optical properties of solids 2nd ed by mark fox

Optical Properties of Solids

Second Edition

Mark Fox

Oxford University Press, 2010

SOLUTIONS TO EXERCISES

These notes contain detailed solutions to the Exercises at the end of eachchapter of the book, for the benefit of class instructors. Please note that figureswithin the solutions are numbered consecutively from the start of the document(e.g. Fig. 1) in order to distinguish them from the figures in the book, whichhave an additional chapter label (e.g. Fig. 1.1). A similar convention applies tothe labels of tables.

The author would be very grateful if mistakes that are discovered in the solu-tions would be communicated to him. He is also very appreciative of commentsabout the text and/or the Exercises. He may be contacted at the followingaddress:

Department of Physics and AstronomyUniversity of Sheffield

Hicks BuildingSheffield, S3 7RHUnited Kingdom.

email: [email protected]

c© Mark Fox 2010

Name: Tapas Banerji Email: [email protected] IP: 129.82.195.78

Address: C-3/144,Vikrant Khand, Gomtinagar, Lucknow, UP 226010, IND

Chapter 1

Introduction

(1.1) Glass is transparent in the visible spectral region and hence we can assumeα = κ = 0. The reflectivity is calculated by inserting n = 1.51 and κ = 0into eqn 1.29 to obtain R = 0.041. The transmission is calculated fromeqn 1.9 with R = 0.041 to obtain T = 92%.

(1.2) From Table 1.4 we read that the refractive indices of fused silica anddense flint glass are 1.46 and 1.746 respectively. The reflectivities are thencalculated from eqn 1.29 to be 0.035 and 0.074 respectively, with κ = 0in both cases because the glass is transparent. We thus find that thereflectivity of dense flint glass is larger than that of fused silica by a factorof 2.1. This is why cut–glass products made from dense flint glass have asparkling appearance.

(1.3) We first use eqns 1.25 and 1.26 to convert εr to n, giving n = 3.01 andκ = 0.38. We then proceed as in Example 1.2. This gives:v = c/n = 9.97× 107 ms−1,α = 4πκ/λ = 9.6× 106 m−1,R = [(n− 1)2 + κ2]/[(n + 1)2 + κ2] = 25.6%.

(1.4) The anti–reflection coating prevents losses at the air–semiconductor in-terface, and 90% of the light is absorbed when exp(−αl) = 0.1 at theoperating wavelength. With α = 1.3 × 105 m−1 at 850 nm, we then findl = 1.8× 10−5 m = 18 µm.

(1.5) We are given n = 3.68 and we can use eqn 1.19 to work out κ = αλ/4π =0.083. We then use eqn 1.29 to find R = 0.328. Since αl = 2.6, we do notneed to consider multiple reflections and we can just use eqn 1.8 to findthe transmission. This gives: T = (1 − 0.328)2 exp(−1.3 × 2) = 0.034.The optical density is calculated from eqn 1.11 as 0.434× 1.3× 2 = 1.1.

(1.6) 99.8% absorption in 10 m means exp(−αl) = 0.002, and hence α = 0.62m−1.We use eqn 1.19 to find κ = αλ/4π = 3.5 × 10−8. We thus have n =1.33 + i 3.5 × 10−8. The real and imaginary parts of εr are found fromeqns 1.23 and 1.24 respectively, and we thus obtain εr = 1.77+i 9.2×10−8.

(1.7) The filter appears yellow and so it must transmit red and green light, butnot blue. The filter must therefore have absorption at blue wavelengths.

(1.8) (a) In the incoherent limit, we just add the intensities of the beams. Theintensities of the beams transmitted after multiple reflections are shown

1



incident light

transmitted light

reflected light

I0I0(1-R1)

I0(1-R1)e-a l

I0(1-R1)R2e-a l

I0(1-R1) (1-R2) e-a l

I0(1-R1) R2R1e-2a l

I0R1I0(1-R1)R2e

-2a l

I0(1-R1)2R2e

-2a l

I0(1-R1) R2R1e-3a l

I0(1-R1) (1-R2) R2R1 e-3a l

I0(1-R1) R22R1e

-3a lI0(1-R1) R2

2R1e-4a l

I0(1-R1)2 R2

2R1 e-4a l

I0(1-R1) R22R1

2e-4a lI0(1-R1) R2

2R12e-5a l

I0(1-R1) (1-R2) R22R1

2 e-5a l

I0(1-R1) R23R1

2e-5a l

R1

R2

e-al

incident light

transmitted light

reflected light

I0I0(1-R1)

I0(1-R1)e-a l

I0(1-R1)R2e-a l

I0(1-R1) (1-R2) e-a l

I0(1-R1) R2R1e-2a l

I0R1I0(1-R1)R2e

-2a l

I0(1-R1)2R2e

-2a l

I0(1-R1) R2R1e-3a l

I0(1-R1) (1-R2) R2R1 e-3a l

I0(1-R1) R22R1e

-3a lI0(1-R1) R2

2R1e-4a l

I0(1-R1)2 R2

2R1 e-4a l

I0(1-R1) R22R1

2e-4a lI0(1-R1) R2

2R12e-5a l

I0(1-R1) (1-R2) R22R1

2 e-5a l

I0(1-R1) R23R1

2e-5a l

R1

R2

e-al

Figure 1: Multiple reflections in the incoherent limit, as considered in Exercise1.8.

in Fig. 1. The transmitted intensity is given by:

It = I0(1−R1)(1−R2)e−αl + I0(1−R1)(1−R2)R1R2e−3αl

+ I0 (1−R1)(1−R2)R21R

22e−5αl + · · ·

= I0(1−R1)(1−R2)e−αl(1 + R1R2e−2αl + (R1R2)2e−4αl + · · · ) ,

= I0(1−R1)(1−R2)e−αl∞∑

k=0

(R1R2e−2αl)k ,

= I0(1−R1)(1−R2)e−αl 11−R1R2e−2αl

,

where we used the identity∑∞

k=0 xk = 1/(1 − x) in the last line. Thetransmissivity is thus:

T =It

I0=

(1−R1)(1−R2)e−αl

1−R1R2e−2αl.

(b) We have an air-medium-air situation, and so it will be the case thatR1 = R2 ≡ R. We need to compare the exact formulae given in eqns 1.6and 1.9 with the approximate one that neglects multiple reflections givenin eqn 1.8.

(i) With α = 0 the extinction coefficient κ will also be zero. We thencalculate the reflectivity from eqn 1.29 to be:

R =(3.4− 1)2

(3.4 + 1)2=

2.42

4.42= 0.30 .

2



Hence the exact transmission from eqn 1.9 is given by:

Texact =1− 0.301 + 0.30

=0.701.30

= 54% .

The approximate result with multiple reflections neglected is found fromeqn 1.8 with α = 0:

Tapprox = (1−R)2 = (1− 0.30)2 = (0.70)2 = 49% .

Thus by neglecting multiple reflections, the transmission is underestimatedby a factor of about 10%.

(ii) If we assume n À κ, we can ignore the terms in κ in the formula forthe reflectivity (eqn 1.29) and so just obtain R = 0.30 as in part (a). Theexact transmission is calculated from eqn 1.6 to be:

Texact =(1− 0.30)2e−1

1− 0.302e−2= 18.2% .

The formula with multiple reflections neglected (eqn 1.8) gives:

Tapprox = (1− 0.30)2e−1 = 18.0% .

The transmission is thus underestimated by a factor of 1%.

(iii) Following the same method as in part (b)(i), we find:

R = 0.772/2.772 = 0.077 ,

Texact = (1− 0.077)/(1 + 0.077) = 85.7% ,

Tapprox = (1− 0.077)2 = 85.2% .

The error is thus −0.6%.

(c) It is intuitively obvious that it is valid to neglect multiple reflectionswhen the absorption is strong (i.e. αl & 1), since the reflected beams willbe very weak. Part (b)(ii) confirms this.

For the case of transparent materials (i.e. α = 0), it will be valid toneglect multiple reflections when R is small, since then it will be validto approximate 1/(1 + R) as (1 − R), and eqn 1.8 with α = 0 will beequivalent to eqn 1.9. The reflectivity will be small when (n− 1) is small(see eqn 1.29), and so we conclude that multiple reflections are insignificantfor small (n− 1). This point is illustrated by comparing parts (b)(i) and(iii).

(1.9) In contrast to the previous exercise, we must now add up the electric fieldamplitudes of the beams and consider their relative phases. Let r and tbe the amplitude reflection and transmission coefficients for going fromair to the medium, and r′ and t′ be the equivalent quantities for goingfrom the medium to air. If the absorption coefficient is α, the amplitudewill decay by a factor of e−αl/2 during a single pass of the medium, sinceI ∝ E2. For simplicity we define x = e−αl/2 eiΦ/2, where Φ = 4πnl/λ isthe round-trip phase shift, so that the complex amplitude changes by a

3



incident

amplitude

transmitted

amplitude

reflected

amplitude

E0

t E0

tx E0

txr' E0

r E0

tx2r' E0

tt'x2r'E0

x = eif/2 e-al / 2

tx2r'2E0 tx3r'2E0

tx3r'3E0tx4r'3 E0

tt'x4r'3E0 tx4r'4 E0tx5r'4 E0

tx5r'5 E0

tt'x E0

tt'x3r'2E0

tt'x5r'4 E0

t

r

t'

r'

incident

amplitude

transmitted

amplitude

reflected

amplitude

E0

t E0

tx E0

txr' E0

r E0

tx2r' E0

tt'x2r'E0

x = eif/2 e-al / 2

tx2r'2E0 tx3r'2E0

tx3r'3E0tx4r'3 E0

tt'x4r'3E0 tx4r'4 E0tx5r'4 E0

tx5r'5 E0

tt'x E0

tt'x3r'2E0

tt'x5r'4 E0

t

r

t

r

t'

r'

t'

r'

Figure 2: Multiple reflections in the coherent limit, as considered in Exercise1.9.

factor x after one pass through the medium.

(a) With the definitions above, the transmitted amplitude Et is given by(see Fig. 2):

Et = tt′xE0 + tt′x3r′2E0 + tt′x5r′4E0 + · · · ,

= tt′xE0(1 + x2r′2 + x4r′4 + · · · ) ,

= tt′xE0

∞∑

k=0

(x2r′2)k ,

= E0tt′x

1− x2r′2,

where we used∑∞

k=0 ak = 1/(1 − a) in the last line. The transmittedintensity is then given by:

It ∝ Et(Et)∗ = |E0|2 |x|2|tt′|2(1− x2r′2)(1− x2r′2)∗

.

Now x2 = e−αl eiΦ, r = −r′ (due to the π phase shift on going from a moredense to a less dense medium), tt′ = 1− r2, and r2 = r′2 = R. Therefore

4



we have:

T =It

I0=|Et|2|E0|2 ,

=(1−R)2 e−αl

(1−Re−αleiΦ)(1−Re−αle−iΦ),

=(1−R)2 e−αl

(1−Re−αl(eiΦ + e−iΦ) + R2e−2αl),

=(1−R)2 e−αl

(1− 2Re−αl cosΦ + R2e−2αl),

where we used eiϕ + e−iϕ = 2 cos ϕ in the last line.

(b) For the reflectivity, we proceed as in part (a). From Fig. 2 it is apparentthat the reflected amplitude Er is given by:

Er = rE0 + tt′x2r′E0 + tt′x4r′3E0 + · · · ,

= E0(r + r′tt′x2[1 + x2r′2 + x4r′4 + · · · ]) ,

= E0

(r +

r′tt′x2

1− x2r′2

),

where we again used∑∞

k=0 ak = 1/(1 − a). On inserting r′ = −r andtt′ = 1− r2 = 1−R, we find:

Er = E0 r

(1− x2(1−R)

1− x2R

)= E0 r

(1− x2

1− x2R

).

Therefore, with x2 = e−αl eiΦ, and again using eiϕ + e−iϕ = 2 cos ϕ, wefind:

reflectivity =|Er|2|E0|2 ,

= R(1− e−αl eiΦ)(1− e−αl e−iΦ)

(1−Re−αl eiΦ)(1−Re−αl e−iΦ),

= R(1− 2e−αl cosΦ + e−2αl)

(1− 2Re−αl cosΦ + R2e−2αl).

(c) In the limit with α = 0, the reflectivity and transmission formulaejust reduce to the standard ones for a Fabry–Perot etalon (see e.g. Hechtchapter 9):

limα→0(T ) =(1−R)2

(1− 2R cosΦ + R2),

limα→0(reflectivity) =2R(1− cosΦ)

(1− 2R cosΦ + R2).

Note that in this α → 0 limit, conservation of energy requires that thetransmission and reflection coefficients must sum to unity. This can be

5



proven as follows. We have It = T × I0 and Ir = reflectivity× I0. Hence:

It + Ir = I0(1−R)2

(1− 2R cosΦ + R2)+ I0

2R(1− cosΦ)(1− 2R cosΦ + R2)

= I0(1−R)2 + 2R(1− cosΦ)

(1− 2R cosΦ + R2).

= I0(1− 2R + R2 + 2R− 2R cosΦ)

(1− 2R cos Φ + R2),

= I0(1 + R2 − 2R cosΦ)(1− 2R cosΦ + R2)

,

= I0 .

(d) If αl À 1, then e−αl ¿ 1. Since R ≤ 1, we can neglect the second andthird terms in the denominator, and so the transmission is just:

T = (1−R)2 e−αl ,

as in eqn 1.8.

(e) For negligible absorption, we expect to observe thin-film fringes. Thetransmission will be as given in part (c):

limα→0(T ) =(1−R)2

(1− 2R cosΦ + R2).

When Φ = 2mπ, where m is an integer, we have constructive interference,with:

T =(1−R)2

(1− 2R + R2)=

(1−R)2

(1−R)2= 1 .

Since Φ = 4πnl/λ, this constructive interference condition occurs when2nl = mλ, that is, the round trip path length is equal to an integernumber of wavelengths. The other extreme occurs when Φ = (2m + 1)π:

T =(1−R)2

(1 + 2R + R2)=

(1−R)2

(1 + R)2.

At this point, we have 2nl = (m + 1/2)λ′, so that the waves from eachround trip are out of phase and interfere destructively. Therefore, asthe wavelength is scanned, the transmission oscillates between unity and(1−R)2/(1 + R)2, as in a Fabry–Perot etalon.

(1.10) This problem is meant to be a rough attempt to model the optical prop-erties of GaAs near its band edge. The band gap is taken to be 1.42 eV,as appropriate for GaAs at room temperature, and the form of the ab-sorption is meant to be rough fit to the data in Fig. 3.9, with C chosen togive the correct absorption around 2 eV, namely ∼ 4× 106 m−1.

The reflectivity and transmissivity have to be calculated from the two for-mulae given in parts (a) and (b) of the previous exercise. In principle, forwavelengths below the band edge, we have to work out the extinction co-efficient from the absorption via eqn 1.19, and then work out the reflection

6



600 700 800 900 10000.0

0.2

0.4

0.6

0.8

1.0

Tra

nsm

issi

on

orR

efle

ctiv

ity

Wavelength (nm)

TransmissionReflectivity

Figure 3: Transmission and reflectivity of a 2µm thick platelet, as consideredin Exercise 1.10.

coefficient from eqn 1.29 with both the terms in n and κ included. How-ever, the maximum value of κ in this exercise is only 0.19, which occursat 600 nm, where α = 4× 106 m−1, and (0.19)2 is negligible compared to(3.5− 1)2. Therefore, in fact we can just work out R from the real part ofthe refractive index and use R = 2.52/4.52 = 0.31 at all wavelengths. Thetransmissivity and reflectivity calculated in this way are plotted in Fig. 3.

For wavelengths below the band edge (i.e. 870–1000 nm), the medium istransparent, and we observe Fabry–Perot fringes. Peaks in the transmis-sion with T = 1 occur at wavelengths that satisfy 2nl = mλ, (m =integer), i.e. at 875 nm (m = 16), 933 nm (m = 15), and 1000 nm(m = 14). The reflectivity at these wavelengths is equal to zero. Peaksin the reflectivity and minima in the transmission are observed in be-tween the transmission maxima. The minimum value of the transmissionis (1 − R)2/(1 + R)2 = 0.28, implying that the maximum value of thereflectivity is 0.72.

For wavelengths below the band edge, the absorption increases rapidly,and the interference fringes are rapidly damped, with multiple reflectionsbecoming negligible. The reflectivity settles to the value determined bythe front surface (i.e. 31%), while the transmission decreases exponentiallyaccording to eqn 1.8. The transmissivity at 600 nm where α = 4×106 m−1

is (0.69)2 exp(−8) = 1.6× 10−4.

(1.11) In the incoherent limit, the transmissivity of a transparent plate is givenby eqn 1.9 as (1 − R)/(1 + R), while the reflectivity is obtained from

7



eqn 1.29, which becomes R = (n− 1)2/(n + 1)2 when κ = 0. Now:

1−R = 1− (n− 1)2

(n + 1)2=

(n + 1)2 − (n− 1)2

(n + 1)2=

4n

(n + 1)2.

Similarly:

1 + R = 1 +(n− 1)2

(n + 1)2=

(n + 1)2 + (n− 1)2

(n + 1)2=

2(n2 + 1)(n + 1)2

.

We therefore have:

T =1−R

1 + R=

4n

(n + 1)2× (n + 1)2

2(n2 + 1)=

2n

(n2 + 1).

(1.12) The reflection coefficient for the interface between two transparent mediawith refractive indices of n1 and n2 is (see eqn A.55):

R =(n1 − n2)2

(n1 + n2)2.

Note that this reduces to R = (n − 1)2/(n + 1)2 for the case of an air-medium interface with n1 = 1 and n2 = n. (cf. eqn 1.29 with κ = 0.) Forthe three cases considered here we have:

• air → film: n1 = 1, n2 = 2.5, so R = (1.5/3.5)2 = 18%.

• film → substrate: n1 = 2.5, n2 = 1.5, so R = (2.5 − 1.5)2/(2.5 +1.5)2 = (1/4)2 = 6%.

• substrate → air: n1 = 1.5, n2 = 1, so R = (0.5/2.5)2 = 4%.

(1.13) For a thick sample it is appropriate to use eqn 1.8 for the transmission.We take the log10 of eqn 1.8 to obtain (using log10 x = loge x/ loge 10):

− log10(T ) = −2 log10(1−R) + αl/ loge 10 .

We can then substitute αl/ loge 10 from eqn 1.11 to obtain the requiredresult.

If the medium is transparent at λ′ then we will have that Tλ′ = (1 −R)/(1 + R) for the incoherent limit, where R is the reflectivity at λ′. Weassume that the reflectivity varies only weakly with wavelength. This isa reasonable assumption for most materials if we choose λ′ sensibly, forexample, just above the absorption edge we are trying to measure. Withthis assumption, we can work out the value of R from a measurementof the transmission at λ′, and then use this value in the formula derivedin the exercise to work out the optical density from a measurement ofthe transmission at λ. Measurements of T (λ) and T (λ′) thus allow theoptical density to be determined. The absorption coefficient can then bedetermined from the optical density by using eqn 1.11.

(1.14) With σ = 6.6 × 107 Ω−1 m, and ω = 1.88 × 1013 rad/s, we find εr =εr + i 3.97 × 105. The imaginary part of εr is very large, and hence the

8



approximation ε2 À ε1 is a good one. In this approximation, we have(with

√i = (1 + i)/

√2):

n =√

εr = (3.97× 105)1/2√

i = 445(1 + i) .

By inserting n = κ = 445 into eqn 1.29, we then obtain R = 99.6%.

(1.15) We use the same formula for the complex dielectric constant as in theprevious exercise. With ω = 1.88× 1013 rad/s, we find

n2 = εr = ε1 + i 2.94× 105 .

Since ε2 À ε1, this implies:

n =√

εr = (2.94× 105)1/2√

i = 383(1 + i) .

We then find from eqn 1.19 that α = 4πκ/λ = 4.8× 107 m−1. Beer’s lawmeans that we set exp(−αl) = 0.5 for a drop in intensity by a factor of 2,giving l = 1.4× 10−8 m = 14 nm.

(1.16) It is apparent from eqn 1.29 that R = 1 when n = 1 and κ = 0. For zeroreflectivity we thus require εr = (n + iκ)2 = 1.

(1.17) (a) We convert wavelengths to photon energies using E = hc/λ to obtainthe energy level scheme shown in Fig. 4. It is thus apparent that 0.294 eVof energy is dissipated during each absorption / emission process.

(b) When the quantum efficiency is 100%, every absorbed photon pro-duces a luminescent photon. The ratio of the light energy emitted to thatabsorbed is then simply given by the ratio of the relevant photon energies.The emitted power is thus (1.165/1.459) × 10 = 8 W, and the dissipatedpower is 2 W.

(c) For a luminescent quantum efficiency of 50% the number of photonsemitted drops by a factor of 2 compared to part (b), and so the lightpower emitted falls to 4 W. The remaining 6 W of the absorbed power isdissipated as heat.

absorption

1.459 eV

emission

1.165 eV

relaxation (0.294 eV)

Figure 4: Energy level scheme for Exercise 1.17.

(1.18) This is an example of Raman scattering, which is discussed in detail inSection 10.5. Conservation of energy in the scattering process is satisfiedwhen

hνout = hνin − hνphonon .

With ν = c/λ, we then find λout = 521 nm.

9



(1.19) The transmission is given by eqn 1.12, with the wavelength dependenceof the scattering cross section given by eqn 1.13. At 850 nm we have 10%transmission, so that Nσsl = 2.30. Since σs ∝ λ−4, the scattering cross-section is 11.1 times larger at 850 nm than at 1550 nm, and so we haveNσsl = 2.30/11.1 = 0.21 at the longer wavelength, implying a transmis-sion of 81 %.

In general, the scattering losses decrease as the wavelength increases, andhence the propagation losses decrease. Longer wavelengths are thereforepreferable for long range communication systems. At the same time, thefibres start to absorb in the infrared due to phonon absorption. 1550 nm isthe longest practical wavelength for silica fibres before phonon absorptionbecomes significant.

(1.20) We again use eqn 1.12 to calculate the transmission, setting exp(−Nσsl) =0.5. This gives Nσsl = 0.69, which implies l = 3.5m for the given valuesof N and σs.

If the wavelength is reduced by a factor of two, Rayleigh’s scattering law(eqn 1.13) implies that σs increases by a factor of 16. The length requiredfor the same transmission is thus smaller by a factor of 16: i.e. l =3.5/16 = 0.22m.

(1.21) Birefringence is an example of optical anisotropy as discussed in Sec-tion 1.5.1, and also in Section 2.5. Ice is a uniaxial crystal, and thereforehas preferential axes, making optical anisotropy possible. Water, by con-trast, is a liquid and has no preferential axes. The optical properties musttherefore be isotropic, making birefringence impossible.

10



Chapter 2

Classical propagation

(2.1) We envisage two displaced masses as shown in Fig. 5. The spring isextended by a distance (x1 − x2) and so the force on the masses are±Ks(x1 − x2). The equations of motion are therefore

m1d2x1

dt2= −Ks(x1 − x2)

and

m2d2x2

dt2= −Ks(x2 − x1) .

Divide the equations by m1 and m2 respectively and subtract them toobtain:

d2

dt2(x1 − x2) = −Ks

(1

m1+

1m2

)(x1 − x2) .

On defining the relative displacement x = x1 − x2 and introducing thereduced mass µ, where 1/µ = 1/m1 + 1/m2, we then have:

µd2x

dt2= −Ksx .

This is the equation of motion of an oscillator of angular frequency (Ks/µ)1/2.

m1 m2

x1 x2

rest

displaced

m1 m2

x1 x2

rest

displaced

Figure 5: Displacement of two masses as described in Exercise 2.1

(2.2) The solution is simpler if complex exponentials are used. We thereforewrite the force as the real part of F0e−iωt, and look for solutions of theform x(t) = x0e−iωt. On substituting into the equation of motion we thenobtain:

m(−ω2 − iωγ + ω20)x0e−iωt = F0e−iωt ,

which implies:

x(t) =F0

m

1(ω2

0 − ω2 − iωγ)e−iωt .

11



The phase factor comes from the factor of [ω20 − ω2 − iωγ]−1. On multi-

plying the top and bottom by the complex conjugate, we find:

1ω2

0 − ω2 − iωγ=

(ω20 − ω2) + iωγ

(ω20 − ω2)2 + (ωγ)2

.

On writing this in the form:

a + ib = reiθ ≡ r(cos θ + i sin θ) ,

we then see that the phase factor θ is given by:

tan θ =ωγ

(ω20 − ω2)

.

This implies that the displacement of the oscillator is of the form:

x(t) ∼ eiθe−iωt = e−i(ωt−θ) ,

which shows that the oscillator has a relative phase lag of:

θ = tan−1[ωγ/(ω20 − ω2)] .

(2.3) By applying the Lorentz oscillator model of Section 2.2.1, we realize thatthe refractive index will have a frequency dependence as shown in Fig. 2.4,with ω0 corresponding to 500 nm. (i.e. ω0 = 3.8 × 1015 rad/s.) Forfrequencies well above the resonance, we will just have the contribution ofthe undoped sapphire crystal:

n∞ ≡ n(ω À ω0) = 1.77 ,

which implies ε∞ = (1.77)2. The refractive index well below the resonancecan be worked out from eqn 2.19. On using the value of N given in theexercise, we find εst − ε∞ = 2.23× 10−3. We thus have:

nst =√

εst = [(1.77)2 + 2.23× 10−3]1/2 .

We thus find nst − n∞ = 6.3× 10−4.

(2.4) We again use the Lorentz oscillator model of Section 2.2. The Exerciseis similar to Example 2.1, because we are dealing with a relatively smallnumber of absorbers and the overall refractive index will be dominatedby the host crystal. We can therefore assume n = 1.39 throughout theExercise. On the other hand, the host crystal is transparent at 405 nm, andso the absorption will be determined by the impurity atoms. The otherfactor we have to include is the low oscillator strength of the transition.We therefore modify the first equation in Example 2.1 to:

κ(ω0) =ε2(ω0)

2n=

Ne2

2nε0m0

1γω0

× f ,

where f = 9 × 10−5 is the oscillator strength. For the absorption linewe have ω0 = 2πc/405 nm = 4.65 × 1015 rad/s, and γ = ∆ω = 2π∆ν =5.15 × 1014 s−1. With N = 2 × 1026 m−3 and n = 1.39, we then findκ(ω0) = 8.6 × 10−6. We finally obtain the absorption at the line centre(405 nm) from eqn 1.19 as 270m−1.

This Exercise is broadly based on the results presented in the paper byIverson and Sibley in J. Luminescence 20, 311 (1979).

12



(2.5) The result of this Exercise works in the limit where the contribution ofthe particular oscillator to the dielectric constant is relatively small, as inExample 2.1 and the previous exercise. In this limit we have ε2 ¿ ε1, andtherefore κ(ω0) = ε2(ω0)/2n, where n =

√ε1, and ε1(ω0) = 1+χ. We then

see from eqn 2.16 that ε2(ω0) = Ne2/ε0m0γω0, so that the absorption is(cf. eqn 1.19):

α(ω0) =4πκ(ω0)

λ= 4π × ε2(ω0)

2n÷ (2πc/ω0) =

Ne2

nε0m0γc.

This shows that it is the linewidth that determines the peak absorptionstrength per oscillator. The oscillator strength is, of course, also impor-tant.

(2.6) The data can be analysed by comparison with Fig. 2.4 with the assumptionthat the oscillator strength is unity.

(a) The low frequency refractive index corresponds to√

εst. With n = 2.43for ω ¿ ω0 from the data, we find εst = 5.9.

(b) The resonant frequency is the mid point of the “wiggle”, i.e. 5.0 ×1012 Hz.

(c) The natural frequency is given by eqn 2.2, which implies Ks = µω20 .

The reduced mass µ is given by:

1/µ = 1/m1 + 1/m2 = 1/23 + 1/35.5 amu−1 ,

which gives µ = 14 amu = 2.33 × 10−26 kg. With ω0 = 2πν0 = 3.1 ×1013 rad/s, we find Ks = 23 kg s−2. The restoring force is given by F =−Ksx, which implies |F | = 23 N for x equal to unity.

(d) The oscillator density can be found from eqn 2.19. εst = 5.9 has beenfound in part (a), and ε∞ can be read from the graph for ω À ω0 asε∞ = n2 = (1.45)2 = 2.10. We thus have εst − ε∞ = 3.8. Using the valuesof ω0 and µ worked out previously, we then find N = 3.0× 1028 m−3.

(e) γ is equal to the shift between the maximum and minimum in the re-fractive index in angular frequency units. We can only make a rough esti-mate of γ because the data does not follow a simple line shape. The damp-ing rate depends strongly on the frequency, which is why the resonanceline is asymmetric. By comparison with Fig. 2.4 we find ∆ν ∼ 1×1012 Hz,and hence γ = 2π∆ν ∼ 6× 1012 s−1.

(f) The result of Exercise 2.5 tells us that α = Ne2/nε0µγc at the linecentre for a weak absorber. This limit does not really apply here, but wecan still use it to get a rough answer. On inserting the values of N , µ andγ found above, and taking n ∼ 2, we find α ∼ 1× 106 m−1.

(2.7) We start by re-writing eqn 2.25 as:

1vg

=dk

dω.

13



We then substitute k = nω/c and v = c/n to obtain:

1vg

=1c

(n + ω

dn

dω

),

=n

c

(1 +

ω

n

dn

dω

),

=1v

(1 +

ω

n

dn

dω

).

The first relationship in eqn 2.26 follows immediately by taking the recip-rocal.

The second relationship in eqn 2.26 is obtained by substituting λ = 2πc/ωso that

dn

dω=

dn

dλ

dλ

dω= − λ2

2πc

dn

dλ,

and hence:

vg = v[1 + (2πc/λn)(−λ2/2πc)dn/dλ]−1 = v[1− (λ/n)dn/dλ]−1 .

(2.8) We consider three separate frequency regions.

(i) ω < ω0: In this frequency region εr is real, and increases with frequency.Since n =

√εr, it is apparent that dn/dω is positive, so that from eqn 2.26

we see that vg < v. Since εr > 1, n > 1, and hence v = c/n < c. Thereforevg < c.

(ii) ω0 < ω < (ω20 +Ne2/ε0m0)1/2: In this frequency region, εr is negative.

The refractive index is purely imaginary and the wave does not propagate.This is an example of the Reststrahlen effect discussed in Section 10.2.3.

(iii) ω > (ω20 + Ne2/ε0m0)1/2: In this region εr is positive and increases

with frequency, approaching unity asymptotically. dn/dω is therefore pos-itive, but we cannot use the same line of argument as part (i) becausen < 1 and therefore v > c. We must therefore work out vg explicitly usingeqn 2.26. With n =

√εr, we obtain:

dn

dω=

ddω

(1 +

Ne2

ε0m0

1ω2

0 − ω2

)1/2

=1n

Ne2

ε0m0

ω

(ω20 − ω2)2

.

Hence

1vg

=n

c+

Ne2

ncε0m0

ω2

(ω20 − ω2)2

,

=1nc

(n2 +

Ne2

ε0m0

ω2

(ω20 − ω2)2

),

=1nc

(εr +

Ne2

ε0m0

ω2

(ω20 − ω2)2

).

On substituting for εr from the exercise, we find:

1vg

=1nc

(1 +

Ne2

ε0m0

ω20

(ω20 − ω2)2

),

14



and hence:

vg = nc

(1 +

Ne2

ε0m0

ω20

(ω20 − ω2)2

)−1

.

The denominator is greater than unity, and n < 1, so vg < c.

(2.9) (a) Consider a dipole p placed at the origin. The electric field generatedat position vector r is given by:

E(r) =3(p · r)r − r2p

4πε0r5.

The electric field generated at the origin by a dipole p at position vectorr is therefore given by:

E(−r) =3(p · (−r))(−r)− r2p

4πε0r5=

3(p · r)r − r2p

4πε0r5.

Consider the ith dipole within the sphere illustrated in Fig. 2.8. Weassume that the dipole is oriented parallel to the z axis so that we canwrite pi = (0, 0, pi). Then, on writing ri = (xi, yi, zi) in the formula forE, we find that the z component of the field at the origin from the ithdipole is:

E i =pi(3z2

i − r2i )

4πε0r5i

.

We now sum over the cubic lattice of dipoles within the sphere. By sym-metry, the x and y components sum to zero, giving a resultant field alongthe z axis of magnitude:

Esphere =1

4πε0

∑

i

pi3z2

i − r2i

r5i

,

as required.

(b) If all the dipoles have the same magnitude p, then the resultant fieldis given by:

Esphere =p

4πε0

∑

i

2z2i − x2

i − y2i

r5i

.

The x, y and z axes are equivalent for the cubic lattice within the sphere,and so we must have:

∑

i

x2i

r5i

=∑

i

y2i

r5i

=∑

i

z2i

r5i

.

It is thus apparent that

∑

i

2z2i − x2

i − y2i

r5i

= 0 .

The net field is therefore zero.

(c) Consider a hollow sphere of radius a placed within a polarized dielectricmedium as illustrated in Fig. 6. (cf. Fig 2.8.) We assume that the polar-ization is parallel to the z axis. The surface charge on the sphere must

15



balance the normal component of the polarization P . With P = (0, 0, P ),the normal component at polar angle θ is equal P cos θ, as shown in Fig. 6.Hence the surface charge density σ at angle θ is equal to −P cos θ. Thecharge contained in a circular element at angle θ subtending an incremen-tal angle dθ as defined in Fig. 6 is then given by:

dq = σ dA = −P cos θ × (2πa sin θ · adθ) = −2πPa2 cos θ sin θ dθ .

The x and y components of the field generated at the origin by this in-cremental charge sum to zero by symmetry, leaving just a z component,with a magnitude given by Coulomb’s law as:

dEz = − dq

4πε0a2cos θ = +

P cos2 θ sin θ dθ

2ε0.

On integrating over θ, we then obtain:

Ez =∫ π

θ=0

dEz =P

2ε0

∫ π

0

cos2 θ sin θ dθ =P

3ε0.

Since P is parallel to the z axis, and the x and y components of E arezero, we therefore have:

E =P

3ε0,

as required for eqn 2.28.

+ +++

++

+

- ----

--

P

q dq

adq

P cosq

+ +++

++

+

+ +++

++

+

- ----

--

P

q dq

adq

P cosq

Figure 6: Definition of angles and charge increment as required for Exercise2.9(c).

(2.10) If εr− 1 is small, the left hand side of the Clausius–Mossotti relationshipbecomes equal to (εr − 1)/3, and we then find:

εr = 1 + Nχa ≡ 1 + χ ,

where χ = Nχa, as in eqn A.4. It is apparent that εr − 1 will be smallif either N is small or χa is small. This means that we either have a lowdensity of absorbing atoms (as in a gas, for example), or we are workingat frequencies far away from any resonances.

16



(2.11) We are working with a gas, and we can therefore forget about Clausius-Mossotti. At s.t.p. we have NA (Avogadro’s constant) molecules in avolume of 22.4 litres. Hence N = 2.69× 1025 m−3. We then find χa from:

χa = (εr − 1)/N = 2.2× 10−29 m3 .

The atomic dipole is worked out from

p = ε0χaE .

The displacement of an electron by 1A produces a dipole of 1.6×10−29 C m.Hence we require a field of 0.8×1011 V/m. The field acting on an electronat a distance r from a proton is given by Coulomb’s law as:

E =e

4πε0r2.

On substituting r = 1 A, we find E = 1.4× 1011 V/m. It is not surprisingthat these two fields are of similar magnitude because the external fieldmust work against the Coulomb forces in the molecule to induce a dipole.

(2.12) We are given that α(E) = α0 for E2 ≥ E ≥ E1, where E is the photonenergy, and α(E) = 0 at all other energies. By using eqn 1.19 we thenhave:

κ(E) = c~α0/2E , E2 ≥ E ≥ E1 ,

κ(E) = 0 , elsewhere .

We can then find the refractive index from eqn 2.36 as follows:

n(E) = 1 +2π

∫ E2/~

E1/~

ω′κ(ω′)ω′2 − ω2

dω′ ,

= 1 +2π

∫ E2

E1

E′κ(E′)E′2 − E2

dE′ .

Note that we do not have to worry about taking principal parts becauseκ(E) = 0 when E′ = E if E < E1. On substituting for κ(E′) we find:

n(E) = 1 +c~α0

π

∫ E2

E1

1E′2 − E2

dE′ .

Now: ∫1

x2 − a2dx =

12a

∫ [1

x− a− 1

x + a

]dx ,

=12a

ln(

x− a

x + a

).

Hence:

n(E) = 1 +c~α0

π

12E

[ln

(E′ − E

E′ + E

)]E2

E1

,

= 1 +c~α0

2πE

[ln

(E2 − E

E2 + E

)− ln

(E1 − E

E1 + E

) ],

= 1 +c~α0

2πEln

((E2 − E)(E1 + E)(E2 + E)(E1 − E)

),

as required.

17



(2.13) (a) If we are far away from resonance frequencies, we can ignore thedamping term, and write eqn 2.24 as:

εr ≡ n2 = 1 +Ne2

ε0m0

∑

j

fj

ω20j − ω2

.

On substituting ω = 2πc/λ, this becomes:

n2 = 1 +Ne2

ε0m0

1(2πc)2

∑

j

fjλ2jλ

2

λ2 − λ2j

,

where λj = 2πc/ω0j . This is of the Sellmeier form if we take:

Aj = Ne2fjλ2j/4π2ε0m0c

2 .

(b) With the approximations stated in the exercise, we have:

n2 = 1 +A1λ

2

λ2 − λ21

= 1 + A1(1− λ21/λ2)−1 .

With x ≡ λ21/λ2 ¿ 1, we can expand this to:

n2 = 1 + A1(1 + x + x2 + · · · ) ,

which implies:

n = [(1 + A1) + A1(x + x2 + · · · )]1/2

= (1 + A1)1/2[1 + A1/(1 + A1)(x + x2 + · · · )]1/2

= (1 + A1)1/2[1 + (1/2)A1/(1 + A1)(x + x2)− (1/8)(A1/(1 + A1))2(x + x2)2 + · · · ]

= (1 + A1)1/2 +A1

2√

1 + A1

x +(

A1

2√

1 + A1

− A21

8(1 + A1)3/2

)x2 + · · ·

On re-substituting for x, we then find:

n = (1 + A1)1/2 +A1

2√

1 + A1

λ21

λ2+

(A1

2√

1 + A1

− A21

8(1 + A1)3/2

)λ4

1

λ4,

which shows that:

C1 = (1 + A1)1/2 ,

C2 = A1λ21/2(1 + A1)1/2 ,

C3 = A1(4 + 3A1)λ41/8(1 + A1)3/2 .

Note that the Cauchy formula generally applies to transparent materials(eg glasses) in the visible spectral region. In this situation, the dispersionis dominated by the electronic absorption in the ultraviolet. We shouldthen take λ1 as the wavelength of the band gap, and the approximationλ2

1/λ2 ¿ 1 will be reasonable, as we are far away from the band gapenergy.

18



(2.14) (a) On neglecting the λ4 term in Cauchy’s formula, we have

n = C1 + C2/λ2 .

On inserting the values of n at 402.6 nm and 706.5 nm and solving, we findC1 = 1.5255 and C2 = 4824.7 nm2, so that we have:

n = 1.5255 + 4824.7/λ2 ,

where λ is measured in nm.

(b) The values are found by substituting into the result found in part (a)to obtain n = 1.5493 at 450 nm and n = 1.5369 at 650 nm.

(c) Referring to the angles defined in Fig. 7, we have

sin θin

sin θ1=

sin θout

sin θ2= n ,

from Snell’s law. Furthermore, for a prism with apex angle α, we haveθ1 + θ2 = α. With θin = 45 and α = 60, we then find θout = 57.17 forn = 1.5493 (450 nm) and θout = 55.91 for n = 1.5369 (650 nm). Hence∆θout = 1.26.

qinqout

q2q1

60°qin

qout

q2q1

60°

Figure 7: Angles required for the solution of Exercise 2.14.

(2.15) The transit time is given by:

τ =L

vg= L

dk

dω,

which, with k = nω/c, becomes:

τ =L

c

(n + ω

dn

dω

).

We introduce the vacuum wavelength λ via ω = 2πc/λ, and write:

dn

dω=

dn

dλ· dλ

dω= − λ2

2πc

dn

dλ.

We then have (cf. eqn 2.26 with τ = L/vg):

τ =L

c

(n− λ

dn

dλ

).

The difference in the transit time for two wavelengths separated by ∆λ,where ∆λ ¿ λ, is given by:

∆τ =dτ

dλ∆λ .

19



On using the result for τ above, we find:

dτ

dλ= −L

cλ

d2n

dλ2,

which implies:

∆τ = L

(−λ

c

d2n

dλ2

)∆λ = L D ∆λ ,

where D is the material dispersion parameter defined in eqn 2.40. Wetherefore finally obtain:

|∆τ | = L

∣∣∣∣−λ

c

d2n

dλ2

∣∣∣∣ ∆λ = L |D| ∆λ .

The time–bandwidth product of eqn 2.39 implies that a pulse of lightcontains a spread of frequencies and therefore a spread of wavelengths.In a dispersive medium, the different wavelengths will travel at differentvelocities, and this will cause pulse broadening. The precise amount ofbroadening depends on the numerical value of the time–bandwidth prod-uct assumed for the pulse. For a 10 ps pulse with ∆ν∆t = 1, we have∆ν = 1011 Hz. With λ = c/ν, we have |∆λ| = (λ2/c)|∆ν|, so that|∆λ| = 0.8 nm in this case. We thus find:

|∆τ | = L|D|∆λ = 1 km× 17 ps km−1 nm−1 × 0.8nm = 14ps .

direction of

propagation

e-ray

polarization

vector

index ellipsoid

ne

no no

ne

n(q )

y

z

q

direction of

propagation

e-ray

polarization

vector

index ellipsoid

ne

no no

ne

n(q )

y

z

q

Figure 8: Index ellipse for the e–ray of a wave propagating at an angle θ to theoptic (z) axis of a uniaxial crystal, as required for Exercise 2.16.

(2.16) It is apparent from eqn 2.50 that ε11/ε0 = ε22/ε0 = n2o and ε33/ε0 = n2

e .Hence we can re-cast the index ellipsoid in the form:

x2

n2o

+y2

n2o

+z2

n2e

= 1 .

Owing to the x-y symmetry about the optic (z) axis, we can choose theaxes of the index ellipsoid so that the x axis coincides with the polarization

20



vector of the o–ray as in Fig 2.13(a). The polarization of the e–ray willthen lie in the y–z plane, as in Fig. 2.13(b). The projection of the indexellipsoid onto the plane that contains the direction of propagation and thepolarization vector of the e–ray thus appears as the ellipse drawn in Fig. 8.The refractive index n(θ) that we require is the distance from the originto the point of the ellipse where the E-vector cuts it. The co-ordinates ofthis point are x = 0, y = n(θ) cos θ and z = n(θ) sin θ. On substitutinginto the equation of the index ellipsoid, we then have:

0 +n(θ)2 cos2 θ

n2o

+n(θ)2 sin2 θ

n2e

= 1 ,

which implies:1

n(θ)2=

cos2 θ

n2o

+sin2 θ

n2e

,

as required.

(2.17) The condition for total internal reflection for a medium–air interface isthat the angle of incidence should exceed the critical angle θc given by:

sin θc = 1/n ,

where n is the refractive index of the medium. We see from Table 2.1that no = 1.658 and ne = 1.486 for calcite. The critical angles for theo- and e-rays are therefore 37.1 and 42.3 respectively. For the polarizerto work, we want the o-ray to suffer total internal reflection, but not thee-ray. With normal incidence as shown in Fig. 2.14, this will occur if theapex angle θ lies in the range 37.1 − 42.3.

optic axis (z)

InputE-vector

front surface of

wave plate

y

q

Ez

Ey

(a)

optic axis (z)

InputE-vector

direction

y

q

Ez

-EyOutputE-vector

(b)

optic axis (z)

InputE-vector

front surface of

wave plate

y

q

Ez

Ey

(a)

optic axis (z)

InputE-vector

direction

y

q

Ez

-EyOutputE-vector

(b)

Figure 9: (a) Input polarization vector for light propagating along the x direc-tion, relative to the crystal axes, as required for the solution of Exercise 2.18(a).(b) Output polarization of the half-wave plate.

(2.18) (a) We resolve the input polarization into two components, one along theoptic axis, and one orthogonal to it, as shown in Fig. 2.15(b). We defineour axes so that z lies along the optic axis, and y lies in the plane of thefront surface of the waveplate, as shown in Fig. 9(a). The x axis is taken

21



to be the direction of propagation. If the input beam has amplitude E0,the input polarization has components:

E inz = E0 cos θ ,

E iny = E0 sin θ .

The component along z is the e-ray and the one along y is the o-ray.

In a half-wave plate, the phase of the o-ray is shifted by π relative to thee-ray at the output of the retarder plate. This means that the outputpolarizations will be given by:

Eoutz = E0 cos θ = E in

z ,

Eouty = E0eiπ sin θ = −E0 sin θ = −E in

y ,

as shown in Fig. 9(b). The resultant E-vector points at an angle −θ withrespect to the optic axis. Hence the output polarization is rotated by anangle 2θ with respect to the input polarization.

(b) If we set θ = 45, the input polarization will be given by:

E inz = E0 cos 45 = E0/

√2 ,

E iny = E0 sin 45 = E0/

√2 .

The quarter-wave plate introduces a phase difference of π/2 between thee-ray and o-rays. The output polarization will therefore be:

Eoutz = E0/

√2 ,

Eouty = (E0/

√2)eiπ/2 = i E0/

√2 .

The output therefore consists of two orthogonal linearly polarized com-ponents of equal amplitude with a phase difference of 90 between them.This is circularly-polarized light.

(c) When θ 6= 45, the amplitudes of the e- and o-rays will be different.The quarter-wave plate still introduces a phase shift of 90 between them,and so the output consists of two orthogonal linearly polarized componentswith unequal amplitude and with a 90 phase difference between them.This is elliptically polarized light.

(2.19) We follow Example 2.4, but set the phase difference ∆φ to be π/2 becausewe have a quarter-wave plate instead of a half-wave plate. We thereforerequire:

∆φ =π

2=

2π|∆n|dλ

,

which implies d = λ/4|∆n|. In this case we have |∆n| = 0.0091 andλ = 500 nm, and so we require d = 1.4× 10−5 m.

In this exercise we have neglected the optical activity of the quartz platebecause the thickness is very small. In a thicker multiple-order waveplate(i.e one with ∆φ = (2πm+π/2), where m is an integer) a small correctionwould have to be introduced to account for the optical activity.

22



(2.20) The crystal will be isotropic if the medium has high symmetry so thatthe x, y and z axes are equivalent. If not, it will be birefringent. For thecrystals listed in the Exercise we have:

Crystal Structure x, y, z equivalent ? Birefringent ?(a) NaCl cubic (fcc) yes no(b) Diamond cubic yes no(c) Graphite hexagonal no yes(d) Wurtzite hexagonal no yes(e) Zinc blende cubic yes no(f) Solid argon cubic (fcc) yes no(g) Sulphur orthorhombic no yes

The two hexagonal crystals are uniaxial, with the optic axis lying along thedirection perpendicular to the hexagons. Sulphur has the lowest symmetryand is the only biaxial crystal included in the list.

(2.21) (a) The birefringent phase shift for a refractive difference of ∆n is givenby eqn 2.46. For a half wavelength shift, this is equal to π. On substitutingfor the Kerr-induced birefringence from eqn 2.51, we obtain:

φ = π =2π|∆n|d

λ=

2π(λKE2)dλ

= 2πKE2d .

On solving for the field, we then find:

Eλ/2 = 1/√

2Kd ,

as required.

(b) We substitute for K and d to find:

Eλ/2 = 1/√

2× (8.7× 10−14)× 0.02 = 1.7× 107 V/m .

The field is dropped across a distance of 5 mm, and so the voltage requiredis Vλ/2 = Eλ/2 × 0.005 = 85 kV.

As Table 2.2 shows, the value of K for the chalcogenide considered hereis large for a glass, but the voltage required to make a Kerr cell is stillvery high. Practical Kerr cells use liquids with larger Kerr constants, forexample, nitrobenzene.

(2.22) (a) Set up axes so that z corresponds to the direction of propagation andx to the input polarization. With these definitions the input polarizationcan be written as:

E in = E0 x ,

where E0 is the amplitude of the light.

Circular polarization consists of two orthogonally polarized waves of equalamplitude with a 90 phase difference between them, and so we can writeleft and right circular light as:

σ+ = E0(x + iy)/√

2 ,

σ− = E0(x− iy)/√

2 ,

23



where the factor of i ≡ eiπ/2 represents the 90 phase shift. This allowsus to re-write the input polarization in terms of the left and right circularlight as:

E in = E0(σ+ + σ−)/√

2 .

An optically-active medium will introduce a phase difference φ betweenthe left- and right-circular components, where:

φ =2π

λ∆nd .

Here, λ is the vacuum wavelength, ∆n = (nR−nL) is the difference of therefractive indices for right- and left-circular light, and d is the thicknessof the medium. The output wave will therefore be of the form:

Eout = E0(σ+ + eiφσ−)/√

2 .

We revert to the linear basis by substituting for σ+ and σ− and re-writingthis as:

Eout = E0

((x + iy) + eiφ(x− iy)

)/2 .

Collecting the terms in x and y, we find:

Eout = E0

((1 + eiφ)x + i(1− eiφ)y)

)/2 ,

= E0eiφ/2((e−iφ/2 + eiφ/2)x + i(e−iφ/2 − eiφ/2)y)

)/2 ,

= E0eiφ/2 (2 cos(φ/2) x + i(−2i sin(φ/2)) y) /2 ,

= E0eiφ/2 (cos(φ/2) x + sin(φ/2) y) .

This shows that the output polarization is linearly polarized at an an-gle φ/2 with respect to the x axis. The polarization rotation angle θ istherefore equal to φ/2, and so we find:

θ =φ

2=

π

λ∆nd =

πd

λ(nR − nL) .

(b) Using the result from part(a), we find that the rotatory power of anoptically-active medium is given by:

RP = |θ/d| = π|∆n|/λ .

For quartz we thus find

RP = π(7.1× 10−5)/589× 10−9 = 379 rad/m = 21.7 /mm .

(2.23) We apply a Lorentzian model according to the theory in Section 2.2. TheLorentzian lineshapes are shifted for σ+ and σ− light due to the Zeemaneffect, which shifts the energy of the transition from ~ω0 to ~ω0 ± µBBaccording to the circular polarization.

(a) The Faraday rotation is caused by the difference of the real part ofthe refractive index for σ+ and σ− light. Figure 10(a) plots the refractiveindices for two Lorentzian lines shifted from each other by 2µBB and their

24



0.6 0.8 1.0 1.2 1.40.96

0.98

1.00

1.02

1.04

Refr

active

inde

xn

Angular frequency (w/w0

)

-0.04

-0.02

0.00

0.02

0.04

Dn

0.6 0.8 1.0 1.2 1.4-0.06

-0.04

-0.02

0.00

0.02

0.04

0.06

Extinction

coeffic

ient

k


)

(a)

(b)

Dk

n+

n-

Dn

Dn

k+

k-

Dk

0.6 0.8 1.0 1.2 1.40.96

0.98

1.00

1.02

1.04

Refr

active

inde

xn


)

-0.04

-0.02

0.00

0.02

0.04

Dn

0.6 0.8 1.0 1.2 1.4-0.06

-0.04

-0.02

0.00

0.02

0.04

0.06

Extinction

coeffic

ient

k


)

(a)

(b)

Dk

n+

n-

Dn

n+

n-

Dn

Dn

k+

k-

Dk

k+

k-

Dk

Figure 10: Lorentzian model calculation of Faraday effect and magnetic circulardichroism, as considered in Exercise 2.23. (a) Faraday rotation. (b) Magneticcircular dichroism.

difference. The lines are plotted against the normalized angular frequency(ω/ω0) for a Lorentzian line with γ = 0.05ω0. The value of µBB/~ is setto be equal to γ. The value of ∆n = Re(n+− n−), and hence the Faradayrotation is negative above and below the line, and positive near ω0. Therotation decays as the frequency is tuned away from resonance.

(b) The magnetic circular polarization is found by calculating the differ-ence ∆κ of the imaginary part of (n+ − n−). This is shown in Fig. 10 forthe same Lorentzian line as in part (a). We see that the magnetic circulardichroism follows a dispersive lineshape, with a negative signal below ω0

that peaks at ω0 − µBB/~, and a positive signal above ω0 that peaks atω0 + µBB/~. The signal precisely at ω0 is zero.

(2.24) We require a Faraday rotation of π/4 at a magnetic field of 0.5 T. TheFaraday rotation is given by eqn 2.53, and so the thickness of glass is given

25



by:

d =θ

V b=

π/49.0× 0.5

= 0.17m = 17 cm .

This long length is impractical, and real Faraday isolators use specialistglasses or crystals with larger Verdet coefficients.

26



Chapter 3

Interband absorption

(3.1) The Born–von Karman boundary conditions are satisfied when:

kxL = nxL ,

kyL = nyL ,

kzL = nzL ,

where nx, ny and nz are integers. The wave vector is therefore of theform:

k = (2π/L)(nx, ny, nz) .

The allowed values of k form a grid as shown in Fig. 11. Each allowed kstate occupies a volume of k space equal to (2π/L)3. This implies thatthe number of states in a unit volume of k space is L3/(2π)3. Hence aunit volume of the material would have 1/(2π)3 states per unit volume ofk space.

In most calculations we actually require the density of states in k space asgiven in eqn 3.15. This is found by calculating the number of states withinthe incremental shell between k vectors of magnitude k and k + dk, asshown in Fig. 11. This volume of the incremental shell is equal to 4πk2dk,and contains 4πk2dk× 1/(2π)3 = k2dk/2π2 states. Hence g(k) = k2/2π2.

kx

ky

2p/L

k

dk

Figure 11: Grid of allowed values of k permitted by the Born–von Karmanboundary conditions, as considered in Exercise 3.1. The points of the grid areseparated from each other by distance 2π/L in all three directions, giving avolume per state of (2π/L)3. Note that the diagram only shows the x-y planeof k space. The incremental shell considered for the derivation of eqn 3.15 isalso shown.

27



(3.2) With E(k) = ~2k2/2m∗, we have:

dE

dk=~2k

m∗ .

On inserting into eqn 3.14–15 and substituting for k, we find:

g(E) = 2k2/2π2

~2k/m∗ =m∗kπ2~2

=m∗

π2~2

(2m∗E~2

)1/2

=1

2π2

(2m∗

~2

)3/2

E1/2 ,

as required.

(3.3) (a) The parity of a wave function is equal to ±1 depending on whetherψ(−r) = ±ψ(r). Atoms are spherically symmetric, and so measurableproperties such as the probability amplitude must possess inversion sym-metry about the origin: i.e. |ψ(−r)|2 = |ψ(r)|2. This is satisfied ifψ(−r) = ±ψ(r). In other words, the wave function must have a defi-nite parity.

(b) r is an odd function, and so the integral over all space will be zerounless the product ψ∗f ψi is also an odd function. This condition is satisfiedif the two wave functions have different parities (parity selection rule).Since the wave function parity is equal to (−1)l, the parity selection ruleimplies that ∆l is an odd number.

(c) In spherical polar co-ordinates (r, θ, φ) we have:

x = r sin θ cosφ = r sin θ (eiφ + e−iφ)/2 ,

y = r sin θ sinφ = r sin θ (eiφ − e−iφ)/2i ,z = r cos θ .

The selection rules on m can be derived by considering the integral overφ. For light polarized along the z axis we have:

M ∝∫ 2π

φ=0

e−im′φ · 1 · eimφ dφ ,

since z is independent of φ. The integral is zero unless m′ = m. Theselection rule for z-polarized light is therefore ∆m = 0.

For x or y polarized light we have:

M ∝∫ 2π

φ=0

e−im′φ (eiφ ± e−iφ) eimφ dφ ,

which is zero unless m′ = m±1. We thus have the selection rule ∆m = ±1for light linearly polarized along the x or y axis.

(d) From eqn A.40 we have that:

E+ = E0(x + iy)/√

2 ,

E− = E0(x− iy)/√

2 ,

28



where E± are the electric fields for σ± polarizations. With x = r sin θ cos φand y = r sin θ sin φ, we therefore find:

E+ = E0 r sin θ(cos φ + i sin φ)/√

2 ∝ e+iφ ,

E− = E0 r sin θ(cos φ− i sin φ)/√

2 ∝ e−iφ .

On inserting these into the matrix element, we have for σ+ light:

M+ ∝∫ 2π

φ=0

e−im′φ e+iφ eimφ dφ ,

which is zero unless m′ = m + 1. Similarly, for σ− light we have:

M− ∝∫ 2π

φ=0

e−im′φ e−iφ eimφ dφ ,

which is zero unless m′ = m− 1. With circularly polarized light we there-fore have ∆m = +1 for σ+ polarization and ∆m = −1 for σ− polarization.

(3.4) The apparatus required is basically the same as for Figs 3.14–15, but withmodifications to take account of the fact that the required energy range of0.3–0.6 eV corresponds to a wavelength range of 2–4µm. This means thata detector with a band gap smaller than 0.3 eV must be used, e.g. InSb.(See Table 3.3.) As InSb array detectors are not available, a scanningmonochromator with a single channel detector would normally be used.A thermal source would suffice as the light source.

Another point to consider is that standard glass lenses do not transmitin this wavelength range, and appropriate infrared lenses would have tobe used, e.g. made from CdSe. (See Fig. 1.4(b).) Also, since the data istaken at room temperature, no cryostat is needed.

Figure 12 gives a diagram of a typical arrangement that could be used.

InAs sample

scanning

monochromator

computer

:

infrared lenses

white light

source

InSb

detector

InAs sample

scanning

monochromator

computer

:

infrared lenses

white light

source

InSb

detector

InSb

detector

Figure 12: Apparatus for measuring infrared absorption spectra in the range2–4 µm, as discussed in Exercise 3.4.

(3.5) The type of band gap can be determined from an analysis of the variationof the absorption coefficient α with photon energy. The material is director indirect depending on whether a graph of α2 or α1/2 against ~ω is a

29



straight line. Other factors to consider are that the absorption is muchstronger in a direct gap material, and that the temperature dependenceof α is expected to be different. In an indirect gap material, phononabsorption mechanisms will freeze out as the temperature is lowered.

(3.6) Plot α2 and α1/2 against ~ω as shown in Fig. 13. In the range 2.2 ≤ ~ω ≤2.7 eV, the graph of α2 is a straight line with an intercept at 2.2 eV. Wethus deduce that GaP has an indirect band gap at 2.2 eV. For ~ω > 2.7 eV,the graph of α1/2 is a straight line with an intercept at ∼ 2.75 eV, whichimplies that GaP has a direct gap at 2.75 eV. The huge difference in thetwo axis scales of Fig. 13 is a further indication of the indirect nature ofthe transitions below 2.75 eV, and their direct nature above 2.75 eV. Wethus conclude that the conduction band has two minima: one away fromk = 0 at 2.2 eV, and another at k = 0 at 2.75 eV.

2.2 2.4 2.6 2.8 3.00

20

40

60

80

a2

(10

12

m-

2

)

Energy (eV)

0

200

400

600

800

1000

a1/2

(m-

1/2

)

2.2 2.4 2.6 2.8 3.00

20

40

60

80

a2

(10

12

m-

2

)

Energy (eV)

0

200

400

600

800

1000

a1/2

(m-

1/2

)

Figure 13: Analysis of GaP absorption data as required for Exercise 3.6.

(3.7) A wavelength of 1200 nm corresponds to a photon energy of 1.03 eV. Thisis above the direct gap of germanium at 0.805 eV, and thus the absorptionwill be given by (cf. eqn 3.25):

α = C(~ω − 0.805)1/2 .

The scaling coefficient C can be determined from the data in Fig. 3.11:α ≈ 6× 105 m−1 at 0.86 eV implies that C ≈ 2.5× 106 m−1eV−1/2. Withthis value of C, we then find α ≈ 1.2× 106 m−1 at 1.03 eV.

(3.8) (a) We consider transitions 1 and 2 in Fig. 3.5. The k vectors for thetransitions can be worked out from eqn 3.23. The appropriate parametersare read from Table D.2 as follows: Eg = 1.424 eV, m∗

e = 0.067me, m∗hh =

0.5me, and m∗lh = 0.08me. For the heavy-hole and light-hole transitions we

find from eqn 3.22 that µhh = 0.059 me and µlh = 0.036 me respectively.Hence for ~ω = 1.60 eV, we find k = 5.3 × 108 m−1 for the heavy holesand k = 4.1× 108 m−1 for the light holes.

(b) The air wavelength λ of the photon is 775 nm. The wavelength insidethe crystal is reduced by a factor n. The photon wave vector inside the

30



crystal is therefore given by:

k =2π

(λ/n)= 3.0× 107 m−1 .

This is more than an order of magnitude smaller than the electron wavevector, and hence the approximation in eqn 3.12 is justified.

(c) Equation 3.24 shows that the joint density of states is proportional toµ3/2. Hence the ratio of the joint density of states for heavy- and light-holetransitions with the same photon energy is equal to:

(µhh/µlh)3/2 = (0.059/0.036)3/2 = 2.1 .

(d) It is apparent from Fig. 3.5 that the lowest energy (i.e. k = 0) split-offhole transition occurs at ~ω = Eg + ∆. Reading a value of ∆ = 0.34 eVfrom Table C.2, we find ~ω = 1.76 eV, which is equivalent to λ = 704 nm.

(3.9) Consider a σ+ transition. By referring to Fig. 3.8, we see that in a heavy-hole transition we start from the |J,MJ〉 = |3/2,−3/2〉 sublevel. Forheavy holes we must therefore insert J = 3/2 and MJ = −3/2 into thematrix element for σ+ transitions. This gives:

|〈J − 1, MJ + 1|σ+|J,MJ〉|2 = 12(J −MJ )(J −MJ − 1)C ,

= 12(3/2− (−3/2))(3/2− (−3/2)− 1)C ,

= 3C.

For light holes we start from the |J,MJ 〉 = |3/2,−1/2〉 sublevel in a σ+

transition, and so the matrix element is given by:

|〈J − 1, MJ + 1|σ+|J,MJ〉|2 = 12(J −MJ )(J −MJ − 1)C ,

= 12(3/2− (−1/2))(3/2− (−1/2)− 1)C ,

= C.

The ratio of the squares of the matrix elements is therefore 3C : C = 3 : 1.

The argument is the same for σ− transitions. For heavy holes we startfrom the |J,MJ 〉 = |3/2, +3/2〉 sublevel, and so the heavy-hole matrixelement is given by:

|〈J − 1,MJ − 1|σ−|J,MJ〉|2 = 12(J + MJ)(J + MJ − 1)C ,

= 12(3/2 + (+3/2))(3/2 + (+3/2)− 1)C ,

= 3C.

For light holes we have |J,MJ 〉 = |3/2,+1/2〉, and hence:

|〈J − 1,MJ − 1|σ−|J,MJ〉|2 = 12(J + MJ)(J + MJ − 1)C ,

= 12(3/2 + (+1/2))(3/2 + (+1/2)− 1)C ,

= C.

The ratio is therefore 3 : 1 again.

31



(3.10) Linearly polarized light can be considered as a superposition of two op-positely circularly polarized waves. (See eqn A.41.) The σ+ part of thewave will create negative spin polarization, while the σ− part will createan identical positive spin polarization. Hence the net spin polarization isalways equal to zero.

(3.11) The main issue to think about is the effect of the split-off hole band.A split-off hole transition with σ+ polarization will generate a spin upelectron, and thus start to cancel the negative spin polarization createdby the heavy-hole transitions. The matrix element squared for σ± split-offholes transitions is twice that of the light-hole transitions, [see eg Meier &Zacharchenya (1984), p.24] so that the combined strength of the (light-hole+ split-off hole) transitions is equal to that of the heavy-hole transition.This means that the net spin polarization will ultimately go to zero whensplit-off hole transitions become significant. Thus we expect that theelectron spin polarization will be constant at ±50% for Eg ≤ ~ω ≤ (Eg +∆), where ∆ is the spin–orbit energy, and then to drop for higher photonenergies. This is in fact observed experimentally.

The spin polarization is effectively zero for ~ω À (Eg + ∆). One way tounderstand this is to realize that the electric field of the light only interactswith the orbital motion of the electron and does not interact directly withits spin. Therefore, if spin polarization is being generated optically, itmust be through the spin–orbit interaction. Hence we expect the effect todisappear when the electron energy is larger than the spin–orbit energy∆.

(3.12) (a) The lowest conduction band states of silicon are p–like at the Γ point(i.e k = 0). The spin–orbit splitting is small, and the J = 1/2 andJ = 3/2 conduction bands states are degenerate at k = 0, but not forfinite k. Electric dipole transitions from the p–like valence band statesare forbidden to these p–like conduction band states at k = 0. The firstdipole allowed transition is to the s–like antibonding state at ∼ 4.1 eV.Hence the direct gap at the Γ point is equal to 4.1 eV.

(b) The discussion of the atomic character of bands given in Section 3.3.1only applies at the Γ point where k = 0 and we are considering stationarystates. This means that electric dipole transitions can be allowed at thezone edges, even though they are forbidden at k = 0.

(3.13) At low temperatures, phonon absorption is impossible, and the indirecttransition must proceed by phonon emission, with a threshold energy ofEind

g + ~Ω. The band structure diagram of germanium given in Fig. 3.10shows that the indirect gap occurs at the L point of the Brillouin zone. Wetherefore need a phonon with a wave vector equal to the k vector at theL point. The energies of these phonons are given in Table 3.1. The lowestenergy energy phonon is the TA phonon with an energy of 0.008 eV. Theabsorption threshold would thus occur at Eind

g + 0.008 eV, i.e. at 0.75 eV,as indeed demonstrated by the data in Fig. 3.11(a).

(3.14) The absorption coefficient with a field applied is given by eqn 3.26. Theabsorption decreases exponentially for ~ω < Eg, and this produces an ex-ponential absorption tail below Eg. Although there is no clear absorption

32



edge as for the case at zero field, a reasonable point to take is when theabsorption has decayed by a factor e−1 from its value at Eg. This occurswhen

4√

2m∗e

3|e|~E (Eg − ~ω)3/2 = 1 .

We are looking for the field at which this condition is satisfied for ~ω =(Eg − 0.01) eV. We thus need to solve:

4√

2m∗e

3|e|~E (0.01 eV)3/2 = 1 .

With m∗e = 0.067me, we find E = 1.8× 106 V/m.

v

B

r

w

F

v

B

r

w

F

Figure 14: Force acting on a particle moving in a magnetic field pointing intothe paper, as required for Exercise 3.15. The charge is assumed to be positive.

(3.15) We consider a particle of charge q, mass m and velocity v moving in amagnetic field B. The particle experiences the Lorentz force F = qv×Bwhich is at right angles both to the velocity and the field, as shown inFig. 14. This perpendicular force produces circular motion at angularfrequency ω with radius r. We equate the central force with the Lorentzforce to obtain, with v = ωr:

mω2r = qωrB ,

which, with |q| = e, implies:

ω = eB/m ,

as required.

With a magnetic field pointing in the z direction, the motion in the x-yplane is quantized, but the motion in the z direction is free. We haveseen above that, in the classical analysis, the field causes circular motion.The quantized motion will therefore correspond to a quantum harmonicoscillator. These quantized states are called Landau levels. The Blochwave functions of eqn 3.7–8 are therefore modified to the form:

ψn(r) ∝ u(r) ϕn(x, y) eikzz ,

where ϕn is a harmonic oscillator function, and n is the quantum numberof the Landau level. The selection rule for transitions between the Landaulevels can be deduced by repeating the derivation in Section 3.2 with the

33



modified wave functions. For x-polarized light, the matrix element is nowgiven by:

M ∝∫

u∗f (r)ϕn′(x, y)e−ik′zz xu∗i (r)ϕn(x, y)e−ikzz d3r ,

∝∫

unit cell

u∗f (r)x u∗i (r) d3r × 〈ϕn′ |ϕn〉 ,

where we assumed k′z = kz as usual in the second line. The electric dipolematrix element is therefore proportional to the overlap of harmonic oscil-lator wave functions with different values of n. Now harmonic oscillatorfunctions form an orthonormal set, and so the wave functions of differingn are orthogonal. Hence the matrix element is zero unless n′ = n, i.e.∆n = 0.

(3.16) (a) Let z be the free direction. Apply Born–von Karman boundary con-ditions as in Exercise 3.1 to show that kz = 2πn/L, where n is an integer.There is therefore one k state in a distance 2π/L, so that the density ofstates in k space is 1/2π per unit length of material. For free motion inthe z direction we have E = ~2k2/2m for a particle of mass m, whichimplies:

dE/dk = ~2k/m = ~√

2E/m .

The density of states in energy space is then worked out from eqn 3.14:

g1D(E) = 2g(k)

dE/dk= 2

1/2π

~√

2E/m= (2m/Eh2)−1/2 .

We thus see that g1D(E) ∝ E−1/2.

(b) The threshold energy will be equal to the band gap Eg of the semi-conductor as for a 3-D material. Fermi’s golden rule indicates that theabsorption is proportional to the density of states. Since g1D(E) ∝ E−1/2,we therefore expect α ∝ (~ω − Eg)−1/2, for ~ω > Eg. See Fig. 15(i).

0 2 4 6 8 100

2

4

6

Abso

rpti

on

(a.u

.)

Energy relative to Eg in arb. units

(i) (ii)

0 1 2 30

2

4

6

Abso

rpti

on

(a.u

.)

Energy in units of hwL relative to Eg

0 2 4 6 8 100

2

4

6

Abso

rpti

on

(a.u

.)


0 2 4 6 8 100

2

4

6

Abso

rpti

on

(a.u

.)


(i) (ii)

0 1 2 30

2

4

6

Abso

rpti

on

(a.u

.)


0 1 2 30

2

4

6

Abso

rpti

on

(a.u

.)


Figure 15: (i) Absorption of a one-dimensional semiconductor as discussed inExercise 3.16(b). (ii) Absorption for a system with quantized Landau levels intwo directions and free motion in the third. ωL is the Landau level angularfrequency, as discussed in Exercise 3.16(c).

(c) The magnetic field quantizes the motion in two dimensions to giveLandau levels, leaving the particle free to move in the third dimension.

34



Optical transitions are possible between Landau levels with the same valueof n. (See Exercise 3.15.) These will occur at energies given by (cf. eqn3.32):

En = Eg + (n + 1/2)~ωL ,

whereωL = eB/m∗

e + eB/m∗h = eB/µ .

Each Landau level transition has a 1-D density of states due to the freemotion parallel to the field. Hence for each Landau level we expect:

α ∝ (~ω − En)−1/2 .

The total absorption is found by adding the absorption for each Landaulevel transition together, as shown in Fig. 15(ii).

When comparing to the experimental data in Fig. 3.7, we expect α(~ω) todiverge each time the frequency crosses the threshold for a new value ofn. These divergences are broadened by scattering. We therefore see dipsin the transmission at each value of ~ω that satisfies eqn 3.32.

(d) Minima in the transmission occur at 0.807 eV, 0.823 eV, 0.832 eV and0.844 eV, with an average separation of 0.012 eV. We equate this separationenergy to e~B/µ, and hence find µ = 0.035me. If m∗

h À m∗e , we will have

µ = m∗e , and hence we deduce m∗

e ≈ 0.035me. The lowest energy transitionoccurs at Eg + (1/2)e~B/µ, which implies Eg = 0.80 eV.

The values we have deduced refer to the Γ point of the Brillouin zone.(See Fig. 3.10.) The indirect transitions for ~ω > 0.66 eV are too weak tobe observed compared to the direct transitions above 0.80 eV.

(3.17) The responsivity is calculated by using eqn 3.39. The device will be mostefficient if it has 100% quantum efficiency, and so we set η = 1, giving:

Responsivitymax =e

~ω(1− e−αl) .

On inserting the values given in the Exercise, we find responsivities of0.46A/W at 1.55 µm and 1.05 A/W at 1.30 µm.

(3.18) (a) The p–i–n diode structure is described in Appendix E. The p- andn-regions are good conductors, whereas the i-region is depleted of freecarriers and therefore acts like an insulator. We thus have two parallelconducting sheets separated by a dielectric medium, as in a parallel platecapacitor.

(b) The capacitance of a parallel-plate capacitor of area A, permittivityεrε0 and plate separation d is given by:

C =Aεrε0

d.

In applying this formula to a p–n junction, we should use the depletionregion thickness for d. In the case of a p–i–n diode, we assume that thedepletion lengths in the highly doped p- and n-regions are much smallerthan the i-region thickness, so that can set d = li. We then find C = 10pFfor a silicon p–i–n diode with A = 10−6 m2, εr = 11.9 and li = 10−5 m.

35



(c) The electric field for an applied bias of −10V can be calculated fromeqn E.3 as:

E =1.1− (−10)

10−5= 1.1× 106 V/m .

The electron and hole velocities can be calculated from the field and therespective mobilities:

v = µE .

This gives v = 1.7 × 105 m/s for the electrons and v = 5 × 104 m/s forthe holes. The drift time is finally calculated from t = li/v, which gives60 ps for the electrons and 200 ps for the holes.

Note that velocity saturation effects have been neglected here. The linearrelationship between the velocity and field breaks down at high fields,and the velocity approaches a limiting velocity called the saturation driftvelocity. The field in this example is quite large, and the transit times willactually be slightly longer than those calculated from the mobility due tothe saturation of the velocity.

(d) With R = 50Ω and C = 10 pF, we find RC = 500 ps. To obtain thesame transit time, we need a velocity of

v = li/t = 10−5/5× 10−10 = 2× 104 m/s .

This velocity occurs for an electric field of v/µe = 1.3 × 105 V/m. Wefinally find the voltage from eqn E.3:

E = 1.3× 105 =|1.1− V |

10−5,

which gives V = −0.2V. We therefore need to apply a reverse bias of0.2V.

The point about this last part of the question is to make the studentsthink about the factors that limit the response time of the photodetector.In most situations, the time constant will be capacitance-limited becausethe transit time is much shorter than the RC time constant. It is only insmall-area, low-capacitance devices that we need to worry about the drifttransit time.

36



Chapter 4

Excitons

(4.1) The Hamiltonian for the hydrogen atom has three terms corresponding tothe kinetic energies of the proton and electron, and the Coulomb attractionbetween them. On writing the position vectors of the electron and protonas r1 and r2, the Hamiltonian thus takes the form:

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 −e2

4πε0|r1 − r2| ,

where m1 = me, m2 = mp, and

∇2i =

∂2

∂x2i

+∂2

∂y2i

+∂2

∂z2i

.

We introduce the relative co-ordinate r and the centre of mass co-ordinateR according to:

r = r1 − r2

R =m1r1 + m2r2

m1 + m2.

Now

∂

∂x1=

∂x

∂x1

∂

∂x+

∂X

∂x1

∂

∂X=

∂

∂x+

m1

m1 + m2

∂

∂X∂

∂x2=

∂x

∂x2

∂

∂x+

∂X

∂x2

∂

∂X= − ∂

∂x+

m2

m1 + m2

∂

∂X,

which implies

∂2ψ

∂x21

=∂2ψ

∂x2+

2m1

m1 + m2

∂2ψ

∂x∂X+

(m1

m1 + m2

)2∂2ψ

∂X2

∂2ψ

∂x21

=∂2ψ

∂x2− 2m2

m1 + m2

∂2ψ

∂x∂X+

(m2

m1 + m2

)2∂2ψ

∂X2.

It is then apparent that:

1m1

∂2ψ

∂x21

+1

m2

∂2ψ

∂x22

=(

1m1

+1

m2

)∂2ψ

∂x2+

1m1 + m2

∂2ψ

∂X2.

Similar results may be derived for the other co-ordinates, so that we have:

1m1

∇21 +

1m2

∇22 =

(1

m1+

1m2

)∇2

r +1

m1 + m2∇2

R

37



On introducing the total mass M and reduced mass µ according to:

M = m1 + m2 ,1µ

=1

m1+

1m2

,

and substituting into the Hamiltonian, we then find

H = − ~2

2M∇2

R −~2

2µ∇2

r −e2

4πε0|r| .

The three terms in the Hamiltonian now represent respectively:

• the kinetic energy of the whole atom,• the kinetic energy due to relative motion of the two particles,• the Coulomb attraction, which depends only on the relative co-ordinate.

The Hamiltonian thus breaks down into two terms:

H = Hwhole atom + Hrelative ,

where:

Hwhole atom = − ~2

2M∇2

R

Hrelative = − ~2

2µ∇2

r −e2

4πε0|r| .

These two terms correspond respectively to the motions of:

• a free particle of mass M moving with the centre of mass co-ordinate,• a particle of mass µ experiencing the Coulomb force and moving

relative to a stationary origin.

The Hamiltonian is thus separable into the free kinetic energy of the atomas a whole and the bound motion of the electron relative to the nucleus.For the latter case, we describe the motion by using the reduced mass µ,rather than the individual electron mass. The reduced mass correctiondoes not make much difference for hydrogen itself, where m2 À m1 andhence µ ≈ m1, but it is very important for excitons, where the electronand hole masses are typically of the same order of magnitude.

(4.2) (a) The Hamiltonian describes the relative motion of the electron and holeas they experience their mutual Coulomb attraction within the semicon-ductor. The kinetic energy of the exciton as a whole is not included. Asexplained in Exercise 4.1, the appropriate mass is the reduced electron-hole mass µ, and the co-ordinate r is the position of the electron relativeto the hole. The first term represents the kinetic energy, and the second isthe Coulomb potential. The inclusion of εr in the Coulomb terms accountsfor the relative permittivity of the semiconductor.

(b) We substitute Ψ into the Schrodinger equation with the ∇2 operatorwritten in spherical polar co-ordinates:

∇2 =1r2

∂

∂r

(r2 ∂

∂r

)+

1r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1r2 sin2 θ

∂2

∂φ2.

38



Since Ψ does not depend on θ or φ, the Schrodinger equation becomes:

− ~2

2µ

1r2

∂

∂r

(r2 ∂Ψ

∂r

)− e2

4πε0εrrΨ = E Ψ .

On substituting Ψ = C exp(−r/a0), we obtain:(− ~2

2µa20

+~2

µa0r− e2

4πε0εrr

)Ψ = E Ψ .

The wave function is therefore a solution if

+~2

µa0r− e2

4πε0εrr= 0 ,

which implies

a0 =4πε0εr~2

µe2≡ εr

m0

µaH ,

where aH is the hydrogen Bohr radius. With this value of a0, we thenfind:

E = − ~2

2µa20

= − µe4

8ε2r ε20h

2≡ − µ

m0

1ε2r

RH ,

where RH is the hydrogen Rydberg energy.

The normalization constant is found by solving:∫ ∞

r=0

∫ π

θ=0

∫ 2π

φ=0

Ψ∗Ψ r2 sin θ drdθdφ = 1 .

This gives:

4πC2

∫ ∞

r=0

r2e−2r/a0 dr = 4πC2 × a30

4= 1 ,

which implies:

C =(

1πa3

0

)1/2

.

In the language of atomic physics, the wave function considered here is a1s state.

(4.3) The radial probability density P (r) is proportional to r2|R(r)|2, whereR(r) is the radial part of the wave function. For the 1s wave function ofExercise 4.2 we then have:

P (r) ∝ r2e−2r/a0 .

On differentiating, we find that this peaks at a0.

The expectation value of r is found from

〈r〉 =∫ ∞

r=0

∫ π

θ=0

∫ 2π

φ=0

Ψ∗rΨ r2 sin θ drdθdφ ,

= 4π

(1

πa30

) ∫ ∞

r=0

r3e−2r/a0 dr ,

= (3/2)a0 .

The peak of the probability density and the expectation value thus differby a factor of 3/2.

39



(4.4) The purpose of this exercise is to familiarize the student with the varia-tional method and demonstrate that it works. This will be useful to uslater for obtaining an estimate of the wave function and binding energyof the excitons in quantum wells. (See Exercise 6.9.)

(a) We require a spherically-symmetric wave function with a functionalforms that makes the probability density peak at some finite radius andthen decay to zero for large values of r. The given wave function satisfiesthese criteria. It is actually a correctly normalized 1s-like atomic wavefunction, but with a variable radius parameter ξ.

(b) We substitute Ψ into the Hamiltonian with the ∇2 written in sphericalpolar co-ordinates, as in Exercise 4.2. Since Ψ again depends only on r,this gives:

HΨ = − ~2

2µ

1r2

∂

∂r

(r2 ∂Ψ

∂r

)− e2

4πε0εrrΨ .

On evaluating the derivatives, we find:

HΨ =[− ~2

2µξ2+

(~2

µξ− e2

4πε0εr

)1r

](1

πξ3

)1/2

e−r/ξ .

The expectation value is then given by:

〈E〉 =∫ ∞

r=0

∫ π

θ=0

∫ 2π

φ=0

Ψ∗HΨ r2 sin θ drdθdφ

= 4π

(1

πξ3

) ∫ ∞

r=0

[− ~2

2µξ2r2 +

(~2

µξ− e2

4πε0εr

)r

]e−2r/ξ dr

=~2

2µξ2− e2

4πεrε0ξ.

(c) On differentiating 〈E〉 with respect to ξ, we find a minimum when ξ =4πε0εr~2/µe2. The value of 〈E〉 at this minimum is 〈E〉 = −µe4/8h2ε20ε

2r .

(d) The value of ξ that minimizes 〈E〉 is equal to a0, and the minimumvalue of 〈E〉 is the same energy as that found in Exercise 4.2.

The variational method gives exactly the right energy and wave functionhere because our ‘guess’ wave function had exactly the right functionalform. In other situations, this will not be the case, and the energy andwave functions obtained by the variational method will only be an approx-imation to the exact ones. The accuracy of the results will depend on howgood a guess we make for the functional form of the trial wave function.

(4.5) (a) The electron performs circular motion around the nucleus with quan-tized angular momentum equal to n~. The orbits are stable, and photonsare only emitted or absorbed when the electron jumps between orbits.

(b) The central force for the circular motion is provided by the Coulombattraction, and the electron velocity v must therefore satisfy:

µv2

r=

e2

4πε0εrr2,

40



where r is the radius of the orbit and µ is the reduced mass. (See Exercises4.1 and 4.2 for a discussion of why it is appropriate to use the reducedmass here.) The Bohr assumption implies that the angular momentum isquantized:

L = µvr = n~ .

On eliminating v from these two equations we find:

r =4πε0εr~2

µe2n2 ≡ m0

µεrn

2aH ,

where aH = 4πε0~2/m0e2 is the hydrogen Bohr radius. The energy is

found from:

E =12µv2 − e2

4πε0εrr.

On solving for v and substituting, we find:

E = − µe4

8h2ε20ε2rn

2≡ − µ

m0

1ε2r

RH

n2,

where RH = m0e4/8h2ε20 is the hydrogen Rydberg energy. These are the

same results as in eqns 4.1 and 4.2.

(c) E is identical to the exact solution of the hydrogen Schrodinger equa-tion.

(d) The radius for n = 1 corresponds to the peak in the radial probabilitydensity for the ground state 1s wave function. For higher atomic shells, theBohr radius corresponds to the peak radial density of the wave functionwith the highest value of the orbital quantum number l, namely l = n−1.

(4.6) The binding energies and radii can be calculated from eqns 4.1 and 4.2respectively. The reduced mass is calculated from eqn 3.22 to be µ =0.179m0, and with εr = 7.9 we then find RX = 39.1meV and aX = 2.3 nm.Hence we obtain E(1) = −39.1meV, E(2) = −9.8 meV, r1 = 2.3 nm andr2 = 9.3 nm.

We expect the excitons to be stable if E(n) > kBT . At room temperaturekBT = 25 meV, so that we would expect the n = 1 exciton to be stable,but not the n = 2 exciton.

(4.7) The reduced mass is calculated from eqn 3.22 to be 0.056m0, and hencewe calculate RX = 4.9meV from eqn 4.1 using εr = 12.4. Equation 4.4gives the wavelengths of the n = 1 and n = 2 excitonic transitions as873.7 nm and 871.4 nm respectively. Hence ∆λ = 2.3 nm.

(4.8) We assume that the exciton has a Lorentzian line with a centre energy of1.5149 eV and a full width at half maximum of 0.6meV. We then expectthe absorption and refractive index to follow a frequency dependence asin Fig. 2.5. This implies that the maximum in the refractive index wouldoccur at ω0 − γ/2, i.e. [1.5149− (0.6/2)] = 1.5146 eV.

The peak value of the refractive index can be worked out from eqns 2.15–21. At line centre, we have from eqns 1.19 and 1.28:

α =4πκ

λ=

4πε22nλ

,

41



which implies ε2(ω0) = 1.37 if we assume that n ≈ 3.5 (i.e. the excitoniccontribution to the refractive index is relatively small.) From eqn 2.21 wehave

ε2(ω0) = 1.37 = (εst − ε∞)ω0

γ= (εst − ε∞)

1.51490.6

,

implying (εst − ε∞) = 5.4 × 10−4. With εst = (3.5)2 = 12.25, we canthen find the dielectric constant at (ω0 − γ/2) using eqns 2.20–21. Thisgives ε(ω0 − γ/2) = 12.93 + 0.685i. We finally find n from eqn 1.25 to be3.60. This justifies the approximation n ≈ 3.5 in the calculation of ε2(ω0)above.

(4.9) The n = 1 and n = 2 excitons have energies of Eg − RX and Eg −RX/4 respectively. Hence the n = 1 → 2 transition occurs at a photonenergy of 3RX/4. We calculate RX = 4.2 meV from eqn 4.1, and henceconclude that the transition energy is 3.15 eV. This occurs at a wavelengthof 394 µm.

(4.10) The magnitude of the electric field between an electron and hole sepa-rated by a distance of r in a medium of relative permittivity εr is givenby Coulomb’s law as:

E =e

4πε0εrr2.

In the Bohr model we have (see e.g. Exercise 4.5):

|E(n)| = µe4

8(ε0εrhn)2

and

rn =4πε0εr~2n2

µe2.

It is thus apparent that for n = 1 we have:

2|E(1)|er1

=2RX

eaX=

e

4πε0εr

(πµe2

ε0εrh2

)2

=e

4πε0εrr2.

Hence E = 2RX/eaX.

(4.11) We use eqn 3.22 to find µ = 0.028m0, and then use eqns 4.1–2 withεr = 16 to calculate E(1) = 1.5 meV and r1 = 31nm. Then, using theresult of the previous exercise, we find that the internal field in the excitonhas a magnitude of 9.7 × 104 V/m. We expect the excitons to be ionizedwhenever the applied field exceeds this value. The voltage at which thisoccurs can be worked out from eqn 4.5. With Vbi = 0.74V and li =2× 10−6 m, we find E = 9.7× 104 V/m for V0 = +0.55V. Hence we needto apply a forward voltage of 0.55V. The excitons will therefore only beobserved for forward bias voltages above about 0.5 V. At zero bias and inreverse bias, the excitons will be ionized.

(4.12) The cyclotron energy is given by eqn 4.6 and the exciton Rydberg byeqn 4.1. The condition ~ωc = RX can thus be written:

e~Bµ

=µRH

m0ε2r,

42



which, on solving for B, gives:

B =µ2RH

m0ε2re~.

On inserting the numbers for GaAs we find B = 1.8T.

(4.13) The magnetic field is related to the vector potential by

B = ∇ × A .

Thus for A = (B/2)(−y, x, 0), we obtain:

B =B

2

∂x

∂y

∂z

×

−y

x0

=

00B

.

In the analysis following eqn B.17, we neglected the term in A2 becausethe magnetic vector potential of a light wave is small. However, we arenow considering the interaction between an exciton and a strong magneticfield, and it is precisely the term in A2 that gives rise to the diamagneticshift. It is then apparent from eqn B.17 that the diamagnetic perturbationfor a vector potential of A = (B/2)(−y, x, 0) is given by:

H ′ =e2A2

2m0=

e2B2

8m0(x2 + y2) .

The diamagnetic energy shift is calculated from

δE = 〈ψ|H ′|ψ〉 =e2B2

8m0〈ψ|(x2 + y2)|ψ〉 .

In the case of an exciton, the wave functions are spherically symmetric sothat:

〈x2〉 = 〈y2〉 = 〈z2〉 = 〈r2〉/3 = r2n/3 .

The total shift is obtained by summing the energy shifts of the electronsand holes to obtain:

δE =e2B2

8m∗e

2r2n

3+

e2B2

8m∗h

2r2n

3=

e2B2r2n

12µ.

(4.14) It is shown in Example 4.1 that the radius of the ground state exciton inGaAs is 13 nm. Hence for µ = 0.05m0 we calculate δE = +4.9× 10−5 eVat B = 1 T. The wavelength shift can be calculated from:

δλ =dλ

dEδE = − hc

E2δE = −0.026 nm .

(4.15) The effective masses given imply that µ = 0.17m0 from eqn 3.22, sothat we can calculate r1 = 3.1 nm and r2 = 12.3 nm from eqn 4.2 forεr = 10. The Mott densities are estimated from eqn 4.8. Hence we obtainNMott = 8.1 × 1024 m−3 and 1.3 × 1023 m−3 for the n = 1 and n = 2excitons respectively.

43



(4.16) The classical theory of equipartition of energy states that we have athermal energy of kBT/2 per degree of freedom. A free particle has threedegrees of freedom corresponding to the three velocity components. Hencethe total thermal kinetic energy is given by:

p2

2m=

32kBT .

This implies a thermal de Broglie wavelength of:

λdeB =h

p=

h√3mkBT

.

The particle density at the Bose–Einstein condensation temperature isgiven by eqn 4.9. On setting N = 1/r3, where r is the average inter-particle distance, we find

1r

= (2.612)1/3

√2πmkBT

h.

Hence:r

λdeB=

1(2.612)1/3

√32π

= 0.50 .

(4.17) Bose–Einstein condensation refers to the quantization of the kinetic en-ergy due to free translational motion. In applying this to excitons, we needto consider the centre of mass motion of the whole exciton, which behavesas a composite boson. The appropriate mass to use to calculate the con-densation temperature is therefore the total mass of the exciton, namely(m∗

e + m∗h) = 1.7m0. On substituting into eqn 4.9 with N = 1× 1024 m−3

and solving for Tc, we find Tc = 17.2 K.

(4.18) The excitonic radii calculated from eqn 4.2 are r1 = 0.85 nm and r2 =3.4 nm. For the n = 1 exciton, the radius is comparable to the unit cellsize a and thus the Wannier model is invalid. On the other hand, then = 2 exciton satisfies the condition r À a, and the Wannier model isvalid.

44



Chapter 5

Luminescence

(5.1) Indirect transitions involve the absorption or emission of a phonon to con-serve momentum. This means that they have a low transition probability,and hence low quantum efficiency. See Section 5.5.2.

(5.2) This occurs because the electrons can relax very rapidly to the bottomof the conduction band by emission of phonons, and similarly for holes.The relaxation processes occur on a much faster timescale (∼ps) thanthe radiative recombination (∼ns). Hence all the carriers have relaxedbefore emission occurs, and it makes no difference where they were initiallyinjected. See Section 5.3.1.

(5.3) In the 2p → 1s transition, the upper 2p level has n = 2, l = 1 and m = −1,0 or +1, while the lower 1s level has n = 1, l = 0 and m = 0. The EinsteinA coefficient is therefore given from eqn B.31 as:

A =e2ω3

3πε0~c3

13

∑m=−1,0,1

|〈2p,m|r|1s〉|2 ,

where the factor of 1/3 accounts for the triple degeneracy of the 2p state.(It is not necessary to consider the spin degeneracies here because theycancel out.) We write:

r = xi + yj + zk ,

so that:|〈r〉|2 = |〈x〉|2 + |〈y〉|2 + |〈z〉|2 .

Now atoms are spherically symmetric, and so it must be the case that:

|〈x〉|2 = |〈y〉|2 = |〈z〉|2 .

We therefore only have to evaluate one of these, and we chose 〈z〉 becausethe mathematics is easier. Written explicitly, with z = r cos θ, we have:

〈z〉 =∫ ∞

r=0

∫ π

θ=0

∫ 2π

φ=0

Ψ∗2p r cos θ Ψ1s r2 sin θ drdθdφ .

This has to be evaluated for each of the three possible m values of the 2pstate. However, since z has no dependence on φ, and the φ dependence ofthe wave functions is determined only by m, the integral is only non-zerofor the m = 0 level of the 2p state. Hence we only have to evaluate oneintegral. On inserting the explicit forms of the wave functions, we thenhave:

〈z〉 =∫ ∞

r=0

R∗21rR10r2dr

∫ π

θ=0

∫ 2π

φ=0

Y ∗1,0 cos θ Y0,0 sin θ dθdφ ,

=1√6a4

H

∫ ∞

r=0

r4 exp(−3r/2aH) dr ×√

34π

∫ π

θ=0

∫ 2π

φ=0

cos2 θ sin θ dθdφ .

45



This gives:

〈z〉 =24√

6

(23

)5

aH × 1√3

= 0.745 aH = 3.94× 10−11 m .

Then, with |〈r〉|2 = 3|〈z〉|2, we obtain:

A =e2ω3

3πε0~c3

13

3|〈z〉|2 =e2ω3

3πε0~c3|〈z〉|2 .

The 2p → 1s transition in hydrogen has an energy of (3/4)RH = 10.2 eV,and therefore ω = 1.55 × 1016 rad/s. Hence we obtain A2p→1s = 6.27 ×108 s−1. We then see from eqn 5.2 that the radiative lifetime τR = 1/A =1.6 ns. This agrees with the experimental value.

(5.4) It follows from eqn 5.4 that:

1τ

=1τR

+1

τNR,

where 1/τNR is the non-radiative recombination rate. Hence the excitedstate lifetime can be shorter than the radiative lifetime. The radiativelifetime τR would normally be independent of T , since it depends only onthe electron wave functions. However, the non-radiative recombinationrate 1/τNR generally increases with T due to increased non-radiative re-combination by phonon emission or thermally-activated traps. Hence τdecreases with T . The quantum efficiency can calculated from eqn 5.5 tobe equal to τ/τR. At 300 K we find ηR = 79 %, while at 350 K we findηR = 56 %.

(5.5) Semiconductors emit light at their band gap energy. We are therefore look-ing for a semiconductor with Eg = 2.3 eV. Inspection of Table C.3 suggesttwo possibilities: ZnTe or GaP. The latter has an indirect band gap andwould therefore not be very efficient. Hence the most likely candidate ma-terial is ZnTe. Note, however, that we could also make an emitter at thiswavelength by using an alloy, for example: InxGa1−xN. (See Fig. 5.11.)Linear extrapolation between the GaN and InN band gaps would suggestthat a composition with x ≈ 0.41 would emit at 540 nm.

(5.6) (a) Consider an incremental beam slice at a position z within the absorbingmaterial as shown in Fig. 16. Let A be the area of the slice, dz its thickness,I the incoming intensity (power per unit area), and δI the loss of intensitydue to absorption within the slice. From Beer’s law (eqn 1.3) we have:

δI = αIdz .

We assume that each absorbed photon generates an electron-hole pair.The number of electron-hole pairs generated within the slice per unit timeis therefore equal to AδI/hν. Hence the carrier generation rate G per unitvolume is given by:

G =AδI/hν

Adz=

AαIdz/hν

Adz=

αI

hν.

46



(b) The rate equation for the carrier density N is

dN

dt= G− N

τ=

Iα

hν− N

τ,

where the first term accounts for carrier generation and the second forcarrier recombination. In steady-state conditions we must have that

dN

dt= 0 .

Hence:N =

Iατ

hν.

(c) We calculate the laser intensity from:

I =P

A=

1mWπ (50 µm)2

= 1.3× 105 Wm−2 .

The sample is antireflection coated, and so this is also the intensity insidethe sample. We can then calculate the carrier density using the resultfrom part (b) with the values of τ and α given in the exercise. Hence wefind N = 6.6 × 1020 m−3 for hν = 2.41 eV. Note that this is the carrierdensity at the front of the sample. The intensity will decay exponentially(cf. eqn 1.4) and the carrier density will follow a similar exponential decay.

Incoming

photon number

N (z)

Outgoing

photon number

N-dN

A

dz

Incoming

intensity

I(z)

Outgoing

intensity

I - dI

z

Incoming

photon number

N (z)

Outgoing

photon number

N-dN

A

dz

Incoming

intensity

I(z)

Outgoing

intensity

I - dI

z

Figure 16: An incremental slice of a laser beam at a position z within anabsorbing material. A is the area of the slice and dz its thickness. In Exercise 5.6we consider a continuous laser beam with an intensity I(z), whereas in Exercise5.7 we consider a pulse with a photon number of N(z).

(5.7) (a) The argument proceeds along similar lines as for the previous exercise.Consider again an incremental beam slice of area A and thickness dz, asshown in Fig. 16. Since we are now dealing with a pulse rather than acontinuous beam, we need to consider the photon number rather than theintensity. Let N be the incoming photon number, and δN the number ofphotons absorbed within the slice. From Beer’s law (eqn 1.3) we have:

δN = αNdz .

We assume that each absorbed photon generates an electron-hole pair.Furthermore, we assume that the pulse is so short that no recombina-tion takes place while the material is being excited. Hence the number

47



of electron-hole pairs generated is just equal to the number of photonsabsorbed. The carrier density generated is then given by:

N =δN

Adz=

αNdz

Adz=

αN

A.

Now the number of photons N in the pulse is just equal to E/hν, whereE is the pulse energy. Hence we obtain:

N =αE

Ahν.

On inserting the relevant numbers from the exercise, we find N = 1.9 ×1024 m−3.

(b) The pulse excites the carrier density calculated in part (a) at timet = 0. The carriers then recombine and the carrier density decreasesaccording to

N(t) = N0 exp(−t/τ) ,

where τ is the total decay rate including both radiative and non-radiativerecombination. (See eqn 5.4.) With τ = (1/τR + 1/τNR)−1 = 0.89 ns, wefind N(t) = N0/2 at t = 0.62 ns.

(c) The quantum efficiency is calculated from eqn 5.5 as ηR = 89%. Eachlaser pulse contains E/hν photons. We are told that the crystal is ‘thick’,and so we can assume that all the laser photons are absorbed in the crystal.The number of photons re-emitted by luminescence is therefore EηR/hν =3.5× 1010 photons.

hnEg

E

kk = 0

Ee

Eh

hnEg

E

kk = 0

Ee

Eh

Figure 17: Definition of energies as required for Exercise 5.8.

(5.8) The emission rate is proportional to the probability that the upper levelis occupied and that the lower level is empty. These probabilities canbe calculated from the Fermi–Dirac functions for the electrons and holes,with fe as the electron occupancy of the upper level and fh as the holeoccupancy of the lower level (i.e. the probability that the lower level isempty.) Hence the emission probability is proportional to fe × fh.

48



In the classical limit, the occupancy factors follow Boltzmann statisticswith:

fe,h ∝ exp(−Ee,h/kBT ) ,

where Ee and Eh are the electron and hole energies within the conductionor valence band, respectively. (See Fig. 17.) Hence:

fe fh ∝ exp(−Ee/kBT ) · exp(−Eh/kBT ) = exp[−(Ee + Eh)/kBT ] .

Now it is apparent from Fig. 17 that

hν = Eg + Ee + Eh ,

and hence that (Ee + Eh) = hν − Eg. We therefore conclude that:

I(hν) ∝ fe fh ∝ exp[−(hν − Eg)/kBT ] .

(5.9) The number of electrons in the conduction band is given by eqn 5.6. In theclassical limit, we have (E − EF) À kBT , so that the electron occupancyfactor is given by:

fe(E) =1

exp[(E − EF)/kBT ] + 1→ exp[(EF − E)/kBT ] .

On using the density of states given in eqn 5.7, we then obtain:

Ne =1

2π2

(2m∗

e

~2

)3/2

exp(

EF

kBT

) ∫ ∞

Eg

(E − Eg)1/2 exp( −E

kBT

)dE .

On introducing the variable x = (E − Eg)/kBT we then obtain:

Ne =1

2π2

(2m∗

ekBT

~2

)3/2

exp(

(EF − Eg)kBT

) ∫ ∞

0

x1/2 exp(−x) dx .

The final result is obtained by setting (EF − Eg) ≡ EcF, where Ec

F is theelectron Fermi energy measured relative to the bottom of the conductionband.

For the case of GaAs at 300 K, we can insert the effective mass andtemperature, and use the definite integral given in the Exercise to obtain:

Ne = exp(

EcF

kBT

)× (4.4× 1023) m−3 .

In part (a) we then find EcF = −0.216 eV ≡ −8.4kBT , whereas in part (b)

we find EcF = +0.021 eV ≡ +0.83kBT . The approximations are therefore

valid for part (a) because we have (E − EF) À kBT for all states in theconduction band, but not for part (b), where the Fermi level comes outabove the conduction band minimum.

The aim of this exercise is to get the students to think about the conditionsunder which the use of Boltzmann statistics is valid.

49



(5.10) At T = 0 we have f(E) = 1 for E < EF and f(E) = 0 for E > EF.We can therefore cut off the integral at the relevant Fermi energy, andreplace the Fermi function by unity up to this energy. With the energiesmeasured relative to the band edge, we then find:

Ne =1

2π2

(2m∗

e

~2

)3/2 ∫ EcF

0

E1/2 dE ,

for the electrons and similarly for the holes.

On evaluating the integral we find:

Ne =1

2π2

(2m∗

e

~2

)3/2

× 23(Ec

F)3/2 ,

and similarly for the holes. Equation 5.13 then follows by simple re-arrangement.

(5.11) We use eqn 5.13 to evaluate the Fermi energies. In part (a) we findEc

F = 0.36meV for the electrons, and EvF = 0.073meV for the holes. In

part (b) we find EcF = 36 meV for the electrons, and Ev

F = 7.3meV for theholes.

For the conditions of degeneracy to apply, we need EF À kBT . In part (a),the electrons will be degenerate for T ¿ 4.2K, while the holes will bedegenerate for T ¿ 0.9 K. In part (b), the electrons will be degenerate forT ¿ 420 K, while the holes will be degenerate for T ¿ 85K. This showsthat it is much easier to obtain degenerate statistics for the electrons thanthe holes because they are lighter.

(5.12) The Fermi wave vector kF is, in general, related to the Fermi energy EF

by:

EF =~2k2

F

2m.

With EF given by eqn 5.13, we then find:

kF = (3π2N)1/3 ,

where N = Ne or Nh as appropriate. In a photoluminescence experiment,we excite equal numbers of electrons and holes, so that Ne = Nh. Hencethe Fermi wave vectors of the electrons and holes are identical. The factthat the mass does not appear in the formula for the Fermi wave vectoris a consequence of the fact that the density of states in k-space does notdepend on the mass. (See eqn 3.15.)

(5.13) (a) The solid angle subtended by an object of area dA at a distance r isgiven by (See Fig 18(a)):

dΩ =dA

r2.

The lens will be positioned at a distance equal to its focal length from thesample, and so we set r = 100 mm in this case. We then find:

dΩ =π(25mm/2)2

(100 mm)2= 0.049 .

50



dA

r

dW

(a)

qinside

qoutside

medium

n

air

n = 1

S

(b)

dA

r

dW

(a)

qinside

qoutside

medium

n

air

n = 1

S

(b)

Figure 18: (a) Definition of solid angle dΩ for an object of area dA at a distanceof r, as required for Exercise 5.13. (b) Emission of a ray at angle θinside from asource S embedded within a medium of refractive index n.

(b) Consider a ray emitted by a source embedded within a medium ofrefractive index n as illustrated in Fig. 18(b). The ray is refracted at thesurface according to Snell’s law, with:

sin θoutside

sin θinside= n .

The light is being collected by a lens of radius 12.5mm positioned 100mmfrom the surface, and therefore only those rays that satisfy θoutside ≤12.5/100 will be collected (assuming small angles). By Snell’s law wededuce that only those rays within a cone with θinside ≤ 0.125/n will becollected.

The source emits uniformly over all 4π steradians within the medium.The fraction F of the photons emitted that can be collected by the lens isworked out by considering a circle of radius rθinside placed at a distanceof r from the source:

F =dA

4πr2=

πr2(θinside)2

4πr2=

(θinside)2

4≈ (θoutside)2

4n2.

Finally, we have to consider that some of the photons will be reflectedat the surface. The reflectivity is calculated from eqn 1.29 to be 21% forn = 2.7. Hence the final fraction collected is 0.79F = 4.2× 10−4.

(c) The semiconductor absorbs photons of energy 2.41 eV and emits lu-minescent photons at the band gap energy of 1.61 eV with a probabilityequal to ηR. Hence the power emitted is equal (1.61/2.41)ηR times thepower absorbed. The incoming laser will be partially reflected at thesurface, and so the maximum power that can be absorbed is equal to(1 − R) × Pincident = 0.79 × 1 mW. Hence the maximum luminescentpower is equal to:

Plum =1.612.41

ηR × 0.79× 1mW = 0.53 ηR mW .

(d) We obtain the luminescent power collected by the lens by multiplying

51



the total luminescent power by the collected fraction:

P collected = 0.53 ηR × 4.2× 10−4 mW = 0.22 ηR µW .

(5.14) (a) The electron Fermi energy is calculated from eqn 5.13 to be 0.14 eV.

(b) The hole Fermi energy is calculated from:

Nh =∫ Ev

F

0

[ghh(E) + glh(E)] dE

=∫ Ev

F

0

12π2

(2~2

)3/2

(m3/2hh + m

3/2lh )E1/2 dE

=1

3π2

(2~2

)3/2

(m3/2hh + m

3/2lh ) (Ev

F)3/2 .

Hence for Nh = 2× 1024 m−3 we obtain EvF = 0.012 eV.

(c) The carriers will be degenerate if EF > kBT . At 180 K we have kBT =0.018 eV, and so the electrons are degenerate, but not the holes.

(d) When both the electrons and holes are degenerate, we expect emissionfrom the band edge up to the Fermi energies as illustrated in Fig. 5.7.We then expect emission from the band gap to Eg + Ec

F + EvF. At T = 0

we would have a sharp cut-off at this energy, but at finite T the edge isbroadened over ∼ kBT . The luminescence would then be expected to fallto 50 % of its peak value at Eg + Ec

F + EvF. On inserting the calculated

Fermi energies, and the value of the band gap, we expect the 50% pointat around 0.95 eV. This can be compared to the experimental value of∼ 0.94 eV. The agreement between the experiment and model is thusgood.

(e) The luminescence spectrum at 250 ps given in Fig. 5.8 is consistentwith a value of the electron Fermi energy of ∼ 0.035 eV. This implies fromeqn 5.13 that the carrier density is Ne ≈ 3× 1023 m−3. The electrons arestill degenerate at this density for T = 55 K. The lifetime is estimatedfrom:

N(t) = N0 exp(−t/τ) ,

which implies N(24)/N(250) = exp(−226/τ), where τ is the lifetime inps. On inserting the two values, we find τ ≈ 120 ps = 0.12 ns.

(5.15) We consider a zinc–blende III–V semiconductor with the band structureshown in Fig. 3.8. As explained in Section 3.3.7, the heavy-hole transi-tions are three times as strong as the light-hole transitions. The selectionrules are ∆MJ = ∓1 for σ± light, as indicated in Fig. 19. Note thatthese selection rules are the opposite way around to those for absorption,because we are now considering emission.

Consider the situation after excitation with σ+ light resonant with theband gap. There will be three times as many electrons in the MJ = −1/2state as the +1/2 state. On the other hand, the holes will be equally dis-tributed between all four possible sub-levels due to fast spin–orbit mixing.The transitions that are possible are indicated in Fig. 19. The intensity

52



E

MJ

J = 3/2

J = 1/2

J = 1/2

-1/2-3/2 +3/2+1/2

valence band

conduction

band

-D

0

Eg

s+

s+

s-

s-

hh hhlh lh

so so

3N N

31

31

E

MJ

J = 3/2

J = 1/2

J = 1/2

-1/2-3/2 +3/2+1/2

valence band

conduction

band

-D

0

Eg

s+

s+

s-

s-

hh hhlh lh

so so

3N N

31

31

Figure 19: Optical transitions after excitation with σ+ photons, as required forExercise 5.15. The relative weights of the transitions are indicated.

will be proportional to the population of the initial electron state and thetransition weight (i.e. |M |2).From the MJ = −1/2 electron state with population 3N we have twotransitions:

• A σ+ transition to the MJ = −3/2 hh level: relative intensity =3× 3 = 9.

• A σ− transition to the MJ = +1/2 lh level: relative intensity =3× 1 = 3.

From the MJ = +1/2 electron state with population N we also have twotransitions:

• A σ+ transition to the MJ = −1/2 lh level: relative intensity =1× 1 = 1.

• A σ− transition to the MJ = +3/2 hh level: relative intensity =1× 3 = 3.

The total relative σ+ intensity I+ is therefore 9 + 1 = 10, while the totalrelative σ− intensity I− is 3 + 3 = 6. The polarization is then given byeqn 5.14 as:

P =I+ − I−

I+ + I−=

10− 610 + 6

=416

= 25% .

The argument works equally well for σ− excitation.

(5.16) We see from eqn 5.16 that P (B)/P (0) = 1/2 when ΩTS = 1. Hence atB1/2 we have (from eqn 5.17):

ΩTS =geµBB1/2TS

~= 1 .

53



Hence (cf eqn 5.18):

1TS

≡ 1τ

+1τS

=geµBB1/2

~.

Now from eqn 5.15 we have that:

P (0)P0

=1

1 + τ/τS,

which implies:1τ

P0

P (0)=

1τ

+1τS

.

Hence:1τ

P0

P (0)=

geµBB1/2

~,

which gives:

τ =~

geµBB1/2

P0

P (0).

We can then work out τS from:

1τS

=geµBB1/2

~− 1

τ,

which implies:

τS =~

geµBB1/2

(1− P (0)

P0

)−1

.

Note that P0 is expected to be 25% for a bulk III-V semiconductor, asshown in Exercise 5.15.

(5.17) We follow Example 5.1. We extrapolate linearly between the direct gapsof GaP and GaAs to find that:

Edirectg (x) = 1.42 + 1.36x eV .

(a) The LED will cease to function efficiently when the gap becomes in-direct, i.e. for x > 0.45, where the conduction band minimum at theBrillouin zone edge drops below the conduction band minimum at k = 0.Hence the minimum wavelength occurs for x = 0.45, where Eg = 2.03 eV.This corresponds to a wavelength of 610 nm.

(b) We require a direct band gap equivalent to 670 nm, namely 1.85 eV.On substituting into the formula for the direct gap, we find x = 0.316.Note that this is less than 45%, so that the alloy will have a direct gapand hence be an efficient emitter.

(5.18) (a) The reflectivity is calculated from eqn 1.29 to be 31% for n = 3.5.

(b) The cavity mode frequency is given by eqn 5.20, and implies a modespacing of ∆ν = c/2nl. With n = 3.5 and l = 1 mm, we find ∆ν =4.3× 1010 Hz.

54



(c) The threshold gain can be calculated from eqn 5.24. We are told toignore background absorption and scattering, and so we set αb = 0. Thecoated facet has a reflectivity of 95%, while the other facet will just havethe natural reflectivity of 31%. Hence we find:

γth = − 12× 10−3 m

ln(0.95× 0.31) = 610 m−1 .

(5.19) (a) The maximum possible power would be obtained if one photon isemitted for each electron that flows through the device. The power isthen equal to the electron flow rate multiplied by the photon energy:

Pmax = hν ×(

100mAe

)= 150 mW .

(b) The electrical power input is equal to IV = 0.1 × 1.9 = 0.19W =190mW. With a power output of 50mW, the power conversion efficiencyis therefore equal to (50/190) = 26 %.

(c) The power output will vary as shown in Fig. 5.15(a). The slope effi-ciency is calculated from eqn 5.26 as (50/(100 − 35) = 0.77 mW/mA =0.77 W/A. The differential quantum efficiency is calculated from eqn 5.26to be 51 %.

(5.20) The incident primary electron flux per unit area is equal to J/e elec-trons m−2. Each primary electron will generate N eh electron hole pairs ina distance Re from the surface, where N eh is given by eqn 5.27 and Re isthe penetration depth. The generation rate per unit volume of crystal isthus given by:

G =J

eRe×N eh =

J

eRe(1− γ)

Ep

Ei.

The rate equation for the number of electron–hole pairs is thus given by:

dN

dt= G− N

τ,

where τ is the lifetime. In steady state conditions we must have N = 0(cf Exercise 5.6), and we thus have:

N = Gτ =Jτ

eRe(1− γ)

Ep

Ei.

55



Chapter 6

Quantum confinement

(6.1) Substitute into eqn 6.3 with ∆x = 10−6 m and m = 0.1m0 to obtainT . 0.01K.

(6.2) The energy difference between the n = 1 and n = 2 levels is worked outfrom eqn 6.13 to be 3~2π2/2m∗d2. On setting this energy difference to beequal to kBT/2, we then derive the required result, namely

d =

√3~2π2

m∗kBT.

On substituting into this formula with T = 300K, we find d = 9.3 nm form∗ = m0, and d = 30 nm for m∗ = 0.1m0.

On comparing with eqn 6.4, it is apparent that d =√

3π2∆x. This showsthat the two criteria used to determine when quantum size effects areimportant give the same answer, apart from a numerical factor of ∼ 5.The differing numerical factor is to be expected, given the approximatenature of the criteria.

kx

ky

(2p/L)2

k

dk

kx

ky

(2p/L)2

k

dk

Figure 20: Grid of allowed k states in a two-dimensional material of area L2,as required for the solution of Exercise 6.3.

(6.3) The x and y components of the k vector must each satisfy the criterionexp(ikL) = 1, which implies k = integer × 2π/L. The possible values ofthe k vector therefore form a regular grid in k-space as shown in Fig. 20,with a grid-spacing of 2π/L. The area per k-state is (2π/L)2, and the

56



density of states for an area L2 is therefore L2/(2π)2. This implies thatthe density of states in k-space for a unit area of crystal is 1/(2π)2.

It is also apparent from Fig. 20 that the area of k-space enclosed by theincrement k → k+dk is 2πkdk. If we call the number of k-states enclosedby this area gL(k)dk, we then find:

gL(k)dk =2πkdk

(2π/L)2=

L2

2πkdk .

Hence for a unit area of crystal we have:

g2D(k)dk =k

2πdk .

The density of states in energy space can be found from eqns 3.13–14. (Thefactor or two accounts for the fact that spin 1/2 particles have two spinstates for each k-state.) With E(k) = ~2k2/2m, we have dE/dk = ~2k/m,and hence

g2D(E) =2g2D(k)dE/dk

=2k/2π

~2k/m=

m

π~2,

as required.

(6.4) In one dimension the k states must all lie on a single axis (say the xaxis). Let L be the length of the wire. The Born–von Karman boundaryconditions require that exp(ikL) = 1, and hence that k = integer× 2π/L.The density of states in k space for a sample of unit length is therefore1/2π. We thus have:

g1D(k)dk =12π

dk .

The density of states in energy space is then found from eqns 3.13–14 withE = ~2k2/2m. This gives:

g1D(E) =2g1D(k)dE/dk

=2/2π

~2k/m=

m

π~2k=

( m

2~2π2

)1/2

E−1/2 ,

as required.

(6.5) The depth of the potential well enters eqn 6.26 via ξ, which is defined ineqn 6.27. The function on the right hand side of eqn 6.26 decreases from(m∗

w/m∗b)1/2

√ξ at x = 0 to zero at x =

√ξ. (See, for example, Fig. 6.5,

for the case with ξ = 13.2.) The function on the RHS of eqn 6.26 willtherefore always cross the x tan x function between 0 and π/2, no matterhow small ξ is.

(6.6) We follow the method of Example 6.1. We have

(m∗

w

m∗b

)1/2

=(

0.340.5

)1/2

= 0.82 ,

and

ξ =0.34m0 × (10−8)2 × 0.15 eV

2~2= 33.5 .

57



0 2 4 60

2

4

6

y

x

y = x tan x

y = 0.82 (33.5-x2)1/2

x=

1.3

00 2 4 6

0

2

4

6

y

x

y = x tan x

y = 0.82 (33.5-x2)1/2

x=

1.3

0

Figure 21: Graphical solution required for Exercise 6.6.

Hence we must solve:

x tan x = 0.82√

33.5− x2 .

The two functions are plotted in Fig. 21, which shows that the solution isx = 1.30. The energy is then found from

E =2~2x2

m∗wd2

= 7.6meV ,

for d = 10nm. The equivalent energy for an infinite well is calculatedfrom eqn 6.13 to be 11 meV.

(6.7) (a) The wave functions of an infinite potential well form a complete or-thonormal basis, with

∫ +∞

−∞ϕ∗nϕn′ dz = δnn′ ,

where δnn′ is the Kronecker delta function:

δnn′ = 1 if n = n′

= 0 if n 6= n′ .

It is apparent from eqn 6.11 that the wave functions depend only on nand are therefore identical for electrons and holes with the same n. Hencethe orthonormality condition applies irrespective of whether ϕn and ϕn′

are electron or hole wave functions, or a mixture. Hence:

Mnn′ = 1 if n = n′

= 0 if n 6= n′ .

This result can also be derived by explicit (and somewhat tedious) substi-tution of the wave functions from eqn 6.11 into the formula for Mnn′ .

(b) This result follows from parity arguments. The wave functions ofa finite well have well-defined parities as a consequence of the inversion

58



symmetry about the centre of the well. Wave functions with odd n haveeven parity, while those with even n have odd parity. Hence the productϕ∗enϕ∗hn′ has even parity if n−n′ is even and has odd parity for odd n−n′.The integral of an odd function from −∞ → +∞ is zero, and so theoverlap integral is zero if ∆n is equal to an odd number.

n Electron Heavy hole Light hole1 225 30 1882 898 120 750

Table 1: Confinement energies in meV calculated for an infinite quantum wellof width 5 nm, as required for Exercise 6.8.

(6.8) The confinement energies of the electrons, heavy holes and light holescalculated from eqn 6.13 are given in Table 1. With an infinite well, weonly need consider ∆n = 0 transitions. The threshold photon energies forthese transitions are given by eqn 6.39, or the equivalent for higher valuesof n or for the light holes. Two transitions fall in the range 1.4 → 2.0 eV:

• Heavy hole 1 → electron 1 at 1.679 eV,

• Light hole 1 → electron 1 at 1.837 eV.

For each transition we expect a step at the threshold energy as in Fig. 6.9.In 2-D materials the joint density of states is proportional to the electron-hole reduced mass µ (see eqn 6.41), and hence the relative height of theheavy-hole and light-hole transition steps is in proportion to their reducedmasses, that is 0.059 : 0.036. Hence the final spectrum would appear as inFig. 22.

1.4 1.6 1.8 2.0

0.0

0.1

Abso

rpti

on

(arb

.unit

s)

Photon energy (eV)

hh1 ® e1

lh1 ® e11.679 eV

1.837 eV

1.4 1.6 1.8 2.0

0.0

0.1

Abso

rpti

on

(arb

.unit

s)

Photon energy (eV)

hh1 ® e1

lh1 ® e11.679 eV

1.837 eV

Figure 22: Absorption spectrum for Exercise 6.8

(6.9) (a) In finite wells the confinement energies are reduced compared to thoseof an infinite well of the same width. Hence the transition energies wouldbe lower. Furthermore, transitions that are forbidden in infinite wellsbecome weakly allowed, such as the hh3 → e1 transition. This transitionwould fall within the observed energy range, as might also some n = 2transitions.

59



(b) Peaks would appear below the steps in the absorption spectrum dueto excitonic absorption. The difference in energy between the peak andthe continuum absorption would be equal to the exciton binding energy.

(6.10) (a) To prove normalization, we must show that∫∫

ψ∗ψ dA = 1. For thegiven wave function we have:

∫ ∞

r=0

∫ 2π

φ=0

Ψ∗Ψ rdrdφ =(

2πξ2

) ∫ ∞

r=0

∫ 2π

φ=0

exp(−2r/ξ) rdrdφ ,

=(

2πξ2

)× 2π ×

∫ ∞

r=0

exp(−2r/ξ) rdr ,

= 1 ,

as required.

(b) We first compute the effect of H on Ψ, using the fact that ∂Ψ/∂φ = 0:

HΨ = − ~2

2µr

ddr

(rdΨdr

)− e2

4πε0εrrΨ ,

= − ~2

2µξ2Ψ +

(~2

2µξ2− e2

4πε0εr

)Ψr

.

We can then calculate 〈E〉var from:

〈E〉var =∫ ∞

r=0

∫ 2π

φ=0

Ψ∗HΨ rdrdφ ,

=∫ ∞

r=0

∫ 2π

φ=0

Ψ∗[− ~2

2µξ2Ψ +

(~2

2µξ2− e2

4πε0εr

)Ψr

]rdrdφ ,

= − ~2

2µξ2

∫ ∞

r=0

∫ 2π

φ=0

Ψ∗Ψ rdrdφ +(~2

2µξ2− e2

4πε0εr

) ∫ ∞

r=0

∫ 2π

φ=0

Ψ∗Ψdrdφ ,

= − ~2

2µξ2× 1 +

(~2

2µξ2− e2

4πε0εr

)× 2

ξ,

= +~2

2µξ2− e2

2πε0εrξ.

(c) On differentiating 〈E〉var with respect to ξ, we find that 〈E〉var achievesits minimum value of Emin = −µe4/8(πε0εr~)2 for ξ = 2π~2ε0εr/µe2. Theminimum energy is four times larger than the bulk exciton binding energyfound in Exercise 4.4.

(d) In part (c) we found ξmin = 2π~2ε0εr/µe2. This can be written interms of the bulk exciton radius aX defined in eqn 4.2 as:

ξmin = aX/2 .

Hence the radius of the exciton in 2-D is half the radius of the bulk.

(6.11) At d = ∞ we have bulk GaAs, while at d = 0 we have bulk AlGaAs.For intermediate values of d, we have a GaAs quantum well exciton withan enhanced binding energy. In an ideal 2–D system we would expect

60



four times the binding energy of bulk GaAs (i.e. 16 meV), but in realisticsystems, the enhancement might be smaller due to the imperfect quantumconfinement of the finite-height AlGaAs barriers. Thus as d is reducedfrom ∞, the binding energy increases from 4 meV, going through a peak,and then dropping to 6 meV as d → 0. The height of the peak mighttypically be around 10 meV.

1.58 1.60 1.62 1.640.0

1.0

Energy (eV)

PL

Ein

ten

sity

hh1 ® e1

exciton

lh1 ® e1

exciton

hh1 ® e1 band edge

lh1 ® e1

band edge

1.58 1.60 1.62 1.640.0

1.0

Energy (eV)

PL

Ein

ten

sity

hh1 ® e1

exciton

lh1 ® e1

exciton

hh1 ® e1 band edge

lh1 ® e1

band edge

Figure 23: Interpretation of the data in Fig. 6.23 as required for Exercise 6.12.

(6.12) (a) See Section 5.3.5.

(b) The interpretation of the principal features in the data is shown inFig. 23. For both heavy- and light-hole transitions, we expect to observe apeak due to excitonic absorption followed by a step at the band edge. Thetwo strong peaks observed in the data correspond to the excitons for thehh1→ e1 and lh1→ e1 transitions, while the flat absorption bands aboveboth excitons correspond to the interband transitions. These interbandabsorption bands are flat because the density of states is independent ofthe energy in 2-D systems (see eqn 6.41.)

(c) The hh1→ e1 interband continuum starts at 1.592 eV. In the infinitewell model, the transition energy is given by eqn 6.42 with n = 1, which,on using Eg = 1.519 eV, m∗

e = 0.067m0 and m∗hh = 0.5m0, implies d =

9.3 nm. The real well width would be smaller, because the infinite wellmodel overestimates the confinement energy.

(d) The binding energies can be read from Fig. 6.23 as the energy gapbetween the exciton peak and the appropriate band edge. With the tran-sitions identified as in Fig. 23, we find binding energies of 11 meV for theheavy holes and 12 meV for light holes. For a perfect 2-D system we wouldexpect 4×RX for GaAs, i.e. 16.8 meV. The experimental values are lowerbecause a real quantum well is not a perfect 2-D system.

(6.13) By considering Fig. 6.12 we would expect three different regimes to apply.

1. For photon energies in the range set by eqn 6.43, that is:

Eg + Ee1 + Ehh1 ≤ ~ω < Eg + Ee1 + Elh1 ,

we can only excite hh1 → e1 transitions. These only create MJ =−1/2 spin down electrons, and so we expect Πe = −100%.

61



2. For photon energies in the range:

Eg + Ee1 + Elh1 ≤ ~ω < Eg + Ee1 + Eso1 ,

where Eso1 is the confinement energy of the first split-off hole band,we can excite both hh1 → e1 and lh1 → e1 transitions, which createoppositely polarized electrons. Hence we expect the magnitude of theelectron spin polarization to decrease. If the matrix elements are thesame as in the bulk, the heavy-hole transitions will be three timesstronger than the light-hole transitions, and this gives an electronspin polarization of −50%. (See Section 3.3.7.)

3. For photon energies in the range:

~ω > Eg + Ee1 + Eso1 ,

we expect that no spin polarization will be be created. The extracontribution of the split-off hole band exactly cancels the net polar-ization created by the stronger weight of the heavy-hole transition.This can be understood in general terms by realizing that the elec-tron spin polarization is created by spin–orbit coupling, so that whenthe excess photon energy exceeds the spin–orbit interaction energy∆, then the effect will be negligible.

Experimental data showing these effects may be found in Vina, J. Phys.:Condens. Matter 11, 5929 (1999), Fig. 2. The basic trends discussedabove are reproduced in the experimental data, although there are someimportant differences. This is because we really should work in the ex-citonic rather than single-particle picture, and also because the relativeweight of the heavy- and light-hole transitions is not exactly 3:1. See thediscussion in Vina’s article for further details.

(6.14) (a) With H ′ = +ezEz, we have:

∆E(1) = eEz

∫ +∞

−∞ϕ∗zϕ dz .

Since z is an odd function, and ϕ∗ϕ ≡ |ϕ|2 is an even one, the integral iszero.

(b) With H ′ = +ezEz, the second-order energy shift is given by:

∆E(2) = e2E2z

∞∑n=2

|〈1|z|n〉|2E1 − En

.

Since the wave functions have parity (−1)n+1, and z is an odd parityoperator, all the terms with odd n are zero. Hence we have:

∆E(2) = e2E2z

|〈1|z|2〉|2E1 − E2

+ e2E2z

|〈1|z|4〉|2E1 − E4

+ · · · .

Now the terms with n ≥ 4 are much smaller than the term with n = 2.(This can be verified by working through the integrals, but it is fairly

62



obvious given the larger denominator.) Hence we only need to considerthe first term:

∆E(2) = e2E2z

|〈1|z|2〉|2E1 − E2

.

On substituting the energies for an infinite well from eqn 6.13, this be-comes:

∆E(2) = −2e2E2zm

∗d2

3~2π2|〈1|z|2〉|2 .

Now, on redefining the origin so that the quantum well runs from z = 0to z = +d, we have:

〈1|z|2〉 =∫ +∞

−∞ϕ∗1 z ϕ2 dz ,

=2d

∫ d

0

sin(πz/d) z sin(2πz/d) dz ,

= −16d

9π2.

Hence we find

∆E(2) = −2e2E2zm

∗d2

3~2π2×

∣∣∣∣−16d

9π2

∣∣∣∣2

= −24(

23π

)6e2E2

zm∗d4

~2,

as required.

(6.15) (a) Figure 6.13 predicts that the energy of the hh1→e1 transition willshift from 1462.4 meV to 1438.5 meV in a 10 nm quantum well on in-creasing the field from zero to 1 × 107 V/m. This corresponds to a redshift of 14.1 nm. This can be compared to the experimental red shift of10.5 nm for a 9 nm well for a voltage change of 10 V. Excitonic effectsare not included in the calculation of Fig. 6.13, but this is not expectedto make a large difference. Two main factors account for the differencebetween the experimental and theoretical results.

• The well width for the experiment was smaller by a factor of 0.9.From eqn 6.47 we would then expect the Stark shift to be smaller bya factor (0.9)4 = 0.66.

• For the experimental data, we can calculate the field from eqn 6.52.From this we find that Ez changes from 0.15 × 107 V/m to 1.15 ×107 V/m, which should be compared to the theoretical data, whichcorresponds to changing Ez from 0 → 1× 107 V/m. The field changeis the same in both cases, but because the Stark shift is quadratic atlow fields, we do not expect the Stark shifts to be the same. If theStark shift remained quadratic at all field strengths, we would expectthe experiment shift to be larger by a factor (1.152−0.152)/(12−02) =1.30.

Putting these two factors together, we expect the experimental Stark shiftto be smaller than the theoretical one by a factor of 0.94×1.3 = 0.85. Theactual ratio is (10.5/14.1) = 0.74. This difference is easily explained by

63



the fact that the Stark shift does not remain quadratic at all fields, andso the factor due to the different fields should be smaller.

(b) We assume that the Stark shift is quadratic, and hence that ∆E ∝E2

z. We work out the field strength from eqn 6.52, which implies Ez =11.5 MV/m at 10V and Ez = 6.5 MV/m at 5V. For small shifts we have∆λ ∝ ∆E, and so we find:

∆λ(5V) =(

6.511.5

)2

∆λ(10V) = 0.32× 10.5 nm = 3.4 nm .

(c) The wavelength red shift of 10.5 nm corresponds to an energy shift of–18meV. This energy shift is related to the net electron-hole displacement〈∆z〉 by:

∆E = −pzEz = −e〈∆z〉Ez .

With Ez = 11.5MV/m, we thus obtain 〈∆z〉 = 1.6 nm.

Sample A Sample BEz (MV/m) ∆Ecalc ∆Eexp ∆Ecalc ∆Eexp

3 1.5 1 15 66 5.9 5 62 279 13 13 139 54

Table 2: Comparison of the calculated and measured Stark shifts (in meV) forthe two samples discussed in Exercise 6.16.

(6.16) The experimental data is taken from Polland et al., Phys. Rev. Lett. 55,2610 (1985). We analyse it by using the energy shift calculated by second-order perturbation theory in Exercise 6.14(b). Since we are consideringan electron-hole transition, we must add together the Stark shifts of theelectrons and holes, giving:

∆E(2) = −24(

23π

)6e2E2

zd4

~2(m∗

e + m∗hh) .

The calculated shifts using m∗e = 0.067m0 and m∗

hh = 0.5m0 are comparedto the experimental ones in Table 2. It is apparent that the model workswell for sample A, but not for sample B. The model breaks down when thesize of the Stark shift becomes comparable to the energy splitting of theunperturbed hh1 and hh2 levels. This is essentially the same criterion asfor the transition from the quadratic to the linear Stark effect in atomicphysics. In sample B, we are in this regime at all the fields considered.

(6.17) At Ez = 0 the quantum well is symmetric about the centre of the well.The electron and hole states therefore have a definite parity with respectto inversion about z = 0. The parity is (−1)(n+1), and the electron–holeoverlap given by eqn 6.36 is zero if ∆n is odd: see Exercise 6.7(b). Atfinite Ez, the inversion symmetry is broken, and the states no longer havea definite parity. Therefore, the selection rule based on parity no longerholds.

64



(6.18) With the confinement energy E ∝ d−2, we have dE/dd ∝ −2/d3, andhence we expect:

∆E

E= −2

∆d

d.

A ±5% change in d is thus expected to give ∆E/E = ±10%. WithE = 0.1 eV, we then expect a full-width broadening of 0.02 eV. This iscomparable to the linewidth observed in the 10 K data shown in Fig. 6.16.

A ±5% variation in d corresponds more or less to a fluctuation of oneatomic layer. Such “monolayer” fluctuations are unavoidable in the crystalgrowth. The linewidth at room temperature is further broadened by thethermal spread of the carriers in the bands.

(6.19) We assume infinite barriers and use eqn 6.53 to calculate the emissionenergy. With ~ω = 0.80 eV, Eg = 0.75 eV, and µ = 0.038m0, we findd = 14nm. In reality, the quantum well would have to be narrower tocompensate for the imperfect confinement of the barriers.

(6.20) (a) z is an odd function with respect to inversion about z = 0. Theintegral from −∞ to +∞ will therefore be zero unless ϕ∗nϕn′ is also anodd function, which requires that the wave functions must have differentparities. Since the wave functions have parities of (−1)n+1, the conditionis satisfied if n is an even number and n′ odd, and vice versa. Hence ∆nmust be equal to an odd number.

(b) The strength of the intersubband transitions is proportional to thesquare of the matrix elements. These matrix elements can be evaluatedby substituting the wave functions from eqn 6.11. On redefining the originso that the quantum well runs from z = 0 to z = +d, we find:

〈1|z|2〉 =2d

∫ d

0

sin(πz/d) z sin(2πz/d) dz = − 169π2

d ,

and

〈1|z|4〉 =2d

∫ d

0

sin(πz/d) z sin(4πz/d) dz = − 445π2

d .

Hence the 1 → 4 transition is weaker than the 1 → 2 transition by a factor[(4/45)/(16/9)]2 = [1/20]2 = 2.5× 10−3.

It is apparent from Fig. 6.17 that the wavelength of the 1 → 2 transitionis given by

hc

λ= E2 −E1 =

3π2~2

2m∗ed

2,

where we used the infinite well energies of eqn 6.13 in the second equality.On inserting m∗

e = 0.067m0 and d = 20 nm, we find hc/λ = 0.042 eV, andhence λ = 29 µm.

(6.21) Consider a ray incident at angle θ to the normal as shown in Fig. 24.The ray will be refracted according to Snell’s law, with

sin θ

sin θ′= n ,

65



q ¢

q

semiconductor, refractive index n

air

n = 1 E

E¢

z

quantum well

q ¢

q

semiconductor, refractive index n

air

n = 1 E

E¢

z

quantum well

Figure 24: Refraction of light entering a semiconductor containing a quantumwell, as discussed in Exercise 6.21.

where θ′ is the angle inside the crystal. For intersubband transitions, weare interested in the z component of the electric field of the light at thequantum well, namely:

Ez = E ′ sin θ′ = E ′ sin θ/n .

If I0 is the incident intensity, and there are no intensity losses at thesurface, then the intensity in the z component at the quantum well isgiven by

Iz = I0 (sin θ/n)2 ,

since the intensity is proportional to E2. Hence the fraction of the powerof the beam in the z polarization at the quantum well is (sin θ/n)2. Thisfraction has a maximum value of 1/n2 for θ = 90. Therefore even if wecompletely absorb all the z polarized light by intersubband transitions,and we use glancing incidence, we can only remove a fraction of 1/n2 ofthe power in the incident beam. This fractional absorption is equal to 9%if n = 3.3.

nx ny nx (n2x + n2

y + n2z) g

1 1 1 3 12 1 1 6 32 2 1 9 33 1 1 11 32 2 2 12 13 2 1 14 63 2 2 17 3

Table 3: Quantum numbers of the energy levels for a cubic quantum dots in or-der of increasing energy, as discussed in Exercise 6.22. g denotes the degeneracyexcluding spin.

(6.22) For a cubic dot, the energies of the quantized levels are given by eqn 6.54with dx = dy = dz = d, implying:

E(nx, ny, nz) =π2~2

2m∗d2(n2

x + n2y + n2

z) ,

where nx, ny and nz are integers with a minimum value of 1. As demon-strated by Table 3, the quantized levels occur at energies of 3, 6, 9, 11,12, 14, 17,. . . in units of h2/8m∗d2.

66



The degeneracy factor g is worked out by finding the number of per-mutations of (nx, ny, nz) that can give the same energy. The groundstate is unique, but for the first excited we can have (2, 1, 1), (1, 2, 1)and (1, 1, 2), and similarly for the levels at 9, 11, and 17 h2/8m∗d2. Thelevel at 14 h2/8m∗d2 has a degeneracy of 6 because there are 3! = 6 waysof arranging the numbers 3, 2 and 1.

(6.23) The radial equation for a spherical dot is given by eqn 6.57. If the con-fining potential is V0, then we can put V (r) = −V0 for r ≤ R0. Therefore,when the particle is inside the dot (i.e. r ≤ R0), the radial wave functionfor states with l = 0 must satisfy:

[− ~2

2m∗1r2

ddr

(r2 d

dr

)− V0

]R(r) = E R(r) .

Now1r2

ddr

(r2 d

dr

)(sin kr

r

)= −k2

(sin kr

r

).

Hence if R(r) = sin kr/r, then substitution into the radial equation gives:

~2k2

2m∗ R(r)− V0R(r) = E R(r) ,

which implies that

E = −V0 +~2k2

2m∗ .

For an infinite well it must be the case the R(r) = 0 for r ≥ R0. Thereforeat r = R0 we require that sin kR0/R0 = 0, which implies that kR0 = nπ,where n is an integer. Hence:

E = −V0 +~2n2π2

2m∗R20

.

The confinement energy relative to the bottom of the potential well at−V0 is thus:

En =~2n2π2

2m∗R20

.

(6.24) For the dots to have the same volume, we require that:

d3 =43πR3

0 ,

where d is the cube size and R0 is the radius of the spherical dot. Thisimplies that (R0/d) = (3/4π)1/3. From Exercise 6.22 we find that theconfinement energy of the first level in a cubic dot with infinite barriersis Ecubic

1 = 3~2π2/2m∗d2. For a spherical dot eqn 6.58 gives Espherical1 =

~2π2/2m∗R20 for C10 = 1, as appropriate for infinite barriers. Hence:

Ecubic1

Espherical1

=3~2π2

2m∗d2÷ ~2π2

2m∗R20

= 3(

R0

d

)2

= 3(

34π

)2/3

= 1.15 .

67



The reason why the cube has the higher energy can be understood byreference to Fig. 25. In the infinite well approximation, the wave functionsare proportional to sine waves with nodes at the edges for both the cubeand the sphere. It is apparent from Fig. 25 that if the cube and spherehave the same volume, then the wavelength of the sine wave in the cubewill be smaller. This corresponds to a larger k vector, and hence largerconfinement energy, since E = ~2k2/2m∗ inside the dot in both cases.

Figure 25: Comparison of cube and sphere of the same volume, as discussed inExercise 6.24. The wave functions have nodes at the edges.

(6.25) (a) In Cartesian co-ordinates, the Schrodinger equation for a 2–D har-monic oscillator takes the form:

[− ~2

2me

(∂2

∂x2+

∂2

∂y2

)+

12meω

20(x2 + y2)

]Ψ(x, y) = EΨ(x, y) .

On writing Ψ(x, y) = X(x)Y (y) and substituting, we find:

− ~2

2me

d2X

dx2Y +

12meω

20x2XY − ~2

2meX

d2Y

dy2+

12meω

20y2XY = EXY .

On dividing by XY and re-arranging this gives:

− ~2

2me

1X

d2X

dx2+

12meω

20x2 = E +

~2

2me

1Y

d2Y

dy2− 1

2meω

20y2 .

The left hand side is a function of x, and the right hand side a functionof y, and so they both must equal a constant. On calling this constant Cand rearranging we have:

− ~2

2me

d2X

dx2+

12meω

20x2X = CX

− ~2

2me

d2Y

dy2+

12meω

20y2Y = (E − C)Y .

Both of these are standard harmonic oscillator equations. The one for ximplies C = (nx + 1/2)~ω0, where nx is an integer, while the one for yimplies that (E−C) = (ny +1/2)~ω0, where ny is another integer. Hence:

E = C + (ny + 1/2)~ω0 = (nx + ny + 1)~ω0 ≡ (n + 1)~ω0 ,

where n = nx + ny.

The degeneracies are found by working out the different possible permu-tations of nx and ny to get the same energy:

68



• Ground state: E = ~ω0, n = 0, (nx, ny) = (0, 0), degeneracy = 1.

• 1st excited state: E = 2~ω0, n = 1, (nx, ny) = (1, 0) or (0,1), degen-eracy = 2.

• 2nd excited state: E = 3~ω0, n = 2, (nx, ny) = (2, 0), (1,1) or (0,2),degeneracy = 3.

• 3rd excited state: E = 4~ω0, n = 3, (nx, ny) = (3, 0), (2,1), (1,2) or(0,3), degeneracy = 4.

• 4th excited state: E = 5~ω0, n = 4, (nx, ny) = (4, 0), (3,1), (2,2),(1,3) or (0,4), degeneracy = 5.

It is thus apparent that the degeneracy is equal to n. This results can alsobe worked out by considering the m states: there are n states from −n to+n in steps of 2.

(b) In polar co-ordinates, the Schrodinger equation for a 2–D harmonicoscillator takes the form:

[− ~2

2me

1r

∂

∂r

(r

∂

∂r

)+

1r2

∂2

∂φ2+

12meω

20r2

]ψ(r, φ) = Eψ(r, φ) .

On writing ψ(r, φ) = R(r)Φ(φ) and substituting, we find:

− ~2

2me

1r

ddr

(rdR

dr

)Φ +

R

r2

d2Φdφ2

+12meω

20r2RΦ = ERΦ .

On multiplying through by r2/RΦ and re-arranging, this gives:

− ~2

2me

r

R

ddr

(rdR

dr

)+

12meω

20r4 − Er2 = − 1

Φd2Φdφ2

.

The left hand side is a function of r, and the right hand side a functionof φ, and so they both must equal a constant. On calling this constant Cand rearranging we have:

d2Φdφ2

= −CΦ ,

which has solutions Φ(φ) = eimφ, where m2 = C. Now the wave functionmust be single valued for each value of φ and so we require:

Φ(φ) = Φ(φ + 2π) ,

which implies that eim2π = 1 and hence that m must be an integer. There-fore, the wave functions are of the form ψ(r, φ) = R(r)eimφ, where m isan integer.

(c) The fact that m is equal to −n to +n in steps of 2 can be provedmathematically, but requires a lot of work. (See, for example, Pauling &Wilson, Introduction to quantum mechanics, 1935.) We thus restrict ourefforts here to justifying this result for the first three levels. From part (a)we know that the wave function for the state with energy (n+1)~ω has theform ψnx(x)ψny (y), where n = nx+ny. We can draw up a correspondence

69



Energy Polar co-ordinates Cartesian co-ordinates~ω n = 0, m = 0 (nx, ny) = (0, 0)2~ω n = 1, m = ±1 (nx, ny) = (1, 0) or (0, 1)3~ω n = 2, m = −2, 0, +2 (nx, ny) = (2, 0), (1, 1) or (0, 2)

Table 4: Equivalence of the states of a 2–D harmonic oscillator in polar andCartesian co-ordinates, as considered in Exercise 6.25(c).

between the Cartesian and polar co-ordinates as shown in Table 4. Weconsider each energy state in turn.

Ground state with energy ~ωThis state is easy. There is only one possibility with (nx, ny) = (0, 0). Thewave function in polar co-ordinates is therefore of the form:

ψ0,0(r, φ) = ψ0(x)ψ0(y) ∝ exp(−αx2/2) · exp(−αy2/2) = exp(−αr2/2) ,

as required.

First excited state with energy 2~ωWe now have two possibilities, and the wave functions must be linearcombinations of each other, that is:

ψ1,m(r, φ) = C1ψ1(x)ψ0(y) + C2ψ0(x)ψ1(y) .

It is apparent that if we take

ψ1,m(r, φ) ∝ ψ1(x)ψ0(y)± iψ0(x)ψ1(y) ,

then we have:

ψ1,m(r, φ) ∝ (x± iy) exp(−αx2/2) · exp(−αy2/2) .

Now x = r cosφ and y = r sin φ. Therefore:

ψ1,m(r, φ) ∝ r(cosφ± i sin φ) exp(−αr2/2) = r e±iφ exp(−αr2/2) .

This shows that m = ±1 for the state with energy 2~ω. Note that theradial form is as given in the Exercise.

Second excited state with energy 3~ωWe now have three possibilities, and the wave functions must be linearcombinations of each other, that is:

ψ2,m(r, φ) = C1ψ2(x)ψ0(y) + C2ψ1(x)ψ1(y) + C3ψ0(x)ψ2(y) .

Consider the combination ψ2(x)ψ0(y) + ψ0(x)ψ2(y). This can be written:

ψ2,m(r, φ) ∝ (1− 2αx2) exp(−αx2/2) · exp(−αy2/2)+ exp(−αx2/2) · (1− 2αy2) exp(−αy2/2) ,

∝ 2(1− αx2 − αy2) exp(−α(x2 + y2)/2) ,

∝ 2(1− αr2) exp(−αr2/2) .

70



This clearly has no dependence on φ and so it must correspond to thestate with m = 0. Note that the radial form is correct for this state. Wecan thus write:

ψ2,0(r, φ) ∝ ψ2(x)ψ0(y) + ψ0(x)ψ2(y) .

Now consider the combination: −ψ2(x)ψ0(y)±√2iψ1(x)ψ1(y)+ψ0(x)ψ2(y).By carefully considering the normalization constants in front of the har-monic oscillator functions, we find that this gives:

ψ2,m(r, φ) ∝ (−1 + 2αx2 ± 4iαxy + (1− 2αy2)) exp(−α(x2 + y2)/2) ,

∝ 2α(x2 − y2 ± 2ixy) exp(−αr2/2) ,

∝ (r2 cos2 φ− r2 sin2 φ± 2ir2 cosφ sin φ) exp(−αr2/2) ,

∝ r2((cos2 φ− sin2 φ)± 2i cos φ sinφ) exp(−αr2/2) ,

∝ r2(cos 2φ± i sin 2φ) exp(−αr2/2) ,

∝ r2e±2iφ exp(−αr2/2) .

By comparing with the radial functions given in the Exercise, we thenconclude that

ψ2,2(r, φ) ∝ −ψ2(x)ψ0(y) +√

2iψ1(x)ψ1(y) + ψ0(x)ψ2(y) ,

ψ2,−2(r, φ) ∝ −ψ2(x)ψ0(y)−√

2iψ1(x)ψ1(y) + ψ0(x)ψ2(y) .

This shows that the allowed values of m for n = 2 are indeed −2, 0 and+2. It can be verified that these three wave functions are orthogonal, i.e.that they satisfy

∫ψ∗ψ d2r = 0.

71



Chapter 7

Free electrons

(7.1) The method for determining the electron Fermi energy was considered inExercise 5.10, and the formula for EF is given in eqn 5.13:

EF =~2

2m(3π2N)2/3,

which implies:

N =1

3π2

(2mEF

~2

)3/2

.

On substituting this form of N into the formula for the plasma frequencyin eqn 7.6, we then obtain

E3F =

9ε20~2

8m

(π~ωp

e

)4

.

(7.2) We expect 100% reflectivity below the plasma frequency and transmissionat higher frequencies. (See Fig. 7.1.) Hence we can set ωp/2π = 3 MHzin this example. On substituting into eqn 7.6 and solving for N , we findN ∼ 1011 m−3. (n.b. The electron density calculated here is just a typicalone. The value of N varies somewhat due to atmospheric conditions.)

(7.3) The formula for the skin depth is given in eqn 7.20. On inserting thevalue of σ for salt water we find δ ∼ 0.5m for ω/2π = 200 kHz. Thisshows that electromagnetic waves of this frequency penetrate less than1m from the surface of the sea. To obtain a skin depth of 30 m as requiredfor communication with a submarine submerged at this depth, we requireω/2π = 70Hz. Even lower frequencies are required for deeper depths. Thedata transmission rate is very low at these small carrier frequencies.

(7.4) From Table 7.1 we read N = 0.91× 1028 m−3 for cesium, while the trans-mission edge at 440 nm implies λp = 440 nm, and hence:

ωp = 2πc/λp = 4.3× 1015 rad/s .

On substituting into eqn 7.6 and solving for m, we find m∗e = 1.4 ×

10−30 kg ≡ 1.6 m0.

(7.5) This Exercise closely follows Example 7.3. We read ωp = 1.36×1016 rad/sfrom Table 7.1, and work out τ = 4.0 × 10−14 s from eqn 7.14 using thevalue of N from Table 7.1 and the value of σ0 given in the exercise. At500 nm we have ω = 3.77 × 1015 rad/s, and the relative permittivity atthis wavelength is then given by eqns 7.16–17 as:

εr = −12.0 + 0.086i .

We finally use eqns 1.25–26 to calculate n = −0.012+3.5i, and substituten = −0.012 and κ = 3.5 into eqn 1.29 to obtain R = 99.6 %.

72



(7.6) We first use eqn 7.14 to find τ = 1.2× 10−13 s, taking N = 5.9× 1028 m−3

from Table 7.1, and then proceed as in Example 7.3 to find the extinctioncoefficient κ. With ωp = 1.37 × 1016 rad/s (cf. Table 7.1), and ω =2πc/λ = 1.88 × 1015 rad/s, we find from eqns 7.16–17 that εr = −52.1 +0.235i, and hence κ = 7.22 (see eqn 1.26). This value of κ implies, througheqn 1.19, an absorption coefficient α = 9.1 × 107 m−1. The transmissionis given by Beer’s law (see eqn 1.4) as exp(−αz) = 0.16 for z = 20nm.

Note that the skin depth approximation in eqns 7.18–20 is not valid herebecause we have ωτ = 230 for λ = 1 µm (ω = 1.9 × 1015 rad/s), andhence we do not satisfy ωτ ¿ 1. The absorption coefficient estimatedfrom eqn 7.19 is wrong by approximately one order of magnitude.

(7.7) Based on the plasma frequency of gold given in Table 7.1, we expect highreflectivity above ∼ 140 nm. The low reflectivity up to 600 nm is caused byinterband transitions as illustrated schematically in Fig 7.4. The energygap between the d bands and the Fermi energy can be read from thedata as the energy equivalent of ∼ 520 nm, i.e. ∼ 2.4 eV. It is apparentfrom the reflectivity spectrum that gold reflects red, orange and yellowlight stronger than green and blue. This accounts for its characteristicyellowish colour.

(7.8) It is apparent from eqn 1.29 that R = 0 when n = 1, and hence εr = 1.The relative permittivity of a doped semiconductor is given by eqn 7.22,and ‘light damping’ implies that we take γ = 0. Hence we shall have zeroreflectivity when: (see eqn 7.24 for ωp):

1 = εopt − Ne2

m∗ε0ω2= εopt

(1− ω2

p

ω2

).

Equation 7.25 is then derived by solving this formula for ω2.

N (1024 m−3) λ(R = 0) (µm) ω(R = 0) (1013 rad/s) m∗e/m0

0.35 33 5.7 0.0200.62 27 7.0 0.0281.2 22 8.6 0.0362.8 16 12 0.0444.0 14 13 0.048

Table 5: Effective masses of n-type InSb calculated from the data in Fig 7.7 asrequired for Exercise 7.9.

(7.9) The easiest way to determine the effective mass from the spectra is to readthe wavelength at which R = 0 from the data. We then use eqns 7.24–25to write:

m∗ =Ne2

ε0(εopt − 1)ω2,

where ω is the angular frequency corresponding to R = 0. The valuesof the effective masses found in this way with εopt = 15.6 are given inTable 5.

73



It is apparent from Table 5 that m∗e increases strongly with N , rising from

0.020 m0 at 3.5 × 1023 m−3 to 0.048 m0 at 4 × 1024 m−3. This increasein the effective mass with carrier density is caused by non-parabolicity inthe conduction band of InSb. The effective mass approximation assumesthat the bands are parabolic in shape as in Fig 3.5. However, a glance atthe band structure of a real III-V semiconductor (eg GaAs, see Fig 3.4)indicates that this approximation only holds near k = 0. To get a goodfit to the energy bands for larger values of k (but still below the maxima),we have to re-write eqn 3.17 as:

Ec(k) = Eg +~2k2

2m∗(k),

where m∗(k) = m∗e for small k, but then increases at larger values of k. As

we dope the semiconductor with more electrons, the Fermi energy increases(see eqn 5.13) and we are probing states of the conduction band with largervalues of E and k. It is therefore to be expected that the effective massshould increase with the doping density. The band non-parabolicity isparticularly strong in some narrow gap semiconductors such as InSb.

(7.10) The free carrier absorption coefficient in the limit ωτ À 1 is given byeqn 7.28. At 10 µm we have ω = 1.9 × 1014 rad/s, and on inserting therelevant values into eqn 7.28 we find τ = 1.0 ps. This implies ωτ ∼ 200,so that the approximations used to derive eqn 7.28 are justified.

(7.11) The laser beam generates free carriers in the conduction and valencebands which can then induce free carrier absorption. The carrier densitygenerated by a continuous laser beam is given by (see Exercise 5.6):

N =Iατ

~ω,

where I is the intensity, α is the absorption coefficient, τ is the carrierlifetime, and ~ω is the photon energy. On inserting the appropriate valuesas given in the Exercise, we find N = 1.9× 1022 m−3. This is the densityof electrons in the conduction band and holes in the valence band, bothof which cause free carrier absorption. At λ = 10.6 µm we have ω =1.8 × 1014 rad/s, and hence ωτ = 36 for the electrons and ωτ = 9 forthe holes. We thus have ωτ À 1 in both cases, and we can thereforecalculate the free carrier absorption coefficients by using eqn 7.28. Thisgives α = 130 m−1 for the electrons and α = 70 m−1 for the holes. Hencethe total free carrier absorption coefficient is 200m−1.

Here are a few points to note about this exercise:

• Students should be careful not to confuse the two different usages ofτ . In the calculation of the carrier density, τ represents the carrierlifetime, whereas in eqn 7.28 it represents the momentum scatteringtime.

• Intervalence band absorption has been neglected here. However, withonly ∼ 1022 m−3 holes, it is probable that intervalence band absorp-tion will be insignificant. (See the next exercise.)

74



• This exercise is actually an example of a nonlinear optical effect: thelight beam at 633 nm induces changes in the optical properties. Themechanism is that the beam creates carriers which then alter theoptical properties. These types of nonlinear effects are called “freecarrier nonlinearities” for obvious reasons. See Chapter 11 for moreinformation on nonlinear optics.

kEF

D

E

min

max

max

min max,min

kF

lhkF

hh

kEF

D

E

min

max

max

min max,min

kF

lhkF

lhkF

hhkF

hh

Figure 26: Intervalence band transitions as required for the solution of Exercise7.12. The Fermi energy is not drawn to scale: with the parameters given in theExercise, the Fermi energy is just above the split-off band. The labels (1), (2)and (3) refer to the lh→hh, SO→lh, and SO→hh transitions as in Fig. 7.9.

(7.12) (a) The holes are distributed between the heavy- and light-hole bands,and so we can find the Fermi energy from (see eqn 3.16):

N =∫ EF

0

(ghh + glh) dE

=23/2

2π2~3(m3/2

hh + m3/2lh )

∫ EF

0

E1/2 dE .

With m∗hh = 0.5 m0 and m∗

lh = 0.08 m0 we then find EF = 0.032 eV forN = 1× 1025 m−3. Note that this is smaller than the spin–orbit energy ∆and so our neglect of the occupancy of the split-off band is justified. Thewave vector at the Fermi energy is worked out from:

EF =~2k2

F

2m∗ .

This give khhF = 6.5× 108 m−1 and klh

F = 2.6× 108 m−1 for the heavy andlight holes respectively. (See Fig. 26.)

75



(b) The energies of the three types of intervalence band transitions indi-cated in Fig. 7.9 can be calculated by using eqns 3.18–20:

(1) lh → hh : ~ω = |Elh(k)− Ehh(k)| = ~2

2

(1

m∗lh

− 1m∗

hh

)k2 ,

(2) SO → lh : ~ω = |Eso(k)− Elh(k)| = ∆ +~2

2

(1

m∗so

− 1m∗

lh

)k2 ,

(3) SO → hh : ~ω = |Eso(k)− Ehh(k)| = ∆ +~2

2

(1

m∗so

− 1m∗

hh

)k2 ,

where k is the wave vector at which the transition occurs. Transitions canonly take place from occupied states to empty ones. The upper and lowerlimits of the transition energies are therefore set by the Fermi wave vectorsas indicated in Fig. 26. The energy limits calculated using the Fermi wavevectors from part (a) are therefore as follows:

1. lh→hh transitions: The lower and upper limits are set by klhF and

khhF respectively. On inserting the into (1) above, we find a transition

range of 0.03 → 0.17 eV.

2. SO→lh transitions: Since the light hole effective mass is smaller thanthe split-off mass, the lower and upper limits correspond to the tran-sitions at klh

F and k = 0 respectively. On inserting into (2) above wefind a range 0.32 → 0.34 eV.

3. SO→hh transitions: The lower limit is set by the transition at k = 0,and the upper limit by the one at khh

F , giving a range of 0.34 →0.42 eV.

(7.13) (a) The energies of the 2p0, 3p0 and 4p0 transitions can be read fromFig. 7.11 as 34.0, 40.1, and 42.4 meV respectively. These energies fit wellto the formula:

hν = R∗(

1− 1n2

),

with R∗ = 45.2meV. This is consistent with eqn 7.30 if we set m∗e =

0.85 m0 for the donor levels.

(b) The energies of the 2p±, 3p±, 4p± and 5p± transitions can be readfrom Fig. 7.11 as 39.2, 42.4, 43.3 and 44.1meV respectively. The np±

transitions correspond to transitions from np hydrogenic states to the 1sstate, with energies given by:

hν = |E1s − Enp± | .

The 1s energy is just equal to the value of R∗ found in the part (a), andso we can identify R∗0 = R∗ = 45.2 meV. The value of R∗± is then foundby fitting the transition energies. A good fit is found with R∗± ≈ 25 meV.

(7.14) These are donor level transitions as shown in Fig. 7.10(a). Their transi-tion energies are given by eqn 7.30. It is apparent that they depend onlyon the effective mass and relative permittivity of the host crystal, andnot on any of the properties of the dopant atom. The reason why this isso is that the donor levels are formed by the interaction between a band

76



electron and an ionized impurity. The donor atoms are chosen so thatthey have a single excess electron, and so they are all singly ionized. Thedonor level energies are therefore hydrogenic with the relevant mass andpermittivity of the band electron.

On comparing with eqn 7.30, we see that we must have R∗ = (m∗e/m0ε

2r )×

RH. On setting this equal to 2.1 meV we find m∗e = 0.036 m0 for εr = 15.2.

conduction band

valence band

acceptor

levels

Eg

conduction band

valence band

acceptor

levels

conduction band

valence band

acceptor

levels

Eg

Figure 27: Acceptor to conduction band transitions in a p-type semiconductoras discussed in Exercise 7.15. Note that this figure is not drawn to scale: theacceptor energies are much smaller than the band gap.

(7.15) Doping with acceptors creates a p-type semiconductor with a series ofacceptor levels just above the valence band as indicated in Fig. 27. Theenergy of the acceptor levels relative to the top of the valence band willbe given by eqn 7.29 with m∗

e replaced by m∗h. Electrons from the valence

band can be thermally excited to fill the acceptor levels, giving rise tothe possibility of acceptor to conduction band transitions as indicated inFig. 27. These transitions occur at a photon energy of Eg −EA. The ab-sorption edge of a p-type semiconductor will therefore decrease on dopingfrom Eg to Eg − EA, where EA is the energy of n = 1 acceptor level. Awavelength shift from 5.26 µm to 5.44 µm corresponds to an energy shiftof 8meV, and hence we deduce EA ∼ 8meV.

This exercise is based on experimental data for zinc acceptors in InSb.See Figure 2 in “Impurity and Exciton Effects on the Infrared AbsorptionEdges of III-V Compounds”, E.J. Johnson H.Y. Fan, Phys. Rev. 139,A1991–A2001 (1965).

(7.16) The peak is caused by Raman scattering from plasmon modes, as shownin Fig. 7.13. The energy of the scattered photons is given by eqn 7.49.In this exercise, we are clearly considering the case of plasmon emissionwhere the ‘–’ sign is appropriate. We thus deduce:

~ωp = ~ωin − ~ωout = (2.410− 2.321) eV = 0.089 eV .

On substituting into eqn 7.24 with εopt = n2 and the given value of m∗, wefind N = 4.2× 1024 m−3. Note that we would also expect Raman signalsfrom optical phonons, but these would occur at smaller energy shifts with∆ν ∼ 300 cm−1 ≡ 37meV. (See Fig. 10.11.)

77



(7.17) We set

ωp =(

Ne2

εoptε0m∗e

)1/2

= ΩLO ,

with εopt = n2 and solve for N . A wave number ν of 297 cm−1 impliesω = 2πcν = 5.6 × 1013 rad/s, and hence we find N = 7.2 × 1023 m−3. Amixed plasmon–phonon mode is formed at this doping density because thetwo longitudinal excitations couple strongly together if their frequenciesare the same.

(7.18) We first calculate ωp using eqn 7.6. With N = 1 × 1029 m−3, we findωp = 1.8× 1016 rad/s. Next, we work out vF from:

EF =12mv2

F .

With EF = ~2(3π2N)2/3/2m (cf. eqn 5.13), we then find:

vF =~m

(3π2N)1/3 ,

which gives vF = 1.7× 106 m/s.

A wave vector of 0.1π/a has a magnitude of 7.9× 108 m−1 if a = 4 A. Wethen have:

ω ≈ ωp

(1 +

3 (1.7× 106)2 (7.9× 108)2

10 (1.8× 1016)2

)= 1.002 ωp .

We thus see that the term in k2 only makes a difference of 0.2%. It cantherefore safely be ignored in most circumstances.

(7.19) In an electron energy loss experiment on a metal, we expect to observepeaks from both the surface and bulk plasmons. On inserting εd = 1 intoeqn 7.61 as appropriate for air, we expect the two plasmons to differ infrequency by a factor of

√2 = 1.41. The data fits this model quite well,

with the experimental ratio being 1.49. The 10.3 eV and 15.3 eV seriesare thus the surface and bulk plasmons respectively. The bulk plasmaenergy of 15.3 eV implies ωp/2π = 3.7× 1015 Hz. It is then apparent fromTable 7.1 that the metal is aluminium.

This Exercise is based on the electron energy loss spectroscopy data shownin Kittel (7th edition), Chapter 10, Figure 8.

(7.20) (a) At this point, we ignore the imaginary part of εm, and put εd = 1and εm = −18 into eqn 7.62. With ω/c = 2π/λ, this gives:

kdz =

2π

600× 10−9

( −12

−18 + 1

)1/2

= 2.5× 106 m−1 ,

kmz =

2π

600× 10−9

(−(−18)2

−18 + 1

)1/2

= 4.6× 107 m−1 .

Hence ldz = (kdz )−1 = 394 nm and lmz = (km

z )−1 = 22nm.

78



(b) The propagation length in the x direction is determined by the imag-inary part of kx and so in this part of the exercise we do have to includethe imaginary part of εm. On substituting εd = 1 and εm = (−18 + i) intoeqn 7.58 with ω/c = 2π/λ, we find:

kx =2π

600× 10−9

(−18 + i−17 + i

)1/2

= 1.05× 107(1.03 + 1.7× 10−3i)m−1 .

On writing the real and imaginary parts of kx as k′ and k′′ respectively,we then obtain k′′ = 1.8× 104 m−1. The wave therefore propagates with:

E(x) = E0 exp[ikxx] = E0 exp[(i(k′ + ik′′)x] = E0 exp(ik′x) exp(−k′′x) .

The intensity is proportional to EE∗, and therefore decays with x accordingto I(x) = I0 exp (−2k′′x). The propagation distance is therefore given by(2k′′)−1 = 28 µm.

(7.21) Equation 7.63 shows that the resonance in the polarizability of a colloidalmetal nanoparticle occurs when εm = −2εd. On setting εd = n2 for thedielectric, and using the standard frequency dependence of εm for a metal(cf. eqn 7.7), we see that this resonance occurs when:

1− ω2p

ω2= −2n2 .

On rearranging, this gives:

(1 + 2n2)ω2 = ω2p ,

and hence:ω =

ωp√1 + 2n2

.

When the dielectric is air, we have n = 1, and hence ω = ωp/√

3.

(7.22) (a) As shown in the discussion of eqn 7.63, the resonance occurs whenεm = −2εd. When εm is complex, as is the case considered here, thisresonance condition is satisfied when the real part of εm, namely ε1, isequal to −2εd. This is because εd is real, and so the denominator ofeqn 7.63 is smallest when ε1 +2εd = 0. For water we have εd = n2 = 1.77,and so we require ε1 = −3.5. A linear fit to the data given in Table 7.3indicates that this condition is satisfied at about 517 nm.

(b) The result derived in Exercise 7.21 predicts a resonance wavelengthgiven by:

λ =2πc

ω=√

1 + 2n2

ωp/2π=

√1 + 2n2λp ,

where λp is the wavelength corresponding to ωp. On using the value of λp

for gold given in Table 7.1, namely 138 nm, we predict λ = 294 nm, whichdiffers significantly from the correct value obtained in part (a).

The reason for the discrepancy is that the undamped Drude formula givenin eqn 7.7 is not a good approximation for gold in the optical frequencyrange. For example, the Drude model would predict high reflectivity up to

79



the plasma wavelength of 138 nm, but this is not observed in experimentalreflectivity data. (See Fig. 7.20.) At optical frequencies the d-band tran-sitions are very strong, and this modifies the relative permittivity verysignificantly.

In fact the situation is more complicated. Equation 7.63 only appliesto spherical nanoparticles. Furthermore, it predicts that the resonancefrequency should be independent of the nanoparticle size, which is notactually the case: see Maier (2007) for details. This becomes apparentwhen attempting to fit the data in Fig. 7.17. With εd = 2.5, we wouldexpect the resonance to occur when ε1 = −5. Table 7.3 gives ε1 = −5 at538 nm for gold, but the peak actually occurs at 580 nm. This shows that amore detailed model is required to get a good fit to the data. Equation 7.63does, however, explain the basic phenomenon, and correctly explains thered shift of the resonance with increasing εd.

80



Chapter 8

Molecular materials

(8.1) This is a standard example covered in all elementary quantum mechanicstexts. On substituting Ψ1 into the Schrodinger equation, we obtain:

− ~2

2m

(x2

a4− 1

a2

)Ψ1 +

12mΩ2x2Ψ1 = E1Ψ1 ,

which can be re-arranged to give:(

12mΩ2 − ~2

2ma4

)x2Ψ1 +

~2

2ma2Ψ1 = E1Ψ1 .

Ψ1 is therefore a solution if we set a = (~/mΩ)1/2 to eliminate the firstterm. We can then deduce:

E1 =~2

2ma2=

12~Ω .

By a similar method we can show that Ψ2 and Ψ3 are solutions withE2 = (3/2)~Ω and E3 = (5/2)~Ω respectively, and with a again equal to(~/mΩ)1/2.

(8.2) The energy levels for a particle of mass m in an infinite potential well ofwidth d are given by eqn 6.13 as:

En =h2n2

8md2.

Selection rules permit transitions with ∆n equal to an odd number. (Seethe discussion of intersubband transitions in Section 6.7.) The lowestenergy transition occurs for n = 1 → 2, with hν = 3h2/8md2. On settingm = m0 for the π-electron, and hν = 2.5 eV, we find d = 6.7 × 10−10 m.This corresponds to about six carbon–carbon bonds.

Note that this model should not be taken too seriously, since it predictsthat the transition energy scales as d−2. The data in Fig. 8.11 show thatthe actual scaling is much weaker, with E ∼ d−0.4.

(8.3) The ratio of the number of molecules with one vibrational quantum excitedcompared to those with none is given by Boltzmann’s law as exp(−hν/kBT ),where ν is the frequency of the vibration. The calculated ratios for thethree modes at 300 K are:

• ν = 2× 1013 Hz: 4× 10−2,

• ν = 4× 1013 Hz: 1.6× 10−3,

• ν = 7× 1013 Hz: 1.4× 10−5.

81



The point of this exercise is to make the student appreciate that it is agood approximation to assume that all the molecules are in the vibrationalground state at room temperature.

(8.4) The energy of isolated hydrogen atoms is given by the standard Rydbergformula with En = −RH/n2. The energy required to promote one of theatoms from a 1s to a 2p state is therefore (3/4)RH = 10.2 eV. This issmaller than the equivalent transition in the H2 molecule. The reason isthat the energy of a diatomic molecule is given by:

Emolecule = Eatom1 + Eatom2 − Ebinding ,

where Eatomi is the energy of the isolated atoms, and Ebinding is thebinding energy of the molecule. Now the ground state of the moleculeis more strongly bound than the excited state, and the transition includesthe difference of the two binding energies. Hence in the molecule thetransition energy is:

hν = (3/4)RH + ∆Ebinding .

We can then account for the molecular transition energy of 11.3 eV if weassume that the difference of the binding energies of the 1s1s and 1s2pmolecular configurations is 1.1 eV.

0.5 1.0 1.5 2.0

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

U(r

)

r

0.5 1.0 1.5 2.0

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

U(r

)

r

Figure 28: Lennard–Jones potential for A = B = 1, as considered in Exercise8.5.

(8.5) (a) The attractive r−6 term is caused by the van der Waals interaction,which is the main attractive force between neutral molecules. This is adipole-dipole interaction. A fluctuating dipole p1 on molecule 1 generatesan electric field of strength E1 ∝ p1/r3 at molecule 2. This induces adipole of magnitude p2 ∝ E1 ∝ p1/r3 on molecule 2, which then generatesa field E2 ∝ p2/r3 ∝ p1/r6 at molecule 1. The interaction energy is then−p1E2 ∝ −(p1)2/r6. Although the time average of p1 will be zero, the timeaverage of (p1)2 is not. Hence the dipole-dipole mechanism generates anattractive potential ∝ r−6.

(b) The Lennard-Jones potential is plotted for the case A = B = 1 inFig. 28. The potential is attractive at large r, but the repulsive termdominates for small r. This gives a minimum at a well-defined value of r,

82



which we call r0. The position of the minimum energy can be calculatedby setting:

dU

dr= −12A

r13+

6B

r7= 0 ,

at r = r0, which implies r0 = (2A/B)1/6.

(c) The Taylor expansion for U(r) near r0 is:

U(r) = U(r0) +(

dU

dr

)(r − r0) +

12

(d2U

dr2

)(r − r0)2 + · · · ,

where the derivatives are evaluated at r = r0. If r0 is the position of theminimum, then the first derivative will be zero, and U(r) reduces to:

U(r) = U(r0) +12

(d2U

dr2

)(r − r0)2 + · · · .

Nowd2U

dr2= +

156A

r14− 42B

r8,

which, with r = r0 = (2A/B)1/6, gives:

d2U

dr2=

18B2

A

(B

2A

)1/3

.

Thus:

U(r) = U(r0) +18B2

A

(B

2A

)1/3

(r − r0)2 + · · · ,

≡ U(r0) +12µΩ2(r − r0)2 .

Hence Ω2 = (18B2/Aµ)(B/2A)1/3.

(8.6) (a) This result follows directly from the Franck–Condon principle. Weare at T = 0 and so the initial state of the system will be the lowestvibrational level of the lower state, i.e. the n = 0 vibrational level of theelectronic ground state. The initial wave function is therefore ϕ0(Q−Q0).Transitions are possible to any of the vibrational levels of the excited state,as shown in Fig. 8.7. The energy of the transition to the nth level is givenby eqn 8.3 and can be written:

~ωn = ~ω0 + n~Ω .

We are at T = 0 and thus expect negligible broadening of the transitions.The probability for the transition is given by the Franck–Condon factor(eqn 8.12), which in this case takes the form:

P (n, 0) =∣∣∣∣∫ ∞

0

ϕ∗n(Q−Q′0) ϕ0(Q−Q0) dQ

∣∣∣∣2

≡ |〈n,Q′0|0, Q0〉|2 .

83



The spectrum will therefore be given by a series of sharp transitions toeach vibrational level, with their relative weight given by P (n, 0). This isconveniently expressed in Dirac delta function notation as:

I(~ω) ∝∞∑

n=0

P (n, 0)δ(~ω − ~ωn)

∝∞∑

n=0

|〈n,Q′0|0, Q0〉|2δ(~ω − ~ω0 − n~Ω) .

as required.

Inte

nm

sity

(arb

un

its)

0 2 4 6 8 10 12 14(hw - hw0) in units of hW

S = 0

S = 1

S = 5

Inte

nm

sity

(arb

un

its)

0 2 4 6 8 10 12 14(hw - hw0) in units of hW

S = 0

S = 1

S = 5

Figure 29: Vibronic spectra, as considered in Exercise 8.6.

(b) We substitute for the Franck–Condon factor to obtain:

I(~ω) ∝∞∑

n=0

exp(−S)Sn

n!δ(~ω − ~ω0 − n~Ω) .

This defines a Poisson distribution. The spectra are shown in Fig.29.(i) If S = 0, then the only non-zero term is the one with n = 0. Thespectrum therefore just consists of the zero-phonon line at ~ω0.(ii) When S = 1, the strongest two transitions are the ones with n = 0and n = 1. The intensity drops sharply with increasing n.(iii) We now have a bell-shaped distribution peaked at n = 5. The halfwidth of the distribution is given approximately by 2

√5 ∼ 5. In the limit

of large S, the spectrum is Gaussian.

More details about how the Huang–Rhys factor determines the vibronicline shape may be found, for example, in Henderson & Imbusch, Opticalspectroscopy of inorganic solids (Oxford, 1989) or Hayes & Stoneham,Defects and defect processes in nonmetallic solids (Wiley, 1985).

84



0 2 4 6

4.6

4.8

5.0

5.2

5.4

Tra

nsi

tion

ener

gy

(eV

)

Quantum number n

0 2 4 6

4.6

4.8

5.0

5.2

5.4

Tra

nsi

tion

ener

gy

(eV

)

Quantum number n

Figure 30: Analysis of transition energies of benzene as considered in Exercise8.7.

(8.7) The absorption spectrum will consist of a series of vibronic lines withenergies given by eqn 8.3. The longest wavelength will correspond tothe transition with n = 0, and the others to increasing values of n. Thetransition energies are plotted against this assignment of n in Fig. 30. Thegood linear fit confirms the assignment. The fitting parameters that comeout of the analysis are ~ω0 = 4.65 eV and ~Ω2 = 0.113 eV. On making thereasonable assumption that ~|Ω2−Ω1| ¿ (E2−E1), we then deduce fromeqn 8.4 that E2 = E1+4.65 eV. We thus identify the energy of the S1 staterelative to the S0 ground state as 4.65 eV, and find Ω/2π = 2.7× 1013 Hz.

E

QQ0

S0

S1

S2

5.7

eV

7.3

eV

n = 5

n = 6

0.13 eV

0.11 eV

E

QQ0

S0

S1

S2

5.7

eV

7.3

eV

n = 5

n = 6

0.13 eV

0.11 eV

Figure 31: Configuration diagram for ammonia deduced from the data of Fig. 8.9as considered in Exercise 8.8.

(8.8) The absorption spectrum shows a progression of vibronic lines obeyingeqn 8.3 with at least two excited electronic states. The first progression

85



starts at ∼ 5.7 eV, and has a vibrational splitting of 0.11 eV. The intensityof the lines peaks for n = 6. The second progression starts at ∼ 7.3 eV, hasa vibrational splitting of 0.13 eV, and has maximum intensity for n = 5.We thus deduce that we have two excited states:

• S1 at energy 5.7 eV with ~Ω = 0.11 eV,

• S2 at energy 7.3 eV with ~Ω = 0.13 eV.

The line with the maximum intensity tells us about the relative values ofQ0 for the states. The intensity peaks when the overlap of the vibronicstates is largest. The wave function of the ground state S0 peaks at Q0,while for the excited states the wave function gradually peaks more andmore at the classical turning points. The potential minima of S1 and S2

therefore occur so that the edge of their sixth and fifth vibronic levels alignrespectively with Q0 for S0. We thus obtain the schematic configurationdiagram given in Fig. 31.

(8.9) The spin–orbit interaction introduces a coupling mechanism between thespin and orbital angular momenta S and L. It is therefore possible to havean interaction between different S states via their spin–orbit interactionwith a common L state. This interaction produces a small mixing of thewave functions, so that triplet states contain a small admixture of singletcharacter. This small singlet admixture gives a finite probability for atriplet-to-singlet transition, which would otherwise be totally forbidden ifthe spin states were pure.

Spin-orbit coupling is now routinely used to increase the intensity of phos-phorescence in organic LEDs. A heavy metal (eg platinum) is introducedinto the molecule, and this increases the spin–orbit interaction, because thespin–orbit coupling generally scales as Z2, where Z is the atomic number.The use of heavy-metal dopants strongly enhances the triplet-to-singlettransition rate, and hence the overall light emission efficiency. This isespecially important in electrically-pumped devices, which are otherwiselimited to a maximum efficiency of 25%. (See Exercises 8.16 and 8.17.)

(8.10) The weak nature of the emission and the long radiative lifetime indicatesthat we are dealing with phosphorescence from a triplet state. The energylevel scheme for pyrromethene 567 would be qualitatively similar to thatfor anthracene shown in Fig. 8.12(b). From the spectra shown in Fig. 8.10we deduce that the S1 level has an energy of 2.3 eV, while the wavelengthof the phosphorescence indicates that the triplet lies at an energy of 1.6 eV.In the case of optical excitation, the phosphorescence could be caused byintersystem crossing from excited singlet states.

(8.11) The assignment of the vibronic peaks of the solution is given in Fig. 8.13.The energies of the vibronic peaks are plotted in Fig. 32 and a goodstraight line is obtained. The linear fit according to eqn 8.3 gives thevibrational energy as 0.16 eV.

In the case of the crystal, we have first to identify the various peaks in theabsorption spectrum. A strong vibronic progression with energies of 3.13,3.30, 3.46 and 3.61 eV is observed in the data. These can be identified as

86



0 2 4

3.2

3.6

4.0

4.4

En

erg

y(e

V)

Vibronic number

crystal

solution

0 2 4

3.2

3.6

4.0

4.4

En

erg

y(e

V)

Vibronic number0 2 4

3.2

3.6

4.0

4.4

En

erg

y(e

V)

Vibronic number

crystal

solution

crystal

solution

Figure 32: Analysis of the vibronic peaks of anthracene shown in Fig. 8.13 asconsidered in Exercise 8.11.

the 0–0, 0–1, 0–2, and 0–3 transitions, and a linear fit according to eqn 8.3gives the vibrational energy as 0.18 eV. (See Fig. 32.)

Two other features can also be identified in the absorption spectrum ofthe crystal at 3.18 and 3.35 eV. These have the same splitting as the otherprogression and therefore involve similar types of vibrations. The mostprobable cause is a splitting of the electronic states in the crystal due toits lower symmetry, eg by the Davydov effect. (See Pope and Swenberg,Electronic processes in organic crystals and polymers, 2nd edn, OxfordUniversity Press, 1999, Section I.D.5, pp 59–66.)

(8.12) When the ‘mirror symmetry’ rule works, we expect the emission spectrumto be a mirror image of the absorption spectrum about the 0–0 transition.(See, for example, Figs 8.10 or 8.17.) We thus expect a broad vibronicband extending downwards from the 0-0 transition at 3.13 eV. The widthof the band will be about 1 eV. A series of vibronic peaks will occur withenergies given by hν ≈ (3.13 − n~Ω), with ~Ω ≈ 0.18 eV. We would thusexpect peaks at 3.13, 2.95, 2.77, 2.59 eV · · · .

(8.13) The S1 absorption band has a 0–0 transition at 1.9 eV and extends to∼ 2.8 eV. The emission band would thus have a 0–0 transition at 1.9 eVand extend down to about 1.0 eV. The 0–1 vibronic peaks occurs at 2.1 eVin the absorption spectrum, which implies a vibrational energy of ∼ 0.2 eV,and hence a 0–1 transition in emission at around 1.7 eV.

(8.14) As discussed for the PDA data in Fig. 8.16, the difference between theabsorption and photoconductivity edges is caused by excitonic effects. Theabsorption edge corresponds to the creation of tightly-bound (Frenkel)excitons. Since excitons are neutral particles, they do not contribute tothe photoconductivity. The photoconductivity edge therefore correspondsto the band edge where free electrons and holes are first created. Thedifference in the two edges gives the exciton binding energy, which worksout to be 1.1 eV.

It is important to realize that this is a different situation to that encoun-tered for weakly-bound (Wannier) excitons. Weakly-bound excitons can

87



be easily ionized to produce free electrons and holes, and hence producea photocurrent. (See, for example, Figs 4.5 and 6.15.)

(8.15) The dominant vibrational frequency can be deduced by analysing the vi-bronic progression of either the absorption or emission spectra in Fig. 8.17according to eqns 8.3 or 8.5 as appropriate. This gives ~Ω ∼ 0.17 eV, im-plying Ω ∼ 2.6 × 1014 rad/s. With ω = 2.98 × 1015 rad/s, we then findωRaman = (ω−Ω) = 2.72×1015 rad/s, which is equivalent to a wavelengthof 693 nm.

Two points could be made here:

• The molecule will have other vibrational modes, and these will giveadditional Raman lines.

• Not all vibrational modes are Raman-active, (see Section 10.5.2) andit is not immediately obvious that the 0.17 eV mode responsible forthe vibronic spectra will show up in the Raman spectrum. In fact,these modes are observed in the experimental Raman spectra, but itrequires a careful analysis by group theory to prove that this is so.

(8.16) Optical excitation creates only singlets because the ground state is asinglet and optical transitions do not change the spin. With only singletstates excited, the recombination of the electrons and holes is opticallyallowed for all the carriers.

Wave function Sz S State↑e↑h +1 1 triplet

(↑e↓h + ↓e↑h)/√

2 0 1 triplet(↑e↓h − ↓e↑h)/

√2 0 0 singlet

↓e↓h −1 1 triplet

Table 6: Possible arrangements of relative electron-hole spins as discussed inExercise 8.16.

On the other hand, with electrical injection there is no control of therelative spin of the electrons and holes. The spins can be either parallel oranti-parallel, and this gives rise to four possible total spin wave functions,as indicated in Table 6. Three of these are triplets and only one is asinglet state. The relative number of triplet and singlet excitons createdby electrical injection is therefore in the ratio 3:1, which implies that only25% of the excitons are in singlet states. The remaining 75% are in tripletstates with very low emission probabilities. Hence the emission is expectedto be weaker than that for optical excitation by a factor of four.

The creation of triplets in electrically-driven organic LEDs is a seriousissue that limits their efficiency. One way to enhance the efficiency isto increase the spin–orbit interaction to encourage inter-system crossing.This is typically done by including a heavy metal atom in the molecule.(See Exercise 8.9.)

(8.17) (a) As we have seen in Exercise 8.16, we expect that 75% of the excitonscreated will be in triplet states with very low emission probabilities. Hence

88



the maximum quantum efficiency that we can expect corresponds to thenumber of singlet excitons that we create, namely 25%.

(b) The number of electrons and holes flowing into a device carrying acurrent i is equal to i/e. The quantum efficiency is defined as the ratio ofphotons out to electrons in, and so the number of photons emitted will beequal to ηi/e. The power emitted is then equal to hν×ηi/e, and we have:

P = 2.25 eV × 25%× 10mA/e = 5.6 mW .

(c) The electrical power consumed by the device is equal to iV = 50 mW.The power conversion efficiency is thus equal to 5.6/50 = 11 %. Theefficiency of a real device would be much lower, mainly due to the difficultyof collecting the photons, which are emitted in all directions (ie over 4πsolid angle). Only a small fraction of these would be collected by theoptics. This latter point is exacerbated by the high refractive index of themolecular material, which tends to limit the effective collection efficiencyeven further. (See Exercise 5.13.)

aa

1

30°

a2

a

(a) (b)

n1 a

1

n 2a 2

c

q

60°

n2 |a2| sin 60°

n2 |a2| cos 60°

aa

1

30°

a2

a

aa

1

30°

a2

a

(a) (b)

n1 a

1

n 2a 2

c

q

60°

n2 |a2| sin 60°

n2 |a2| cos 60°

n1 a

1

n 2a 2

c

q

60°

n2 |a2| sin 60°

n2 |a2| cos 60°

Figure 33: (a) Analysis of the lattice vectors of graphene, as considered in inExercise 8.18(a). (b) Evaluation of the chiral angle θ, as required in Exer-cise 8.18(c).

(8.18) (a) The unit cell of graphene and its lattice vectors are shown in Fig. 33(a).From inspection of this figure, we see that:

|a1| = 2× a cos 30 = 2a√

3/2 =√

3a ,

where a is the length of the hexagon edge. The result for |a2| is identicalby symmetry.

(b) We first find the length of the chiral vector c defined in eqn 8.14 byevaluating its scalar product:

c2 ≡ c · c = (n1a1 + n2a2) · (n1a1 + n2a2) ,

= n21 a1 · a1 + 2n1n2 a1 · a2 + n2

2 a2 · a2 .

Now a1 · a1 = a2 · a2 = a20 and a1 · a2 = |a1| |a2| cos 60 = a2

0/2. Hence

c2 = a20(n

21 + n1n2 + n2

2) .

89



Since c is the circumference, the tube diameter d is then given by:

d =c

π=

a0

π

√n2

1 + n1n2 + n22 ,

as required.

(c) The chiral angle for a nanotube with the chiral vector defined ineqn 8.14 can be evaluated by reference to Fig. 33(b). It is apparent that:

tan θ =n2|a2| sin 60

n1|a1|+ n2|a2| cos 60.

On noting that |a1| = |a2|, this simplifies to:

tan θ =n2 sin 60

n1 + n2 cos 60=

√3n2/2

n1 + n2/2=

√3n2

2n1 + n2.

Equation 8.16 follows immediately.

(8.19) (a) Symmetry requires that the electron wave function should be single-valued on rotating the tube by 2π. The phase change of the electron wavefunction for a rotation through 2π is given by:

∆φ = k⊥|c| ≡ k · c

where k⊥ is the component of the wave vector in the direction perpen-dicular to the tube axis and c is the chiral vector. This must be equalto an integer multiple of 2π in order that the electron wave function issingle-valued. Hence we require:

k · c = 2πm ,

where m is an integer.

(b) The tube will be metallic if the k vector corresponding to the K pointof the Brillouin zone is one of the allowed wave vectors of the nanotube.This will be the case if the condition derived in part (a) is satisfied whenk corresponds to the K point, that is if:

K · c = 2πm ,

where K = (k1 − k2)/3. Now the reciprocal lattice vectors are definedby:

a1 · k1 = a2 · k2 = 2π ,

a2 · k1 = a1 · k2 = 0 .

Hence

K · c = (k1 − k2) · (n1a1 + n2a2)/3 = (2πn1 − 2πn2)/3 ,

and therefore2π(n1 − n2)/3 = 2πm ,

which implies n1 − n2 = 3m.

90



(8.20) We treat the nanotube as a 1–D quantum wire of length L with its axisalong the z direction, and apply periodic boundary conditions so that

Ψ(x, y, z) = Ψ(x, y, z + L) .

The electrons are assumed to have free motion along the z axis and sotheir wave function is of the form:

Ψ(x, y, z) = ψ(x, y) exp ikz .

The periodic boundary condition requires that:

exp ik(z + L) = exp ikz .

This means that exp ikL = 1, and hence that k = 2πm/L, where m is aninteger. The density of states in k space is therefore L/2π. We can thuswrite the density of states per unit length in k space as

g(k) = 1/2π .

The density of states per unit length in energy space is worked out from:

g(E) = 2× 2g(k)dE/dk

.

The additional factor of two here compared to eqn 3.14 comes from thefact that the +k and −k velocity states are degenerate. We assume thatwe have a free electron moving in a band with energy En associated withthe quantized motion in the (x, y) directions. The total energy is thengiven by:

E(k) = En +~2k2

2m∗ ,

so thatdE

dk=~2k

m∗ =(

2~2

m∗

)1/2

(E − En)1/2 .

The density of states is thus given by:

g(E) = 412π

(m∗

2~2

)1/2

(E − En)−1/2 =√

2m∗

π~(E − En)−1/2 ,

as required for eqn 8.20.

(8.21) The radiative quantum efficiency is given by eqn 5.5. The decay routevia intersystem crossing to the T1 level is non-radiative, and so we cantake τNR = 1.2 ns. The radiative lifetime τR is given as 1.8 µs. Hence:

ηR =1

1 + τR/τNR=

11 + (1.8× 10−6/1.2× 10−9)

= 6.7× 10−4 .

The radiative quantum efficiency is therefore only 0.07%.

91



Chapter 9

Luminescence centres

(9.1) The solution for a one-dimensional potential well with infinite barriersis given in Section 6.3.2. In a cube the motion is quantized in all threedimensions, and the energies for the x, y and z directions just add together.For each direction, we have (cf eqn 6.13):

E =~2π2

2m∗(2a)2n2

where n is the quantum number. The quantum numbers for the threedegrees of freedom are independent of each other, and we derive eqn 9.4by adding the quantized energies for the x, y and z directions together.Note that the similar case of a quantum dot was considered previously inSection 6.8.1. See especially eqn 6.54.

(9.2) Equation 9.5 predicts E = 0.28/a2. The experimental energies are lowerbecause a real F-centre is not a rigid cubic box, and hence it would bemore appropriate to use a finite rather than infinite potential well model.As discussed in Section 6.3.3, the quantization energies of finite potentialwells are smaller than those of infinite wells of the same dimension.

(9.3) We can set a = 0.33 nm and calculate hν = 2.6 eV from eqn 9.5. Alterna-tively, we can just read hν ≈ 2 eV from Fig. 9.4.

2a

- +

+ -- +

2a

+

+ -- +

- +

+ -- +

- +

+ -- +

- +

+ -- +

- +

+ -+

- +

+ -- +

-+

-

+

-+

F2+

2b

b

b

(a) (b)

2a

- +

+ -- +

2a

+

+ -- +

- +

+ -- +

- +

+ -- +

- +

+ -- +

- +

+ -+

- +

+ -- +

-+

-

+

-+

F2+

2a

- +

+ -- +

- +

+ -- +

2a

+

+ -- +

- +

+ -- +

- +

+ -- +

- +

+ -- +

- +

+ -- +

- +

+ -- +

- +

+ -- +

- +

+ -+

- +

+ -- +

- +

+ -- +

-+

-

-+

-

+

-+

+

-+

F2+F2+

2b

b

b

2b

b

b

(a) (b)

Figure 34: (a) An F+2 centre in an alkali halide with cell size 2a, as considered

in Exercise 9.4. The colour centre is modelled as a rectangular box with squarecross-section as shown in part (b).

(9.4) The energy of the electron will be given by eqn 6.54 with dx = dy = b and

92



dz = 2b. We therefore have:

E =~2π2

2m0

(n2

x

b2+

n2y

b2+

n2z

4b2

).

We can apply this model to the F+2 centre by taking the appropriate value

of b. The rectangle equivalent to the F+2 centre is sketched in Fig. 34. If the

cubic unit cell size is 2a, then the square end of the box has dimension√

2a,and the longer dimension is 2

√2a. We therefore need to take b =

√2a for

an F+2 centre. The lowest energy transition is the nz = 1 → 2 transition,

which has an energy of hν = 3h2/32m0b2. With b =

√2a, this gives

hν = 3h2/64m0a2, which is half the value given in eqn 9.5.

The experimental absorption peak of KF:F+2 occurs at 1.1 eV (see Fig. 9.5),

while that of KF:F occurs at 2.9 eV (see Fig. 9.4.) The experimental ra-tio is thus ∼ 0.4, which is close to the predicted value of 0.5. This isremarkably good agreement considering the simplicity of the model.

(9.5) (a) By applying Boltzmann’s law to the two levels, we see that the occu-pancy of the MS = ±1 upper level relative to the MS = 0 lower level isgiven by:

N2

N1=

g2

g1exp

(∆E

kBT

),

where g2 and g1 are the level degeneracies (i.e. g2 = 2, g1 = 1), and ∆E/his given in Fig 7.9(a) as 2.9 GHz. We thus find N2/N1 = 1.87 for T = 2 K,which means that 1/2.87 = 35% of the centres are in the MS = 0 level,and 65% of them are in the MS = ±1 level.

(b) We require N1/(N1 + N2) = 80%, which implies N2/N1 = 0.25. Wethus solve:

N2

N1=

g2

g1exp

(∆E

kBT

)= 0.25 ,

for T . This gives T = 0.07K.

(9.6) The emission spectrum shows four clear peaks within the vibronic band:the zero-phonon line at 638 nm, and three phonon sidebands at 660, 685,and 708 nm. These are assigned to multiple-phonon emission processes.A fit of these energies to eqn 9.2 is shown in Fig. 35. The phonon energyis 0.064 eV.

(9.7) (a) For the 3d electron we have n = 3 and l = 2, which implies 〈r〉3d =(21/2)aH/Z, while for the 4f electron we have n = 4 and l = 3 and hence〈r〉4f = 18 aH/Z. We thus find:

〈r〉3d〈r〉4f =

712

Z4f

Z3d.

As a crude estimate we use average values for the atomic numbers of therelevant series: i.e. we take Z = 25 for the transition metals, and Z = 64for the rare earths. This implies

〈r〉3d〈r〉4f ∼

712× 64

25∼ 1.5 .

93



0 1 2 3

1.75

1.80

1.85

1.90

1.95

Sid

eban

den

erg

yE

(eV

)

Sideband number n

E = 1.941 - 0.0636 n

0 1 2 3

1.75

1.80

1.85

1.90

1.95

Sid

eban

den

erg

yE

(eV

)

Sideband number n

E = 1.941 - 0.0636 n

Figure 35: Fit to the phonon sidebands in Fig. 9.7(b) using eqn 9.2, as requiredfor Exercise 9.6.

This shows that the 3d series is expected to have a larger radius by afactor of ∼ 1.5.

It is instructive to compare the radial probability densities (i.e. r2|R(r)|2)derived from the radial wave functions:

R3d(r) =Z7/2

a3/2H

481√

30

(r

aH

)2

exp(− Zr

3aH

),

R4f(r) =Z9/2

a3/2H

1768

√35

(r

aH

)3

exp(− Zr

4aH

).

Figure 36 compares r2R(r)2 for a 3d wave function with Z = 25 with thatfor a 4f wave function with Z = 64. It is apparent that the 3d orbital hasthe larger average radius.

0.0 0.2 0.4 0.6 0.8 1.0 1.20

1

2

3

4

5

Rad

ial

pro

bab

ilit

yd

ensi

ty(a

.u.)

r / aH

3d (Z = 25)

4f (Z = 64)

0.0 0.2 0.4 0.6 0.8 1.0 1.20

1

2

3

4

5

Rad

ial

pro

bab

ilit

yd

ensi

ty(a

.u.)

r / aH

3d (Z = 25)

4f (Z = 64)

Figure 36: Radial probability densities for the 3d orbitals of a transition-metalion and the 4f orbitals of a rare-earth ion, as considered in Exercise 9.7.

It should be stressed that these arguments are not totally convincing be-cause they take no account of the screening effect of the inner shells, which

94



will reduce the effective value of Z experienced by the electron. However,the more accurate calculations of the wave functions do confirm the gen-eral trends predicted by this argument.

(b) The 3d transition metal ions have lost the 4s electrons, and so the 3delectrons are the outermost orbitals. By contrast, the 4f orbitals in therare earths are inside the filled 5s and 5p orbitals. The 5s and 5p orbitalshave n = 5 and l = 0 & 1 respectively. They therefore have larger radiithan the 4f shell on account of the fact that 〈r〉 is proportional to n2 anddecreases with l.

z

x, y y

x

z

xz

x, y

(a) (b) (c) (d)

smaller

distancez

x, y

z

x, y y

x

y

x

z

x

z

xz

x, y

z

x, y

(a) (b) (c) (d)

smaller

distance

Figure 37: Discussion of p orbitals as required for Exercise 9.8. (a) A pz orbital.(b) pz, px and py orbitals in an octahedral lattice, as seen in the x-z and x-yplanes. (c) pz, px and py orbitals in a uniaxially-distorted lattice, as seen in thex-z and x-y planes. (d) Splitting of the p orbitals in the distorted lattice, withnegative nearest neighbours.

(9.8) (a) The three p orbitals are dumb-bell shaped as shown in Fig. 37(a) forthe case of the pz orbital. In an octahedral lattice, the x, y and z directionsare all equivalent. This implies that the px, py and pz orbitals must allexperience the same interaction with the crystal. This is apparent fromFig. 37(b), which shows that the distance from the electron cloud to theions is the same for the pz, px and py orbitals. Hence they will experienceidentical Coulomb interactions.

(b) In a uniaxial crystal, the octahedral symmetry is lost and the z direc-tion is now different. This means that the pz orbitals are closer to the ionsthan the px or py orbitals. (See Fig. 37(c).) The Coulomb interactionsbetween the electron cloud will now be different for the pz orbital and theother two, and so its energy will be different. On the other hand, the xand y directions remain equivalent (see lower half of Fig. 37(c)), and sothe px and py orbitals are still degenerate. Hence the triplet p state splitsinto a singlet and a doublet.

(c) If the nearest neighbour ions are negative, the pz electrons will ex-perience a stronger repulsive interaction with the lattice because of thesmaller distance to the ion. Hence the pz states will have a larger energy

95



than the px and py states. This will give a splitting as shown in Fig. 37(d),with the singlet at higher energy.

(9.9) (a) We write the potential in the form:

V (r, θ, φ) =6α

a+ f(r)[f1(θ) + f2(θ)(e+i4φ + e−i4φ)] ,

and evaluate the matrix element from:

〈m|V |m′〉 =∫∫∫

ψ∗m V ψm′ r2 sin θ dr dθ dφ .

It is immediately obvious that the constant term 6α/a can only give non-zero terms if m′ = m by wave function orthogonality. Consider the integralover φ for the remaining terms. This is of the form:

I =∫ 2π

0

e−imφ [f1(θ) + f2(θ)(e+i4φ + e−i4φ)] eim′φ dφ .

Now it is apparent that∫ 2π

0

e−imφeim′′φeim′φ dφ = 2πδm,(m′+m′′) ,

where δi,j is the Kronecker delta function. Hence the only non-zero matrixelements are those with m′ = m, m′ = m− 4 or m′ = m + 4.

(b) A general state of the system may be written as Ψ =∑2

i=−2 ci|i〉,which may be expressed as the column vector:

Ψ =

c−2

c−1

c0

c1

c2

.

In this basis, and with the notation introduced in the Exercise, the crystal-field Hamiltonian matrix takes the form:

Hcf =

A 0 0 0 D0 B 0 0 00 0 C 0 00 0 0 B 0D 0 0 0 A

.

The eigenenergies and states are found by diagonalizing the Hamiltonian.It is apparent that the central part of the matrix is already diagonal, andso immediately we find three eigenstates, namely:

• | − 1〉 with energy B,

• |0〉 with energy C,

• |+ 1〉, also with energy B.

96



The states | − 2〉 and |+ 2〉 separate into a submatrix, with:(

A DD A

)(c−2

c2

)= λ

(c−2

c2

).

The solutions are: c2 = ±c−2, λ = A ± D. We thus find the final twoeigenstates, namely:

• (|2〉+ | − 2〉)/√2 with energy A + D,• (|2〉 − | − 2〉)/√2 with energy A−D.

(c) The explicit calculation of the matrix elements A, B, C and D, andhence the demonstration that A + D = C and A−D = B, may be foundin Henderson & Imbusch (1989), §2.6.

If A+D = C, then the eigenstates |0〉 and (|2〉+ |−2〉)/√2 are degenerate,and if A−D = B, then the eigenstates | − 1〉, |+ 1〉 and (|2〉 − | − 2〉)/√2are degenerate. We thus identify the two dγ states with the |0〉 and (|2〉+| − 2〉)/√2 states, or a linear combination of them. Similarly, we identifythe three dε states with the eigenstates |− 1〉, |+1〉 and (|2〉− |− 2〉)/√2,or a linear combination of them.

It is simplest to see the connection between these states and those givenin the Exercise by working backwards from the Cartesian versions, andrelating them to the spherical harmonic functions given in part (a) of theexercise.

2z2 − x2 − y2 = 2r2 cos2 θ − r2 sin2 θ cos2 φ− r2 sin2 θ sin2 φ ,

= r2(2 cos2 θ − sin2 θ[cos2 φ + sin2 φ]) ,

= r2(2 cos2 θ − sin2 θ) ,

= r2(3 cos2 θ − 1) ,

∝ C2,0(θ, φ) ,

∝ |0〉 ;

x2 − y2 = r2 sin2 θ cos2 φ− r2 sin2 θ sin2 φ ,

= r2 sin2 θ(cos2 φ− sin2 φ) ,

= r2 sin2 θ cos 2φ ,

= r2 sin2 θ (ei2φ + e−i2φ)/2 ,

∝ (C2,2 + C2,−2) ,

∝ (|2〉+ | − 2〉)/√

2 .

We thus identify the wave functions 2z2 − x2 − y2 and x2 − y2 with thetwo dγ states.

Similarly:

xy = r2 sin2 θ cos φ sin φ ,

= r2 sin2 θ sin 2φ/2 ,

= r2 sin2 θ (ei2φ − e−i2φ)/4i ,∝ (C2,2 − C2,−2) ,

∝ (|2〉 − | − 2〉)/√

2 ;

97



yz = r2 sin θ sin φ cos θ ,

= r2 cos θ sin θ(eiφ − e−iφ)/2i ,∝ (C2,1 − C2,−1) ,

∝ (|1〉 − | − 1〉)/√

2 ;

zx = r2 cos θ sin θ cosφ ,

= r2 cos θ sin θ(eiφ + e−iφ)/2 ,

∝ (C2,1 + C2,−1) ,

∝ (|1〉+ | − 1〉)/√

2 .

We thus identify the wave functions xy, yz and zx with the three dε states.

x

y

xx

y z

2 2dx y-

dxy 2d

z

q = -Ze

(a) (b) (c)

x

y

xx

y z

2 2dx y-

dxy 2d

z

q = -Ze

(a) (b) (c)

Figure 38: Probability densities for the dε and dγ states, as considered in Exer-cise 9.9(d): (a) dxy, (b) dx2−y2 , and (c) dz2 . The circles on the axes representthe anions, which have charge −Ze. The probability densities for the dyz anddzx states are obtained by exchanging the axis labels.

(d) The probability densities for the dγ and dε states are plotted in Fig. 38.The dγ state with wave function 2z2 − x2 − y2 is labelled dz2 . The dxy,dyz and dzx states have zero probability along the axes, and have highprobability between the axes. On the other hand, the dx2−y2 and dz2

states have high probability densities along the axes, but not along thediagonal directions.

For a d1 configuration the ion has a single d electron. In the dγ statesthe electron probability density is high along the crystalline axes, and sothe electron experiences a strong repulsion from the negatively-chargedanions. The dε states, by contrast, have a smaller probability densityclose to the anions and thus experience a smaller repulsion. The two dγstates therefore have the higher energy, and hence the doublet lies abovethe triplet.

The argument is reversed for a d9 configuration, which can be consideredas a single hole in the filled d shell. The charge cloud is now positive ratherthan negative, and the two dγ states experience a stronger attractionrather than repulsion. They therefore have the lower energy.

98



(9.10) The 1064 nm line in a Nd:YAG crystal corresponds to a transition fromthe 11 502 cm−1 level as shown in Fig. 9.9(b). The relative populations ofthe 11 502 cm−1 and 11 414 cm−1 levels of the 4F3/2 term are proportionalto the Boltzmann factor:

N(11502 cm−1)N(11414 cm−1)

= exp(−∆E

kBT

),

where ∆E = 88 cm−1. This ratio equals 0.19 at 77 K and 0.66 at 300 K.The spontaneous emission rate increases in proportion to these factors,and therefore the relative intensity of the 1064 nm line increases with T .

(9.11) The transition rates between the two levels are governed by the Einsteincoefficients, the populations of the levels, and the light energy density.(See Section B.1 in Appendix B.) Three types of transitions are possible,namely spontaneous emission, stimulated emission, and absorption. Forgain, we need that the stimulated emission rate should exceed the absorp-tion rate. (Spontaneous emission is negligible at high light intensities.)The condition for this to occur is (see eqns B.5 and B.6):

B21N2u(ν) > B12N1u(ν) ,

which implies:N2

N1>

B12

B21.

On substituting from eqn B.10, we derive the condition for net gain:

N2

N1>

g2

g1,

where g2 and g1 are the degeneracies of the two levels. This condition iscalled population inversion.

(9.12) (a) When the pump is turned off, all of the atoms will be in the groundstate, so that:

N0 = N ,

N2 = 0 ,

where N is the total number of atoms. When the pump is turned on, ∆Natoms will be pumped to the upper laser level via level 1, so that:

N0 = N −∆N .

N2 = ∆N .

For population inversion, we require N2 > N0, which implies ∆N > (N −∆N), i.e. ∆N > N/2. Population inversion is therefore only achievedwhen more than 50% of the atoms are pumped to the excited state.

(b) In this particular example, we have N2 = 0.6N at t = 0, and wehave seen in part (a) that the laser will stop oscillating when N2 = 0.5N .The sequence of events is therefore as shown in Fig. 39. The number ofatoms that make stimulated radiative transitions (and hence the number

99



of photons emitted) during the laser pulse is therefore 0.1N . This gives apulse energy of 0.1N × hν, where hν is the laser photon energy, namely1.79 eV. On inserting the appropriate numbers we find:

Epulse = 0.1× (1025 × 10−6)× 1.79 eV = 0.3 J .

0

2

t = 0

60%

40%0

2

after pulse

50%

50%

laser pulse

emitted

0

2

t = 0

60%

40%0

2

t = 0

60%

40%0

2

after pulse

50%

50%

laser pulse

emitted

Figure 39: Sequence of events for the ruby laser considered in Exercise 9.12(b).

(9.13) (a) The optical intensity I(t) is proportional to |E(t)|2, and hence if wehave a Gaussian pulse as defined in the exercise, we have a time-varyingelectric field of the form (note the extra factor of two):

E(t) = E0 exp(−t2/2τ2) e−iω0t ,

where ω0 is the centre angular frequency. The spectrum of the pulse isfound by first taking the Fourier transform of E(t):

E(ω) =1√2π

∫ +∞

−∞E(t) eiωt dt ,

=E0√2π

∫ +∞

−∞exp(−t2/2τ2) ei(ω−ω0)t dt ,

= E0τ exp[−τ2(ω − ω0)2/2] ,

where we used the standard integral:

F (Ω) =∫ +∞

−∞exp(−t2/σ2)e−iΩt dt =

√πσ exp(−σ2Ω2/4) ,

in the last line. The final result is obtained by using:

I(ω) ∝ |E(ω)|2 ∝ exp[−τ2(ω − ω0)2] .

(b) The full width at half maximum (FWHM) of the pulse in the timedomain is found by finding the times for which I(t) = I0/2, i.e. by solving:

exp(−t2/τ2) = 0.5 .

This gives t = ±√

ln 2 τ , so that ∆t = 2√

ln 2 τ . The FWHM of the pulsein the frequency domain is likewise found by solving:

exp[−τ2(ω − ω0)2] = 0.5 ,

100



which gives (ω − ω0) = ±√

ln 2/τ , and hence ∆ω = 2√

ln 2/τ . We thenobtain:

∆ν∆t = ∆ω∆t/2π = 2 ln 2/π = 0.441 .

(9.14) The crystal-field shifts of the energy levels depend on the local environ-ment of the ion. For example, small perturbations to the atomic positionsaffect the energy levels of the ion through the alterations to the local elec-tric field the ion experiences. There will be much larger inhomogeneity inthe local environment in a glass than in a crystal, due to the lack of long-range order. Hence we expect much larger inhomogeneous broadening ofthe crystal-field split transitions in a glass than in a crystal.

If we assume a Gaussian pulse, we expect a time-bandwidth product of0.441, and hence ∆tmin = 60 fs. Other pulse shapes would give comparableminimum pulse durations.

(9.15) The subscript ‘g’ stands for gerade and implies even parity. A g→gtransition therefore involves no parity change and is forbidden for electric-dipole transitions. (See Section B.3.) The transition will therefore havea lower probability than for an electric-dipole allowed process, and hencehave a long excited state lifetime (∼ 4 µs). Since the lifetime is long, theradiation would be classified as phosphorescence rather than fluorescence.(See Section B.3.)

(9.16) The excited state lifetime is determined by both radiative and non-radiative processes. It follows from eqn 5.4 that:

1τ

=1τR

+1

τNR.

The radiative lifetime is governed by atomic transition probabilities andis not expected to vary significantly with the temperature. On the otherhand, the non-radiative transition rate is governed by phonon-assistedprocesses and is expected to increase strongly with T . On substitutingτR = 1.8ms into the equation above, we find τNR = 6.3 ms at 77Kand 0.062 ms at 300 K. This implies, through eqn 5.5, that the radiativequantum efficiency is 78% at 77 K and 3% at 300 K. The radiative efficiencyis too low at 300K to allow lasing.

(9.17) The level scheme for Ti:sapphire is shown in Fig. 9.13. If we assume thatthe laser threshold is low, and that the slope efficiency is 100% (i.e. onelaser photon emitted for each pump photon absorbed), then the ratio ofthe output power to the input power would just be proportional to theratio of the respective photon energies, which implies:

Pout =hνout

hνinPin =

1/8001/514

Pin = 3.2W .

In this case, the remaining 1.8 W of power produces phonons (i.e. heat)in the crystal. Note that a substantial amount of heat is generated inthe crystal even for the ideal case of 100% quantum efficiency due to thedifference in the photon energies of the argon and Ti:sapphire lasers.

101



In reality, a Ti:sapphire laser typically gives about 1 W for a pump powerof 5W at ∼ 500 nm. This reduction of the power from the ideal value iscaused by a number of factors:

• The laser threshold will be significant, and the output power is pro-portional to (Pin − Pth) rather than to just Pin (cf. Fig 5.15(a));

• the quantum efficiency is less than 100% due to significant non-radiative decay;1

• the absorption of the pump laser in the crystal is not perfect, and sonot all of the pump power is absorbed;

• there might be optical losses within the cavity.

(9.18) The energy conversion efficiency is calculated by assuming that we obtainone emitted photon from the phosphor for each photon absorbed from theLED. The intrinsic energy conversion efficiency is then just determined bythe ratio of the photon energies:

η =hνout

hνin=

λin

λout,

where λin and λout are the wavelengths of the photons absorbed and emit-ted, respectively. In this case we have λout = 650 nm. Hence:(a) η = 54% for λin = 350 nm;(b) η = 69% for λin = 450 nm.

1The operating temperature of the laser crystal will be above room temperature due tothe heat generated within it, and this further increases the non-radiative decay rate. Coolingof the laser rod is therefore essential.

102



Chapter 10

Phonons

(10.1) The phonon modes of purely covalent crystals do not give rise to infraredabsorption because the atoms are neutral and do not interact with theelectric field of the light wave. Of the five materials listed, germaniumand argon are elemental, and must therefore have neutral atoms with non-polar bonds, and hence no infrared absorption. The other three, namelyice, ZnSe and SiC, are polar, and would therefore have some infrared-activephonons.

(10.2) It is apparent from eqn 1.29 that R = 0 when n = 1, and hence εr = 1.For an undamped oscillator, εr(ν) is given by eqn 10.15. We thus solve:

εr(ν) = ε∞ + (εst − ε∞)ν2TO

(ν2TO − ν2)

= 1 ,

for ν. Rearrangement gives:

ν2 =(

εst − 1ε∞ − 1

)ν2TO ,

leading to the result quoted in the exercise.

(10.3) The exercise follows Example 10.1(a). We calculate νLO = 20THz fromthe LST relationship, and hence that the Reststrahl band runs from 9.2to 20THz, i.e. 15 µm to 33 µm.

(10.4) The exercise closely follows Example 10.1(b). The relative permittivityis given by eqn 10.10 as:

εr(ν) = 10 +210

100− ν2 − iγ′ν/2π,

where ν is measured in THz and γ′ = γ/1012. We calculate νLO = 11 THzfrom the LST relationship, so that the Reststrahl band runs from 10 to11THz. We thus need to evaluate εr at ν = 10.5 THz.

(a) With γ = 1011 s−1, we find εr = −10.48+0.336i at 10.5 THz, and hencethat n = 0.0519 and κ = 3.238 from eqns 1.25–26. Then from eqn 1.29 wefind R = 0.98.

(b) With γ = 1012 s−1, we find εr = −9.958+0.252i and n = 0.509+3.196iat 10.5THz, and hence that R = 0.84.

(10.5) (a) We identify the Reststrahl band from the region of high reflectivityfrom 30–32µm. (See Fig. 40(a).) On equating the upper and lower wave-length limits with νTO and νLO respectively, we find νTO ≈ 9.5 THz andνLO ≈ 10 THz.

103



16 20 24 28 32 36 400.0

0.2

0.4

0.6

0.8

1.0R

efle

ctiv

ity

Wavelength (mm)

AlSb

nLO nTO

R(¥)R(0)

(a)

0.0 0.4 0.8 1.2 1.6

0.84

0.88

0.92

0.96

1.00

Ref

lect

ivit

y

g (1012 s-1)

(b)

16 20 24 28 32 36 400.0

0.2

0.4

0.6

0.8

1.0R

efle

ctiv

ity

Wavelength (mm)

AlSb

nLO nTO

R(¥)R(0)

(a)

16 20 24 28 32 36 400.0

0.2

0.4

0.6

0.8

1.0R

efle

ctiv

ity

Wavelength (mm)

AlSb

nLO nTO

R(¥)R(0)

(a)

0.0 0.4 0.8 1.2 1.6

0.84

0.88

0.92

0.96

1.00

Ref

lect

ivit

y

g (1012 s-1)

(b)

0.0 0.4 0.8 1.2 1.6

0.84

0.88

0.92

0.96

1.00

Ref

lect

ivit

y

g (1012 s-1)

(b)

Figure 40: (a) Interpretation of the data in Fig. 10.14 as required for Exercise10.5(a) and (b). (b) Calculated reflectivity versus damping constant, as requiredfor Exercise 10.5(c).

(b) The high- and low-frequency permittivities can be deduced from theasymptotic reflectivities. (See Fig. 40(a).) The low-frequency limit givesεst from (see eqn 1.29, with n =

√εr):

R(0) =(√

εst − 1√εst + 1

)2

.

At ω → 0, there is no absorption, and so εr will be real. On readingR ≈ 0.30 ≡ R(0) at long wavelengths, we deduce εst ≈ 12. On similarlyequating the short wavelength limit of R, namely 26%, with R(∞), wededuce ε∞ ≈ 9.5.

(c) The peak reflectivity is about 90%, and is limited by γ, which in turnis determined by the lifetime of the TO phonons. The middle of theReststrahl band occurs at 9.75 THz. We thus need to evaluate εr from(see eqn 10.10, with ν measured in THz and γ′ = γ/1012):

εr(ν) = ε∞ + (εst − ε∞)ν2TO

(ν2TO − ν2)− iγν/2π

= 9.5 +225

90− ν2 − iγ′ν/2π,

at ν = 9.75THz. We split this into the real and imaginary parts, computen and κ from eqns 1.25–26, and R from eqn 1.29. The reflectivity calcu-lated in this way is plotted as a function of γ in Fig. 40(b). It is apparentthat we have R = 0.9 for γ = 8.6× 1011 s−1. This implies, from γ = 1/τ ,that τ = 12ps.

The values of νLO and νTO found in part (a) can be compared to theLyddane–Sachs–Teller relationship, which predicts νLO/νTO = 1.11. Theexperimental ratio is slightly smaller. The values given here are onlyapproximate, and depend on how exactly they are extracted from thedata. The departure from LST is therefore not very significant.

(10.6) The exercise closely follows Example 10.1(c). We first use eqn 10.17to find εr at νTO, which gives εr = 10 + 132i for γ = 1012 s−1 andεr = 10 + 1320i for γ = 1011 s−1. We then use eqn 1.26 to find κ and

104



eqn 1.19 to find α.(a) For γ = 1012 s−1 we find κ = 7.8 and α = 3.4× 106 m−1.(b) For γ = 1011 s−1 we find κ = 26 and α = 1.1× 107 m−1.Note that the peak absorption increases for the smaller value of the damp-ing, as normal for a damped oscillator.

(10.7) The peak reflectivity is governed by the damping constant γ. (See, forexample, Fig. 40(b) above.) As the temperature increases, we expect γto increase, and hence R to decrease, due to the increased probabilityof anharmonic decay processes of the type illustrated in Fig. 10.13. Thereason why anharmonic phonon decay increases with T is that phonons arebosons, and the probability for phonon emission increases as the thermalpopulation of the final state increases. (n.b. This contrasts with fermions,for which the transition probabilities decrease with increasing occupancyof the final state.)

(10.8) With negligible damping, we can use eqn 10.15 to calculate εr = 21.5 at8THz. We then substitute this value of εr into eqn 10.18 to compute thewave vector. This gives:

q =√

εrω

c=√

21.5× 2π × 8× 1012

3× 108= 7.8× 105 m−1 .

(10.9) (a) The condition for cyclotron resonance is given in eqn 10.24. In apolar material, the mass that is measured is the polaron mass m∗∗. Wethus obtain:

m∗∗ =eλB

2πc= 0.097 m0 .

(b) The rigid lattice mass m∗ can be calculated from eqn 10.21. Oninserting the relevant values into eqn 10.19, we find αep ≈ 0.33 for CdTe.Then from eqn 10.21 we find that m∗∗ = 0.097 m0 implies m∗ = 0.092 m0.

(10.10) The Raman spectra for a number of III-V crystals are shown in Fig. 10.11.In each case we observe two peaks: one for the TO phonons and the otherfor the LO phonons. These two phonon modes have different frequenciesbecause III-V compounds have polar bonds with partially charged atoms.They therefore interact with light, and obey the LST relationship. Thesituation in diamond is different because it is a purely covalent crystal,with neutral atoms that do not interact with the light. The LST analysisdoes not apply, and the optical phonons are degenerate at q = 0. TheRaman spectrum therefore has only one peak for the optical phonons.

(10.11) Silicon, like diamond in the previous exercise, is covalent, and its LOand TO phonons are degenerate at q = 0. The two peaks correspond tothe Stokes and anti-Stokes lines from these degenerate optical phonons.The line at 501.2 nm is shifted up in frequency compared to the laser andis thus the anti-Stokes line, while that at 528.6 nm is the Stokes line. Thephonon frequency can be worked out from eqn 10.27, which gives, for thecase of the Stokes line:

Ω/2π =c

514.5 nm− c

528.6 nm= 15.5THz .

105



The relative intensities of the lines are given by eqn 10.29:

I(501.2 nm)I(528.6 nm)

= exp(

h× 15.5× 1012

kBT

)= 0.08 .

(10.12) NaCl has an inversion centre and so the mutual exclusion rule applies. Itis apparent from Fig. 41 that the TO mode has odd parity under inversion.The TO mode is therefore IR active but not Raman active.

invertinvert

Figure 41: Inversion of a TO mode in an ionic crystal, as considered in Exercise10.12. The inversion centre is circled.

(10.13) The energies of the phonon modes can be deduced directly from theRaman spectra by applying eqn 10.27, with the + sign as appropriate fora Stokes shift. This shows that the Raman shift is exactly equal to thephonon frequency. For each crystal, two lines are observed. The lowerfrequency line comes from the TO phonons, while the higher frequencyline originates from the LO phonons. The Raman shifts in cm−1 fromFig. 10.11 are given in Table 7, together with the energies deduced ac-cording to:

E (meV) = 0.124× Raman shift (cm−1) .

Crystal Raman line 1 Raman line 2 TO phonon energy LO phonon energycm−1 cm−1 meV meV

GaAs 262 286 32.5 35.5InP 299 341 37.1 42.3AlSb 312 332 38.7 41.1GaP 364 403 45.1 50.0

Table 7: Raman shifts deduced from the data in Fig. 10.11, as considered inExercise 10.13.

On comparing the frequencies of the TO and LO phonons of GaAs inTable 7 with those deduced from the infrared reflectivity data in Fig. 10.5,we see that there is a small shift of a few wave numbers between the twosets of data. This is caused by the slight decrease of the optical phononfrequencies between 4 K and 300 K.

(10.14) Equation 10.28 implies that momentum is conserved during the Ra-man scattering process so that the vectors form a triangle as depicted inFig. 42(a). In the case of inelastic scattering by acoustic phonons, thefrequency shift of the photon is very small because ω À Ω. This impliesthat the magnitude of the photon wave vector hardly changes, so that wecan approximate:

|k1| = |k2| ≡ k =nω

c.

106



k1

k2

q

q/2

k1k2

q

(a) (b)

k1

k2

q

q/2

k1k2

q

(a) (b)

Figure 42: (a) Conservation of momentum during Raman scattering by a phononof wave vector q, as as considered in Exercise 10.14. (b) Back-scattering geom-etry.

It is then apparent from Fig. 42(a) that:

q

2= k sin

θ

2=

nω

csin

θ

2.

On writing q = Ω/vs as appropriate for acoustic phonons, we derive theresult in the exercise. Equation 10.32 then follows by writing:

δω = |ω2 − ω1| = Ω .

(10.15) In back-scattering geometry, we have θ = 180, so that q = 2k. (SeeFig. 42(b) with |k1| = |k2| = k.) It then follows from eqn 10.32 withsin(θ/2) = 1 that:

vs =cδω

2nω=

λδν

2n.

On inserting the data given in the exercise, we find vs = 813 m s−1.

(10.16) (a) The negative term in r−1 is the attractive potential due to theCoulomb interaction between the ions. The Madelung constant α accountsfor the summation of the contributions of the positive and negative ionsover the whole crystal. The positive term in r−12 represents the shortrange repulsive force due to the Pauli exclusion principle when the electronwave functions overlap.

(b) The graph of U(r) is qualitatively similar to that for the Lennard–Jonespotential, being attractive for large r, repulsive for small r, and with aminimum at some intermediate value of r, labelled r0. (cf. Fig. 28.) Thevalue of r0 is found by differentiating U(r):

dU

dr= −12β

r13+

αe2

4πε0r2,

and finding the value of r for which dU/dr = 0, namely:

12β

r130

=αe2

4πε0r20

,

which implies r110 = 12β × 4πε0/αe2.

(c) The Taylor series for U(r) expanded about r0 is:

U(r) = U(r0)+dU

dr r=r0

(r−r0)+12

d2U

dr2 r=r0

(r−r0)2+16

d3U

dr3 r=r0

(r−r0)3+· · · .

107



Now U(r) has a minimum at r0, and the first derivative is therefore zero,so that:

U(r) = U(r0) +12

d2U

dr2 r=r0

(r − r0)2 +16

d3U

dr3 r=r0

(r − r0)3 + · · · .

We can reconcile this with eqn 10.33 by taking x = r − r0 and U(x) =U(r)− U(r0). It is then apparent that:

C2 =12

d2U

dr2 r=r0

,

C3 =16

d3U

dr3 r=r0

.

At r = r0 we have

d3U

dr3= −2184β

r150

+6αe2

4πε0r40

,

=(

6αe2

4πε0− 2184β

r110

)1r40

,

= −176αe2

4πε0r40

,

where we used the result of part (b) to derive the last line. Hence:

C3 = −22αe2/3πε0r40 .

(10.17) If the Raman spectrum is lifetime-broadened, we shall have a Lorentzianline shape with:

∆ν ∆t =12π

.

Hence with ∆t = τ , and ∆ν = c∆ν, we have:

τ =1

2πc∆ν.

On inserting the data given in the exercise, we find τ = 6 ps. This valueagrees with the lifetime measured by time-resolved Raman scattering. See:von der Linde et al., Phys. Rev. Lett. 44, 1505 (1980).

108



Chapter 11

Nonlinear optics

(11.1) In the Bohr model for hydrogen, the radius of the electron in the nthquantum level is given by:

rn =4πε0~2

me2

n2

Z=

n2

ZaH .

The magnitude of the electric field is given by the standard Coulombformula:

E =Ze

4πε0r2,

which, on inserting rn from the Bohr formula, gives:

E =Ze

4πε0r2n

=e

4πε0a2H

Z3

n4.

For the outer 3s and 3p electrons in atomic silicon, use n = 3 and aneffective nuclear charge Z ∼ 4.2 We then obtain a value of E = 5 ×1011 Vm−1. The field for the conduction electrons in crystalline siliconwould, of course, be different due to the high relative permittivity and thelow effective mass.

(11.2) We can relate the optical intensity to the electric field by using eqn A.44.(a) The optical intensity is found from:

I =P

A=

Epulse/τpulse

πr2=

1÷ 10−8

π(2.5× 10−3)2= 5.1× 1012 W/m2

.

Hence with n = 1 for air, we find from eqn A.44 that E = 6.2×107 V m−1.(b) The optical intensity is found from:

I =P

A=

10−3

20× 10−12= 5× 107 W/m2

.

Then with n = 1.45 as appropriate for the fibre, we find from eqn A.44that E = 1.6× 105 Vm−1.

(11.3) With no external field applied, the gas is isotropic and therefore possessesinversion symmetry. Hence χ(2) = 0, and no frequency doubling will occur.

With the electric field applied, the gas is no longer isotropic as the electronclouds of the atoms will be distorted along the axis defined by the field.This means that inversion symmetry no longer holds, so that χ(2) 6= 0 andfrequency doubling can, in principle, occur. However, it would give a veryweak signal due to the low density of atoms.

2Zeff is the difference between the nuclear charge and the total number of inner shellelectrons that screen the valence electrons from the nucleus. i.e. Zeff = 14 − 10, where 10 isthe total number of electrons in the 1s, 2s and 2p shells.

109



(11.4) The second-order nonlinear susceptibility is zero if the material has aninversion centre. We must therefore consider the microscopic structure tosee if the material has inversion symmetry or not.(a) NaCl is a face-centred cubic crystal with inversion symmetry: χ(2) = 0.(b) GaAs has the zinc-blende structure, which is similar to the diamondstructure except that the bonds are asymmetric. It does not possess in-version symmetry and therefore χ(2) 6= 0.(c) Water is a liquid and is therefore isotropic; hence inversion symmetryapplies and χ(2) = 0.(d) Glass is amorphous and has no preferred axes; inversion symmetryapplies and χ(2) = 0.(e) Crystalline quartz is a uniaxial crystal with the trigonal 3m structure.It does not possess inversion symmetry, so that χ(2) 6= 0.(f) ZnS has the hexagonal wurtzite structure (6mm), without inversionsymmetry. Hence χ(2) 6= 0.

(11.5) (a) Consider the absorption and stimulated emission transitions as indi-cated in Fig. 11.2. (Spontaneous emission can be neglected if uν is suf-ficiently large.) The absorption and stimulated emission rates are equalto N1B12uνg(ν) and N2B21uνg(ν) respectively. (See eqns B.5–6 with theadditional factor of g(ν) explained in the discussion of eqn 11.30.) If thelevels are non-degenerate, then eqn B.10 tells us that B12 = B21. At t = 0,all the atoms are in level 1, and there is net absorption, which increasesN2 and decreases N1. As the atoms are pumped to level 2, the stimu-lated emission rate becomes increasingly significant. Eventually, we reacha stage where N1 = N2 = N0/2, and the stimulated emission and absorp-tion rates are identical. There is therefore no net absorption or emission,and N2 cannot increase further. The maximum value of N2 that can beachieved is therefore N0/2.3

(b) If we only have two levels and we neglect spontaneous emission, thenthe rate equations for N1 and N2 are:

dN1

dt= −B12N1uνg(ν) + B21N2uνg(ν) ,

dN2

dt= +B12N1uνg(ν)−B21N2uνg(ν) .

On setting B12 = B21 as appropriate for non-degenerate levels, and sub-tracting, we find:

ddt

(N1 −N2) =d∆N

dt= −2B12uνg(ν)∆N ,

where ∆N = N1 −N2. Integration yields:

∆N(t) = ∆N(0) exp(−2B12uνg(ν)t) ,

which, with ∆N(0) = N0, gives the required result.

3This shows that it is not possible to achieve population inversion (i.e. N2 > N1) in atwo-level system: three or more levels are required. This is why lasers, in which populationinversion is essential, are always classified as either three or four-level systems.

110



The result quantifies the way the laser pumps atoms from level 1 to level2, and hence reduces the net absorption. The equation implies that thepopulations will eventually equalize no matter how weak the laser beamis. This unphysical result arises from neglecting spontaneous emission andtransitions to other levels that occur in real atoms.

direction of propagation

x

y

E

q

direction of propagation

x

y

E

q

Figure 43: Propagation and polarization vectors for the light wave consideredin Exercise 11.6.

(11.6) With the beam propagating in the x-y plane and with its polarization inthe same plane, the light must be linearly polarized as shown in Fig. 43.The z-component of the electric field is therefore zero. The nonlinearpolarization, for the given nonlinear optical tensor, is found from eqn 11.43to be:

P(2)x

P(2)y

P(2)z

=

0 0 0 d14 0 00 0 0 0 d25 00 0 0 0 0 d36

ExEx

EyEy

EzEz

2EyEz

2EzEx

2ExEy

=

2d14EyEz

2d25EzEx

2d14EyEz

.

On setting Ez = 0, we obtain:

P(2)x

P(2)y

P(2)z

=

00

2d14EyEz

.

Only P(2)z is non-zero, and therefore the second harmonic beam must be

polarized along z.

We assume that the direction of propagation makes an angle θ with respectto the x axis as shown in Fig. 43 and that the light has an electric fieldof magnitude E0. It will then be the case that Ex = E0 cos θ and Ey =E0 sin θ, and hence that:

P (2)z = 2d36E2

0 cos θ sin θ = d36E20 sin 2θ .

This is maximized when 2θ = 90: i.e. θ = 45.

(11.7) This exercise closely follows Example 11.2, and the phase-matching angleis found by substituting the appropriate refractive indices into eqn 11.52.

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With the data given in the exercise, this gives:

1(1.506)2

=sin2 θ

(1.490)2+

cos2 θ

(1.534)2.

On using cos2 θ = 1 − sin2 θ and re-arranging, we find sin2 θ = 0.626 andhence θ = 52.3.

(11.8) (a) In the absence of the field, the index ellipsoid given in eqn 11.54 isof the form:

x2 + y2

n2o

+z2

n2e

= 1 .

On substituting into eqn 11.60 with Ex = Ey = 0 and Ez = E , we findthat the only non-zero changes to the index ellipsoid induced by the fieldare as follows:

∆(

1n2

)

1

= ∆(

1n2

)

2

= r13E ,

∆(

1n2

)

3

= r33E .

From eqn 11.56 we then see that the modified index ellipsoid is:(

1n2

o

+ r13E)

x2 +(

1n2

o

+ r13E)

y2 +(

1n2

e

+ r33E)

z2 = 1 ,

which can be written in the form :

x2 + y2

no(E)2+

z2

ne(E)2= 1 ,

where:

1n2

o(E)=

1n2

o

+ r13E ,

1n2

e(E)=

1n2

e

+ r33E .

This implies:

no(E) = no(1 + n2or13E)−1/2 ,

ne(E) = ne(1 + n2er33E)−1/2 .

We assume that the field-induced changes are small and use the approxi-mation (1 + x)−1/2 = (1− x/2) for small x to obtain the final result:

no(E) = no − 12n3

or13E ,

ne(E) = ne − 12n3

er33E .

(b) The geometry of the crystal is shown in Fig. 44. If the light is prop-agating along the y axis, and is polarized along the z direction, then it

112



x

y

z

L

V x

y

z

x

y

z

L

V

Figure 44: Experimental geometry of the electro–optic phase modulator consid-ered in Exercise 11.8.

will experience a refractive index of ne(E). From part (a) we know thatne(E) = ne − n3

er33E/2, and hence the refractive index change when thefield is applied. The phase change associated with the change in ne istherefore given by

∆φ(E) =2π

λ∆neL = −πn3

er33EL/λ .

The argument would work equally well if the light were polarized along xinstead of z. In this case the phase change would be given by:

∆φ(E) =2π

λ∆noL = −πn3

er13EL/λ .

In the case of LiNbO3, r33 ∼ 3r13 (see Table 11.3), and so it is convenientto use the z polarization, but this is not necessarily true for other crystals.

(c) It is apparent from part (b) that the electric field produces a linearmodulation of the phase. Hence the application of a voltage modulatesthe phase in proportion to the voltage. This is a phase modulator.

y

z

x

y¢x¢

L

V

y

z

x

y¢x¢

L

V

Figure 45: Experimental geometry of the electro-optic crystal considered inExercise 11.9.

(11.9) We consider an electro-optic crystal with axes as defined in Fig. 45. Thevoltage is applied so as to produce an electric field of magnitude Ez alongz axis.

(a) If the light propagates along the z axis, the polarization vector willlie in the x-y plane, or equivalently, in the x′-y′ plane. We resolve the

113



light polarization vector into its components along the x′ and y′ axes.The phase shift induced by a refractive index change ∆n in a medium oflength L is, in general, given by:

∆φ =2π

λ∆nL ,

where λ is the vacuum wavelength. On applying this to the two compo-nents along the x′ and y′ axes, we then have:

∆Φx′ =2πL

λ∆nx′ =

πLn30r41

λEz

∆Φy′ =2πL

λ∆ny′ = −πLn3

0r41

λEz ,

where ∆nx′ and ∆ny′ are the field-induced refractive index changes alongthe x′ and y′ axes as given in the exercise. The phase difference ∆Φ isthus given by (with Ez = V/L):

∆Φ = ∆Φx′ −∆Φy′ =2πLn3

0r41Ez

λ=

2πn30r41V

λ.

Note that the phase change is independent of the length in this longitudinalgeometry.

(b) On setting ∆Φ = π, we find:

Vπ =λ

2n30r41

.

With the appropriate figures for CdTe given in the exercise, we obtainVπ = 44 kV.

(11.10) The phase change is given by eqn 11.62. For a crystal of length L, thefield strength at voltage V is equal to V/L. Hence the phase change is:

∆φ =2π

λn3

or63(V/L)L = 2πn3or63V/λ .

The transmission will be equal to 50% when ∆φ = π/2, since this meansthat the crystal acts like a quarter wave plate. In these circumstances, theoutput of the crystal is circularly polarized, so that half of the intensity istransmitted through the second polarizer. This occurs when:

V = λ∆φ/2πn3or63 = λ/4n3

or63 .

On inserting the values of no and r63 for λ = 633 nm, we find V = 4.2 kV.

(11.11) In a third-order nonlinear medium, the change in the relative permit-tivity is given by (see eqn 11.69):

∆εr = ∆ε1 + i∆ε2 = χ(3)E2 .

We therefore have ∆ε2 = Im(χ(3))E2 and hence that ∆ε2 ∝ Im(χ(3))Ibecause I ∝ E2. It follows from eqns 1.19 and 1.24 that ∆α ∝ ∆ε2.Hence, in a medium with Im(χ(3)) 6= 0, we expect

∆α ∝ Im(χ(3))I .

114



A saturable absorber has an absorption coefficient that obeys eqn 11.37.In the limit of small intensities, this is of the form:

α(I) = α0 − α0I/Is = α0 −∆α ,

where ∆α = α2I and α2 = α0/Is. We thus have an intensity dependenceexactly as described above, and we therefore conclude that the saturableabsorber must have Im(χ(3)) 6= 0.

(11.12) We can choose our axes as we please in an isotropic medium. Thereforechoose z as the direction of propagation, and x as the polarization vector,so that the electric field is given by:

E = (Ex, 0, 0) .

The third-order nonlinear polarization is given by eqn 11.11, and, withEy = Ez = 0, the nonlinear polarization is of the form:

P(3)x

P(3)y

P(3)z

= ε0E3

x

χxxxx

χyxxx

χzxxx

.

However, we see from Table 11.6 that χyxxx = χzxxx = 0. Hence we find

P = ε0χxxxxE3x(1, 0, 0) ,

which means that P is parallel to E.

(11.13) This exercise is very similar to Example 11.4. From eqn 11.76, werequire that:

∆Φnonlinear =2π

λn2IL = π ,

which implies:

I =λ

2n2L=

1.55× 10−6

2 · 2× 10−20 · 10= 3.9× 1012 W m−2 .

The optical power to produce this intensity is given by:

P = IA = 3.9× 1012 × π(2.5× 10−6)2 = 76 W .

This is a large power level for a continuous-wave laser, but not for a pulsedlaser. Consider, for example, a mode-locked laser with an average powerof Pav, pulse repetition rate f , and pulse width tp. The peak power isgiven by:

Ppk =Epulse

tp=

Pav

f tp.

On inserting typical values, namely Pav = 10 mW, f = 100 MHz, andtp = 1 ps, we find Ppk = 100W.

115



(11.14) It is apparent from Fig. 11.10 that the presence of electrons causes thestates in the conduction band up to Ec

F to be filled up, and likewise forthe holes in the valence band. The absorption between Eg and (Eg +Ec

F +Ev

F) will therefore be blocked, and the new absorption edge will occur at(Eg + Ec

F + EvF). The shift in the absorption edge is therefore (Ec

F + EvF).

The Fermi energy in the conduction band can be calculated from eqn 5.13.On inserting m∗

e = 0.067m0, we find EcF = 0.054 eV. In the case of the

valence band, we must consider the occupancy of both the heavy and lighthole bands. On using the result of Exercise 5.14(b), we have:

Nh =1

3π2

(2~2

)3/2

(m3/2hh + m

3/2lh ) (Ev

F)3/2 ,

which gives EvF = 0.007 eV for m∗

hh = 0.5m0 and m∗lh = 0.08m0. Hence the

absorption edge will shift to higher energy by 0.054 + 0.007 = 0.061 eV.4

(11.15) If we treat the exciton as a classical oscillator, we can use the resultsderived in Chapter 2. If we assume that the contribution of the exciton tothe refractive index is small compared to the non-resonant value, (whichis indeed the case, as we shall show below,) then we expect a refractiveindex variation as in Fig. 2.5. The refractive index will thus have localmaxima and minima just below and above the centre of the absorptionline. It is this extra contribution that we are considering in this exercise.

The magnitude of the excitonic contribution can be calculated by followingthe method of Example 2.1. In part (a) of Example 2.1 it is shown that

κmax =Ne2

2nε0m0γω0,

where κmax is the extinction coefficient at the line centre, while in part (c)it is shown that:

nmax =(

ε∞ +Ne2

2ε0m0γω0

)1/2

.

It then follows, with ε∞ = n2, that:

nmax = n(1 +

κmax

n

)1/2

,

where nmax is the maximum value of the refractive index. If we assume,as is demonstrated below, that κmax ¿ n, we then find:

nmax = n + κmax/2 .

Now we know from eqn 1.19 and the data given in the exercise that

κmax =λαmax

4π=

847× 10−9 · 8× 105

4π= 0.054 .

4In a real experiment, the behaviour would be more complicated due to many-body effectssuch as band gap renormalization. This causes a shift of Eg ∝ −N1/3, which reduces the blueshift of the absorption edge.

116



We thus deduce that there is a local maximum in the refractive indexjust below the absorption line with (nmax − n) = 0.027. If the excitonabsorption is saturated, this local maximum will disappear. Hence themaximum change in the refractive index is 0.027.

(11.16) (a) The saturation density for the excitons is equal to the Mott densitygiven in eqn 4.8. We consider here only the n = 1 exciton, since this isthe ground state of the system and has the highest stability. For InP wehave:

µ = (1/m∗e + 1/m∗

h)−1 ≈ 0.06m0

for m∗e = 0.077m0 and m∗

h ∼ 0.3m0 (i.e. a mean of m∗hh and m∗

lh).Equation 4.2 then gives r1 = 12.5 aH/0.06 = 11 nm, and we hence findNMott ≈ 1.8× 1023 m−3.

(b) We first calculate the energy of the n = 1 exciton. We see from eqn 4.4that the n = 1 exciton absorption line will occur at

hν = Eg −RX ,

where Eg is given in Table D.2 as 1.42 eV. With µ = 0.06m0 and εr = 12.5,we find from eqn 4.1 that RX = (0.06/12.52)RH = 5.2meV. The excitonenergy is thus 1.41 eV at low temperatures. (We consider low temperatureshere because the exciton would be ionized at room temperature.)

The saturation intensity Is is the optical intensity required to produce theMott density worked out in part (a). By using the result of Exercise 5.6(b),namely:

N =Iατ

hν,

we then find

Is =hν

ατNMott =

1.41 eV106 · 10−9

× (1.8× 1023) ≈ 4× 107 W m−2 .

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optical properties of solids 2nd ed by mark fox

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