ops solutions manual
TRANSCRIPT
Optical Properties of Solids
Mark Fox
Oxford University Press, 2001
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 2006
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 1.26 to obtain R = 0.041. The transmission is calculated fromeqn 1.6 with R = 0.041 and α = 0 to obtain T = 0.92.
(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 1.26 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 1.22 and 1.23 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 1.16 to work out κ = αλ/4π =0.083. We then use eqn 1.26 to find R = 0.328. The transmission coeffi-cient is calculated from eqn 1.6 as T = (1−0.328)2 exp(−1.3×2) = 0.034.The optical density is calculated from eqn 1.8 as 0.434×1.3×2 = 1.1. (n.b.In principle we should take account of multiple reflections as in Fig 1.10.However, we can neglect these effects here because the absorption is veryhigh: see Exercise 1.9.)
(1.6) 99.8% absorption in 10 m means exp(−αl) = 0.002, and hence α = 0.62m−1.We use eqn 1.16 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 1.20 and 1.21 respectively, and thus we 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) (i) The beam that has suffered no reflections from the back surface passesthrough the first interface, then propagates through the material, andfinally passes through the second interface. Its intensity is thus given by:
I1 = (1−R) · e−αl · (1−R) .
1
The intensity of the beam that is reflected once from the back surfaceis found by following its path through the medium as it is reflected offthe front and back surfaces and propagates through the medium. Theintensity is thus given by:
I2 = (1−R) · e−αl ·R · e−αl ·R · e−αl · (1−R) .
The ratio of the two intensities is thus given by R2 exp(−2αl).
(ii) The window is transparent and hence non–absorbing, implying α =κ = 0. The reflectivity is calculated from eqn 1.26 to be 0.04, and theratio is thus R2 = 1.6× 10−3.
(iii) The intensity is proportional to the square of the field. (cf. eqn A.40.)The field ratio is thus [R2 exp(−2αl)]1/2 = 0.04.
(iv) The field ratio is important at normal or near-normal incidence be-cause multiple beam interference can occur for the exiting and reflectedbeams. The plate would then behave as a Fabry–Perot etalon. (See opticstexts e.g. Hecht (1998) for further details.) The effects of multiple beaminterference are a nuisance when it comes to extracting reliable values ofthe optical constants, but their effects can be assumed to be small in limitwhere the reflectivity is small (as in this Exercise), and when the sampleis strongly absorbing (as in the next).
(1.9) In Exercise 1.5 we worked out R = 0.328. With α = 1.3 × 106 m−1 andl = 2× 10−6 m, the intensity ratio is equal to R2 exp(−2αl) = 5.9× 10−4,while the field ratio is its square root, namely 0.024.
(1.10) We take the log10 of eqn 1.6 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 1.8 to obtain the requiredresult.
If the medium is transparent at λ′ then we will have that Tλ′ = (1 −R)2, where R is the reflectivity at λ′. We assume that the reflectivityvaries only weakly with wavelength. This is a reasonable assumption formost materials if we choose λ′ sensibly. For example, we would chooseλ′ just above the absorption edge we are trying to measure. With thisassumption, the optical density at λ is then given by
O.D.(λ) = − log10(Tλ) + log10(Tλ′) .
Measurements of T (λ) and T (λ′) thus allow the optical density to bedetermined. If l is known, the absorption coefficient can then be foundfrom eqn 1.8.
(1.11) The propagation time is equal to l/v = ln/c. The time difference isthus l(n1 − n2)/c = 27 ns. The pulses at the wavelength with the smallerrefractive index (i.e. 1500 nm) take the shorter time.
(1.12) With σ = 6.6 × 107 Ω−1 m, and ω = 1.88 × 1013 rad/s, we find εr =εr + 3.97 × 105i. The imaginary part of εr is very large, and hence the
2
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 1.26, we then obtain R = 99.6%.
(1.13) 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 + 2.94× 105i ,
which implies:
n =√
εr = (2.94× 105)1/2√
i = 383(1 + i) .
We then find from eqn 1.16 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.
(1.14) It is apparent from eqn 1.26 that R = 1 when n = 1 and κ = 0. For zeroreflectivity we thus require εr = (n + iκ)2 = 1.
(1.15) (i) We convert wavelengths to photon energies using E = hc/λ to obtainthe energy level scheme shown in Fig. 1 of this solutions manual. It is thusapparent that 0.294 eV of energy is dissipated during each absorption /emission process.
(ii) 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.
(iii) For a luminescent quantum efficiency of 50% the number of photonsemitted drops by a factor of 2 compared to part (ii), 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 1: Energy levels scheme for Exercise 1.15.
(1.16) 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.
3
(1.17) The transmission is given by eqn 1.9, with the wavelength dependenceof the scattering cross section given by eqn 1.10. At 850 nm we have10% transmission, so that Nσsl = 2.30. The scattering cross-section is11.1 times larger at 850 nm than at 1550 nm, and so we have Nσsl =2.30/11.1 = 0.21 at the longer wavelength, implying a transmission of81%.
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.
(1.18) We again use eqn 1.9 to calculate the transmission, setting exp(−Nσsl) =0.5. This gives Nσsl = 0.69, which implies l = 3.5m for the given valueson N and σs.
If the wavelength is reduced by a factor of two, Rayleigh’s scattering law(eqn 1.10) 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.
(1.19) Birefringence is an example of optical anisotropy as discussed in Sec-tion 1.5.1, and also in Section 2.4. 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.
4
Chapter 2
Classical propagation
(2.1) We envisage two displaced masses as shown in Fig. 2. 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 frequency (Ks/µ)1/2.
m1 m2
x1 x2
rest
displaced
m1 m2
x1 x2
rest
displaced
Figure 2: 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
1ω2
0 − ω2 − iωγe−iωt .
5
The phase factor comes from the middle term, namely [ω20 − ω2 − iωγ]−1.
On multiplying 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. Using the value of N given in thequestion, 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 1.16 as 270m−1.
This Exercise is broadly based on the results presented in the paper byIverson and Sibley in J. Luminescence 20, 311 (1979).
6
(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 1.16):
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.
(i) The low frequency refractive index corresponds to√
εst. With n(ω ¿ω0) = 2.43 from the data, we find εst = 5.9.
(ii) The resonant frequency is the mid point of the “wiggle”, i.e. 5.0 ×1012 Hz.
(iii) The natural frequency is given by eqn 2.2, which implies Ks = µω20 .
The reduced mass µ is given by eqn 2.1:
1/µ = 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.
(iv) The oscillator density can be found from eqn 2.19. εst = 5.9 has beenfound in part (i), and ε∞ can be read from the graph as ε∞ = [n(ω Àω0)]2 = (1.45)2 = 2.10. We thus have εst − ε∞ = 3.8. Using the values ofω0 and µ worked out previously, we then find N = 3.0× 1028 m−3.
(v) 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.
(vi) 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) Use ω = ck/n in eqn 2.25 to obtain:
vg =dω
dk= c
ddk
(k/n) =c
n− ck
n2
dn
dk.
Then substitute v = c/n to obtain eqn 2.26.
7
(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 because k follows ω. Since εr > 1, n > 1, and hencev = c/n < c. Therefore vg < 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 2.25. It is easier to take dn/dω than dn/dk, and we can use k = nω/cto find vg from:
1vg
=dk
dω=
n
c+
ω
c
dn
dω.
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
),
=1nc
(1 +
Ne2
ε0m0
ω20
(ω20 − ω2)2
).
Hence we find:
vg = nc
(1 +
Ne2
ε0m0
ω20
(ω20 − ω2)2
)−1
.
The denominator is greater than unity, and n < 1, so vg < c.
(2.9) (i) 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.
8
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.
(ii) 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.
(iii) Consider a hollow sphere of radius a placed within a polarized dielec-tric medium as illustrated in Fig. 3. (cf. Fig 2.8.) We assume that thepolarization is parallel to the z axis. The surface charge on the sphere mustbalance 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. 3.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. 3 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.
9
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 3: Definition of angles and charge increment as required for Exercise2.9(iii).
(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.
(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.
10
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) (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.
11
(2.13) (i) 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.
(ii) The values are found by substituting into the result found in part (i)to obtain n = 1.5493 at 450 nm and n = 1.5369 at 650 nm.
(iii) Referring to the angles defined in Fig. 4, 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.
qin
qout
q2
q1
60°
qin
qout
q2
q1
60°
Figure 4: Angles required for the solution of Exercise 2.13.
(2.14) 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, so that we then have:
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 .
12
On using the result above, we find
dτ
dλ= −L
cλ
d2n
dλ2,
which implies:
|∆τ | = L
cλ
d2n
dλ2∆λ =
L
cλ2 d2n
dλ2
∆λ
λ,
as required.
The time–bandwidth product of eqn 2.38 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. With λ = c/ν, we have|∆λ| = (λ2/c)∆ν, so that:
|∆τ | = L
c
(λ2 d2n
dλ2
)λ
c∆ν .
The precise amount of broadening depends on the numerical value of thetime–bandwidth product assumed for the pulse. For a 1 ps pulse with∆ν∆t = 1, we have ∆ν = 1012 Hz, and hence |∆τ | = 0.17 ps for theparameters given in the exercise.
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 5: 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 2.15.
(2.15) It is apparent from eqn 2.46 that ε11 = ε22 = n2o and ε33 = n2
e . Hence wecan re-cast the index ellipsoid in the form:
x2
n2o
+y2
n2o
+z2
n2e
= 1 .
Owing to the spherical symmetry about the optic (z) axis, we can choosethe axes of the index ellipsoid so that the x axis coincides with the po-larization vector of the o–ray as in Fig 2.12(a). The polarization of the
13
e–ray will then lie in the y–z plane, as in Fig. 2.12(b). The projection ofthe index ellipsoid onto the plane that contains the direction of propaga-tion and the polarization vector of the e–ray thus appears as the ellipsedrawn in Fig. 5. The refractive index n(θ) that we require is the distancefrom the origin to the point of the ellipse where the E-vector cuts it. Theco-ordinates of this point are x = 0, y = n(θ) cos θ and z = n(θ) sin θ. Onsubstituting into 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
= 1 ,
as required.
optic axis (z)
E-vector
45°
front surface of
wave plate
y
optic axis (z)
E-vector
45°
front surface of
wave plate
y
Figure 6: Axes of the wave plate relative to the polarization vector for lightpropagating along the x direction, as required for the solution of Exercise 2.16.
(2.16) We assume that the optic axis of the wave plate is along the z axis, asshown in Fig. 6, and assume that the light is propagating along the xdirection. Light with its polarization vector at 45 to the optic axis willthen have its E-vector as shown in the figure. We resolve the E-vectorinto two components of equal amplitude, one along the z axis and theother along the y axis. The components along the z and y axes experiencerefractive indices of ne and no respectively. The phase difference betweenthe two components at the rear surface of the wave plate is thus ∆φ =2πL∆n/λ, where L is the thickness, ∆n = ne − no and λ is the vacuumwavelength of the light.
The wave plate operates at a quarter wave plate when ∆φ = π/2. Inthis situation, the output is two orthogonally polarized waves of equalamplitude but with a π/2 phase difference between them, i.e. circularlypolarized light. The condition for this is L = λ/4∆n, which is equal to14 µm for the parameters given.
(2.17) 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 the
14
crystals 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.
15
Chapter 3
Interband absorption
(3.1) The Born–von Karmen boundary conditions are satisfied when:
kxL = nxL ,
kyL = nyL ,
kzL = nzL ,
where nx, ny and nz are integers. The wave vector is therefore of the form:
k = (2π/L)(nx, ny, nz) .
The allowed values of k form a grid as shown in Fig. 7. Each allowedk-state 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.
The density of states in k-space is found by calculating the number ofstates within the incremental shell between k-vectors of magnitude k andk + dk, as shown in Fig. 7. This volume of the incremental shell is equalto 4πk2dk, and contains 4πk2dk× 1/(2π)3 = k2dk/2π2 states, as given ineqn 3.15.
kx
ky
2p/L
k
dk
Figure 7: Grid of allowed values of k permitted by the Born–von Karmen bound-ary conditions, as considered in Exercise 3.1. The points of the grid are sepa-rated from each other by distance 2π/L in all three directions, giving a volumeper state of (2π/L)3. Note that the diagram only shows the x-y plane of k-space. The incremental shell considered for the derivation of eqn 3.15 is alsoshown.
16
(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) (i) 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.
(ii) 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.
(iii) 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
eim′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
eim′φ (eiφ ± e−iφ) eimφ dφ ,
which is zero unless m′ = m ± 1. We thus have the selection rule ∆m =±1 for light linearly polarized along the x or y axis. With circularlypolarized light we have ∆m = +1 for σ+ polarization and ∆m = −1 forσ− polarization.
(3.4) The apparatus required is basically the same as for Fig. 3.13, 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 is below theband gap of silicon, and the spectrograph/silicon CCD array arrangement
17
shown in Fig. 3.13 is not appropriate. Instead, a detector with a band gapsmaller than 0.3 eV must be used, e.g. InSb. (See Table 3.2.) As InSbarray detectors are not available, a scanning monochromator with a singlechannel detector would normally be used. A thermal source would sufficeas 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 8 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 8: 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 astraight line. Other factors to consider are that the absorption is muchstronger in a direct-gap material, and that the temperature dependence ofα is expected to be different. In an indirect gap material, phonon-assistedabsorption mechanisms will freeze out as the temperature is lowered.
(3.6) Plot α2 and α1/2 against ~ω as shown in Fig. 9. 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. (Notethe huge difference in the two axis scales, which is a further indicationof the indirect nature of the transitions below 2.75 eV, and their directnature above 2.75 eV.) We thus deduce that GaP has a direct gap at2.75 eV. Hence we conclude that the conduction band has two minima:one away from k = 0 at 2.2 eV, and another at k = 0 at 2.75 eV.
(3.7) A wavelength of 1200 nm corresponds to a photon energy of 1.03 eV. Thisis above the direct gap of germanium at 0.80 eV, and thus the absorptionwill be given by (cf. eqn 3.25):
α = C(~ω − 0.80)1/2 .
18
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 9: Analysis of GaP absorption data as required for Exercise 3.6.
The scaling coefficient C can be determined from the data in Fig. 3.10:α2 ≈ 0.5 × 1012 m−2 at 0.90 eV implies C ≈ 2.2 × 106 m−1eV−1/2. Withthis value of C, we then find α ≈ 1× 106 m−1 at 1.02 eV.
(3.8) (i) 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 C.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.
(ii) 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 thecrystal 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.
(iii) Equation 3.24 shows that the joint density of states is proportionalto µ3/2. Hence the ratio of the joint density of states for heavy and lighthole transitions with the same photon energy is equal to:
(µhh/µlh)3/2 = (0.059/0.036)3/2 = 2.1 .
(iv) 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) (i) 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 and j = 3/2conduction bands states are degenerate at k = 0, but not for finite k.
19
Electric-dipole transitions from the p-like valence band states are forbiddento these p-like conduction band states at k = 0. The first dipole-allowedtransition is to the s-like antibonding state at ∼ 4.1 eV. Hence the directgap at the Γ point is equal to 4.1 eV.
(ii) 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.
(3.10) 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. 3.9shows 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.
(3.11) 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 absorptionedge 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 10: Force acting on a particle moving in a magnetic field pointing intothe paper, as required for Exercise 3.12. The charge is assumed to be positive.
(3.12) 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. 10. This perpendicular force produces circular motion at angular
20
frequency 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 themodified 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)xu∗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.
(3.13) (i) Let z be the free direction. Apply Born–von Karmen 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.
(ii) 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 the
21
0 2 4 6 8 100
2
4
6
Ab
sorp
tion
(a.u
.)
Energy relative to Eg in arb. units
(ii) (iii)
0 1 2 30
2
4
6
Ab
sorp
tion
(a.u
.)
Energy in units of hwL relative to Eg
0 2 4 6 8 100
2
4
6
Ab
sorp
tion
(a.u
.)
Energy relative to Eg in arb. units
0 2 4 6 8 100
2
4
6
Ab
sorp
tion
(a.u
.)
Energy relative to Eg in arb. units
(ii) (iii)
0 1 2 30
2
4
6
Ab
sorp
tion
(a.u
.)
Energy in units of hwL relative to Eg
0 1 2 30
2
4
6
Ab
sorp
tion
(a.u
.)
Energy in units of hwL relative to Eg
Figure 11: (ii) Absorption of a one-dimensional semiconductor as discussed inExercise 3.13. (iii) Absorption for a system with quantized landau levels intwo directions and free motion in the third. ωL is the Landau level angularfrequency.
absorption 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. 11(ii).
(iii) The magnetic field quantizes the motion in two dimensions to giveLandau levels, leaving the particle free to move in the third dimension.Optical transitions are possible between Landau levels with the same valueof n. (See Exercise 3.12.) These will occur at energies given by (c.f. 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. 11(iii).
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.
(iv) 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. 3.9.) The indirect transitions for ~ω > 0.66 eV are too weak tobe observed compared to the direct transitions above 0.80 eV.
(3.14) The responsivity is calculated using eqn 3.38. The device will be mostefficient if it has 100% quantum efficiency, and so we set η = 1, giving:
Responsivitymax =e
~ω(1− e−αl) .
22
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.
(3.15) (i) The p-i-n diode structure is described in Appendix D. 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.
(ii) 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.
(iii) The electric field for an applied bias of −10V can be calculated fromeqn D.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.
(iv) With R = 50 Ω and C = 10pF, 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 D.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.
23
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.
24
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
25
On introducing the total mass M and reduced mass m according to:
M = m1 + m2 ,1m
=1
m1+
1m2
,
and substituting into the Hamiltonian, we then find
H = − ~2
2M∇2
R −~2
2m∇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
2m∇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 m 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 m,rather than the individual electron mass. The reduced mass correctiondoes not make much difference for hydrogen itself, where m2 À m1 andhence m ≈ m1, but it is very important for excitons, where the electronand hole masses are typically of the same order of magnitude.
(4.2) (i) 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.
(ii) 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.
26
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 infact a 1s 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 valueof 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.
27
(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.)
(i) 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.
(ii) 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.
(iii) 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 .
(iv) 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) (i) 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.
(ii) The central force for the circular motion is provided by the Coulombattraction, and the electron velocity v must therefore satisfy:
µv2
r=
e2
4πε0εrr2,
28
where r is the radius of the orbital and µ is the reduced mass. (SeeExercise 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.
(iii) E is identical to the exact solution of the hydrogen Schrodinger equa-tion.
(iv) 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 hence wecalculate RX = 4.9 meV from eqn 4.1 using εr = 12.4. Equation 4.4 givesthe wavelengths of the n = 1 and n = 2 excitonic transitions as 873.7 nmand 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 2.17–21. At line centre, we have from eqns 1.16 and 1.22:
α =4πκ
λ=
4πε22nλ
,
29
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 1.22 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/4respectively. Hence the n = 1 → 2 transition occurs at a photon energyof 3RX/4. We calculate RX = 4.2 meV from eqn 4.1, and hence con-clude that the transition energy is 3.15 eV. This occurs at a wavelength of394 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.
(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,
30
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 = (B0/2)(−y, x, 0), we obtain:
B =B0
2
∂x
∂y
∂z
×
−y
x0
=
00
B0
.
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 = (B0/2)(−y, x, 0) is given by:
H ′ =e2A2
2m0=
e2B20
8m0(x2 + y2) .
The diamagnetic energy shift is calculated from
δE = 〈ψ|H ′|ψ〉 =e2B2
0
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
0
8m∗e
2r2n
3+
e2B20
8m∗h
2r2n
3=
e2B20r2
n
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 given effective masses imply µ = 0.17m0 from eqn 3.22, so that wecan calculate r1 = 3.1 nm and r2 = 12.3 nm from eqn 4.2 using εr = 10.The Mott densities are estimated from eqn 4.8. Hence we obtain NMott =8.1 × 1024 m−3 and 1.3 × 1023 m−3 for the n = 1 and n = 2 excitonsrespectively.
31
(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, the radius is comparable to the unit cell size a andthus the Wannier model is invalid. On the other hand, the n = 2 excitonsatisfies the condition r À a, and the Wannier model is valid.
32
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.29 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 that it must be the casethat:
|〈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 .
33
This gives:
〈z〉 =24√
6
(23
)5
aH × 1√3
= 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 concurs 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. τR would normallybe independent of T , but the non-radiative recombination rate 1/τNR gen-erally increases with T due to increased non-radiative recombination byphonon emission. Hence τ decreases with T . The quantum efficiency cancalculated from eqn 5.5 to be equal to τ/τR. At 300K we find ηR = 79 %,while at 350 K we find ηR = 56 %.
(5.5) Semiconductors emit light at their band gap energy. We are thereforelooking for a semiconductor with Eg = 2.3 eV. Inspection of Table C.3suggest two possibilities: ZnTe or GaP. The latter has an indirect bandgap and would therefore not be very inefficient. Hence the most likelycandidate material is ZnTe. Note, however, that we could also make anemitter at this wavelength by using an alloy, for example: GaxIn1−xN.(See Fig. 5.11 and Exercise 5.15.) Linear extrapolation between the GaNand InN band gaps would suggest that a composition with x ≈ 0.26 wouldemit at 540 nm.
(5.6) (i) Consider an incremental beam slice at a position z within the absorbingmaterial as shown in Fig. 12. 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.
34
(ii) 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.
(iii) 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 (ii) 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 12: 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) (i) 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. 12. 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
35
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.
(ii) The pulse excites the carrier density calculated in part (i) at time t = 0.The carriers then recombine and the carrier density decreases accordingto
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.
(iii) The quantum efficiency is calculated from eqn 5.5 as η = 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 Eη/hν =3.5× 1010 photons.
hnEg
E
kk = 0
Ee
Eh
hnEg
E
kk = 0
Ee
Eh
Figure 13: 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.
36
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. 13.) Hence:
fe fh ∝ exp(−Ee/kBT ) · exp(−Eh/kBT ) = exp[−(Ee + Eh)/kBT ] .
Now it is apparent from Fig. 13 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 Boltzmann statistics is valid.
37
(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 willbe degenerate for T ¿ 0.9K. In part (b), the electrons will be degeneratefor T ¿ 420 K, while the holes will be degenerate for T ¿ 85K. Thisshows that it is much easier to obtain degenerate statistics for the lighterelectrons than for the holes.
(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.
(5.13) (i) The solid angle subtended by an object of area dA at a distance r isgiven by (See Fig 14(a)):
dΩ =dA
4πr2.
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
4π(100mm)2= 0.049 .
38
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 14: (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.
(n.b. The answer given in some of the printed versions of the book isincorrect.)
(ii) Consider a ray emitted by a source embedded within a medium ofrefractive index n as illustrated in Fig. 14. 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 1.26 to be 21% forn = 2.7. Hence the final fraction collected is 0.79F = 4.2× 10−4.
(iii) 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 .
39
(iv) We obtain the luminescent power collected by the lens by multiplyingthe total luminescent power by the collected fraction:
P collected = 0.53 ηR × 4.2× 10−4 mW = 0.22 ηR µW .
(5.14) (i) The electron Fermi energy is calculated from eqn 5.13 to be 0.14 eV.
(ii) 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.
(iii) The carriers will be degenerate if EF > kBT . At 180K we havekBT = 0.018 eV, and so the electrons are degenerate, but not the holes.
(iv) 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. Wethen 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.
(v) 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(250)/N(24) = exp(−226/τ), where τ is the lifetime inps. On inserting the two values, we find τ ≈ 0.13 ns.
(5.15) We follow Example 5.1(i). We require a band gap equivalent to 500 nm,namely 2.48 eV. Linear extrapolation between the band gaps of GaN andInN implies that an alloy with x = 0.39 would have the required bandgap.
(5.16) (i) The reflectivity is calculated from eqn 1.26 to be 31% for n = 3.5.
(ii) The cavity mode frequency is given by eqn 5.15, and implies a modespacing of ∆ν = c/2nl. With n = 3.5 and l = 1 mm, we find ∆ν =4.3× 1010 Hz.
(iii) The threshold gain can be calculated from eqn 5.19. We are told toignore background absorption and scattering, and so we set αb = 0. The
40
coated 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 .
(5.17) (i) 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 .
(ii) 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 %.
(iii) The power output will vary as shown in Fig. 5.15(a). The slopeefficiency is calculated from eqn 5.21 as (50/(100− 35) = 0.77 mW/mA =0.77 W/A. The differential quantum efficiency is calculated from eqn 5.21to be 51 %.
41
Chapter 6
Semiconductor quantumwells
(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 15: 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. 15,
42
with a grid-spacing of 2π/L. The area per k-state is (2π/L)2, and thedensity 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. 15 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:
g(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
g(E) =2g(k)dE/dk
=2k/2π
~2k/m=
m
π~2,
as required.
(6.4) 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+∞ at x = 0 to zero at x =
√ξ. (See, for example, Fig. 6.4, for the case
with ξ = 13.2.) The function on the RHS of eqn 6.26 will therefore alwayscross the tan x function between 0 and π/2, no matter how small ξ is.
0.0 0.4 0.8 1.20
2
4
6
8
10
y
x
y = tan(x)
y = 0.82 (33.5-x2)1/2/x
x = 1.30
0.0 0.4 0.8 1.20
2
4
6
8
10
y
x
y = tan(x)
y = 0.82 (33.5-x2)1/2/x
x = 1.30
Figure 16: Graphical solution required for Exercise 6.5.
(6.5) 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 .
43
Hence we must solve:
tan x = 0.82(
33.5− x2
x
).
The two functions are plotted in Fig. 16, which shows that the solution isx = 1.30. The energy is then found from
E =2~2x2
m∗wd2
= 7.6meV ,
where d = 10nm. The equivalent energy for an infinite well is calculatedfrom eqn 6.13 to be 11 meV.
(6.6) (i) The wave functions of an infinite potential well form a complete or-thonormal basis, with
∫ +∞
−∞ϕ∗nϕn′ dz = δn,n′ ,
where δn,n′ is the Kronecker delta function:
δn,n′ = 1 if n = n′
= 0 if n 6= n′ .
It is apparent from eqn 6.11 that the wave functions depend only on n andare therefore identical for electrons and holes with the same n. Hence theorthonormality condition applies irrespective of whether ϕn and ϕn′ areelectron or hole wave functions, or a mixture. Hence:
Mn,n′ = 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 6.13 into the formula for Mn,n′ .
(ii) This result follows from parity arguments. The wave functions ofa finite well have well-defined parities as a consequence of the inversionsymmetry 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 6.7.
(6.7) 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. Two transitions fall in the range1.4 → 2.0 eV:
44
• 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. 6.8.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 asin Fig. 17.
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 17: Absorption spectrum for Exercise 6.7
(6.8) (i) 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.
(ii) Peaks would appear below the steps in the absorption spectrum dueto exciton absorption. The difference in energy between the peak and thecontinuum absorption is the exciton binding energy.
(6.9) (i) To prove normalization, we must show that∫ ∫
ψ∗ψ dA = 1. For thegiven wave function we have:
∫ ∞
r=0
∫ 2π
θ=0
Ψ∗Ψ rdrdθ =(
2πξ2
) ∫ ∞
r=0
∫ 2π
θ=0
exp(−2r/ξ) rdrdθ ,
=(
2πξ2
)× 2π ×
∫ ∞
r=0
exp(−2r/ξ) rdr ,
= 1 ,
as required.
(ii) We first compute the effect of H on Ψ, using the fact that ∂Ψ/∂θ = 0:
HΨ = − ~2
2µr
ddr
(rdΨdr
)− e2
4πε0εrrΨ ,
= − ~2
2µξ2Ψ +
(~2
2µξ2− e2
4πε0εr
)Ψr
.
45
We can then calculate 〈E〉var from:
〈E〉var = − ~2
2µξ2
∫ ∞
r=0
∫ 2π
θ=0
Ψ∗Ψ rdrdθ +(~2
2µξ2− e2
4πε0εr
) ∫ ∞
r=0
∫ 2π
θ=0
Ψ∗Ψdrdθ ,
= − ~2
2µξ2× 1 +
(~2
2µξ2− e2
4πε0εr
)× 2
ξ,
= +~2
2µξ2− e2
2πε0εrξ.
(iii) 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.
(iv) In part (iii) 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.
(6.10) 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 expectfour 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 18: Interpretation of the data in Fig. 6.17 as required for Exercise 6.11.
(6.11) (i) See Section 5.3.4.
(ii) The interpretation of the principal features in the data is shown inFig. 18. 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 the
46
hh1→ 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.)
(iii) 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.
(iv) The binding energies can be read from Fig. 6.17 as the energy gapbetween the exciton peak and the appropriate band edge. With the tran-sitions identified as in Fig. 18, 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.
(6.12) (i) The interaction between an electric dipole and an external electricfield is of the form:
H ′ = −p · E .
With the field aligned along the z axis, this reduces to H ′ = −pzEz. If theposition vector of the electron relative to the origin is r, then the electrondipole moment is −er, with z component pz = −ez. Hence H ′ = +ezEz.Note that the choice of origin does not matter in the final result, becausewe only calculate the shift of the electron relative to its original position.
(ii) 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.
(iii) 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 fairlyobvious given the larger denominator.) Hence we only need to considerthe first term:
∆E(2) = e2E2z
|〈1|z|2〉|2E1 − E2
.
47
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.
(6.13) (i) From the result in the previous Exercise, we would expect ∆E ∝E2
z. We work out the field strength from eqn 6.44, 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 .
(ii) 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 6.13.
(6.14) 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 6.12(iii). 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) .
48
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 quoted.
(6.15) 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 6.6(ii). Atfinite Ez, the inversion symmetry is broken, and the states no longer havea definite parity. Therefore, the selection rule based on parity no longerholds.
(6.16) 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. 6.13.
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.
(6.17) We assume infinite barriers and use eqn 6.42 to calculate the transitionenergy with n = 1. With ~ω = 0.80 eV, Eg = 0.75 eV, and µ = 0.038m0,we find d = 14 nm. In reality, the quantum well would have to be narrowerto compensate for the imperfect confinement of the barriers.
(6.18) (i) 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.
(ii) 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 .
49
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. 6.14 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.
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 19: Refraction of light entering a semiconductor containing a quantumwell, as discussed in Exercise 6.19.
(6.19) Consider a ray incident at angle θ to the normal as shown in Fig. 19.The ray will be refracted according to Snell’s law, with
sin θ
sin θ′= n ,
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.
(6.20) For a cubic dot, the energies of the quantized levels are given by eqn 6.47with dx = dy = dz = d, implying:
E(nx, ny, nz) =π2~2
2m∗d2(n2
x + n2y + n2
z) ,
50
nx ny nx (n2x + n2
y + n2z)
1 1 1 32 1 1 62 2 1 93 1 1 112 2 2 123 2 1 143 2 2 17
Table 3: Quantum numbers of the energy levels for a cubic quantum dots inorder of increasing energy, as discussed in Exercise 6.20.
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.
51
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 then find δ ∼ 0.5 m for ω/2π = 200 kHz. Thisshows that electromagnetic waves of this frequency penetrate less than 1mfrom 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 ≡ 1.6m0.
(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 1.22–23 to calculate n = −0.012+3.5i, and substituten = −0.012 and κ = 3.5 into eqn 1.26 to obtain R = 99.6 %.
52
(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 1.23). This value of κ implies, througheqn 1.16, 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.
(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 1.26 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 4: 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 4.
It is apparent from Table 4 that m∗e increases strongly with N , rising from
0.020m0 at 3.5 × 1023 m−3 to 0.048m0 at 4 × 1024 m−3. This increase inthe effective mass with carrier density is caused by non-parabolicity in
53
the 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 energyincreases (see eqn 5.13) and we are probing states of the conduction bandwith larger values of E and k. It is therefore to be expected that theeffective mass should increase with the doping density.
(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, both ofwhich cause free carrier absorption. The free carrier absorption coefficientat 10.6 µm can be calculated separately for the electrons and holes usingeqn 7.28. With ω = 1.8 × 1014 rad/s, we obtain α = 130m−1 for theelectrons and α = 70 m−1 for the holes. Hence the total free carrierabsorption coefficient is 200 m−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.)
• 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.
54
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 20: 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) (i) 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 = 0.5m0 and m∗
lh = 0.08m0 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. 20.)
(ii) 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 ,
55
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. 20. The energy limits calculated using the Fermi wavevectors from part (i) 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) (i) 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 =
0.85m0 for the donor levels.
(ii) 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 (i), 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 sketched in Fig. 7.10(a). Their tran-sition 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 bandelectron 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 = 0.036m0 for εr = 15.2.
(7.15) Doping with acceptors creates a p-type semiconductor with a series ofacceptor levels just above the valence band as indicated in Fig. 21. Theenergy of the acceptor levels relative to the top of the valence band will
56
conduction band
valence band
acceptor
levels
Eg
conduction band
valence band
acceptor
levels
conduction band
valence band
acceptor
levels
Eg
Figure 21: 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.
be given by eqn 7.29 with m∗e replaced by m∗
h. Electrons from the valenceband can be thermally excited to fill the acceptor levels, giving rise tothe possibility of acceptor to conduction band transitions as indicated inFig. 21. 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.
Students might get confused about the stipulation of 10 K for the temper-ature in this exercise. At low temperatures we might expect the acceptorsto be ‘frozen out’, with holes in the acceptor levels rather than in thevalence band. The point is that the acceptors here are rather shallow anda significant fraction are ionized even at 10 K. It would not be possible toobserve the acceptor to conduction band transitions at higher tempera-tures due to the thermal broadening of the edge. The exercise is in factbased on real experimental data. See Figure 2 in Johnson, E.J. and Fan,H.Y. (1965): “Impurity and Exciton Effects on the Infrared AbsorptionEdges of III-V Compounds”, Phys. Rev. 139, A1991–A2001.
(7.16) The peak is caused by Raman scattering from plasmon modes. Theenergy of the scattered photons is given by eqn 7.34. In this exercise, weare clearly considering the case of plasmon emission where the ‘–’ sign isappropriate. 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. (See Fig. 10.11.)
(7.17) We set
ωp =(
Ne2
εoptε0m∗e
)1/2
= ΩLO ,
57
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.
58
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 = EΨ1 ,
which can be re-arranged to give:(
12mΩ2 − ~2
2ma4
)x2Ψ1 +
~2
2ma2Ψ1 = EΨ1 .
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 seven carbon–carbon bonds.
(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.
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.
59
(8.4) The energy of isolated hydrogen atoms is just given by the standard Ry-dberg formula with En = −RH/n2. The energy required to promote oneof the atoms from a 1s to a 2p state is therefore (3/4)RH = 10.2 eV. Thisis smaller than the equivalent transition in the H2 molecule. The reasonis that 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 1.1eV.
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 22: Lennard–Jones potential for A = B = 1, as considered in Exercise8.5.
(8.5) (i) 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.
(ii) The Lennard-Jones potential is plotted for the case A = B = 1 inFig. 22. 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.The position of the minimum r0 can be calculated by setting:
dU
dr= −12A
r13+
6B
r7= 0 ,
at r = r0, which implies r0 = (2A/B)1/6.
60
(iii) 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.
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 23: Analysis of transition energies in Exercise 8.6.
(8.6) 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. 23. Thegood linear fit confirms the assignment. The fitting parameters that comeout of the analysis are ~ω0 = 4.65 eV and ~Ω2 = 0.113 eV. We thus identifythe energy of the S1 state as 4.65 eV, and find Ω/2π = 2.7× 1013 Hz.
61
E
0
S0
S1
S2
5.7
eV
7.4
eV
n = 5
n = 6
0.13 eV
0.11 eV
E
0
S0
S1
S2
5.7
eV
7.4
eV
n = 5
n = 6
0.13 eV
0.11 eV
Figure 24: Configuration diagram deduced from the data of Fig. 8.9 as consid-ered in Exercise 8.7.
(8.7) The absorption spectrum shows a progression of vibronic lines obeyingeqn 8.3 with at least two excited electronic states. The first progressionstarts 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. 24.
(8.8) 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.
62
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 8.15 and 8.16.)
(8.9) 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 thatof anthracene shown in Fig. 8.12. From the spectra shown in Fig. 8.10 wededuce that the S1 level has an energy of 2.3 eV, while the wavelength ofthe 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.
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 25: Analysis of the vibronic peaks of anthracene shown in Fig. 8.13 asconsidered in Exercise 8.10.
(8.10) 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. 25 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 asthe 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. 25.)
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 lower symme-try of the crystal, 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.)
(8.11) 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 vibronic
63
band 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 · · · .
(8.12) 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.
(8.13) The difference between the absorption and photoconductivity edges iscaused by excitonic effects. The absorption edge corresponds to the cre-ation of tightly-bound (Frenkel) excitons. Since excitons are neutral par-ticles, they do not contribute to the photoconductivity. The photoconduc-tivity edge therefore corresponds to the band edge where free electrons andholes are first created. The difference in the two edges gives the excitonbinding energy, which works out 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 canbe easily ionized to produce free electrons and holes, and hence producea photocurrent. (See, for example, Figs 4.5 and 6.12.)
(8.14) 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, butit requires a careful analysis by group theory to demonstrate thistheoretically.
(8.15) 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.
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 5. 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 only
64
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 5: Possible arrangements of relative electron-hole spins as discussed inExercise 8.15.
25% 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 8.8.)
(8.16) (i) As we have seen in Exercise 8.15, we expect that 75% of the excitonscreated will be in triplet states with very low emission probabilities. Hencethe maximum quantum efficiency that we can expect corresponds to thenumber of singlet excitons that we create, namely 25%.
(ii) 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 .
(iii) 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.)
65
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 6.9.
(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. 9.5.
2a
- +
+ -- +
2a
+
+ -- +
- +
+ -- +
- +
+ -- +
- +
+ -- +
- +
+ -+
- +
+ -- +
-+
-
+
-+
F2+
2b
b
b
(a) (b)
2a
- +
+ -- +
2a
+
+ -- +
- +
+ -- +
- +
+ -- +
- +
+ -- +
- +
+ -+
- +
+ -- +
-+
-
+
-+
F2+
2a
- +
+ -- +
- +
+ -- +
2a
+
+ -- +
- +
+ -- +
- +
+ -- +
- +
+ -- +
- +
+ -- +
- +
+ -- +
- +
+ -- +
- +
+ -+
- +
+ -- +
- +
+ -- +
-+
-
-+
-
+
-+
+
-+
F2+F2+
2b
b
b
2b
b
b
(a) (b)
Figure 26: (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 6.47 with dx = dy = b and
66
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. 26. 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. 9.6),
while that of KF:F occurs at 2.9 eV (see Fig. 9.5.) 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) We use average values for the atomic numbers of the relevant series:i.e. we take n = 3 and Z = 25 for the 3d series, and n = 4 and Z = 64 forthe 4f series. We thus obtain a crude estimate of the ratio of their radiifrom:
r3d
r4f∼ (32/25)
(42/64)∼ 1.4 .
This shows that the 3d series is expected to have a larger radius by afactor of ∼ 1.4.
(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 therefore have larger radii than the 4f shell due to the n2
dependence of r.
(9.6) (i) The three p orbitals are dumb-bell shaped as shown in Fig. 27(a) for thecase 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 energy with the crystal. This is apparentfrom Fig. 27(b), which shows that the distance from the electron cloud tothe ions is the same for the the pz, px and py orbitals. Hence they willexperience identical Coulomb interactions.
(ii) 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. 27(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. 27(c)), and sothe px and py orbitals are still degenerate. Hence the triplet p state splitsinto a singlet and a doublet.
(iii) 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
67
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 27: Discussion of p orbitals as required for Exercise 9.6. (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.
than the px and py states. This will give a splitting as shown in Fig. 27(d),with the singlet at higher energy.
(9.7) The 1064 nm line in Nd:YAG corresponds to a transition from the 11 502 cm−1
level as shown in Fig. 9.8(b). The relative populations of the 11 502 cm−1
and 11 414 cm−1 levels of the 4F3/2 term are proportional to the Boltz-mann 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 .
(9.8) 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,
68
where g2 and g1 are the degeneracies of the two levels. This condition iscalled population inversion.
0
1
2
PUMP
LASER EMISSION
694.3 nm
rapid decay
ground state
PUMPING BANDS
0
1
2
PUMP
LASER EMISSION
694.3 nm
rapid decay
ground state
PUMPING BANDS
Figure 28: Three-level laser scheme for ruby, as required for Exercise 9.9.
(9.9) Ruby is a three-level laser, with a level scheme as shown in Fig. 28. Thelower laser level (level 0) is the ground state, and the upper level is anexcited state (level 2). Ruby has strong absorption bands in the green/bluespectral regions (see Fig. 1.7), and these are used as intermediate pumpingbands (level 1) to produce population inversion after rapid decay to theupper lasing level.
When the pump is turned off, all the atoms will be in the ground state,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, so that:
N0 = N −∆N .
N2 = ∆N .
For population inversion, we require N2 > N0, which implies ∆N > N/2.Population inversion is therefore only achieved when more than 50% ofthe atoms are pumped to the excited state.
In this particular example, we have N2 = 0.6N at t = 0, and we have seenabove that the laser will stop oscillating when N2 = 0.5N . The number ofatoms that make stimulated radiative transitions (and hence the numberof 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 .
Please note: students might have difficulty with the exercise if they havenot already done a separate laser physics course: the information provided
69
in the text is insufficient for them to be able to answer it properly. I shallendeavour to correct this in the second edition, should there ever be one.
(9.10) (i) The optical intensity I(t) is proportional to |E(t)|2, and hence if wehave Gaussian intensity pulse as defined in the exercise, we have a time-varying electric 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] .
(ii) 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 ,
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 .
(9.11) The crystal-field shifts of the energy levels depend on the local environ-ment of the ion. (e.g. small perturbations to the atomic positions affectthe energy levels of the ion through the alterations to the local electricfields 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.
70
(9.12) The subscript ‘g’ stands for gerade and implies even parity. A g→g transi-tion therefore involves no parity change and is forbidden for electric-dipoletransitions. (See Section B.3.) The transition must therefore take placeby a higher-order process, e.g. the magnetic-dipole or electric-quadrupoleinteraction. These have much lower probabilities than electric-dipole pro-cesses, and hence give long excited state lifetimes in the microsecond ormillisecond range. Since the lifetime is long, the radiation would be clas-sified as phosphorescence rather than fluorescence. (See Section B.3.)
(9.13) 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 300 K to allow lasing.
(9.14) The level scheme for Ti:sapphire is shown in Fig. 9.13. If we assume thatthe radiative quantum efficiency is 100% (i.e. one laser photon emittedfor each pump photon absorbed), then the ratio of the output power tothe input power would just be proportional to the ratio of the respectivephoton 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.
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:
• less than unity quantum efficiency due to significant non-radiativedecay;1
• imperfect absorption of the pump laser in the crystal;
• optical losses within the cavity.
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.
71
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 1.26 that R = 0 when n = 1, and hence εr = 1.For an undamped oscillator, εr(ν) is given by eqn 10.16. 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(i). We calculate νLO = 20THz fromthe LST relationship, and hence that the reststrahlen band runs from 9.2to 20THz, i.e. 15 µm to 33 µm.
(10.4) The exercise closely follows example 10.1(ii). 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 reststrahlen 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 1.22–23. Then from eqn 1.26 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) (i) We identify the reststrahlen band from the region of high reflectivityfrom 30–32µm. (See Fig. 29(a).) On equating the upper and lower wave-length limits with νTO and νLO respectively, we find νTO ≈ 9.5 THz andνLO ≈ 10 THz.
72
16 20 24 28 32 36 400.0
0.2
0.4
0.6
0.8
1.0R
efle
ctiv
ity
Wavelength (mm)
AlSb
nTO nLO
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
nTO nLO
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
nTO nLO
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 29: (a) Interpretation of the data in Fig. 10.14 as required for Exercise10.5(i) and (ii). (b) Calculated reflectivity versus damping constant, as requiredfor Exercise 10.5(iii).
(ii) The high and low frequency permittivities can be deduced from theasymptotic reflectivities. (See Fig. 29(a).) The low frequency limit givesεst from (see eqn 1.26, 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.
(iii) The peak reflectivity is about 90%, and is limited by γ, which inturn is determined by the lifetime of the TO phonons. The middle of thereststrahlen 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 1.22–23, and R from eqn 1.26. The reflectivity calcu-lated in this way is plotted as a function of γ in Fig. 29(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 (i) 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(iii). We first use eqn 10.18to 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 1.23 to find κ and
73
eqn 1.16 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. 29(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 10.16 to calculate εr = 21.5 at8THz. We then substitute this value of εr into eqn 10.19 to compute thewave vector. This gives:
q =√
εrω
c=√
21.5× 2π × 8× 1012
3× 108= 7.8× 105 m−1 .
(10.9) (i) The condition for cyclotron resonance is given in eqn 10.24. In a polarmaterial, the mass that is measured is the polaron mass m∗∗. We thusobtain:
m∗∗ =eλB
2πc= 0.097 m0 .
(ii) The rigid lattice mass m∗ can be calculated from eqn 10.25. Oninserting the relevant values into eqn 10.20, we find αep ≈ 0.33 for CdTe.Then from eqn 10.25 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 10.28, which gives, for thecase of the Stokes line:
Ω/2π =c
514.5 nm− c
528.6 nm= 15.5THz .
74
The relative intensities of the lines are given by eqn 10.30:
I(501.2 nm)I(528.6 nm)
= exp(
h× 15.5× 1012
kBT
)= 0.08 .
invertinvert
Figure 30: Inversion of a TO mode in an ionic crystal, as considered in Exercise10.12. The inversion centre is circled.
(10.12) NaCl has an inversion centre and so the mutual exclusion rule applies. Itis apparent from Fig. 30 that the TO mode has odd parity under inversion.The TO mode is therefore IR active but not Raman active.
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 6: Raman shifts deduced from the data in Fig. 10.11, as considered inExercise 10.13.
(10.13) The energies of the phonon modes can be deduced directly from theRaman spectra by applying eqn 10.28, 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 6, together with the energies deduced ac-cording to:
E (meV) = 0.124× Raman shift (cm−1) .
On comparing the frequencies of the TO and LO phonons of GaAs inTable 6 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 10.29 implies that momentum is conserved during the Ra-man scattering process so that the vectors form a triangle as depicted inFig. 31(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.
75
k1
k2
q
q/2k1k2
q
(a) (b)
k1
k2
q
q/2k1k2
q
(a) (b)
Figure 31: (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. 31(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 10.33 then follows by writing:
δω = |ω2 − ω1| = Ω .
In back-scattering geometry, we have θ = 180, so that q = 2k. (SeeFig. 31(b) with |k1| = |k2| = k.) It then follows from eqn 10.33 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.
(10.15) (i) The negative term in r−1 is the attractive potential due to theCoulomb interaction between the ions. The Madelung constant α ac-counts for the summation of the contributions of the positive and nega-tive ions over the whole crystal. The positive term in r−12 represents theshort range repulsive force due to the Pauli exclusion principle when theelectron wave functions overlap.
(ii) 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. 22.) 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.
(iii) 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+· · · .
76
Now U(r) is 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 10.34 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 (ii) to derive the last line. Hence:
C3 = −22αe2/3πε0r40 .
(10.16) 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).
77
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.40.(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.40 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.40that 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.
78
(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) (i) 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 N1B12uν and N2B21uν respectively (see eqns B.5–6.). If the levels arenon-degenerate, then eqn B.12 tells us that B12 = B21. At t = 0, all theatoms are in level 1, and there is net absorption, which increases N2 anddecreases N1. As the atoms are pumped to level 2, the stimulated emis-sion rate becomes increasingly significant. Eventually, we reach a stagewhere N1 = N2 = N0/2, and the stimulated emission and absorption ratesare identical. There is therefore no net absorption or emission, and N2
cannot increase further. The maximum value of N2 that can be achievedis therefore N0/2.3
(ii) If we only have two levels and we neglect spontaneous emission, thenthe rate equations for N1 and N2 are:
dN1
dt= −B12N1uν + B21N2uν ,
dN2
dt= +B12N1uν −B21N2uν .
On setting B12 = B21 as appropriate for non-degenerate levels, and sub-tracting, we find:
ddt
(N1 −N2) =d∆N
dt= −2B12uν∆N ,
where ∆N = N1 −N2. Integration yields:
∆N(t) = ∆N(0) exp(−2B12uνt) ,
which, with ∆N(0) = N0, gives the required result.
The result quantifies the way the laser pumps atoms from level 1 to level2, and hence reduces the net absorption. The equation implies that the
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.
79
populations 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 32: 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. 32.The z-component of the electric field is therefore zero. The nonlinearpolarization, for the given nonlinear optical tensor, is found from eqn 11.46to 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. 32 and that the light has an electric fieldmagnitude of 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) We consider an electro-optic crystal with axes as defined in Fig. 33. Thevoltage is applied so as to produce an electric field of magnitude Ez alongz axis.
(i) 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
80
yz
x
y¢x¢
L
V
yz
x
y¢x¢
L
V
Figure 33: Experimental geometry of the electro-optic crystal considered inExercise 11.7.
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 .
On applying this to the two components along the x′ and y′ axes, we thenhave:
∆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.
(n.b. The formula given in the text is wrong by a factor of two.)
(ii) On setting ∆Φ = π, we find:
Vπ =λ
2n30r41
.
With the appropriate figures for CdTe given in the exercise, we obtainVπ = 44 kV.
(11.8) This exercise closely follows Example 11.2, and the phase-matching angleis found by substituting the appropriate refractive indices into eqn 11.56.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.
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(11.9) In a third-order nonlinear medium, the change in the relative permittivityis given by (see eqn 11.62):
∆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 1.16 and 1.21 that ∆α ∝ ∆ε2.Hence, in a medium with Im(χ(3)) 6= 0, we expect
∆α ∝ Im(χ(3))I .
A saturable absorber has an absorption coefficient which obeys eqn 11.40.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.
(11.10) 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 11.12, 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 11.3 that χyxxx = χzxxx = 0. Hence we find
P = ε0χxxxxE3x(1, 0, 0) ,
which means that P is parallel to E.
(11.11) This exercise is very similar to Example 11.3. Form eqn 11.70, werequire that:
∆Φnonlinear =2π
λn2Il = π ,
which implies:
I =λ
2n2l=
1.55× 10−6
2 · 3× 10−20 · 10= 2.6× 1012 W m−2 .
The optical power to produce this intensity is given by:
P = IA = 2.6× 1012 × π(2.5× 10−6)2 = 50 W .
82
(11.12) It is apparent from Fig. 11.8 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(ii), 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
(11.13) 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 (i) of Example 2.1 it is shown that
κmax =Ne2
2nε0m0γω0,
where κmax is the extinction coefficient at the line centre, while in part (iii)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 1.16 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.
83
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.
(11.14) We see from eqn 4.4 that the n = 1 exciton absorption line will occurat an energy of
hν = Eg −RX .
For InP we have:
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). Hencewe see from eqn 4.1 that RX = (0.06/12.52)RH = 5.2 meV. The excitonenergy is thus 1.34 eV at low temperatures. (We consider low temperatureshere because the exciton would be ionized at room temperature.)
The saturation density for excitons is given by the Mott density of eqn 4.8.We find from eqn 4.2 that r = 11nm for the n = 1 exciton, and henceNMott ≈ 1.8 × 1023 m−3. The saturation intensity Is is the optical in-tensity required to produce this carrier density. By using the result ofExercise 5.6(ii), namely:
N =Iατ
hν,
we find
Is =hν
ατNMott =
1.34 eV106 · 10−9
× (1.8× 1023) ≈ 4× 107 W m−2 .
84