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 each chapter of the book, for the benefit of class instructors. Please note that figures within 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, which have an additional chapter label (e.g. Fig. 1.1). A similar convention applies to the 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 comments about the text and/or the Exercises. He may be contacted at the following address:
Department of Physics and Astronomy University of Sheffield Hicks Building Sheffield, S3 7RH United 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 assume α = κ = 0. The reflectivity is calculated by inserting n = 1.51 and κ = 0 into eqn 1.26 to obtain R = 0.041. The transmission is calculated from eqn 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 and dense flint glass are 1.46 and 1.746 respectively. The reflectivities are then calculated from eqn 1.26 to be 0.035 and 0.074 respectively, with κ = 0 in both cases because the glass is transparent. We thus find that the reflectivity of dense flint glass is larger than that of fused silica by a factor of 2.1. This is why cut–glass products made from dense flint glass have a sparkling appearance.
(1.3) We first use eqns 1.22 and 1.23 to convert ˜²r to ˜n, giving n = 3.01 and κ = 0.38. We then proceed as in Example 1.2. This gives:
v = c/n = 9.97 × 107m s−1, α = 4πκ/λ = 9.6 × 106m−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 the operating wavelength. With α = 1.3 × 105m−1 at 850 nm, we then find l = 1.8 × 10−5m = 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 very high: see Exercise 1.9.)
(1.6) 99.8% absorption in 10 m means exp(−αl) = 0.002, and hence α = 0.62 m−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 ˜²
r are found from eqns 1.20 and 1.21 respectively, and thus we obtain ˜²r= 1.77+i 9.2×10−8. (1.7) The filter appears yellow and so it must transmit red and green light, but not blue. The filter must therefore have absorption at blue wavelengths. (1.8) (i) The beam that has suffered no reflections from the back surface passes
through the first interface, then propagates through the material, and finally passes through the second interface. Its intensity is thus given by:
The intensity of the beam that is reflected once from the back surface is found by following its path through the medium as it is reflected off the front and back surfaces and propagates through the medium. The intensity 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 R2exp(−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 the ratio 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 [R2exp(−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 reflected beams. The plate would then behave as a Fabry–Perot etalon. (See optics texts e.g. Hecht (1998) for further details.) The effects of multiple beam interference are a nuisance when it comes to extracting reliable values of the optical constants, but their effects can be assumed to be small in limit where the reflectivity is small (as in this Exercise), and when the sample is strongly absorbing (as in the next).
(1.9) In Exercise 1.5 we worked out R = 0.328. With α = 1.3 × 106m−1 and l = 2 × 10−6m, the intensity ratio is equal to R2exp(−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 log10x = logex/ loge10): − log10(T ) = −2 log10(1 − R) + αl/ loge10 .
We can then substitute αl/ loge10 from eqn 1.8 to obtain the required result.
If the medium is transparent at λ0 then we will have that T
λ0 = (1 − R)2, where R is the reflectivity at λ0. We assume that the reflectivity
varies only weakly with wavelength. This is a reasonable assumption for most materials if we choose λ0 sensibly. For example, we would choose λ0 just above the absorption edge we are trying to measure. With this
assumption, the optical density at λ is then given by O.D.(λ) = − log10(Tλ) + log10(Tλ0) .
Measurements of T (λ) and T (λ0) thus allow the optical density to be
determined. If l is known, the absorption coefficient can then be found from eqn 1.8.
(1.11) The propagation time is equal to l/v = ln/c. The time difference is thus l(n1− n2)/c = 27 ns. The pulses at the wavelength with the smaller refractive index (i.e. 1500 nm) take the shorter time.
(1.12) With σ = 6.6 × 107Ω−1m, and ω = 1.88 × 1013rad/s, we find ˜² r = ²r+ 3.97 × 105i. The imaginary part of ˜²r is very large, and hence the
approximation ²2 À ²1 is a good one. In this approximation, we have (with√i = (1 + i)/√2):
˜
n =p˜²r= (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 the
previous exercise. With ω = 1.88 × 1013 rad/s, we find ˜ n2= ˜² r= ²1+ 2.94 × 105i , which implies: ˜ n =p˜²r= (2.94 × 105)1/2√i = 383(1 + i) .
We then find from eqn 1.16 that α = 4πκ/λ = 4.8 × 107m−1. Beer’s law
means that we set exp(−αl) = 0.5 for a drop in intensity by a factor of 2, giving l = 1.4 × 10−8m = 14 nm.
(1.14) It is apparent from eqn 1.26 that R = 1 when n = 1 and κ = 0. For zero reflectivity we thus require ˜²r= (n + iκ)2= 1.
(1.15) (i) We convert wavelengths to photon energies using E = hc/λ to obtain the energy level scheme shown in Fig. 1 of this solutions manual. It is thus apparent 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 that absorbed 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 dissipated power is 2 W.
(iii) For a luminescent quantum efficiency of 50% the number of photons emitted drops by a factor of 2 compared to part (ii), and so the light power emitted falls to 4 W. The remaining 6 W of the absorbed power is dissipated 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 in Section 10.5. Conservation of energy in the scattering process is satisfied when
hνout= hνin− hνphonon. With ν = c/λ, we then find λout= 521 nm.
(1.17) The transmission is given by eqn 1.9, with the wavelength dependence of the scattering cross section given by eqn 1.10. At 850 nm we have 10% transmission, so that N σsl = 2.30. The scattering cross-section is 11.1 times larger at 850 nm than at 1550 nm, and so we have N σsl = 2.30/11.1 = 0.21 at the longer wavelength, implying a transmission of 81 %.
In general, the scattering losses decrease as the wavelength increases, and hence the propagation losses decrease. Longer wavelengths are therefore preferable for long range communication systems. At the same time, the fibres start to absorb in the infrared due to phonon absorption. 1550 nm is the longest practical wavelength for silica fibres before phonon absorption becomes significant.
(1.18) We again use eqn 1.9 to calculate the transmission, setting exp(−N σsl) = 0.5. This gives N σsl = 0.69, which implies l = 3.5 m for the given values on N and σs.
If the wavelength is reduced by a factor of two, Rayleigh’s scattering law (eqn 1.10) implies that σsincreases by a factor of 16. The length required for the same transmission is thus smaller by a factor of 16: i.e. l = 3.5/16 = 0.22 m.
(1.19) Birefringence is an example of optical anisotropy as discussed in Sec-tion 1.5.1, and also in SecSec-tion 2.4. Ice is a uniaxial crystal, and therefore has preferential axes, making optical anisotropy possible. Water, by con-trast, is a liquid and has no preferential axes. The optical properties must therefore be isotropic, making birefringence impossible.
Chapter 2
Classical propagation
(2.1) We envisage two displaced masses as shown in Fig. 2. The spring is extended by a distance (x1− x2) and so the force on the masses are ±Ks(x1− x2). The equations of motion are therefore
m1d 2x 1 dt2 = −Ks(x1− x2) and m2 d2x 2 dt2 = −Ks(x2− x1) .
Divide the equations by m1 and m2 respectively and subtract them to obtain: d2 dt2(x1− x2) = −Ks µ 1 m1 + 1 m2 ¶ (x1− x2) .
On defining the relative displacement x = x1− x2 and introducing the reduced mass µ, where 1/µ = 1/m1+ 1/m2, we then have:
µd 2x
dt2 = −Ksx .
This is the equation of motion of an oscillator of frequency (Ks/µ)1/2.
m
1m
2x
1x
2rest
displaced
m
1m
2x
1x
2rest
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 therefore
write the force as the real part of F0e−iωt, and look for solutions of the
form x(t) = x0e−iωt. On substituting into the equation of motion we then
obtain:
m(−ω2− iωγ + ω2
0)x0e−iωt= F0e−iωt, which implies: x(t) = F0 m 1 ω2 0− ω2− iωγ e−iωt.
The phase factor comes from the middle term, namely [ω2
0− ω2− iωγ]−1. On multiplying top and bottom by the complex conjugate, we find:
1 ω2 0− ω2− iωγ = (ω 2 0− ω2) + iωγ (ω2 0− 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 (ω2
0− 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/(ω02− ω2)] .
(2.3) By applying the Lorentz oscillator model of Section 2.2.1, we realize that the refractive index will have a frequency dependence as shown in Fig. 2.4, with ω0 corresponding to 500 nm. (i.e. ω0 = 3.8 × 1015rad/s.) For frequencies well above the resonance, we will just have the contribution of the undoped sapphire crystal:
n∞≡ n(ω À ω0) = 1.77 ,
which implies ²∞= (1.77)2. The refractive index well below the resonance
can be worked out from eqn 2.19. Using the value of N given in the question, we find ²st− ²∞= 2.23 × 10−3. We thus have:
nst = √
²st= [(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 Exercise is similar to Example 2.1, because we are dealing with a relatively small number of absorbers and the overall refractive index will be dominated by the host crystal. We can therefore assume n = 1.39 throughout the Exercise. On the other hand, the host crystal is transparent at 405 nm, and so the absorption will be determined by the impurity atoms. The other factor 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) =²2(ω0) 2n = N e2 2n²0m0 1 γω0 × f ,
where f = 9 × 10−5 is the oscillator strength. For the absorption line
we have ω0 = 2πc/405 nm = 4.65 × 1015rad/s, and γ = ∆ω = 2π∆ν = 5.15 × 1014s−1. With N = 2 × 1026m−3 and n = 1.39, we then find κ(ω0) = 8.6 × 10−6. We finally obtain the absorption at the line centre
(405 nm) from eqn 1.16 as 270 m−1.
This Exercise is broadly based on the results presented in the paper by Iverson and Sibley in J. Luminescence 20, 311 (1979).
(2.5) The result of this Exercise works in the limit where the contribution of the particular oscillator to the dielectric constant is relatively small, as in Example 2.1 and the previous exercise. In this limit we have ²2¿ ²1, and therefore κ(ω0) = ²2(ω0)/2n, where n =√²1, and ²1(ω0) = 1+χ. We then see from eqn 2.16 that ²2(ω0) = N e2/²
0m0γω0, so that the absorption is (cf. eqn 1.16): α(ω0) = 4πκ(ω0) λ = 4π × ²2(ω0) 2n ÷ (2πc/ω0) = N e2 n²0m0γc.
This shows that it is the linewidth that determines the peak absorption strength 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√²st. With n(ω ¿ ω0) = 2.43 from the data, we find ²st= 5.9.
(ii) The resonant frequency is the mid point of the “wiggle”, i.e. 5.0 × 1012Hz.
(iii) The natural frequency is given by eqn 2.2, which implies Ks = µω2 0. 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 × 10−26kg. With ω0 = 2πν0 = 3.1 ×
1013rad/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. ²st = 5.9 has been found in part (i), and ²∞ can be read from the graph as ²∞ = [n(ω À ω0)]2= (1.45)2= 2.10. We thus have ²st− ²
∞= 3.8. Using the values of ω0 and µ worked out previously, we then find N = 3.0 × 1028m−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 resonance line is asymmetric. By comparison with Fig. 2.4 we find ∆ν ∼ 1×1012Hz, and hence γ = 2π∆ν ∼ 6 × 1012s−1.
(vi) The result of Exercise 2.5 tells us that α = N e2/n²
0µγc at the line centre for a weak absorber. This limit does not really apply here, but we can still use it to get a rough answer. On inserting the values of N , µ and γ found above, and taking n ∼ 2, we find α ∼ 1 × 106m−1.
(2.7) Use ω = ck/n in eqn 2.25 to obtain: vg= dω dk = c d dk(k/n) = c n− ck n2 dn dk . Then substitute v = c/n to obtain eqn 2.26.
(2.8) We consider three separate frequency regions.
(i) ω < ω0: In this frequency region ²ris real, and increases with frequency. Since n =√²r, it is apparent that dn/dω is positive, so that from eqn 2.26 we see that vg < v because k follows ω. Since ²r> 1, n > 1, and hence v = c/n < c. Therefore vg< c.
(ii) ω0< ω < (ω2
0+N e2/²0m0)1/2: In this frequency region, ²ris 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 > (ω2
0+ N e2/²0m0)1/2: In this region ²r 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) because n < 1 and therefore v > c. We must therefore work out vg explicitly using eqn 2.25. It is easier to take dn/dω than dn/dk, and we can use k = nω/c to find vg from: 1 vg = dk dω = n c + ω c dn dω. With n =√²r, we obtain: dn dω = d dω µ 1 + N e2 ²0m0 1 ω2 0− ω2 ¶1/2 = 1 n N e2 ²0m0 ω (ω2 0− ω2)2 . Hence 1 vg = n c + N e2 nc²0m0 ω2 (ω2 0− ω2)2 , = 1 nc µ n2+ N e2 ²0m0 ω2 (ω2 0− ω2)2 ¶ , = 1 nc µ ²r+ N e 2 ²0m0 ω2 (ω2 0− ω2)2 ¶ , = 1 nc µ 1 + N e2 ²0m0 ω2 0 (ω2 0− ω2)2 ¶ . Hence we find: vg= nc µ 1 + N e 2 ²0m0 ω2 0 (ω2 0− 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 generated at position vector r is given by:
E(r) = 3(p · r)r − r 2p 4π²0r5 .
The electric field generated at the origin by a dipole p at position vector r is therefore given by:
E(−r) = 3(p · (−r))(−r) − r 2p
4π²0r5 =
3(p · r)r − r2p 4π²0r5 .
Consider the ith dipole within the sphere illustrated in Fig. 2.8. We assume that the dipole is oriented parallel to the z axis so that we can write pi = (0, 0, pi). Then, on writing ri = (xi, yi, zi) in the formula for E, we find that the z component of the field at the origin from the ith dipole is: Ei=pi(3z 2 i − ri2) 4π²0r5 i .
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 along the z axis of magnitude:
Esphere= 1 4π²0 X i pi3z 2 i − r2i r5 i , as required.
(ii) If all the dipoles have the same magnitude p, then the resultant field is given by: Esphere = p 4π²0 X i 2z2 i − x2i − yi2 r5 i .
The x, y and z axes are equivalent for the cubic lattice within the sphere, and so we must have:
X i x2 i r5 i =X i y2 i r5 i =X i z2 i r5 i . It is thus apparent that
X i 2z2 i − x2i − y2i r5 i = 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 the polarization is parallel to the z axis. The surface charge on the sphere must balance the normal component of the polarization P . With P = (0, 0, P ), the normal component at polar angle θ is equal P cos θ, as shown in Fig. 3. Hence the surface charge density σ at angle θ is equal to −P cos θ. The charge 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 · a dθ) = −2πP a2cos 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
4π²0a2cos θ = +
P cos2θ sin θ dθ
On integrating over θ, we then obtain: Ez= Z π θ=0 dEz= P 2²0 Z π 0 cos2θ sin θ dθ = P 3²0.
Since P is parallel to the z axis, and the x and y components of E are zero, we therefore have:
E = P
3²0, as required for eqn 2.28.
+ + + + + ++
-
--
-P
q
dq
a
dqP cosq
+ + + + + ++ + + + + + ++-
--
-P
q
dq
a
dqP cosq
Figure 3: Definition of angles and charge increment as required for Exercise 2.9(iii).
(2.10) If ²r− 1 is small, the left hand side of the Clausius–Mossotti relationship becomes equal to (²r− 1)/3, and we then find:
²r= 1 + N χa≡ 1 + χ ,
where χ = N χa, as in eqn A.4. It is apparent that ²r− 1 will be small if either N is small or χa is small. This means that we either have a low density of absorbing atoms (as in a gas, for example), or we are working at 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 a volume of 22.4 litres. Hence N = 2.69 × 1025m−3. We then find χafrom:
χa= (²r− 1)/N = 2.2 × 10−29m3. The atomic dipole is worked out from
p = ²0χaE .
The displacement of an electron by 1˚A produces a dipole of 1.6×10−29C m.
Hence we require a field of 0.8×1011 V/m. The field acting on an electron at a distance r from a proton is given by Coulomb’s law as:
E = e
On substituting r = 1 ˚A, we find E = 1.4 × 1011 V/m. It is not surprising that these two fields are of similar magnitude because the external field must 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 the
damping term, and write eqn 2.24 as: ²r≡ n2= 1 + N e 2 ²0m0 X j fj ω2 0j− ω2 . On substituting ω = 2πc/λ, this becomes:
n2= 1 + N e2 ²0m0 1 (2πc)2 X j fjλ2jλ2 λ2− λ2 j , where λj = 2πc/ω0j. This is of the Sellmeier form if we take:
Aj = N e2fjλ2j/4π2²0m0c2.
(b) With the approximations stated in the exercise, we have: n2= 1 + A1λ 2 λ2− λ2 1 = 1 + A1(1 − λ21/λ2)−1. With x ≡ λ2
1/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 + A1x + µ A1 2√1 + A1 − A2 1 8(1 + A1)3/2 ¶ x2+ · · · On re-substituting for x, we then find:
n = (1 + A1)1/2+ A1 2√1 + A1 λ2 1 λ2 + µ A1 2√1 + A1 − A2 1 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 dispersion is dominated by the electronic absorption in the ultraviolet. We should then take λ1 as the wavelength of the band gap, and the approximation λ2
1/λ2 ¿ 1 will be reasonable, as we are far away from the band gap energy.
(2.13) (i) On neglecting the λ4term 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 find C1= 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 have θ1+ θ2= α. With θin= 45◦ and α = 60◦, we then find θout = 57.17◦ for n = 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 µ n + ωdn 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 µ n − λdn dλ ¶ .
The difference in the transit time for two wavelengths separated by ∆λ, where ∆λ ¿ λ, is given by:
∆τ = dτ dλ∆λ .
On using the result above, we find dτ dλ = − L c λ d2n dλ2, which implies: |∆τ | = L c λ d2n dλ2 ∆λ = L c λ 2d2n dλ2 ∆λ λ , as required.
The time–bandwidth product of eqn 2.38 implies that a pulse of light contains a spread of frequencies and therefore a spread of wavelengths. In a dispersive medium, the different wavelengths will travel at different velocities, and this will cause pulse broadening. With λ = c/ν, we have |∆λ| = (λ2/c)∆ν, so that: |∆τ | = L c µ λ2d2n dλ2 ¶ λ c ∆ν .
The precise amount of broadening depends on the numerical value of the time–bandwidth product assumed for the pulse. For a 1 ps pulse with ∆ν∆t = 1, we have ∆ν = 1012Hz, and hence |∆τ | = 0.17 ps for the parameters 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 the optic (z) axis of a uniaxial crystal, as required for Exercise 2.15.
(2.15) It is apparent from eqn 2.46 that ²11= ²22= n2
oand ²33= n2e. Hence we can re-cast the index ellipsoid in the form:
x2 n2 o +y 2 n2 o +z 2 n2 e = 1 .
Owing to the spherical symmetry about the optic (z) axis, we can choose the 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 popo-larization of the
e–ray will then lie in the y–z plane, as in Fig. 2.12(b). The projection of the index ellipsoid onto the plane that contains the direction of propaga-tion and the polarizapropaga-tion vector of the e–ray thus appears as the ellipse drawn in Fig. 5. The refractive index n(θ) that we require is the distance from the origin to the point of the ellipse where the E-vector cuts it. The co-ordinates of this point are x = 0, y = n(θ) cos θ and z = n(θ) sin θ. On substituting into the equation of the index ellipsoid, we then have:
0 + n(θ) 2cos2θ n2 o +n(θ) 2sin2θ n2 e = 1 , which implies: 1 n(θ)2 = cos2θ n2 o +sin 2θ n2 e = 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 light propagating 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, as shown in Fig. 6, and assume that the light is propagating along the x direction. Light with its polarization vector at 45◦ to the optic axis will
then have its E-vector as shown in the figure. We resolve the E-vector into two components of equal amplitude, one along the z axis and the other along the y axis. The components along the z and y axes experience refractive indices of ne and norespectively. The phase difference between the 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 vacuum wavelength of the light.
The wave plate operates at a quarter wave plate when ∆φ = π/2. In this situation, the output is two orthogonally polarized waves of equal amplitude but with a π/2 phase difference between them, i.e. circularly polarized light. The condition for this is L = λ/4∆n, which is equal to 14 µm for the parameters given.
(2.17) The crystal will be isotropic if the medium has high symmetry so that the x, y and z axes are equivalent. If not, it will be birefringent. For the
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 the direction perpendicular to the hexagons. Sulphur has the lowest symmetry and is the only biaxial crystal included in the list.
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 nzare 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 allowed k-state occupies a volume of k-space equal to (2π/L)3. This implies that the number of states in a unit volume of k-space is L3/(2π)3. Hence a unit volume of the material would have 1/(2π)3states per unit volume of k-space.
The density of states in k-space is found by calculating the number of states within the incremental shell between k-vectors of magnitude k and k + dk, as shown in Fig. 7. This volume of the incremental shell is equal to 4πk2dk, and contains 4πk2dk × 1/(2π)3= k2dk/2π2states, as given in eqn 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 volume per 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 also shown.
(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) = 2k 2/2π2 ~2k/m∗ = m∗k π2~2 = m∗ π2~2 µ 2m∗E ~2 ¶1/2 = 1 2π2 µ 2m∗ ~2 ¶3/2 E1/2, as required.
(3.3) (i) The parity of a wave function is equal to ±1 depending on whether ψ(−r) = ±ψ(r). Atoms are spherically symmetric, and so measurable properties such as the probability amplitude must possess inversion sym-metry about the origin: i.e. |ψ(−r)|2 = |ψ(r)|2. This is satisfied if ψ(−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 zero unless the product ψ∗
fψiis also an odd function. This condition is satisfied if the two wave functions have different parities (parity selection rule). Since the wave function parity is equal to (−1)l, the parity selection rule implies that ∆l is an odd number.
(iii) In spherical polar co-ordinates (r, θ, φ) we have: x = r sin θ cos φ = r sin θ (eiφ+ e−iφ)/2 , y = r sin θ sin φ = r sin θ (eiφ− e−iφ)/2i , z = r cos θ .
The selection rules on m can be derived by considering the integral over φ. For light polarized along the z axis we have:
M ∝ Z 2π
φ=0
eim0φ· 1 · eimφdφ ,
since z is independent of φ. The integral is zero unless m0 = m. The
selection rule for z-polarized light is therefore ∆m = 0. For x or y polarized light we have:
M ∝ Z 2π
φ=0
eim0φ(eiφ± e−iφ) eimφdφ ,
which is zero unless m0 = m ± 1. We thus have the selection rule ∆m = ±1 for light linearly polarized along the x or y axis. With circularly polarized light we have ∆m = +1 for σ+ polarization and ∆m = −1 for σ− polarization.
(3.4) The apparatus required is basically the same as for Fig. 3.13, but with modifications to take account of the fact that the required energy range of 0.3–0.6 eV corresponds to a wavelength range of 2–4 µm. This is below the band gap of silicon, and the spectrograph/silicon CCD array arrangement
shown in Fig. 3.13 is not appropriate. Instead, a detector with a band gap smaller than 0.3 eV must be used, e.g. InSb. (See Table 3.2.) As InSb array detectors are not available, a scanning monochromator with a single channel detector would normally be used. A thermal source would suffice as the light source.
Another point to consider is that standard glass lenses do not transmit in this wavelength range, and appropriate infrared lenses would have to be used, e.g. made from CdSe. (See Fig. 1.4(b).) Also, since the data is taken 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 detectorFigure 8: Apparatus for measuring infrared absorption spectra in the range 2–4 µm, as discussed in Exercise 3.4.
(3.5) The type of band gap can be determined from an analysis of the variation of the absorption coefficient α with photon energy. The material is direct or indirect depending on whether a graph of α2 or α1/2 against ~ω is a straight line. Other factors to consider are that the absorption is much stronger in a direct-gap material, and that the temperature dependence of α is expected to be different. In an indirect gap material, phonon-assisted absorption mechanisms will freeze out as the temperature is lowered. (3.6) Plot α2and α1/2against ~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. We thus deduce that GaP has an indirect band gap at 2.2 eV. For ~ω > 2.7 eV, the graph of α1/2is a straight line with an intercept at ∼ 2.75 eV. (Note the huge difference in the two axis scales, which is a further indication of the indirect nature of the transitions below 2.75 eV, and their direct nature above 2.75 eV.) We thus deduce that GaP has a direct gap at 2.75 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. This is above the direct gap of germanium at 0.80 eV, and thus the absorption will be given by (cf. eqn 3.25):
2.2 2.4 2.6 2.8 3.0 0 20 40 60 80 a 2(10 12 m -2) Energy (eV) 0 200 400 600 800 1000 a 1/ 2(m -1/2 ) 2.2 2.4 2.6 2.8 3.0 0 20 40 60 80 a 2(10 12 m -2) Energy (eV) 0 200 400 600 800 1000 a 1/ 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 × 1012m−2 at 0.90 eV implies C ≈ 2.2 × 106m−1eV−1/2. With
this value of C, we then find α ≈ 1 × 106m−1 at 1.02 eV.
(3.8) (i) We consider transitions 1 and 2 in Fig. 3.5. The k vectors for the transitions can be worked out from eqn 3.23. The appropriate parameters are 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 × 108m−1 for the heavy holes
and k = 4.1 × 108m−1 for the light holes.
(ii) The air wavelength λ of the photon is 775 nm. The wavelength inside the crystal is reduced by a factor n. The photon wave vector inside the crystal is therefore given by:
k = 2π
(λ/n) = 3.0 × 10 7m−1.
This is more than an order of magnitude smaller than the electron wave vector, and hence the approximation in eqn 3.12 is justified.
(iii) Equation 3.24 shows that the joint density of states is proportional to µ3/2. Hence the ratio of the joint density of states for heavy and light hole 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–off hole transition occurs at ~ω = Eg+ ∆. Reading a value of ∆ = 0.34 eV from 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 = 3/2 conduction bands states are degenerate at k = 0, but not for finite k.
Electric-dipole transitions from the p-like valence band states are forbidden to these p-like conduction band states at k = 0. The first dipole-allowed transition is to the s-like antibonding state at ∼ 4.1 eV. Hence the direct gap at the Γ point is equal to 4.1 eV.
(ii) The discussion of the atomic character of bands given in Section 3.3.1 only applies at the Γ point where k = 0 and we are considering stationary states. This means that electric-dipole transitions can be allowed at the zone edges, even though they are forbidden at k = 0.
(3.10) At low temperatures, phonon absorption is impossible, and the indirect transition must proceed by phonon emission, with a threshold energy of Eind
g + ~Ω. The band structure diagram of germanium given in Fig. 3.9 shows that the indirect gap occurs at the L-point of the Brillouin zone. We therefore need a phonon with a wave vector equal to the k vector at the L-point. The energies of these phonons are given in Table 3.1. The lowest energy energy phonon is the TA phonon with an energy of 0.008 eV. The absorption 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. The absorption decreases exponentially for ~ω < Eg, and this produces an ex-ponential absorption tail below Eg. Although there is no clear absorption edge as for the case at zero field, a reasonable point to take is when the absorption has decayed by a factor e−1 from its value at E
g. This occurs when 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 × 106V/m. v B r w F v B r w F
Figure 10: Force acting on a particle moving in a magnetic field pointing into the 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 a magnetic field B. The particle experiences the Lorentz force F = qv × B which is at right angles both to the velocity and the field, as shown in Fig. 10. This perpendicular force produces circular motion at angular
frequency ω with radius r. We equate the central force with the Lorentz force 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-y plane is quantized, but the motion in the z direction is free. We have seen above that, in the classical analysis, the field causes circular motion. The quantized motion will therefore correspond to a quantum harmonic oscillator. These quantized states are called Landau levels. The Bloch wave 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 number of the Landau level. The selection rule for transitions between the Landau levels can be deduced by repeating the derivation in Section 3.2 with the modified wave functions. For x-polarized light, the matrix element is now given by: M ∝ Z u∗ f(r)ϕn0(x, y)e−ik 0 zzx u∗ i(r)ϕn(x, y)e−ikzzd3r , ∝ Z unit cell u∗ f(r) x u∗i(r) d3r × hϕn0|ϕni , where we assumed k0
z= kzas usual in the second line. The electric–dipole
matrix element is therefore proportional to the overlap of harmonic oscil-lator wave functions with different values of n. Now harmonic osciloscil-lator functions form an orthonormal set, and so the wave functions of differing n are orthogonal. Hence the matrix element is zero unless n0 = 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 of states in k-space is 1/2π per unit length of material. For free motion in the z direction we have E = ~2k2/2m for a particle of mass m, which implies:
dE/dk = ~2k/m = ~p2E/m .
The density of states in energy space is then worked out from eqn 3.14: g1D(E) = 2 g(k)
dE/dk = 2 1/2π
~p2E/m = (2m/Eh 2)−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
0 2 4 6 8 10 0 2 4 6 A bs o rpt io n (a .u.)
Energy relative to Egin arb. units
(ii) (iii) 0 1 2 3 0 2 4 6 A bs o rpt io n (a .u.)
Energy in units of hwLrelative to Eg 0 2 4 6 8 10 0 2 4 6 A bs o rpt io n (a .u.)
Energy relative to Egin arb. units 0 2 4 6 8 10 0 2 4 6 A bs o rpt io n (a .u.)
Energy relative to Egin arb. units
(ii) (iii) 0 1 2 3 0 2 4 6 A bs o rpt io n (a .u.)
Energy in units of hwLrelative to Eg 0 1 2 3 0 2 4 6 A bs o rpt io n (a .u.)
Energy in units of hwLrelative to Eg
Figure 11: (ii) Absorption of a one-dimensional semiconductor as discussed in Exercise 3.13. (iii) Absorption for a system with quantized landau levels in two directions and free motion in the third. ωL is the Landau level angular frequency.
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 give Landau levels, leaving the particle free to move in the third dimension. Optical transitions are possible between Landau levels with the same value of n. (See Exercise 3.12.) These will occur at energies given by (c.f. eqn 3.32):
En= Eg+ (n + 1/2)~ωL, where
ωL = eB/m∗e+ eB/m∗h= eB/µ .
Each Landau level transition has a 1-D density of states due to the free motion 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 Landau level transition together, as shown in Fig. 11(iii).
When comparing to the experimental data in Fig. 3.7, we expect α(~ω) to diverge each time the frequency crosses the threshold for a new value of n. These divergences are broadened by scattering. We therefore see dips in 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 and 0.844 eV, with an average separation of 0.012 eV. We equate this separation energy to e~B/µ, and hence find µ = 0.035me. If m∗
hÀ m∗e, we will have µ = m∗
e, and hence we deduce m∗e ≈ 0.035me. The lowest energy transition occurs 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 to be observed compared to the direct transitions above 0.80 eV.
(3.14) The responsivity is calculated using eqn 3.38. The device will be most efficient if it has 100% quantum efficiency, and so we set η = 1, giving:
Responsivitymax= e ~ω(1 − e
On inserting the values given in the Exercise, we find responsivities of 0.46 A/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- and n-regions are good conductors, whereas the i-region is depleted of free carriers and therefore acts like an insulator. We thus have two parallel conducting sheets separated by a dielectric medium, as in a parallel-plate capacitor.
(ii) The capacitance of a parallel-plate capacitor of area A, permittivity ²r²0 and plate separation d is given by:
C =A²r²0
d .
In applying this formula to a p-n junction, we should use the depletion region thickness for d. In the case of a p-i-n diode, we assume that the depletion lengths in the highly doped p- and n-regions are much smaller than the i-region thickness, so that can set d = li. We then find C = 10 pF for a silicon p-i-n diode with A = 10−6m2, ²r= 11.9 and li= 10−5m.
(iii) The electric field for an applied bias of −10 V can be calculated from eqn D.3 as:
E = 1.1 − (−10)
10−5 = 1.1 × 10
6V/m .
The electron and hole velocities can be calculated from the field and the respective mobilities:
v = µE .
This gives v = 1.7 × 105 m/s for the electrons and v = 5 × 104 m/s for the holes. The drift time is finally calculated from t = li/v, which gives 60 ps for the electrons and 200 ps for the holes.
Note that velocity saturation effects have been neglected here. The linear relationship between the velocity and field breaks down at high fields, and the velocity approaches a limiting velocity called the saturation drift velocity. The field in this example is quite large, and the transit times will actually be slightly longer than those calculated from the mobility due to the saturation of the velocity.
(iv) With R = 50 Ω and C = 10 pF, we find RC = 500 ps. To obtain the same transit time, we need a velocity of
v = li/t = 10−5/5 × 10−10= 2 × 104m/s .
This velocity occurs for an electric field of v/µe = 1.3 × 105V/m. We finally find the voltage from eqn D.3:
E = 1.3 × 105= |1.1 − V | 10−5 ,
which gives V = −0.2 V. We therefore need to apply a reverse bias of 0.2 V.
The point about this last part of the question is to make the students think about the factors that limit the response time of the photodetector.
In most situations, the time constant will be capacitance-limited because the transit time is much shorter than the RC time constant. It is only in small-area low-capacitance devices that we need to worry about the drift transit time.
Chapter 4
Excitons
(4.1) The Hamiltonian for the hydrogen atom has three terms corresponding to the kinetic energies of the proton and electron, and the Coulomb attraction between them. On writing the position vectors of the electron and proton as r1 and r2, the Hamiltonian thus takes the form:
ˆ H = − ~2 2m1∇ 2 1− ~2 2m2∇ 2 2− e2 4π²0|r1− r2| , where m1= me, m2= mp, and
∇2 i = ∂2 ∂x2 i + ∂2 ∂y2 i + ∂2 ∂z2 i .
We introduce the relative co-ordinate r and the centre of mass co-ordinate R 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 ∂x2 1 = ∂ 2ψ ∂x2 + 2m1 m1+ m2 ∂2ψ ∂x∂X + µ m1 m1+ m2 ¶2 ∂2ψ ∂X2 ∂2ψ ∂x2 1 = ∂ 2ψ ∂x2 − 2m2 m1+ m2 ∂2ψ ∂x∂X + µ m2 m1+ m2 ¶2 ∂2ψ ∂X2. It is then apparent that:
1 m1 ∂2ψ ∂x2 1 + 1 m2 ∂2ψ ∂x2 2 = µ 1 m1 + 1 m2 ¶ ∂2ψ ∂x2 + 1 m1+ m2 ∂2ψ ∂X2. Similar results may be derived for the other co-ordinates, so that we have:
1 m1 ∇2 1+ 1 m2 ∇2 2= µ 1 m1 + 1 m2 ¶ ∇2 r+ 1 m1+ m2 ∇2 R
On introducing the total mass M and reduced mass m according to: M = m1+ m2, 1 m = 1 m1 + 1 m2 , and substituting into the Hamiltonian, we then find
ˆ H = − ~ 2 2M∇ 2 R− ~2 2m∇ 2 r− e2 4π²0|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:
ˆ
H = ˆHwhole atom+ ˆHrelative, where: ˆ Hwhole atom = − ~ 2 2M∇ 2 R ˆ Hrelative = − ~2 2m∇ 2 r− e2 4π²0|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 atom as 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 correction does not make much difference for hydrogen itself, where m2 À m1 and hence m ≈ m1, but it is very important for excitons, where the electron and hole masses are typically of the same order of magnitude.
(4.2) (i) The Hamiltonian describes the relative motion of the electron and hole as they experience their mutual Coulomb attraction within the semicon-ductor. The kinetic energy of the exciton as a whole is not included. As explained 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 relative to the hole. The first term represents the kinetic energy, and the second is the Coulomb potential. The inclusion of ²rin the Coulomb terms accounts for the relative permittivity of the semiconductor.
(ii) We substitute Ψ into the Schr¨odinger equation with the ∇2operator written in spherical polar co-ordinates:
∇2= 1 r2 ∂ ∂r µ r2 ∂ ∂r ¶ + 1 r2sin θ ∂ ∂θ µ sin θ ∂ ∂θ ¶ + 1 r2sin2θ ∂2 ∂φ2.
Since Ψ does not depend on θ or φ, the Schr¨odinger equation becomes: −~2 2µ 1 r2 ∂ ∂r µ r2∂Ψ ∂r ¶ − e2 4π²0²rr Ψ = E Ψ . On substituting Ψ = C exp(−r/a0), we obtain:
µ − ~2 2µa2 0 + ~2 µa0r − e2 4π²0²rr ¶ Ψ = E Ψ . The wave function is therefore a solution if
+ ~ 2 µa0r− e2 4π²0²rr = 0 , which implies a0=4π²0²r~ 2 µe2 ≡ ²r m0 µ aH,
where aH is the hydrogen Bohr radius. With this value of a0, we then find: E = − ~ 2 2µa2 0 = − µe 4 8²2 r²20h2 ≡ − µ m0 1 ²2 r RH, where RH is the hydrogen Rydberg energy.
The normalization constant is found by solving: Z ∞ r=0 Z π θ=0 Z 2π φ=0 Ψ∗Ψ r2sin θ drdθdφ = 1 . This gives: 4πC2 Z ∞ r=0 r2e−2r/a0 dr = 4πC2×a 3 0 4 = 1 , which implies: C = µ 1 πa3 0 ¶1/2 .
In the language of atomic physics, the wave function considered here is in fact a 1s state.
(4.3) The radial probability density P (r) is proportional to r2|R(r)|2, where R(r) is the radial part of the wave function. For the 1s wave function of Exercise 4.2 we then have:
P (r) ∝ r2e−2r/a0.
On differentiating, we find that this peaks at a0. The expectation value of r is found from hri = Z ∞ r=0 Z π θ=0 Z 2π φ=0 Ψ∗rΨ r2sin θ drdθdφ , = 4π µ 1 πa3 0 ¶ Z ∞ r=0 r3e−2r/a0 dr , = (3/2)a0.
The peak of the probability density and the expectation value thus differ by a factor of 3/2.
(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 us later for obtaining an estimate of the wave function and binding energy of the excitons in quantum wells. (See Exercise 6.9.)
(i) We require a spherically symmetric wave function with a functional forms that makes the probability density peak at some finite radius and then decay to zero for large values of r. The given wave function satisfies these criteria. It is actually a correctly normalized 1s-like atomic wave function, but with a variable radius parameter ξ.
(ii) We substitute Ψ into the Hamiltonian with the ∇2written in spherical polar co-ordinates, as in Exercise 4.2. Since Ψ again depends only on r, this gives: ˆ HΨ = −~ 2 2µ 1 r2 ∂ ∂r µ r2∂Ψ ∂r ¶ − e 2 4π²0²rr Ψ . On evaluating the derivatives, we find:
ˆ HΨ = · − ~ 2 2µξ2 + µ ~2 µξ − e2 4π²0²r ¶ 1 r ¸ µ 1 πξ3 ¶1/2 e−r/ξ.
The expectation value is then given by: hEi = Z ∞ r=0 Z π θ=0 Z 2π φ=0 Ψ∗HΨ rˆ 2sin θ drdθdφ = 4π µ 1 πξ3 ¶ Z ∞ r=0 · − ~ 2 2µξ2r 2+ µ ~2 µξ− e2 4π²0²r ¶ r ¸ e−2r/ξdr = ~ 2 2µξ2 − e2 4π²r²0ξ.
(iii) On differentiating hEi with respect to ξ, we find a minimum when ξ = 4π²0²r~2/µe2. The value of hEi at this minimum is hEi = −µe4/8h2²2
0²2r. (iv) The value of ξ that minimizes hEi is equal to a0, and the minimum value of hEi is the same energy as that found in Exercise 4.2.
The variational method gives exactly the right energy and wave function here because our ‘guess’ wave function had exactly the right functional form. In other situations, this will not be the case, and the energy and wave functions obtained by the variational method will only be an approx-imation to the exact ones. The accuracy of the results will depend on how good 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 photons are only emitted or absorbed when the electron jumps between orbits. (ii) The central force for the circular motion is provided by the Coulomb attraction, and the electron velocity v must therefore satisfy:
µv2
r =
e2 4π²0²rr2
where r is the radius of the orbital and µ is the reduced mass. (See Exercise 4.2 for a discussion of why it is appropriate to use the reduced mass here.) The Bohr assumption implies that the angular momentum is quantized:
L = µvr = n~ . On eliminating v from these two equations we find:
r = 4π²0²r~ 2 µe2 n 2≡ m0 µ ²rn 2a H, where aH = 4π²0~2/m
0e2 is the hydrogen Bohr radius. The energy is found from:
E = 1 2µv
2− e2 4π²0²rr. On solving for v and substituting, we find:
E = − µe 4 8h2²2 0²2rn2 ≡ − µ m0 1 ²2 r RH n2 , where RH = m0e4/8h2²2
0 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 Schr¨odinger equa-tion.
(iv) The radius for n = 1 corresponds to the peak in the radial probability density for the ground state 1s wave function. For higher atomic shells, the Bohr radius corresponds to the peak radial density of the wave function with 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 4.2 respectively. The reduced mass is calculated from eqn 3.22 to be µ = 0.179m0, and with ²r= 7.9 we then find RX= 39.1 meV and aX= 2.3 nm. Hence we obtain E(1) = −39.1 meV, E(2) = −9.8 meV, r1 = 2.3 nm and r2= 9.3 nm.
We expect the excitons to be stable if E(n) > kBT . At room temperature kBT = 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 we calculate RX= 4.9 meV from eqn 4.1 using ²r= 12.4. Equation 4.4 gives the wavelengths of the n = 1 and n = 2 excitonic transitions as 873.7 nm and 871.4 nm respectively. Hence ∆λ = 2.3 nm.
(4.8) We assume that the exciton has a Lorentzian line with a centre energy of 1.5149 eV and a full width at half maximum of 0.6 meV. We then expect the absorption and refractive index to follow a frequency dependence as in Fig. 2.5. This implies that the maximum in the refractive index would occur 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 =
4π²2 2nλ ,
which implies ²2(ω0) = 1.37 if we assume that n ≈ 3.5 (i.e. the excitonic contribution to the refractive index is relatively small.) From eqn 2.21 we have
²2(ω0) = 1.37 = (²st− ²∞)ω0
γ = (²st− ²∞) 1.5149
0.6 ,
implying (²st − ²∞) = 5.4 × 10−4. With ²st = (3.5)2 = 12.25, we can
then find the dielectric constant at (ω0− γ/2) using eqns 2.20–21. This gives ²(ω0− γ/2) = 12.93 + 0.685i. We finally find n from eqn 1.22 to be 3.60. This justifies the approximation n ≈ 3.5 in the calculation of ²2(ω0) above.
(4.9) The n = 1 and n = 2 excitons have energies of Eg− RX and Eg− RX/4 respectively. Hence the n = 1 → 2 transition occurs at a photon energy of 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 of 394 µm.
(4.10) The magnitude of the electric field between an electron and hole sepa-rated by a distance of r in a medium of relative permittivity ²r is given by Coulomb’s law as:
E = e
4π²0²rr2. In the Bohr model we have (see e.g. Exercise 4.5):
|E(n)| = µe4 8(²0²rhn)2 and rn= 4π²0²r~ 2n2 µe2 . It is thus apparent that for n = 1 we have:
2|E(1)| er1 = 2RX eaX = e 4π²0²r µ πµe2 ²0²rh2 ¶2 = e 4π²0²rr2. Hence E = 2RX/eaX.
(4.11) We use eqn 3.22 to find µ = 0.028m0, and then use eqns 4.1–2 with ²r = 16 to calculate E(1) = 1.5 meV and r1 = 31 nm. Then, using the result of the previous exercise, we find that the internal field in the exciton has a magnitude of 9.7 × 104V/m. We expect the excitons to be ionized whenever the applied field exceeds this value. The voltage at which this occurs can be worked out from eqn 4.5. With Vbi = 0.74 V and li = 2 × 10−6m, we find E = 9.7 × 104V/m for V0= +0.55 V. Hence we need to apply a forward voltage of 0.55 V.
(4.12) The cyclotron energy is given by eqn 4.6 and the exciton Rydberg by eqn 4.1. The condition ~ωc= RX can thus be written:
e~B
µ =
µRH m0²2r ,
which, on solving for B, gives: B = µ 2R H m0²2re~ .
On inserting the numbers for GaAs we find B = 1.8 T. (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 ∂∂xy ∂z × −yx 0 = 00 B0 .
In the analysis following eqn B.17, we neglected the term in A2 because the magnetic vector potential of a light wave is small. However, we are now considering the interaction between an exciton and a strong magnetic field, and it is precisely the term in A2 that gives rise to the diamagnetic shift. It is then apparent from eqn B.17 that the diamagnetic perturbation for a vector potential of A = (B0/2)(−y, x, 0) is given by:
ˆ H0= e2A2 2m0 = e2B2 0 8m0(x 2+ y2) .
The diamagnetic energy shift is calculated from δE = hψ| ˆH0|ψi = e2B20
8m0hψ|(x
2+ y2)|ψi .
In the case of an exciton, the wave functions are spherically symmetric so that:
hx2i = hy2i = hz2i = hr2i/3 = r2 n/3 .
The total shift is obtained by summing the energy shifts of the electrons and holes to obtain:
δE = e2B20 8m∗ e 2r2 n 3 + e2B2 0 8m∗ h 2r2 n 3 = e2B2 0r2n 12µ .
(4.14) It is shown in Example 4.1 that the radius of the ground state exciton in GaAs is 13 nm. Hence for µ = 0.05m0 we calculate δE = +4.9 × 10−5eV
at B = 1 T. The wavelength shift can be calculated from: δλ = dλ
dE δE = − hc
E2 δE = −0.026 nm .
(4.15) The given effective masses imply µ = 0.17m0 from eqn 3.22, so that we can calculate r1 = 3.1 nm and r2 = 12.3 nm from eqn 4.2 using ²r = 10. The Mott densities are estimated from eqn 4.8. Hence we obtain NMott= 8.1 × 1024m−3 and 1.3 × 1023m−3 for the n = 1 and n = 2 excitons
