to be fulﬁlled:

∇ · D = ∇ · (E) = ∇ · E = 0, (B.51)

i.e., ∇ · E = 0. By substituting Eq. (B.50) into Eq. (B.48) we obtain the wave equation

for the electric ﬁeld, E:

∇

E =

µ

∂

∂t

. (B.52)

Analogously, we can obtain the wave equation for the magnetic ﬁeld intensity, H:

∇

H =

µ

∂

∂t

. (B.53)

Here we introduced the quantity c:

c =

√



= 3 × 10

−1

, (B.54)

the magnitude and dimension of which coincide with those of the speed of light in vacuum.

The parameter

v =

√

µ

(B.55)

B.2 Electromagnetic waves 347

is the phase velocity of the electromagnetic ﬁeld propagating in a medium. For vacuum

 = µ = 1, and the phase velocity, v, of the electromagnetic wave coincides with the

speed of light, c.

To deﬁne the ﬁelds associated with an electromagnetic wave, let us write the solutions

for Eqs. (B.52)and(B.53) and for the ﬁeld intensities of electric and magnetic ﬁelds in

the form of plane monochromatic traveling waves:

E = E

cos(ωt − k · r), (B.56)

H = H

cos(ωt − k · r), (B.57)

where E

and H

are the amplitudes of the corresponding ﬁelds, ω is the frequency of the

oscillations of ﬁeld intensities E and H in the wave, and k is the wavevector.

The magnitude of the wavevector, k, is equal to

k =

2π

, (B.58)

where λ = vT is the wavelength, which is equal to the distance covered by the wave with

phase velocity v during the period of oscillations T = 2π/ω.

The direction of wave propagation coincides with the direction of the wavevector k.

Indeed, if we assume that the argument in Eqs. (B.56)and(B.57) is constant, it leads to

the equation

k ·r −ωt = constant, (B.59)

which deﬁnes the space planes of equal wave phases propagating with velocity v in the

direction of the wavevector k. The scalar product k ·r can be written in terms of the

projections on the Cartesian coordinates

k ·r = k

x + k

y + k

z, (B.60)

where k

= k cos α, k

= k cos β,andk

= k cos γ ; α, β,andγ are the angles between

the wavevector, k, and the corresponding coordinate axis. The constants cosα,cosβ,and

cos γ are called the directional cosines of the vector k (see Fig. B.2). Thus, the spatial

change of the wave’s phase takes place in the direction of the vector k, namely this vector

deﬁnes the direction of wave propagation in an isotropic medium.

Following substitution of solutions (B.56)and(B.57) into Eqs. (B.52)and(B.53), we

can obtain the relation between the frequency, ω, and the wavenumber, k.Asaresult

we will obtain the dispersion equation for the electromagnetic wave in an isotropic non-

absorbing medium,

µ

, (B.61)

ω = vk. (B.62)

348 Appendix B. Electromagnetic ﬁelds and waves

Figure B.2 The orientation

of the wavevector k in

terms of the angles of the

directional cosines α, β,

and γ .

If the parameters of the medium  and µ depend on the frequency, then the phase

velocity, v, also depends on the frequency, i.e., such a medium is dispersive. If  and µ

do not depend on the frequency (this happens for real media in certain frequency ranges),

then such a medium is not dispersive (in this case the phase and group velocities coincide).

Let us substitute Eqs. (B.56)and(B.57) into Eqs. (B.45)and(B.46), then we get

k ×H =−

ωE, (B.63)

k ×E = µ

µωH. (B.64)

According to the deﬁnition, the directions of the three vectors that compose the scalar

triple product, E, H,andk, are perpendicular to each other and constitute in this order a

right-handed system.

Vectors E and H in a monochromatic wave oscillate in orthogonal planes; the vector,

k, is directed along the line of intersection of these planes in the direction of wave

propagation (see Fig. B.3). Such a wave is called linearly polarized. There exist waves for

which vectors E and H, while staying perpendicular to each other during wave propagation,

rotate clockwise or counter-clockwise. These waves are called circularly polarized waves

or elliptically polarized waves, depending on the trajectory of the end of the vector E.

B.2.2 Energy ﬂux and wave momentum

Let us scalarly multiply Eq. (B.63)byE and Eq. (B.64)byH, and then subtract the ﬁrst

result from the second:

H ·(k ×E) −E ·(k × H) = ω(µ

µH

+ 

 E

). (B.65)

B.2 Electromagnetic waves 349

Figure B.3 The

propagation of an

electromagnetic wave

with wavelength λ; E, H,

and k constitute a

right-handed system.

The expression in parentheses on the right-hand side of Eq. (B.65) is equal to twice the

energy density of the electromagnetic ﬁeld, i.e.,

µH

+ 

 E

= 2(w

+ w

) = 2w. (B.66)

The left-hand side of Eq. (B.65), according to cyclic permutation for any scalar triple

product

A · (B × C) = C · (A × B) = B ·(C ×A), (B.67)

will transform to

H · (k ×E) −E ·(k × H) = k · (E ×H) −k ·(H ×E) = 2k · (E ×H). (B.68)

Let us introduce the vector P:

P = E × H. (B.69)

P is known as the Poynting vector. This vector deﬁnes the density of energy ﬂux of

the electromagnetic ﬁeld and has the dimension [J m

−2

−1

]. In an isotropic medium

this vector is parallel to the vector k. Now, taking into account Eqs. (B.61)and(B.65),

Eq. (B.69) can be rewritten as follows:

P = vwe

, (B.70)

P = vw. (B.71)

Here, e

is a unit vector oriented in the direction of the vector k. Taking into account

the orthogonality of vectors E, H,andk, and using Eqs. (B.63)and(B.64), we can write

350 Appendix B. Electromagnetic ﬁelds and waves

the relation between the moduli of vectors E and H:

√

µH =

√



 E. (B.72)

Taking into account this relation, it can be shown that, in traveling waves described by

Eqs. (B.56)and(B.57), the energy density of the electric ﬁeld is equal to the energy

density of the magnetic ﬁeld, and the total energy density is

w = 

 E

= µ

µH

√



µEH. (B.73)

The magnitude of the energy transferred by the electromagnetic wave is deﬁned by the

intensity of the electromagnetic wave, I , which is equal, according to the deﬁnition, to the

average value of the density of the energy ﬂux during a period T . For the monochromatic

waves described by Eqs. (B.56)and(B.57), taking into account that P = vw, we obtain

I =







P(t)dt =

√

µ



 E



cos

(ωt − kr)







, (B.74)

which shows that the intensity of the wave is proportional to the square of its ﬁeld’s

amplitude. The dimension of intensity is the same as that for the density of energy ﬂux,

namely [J s

−1

−2

] =[W m

−2

An electromagnetic wave, in addition to energy, also carries linear momentum. Accord-

ing to Einstein’s relativistic theory, an object that moves with the speed of light must have

mass at rest equal to zero. At the same time the momentum, p, of such an object is

equal to

p =

, (B.75)

where W is the object’s energy. The object in the case considered here is the electromag-

netic wave (later we will show that an electromagnetic wave can be considered as a ﬂux

of photons moving with the speed of light and having mass at rest equal to zero). For the

momentum density of an electromagnetic wave, π, we can write

π =

. (B.76)

If we take into account the density of energy ﬂux in vacuum (Eq. (B.71)), then Eq. (B.76)

can be rewritten in vector form in terms of the Poynting vector, P:

π =

E × H. (B.77)

Because the electromagnetic wave has momentum, when it is incident upon a surface it

exerts pressure on it. Let a plane wave be incident perpendicularly upon the plane surface

of a body, which absorbs this wave. Let us calculate the momentum that the wave transmits

to the part of the body’s surface S during the time t. It must be equal to the volume of

B.2 Electromagnetic waves 351

c t

Figure B.4 The pressure,

P = F/ S= πc, exerted by

an electromagnetic wave

on a plane surface of

cross-section S. The

momentum density of the

electromagnetic wave is

π.

the cylinder, Sc t (see Fig. B.4) multiplied by the momentum density π:

p = π Sc t. (B.78)

The pressure, P,isdeﬁnedbytheforce,F, acting on the surface, divided by the cross-

section S, i.e.,

P =

p

t

= π c = w. (B.79)

In the case of the surface of a mirror that completely reﬂects the incident wave, the

momentum transmitted to the surface during the same time is twice as large, and thus the

pressure in this case is

P = 2w. (B.80)

Let us estimate the magnitude of this pressure for a laser beam, for which the electromag-

netic wave has an amplitude of the electric ﬁeld intensity of E

= 10

−1

P = 2

= 2 × 8.85 ×10

−12

× 10

−1

· V

−2

 1.8 × 10

−5

Pa,

which is 10

times smaller than standard atmospheric pressure (P

atm

≈ 10

Pa). Nev-

ertheless, the pressure of an electromagnetic wave can nowadays easily be detected and

measured with high precision.

Example B.3. For wave ﬁelds as well as for static ﬁelds the superposition principle

(Eq. (B.5)) is valid. Using this principle, estimate the intensity of the total wave ﬁeld

created by two linearly polarized plane waves

= e

cos(ωt − k · r),

= e

cos(ωt − k · r),

(B.81)

propagating in vacuum in the same direction. The unit vectors e

and e

deﬁne the

directions of the wave polarizations, which in general are different.

352 Appendix B. Electromagnetic ﬁelds and waves

Reasoning. In accordance with the superposition principle, the total electric ﬁeld E is

equal to the sum of two ﬁelds E

and E

E = E

+ E

. (B.82)

Then, the intensity, I, of the total electric ﬁeld is deﬁned as

I =















+ E

+ 2E

· E



. (B.83)

Because





= E

/2,





= E

/2, and 2



· E



= E

· e

), after averaging and

summation in Eq. (B.83) we obtain

I =





(1 + e

· e

) = 2I

(1 +e

· e

), (B.84)

where





(B.85)

is the intensity of an individual wave. The maximal magnitude of the total intensity,

max

= 4I

, will occur in the case of parallel vectors E

and E

, i.e., when e

· e

= 1. In

this case the amplitudes of the ﬁelds are added to each other, so we can say that the waves

reinforce each other. The minimal magnitude of the intensity of the total wave, I

min

= 0,

occurs in the case of anti-parallel vectors E

and E

, i.e., when e

· e

=−1andthewaves

cancel each other out. If the polarizations of two waves are orthogonal to each other, i.e.,

· e

= 0, then I = 2I

, and the waves propagate independently from each other. The

example considered here illustrates the superposition principle with respect to the wave

ﬁelds.

Example B.4. Consider the harmonic-in-time wave process when only the averaged

magnitude of the density of energy ﬂux P over the time period has physical meaning. Find





by presenting ﬁelds E and H in complex form.

Reasoning. Using the complex form of ﬁelds E and H

E = E

i(ωt−k·r)

H = H

i(ωt−k·r)

(B.86)

it is convenient to substitute into the expressions for the density of energy ﬂux real values

of ﬁelds, i.e.,

E + E

∗

and

H +H

∗